PROBLEM 1.1
Heat is removed from a rectangular surface by
convection to an ambient fluid at Tf . The heat transfer
coefficient is h. Surface temperature is given by
A
Ts = 1 / 2
x
L
0
W
x
where A is constant. Determine the steady state heat
transfer rate from the plate.
(1) Observations. (i) Heat is removed from the surface
by convection. Therefore, Newton's law of cooling is
applicable. (ii) Ambient temperature and heat transfer
coefficient are uniform. (iii) Surface temperature varies
along the rectangle.
L
0
dq s
x
W
dx
(2) Problem Definition. Find the total heat transfer rate by convection from the surface of a
plate with a variable surface area and heat transfer coefficient.
(3) Solution Plan. Newton's law of cooling gives the rate of heat transfer by convection.
However, in this problem surface temperature is not uniform. This means that the rate of heat
transfer varies along the surface. Thus, Newton’s law should be applied to an infinitesimal area
dAs and integrated over the entire surface to obtain the total heat transfer.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) negligible radiation, (3) uniform heat transfer
coefficient and (4) uniform ambient fluid temperature.
(ii) Analysis. Newton's law of cooling states that
q s = h As (Ts - Tf)
(a)
where
As = surface area, m2
h = heat transfer coefficient, W/m2-oC
q s = rate of surface heat transfer by convection, W
Ts = surface temperature, oC
Tf = ambient temperature, oC
Applying (a) to an infinitesimal area dAs
d q s = h (Ts - Tf) dAs
(b)
The next step is to express Ts ( x) in terms of distance x along the triangle. Ts ( x) is specified as
A
Ts = 1 / 2
(c)
x
PROBLEM 1.1 (continued)
The infinitesimal area dAs is given by
dAs = W dx
(d)
where
x = axial distance, m
W = width, m
Substituting (c) and into (b)
d q s = h(
A
x
- Tf) Wdx
1/ 2
(e)
Integration of (f) gives q s
L
³
q s = dqs = hW ( Ax 1/ 2 Tf )dx
³
(f)
0
Evaluating the integral in (f)
>
@
qs
hW 2 AL1/ 2 LTf
qs
hWL 2 AL1/ 2 Tf
Rewrite the above
>
@
(g)
Note that at x = L surface temperature Ts (L) is given by (c) as
Ts ( L)
(h) into (g)
qs
AL1/ 2
hWL >2Ts ( L) Tf @
(h)
(i)
(iii) Checking. Dimensional check: According to (c) units of C are o C/m1/ 2 . Therefore units
q s in (g) are W.
Limiting checks: If h = 0 then q s = 0. Similarly, if W = 0 or L = 0 then q s = 0. Equation (i)
satisfies these limiting cases.
(5) Comments. Integration is necessary because surface temperature is variable.. The same
procedure can be followed if the ambient temperature or heat transfer coefficient is non-uniform.
PROBLEM 1.2
A right angle triangle is at a uniform surface temperature Ts. Heat is removed by convection to
an ambient fluid at Tf . The heat transfer coefficient h varies along the surface according to
h=
C
x1 / 2
where C is constant and x is distance along the base measured from the apex. Determine the
total heat transfer rate from the triangle.
(1) Observations. (i) Heat is removed from the surface by convection. Therefore, Newton's
law of cooling may be helpful. (ii) Ambient temperature and surface temperature are uniform.
(iii) Surface area and heat transfer coefficient vary along the triangle.
(2) Problem Definition. Find the total heat transfer rate by convection from the surface of a
plate with a variable surface area and heat transfer coefficient.
(3) Solution Plan. Newton's law of cooling gives the rate of
heat transfer by convection. However, in this problem surface
area and heat transfer coefficient are not uniform. This means
that the rate of heat transfer varies along the surface. Thus,
Newton’s law should be applied to an infinitesimal area dAs
and integrated over the entire surface to obtain the total heat
transfer.
dqs
x
W
dx
L
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) negligible radiation and (3) uniform ambient fluid
temperature.
(ii) Analysis. Newton's law of cooling states that
q s = h As (Ts - Tf)
(a)
where
As = surface area, m2
h = heat transfer coefficient, W/m2-oC
q s = rate of surface heat transfer by convection, W
Ts = surface temperature, oC
Tf = ambient temperature, oC
Applying (a) to an infinitesimal area dAs
d q s = h (Ts - Tf) dAs
(b)
The next step is to express h and dAs in terms of distance x along the triangle. The heat transfer
coefficient h is given by
h=
The infinitesimal area dAs is given by
C
x1 / 2
(c)
PROBLEM 1.2 (continued)
dAs = y(x) dx
(d)
where
x = distance along base of triangle, m
y(x) = height of the element dAs, m
Similarity of triangles give
y(x) =
W
x
L
(e)
where
L = base of triangle, m
W = height of triangle, m
Substituting (c), (d) and (e) into (b)
d qs =
C
W
(Ts - Tf) x dx
1/ 2
L
x
(f)
Integration of (f) gives qs. Keeping in mind that C, L, W, Ts and Tf are constants, (f) gives
³
q s = dqs =
CW
(Ts Tf )
L
L
³
0
x
x1 / 2
dx
(g)
Evaluating the integral in (g)
qs =
2
C W L1/2 (Ts - Tf)
3
(h)
(iii) Checking. Dimensional check: According to (c) units of C are W/m3/2-oC. Therefore
units of q s in (h) are
q s = C(W/m3/2-oC) W(m) L1/2(m1/2) (Ts - Tf)(oC) = W
Limiting checks: If h = 0 (that is C = 0) then q s = 0. Similarly, if W = 0 or L = 0 or Ts = Tf
then q s = 0. Equation (h) satisfies these limiting cases.
(5) Comments. Integration was necessary because both area and heat transfer coefficient vary
with distance along the triangle. The same procedure can be followed if the ambient temperature
or surface temperature is non-uniform.
PROBLEM 1.3
A high intensity light bulb with surface heat flux (q / A) s is cooled by a fluid at Tf . Sketch the
fluid temperature profiles for three values of the heat transfer coefficients: h1, h2, and h3, where
h1 < h2 < h3.
(1) Observations. (i) Heat flux leaving the surface is specified (fixed). (ii) Heat loss from the
surface is by convection and radiation. (iii) Convection is described by Newton's law of cooling.
(iv). Changing the heat transfer coefficient affects temperature distribution. (v). Surface
temperature decreases as the heat transfer coefficient is increased. (vi) Surface temperature
gradient is described by Fourier’s law.(vii) Ambient temperature is constant.
(2) Problem Definition. Determine effect of heat transfer coefficient on surface temperature and
surface gradient..
(3) Solution Plan. (i) Apply Newton's law of cooling to examine surface temperature. (ii) Apply
Fourier’s law to determine temperature gradient at the surface.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) no radiation ,(3) uniform ambient fluid temperature
and (4) constant properties.
(ii) Analysis. Newton’s law of cooling
q/ A
s
h(Ts Tf )
(a)
Solve for Ts
(q / A) s
(b)
h
This result shows that for constant (q / A) s surface temperature decreases as h is increased.
Apply Fourier’s law
y
§ wT ·
q / A s k ¨¨ ¸¸
(c)
© wy ¹ y 0
Tf Ts
h1
where y is the distance normal to the
surface. Rewrite (c)
h2
h3
Tf
q/ A w
§ wT ·
¨¨
¸¸
(d)
k
© wy ¹ y 0
This shows that temperature gradient at
the surface remains constant independent
of h. Based on (b) and (d) the temperature
profiles corresponding to three values of
h are shown in the sketch.
Ts
( q / A) s
T
(iii) Checking. Dimensional check: (1) Each term in (b) has units of temperature
Ts ( o C)
Tf ( o C) (q / A) s ( w/m 2 )
2
o
h( w/m C)
o
C
PROBLEM 1.3 (continued)
(2) Each term in (d) has units of
§ wT ·
¨¨
¸¸ ( o C/m)
© wy ¹ y 0
o
C/m
q / A w ( o C/m 2 )
o
o
C/m
k ( W/m- C)
Limiting check: (i) for h = 0 (no heat leaves the surface), surface temperature is infinite. Set h = 0
in (b) gives Ts f.
(5) Comments. Temperature gradient at the surface is the same for all values of h as long as the
thermal conductivity of the fluid is constant and radiation is neglected.
PROBLEM 1.4
Explain why fanning gives a cool sensation.
y
Tf
(1) Observations. (i) Metabolic heat leaves
body at the skin by convection and radiation.
(ii) Convection is described by Newton's law
of cooling. (iii). Fanning increases the heat
transfer coefficient and affects temperature
distribution, including surface temperature.
(iv). Surface temperature decreases as the
heat transfer coefficient is increased. (v)
Surface temperature is described by Newton’s
law of cooling. (vi) Ambient temperature is
constant.
no fan
fan
Ts
skin
T
qcsc
(2) Problem Definition. Determine effect of heat transfer coefficient on surface temperature.
(3) Solution Plan. Apply Newton's law of cooling to examine surface temperature.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) no radiation ,(3) uniform ambient fluid temperature,
(4) constant surface heat flux and (5) constant properties.
(ii) Analysis. Newton’s law of cooling
q csc
h(Ts Tf )
(a)
q csc
h
(b)
where
h = heat transfer coefficient, W/m 2 o C
q csc surface heat flux, W/m 2
Ts = surface temperature, o C
Tf =ambient temperature, o C
Solve (a) for Ts
Ts
Tf This result shows that for constant q csc , surface temperature decreases as h is increased. Since
fanning increases h it follows that it lowers surface temperature and gives a cooling sensation.
(iii) Checking. Dimensional check: Each term in (b) has units of temperature
Ts ( o C)
Tf ( o C) q csc ( w/m 2 )
2
o
h( w/m C)
o
C
PROBLEM 1.4 (continued)
Limiting check: for h = 0 (no heat leaves the surface), surface temperature is infinite. Set h = 0 in
(b) gives Ts f.
(5) Comments. (i) The analysis is based on the assumption that surface heat flux remains
constant. (ii) Although surface temperature decreases with fanning, temperature gradient at the
surface remains constant. This follows from the application of Fourier’s law at the surface
q csc
§ wT ·
¸¸
k ¨¨
© wy ¹ s
Solving for (wT / wy ) s
§ wT ·
¸¸
¨¨
© wy ¹ s
q csc
k
constant
PROBLEM 1.5
A block of ice is submerged in water at the melting temperature. Explain why stirring the water
accelerates the melting rate.
y
(1) Observations. (i) Melting rate of ice depends on
no stirring
the rate of heat added at the surface. (ii) Heat is added
stirring
to the ice from the water by convection. (iii) Newton's
law of cooling is applicable. (iv). Stirring increases
water
surface temperature gradient and the heat transfer
coefficient. An increase in gradient or h increases the
qcsc T
rate of heat transfer. (v) Surface temperature remains
constant equal to the melting temperature of ice. (vi)
0
Ts
water temperature is constant.
ice
ice
(2) Problem Definition. Determine effect of stirring
on surface heat flux.
(3) Solution Plan. Apply Newton's law of cooling to examine surface heat flux.
(4) Plan Execution.
(i) Assumptions. (1) no radiation ,(2) uniform water temperature, (3) constant melting
(surface) temperature.
(ii) Analysis. Newton’s law of cooling
q csc
h(Ts Tf )
(a)
where
h = heat transfer coefficient, W/m 2 o C
q csc surface heat flux, W/m 2
Ts = surface temperature, o C
Tf =ambient water temperature, o C
Stirring increases h . Thus, according to (a) surface heat flux increases with stirring. This will
accelerate melting.
(iii) Checking. Dimensional check: Each term in (a) has units of heat flux.
Limiting check: For Tf Ts (water and ice are at the same temperature), no heat will be added to
the ice. Set Tf Ts in (a) gives q csc 0.
(5) Comments. An increase in h is a consequence of an increase in surface temperature gradient.
Application of Fourier’s law at the surface gives
q csc
§ wT ·
¸¸
k ¨¨
© wy ¹ s
(b)
PROBLEM 1.5 (continued)
Combining (a) and (b)
h
§ wT ·
¸¸
k ¨¨
© wy ¹ s
Ts Tf
According to (c), for constant Ts and Tf , increasing surface temperature gradient increases h.
(c)
PROBLEM 1.6
Consider steady state, incompressible, axisymmetric parallel flow in a tube of radius ro . The
axial velocity distribution for this flow is given by
r2
u u (1 2 )
ro
where u is the mean or average axial velocity. Determine the three components of the total
acceleration for this flow.
(1) Observations. (i) This problem is described by cylindrical coordinates. (ii) For parallel
streamlines v r v T 0 . (iii) Axial velocity is independent of axial and angular distance.
(2) Problem Definition. Determine the total acceleration in the r, T and z directions.
(3) Solution Plan. Apply total derivative in cylindrical coordinates.
(4) Plan Execution.
(ii) Assumptions. (1) Constant radius tube, (2) constant density and (3) streamlines are
parallel to surface.
(ii) Analysis. Total acceleration in cylindrical coordinates is given by
dv r
dt
dv T
dt
Dv r
Dt
wv
wv
wv r v T wv T v T2
vz r r
vr
r wT
r
wt
wz
wr
v wv T v r v T
Dv T
wv
wv
wv
vr T T
vz T T
Dt
wr
r wT
r
wz
wt
dv z Dv z
wv z v T wv z
wv z wv z
vz
vr
dt
Dt
wr
r wT
wz
wt
(1.23a)
(1.23b)
(1.23c)
For streamlines parallel to surface
vr
The axial velocity u
vT
(a)
0
v z is given by
vz
u
u (1 r2
)
ro2
(b)
From (b) it follows that
wv z
wz
Substituting into (1.23a), (1.23b) and (1.23c)
wv z
0
wt
(c)
Radial acceleration:
dv r
dt
Dv r
Dt
0
Angular acceleration;
dv T
dt
Dv T
Dt
0
PROBLEM 1.6 (continued)
Axial acceleration:
dv z
dt
Dv z
Dt
0
(5) Comments. All three acceleration components vanish for this flow.
PROBLEM 1.7
Consider transient flow in the neighborhood of a vortex line where the
velocity is in the tangential direction given by
V (r , t )
§ r 2 ·º
*o ª
¸»
«1 exp¨¨ ¸
2 S r ¬«
© 4ǎ t ¹¼»
V
r
Here r is the radial coordinate, t is time, * o is circulation
(constant) ǎ is kinematic viscosity. Determine the three components
of total acceleration.
(1) Observations. (i) This problem is described by cylindrical coordinates. (ii) streamlines are
concentric circles. Thus the velocity component in the radial direction vanishes ( v r 0 ). (iii)
For one-dimensional flow there is no motion in the z-direction ( v z 0 ). (iv) The T -velocity
component, v T , depends on distance r and time t.
(2) Problem Definition. Determine the total acceleration in the r, T and z directions.
(3) Solution Plan. Apply total derivative in cylindrical coordinates.
(4) Plan Execution.
(ii) Assumptions. (1) streamlines are concentric circles (2) no motion in the z-direction.
(ii) Analysis. Total acceleration in cylindrical coordinates is given by
The three components of the total acceleration in the cylindrical coordinates r ,T , z are
dv r
dt
dv T
dt
dv z
dt
Dv r
Dt
vr
wv
wv
wv r v T wv r v T2
vz r r
r wT
r
wt
wz
wr
(1.23a)
Dv T
Dt
vr
wv T v T wv T v r v T
wv
wv
vz T T
wr
r wT
r
wz
wt
(1.23b)
wv
wv z v T wv z
wv
vz z z
wr
r wT
wz
wt
(1.23c)
Dv z
Dt
vr
For the flow under consideration the three velocity component, v r , v T and v z are
vr
vT (r , t )
0
§ r 2 ·º
*o ª
¨
¸
1
exp
«
¨ 4ǎt ¸»
2 S r «¬
©
¹»¼
vz
Radial acceleration: (a) and (c) into (1.23a)
0
(a)
(b)
(c)
PROBLEM 1.7 (continued)
Dv r
Dt
v T2
r
(d)
(b) into (d)
Dv r
Dt
* o )2 ª
§ r 2 ·º
¨
¸»
1
exp
«
2 3
¨
¸
t
4
Q
4 S r ¬«
©
¹¼»
2
(e)
Tangential acceleration: (a) and (c) into (1.23b)
Dv T
Dt
wv T
wt
(f)
(b) into (f)
DvT
Dt
* o § r2 ·1
r2
¨
¸ exp
2S r ¨© 4ǎt ¸¹ t
4ǎt
(g)
Axial acceleration: (a) and into (1.23c)
Dv z
0
(h)
Dt
(iii) Checking. Dimensional check: Units of acceleration in (e) and (g) are m/s 2 . Note that
according to (b), units of * o are m 2 /s and the exponent of the exponential is dimensionless.
Thus units of (e) are
2
Dv r
Dt
§ r 2 ·º
* o ) 2 (m 4 /s 2 ) ª
¸» = m/s 2
«1 exp¨¨ 2 3
3
¸
4 S r (m ) ¬«
© 4ǎ t ¹¼»
Units of (g) are
Dvș
Dt
* o (m 2 /s) § r 2 (m 2 ) · 1
r2
¨
¸
= m/s 2
exp
2
¨
¸
2S r (m) © 4ǎ (m /s)t (s) ¹ t (s)
4ǎ t
Limiting check: (1) For * o
Dv r Dv T
(g) gives
0
Dt
Dt
0 , all acceleration components vanish. Setting * o
0 in (e) and
f , the tangential velocity vanishes ( v T = 0). Thus all acceleration
Dv r Dv T
components should vanish. Setting t f in (e) and (g) gives
0.
Dt
Dt
(2) According to (b) at t
(5) Comments. The three velocity components must be known to determine the three
acceleration components.
PROBLEM 1.8
An infinitely large plate is suddenly moved parallel to its surface with a velocity U o . The
resulting transient velocity distribution of the surrounding fluid is given by
u
ª
U o «1 (2 / S )
¬
K
³
0
º
exp(K 2 )dK »
¼
y
where the variable K is defined as
K ( x, t )
y
plate
ǎt
2
0
Uo
x
Here t is time, y is the vertical coordinate and ǎ is kinematic viscosity. Note that streamlines
for this flow are parallel to the plate. Determine the three components of total acceleration.
(1) Observations. (i) This problem is described by Cartesian coordinates. (ii) For parallel
streamlines the y-velocity component v 0 . (iii) For one-dimensional flow there is no motion in
the z-direction (w = 0). The x-velocity component depends on distance y and time t.
(2) Problem Definition. Determine the total acceleration in the x, y and z directions.
(3) Solution Plan. Apply total derivative in Cartesian coordinates.
(4) Plan Execution.
(ii) Assumptions.
direction.
(1) streamlines are parallel to surface and (2) no motion in the z-
(ii) Analysis. Total acceleration in Cartesian coordinates is given by
df
dt
Df
Dt
u
wf
wf
wf wf
v
w wx
wy
wz wt
(1.21)
where f represents any of the three velocity components u, v or w. The x-velocity component u is
given by
u
ª
U o «1 (2 / S )
¬
K
º
exp(K 2 )dK »
¼
³
0
(a)
where
K ( x, t )
y
2
ǎt
(b)
Note that u depends on y and t only. For one-dimensional parallel flow
v w 0
Total acceleration in the x-direction, a x . Set f = u in (1.21)
ax
du
dt
Du
Dt
u
wu
wu
wu wu
v
w wx
wy
wz wt
(c)
(d)
PROBLEM 1.8 (continued)
Since u depends on y and t only, it follows that
wu
wx
0
(e)
ax
wu
wt
(f)
wu
wt
du wK
dK wt
(g)
2U o
exp(K 2 )
(h)
Substitute (c) and (e) into (d)
This derivative is obtained using the chain rule
ax
Using (a)
du
dK
y
t 3 / 2
S
Using (b)
wK
wt
2
ǎ
y
1
4 ǎt t
K
(i)
4t
Substitute (h) and (i) into (g)
ax
wu
wt
U o K exp(K 2 )
t
2 S
(g)
Total acceleration in the y-direction, a y . Set f = vȱin (1.21)
ay
dv
dt
Dv
Dt
u
wv
wv wv
wv
v
w wx
wy
wz wt
(h)
Apply (c) to (h)
ay
(i)
0
Total acceleration in the z-direction, a z . Set f = wȱin (1.21)
aw
dw
dt
Dw
Dt
u
ww
ww
ww ww
v
w
wx
wy
wz wt
(j)
Apply (c) to (h)
az
(k)
0
2
(iii) Checking. Dimensional check: Units of acceleration in (g) are m /s. Note that K is
dimensionless. Thus units of (g) are
ax
U o (m/s) K exp(K 2 )
t (s)
2 S
Limiting check: (1) For U o
(2) According to (b) at t
m 2 /s
0 , the acceleration a x
f , K ( y, f)
0. Setting U o
0. Evaluation (a) at K ( y, f)
0 in (g) gives a x
0 gives
0.
PROBLEM 1.8 (continued)
u ( y, f) U o
Since u is constant every where it follows that the a x must be zero. Setting K
(g) gives a x 0.
(l)
0 and t
f in
(5) Comments. The three velocity components must be known to determine the three
acceleration components.
PROBLEM 1.9
Consider two parallel plates with the lower plate stationary and the upper plate moving with a
velocity U o . The lower plate is maintained at temperature T1 and the upper plate at To . The
axial velocity of the fluid for steady state and parallel streamlines is given by
u
Uo
y
y
H
To
Uo
where H is the distance between the two plates.
Temperature distribution is given by
T
PU o2 ª
y2 º
y
y
«
» (To T1 ) T1
H¼
2kH ¬
H
0
T1
x
where k is thermal conductivity and P is viscosity. Determine the total temperature derivative.
(1) Observations. (i) This problem is described by Cartesian coordinates. (ii) For parallel
streamlines the y-velocity component v 0 . (iii) For one-dimensional flow there is no motion in
the z-direction (w = 0). The x-velocity component depends on distance y only.
(2) Problem Definition. Determine the total temperature derivative.
(3) Solution Plan. Apply total derivative in Cartesian coordinates.
(4) Plan Execution.
(ii) Assumptions. (1) streamlines are parallel to surface, (2) no motion in the z-direction
and (3) temperature distribution s one dimensional, T T ( y ).
(ii) Analysis. Total acceleration in Cartesian coordinates is given by
df
dt
Df
Dt
wf
wf wf
wf
v
w wx
wy
wz wt
u
(1.21)
where f represents temperature. Let f = T in (1.21)
dT
dt
DT
Dt
u
wT
wT
wT wT
v
w
wx
wy
wz wt
(a)
where
y
H
(b)
v=w=0
(c)
u
Uo
and
Temperature distribution is given by
T
Using (d)
PU o2 ª
y2 º
y
y
«
» (To T1 ) T1
2kH ¬
H¼
H
(d)
PROBLEM 1.9 (continued)
wT
wx
wT
wt
dT
dt
DT
Dt
0
(e)
Substituting (b), (c) and (e) into
0
(f)
(iii) Checking. Dimensional check: Each term in (d) has units of temperature.
(5) Comments. Velocity and temperature distribution must be know in order to determine the
total derivative of temperature.
PROBLEM 1.10
One side of a thin plate is heated electrically such
Vf y
that surface heat flux is uniform. The opposite side
of the plate is cooled by convection. The upstream Tf
x
velocity is Vf and temperature is Tf . Experiments
qocc
were carried out at two upstream velocities, Vf1
and Vf 2 where Vf 2 ! Vf1 . All other conditions were unchanged. The heat transfer coefficient
was found to increase as the free stream velocity is increased. Sketch the temperature profile
T(y) of the fluid corresponding to the two velocities.
(1) Observations. (i) Heat flux leaving the surface is specified (fixed). (ii) Heat loss from the
surface is by convection and radiation. (iii) Convection is described by Newton's law of cooling.
(iv). Changing the heat transfer coefficient affects temperature distribution. (v). Surface
temperature decreases as the heat transfer coefficient is increased. (vi) Surface temperature
gradient is described by Fourier’s law.(vii) Ambient temperature is constant.
(2) Problem Definition. Determine effect of heat transfer coefficient on surface temperature and
surface gradient..
(3) Solution Plan. (i) Apply Newton's law of cooling to examine surface temperature. (ii) Apply
Fourier’s law to determine temperature gradient at the surface.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) no radiation ,(3) uniform ambient fluid temperature
and (4) constant properties.
(ii) Analysis. Newton’s law of cooling
q occ
h(Ts Tf )
(a)
Solve for Ts
q occ
(b)
h
This result shows that for constant qocc , surface temperature decreases as h is increased. Apply
Fourier’s law
§ wT ·
y
¸¸
(c)
qocc k ¨¨
© wy ¹ y 0
Ts
Tf Tf
where y is the distance normal to the surface.
Rewrite (c)
q cc
§ wT ·
¨¨
¸¸
o
k
© wy ¹ y 0
h1
(d)
This shows that temperature gradient at the surface
remains constant independent of h. Based on (b) and
h2
T
Ts 2
qocc
PROBLEM 1.10 (continued)
(d) the temperature profiles corresponding to two values of h are shown in the sketch.
(iii) Checking. Dimensional check: (1) Each term in (b) has units of temperature
Ts ( o C)
Tf ( o C) (q / A) w ( w/m 2 )
2
h( w/m C)
(2) Each term in (d) has units of
§ wT ·
¨¨
¸¸ ( o C/m)
© wy ¹ y 0
o
o
o
C
C/m
q / A w ( o C/m 2 )
o
o
C/m
k ( W/m- C)
Limiting check: (i) for h = 0 (no heat leaves the surface), surface temperature is infinite. Set h = 0
in (b) gives Ts f.
(5) Comments. Temperature gradient at the surface is the same for all values of h as long as the
thermal conductivity of the fluid is constant and radiation is neglected.
PROBLEM 1.11
Heat is removed from an L-shaped area by convection. The heat
transfer coefficient is h and the ambient temperature is Tf . Surface
temperature varies according to
Ts ( x)
To e
0
a
cx
x 2a
2a
a
a
where c and To are constants. Determine the rate of heat transfer
from the area.
a
(1) Observations. (i) Heat is removed from the surface by convection. Therefore, Newton's
law of cooling is applicable. (ii) Ambient temperature and heat transfer coefficient are uniform.
(iii) Surface temperature varies along the area. (iv) The area varies with distance x.
(2) Problem Definition. Find the total heat transfer rate by convection from the surface of a
plate with a variable surface area and heat transfer coefficient.
(3) Solution Plan. Newton's law of cooling gives the rate of heat transfer by convection.
However, in this problem surface temperature is not uniform. This means that the rate of heat
transfer varies along the surface. Thus, Newton’s law should be applied to an infinitesimal area
dAs and integrated over the entire surface to obtain the total heat transfer.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) negligible radiation, (3) uniform heat transfer
coefficient and (4) uniform ambient fluid temperature.
(ii) Analysis. Newton's law of cooling states that
q s = h As (Ts - Tf)
(a)
where
0
As = surface area, m2
h = heat transfer coefficient, W/m2-oC
q s = rate of surface heat transfer by convection, W
Ts = surface temperature, oC
Tf = ambient temperature, oC
a
h(Ts Tf )dAs1
1
dx
dqs
2
2a
a
Since the L-shaped area varies with distance x, it is divided
into two parts, 1 and 2, each having constant width. Applying
(a) to an infinitesimal area dAs1
d q s1
dx
x
(b)
Integration of (b) from x = 0 to x = a gives the total heat from area 1
a
PROBLEM 1.11 (continued)
a
³
(c)
h(Ts Tf )dAs 2
(d)
h (Ts Tf )dAs1
q s1
0
Similarly, for area 2
d qs2
Integration form x = a to x = 2a gives the total heat from area 2
2a
³
h (Ts Tf )dAs 2
qs2
(e)
a
dAs1 and dAs 2 are given by
dAs1
a dx
(f)
dAs 2
2a dx
(g)
Surface temperature Ts ( x) is specified as
To e
Ts ( x)
cx
(h)
Substitute (f) and (h) into (c)
a
q s1
ha
³
(To e
cx
Tf ) dx
0
Evaluate the integral
q s1
ªT
º
h a « o (e ca 1) aTf »
¬c
¼
(i)
Similarly, (g) and (h) into (e)
2a
qs2
2h a
³
(To e
a
cx
Tf ) dx
Evaluate the integral
qs2
ªT
º
2h a « o (e 2ca e ca ) aTf »
¬c
¼
(j)
The total heat transfer from the L-shaped area is
qs
q s1 q s 2
q s1
ªT
º
ªT
º
h a « o (e ca 1) aTf » 2h a « o (e 2ca e ca ) aTf »
¬c
¼
¬c
¼
Rearrange
qs
ª
º
T
a
hTo «2e 2ca e ca ac f 1»
c
To
¬
¼
(k)
(iii) Checking. Dimensional check: According to (a) units of c are 1/m . Therefore units q s
each term in the bracket of (k) is dimensionless and the coefficient has units of W.
PROBLEM 1.11 (continued)
Limiting checks: (1) If h = 0 then q s = 0. Similarly, if a = 0 the area vanishes and q s = 0.
Equation (i) satisfies these limiting cases.
(2) If To 0 , the entire surface is at uniform temperature Ts
of cooling (a) gives
q s 3a 2 hTf
Setting To
0. Application of Newton’s law
(l)
0 in (k) gives same result.
(5) Comments. Integration is necessary because surface temperature is variable. The same
procedure can be followed if the ambient temperature or heat transfer coefficient is non-uniform.
PROBLEM 2.1
[a] Consider transient (unsteady), incompressible, three dimensional flow. Write the continuity
equation in Cartesian coordinates for this flow.
[b] Repeat [a] for steady state.
The continuity equation in Cartesian coordinates is
wU w
w
w
Uu Uv Uw
w t wx
wy
wz
0
(2.2a)
[a] For incompressible flow the density is constant. Thus the U can be taken out of the
differentiation sign. In addition, for constant density
wU
wt
0
Equation (2.2a) becomes
wu wv ww
wx wy wz
[b] Equation (a) holds for steady state as well.
0
(a)
PROBLEM 2.2
Far away from the inlet of a tube, entrance
effects diminish and stream lines become
parallel and the flow is referred to as fully
developed. Write the continuity equation in
the fully developed region for incompressible
fluid.
r
r
z
fully developed
(1) Observations. (i) The fluid is incompressible. (ii) Radial and tangential velocity
components are zero. (iii) Streamlines are parallel. (iv) Cylindrical geometry.
(2) Problem Definition. Simplify the continuity equation for this flow.
(3) Solution plan. Apply continuity equation in cylindrical coordinates.
(4) Plan Execution.
(i) Assumptions. (1) Incompressible fluid and (2) radial and tangential velocity
components are zero.
(ii) Analysis. The continuity equation in cylindrical coordinates is given by (2.4)
wU 1 w
w
1 w
U rv r U vT Uv z
r wT
wt r wr
wz
0
(2.4)
This equation is simplified for:
wU
0
wt
Parallel streamlines (no radial velocity): v r 0
Incompressible fluid: U is constant,
Parallel streamlines (no tangential velocity): v T
0
Introducing the above simplifications into (2.4), gives
wv z
wz
0
(a)
this result shows that the axial velocity component is invariant with z.
(iii) Checking. Dimensional check: Each term in (2.4) has units of density per unit time.
(5) Comments. (i) The axial velocity varies with radial distance only. (ii) Equation (a) holds
for unsteady state as well. The reason is because for incompressible flow steady or unsteady
the following applies
wU
0
(b)
wt
PROBLEM 2.3
Consider incompressible flow between parallel
plates. Far away from the entrance the axial
velocity component does not vary with the axial
distance.
[a] Determine the velocity component in the ydirection.
[b] Does your result in [a] hold for steady as well as unsteady flow? Explain.
(1) Observations. (i) The fluid is incompressible. (ii) axial velocity is invariant with axial
distance. (iii) Plates are parallel. (iv) Cartesian geometry.
(2) Problem Definition. Determine the velocity component v in the y-direction.
(3) Solution plan. Apply continuity equation.
(4) Plan Execution.
(i) Assumptions. (1) Incompressible fluid, (2) axial velocity is invariant with axial
distance and (3) two-dimensional flow.
(ii) Analysis. The continuity equation in Cartesian coordinates is
wU w
w
w
(2.2a)
Uu Uv Uw 0
w t wx
wy
wz
For incompressible flow the density is constant. Thus the U can be taken out of the
differentiation sign. In addition, for constant density
wU
wt
0
(a)
Since the axial velocity u is invariant with axial distance x, it follows that
wu
wx
0
(b)
w
wz
0
(c)
For two-dimensional flow
(a)-(c) into (2.2a)
wv
wy
0
(d)
f ( x, t )
(e)
Integrating (d)
v
PROBLEM 2.3 (continued)
At the wall the velocity must vanish. Thus
v
0 everywhere in the flow field
[b] Equation (f) holds for steady state as well.
(f)
PROBLEM 2.4
The radial and tangential velocity components for
incompressible flow through a tube are zero. Show
that the axial velocity does not change in the flow
direction. Is this valid for steady as well as
transient flow?
(1) Observations. (i) The fluid is incompressible. (ii) Radial and tangential velocity
components are zero. (iii) Streamlines are parallel. (iv) Cylindrical geometry.
wv z
0.
wz
(3) Solution plan. Apply continuity equation in cylindrical coordinates.
(2) Problem Definition. Show that
(4) Plan Execution.
(i) Assumptions. (1) Incompressible fluid and (2) radial and tangential velocity
components are zero.
(ii) Analysis. The continuity equation in cylindrical coordinates is given by (2.4)
wU 1 w
w
1 w
U rv r U vT Uv z
r wT
wt r wr
wz
0
(2.4)
This equation is simplified for:
Incompressible fluid: U is constant,
No radial velocity: v r
0
No tangential velocity
vT
wU
wt
0
0
Introducing the above simplifications into (2.4), gives
wv z
wz
0
(a)
this result shows that the axial velocity component is invariant with z.
Equation (a) holds for unsteady state as well. The reason is because for incompressible flow
steady or unsteady the following applies
wU
wt
0
(b)
(iii) Checking. Dimensional check: Each term in (2.4) has units of density per unit time.
(5) Comments. (i) Since the radial and tangential velocity component vanishes everywhere
in the flow field, it follows that the streamlines are parallel to the surface. (ii) The axial
velocity varies with radial distance only.
PROBLEM 2.5
Show that W xy
W yx
(1) Observations. (i) Shearing stresses are tangential surface forces. (ii) W xy and W yx are
shearing stresses in a Cartesian coordinate system.(iii) Tangential forces on an element result in
angular rotation of the element. (iv) If the net external torque on an element is zero its angular
acceleration will vanish.
(2) Problem Definition. Find the relationship between W xy and W yx acting on an element.
(3) Solution Plan. Apply Newton’s law of angular motion to an element dx u dy .
(4) Plan Execution.
(i) Assumptions. Continuum,
(ii) Analysis. Consider an element dx u dy with tangential shearing stresses acting on its four
sides. The depth of the element is unity. Apply Newton’s law of angular motion
¦W 0
ID
(a)
W yx y
wW yx
wy
dy
where
I = moment of inertia about 0,
Kg m 2
D = angular acceleration, rad/s 2
W 0 = torque about center 0, N - m
W xy
dy
W xy 0
Note that normal forces (pressure and
dx
normal stress, not shown) exert no torque
on the element since their resultants pass
W yx
through the center 0. The moment of
inertia of the element dx u dy about the
center 0 is
(dx) 2 (dy ) 2
I
U dxdy
12
wW xy
wx
dx
x
(b)
The sum of all external torques acting on the element due to shearing stresses is
¦W 0
W yx dx
wW yx
wW xy
dy
dx
dy
dx
dy ]dx W xy dy [W xy dx]dy
[W yx wy
wx
2
2
2
2
(c)
Note that in the above each shearing stress is multiplied by area to obtain force and by the arm to
give torque. Equation (c) is simplified by neglecting third order
¦W 0
W yx dx
dy
dy
W yx dx
2
2
W xy dy dx W xy dy
2
dx
2
(W yx W xy )dxdy
(d)
PROBLEM 2.5 (continued)
Substituting (b) and (d) into (a)
(W yx W xy )dxdy D
(dx) 2 (dy ) 2
U dxdy
12
Simplify
(W yx W xy ) D U
(dx) 2 (dy ) 2
12
(e)
The right hand side of (e) is of higher order and thus can be neglected to give
(W yx W xy )
0
Thus
W yx
W xy
(f)
(iii) Checking. Dimensional check: Units of (a);
W 0 ( N m) I (Kg m 2 )D (
rad
)
s2
This gives
N
Kg m
s2
which is the correct units for Newton.
(5) Comments. It is incorrect to conclude that W yx
equilibrium.
W xy because the element is in static
PROBLEM 2.6
A fluid flows axially between parallel plates.
Assume: Newtonian fluid, steady state, constant
density, constant viscosity, negligible gravity and
parallel streamlines. Write the three components of
the momentum equations for this flow.
(1) Observations. (i) Properties are constant. (ii) Cartesian coordinates. (iii) Parallel streamlines:
no velocity component in the y-direction. (iv) Axial flow: no velocity component in the zdirection. (v) The Navier-Stokes equations give the three momentum equations.
(2) Problem Definition. Determining the three momentum equations for the flow under
consideration.
(3) Solution Plan. Apply the Navier-Stokes equations of motion in Cartesian coordinates.
Simplify according to the conditions of the problem.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) constant density and viscosity, (3) steady state, (4)
negligible gravity, (5) streamlines are parallel to the plates (no motion in the y-direction) and (6)
axial flow (no motion in the z-direction).
(ii) Analysis. The Navier-Stokes for constant properties are
Ug x § w 2u w 2u w 2u ·
wp
P ¨¨ 2 2 2 ¸¸
wx
wy
wz ¹
© wx
(2.10x)
§ wv
wv
wv
wv ·
u
v
w ¸¸
wx
wy
wz ¹
© wt
Ug y § w 2v w 2v w 2v ·
wp
P ¨¨ 2 2 2 ¸¸
wy
wy
wz ¹
© wx
(2.10y)
§ ww
ww
ww
ww ·
u
v
w ¸¸
wx
wy
wz ¹
© wt
Ug z § w2w w2w w2w ·
wp
P ¨¨ 2 2 2 ¸¸
wz
wz ¹
wy
© wx
(2.10z)
§ wu
wu
wu
wu ·
u
v
w ¸¸
wx
wy
wz ¹
© wt
x-direction:
U ¨¨
y-direction:
U ¨¨
z-direction:
U ¨¨
These equations are simplfied as follows:
Steady state:
w
wt
0
No gravity: g = 0
Parallel streamlines: v
0
Axial flow: w = 0
Substituting these simplifications inot (2.10)
x-direction:
Uu
wu
wx
§ w 2u w 2u ·
wp
P¨ 2 2 ¸
¨ wx
wx
wy ¸¹
©
(a)
PROBLEM 2.6 (continued)
However, continuity equation gives
wU
wU
wU
wU
ª wu w v w w º
u
v
w
U« » 0
wt
wx
wy
wz
¬ wx wx wx ¼
(2.2b)
For incompressible flow this simplifies to
wu w v w w
wx wy wz
0
(b)
This simplifies to
wu
wx
0
(c)
Substituting (c) into (a)
wp
wx
x-direction:
P
w 2u
wy 2
(d)
Equations (2.10y) and (2,10z) simplify to
y-direction:
wp
wy
0
(e)
z-direction:
wp
wz
0
(f)
(iii) Checking. Dimensional check: units of (d)
wp N
(
)
wx m 2 m
N
P(
kg w 2 u m
)
(
)
s m wy 2 sm 2
kg - m
s2
Units of (d) are correct.
(5) Comments. The continuity equation provides additional simplification of the Navier-Stokes
equations.
PROBLEM 2.7
A fluid flows axially (z-direction) through a tube.
Assume: Newtonian fluid, steady state, constant
density, constant viscosity, negligible gravity and
parallel streamlines. Write the three components of
the momentum equations for this flow.
(1) Observations. (i) Properties are constant. (ii) Cylindrical coordinates. (iii) Parallel
streamlines: no velocity component in the r-direction. (iv) Axial flow: no velocity component in
the T -direction. (v) No variation in the T -direction. The Navier-Stokes equations give the three
momentum equations.
(2) Problem Definition. Determining the three momentum equations for the flow under
consideration.
(3) Solution Plan. Apply the Navier-Stokes equations of motion in cylindrical coordinates.
Simplify according to the conditions of the problem.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) constant density and viscosity, (3) steady state, (4)
negligible gravity, (5) streamlines are parallel to surface (no motion in the r-direction) and (6)
axial flow (no motion in the T -direction).
(ii) Analysis. The Navier-Stokes for constant properties are
wv r v T wv r v T 2
wv
wv ·
vz r r ¸
r wT
r
wr
wz
wt ¸¹
§
¨
©
U¨ v r
r-direction:
(2.11r)
2
ª w §1 w
2 wv T w 2 v r º
wp
· 1 w vr
(rv r ) ¸ 2
Ug r P « ¨
»
wr
¹ r wT 2 r 2 wT
wz 2 »¼
«¬ wr © r wr
§
©
U¨vr
T -direction:
wv T v T wv T v r vT
wv
wv ·
vz T T ¸
r wT
r
wr
wz
wt ¹
2
ª w §1 w
1 wp
2 wv r w 2 v T º
· 1 w vT
UgT ( rvT ) ¸ 2
P« ¨
»
2
r wT
r 2 wT
wz 2 ¼
¹ r wT
¬ wr © r wr
§
U¨ v r
©
z-direction:
wv z v T wv z
wv
wv ·
vz z z ¸
wr
wz
wt ¹
r wT
ª 1 w § wv z · 1 w 2 v z w 2 v z º
wp
Ug z P «
¨r
¸ 2
»
2
wz
wz 2 »¼
«¬ r wr © wr ¹ r wT
These equations are simplfied as follows:
Steady state:
w
wt
0
Parallel streamlines: v r
0
(2.11 T )
(2.11z)
PROBLEM 2.7 (continued)
Axial flow: vT = 0
No gravity: g r
gT
gz
0
No variation in the T -direction,
w
wT
0
Substituting these simplifications into (2.11)
wp
wr
wp
wT
r-direction
T -direction:
Uv z
z-direction:
wv z
wz
0
(a)
0
(b)
ª 1 w § wv z
wp
P«
¨r
wz
¬ r wr © wr
2
· w vz º
¸
2 »
¹ wz ¼
(c)
Equation (c) is simplified further using the continuity equation in cylindrical coordinates
w
wU 1 w
1 w
U rv r U vT Uvz
r wT
w t r wr
wz
0
(2.4)
This equation is simplified to
wv z
wz
(d)
0
(d) into (c)
wp
wz
z-direction:
P w § wv z ·
(e)
¨r
¸
r wr © wr ¹
(iii) Checking. Dimensional check: units of (e)
wp
( N/m 3 )
wz
P ( kg/s m)
w § wv ·
1
) ¨ r z ¸( m 2 /m 2 s)
r ( m) wr © wr ¹
kg/s2 m 2
N/m 3
Thus units of (e) are correct.
(5) Comments. The continuity equation provides additional simplification of the Navier-Stokes
equations.
PROBLEM 2.8
Consider two-dimensional flow (x,y) between parallel
plates. Assume: Newtonian fluid, constant density and
viscosity. Write the two components of the momentum
equations for this flow. How many unknown do the
equations have? Can they be solved for the unknowns? If
not what other equation(s) is needed to obtain a solution?
(1) Observations. (i) Properties are constant. (ii) Cartesian coordinates. (iii) Two dimensional
flow (no velocity component in the z-direction. (iv) The Navier-Stokes equations give two
momentum equations.
(2) Problem Definition. Determining the two momentum equations for the flow under
consideration.
(3) Solution Plan. Apply the Navier-Stokes equations of motion in Cartesian coordinates.
Simplify according to the conditions of the problem and count the unknown dependent variables.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) constant density and viscosity, (3) steady state and
(4) two-dimensional flow (no motion in the z-direction).
(ii) Analysis. The Navier-Stokes for constant properties are
§ wu
wu
wu
wu ·
v
w ¸¸
u
wx
wy
wz ¹
© wt
Ug x § w 2u w 2u w 2u ·
wp
P ¨¨ 2 2 2 ¸¸
wx
wz ¹
wy
© wx
(2.10x)
§ wv
wv
wv
wv ·
v
w ¸¸
u
wx
wy
wz ¹
© wt
Ug y § w 2v w 2v w 2v ·
wp
P ¨¨ 2 2 2 ¸¸
wy
wz ¹
wy
© wx
(2.10y)
x-direction:
U ¨¨
y-direction:
U ¨¨
These equations are simplfied as follows:
Steady state:
w
wt
0
Two dimensional flow:
w
wz
w=0
Substituting these simplifications inot (2.10)
x-direction:
§ wu
wu ·
v ¸¸
wy ¹
© wx
Ug x § w 2u w 2u ·
wp
P ¨¨ 2 2 ¸¸
wx
wy ¹
© wx
§ wv
wv ·
v ¸¸
wy ¹
© wx
Ug y § w 2v w 2v ·
wp
P ¨¨ 2 2 ¸¸
wy
wy ¹
© wx
U ¨¨ u
(a)
y-direction:
U ¨¨ u
(b)
These two equations contains three unknowns: u, v and p. A third equation is needed to obtain a
solution. This equation is continuity
PROBLEM 2.8 (continued)
wU
wU
wU
wU
ª wu w v w w º
u
v
w
U« wt
wx
wy
wz
¬ wx wx wx »¼
(2.2b)
0
For incompressible two-dimensional flow this simplifies to
wu wv
wx wx
(c)
0
(iii) Checking. Dimensional check: Each term in (a) and (b) has units of
kg
.
s m2
2
(5) Comments. It is not surprising that continuity is needed to obtain a solution to the flow
field. Conservation of mass (continuity) and momentum (Navier-Stokes equations) must be
satisfied.
PROBLEM 2.9
Consider Two-dimensional (r,z) flow through a tube.
Assume: Newtonian, constant density and viscosity.
Write the two components of the momentum equations for
this flow. How many unknowns do the equations have?
Can the equations be solved for the unknowns? If not
what other equation(s) is needed to obtain a solution?
r
r
z
(1) Observations. (i) Properties are constant. (ii) Cylindrical coordinates. (iii) Two dimensional
flow (no velocity component in the T -direction. (iv) The Navier-Stokes equations give two
momentum equations.
(2) Problem Definition. Determining the two momentum equations for the flow under
consideration.
(3) Solution Plan. Apply the Navier-Stokes equations of motion in cylindrical coordinates.
Simplify according to the conditions of the problem and count the unknown dependent variables.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) constant density and viscosity, (3) steady state and
(4) two-dimensional flow (no motion in the T -direction).
(ii) Analysis. The Navier-Stokes in the r and z directions for constant properties are
§
¨
©
U¨ v r
r-direction:
wv ·
wv
wv r v T wv r v T 2
vz r r ¸
r wT
r
wt ¸¹
wz
wr
(2.11r)
2
ª w §1 w
2 wv T w 2 v r º
wp
· 1 w vr
Ug r P « ¨
( rv r ) ¸ 2
»
wr
¹ r wT 2 r 2 wT
wz 2 »¼
«¬ wr © r wr
§
U¨ v r
z-direction:
©
wv ·
wv z v T wv z
wv
vz z z ¸
wz
wt ¹
wr
r wT
ª 1 w § wv z · 1 w 2 v z w 2 v z º
wp
Ug z P «
¨r
¸ 2
»
2
wz
wz 2 »¼
«¬ r wr © wr ¹ r wT
These equations are simplfied as follows:
Steady state:
w
wt
0
Two dimensional flow:
w
wT
vT = 0
Substituting these simplifications into (2.11)
(2.11z)
PROBLEM 2.9 (continued)
r-direction:
§
U¨ v r
©
z-direction:
§
U¨ v r
©
wv r
wv ·
vz r ¸
wr
wz ¹
Ug r wv z
wv z ·
vz
¸
wr
wz ¹
Ug z 2
ª w §1 w
wp
· w vr º
(rv r ) ¸ P« ¨
»
wr
¹ wz 2 ¼»
«¬ wr © r wr
(a)
ª 1 w § wv z · w 2 v z º
wp
P«
¨r
¸
»
2
wz
«¬ r wr © wr ¹ wz ¼»
(b)
These two equations contains three unknowns: v r , v z and p. A third equation is needed to obtain
a solution. This equation is continuity in cylindrical coordinates
The continuity equation in cylindrical coordinates is given by (2.4)
wU 1 w
w
1 w
U rv r U vT Uv z
r wT
wt r wr
wz
(2.4)
0
This equation is simplified for:
Incompressible fluid: U is constant,
wU
wt
0
Equation (2.4) becomes
wv
1 w
rv r z
r wr
wz
(c)
0
(iii) Checking. Dimensional check: Each term in (a) and (b) has units of
kg
.
s m2
2
(5) Comments. It is not surprising that continuity is needed to obtain a solution to the flow
field. Conservation of mass (continuity) and momentum (Navier-Stokes equations) must be
satisfied.
PROBLEM 2.10
In Chapter 1 it is stated that fluid motion and fluid nature play a role in convection heat transfer.
Does the energy equation substantiate this observation? Explain.
(1) Observations. (i) Motion in energy consideration is represented by velocity components. (ii)
Fluid nature is represented by fluid properties.
(2) Problem Definition. Determine if the energy equation depends on velocity and fluid
properties.
(3) Solution Plan. Write the energy equation and determine if it depends on velocity and
properties.
(4) Plan Execution.
(i) Assumptions. (1) Continuum, (2) Newtonian fluid and (3) negligible nuclear, radiation
and electromagnetic energy transfer.
(ii) Analysis. The energy equation in given by
U cp
DT
Dt
’ ˜ k’T E T
Dp
P)
Dt
(2.15)
where
c p specific heat at constant pressure
k thermal conductivity
p pressure
E coefficient of thermal expansion or compressibility
) = dissipation function
The coefficient of thermal expansion E is a property of material defined as
E
1 ª wU º
U «¬ wT »¼ p
(2.16)
The dissipation function ) is associated with energy dissipation due to friction. It is important
in high speed flow and for very viscous fluids. In Cartesian coordinates ) is given by
)
ª
2 § wv · 2
2 º ª§ wu wv · 2 § wv ww · 2
2º
§ ww ·
§ ww wu · »
§ wu ·
¸¸ ¨
¸ 2 «¨ ¸ ¨¨ ¸¸ ¨ ¸ » «¨¨ ¸¸ ¨¨ «© wx ¹
© wz ¹ » «© wy wx ¹
© wx wz ¹ »
© wy ¹
© wz wy ¹
¬
¼ ¬
¼
2
2 § wu wv ww ·
¸
¨ 3 ¨© wx wy wz ¸¹
(2.17)
Note that the total temperature derivative in (2.15) is defined as
DT
Dt
u
wT
wT
wT wT
v
w
wx
wy
wz wt
(a)
PROBLEM 2.10 (continued)
Examination of the above equations shows that energy equation (2.15) depends on the velocity
components u, v and w. In addition, (2.15) depends on U , c p , E , k and P. These are properties
of fluid.
(5) Comments. To determine temperature distribution it is necessary to know the velocity
distribution.
PROBLEM 2.11
A fluid flows axially (x-direction) between parallel plates.
Assume: Newtonian fluid, steady state, constant density,
constant viscosity, constant conductivity, negligible
gravity and parallel streamlines. Write the energy
equation for this flow.
(1) Observations. (i) Properties are constant. (ii) Cartesian coordinates. (iii) Parallel streamlines:
no velocity component in the y-direction. (iv) Axial flow: no velocity component in the zdirection.
(2) Problem Definition. Determining the energy equations for the flow under consideration.
(3) Solution Plan. Apply the energy equations in Cartesian coordinates. Simplify according to
the conditions of the problem.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) constant density, viscosity and conductivity, (3)
steady state, (4) negligible gravity, (5) streamlines are parallel to the plates (no motion in the ydirection), (6) axial flow (no motion in the z-direction) and (7) negligible nuclear, radiation and
electromagnetic energy transfer.
((4) Plan Execution.
(i) Assumptions. (1) Continuum, (2) Newtonian fluid, (3) no energy generation and (4)
constant properties.
(ii) Analysis. The energy equation for this case is given by
§ wT
wT
wT
wT ·
¸
U c p ¨¨
u
v
w
wz ¸¹
wx
wy
© wt
§ w 2 T w 2T w 2 T ·
k ¨ 2 2 2 ¸ P)
¨ wx
wz ¸¹
wy
©
(2.19b)
where
cp
specific heat at constant pressure
k thermal conductivity
p pressure
T temperature
U density
) = dissipation function
The dissipation function ) in Cartesian coordinates is given by
)
2
2
ª
2 § wv · 2
2º ª
2º
§ wv ww ·
§ ww · » «§ wu wv ·
§ ww wu · »
§ wu ·
«
¸¸ ¨
¸ 2 ¨ ¸ ¨¨ ¸¸ ¨ ¸ ¨¨ ¸¸ ¨¨ «© wx ¹
© wz ¹ » «© wy wx ¹
© wx wz ¹ »
© wy ¹
© wz wy ¹
¬
¼ ¬
¼
2
2 § wu wv ww ·
¸
¨ 3 ¨© wx wy wz ¸¹
These equations are simplfied as follows:
(2.17)
PROBLEM 2.11 (continued)
Steady state:
w
wt
0
No gravity: g = 0
Parallel streamlines: v
Axial flow: w
w
wz
0
0
Substituting these simplifications into (2.19b)
§ w 2T w 2T ·
k ¨ 2 2 ¸ P)
¨ wx
wy ¸¹
©
(a)
2 § ·2
2
wu
2 § wu ·
§ wu ·
2¨ ¸ ¨¨ ¸¸ ¨ ¸
3 © wx ¹
© wx ¹
© wy ¹
(b)
Uc pu
wT
wx
Similarly (2.17) simplifies to
)
Further simplifications are obtained using continuity equation (2.2b)
wU
wU
wU
wU
ª wu w v w w º
u
v
w
U« » 0
wt
wx
wy
wz
¬ wx wx wx ¼
(2.2b)
For incompressible parallel flow this becomes
wu
wx
(c)
0
(c) into (b)
)
§ wu ·
¨¨ ¸¸
© wy ¹
2
(d)
Substitute (d) into (a) gives the energy equation for this flow
Uc pu
wT
wx
2
§ w 2T w 2 T ·
§ wu ·
k ¨ 2 2 ¸ P ¨¨ ¸¸
¨ wx
wy ¸¹
© wy ¹
©
(e)
(iii) Checking. Dimensional check: Each term in (e) has units of W/m 3 .
(5) Comments. The continuity equation provides additional simplification of the dissipation
function.
PROBLEM 2.12
An ideal gas flows axially (x-direction) between parallel
plates. Assume: Newtonian fluid, steady state, constant
viscosity, constant conductivity, negligible gravity and
parallel streamlines. Write the energy equation for this
flow.
(1) Observations. (i) Properties are constant. (ii) Cartesian coordinates. (iii) Parallel streamlines:
no velocity component in the y-direction. (iv) Axial flow: no velocity component in the zdirection. (v) The fluid is an ideal gas.
(2) Problem Definition. Determining the energy equations for the flow under consideration.
(3) Solution Plan. Apply the energy equations in Cartesian coordinates. Simplify according to
the conditions of the problem.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) constant density, viscosity and conductivity, (3)
steady state, (4) negligible gravity, (5) streamlines are parallel to the plates (no motion in the ydirection), (6) axial flow (no motion in the z-direction), (7) negligible nuclear, radiation and
electromagnetic energy transfer and (8) Ideal gas.
(ii) Analysis. The energy equation for this case in given by
&
DT
U cv
’ ˜ k’T p’ ˜ V P)
Dt
(2.23)
Rewriting and noting that k is constant
&
§ wT
wT
wT wT · §¨ w 2T w 2T w 2T ·¸
¸¸ k
¨
U cv ¨ u
’
˜
P)
v
w
p
V
wy
wz wt ¹ ¨© wx 2 wy 2 wz 2 ¸¹
© wx
(a)
where
cv specific heat at constant pressure
k thermal conductivity
p pressure
T temperature
u, v , w velocity components in x, y and z directions
U density
) = dissipation function
The dissipation function ) in Cartesian coordinates is given by
)
ª
2 § wv · 2
2 º ª§ wu wv · 2 § wv ww · 2
2º
§ ww ·
§ ww wu · »
§ wu ·
¸¸ ¨
¸ 2 «¨ ¸ ¨¨ ¸¸ ¨ ¸ » «¨¨ ¸¸ ¨¨ «© wx ¹
© wz ¹ » «© wy wx ¹
© wx wz ¹ »
© wy ¹
© wz wy ¹
¬
¼ ¬
¼
2
2 § wu wv ww ·
¨ ¸
3 ¨© wx wy wz ¸¹
(2.17)
PROBLEM 2.12 (continued)
These equations are simplfied as follows:
Constant k
w
0
wt
Parallel streamlines: v
Steady state:
0
w
0
wz
Incompressible fluid: ’ ˜ V
Axial flow: w
0
Substituting these simplifications into (a)
U cv u
wT
wx
§ w 2T w 2 T ·
k ¨ 2 2 ¸ P)
¨ wx
wy ¸¹
©
(b)
Similarly (2.17) simplifies to
)
2 § ·2
2
wu
2 § wu ·
§ wu ·
2¨ ¸ ¨¨ ¸¸ ¨ ¸
3 © wx ¹
© wx ¹
© wy ¹
(b)
Further simplifications are obtained using continuity equation (2.2b)
wU
wU
wU
wU
ª wu w v w w º
u
v
w
U« » 0
wt
wx
wy
wz
¬ wx wx wx ¼
(2.2b)
For incompressible parallel flow this becomes
wu
wx
(c)
0
(c) into (b)
)
§ wu ·
¨¨ ¸¸
© wy ¹
2
(d)
Substitute (d) into (a) gives the energy equation for this flow
wT
U cv u
wx
2
§ w 2 T w 2T ·
§ wu ·
¸
¨
P ¨¨ ¸¸
k
¨ wx 2 wy 2 ¸
© wy ¹
¹
©
(e)
(iii) Checking. Dimensional check: Each term in (e) has units of W/m 3 .
(5) Comments. The continuity equation provides additional simplification of the dissipation
function.
PROBLEM 2.13
Consider two-dimensional free convection over a vertical plate. Assume:
Newtonian fluid, steady state, constant viscosity, Boussinesq approximation and
negligible dissipation. Write the governing equations for this case. Can the flow
field be determined independently of the temperature field?
(1) Observations. (i) This is a two-dimensional free convection problem. (ii)
The flow is due to gravity. (iii) The flow is governed by the momentum and
energy equations. Thus the governing equations are the Navier-Stokes equations
of motion and the energy equation. (iv) The geometry is Cartesian.
u
g
x
y
(2) Problem Definition. Determine: the x and y components of the Navier-Stokes equations of
motion, and the energy equation for the flow under consideration .
(3) Solution Plan. Start with the Cartesian coordinates Navier-Stokes equations of motion and
energy equation for constant properties. Simplify them for this special case.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4)
Boussinesq approximations and (5) negligible nuclear, electromagnetic and radiation energy
transfer.
(ii) Analysis. Momentum equations. The Navier Stokes equations of motion for free
convection are given in (2.29)
&
&
&
1
DV
(2.29)
E g T Tf ’ p p f v ’ 2V
Uf
Dt
This vector equation gives the x and y components
§ w 2u w 2u w 2u ·
§ wu
wu
wu
wu ·
1 wp
u
v
w ¸¸ Eg (T Tf ) Q ¨ 2 2 2 ¸
¨ wx
wx
wy
wz ¹
U f wx
wy
wz ¸¹
© wt
©
U ¨¨
§ w 2v w 2v w 2v ·
§ wv
wp
wv
wv
wv ·
P¨ 2 2 2 ¸
w ¸¸ u
v
¨ wx
wy
wx
wy
wz ¹
wy
wz ¸¹
© wt
©
U ¨¨
(a)
(b)
Gravity is assumed to point in the negative x-direction. The Cartesian coordinates energy
equation for incompressible constant conductivity fluid is given by equation (2.19b)
§ wT
wT
wT
wT ·
¸
u
v
w
wx
wy
wz ¸¹
© wt
U c 5 ¨¨
§ w 2 T w 2T w 2 T ·
k ¨¨ 2 2 2 ¸¸ P)
wy
wz ¹
© wx
(2.19b)
where ) is the dissipation function. These equations are simplified based on the following
assumptions
w
Steady state:
0
wt
Axial flow: w
w
wz
0
PROBLEM 2.13 (continued)
No dissipation: )
0
(a), (b) and (2.19b) become
wu
wu
u
v
wx
wy
§ w 2u w 2u ·
1 wp
Eg (T Tf ) Q ¨ 2 2 ¸
¨ wx
U f wx
wy ¸¹
©
wv
wv
u
v
wx
wy
§ w 2v w 2v ·
1 wp
Q ¨ 2 2 ¸
¨ wx
U f wy
wy ¸¹
©
§ w 2T w 2 T ·
§ wT
wT ·
¸¸ k ¨
2¸
v
2
¨
x
y
w
w
x
w
wy ¸¹
¹
©
©
U c p ¨¨ u
(c)
(d)
(e)
Equations (c), (d) and (e) are the governing equations for this flow. Examination of momentum
equations (c) and (d) shows that they contain the unknown temperature variable T. Thus these
equations can not be solved for the flow field without invoking the energy equation. Note that
the three equations contain four unknowns: u, v , p and T. Continuity provides the fourth
equation.
(iii) Checking. Dimensional check: Each term of momentum equations (c) and (d) has units of
m/s 2 . Each term in (e) has units of W/m 3 .
Limiting check: If the fluid is not moving, the energy equation should reduce to pure conduction.
Setting u 0 in (e) gives
w 2T
wx 2
w 2T
wy 2
0
This is the correct equation for this limiting case.
(5) Comments. (i) Governing equations (c), (d) and (e) are coupled. Thus they must be solved,
together with continuity, for the flow field and temperature field. (ii) In energy equation (e),
properties c p , k and U represent fluid nature. Velocity components u and v represent fluid
motion. This confirms the observation made in Chapter 1 that fluid motion and nature play a role
in convection heat transfer (temperature distribution).
PROBLEM 2.14
Discuss the condition(s) under which the Navier-Stokes equations of motion can be solved
independently of the energy equation.
Solution
Examination of the smallest rectangle in Table 2.1 shows that for constant properties (density
and viscosity), continuity and momentum (4 equations) contain the four flow field unknowns u,
v, w and p. Thus for constant properties the Navier-Stokes equations and continuity can be
solved for the flow field independently of the energy equation.
PROBLEM 2.15
Consider a thin film of liquid condensate which is falling over a flat surface by
virtue of gravity. Neglecting variations in the z-direction and assuming
Newtonian fluid, steady state, constant properties and parallel streamlines.
[a] Write the momentum equation(s) for this flow.
[b] Write the energy equation including dissipation effect
g
x
(1) Observations. (i) The flow is due to gravity. (ii) For parallel streamlines the
velocity component v = 0 in the y-direction. (iii) Pressure at the free surface is
uniform (atmospheric). (iv) Properties are constant. (v) The geometry is Cartesian.
y
(2) Problem Definition. Determine: [a] the x and y components of the Navier-Stokes equations
of motion, and [b] the energy equation for the flow under consideration .
(3) Solution Plan. Start with the Cartesian coordinates Navier-Stokes equations of motion and
energy equation for constant properties. Simplify them for this special case.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) flow is in the x-direction only, (4)
constant properties, (5) uniform ambient pressure, (6) parallel streamlines. (7) negligible nuclear,
electromagnetic and radiation energy transfer.
(ii) Analysis. [a] Momentum equations. The Navier Stokes equations of motion in Cartesian
coordinates for constant properties are given in equations (2.10x ) and (2.10y)
§ wu
wu
wu
wu ·
u
v
w ¸¸
wx
wy
wz ¹
© wt
U ¨¨
§ wv
wv
wv
wv ·
u
v
w ¸¸
wx
wy
wz ¹
© wt
U ¨¨
Ug x § w 2u w 2u w 2u ·
wp
P ¨¨ 2 2 2 ¸¸
wx
wy
wz ¹
© wx
(2.10x)
Ug y § w 2v w 2v w 2v ·
wp
P ¨¨ 2 2 2 ¸¸
wy
wy
wz ¹
© wx
(2.10y)
The gravitational components are
gx
gy
g,
0
(a)
Based on the above assumptions, equations (2.10) are simplified as follows:
Steady state:
Axial flow (x-direction only):
Parallel flow:
wu
wt
wv
wt
0
(b)
w
w
wz
0
(c)
0
(d)
§ w 2u
wp
wu
w 2 u ·¸
Ug P¨
¨ wx2 w y2 ¸
wx
wx
©
¹
(e)
v
Substituting (a)-(d) into (2.10x) and (2.10y), gives
Uu
and
Problem 2.15 (continued)
wp
0
(f)
wy
The x-component (e) can be simplified further using the continuity equation for incompressible
flow, equation (2.3)
& wu w v w w
’ ˜V
0
(g)
wx w y wz
Substituting (c) and (d) into (g), gives
wu
0
(h)
wx
Using (h) into (e) gives the x-component
wp
w 2u
Ug (i)
P 2 =0
wx
wy
Integrating (f) with respect to y
p f (x)
(j)
where f(x) is the constant of integration. At the free surface, y
equal to p f . Therefore, setting y H in (j) gives
f ( x)
H , the pressure is uniform
(k)
pf
Substituting (k) into (j) gives the pressure solution
p
(l)
pf
wp
=0
wx
Substituting (m) into (i) gives the x-component of the Navier-Stokes equations
Differentiating (k) with respect to x gives
Ug P
d 2u
d y2
(m)
0
(n)
[b] Energy equation. The Cartesian coordinates energy equation for incompressible constant
conductivity fluid is given by equation (2.19b)
§ wT
wT
wT
wT ·
¸
u
v
w
U c 5 ¨¨
wx
wy
wz ¸¹
© wt
§ w 2 T w 2T w 2 T ·
k ¨¨ 2 2 2 ¸¸ P)
wy
wz ¹
© wx
(2.19b)
where the dissipation function in Cartesian coordinates is given by equation (2.17)
)
2
2
ª
2 § wv · 2
2º ª
2º
§ wv ww ·
§ ww · » «§ wu wv ·
§ ww wu · »
§ wu ·
«
¸¸ ¨
¸ 2 ¨ ¸ ¨¨ ¸¸ ¨ ¸ ¨¨ ¸¸ ¨¨ «© wx ¹
© wz ¹ » «© wy wx ¹
© wx wz ¹ »
© wy ¹
© wz wy ¹
¬
¼ ¬
¼
2
2 § wu wv ww ·
¸¸
¨¨ 3 © wx wy wz ¹
(2.17)
Problem 2.15 (continued)
Based on the above assumptions, these equations are simplified as follows:
wT
wt
Steady state:
0
(o)
Substituting (c), (d) and (o) into (2.19b), gives
Uc pu
§ w 2T w 2T ·
k ¨ 2 2 ¸ P)
¨ wx
wy ¸¹
©
wT
wx
(p)
The dissipation function (2.17) is simplified using (c), (d) and (h)
§ wu ·
) ¨¨ ¸¸
© wy ¹
2
(q)
Substituting (q) into (p) gives the energy equation
wT
U c5 u
wx
§ w 2T w 2T ·
§ wu ·
k ¨¨ 2 2 ¸¸ P ¨¨ ¸¸
wy ¹
© wy ¹
© wx
2
(r)
(iii) Checking. Dimensional check: Each term of the x-component equation (n) must have
the same units
Ug = kg/m2-s2
P
d 2u
dy
2
= (kg/m-s)
m/s
= kg/m2-s2
m2
Each term in (r) has the same units of W/m 3 .
Limiting check: If the fluid is not moving, the energy equation should reduce to pure conduction.
Setting u 0 in (i) gives
w 2T
wx 2
w 2T
wy 2
0
This is the correct equation for this limiting case.
(5) Comments. (i) For two-dimensional incompressible parallel flow, the momentum equations
are considerably simplified because the vertical velocity component v = 0.
(ii) The flow is in fact one-dimensional since u does not change with x and is a function of y
only.
(iii) In energy equation (r), properties c p , k , U and P represent fluid nature. The velocity u
represents fluid motion. This confirms the observation made in Chapter 1 that fluid motion and
nature play a role in convection heat transfer (temperature distribution).
(ii) The last term in energy equation (r) represents dissipation.
PROBLEM 2.16
A wedge is maintained at T1 along one side and T2 along
the opposite side. A solution for the flow field is obtained
based on Newtonian fluid and constant properties. The
fluid approaches the wedge with uniform velocity and
temperature. Examination of the solution shows that the
velocity distribution is not symmetrical with respect to the
x-axis. You are asked to support the argument that the
solution is incorrect.
y
Vf
Tf
D
D
x
T1
x
T
x2
(1) Observations. (i) This is a forced convection problem. (ii) Flow properties (density and
viscosity) are constant. (iii) Upstream conditions are uniform (symmetrical) (iv) The velocity
vanishes at both wedge surfaces (symmetrical). (v) Surface temperature is asymmetric. (vi) Flow
field for constant property fluids is governed by the Navier-Stokes and continuity equations. (vii)
If the governing equations are independent of temperature, the velocity distribution over the
wedge should be symmetrical with respect to x. (viii) The geometry is Cartesian.
(2) Problem Definition. Determine if the governing equations for the velocity distribution is
independent of temperature.
(3) Solution Plan. Examine the Navier-Stokes and continuity equations in Cartesian coordinates
for dependency on temperature.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) two-dimensional (x and y), (3) constant properties
and (4) uniform upstream conditions.
(ii) Analysis. The Navier Stokes equations of motion in Cartesian coordinates for constant
properties are given in equations (2.10x ) and (2.10y)
§ wu
wu
wu
wu ·
u
v
w ¸¸
wx
wy
wz ¹
© wt
U ¨¨
§ wv
wv
wv
wv ·
u
v
w ¸¸
wx
wy
wz ¹
© wt
U ¨¨
Ug x § w 2u w 2u w 2u ·
wp
P ¨¨ 2 2 2 ¸¸
wx
wy
wz ¹
© wx
(2.10x)
Ug y § w 2v w 2v w 2v ·
wp
P ¨¨ 2 2 2 ¸¸
wy
wy
wz ¹
© wx
(2.10y)
For two-dimensional conditions (w = 0) these equations become
§ wu
wu
wu ·
U ¨¨ u v ¸¸
wx
wy ¹
© wt
§ wv
wv
wv ·
u
v ¸¸
wx
wy ¹
© wt
U ¨¨
§ w 2u w 2u ·
wp
Ug x P¨ 2 2 ¸
¨ wx
wx
wy ¸¹
©
(a)
§ w 2v w 2v ·
wp
P¨ 2 2 ¸
¨ wx
wy
wy ¸¹
©
(b)
Ug y These equations contain three unknowns: u, v and p. Continuity provides the fourth equation
wU w
w
w
Uu Uv Uw
w t wx
wy
wz
For two-dimensional constant density (2.2a) simplifies to
0
(2.2a)
PROBLEM 2.16 (continued)
wu wv
wx wy
0
(b)
Since properties are constant, U and P are constant. Thus (a), (b) and (c) are independent of
temperature. It follows that the solution to these equations for u, v and p is independent of the
asymmetry of the boundary temperature. A solution based on the assumption of constant
property that give asymmetrical velocity distribution must be incorrect.
(iii) Checking. Dimensional check: Each term in (a) and (b) has units of kg/m2-s2
(5) Comments. (i) Although the flow was assumed two-dimensional, the same conclusion
applies to three-dimensional flow as long as the geometry is symmetrical about the x-axis and
upstream conditions are uniform. (ii) Examination of the smallest rectangle in Table 2.1 shows
that for constant properties (density and viscosity), continuity and momentum (4 equations)
contain the four flow field unknowns u, v, w and p. Thus for constant properties the NavierStokes equations and continuity can be solved for the flow field independently of the energy
equation. This is valid for steady as well as transient flow.
PROBLEM 2.18
Consider two-dimensional (x and y), steady, constant
properties, parallel flow between two plates
separated by a distance H. The lower plate is
stationary while the upper plate moves axially with a
velocity U o . The upper plate is maintained at
uniform temperature To and the lower plate is cooled
with a flux q occ . Taking into consideration dissipation, write the Navier-Stokes equations of
motion, energy equation and boundary conditions at the two plates.
(1) Observations. (i) The geometry is Cartesian. (ii) Properties are constant. (ii) Axial flow (no
motion in the z-direction), (iv) Parallel streamlines means that the normal velocity component is
zero. (v) Specified flux at the lower plate and specified temperature at the upper plate.
(2) Problem Definition. Determine: [a] the x and y components of the Navier-Stokes equations
of motion, [b] the energy equation for the flow under consideration, and [c] velocity and
temperature boundary conditions at the lower and upper plates.
(3) Solution Plan. Start with the Cartesian coordinates Navier-Stokes equations of motion and
energy equation for constant properties. Simplify them for this special case.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) flow is in the x -direction, (3) constant properties
and (4) negligible nuclear, electromagnetic and radiation energy transfer.
(ii) Analysis. The Navier Stokes equations of motion in Cartesian coordinates for constant
properties are given in equations (2.10x ) and (2.10y)
§ wu
wu
wu
wu ·
v
w ¸¸
u
wx
wy
wz ¹
© wt
Ug x § w 2u w 2u w 2u ·
wp
P ¨¨ 2 2 2 ¸¸
wx
wy
wz ¹
© wx
(2.10x)
§ wv
wv
wv
wv ·
u
v
w ¸¸
wx
wy
wz ¹
© wt
Ug y § w 2v w 2v w 2v ·
wp
P ¨¨ 2 2 2 ¸¸
wy
wy
wz ¹
© wx
(2.10y
U ¨¨
U ¨¨
The energy equation given by (2.19b)
§ wT
wT
wT
wT ·
¸
w
u
v
wx
wy
wz ¸¹
© wt
U c p ¨¨
§ w 2 T w 2T w 2 T ·
k ¨ 2 2 2 ¸ P)
¨ wx
wz ¸¹
wy
©
(2.19b)
The dissipation function ) in Cartesian coordinates is given by (2.17)
)
2
2
2
ª
2 § ·2
2 º ª§
2º
§ wv ww ·
wu wv ·
wv
§ ww ·
§ ww wu · » 2 § wu wv ww · (2.17)
§ wu ·
¸¸
¸¸ ¨
2 «¨ ¸ ¨¨ ¸¸ ¨ ¸ » «¨¨ ¸¸ ¨¨ ¸ ¨¨ «© wx ¹
© wz ¹ » «© wy wx ¹
© wx wz ¹ » 3 © wx wy wz ¹
© wy ¹
© wz wy ¹
¬
¼ ¬
¼
These equations are simplfied as follows:
Parallel streamlines: v
0
PROBLEM 2.18 (continued)
Axial flow: w
w
wz
0
Further simplifications are obtained using continuity equation (2.2b)
ª wu w v w w º
wU
wU
wU
wU
u
v
w
U« » 0
wt
wx
wy
wz
¬ wx wy wz ¼
(2.2b)
For incompressible parallel flow this reduces to
wu
wx
(a)
0
Substituting these simplifications into (2.10x), (2.10y), (2.19b) and (2.17)
U
wu
wt
Ug x Ug y wp
w 2u
P 2
wx
wy
wp
wy
0
§ w 2T w 2T ·
k ¨ 2 2 ¸ P)
¨ wx
wy ¸¹
©
wT ·
§ wT
u
¸
wx ¹
© wt
Uc p ¨
§ wu ·
¨¨ ¸¸
© wy ¹
)
(b)
(c)
(d)
2
(e)
The boundary conditions on velocity components are
(1)
(2)
(3)
(4)
u ( x,0) 0
v ( x,0) 0
u ( x, H ) U o
v ( x, H ) 0
The boundary conditions on temperature are
(1) k
wT ( x,0)
wy
(2) T ( x, H )
qocc
To
(iii) Checking. Each term in momentum equations (b) and (c) has units of kg/m 2 s 2 . Each
term in energy equation (d) has units of W/m 3 .
Limiting check: If the upper plate is stationary and there is no axial pressure gradient, the
problem reduces to one-dimensional transient conduction. Since u v ) 0 , (d) becomes
Uc p
wT
wt
§ w 2T w 2T ·
k¨ 2 2 ¸
¨ wx
wy ¸¹
©
(f)
However, since the boundary conditions on temperature are independent of x, axial temperature
PROBLEM 2.18 (continued)
gradient vanishes and (f) simplifies further to
Uc p
wT
wt
k
w 2T
wy 2
(g)
This is the one-dimensional transient conduction equation.
(5) Comments. (i) The continuity equation provides important simplifications in the momentum
and energy equations. (ii) For steady state set wT / wt is set equal to zero.
PROBLEM 2.19
A shaft or radius r1 rotates concentrically inside a sleeve of inner
radius r2 . Lubrication oil fills the clearance between the shaft and
the sleeve. The sleeve is maintained at uniform temperature To .
Neglecting axial variation and taking into consideration dissipation,
write the Navier-Stokes equations of motion, energy equation and
boundary conditions for this flow. Assume constant properties.
r1
r2
shaft
r
T
To
Z
(1) Observations. (i) The geometry is cylindrical. (ii) No variation in
the axial and angular directions. (iii) Properties are constant.
(2) Problem Definition. Determine: [a] the r and T components of the Navier-Stokes equations
of motion, [b] the energy equation and [c] boundary conditions for the flow under consideration.
(3) Solution Plan. Start with the cylindrical coordinates Navier-Stokes equations of motion and
energy equation for constant properties. Simplify them for this special case.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) flow is in the T -direction, (4)
constant properties, (5) negligible nuclear, electromagnetic and radiation energy transfer and (6)
negligible gravity.
(ii) Analysis. [a]. The Navier Stokes equations of motion in cylindrical coordinates for
constant properties are given in equations (2.11r) and (2.11 T )
wv r v T wv r v T 2
wv
wv ·
vz r r ¸
wr
wz
wt ¸¹
r wT
r
§
¨
©
U¨ v r
r-direction:
(2.11r)
2
ª w §1 w
wp
2 wv T w 2 v r º
· 1 w vr
(rv r ) ¸ 2
Ug r P « ¨
»
wr
¹ r wT 2 r 2 wT
wz 2 »¼
«¬ wr © r wr
§
U¨ v r
T -direction:
©
wv ·
wv T v T wv T v r v T
wv
vz T T ¸
wr
wz
wt ¹
r wT
r
2
ª w §1 w
1 wp
2 wv r w 2 v T º
· 1 w vT
P« ¨
2
( rv T ) ¸ 2
UgT »
r wT
¹ r wT 2
r wT
wz 2 »¼
¬« wr © r wr
These equations are simplfied as follows:
Steady state:
w
wt
No gravity: g r
No axial flow:
0
gT
w
wz
0
vz
0
(2.11 T )
PROBLEM 2.19 (continued)
w
Symmetry:
wT
0
Substituting these simplifications inot (2.11r)
§
¨
©
U¨ v r
wv r v T 2 ·¸
ª w §1 w
wp
·º
P« ¨
(rv r ) ¸»
r ¸¹
wr
wr
¹¼
¬ wr © r wr
(a)
wv T v r v T ·
w §1 w
·
(rv T ) ¸
¸ P ¨
wr © r wr
r ¹
wr
¹
(b)
Similarly, (2.11 T ) become
§
U¨ v r
©
However, continuity equation gives
wU 1 w
1 w
w
U rv r U vT Uv z
r wT
wt r wr
wz
For incompressible fluid and
w
wz
w
wT
0
(2.4)
0 , this simplifies to
w
rv r
wr
(c)
0
Integrating
rv r
C
where C is constant of integration. Since v r (r1 ) 0 it follows that C = 0. Therefore
vr
0
(d)
wp
wr
(e)
Introducing (d) into (a) and (b)
U
r-direction:
vT 2
r
w §1 w
·
( rv T ) ¸
¨
wr © r wr
¹
T -direction
0
(f)
[b] Energy equation. For constant properties the energy equation is given by
wT v T wT
wT ·
§ wT
vr
vz
¸
wr
wz ¹
r wT
© wt
Uc p ¨
where the dissipation function is
ª 1 w § wT · 1 w 2T w 2T º
2 » P)
k«
¨r
¸ 2
2
wz ¼»
¬« r wr © wr ¹ r w 0
(2.24)
PROBLEM 2.19 (continued)
2
2
2
2
v
1 wv r ·
§ 1 wv T v r ·
§ wv
§ wv ·
§ wv ·
2¨ r ¸ 2¨
¸ 2¨ z ¸ ¨ T ¸ r ¹
r
r w0 ¹
© wr ¹
© wz ¹
© r wT
© wr
)
2
wv ·
§ 1 wv z wv ·
§ wv
¨
¸ ¨ r z ¸
wz ¹
wr ¹
© wz
© r w0
2
(2.25)
Equations (2.24) and (2.25) are simplified for the conditions of this problem
0 k
)
1 w § wT ·
¨r
¸ P)
r wr © wr ¹
§ wv 0 v T ·
¨
¸
r ¹
© wr
(g)
2
(h)
[c] Boundary conditions. The boundary conditions for Navier-Stokes equations (e) and (f) are
(1) v T (r1 ) Z r1
(2) v T (r2 )
0
The boundary conditions on energy equation (g) are
(1) Insulated surface at r1 :
wT (r1 )
wr
0
(2) Specified temperature at r2 : T (r2 )
To
(iii) Checking. Dimensional check: Units of (e)
U(
kg v T2 m 2
)
(
)
m3 r s 2m
kg
2 2
m s
wp N
(
)
wr m 2 m
N
kg
2
m 2s 2
m m
Units of each term in (g)
k(
w §
wT o
1
·
r
(
m
)
( C/m) ¸
¨
o
wr
¹
m C r (m) wr (m) ©
P(
W
)
kg
)) (1 / s 2 )
sm
W
kg
m3
s 3 m
kg
3
s m
(5) Comments. The continuity equation provides additional simplification of the Navier-Stokes
and energy equations.
PROBLEM 2.20
A rod of radius ri moves axially with velocity
U o inside a concentric tube of radius ro . A
fluid having constant properties fills the space
between the shaft and tube. The tube surface is
maintained at uniform temperature To . Write
the Navier-Stokes equations of motion, energy
equation and surface boundary conditions
Taking into consideration dissipation. Assume that the streamlines are parallel to the surface.
(1) Observations. (i) The geometry is cylindrical. (ii) No variation in the angular direction. (iii)
Properties are constant. (iv) Parallel streamlines means that the radial velocity component is
zero.
(2) Problem Definition. Determine: [a] the r and z components of the Navier-Stokes equations
of motion, [b] the energy equation for the flow under consideration and [c] boundary conditions.
(3) Solution Plan. Start with the cylindrical coordinates Navier-Stokes equations of motion and
energy equation for constant properties. Simplify them for this special case.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) flow is in the z -direction, (4)
constant properties, (5) negligible nuclear, electromagnetic and radiation energy transfer and (6)
negligible gravity.
(ii) Analysis. [a]. The Navier Stokes equations of motion in cylindrical coordinates for
constant properties are given in equations (2.11r) and (2.11 z )
r-direction:
§ wv r v T wv r v T 2
wv
wv ·
vz r r ¸
U¨ v r
¨
wr
wz
wt ¸¹
r wT
r
©
2
ª w §1 w
2 wv T w 2 v r º
wp
· 1 w vr
Ug r P « ¨
( rv r ) ¸ 2
»
wr
¹ r wT 2 r 2 wT
wz 2 ¼»
¬« wr © r wr
§
U¨ v r
z-direction:
©
wv z v T wv z
wv
wv ·
vz z z ¸
wr
wz
wt ¹
r wT
ª 1 w § wv z · 1 w 2 v z w 2 v z º
wp
Ug z P «
¨r
¸ 2
»
2
wz
wz 2 ¼»
¬« r wr © wr ¹ r wT
These equations are simplfied as follows:
w
Steady state:
0
wt
No gravity: g r g z 0
No tangential flow: v T 0
(2.11r)
(2.11z)
PROBLEM 2.20 (continued)
w
wT
Symmetry:
0
Parallel streamlines:
vr
0
Substituting these simplifications inot (2.11r)
r-direction
0
wp
wr
(a)
Similarly, (2.11 z ) become
Uv z
:
wv z
wz
ª 1 w § wv z · w 2 v z º
wp
P«
¸
¨r
2 »
wz
«¬ r wr © wr ¹ wz ¼»
(b)
However, continuity equation gives
wU 1 w
1 w
w
U rv r U vT Uv z
wt r wr
wz
r wT
For incompressible fluid and v r
w
wT
(2.4)
0
0 , this simplifies to
wv z
wz
(c)
0
Introducing (d) into (b)
z-direction
0 ª 1 w § wv z
wp
P«
¨r
wz
¬ r wr © wr
·º
¸»
¹¼
(d)
[b] Energy equation. For constant properties the energy equation is given by
ª 1 w § wT · 1 w 2T w 2T º
wT v T wT
wT ·
§ wT
vr
vz
¸ k«
¨r
¸
» P)
r wT
wr
wz ¹
© wt
«¬ r wr © wr ¹ r 2 wT 2 wz 2 »¼
Uc p ¨
(2.24)
where the dissipation function is
2
2
)
2
2
v
1 wv r ·
§ 1 wv T v r ·
§ wv
§ wv ·
§ wv ·
2¨ r ¸ 2¨
¸ 2¨ z ¸ ¨ T ¸ r ¹
r
r wT ¹
© wr ¹
© wz ¹
© r wT
© wr
2
wv ·
§ 1 wv z wv ·
§ wv
¨
¸ ¨ r z ¸
wz ¹
wr ¹
© wz
© r wT
2
(2.25)
Equations (2.24) and (2.25) are simplified for the conditions of this problems
Uc pv z
wT
wz
ª 1 w § wT · w 2T º
k«
¨r
¸
» P)
«¬ r wr © wr ¹ wz 2 »¼
)
§ wv z ·
¨
¸
© wr ¹
(g)
2
(h)
PROBLEM 2.20 (continued)
[c] Boundary conditions. The boundary conditions for Navier-Stokes equations (e) and (f) are
(1) v z (ri ) U o
(2) v z (ro )
0
The boundary conditions on energy equation (g) are
(1) Equality of temperature at ri : T (ri , z )
(2) Equality of flux at ri : k
wT (ri , z )
wr
Tr (ri , z )
kr
wTr (ri , z )
wr
(3) Specified temperature at ro : T (ro , z ))
To
where the subscript r refers to the rod.
(iii) Checking. Dimensional check: United of each term in (d)
wp N
(
)
wz m 2 m
P(
N
m3
wv (m/s) ·º
kg ª 1
w §
¸»
¨¨ r (m) z
)«
s m ¬ r (m) wr (m) ©
wr (m) ¸¹¼
kg
2
s m
N
2
m3
Units of each term in (h)
Uc pv z
k
w 2T
wz 2
P(
m oC
m 3 kg o C s m
wT
wz
kg
W
J
o
C
W
m o C m 2
m3
kg
)) (1 / s 2 )
sm
W
m3
kg
s 3 m
(5) Comments. (i) The continuity equation provides additional simplification of the NavierStokes and energy equations. (ii) The temperature distribution is two-dimensional. (iii) To solve
for the temperature distribution it is necessary to write a heat equation for the rod as well as
thermal boundary conditions at two axial locations.
PROBLEM 2.21
A rod or radius ri rotates concentrically inside a tube of inner radius ro .
Lubrication oil fills the clearance between the shaft and the tube. Tube
surface is maintained at uniform temperature To . The rod generates heat
volumetrically at uniform rate q ccc . Neglecting axial variation and taking
into consideration dissipation, write the Navier-Stokes equations of
motion, energy equation and boundary conditions for this flow. Assume
constant properties.
ro
To
ri
0
Z
qocc
(1) Observations. (i) The geometry is cylindrical. (ii) No variation in
the axial and angular directions. (iii) Properties are constant.
(2) Problem Definition. Determine: [a] the r and T components of the Navier-Stokes equations
of motion, [b] the energy equation and [c] boundary conditions for the flow under consideration.
(3) Solution Plan. Start with the cylindrical coordinates Navier-Stokes equations of motion and
energy equation for constant properties. Simplify them for this special case.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) flow is in the T -direction, (4)
constant properties, (5) negligible nuclear, electromagnetic and radiation energy transfer and (6)
negligible gravity.
(ii) Analysis. [a]. The Navier Stokes equations of motion in cylindrical coordinates for
constant properties are given in equations (2.11r) and (2.11 T )
r-direction:
§ wv r v T wv r v T 2
wv
wv ·
vz r r ¸
U¨ v r
¨
wr
wz
wt ¸¹
r wT
r
©
(2.11r)
2
ª w §1 w
wp
2 wv T w 2 v r º
· 1 w vr
(rv r ) ¸ 2
Ug r P « ¨
»
wr
¹ r wT 2 r 2 wT
wz 2 ¼»
¬« wr © r wr
§
U¨ v r
T -direction:
©
wv T v T wv T v r v T
wv
wv ·
vz T T ¸
wr
wz
wt ¹
r wT
r
2
ª w §1 w
1 wp
2 wv r w 2 v T
· 1 w vT
P« ¨
( rv T ) ¸ 2
UgT r wT
¹ r wT 2
r 2 wT
wz 2
«¬ wr © r wr
These equations are simplfied as follows:
Steady state:
w
wt
0
No gravity: g r
gT 0
w
vz
No axial flow:
wz
0
º
»
»¼
(2.11 T )
PROBLEM 2.21 (continued)
w
Symmetry:
wT
0
Substituting these simplifications inot (2.11r)
§
¨
©
U¨ v r
wv r v T 2 ·¸
ª w §1 w
wp
·º
P« ¨
(rv r ) ¸»
¸
r ¹
wr
wr
¹¼
¬ wr © r wr
(a)
wv T v r v T ·
w §1 w
·
(rv T ) ¸
¸ P ¨
wr © r wr
r ¹
wr
¹
(b)
Similarly, (2.11 T ) become
§
U¨ v r
©
However, continuity equation gives
wU 1 w
1 w
w
U rv r U vT Uv z
r wT
wt r wr
wz
For incompressible fluid and
w
wz
w
wT
(2.4)
0
0 , this simplifies to
w
rv r
wr
(c)
0
Integrating
rv r
C
where C is constant of integration. Since v r (ro )
vr
0 it follows that C = 0. Therefore
0
(d)
wp
wr
(e)
Introducing (d) into to (a) and (b)
U
r-direction:
vT 2
r
w §1 w
·
( rv T ) ¸
¨
wr © r wr
¹
T -direction
(f)
0
[b] Energy equation. For constant properties the energy equation is given by
wT v T wT
wT ·
§ wT
vr
vz
¸
wr
wz ¹
r wT
© wt
Uc p ¨
ª 1 w § wT · 1 w 2T w 2T º
2 » P)
k«
¨r
¸ 2
2
wz »¼
«¬ r wr © wr ¹ r w 0
(2.24)
where the dissipation function is
2
)
2
2
2
v
1 wv r ·
§ 1 wv T v r ·
§ wv
§ wv ·
§ wv ·
2¨ r ¸ 2¨
¸ 2¨ z ¸ ¨ T ¸ r ¹
r
r w0 ¹
© wr ¹
© wz ¹
© r wT
© wr
2
wv ·
§ 1 wv z wv ·
§ wv
¨
¸ ¨ r z ¸
wz ¹
wr ¹
© wz
© r w0
Equations (2.24) and (2.25) are simplified for the conditions of this problems
2
(2.25)
PROBLEM 2.21 (continued)
0 k
1 w § wT ·
¸ P)
¨r
r wr © wr ¹
§ wv 0 v T ·
¨
¸
r ¹
© wr
)
(g)
2
(h)
[c] Boundary conditions. The boundary conditions for Navier-Stokes equations (e) and (f) are
(1) v T (ri ) Z ri
(2) v T (ro ) 0
The boundary conditions on energy equation (g) are
(1) Specified flux at ri : k
wT (ri )
wr
qicc
(2) Specified temperature at ro : T (ro )
To
Conservation of energy for the rod gives the flux qicc :
Energy generated in rod = energy leaving surface at ri
q cccSri2
2Sri qicc
q cccri
2
qicc
(i)
(iii) Checking. Dimensional check: United of (e)
kg v T2 m 2
U( 3 ) ( 2 )
r s m
m
kg
2 2
m s
wp N
(
)
wr m 2 m
N
kg
2
m 2s 2
m m
Units of each term in (g)
k(
w §
wT o
1
·
¨ r (m) ( C/m) ¸
wr
¹
m C r (m) wr (m) ©
P(
W
o
)
kg
)) (1 / s 2 )
sm
W
m
3
kg
3
s m
kg
3
s m
(5) Comments. The continuity equation provides additional simplification of the Navier-Stokes
and energy equations.
PROBLEM 2.22
Air flows over the two spheres. The radius of sphere 2 is double
that of sphere 1. However, the free stream velocity for sphere 1 is
double that for sphere 2. Determine the ratio of the average heat
transfer coefficients h1 / h2 for the two spheres.
1
Vf1
2
V
f2
(1) Observations. (i) This is a forced convection problem. (ii)
The same fluid flows over both spheres. (iii) Sphere diameter and free stream velocity affect
the Reynolds number which in turn affect the heat transfer coefficient.
(2) Problem Definition. Since the average heat transfer coefficient h is correlated in terms
of the Nusselt number, the problem becomes one of determining the Nusselt number for each
sphere and taking their ratio.
(3) Solution Plan. Use the results of dimensional analysis to obtain a relationship between the
Nusselt number and the significant parameters in forced convection.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state and (3) constant properties.
(ii) Analysis. Non-dimensional form of the governing equations for convection gives
Nu D =
hD
= f( Re D , Pr, GrD , E)
k
(a)
where
D = diameter of sphere, m
E = Eckert number
GrD = Grashof number
2 o
h = average heat transfer coefficient, W/m - C
k = thermal conductivity of fluid, W/m-oC
Nu D = average Nusselt number
Pr = Prandtl number
Re D = Reynolds number
Assume that free convection is negligible compared to forced convection. This eliminates the
Grashof number. Furthermore, neglect dissipation effects. This eliminates the Eckert number.
Thus (a) is simplified to
hD
Nu D =
= f( Re D , Pr)
(b)
k
Since the same fluid flows over both spheres, it follows that the Prandtl number is the same for
both. Thus (b) becomes
hD
Nu D =
= f( Re D )
(c)
k
The Reynolds number is defined as
PROBLEM 2.22 (continued)
Re D =
Vf D
(d)
Q
where
Vf free stream velocity, m/s
Q = kinematic viscosity, m2/s
Solving equation (c) for h
h=
k
f( Re D )
D
(e)
According to (e), to calculate h for each sphere it is necessary to determine: (1) the exact form
of the function f( Re D ), (2) the diameter, (3) the thermal conductivity and (4) the Reynolds
number. However, of interest is determining the ratio of the heat transfer coefficients for two
spheres. Applying (e) to the two spheres and taking the ratio of the resulting equation
D f Re D1
h1
= 2
h2
D1 f Re D 2
(f)
where the subscripts 1 and 2 refer to sphere 1 and 2, respectively. Using (d) to determine the two
Reynolds numbers
Vf1 D1
Re D1
(g)
Q
and
Re D 2
Vf 2 D 2
(h)
Q
However
D2 = 2 D1
(i)
and
Vf 2
(j)
Vf1 / 2
Substituting (i) and (j) into (h)
Re D 2
Vf1 2 D1
V D
= f1 1
2Q
Q
Re D1
(k)
Thus the Reynolds number is the same for both spheres. It follows that
Re D 2
f( Re D 2 ) = f( Re D1 )
(l)
h1
D
= 2 =2
h2
D1
(m)
Substituting this result into (f) gives
(iii) Checking. Qualitative check: From (e) one concludes that for the same fluid (same k)
and Reynolds number, the heat transfer coefficient is inversely proportional to the diameter.
This confirms the result in (m).
PROBLEM 2.22 (continued)
(5) Comments. (i) For constant Reynolds and Prandtl numbers the heat transfer coefficient
increases as the diameter decreases. (ii) The ratio of the total heat transfer rate from the two
spheres is obtained from Newton's law of cooling
q1
q2
h1 (S D12 )(Ts Tf )
D2 D12
h2 (S D22 )(Ts
D1 D22
Tf )
D1
D2
1
2
Thus, although the heat transfer coefficient for the small sphere is greater than that of the large
sphere, its total heat transfer rate is smaller by a factor of two.
PROBLEM 2.23
The average Nusselt number for laminar free convection over an isothermal vertical plate is
determined analytically and is given by
4 ª GrL º
3 «¬ 4 »¼
hL
k
Nu L
1/ 4
f ( Pr )
where GrL is the Grashof number based on the length of the plate L and f(Pr) is a function of the
Prandtl number. Determine the percent change in the average heat transfer coefficient if the
length of the plate is doubled.
(1) Observations. (i) This is a free convection problem. (ii) The average heat transfer
coefficient h depends on the vertical length L of the plate. (iii) L appears in the Nusselt number
as well as the Grashof number.
(2) Problem Definition. Derive a relationship between the average heat transfer coefficient h
and the length of a vertical plate L.
(3) Solution Plan.
length L.
Solve the given Nusselt number correlation equation for h in terms of
(4) Plan Execution.
(i) Assumptions. (1) Laminar flow and (2) given correlation equation for Nusselt number
applies to both plates.
(ii) Analysis. The percent change in h is given by
% change in h = 100
h2 h1
h1
ªh
º
100« 2 1»
¬ h1
¼
(a)
where the subscripts 1 and 2 refer to plates of length L and 2L, respectively and h is the average
heat transfer coefficient. The average Nusselt number Nu L is given by
Nu L
hL
k
4 ª GrL º
3 «¬ 4 »¼
1/ 4
f(Pr)
(b)
where
f(Pr) = function of Prandtl number
GrL = Grashof number
2 o
h = average heat transfer coefficient, W/m - C
k = thermal conductivity, W/m-oC
L = plate length, m
Nu L = average Nusselt number
Pr = Prandtl number
The Grashof number is defined as
GrL =
where
E g (Ts Tf ) L3
Q2
(c)
PROBLEM 2.23 (continued)
g = gravitational acceleration, m/s2
Ts = surface temperature, oC
Tf = ambient temperature, oC
E = coefficient of thermal expansion, 1/K (or 1/oC)
Q = kinematic viscosity, m2/s
Substituting (c) into (b) and solving for h
4 ª E g (Ts Tf ) º
h = k«
»
3 ¬
4Q 2
¼
Applying (d) to the two plates
4 ª E g (Ts Tf ) º
h1 = k «
»
3 ¬
4Q 2
¼
1/ 4
f ( Pr )
1
L
1/ 4
1
f ( Pr )
(e)
1/ 4
L1
and
4 ª E g (Ts Tf ) º
h2 = k «
»
3 ¬
4Q 2
¼
(d)
1/ 4
1/ 4
f Pr
1
L2
(f)
1/ 4
Taking the ratio of (e) and (f)
h2
h1
ª L1 º
« »
¬ L2 ¼
Substituting (g) into (a)
1/ 4
(g)
>
@
% change in h = 100 ( L1 / L2 )1 / 4 1
(h)
(iii) Computations. For the case where the length L2 = 2L1, equation (h) gives
% change in h = 100 [(1/2)1/4 1 ] = 15.9 %
(iv) Checking. Dimensional check: Units of h in (d) should be W/m2-oC:
ª E (1/ o C) g (m/s 2 )(Ts Tf )( o C) º
4
o
h = k ( W/m C) «
»
3
4Q 2 (m 2 /s) 2
¬«
¼»
1/ 4
f(Pr)
1
1/ 4
L
( m)
1/ 4
= W/m2-oC
Qualitative check: According to (d) the average heat transfer coefficient is inversely proportional
to L1/4. Thus increasing L, decreases h . This is consistent with the negative result obtained
( 15.9 %) which indicates a decrease in h .
(5) Comments. Although h decreases as the length of the plate is increased, the total heat
transfer rate increases. Newton's law of cooling gives
q2
q1
h 2 L 2 T s Tf
h2 L 2
h1 L1 Ts Tf
h1 L1
§ L1
¨¨
© L2
·
¸¸
¹
1/ 4
L2
L1
§ L2
¨¨
© L1
·
¸¸
¹
3/ 4
= (2)3/4 = 1.68
PROBLEM 2.24
An experiment was performed to determine the average heat transfer coefficient for forced
convection over spheres. In the experiment a sphere of diameter 3.2 cm is maintained at uniform
surface temperature. The free stream velocity of the fluid is 23.4 m/s. Measurements showed that
the average heat transfer coefficient is 62 W/m 2 o C .
[a] Predict the average heat transfer coefficient for the same fluid which is at the same free
stream temperature flowing over a sphere of diameter 6.4 cm which is maintained at the same
surface temperature. The free stream velocity is 11.7 m/s.
[b] Which sphere transfers more heat?
(1) Observations. (i) This is a forced convection problem. (ii) The same fluid flows over
both spheres. (iii) Sphere diameter and free stream velocity affect the Reynolds number which
in turn affect the heat transfer coefficient. (iv) Newton’s law of cooling gives the heat transfer
rate.
(2) Problem Definition. Since the average heat transfer coefficient h is expressed in terms
of the Nusselt number, the problem becomes one of determining the Nusselt number for each
sphere and taking their ratio.
(3) Solution Plan. Use the results of dimensional analysis to obtain a relationship between the
Nusselt number and the significant parameters in forced convection. Apply Newton’s law of
cooling to determine heat transfer rate.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state and (3) constant properties.
(ii) Analysis and computations. Non-dimensional form of the governing equations for
forced convection gives
Nu D =
hD
= f( Re D , Pr, E)
k
(a)
where
D = diameter of sphere, m
E = Eckert number
2 o
h = average heat transfer coefficient, W/m - C
o
k = thermal conductivity of fluid, W/m- C
Nu D = average Nusselt number
Pr = Prandtl number
Re D = Reynolds number
Assume that dissipation is negligible, equation (a) is simplified to
Nu D =
hD
= f( Re D , Pr)
k
(b)
Since the same fluid flows over both spheres, it follows that the Prandtl number is the same for
both. Thus (b) becomes
PROBLEM 2.24 (continued)
hD
= f( Re D )
k
Nu D =
(c)
The Reynolds number is defined as
Re D
Vf D
(d)
Q
where
Vf free stream velocity, m/s
Q = kinematic viscosity, m2/s
Solving equation (c) for h
k
f( Re D )
D
h=
(e)
Applying (2) to the two spheres and taking the ratio to eliminate k
D f ( Re D1 )
h1
= 2
h2
D1 f ( Re D 2 )
(f)
where the subscripts 1 and 2 refer to sphere 1 and 2, respectively. Using (d) to determine the two
Reynolds numbers
V D
Re D1 = f1 1
(g)
Q
and
Re D 2
Vf 2 D 2
(h)
Q
The total heat transfer rate is determined using Newton’s law of cooling
qT
h A(Ts Tf )
(i)
A SD 2
(j)
where A is surface area of sphere given by
(j) into (i)
q
S h D (T s T f )
(k)
Applying (k) to the two spheres and taking their ratio
q1
q2
S h1 D1 (Ts Tf )
S h 2 D2 (Ts Tf )
h1 D1
h 2 D2
(iii) Computations. Substituting numerical values in (g) and (h)
Re D1
and
23.4(m/s)3.2(m)
Q (m 2 /s)
74.88(m 2 /s)
Q (m 2 /s)
(l)
PROBLEM 2.24 (continued)
11.7(m/s)6.4(m)
Q (m 2 /s)
Re D 2
74.88(m 2 /s)
Q (m 2 /s)
Thus the two Reynolds numbers are identical. It follows that
f ( Re D1 )
f ( Re D 2 )
(i)
Substituting (i) into (f)
h1
D
= 2
h2
D1
(j)
Solving (j) for h2
D1
3.2(m)
62( W/m 2 o C)
6.4(m)
D2
Substitute (j) into (l)
q1 D2 D1
1
q 2 D1 D2
h2
h1
31 W/m 2 o C
Checking.
Dimensional check: computations showed that the Reynolds number is
dimensionless and that units of h are correct.
Qualitative check: From (e) one concludes that for the same fluid (same k) and Reynolds
number, the heat transfer coefficient is inversely proportional to the diameter. Results show that
increasing the diameter by a factor of 2 reduces the heat transfer coefficient by the same factor.
(5) Comments. For constant Reynolds and Prandtl numbers the heat transfer coefficient
increases as the diameter decreases.
PROBLEM 2.25
Atmospheric air flows between parallel plates with a mean velocity of 10 m/s . One plate is
maintained at 25 o C while the other at 115 o C.
[a] Calculate the Eckert number. Can dissipation be neglected?
[b] Use scale analysis to compare the magnitude of normal conduction, k w 2T / wy 2 , with
dissipation, P (wu / wy ) 2 . Is dissipation negligible compared to conduction?
(1) Observations. (i) Dissipation is important when the Eckert number is high compared to
unity. (ii) If the ratio of dissipation to conduction is small compared to unity, it can be neglected.
(2) Problem Definition. [a] Compute the Eckert number. [b] Estimate normal conduction and
dissipation using scaling.
(3) Solution Plan. [a] Using the definition of the Eckert number, compute its value for the given
data. [b] Use scaling to estimate the ratio of dissipation to normal conduction.
(4) Plan Execution.
(i) Assumption. (1) Newtonian fluid and (2) continuum.
(ii) Analysis. The Eckert number is defined in equation (2.43) as
(
u2
c p (Ts1 Ts 2 )
(a)
where
cp
specific heat, J/kg o C
Ts1
temperature of plate 1 = 25 o C
Ts 2
H
u
y
o
Ts 2 temperature of plate 2 = 115 C
u mean axial velocity = 10 m/s
0
Ts1
Dissipation is given by
ª wu º
dissipation = P « »
¬ wy ¼
2
(b)
where
u axial velocity, m/s
y normal coordinate, m
P viscosity, kg/s-m
Normal conduction is given by
conduction = k
w 2T
wy 2
To scale dissipation and conduction, the following scales are introduced
(c)
PROBLEM 2.25 (continued)
Scale for 'T : 'T a (Ts1 Ts 2 ) Scale for ' u : ' u a u
Scale for ' y : ' y a H
where H is spacing between plates. Rewrite (b) and (c)
ª 'u º
dissipation = P « »
¬ 'y ¼
conduction = k
2
(d)
' ('T )
('y ) 2
(e)
u2
H2
(f)
Use the above scales to estimate (d) and (e)
dissipation = P
conduction = k
(Ts1 Ts 2 )
(g)
H2
Taking the ratio of (f) and (g)
dissipation
conduction
Pu2
(h)
k (Ts1 Ts 2 )
(iii) Computations. Properties of air are determined at the average temperatureT
T
cp
Ts1 Ts 2
2
(25 115)( o C)
2
1008.7 J/kg
o
70 o C
C
0.02922 W/m o C
P 20.47 u 10 6 kg/s m
k
Substitute into (a)
E
(10) 2 (m 2 /s 2 )
1008.7(J/kg o C)(115 - 25)( o C)
0.0011
kg m 2
J s2
0.0011
Substitute into (h)
dissipation
conduction
20.47 u 10 6 (kg/s m)(10) 2 (m/s) 2
0.02922(W/m- o C)(115 - 25)( o C)
0.00078
kg - m 2
W - s3
0.00078
W
W
0.00078
Computations show that the Eckert number is small compared to unity. Thus dissipation can be
neglected. Computations also show that dissipation is small compared to normal conduction.
Thus it can be neglected.
(iv) Checking. Dimensional check: Computations show that units for dissipation and
conduction are correct.
PROBLEM 2.25 (continued)
(5) Comments. (i) The spacing between the two plates, H, need not be specified to compare
dissipation with conduction. Their ratio in (h) is independent of H. (ii) The Eckert number is a
measure of the importance of dissipation.
PROBLEM 2.26
An infinitely large plate is immersed in an infinite fluid. The plate is suddenly moved along its
plane with velocity U o . Neglect gravity and assume constant properties.
[a] Show that the axial Navier-Stokes equation is given by
wu
w 2u
U
P 2
wt
wy
[b] Due to viscous forces, the effect of plate motion
penetrates into the fluid. The penetration depth G (t )
increases with time. Use scaling to derive an expression
for G (t ) .
(1) Observations. (i) The plate is infinite. (ii) No changes take place in the axial direction
(infinite plate). (iii) This is a transient problem. (iv) Constant properties. (v) Cartesian
coordinates.
(2) Problem Definition. [a] Determine the equation of motion for resulting from a suddenly
accelerated plate. [b] Use scaling to estimate G (t ).
(3) Solution Plan. [a] Apply the Navier-Stokes equations of motion. Introduce continuity to
identify simplifying conditions. [b] Assign scales to each variable in the governing equation to
estimate G (t ).
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) constant properties, (3) no motion in the zdirection and (4) negligible gravity.
(ii) Analysis. [a] The Navier-Stokes for two-dimensional constant properties are
§ wu
wu
wu
wu ·
u
v
w ¸¸
wx
wy
wz ¹
© wt
x-direction:
U ¨¨
y-direction:
U ¨¨
§ wv
wv
wv
wv ·
u
v
w ¸¸
wx
wy
wz ¹
© wt
These equations are simplfied as follows:
No gravity: g = 0
No axial variation:
w
wx
0
No motion in the z-direction: w
w
wz
0
Substituting these simplifications into (2.10)
Ug x § w 2u w 2u w 2u ·
wp
P ¨¨ 2 2 2 ¸¸
wx
wy
wz ¹
© wx
(2.10x)
Ug y § w 2v w 2v w 2v ·
wp
P ¨¨ 2 2 2 ¸¸
wy
wz ¹
wy
© wx
(2.10y)
PROBLEM 2.26 (continued)
§ wu
wu ·
v ¸¸
wy ¹
© wt
U ¨¨
x-direction:
§ wv
wv ·
v ¸¸
wy ¹
© wt
U ¨¨
y-direction:
P
w 2u
wy 2
(a)
§ w 2v ·
wp
P ¨¨ 2 ¸¸
wy
© wy ¹
(b)
However, continuity equation gives
ª wu w v w w º
wU
wU
wU
wU
u
v
w
U« »
wy
wz
wt
wx
¬ wx wy wz ¼
0
(2.2b)
For two-dimensional incompressible flow this simplifies to
wu wv
wx wy
0
(c)
This simplifies to
wv
wy
0
(d)
f (t )
(e)
Integration of (d) gives
v
where f(t) is “constant” of integration. This time function is determined from the no slip
boundary condition on v
v ( x,0) 0
(f)
Applying (f) to (e) gives
(g)
f (t ) 0
(g) into (e)
(h)
v 0
Substitute (h) into (a) and (b)
U
x-direction:
wu
wt
P
w 2u
wy 2
wp
wy
[b] To obtain scaling estimate of G (t ) rewrite (i)
y-direction
0
U
Scale for ' u : ' u a (U 0) Scale for ' u : ' t a (t 0)
Scale for ' y : ' y a (G 0)
'u
't
P
' ª'u º
' y «¬ ' y »¼
(i)
(j)
(k)
PROBLEM 2.26 (continued)
Substitute into (k)
U
U
U
a P 2 l t
G
Introduce the definition of kinematic viscosity Q
Q
P
U
(m)
Substitute (m) into (l) and solve for G
G (t ) a Q t n
(iii) Checking. Dimensional check: (i) Each term in (i) has units of kg/s2 m 2 .
w 2u
wu
2
2
( m/s2 ) P ( kg/s m) 2 ( m/sm 2 ) = kg/s m
wt
wy
(ii) Units of (n) should be length:
U (kg/m3 )
G (t )
Q (m 2 /s)t (s)
m
Limiting check: Initially the penetration thickness is zero. Setting t = 0 in (n) gives G (0)
which is the correct result.
0,
(5) Comments. (i) A major simplification of this problem is due to the assumption of infinite
plate. Due to this assumption all derivatives with respect to x vanish. (ii) The same governing
equation (i) applies to an oscillating plate moving in a plane normal to y. (iii) Scaling estimate of
the penetration thickness G (t ) is independent of plate velocity U.
PROBLEM 2.27
An infinitely large plate is immersed in an infinite fluid at
uniform temperature Ti . The plate is suddenly maintained at
temperature To . Assume constant properties and neglect gravity.
[a] Show that the energy equation is given by
wT
wt
D
w 2T
wy 2
[b] Due to conduction, the effect of plate temperature propagates into the fluid. The penetration
depth G (t ) increases with time. Use scaling to derive an expression for G (t ) .
(1) Observations. (i) The plate is infinite. (ii) No changes take place in the axial direction
(infinite plate). (iii) This is a transient problem. (iv) Constant properties. (v) Cartesian
coordinates. (vi) Gravity is neglected. Thus there is no free convection. (vii) The fluid is
stationary.
(2) Problem Definition. [a] Determine the energy equation resulting from a step change in
surface temperature. [b] Use scaling to estimate the thermal penetration thickness G (t ).
(3) Solution Plan. [a] Apply energy equation and simplify it for the conditions of the problem.
[b] Assign scales to each variable in the governing equation for temperature distribution.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) constant properties, (3) stationary fluid, and (4)
negligible gravity.
(ii) Analysis. (i) Assumptions. (1) Continuum, (2) Newtonian fluid, (3) constant properties
and (4) negligible nuclear, radiation and electromagnetic energy transfer.
(ii) Analysis. The energy equation for this case is given by
§ wT
wT
wT
wT ·
¸
v
w
u
wx
wy
wz ¸¹
© wt
U c p ¨¨
where
cp
specific heat at constant pressure
k thermal conductivity
p pressure
T temperature
U density
) = dissipation function
Stationary fluid: u
v
w )
0
§ w 2 T w 2T w 2 T ·
k ¨ 2 2 2 ¸ P)
¨ wx
wy
wz ¸¹
©
(2.19b)
PROBLEM 2.27 (continued)
No axial variation:
w
wx
0
No variation in the z-direction:
w
wz
0
Substituting these simplifications into (2.19.b)
Uc p
wT
wt
w 2T
k
wy 2
(a)
Introduce the definition of thermal diffusivity D
D
cpP
k
(b)
(b) into (a)
wT
wt
D
w 2T
wy 2
(c)
[b] To obtain scaling estimate of G (t ) rewrite (i)
'T
't
D
' ª' T º
' y «¬ ' y »¼
(d)
Scale for ' T : ' T a (To Ti ) Scale for ' u : ' t a (t 0)
Scale for ' y : ' y a (G 0)
Substitute into (d)
To Ti
T T
a D o 2 i e t
G
Solve for G
G (t ) a D t f
(iii) Checking. Dimensional check: (i) Each term in (i) has units of. o C/s :
wT o
w 2T o
2
( C/s) D (m /s) 2 ( C/m 2 )
wt
wy
o
C/s
(ii) Units of (f) should be length:
G (t )
D (m 2 /s)t (s)
m
Limiting check: Initially the penetration thickness is zero. Setting t = 0 in (f) gives G (0)
which is the correct result.
0,
(5) Comments. (i) A major simplification of this problem is due to the assumption of infinite
plate. Consequently, all derivatives with respect to x vanish. (ii) Since the fluid is stationary, the
problem is one of pure conduction.
PROBLEM 3.1
A large plate moves with constant velocity U o parallel to a stationary plate separated by a
distance H. An incompressible fluid fills the channel formed by the plates. The stationary plate
is at temperature T1 and the moving plate is at temperature To . Taking into consideration
dissipation, determine the maximum temperature and
the heat flux at the moving plate. Assume laminar
flow and neglect gravity effect and pressure
variation in the channel.
(1) Observations. (i) Moving plate sets fluid in
motion in the x-direction. (ii) Since plates are infinite the flow field does not vary in the axial
direction x. (iii) The effect of pressure gradient is negligible. (iv) The fluid is incompressible
(constant density). (v) Use Cartesian coordinates.
(2) Problem Definition. Determine the velocity and temperature distribution.
(3) Solution Plan. Apply continuity and Navier-Stokes equations to determine the flow field.
Apply the energy equation to determine the temperature distribution.
(4) Plan Execution.
(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) constant properties (density, viscosity
and conductivity), (iv) infinite plates, (v) no end effects, (vi) uniform pressure (vi) negligible
gravitational effect.
(ii) Analysis. Since the objective is the determination of temperature distribution and heat
transfer rate, it is logical to begin the analysis with the energy equation. The energy equation for
constant properties is given by (2.19b)
§ wT
wT
wT
wT ·
¸
U c 5 ¨¨
u
v
w
wx
wy
wz ¸¹
© wt
§ w 2 T w 2T w 2 T ·
k ¨¨ 2 2 2 ¸¸ P)
wz ¹
wy
© wx
(2.19b)
where the dissipation function ) is given by (2.17)
)
ª
2 § wv · 2
2 º ª§ wu wv · 2 § wv ww · 2
2º
§ ww ·
§ ww wu · »
§ wu ·
¸¸ ¨
¸ 2«¨ ¸ ¨¨ ¸¸ ¨ ¸ » «¨¨ ¸¸ ¨¨ «© wx ¹
© wz ¹ » «© wy wx ¹
© wx wz ¹ »
© wy ¹
© wz wy ¹
¬
¼ ¬
¼
2
2 § wu wv ww ·
¸
¨ 3 ¨© wx wy wz ¸¹
(2.17)
Thus it is clear from (2.19b) and (2.17) that the determination of temperature distribution
requires the determination of the velocity components u , v and w. This is accomplished by
applying continuity and the Navier-Stokes equations. We begin with the continuity equation in
Cartesian coordinates
ª wu w v w w º
wU
wU
wU
wU
u
v
w
U« » 0
wt
wx
wy
wz
¬ wx wy wz ¼
For constant density
(2.2b)
PROBLEM 3.1 (continued)
wU
wt
wU
wx
w
wx
w
wz
wU
wy
wU
wz
0
(a)
Since plates are infinite
w 0
(b)
Substituting (a) and (b) into (2.2b), gives
wv
0
wy
Integrating (c)
(c)
v
f (x)
(d)
To determine the “constant” of integration f (x ) we apply the no-slip boundary condition at the
lower plate
v ( x,0) 0
(e)
Equations (d) and (e) give
f ( x) 0
Substituting into (d)
v 0
(f)
Since the vertical component v vanishes everywhere, it follows that the streamlines are parallel.
To determine the horizontal component u we apply the Navier-Stokes equation in the x-direction,
(2.10x)
§ wu
wu ·
wu
wu
w ¸¸
v
u
wz ¹
wy
wx
© wt
U ¨¨
Ug x § w 2u w 2u w 2u ·
wp
P¨ 2 2 2 ¸
¨ wx
wx
wy
wz ¸¹
©
(2.10x)
This equation is simplified as follows: Steady state
wu
wt
0
(g)
gx
0
(h)
wp
wx
0
(i)
0
(j)
Negligible gravity effect
Negligible axial pressure variation
Substituting (b) and (f)-(i) into (2.10x) gives
d 2u
dy 2
The solution to (j) is
u C1 y C 2
where C1 and C 2 are constants of integration. The two boundary conditions on u are:
u (0)
These conditions give
0 and u ( H ) U o
(k)
(l)
PROBLEM 3.1 (continued)
Uo
and C 2
H
C1
(m)
0
Substituting (m) into (k)
u
Uo
y
H
(3.8)
With the velocity distribution determined, we return to the dissipation function and energy
equation. Substituting (b) and (f) into (2.17) gives
§ wu ·
¨¨ ¸¸
© wy ¹
)
2
(n)
Using solution (3.8) into (n) gives
U o2
)
Noting that for steady state wT / wt
simplifies to
(o)
H2
0 and using (b), (f) and (o), the energy equation (2.10b)
k
d 2T
dy 2
P
U o2
H2
(p)
0
In arriving at (p) axial temperature variation was neglected. This is valid for infinite plates at
uniform surface temperature. Equation (p) is solved by direct integration
T
PU o2
2kH
2
y 2 C3 y C 4
(q)
where C3 and C 4 are constants of integration. The two boundary conditions on (q) are
T (0)
T1 and T ( H )
To
(r)
These boundary conditions and solution (q) give
C3
To T1 P U o2
and C 4
H
2 Hk
T1
(s)
Substituting (s) into (q) and rearranging the result in dimensionless form, give
T T1
To T1
PU o2
y
y· y
§
¨1 ¸
H 2k (To T1 ) ©
H¹H
(t)
This can be written in terms of the Eckert and Prandtl numbers as
T T1
To T1
cpP §
U o2
y
y· y
¨1 ¸
H 2c p (To T1 ) k ©
H¹H
T T1
To T1
where
y EPr §
y· y
¨1 ¸
H
2 © H¹H
(u)
PROBLEM 3.1 (continued)
E
U o2
c p (To T1 )
and Pr
cpP
(v)
k
The maximum temperature occurs where the temperature gradient is zero. Differentiating (u),
setting the result equal to zero and solving for position of maximum temperature y m
ym 1
1
(w)
H
2 EPr
Substituting (w) into (u) gives the maximum temperature Tm
Tm T1
To T1
EPr
1
1
2 2 EPr
8
The heat flux at the moving surface is determined by applying Fourier’s law at y
q cc( H )
k
(x)
H
dT ( H )
dy
Using (u) into the above
q cc( H )
k (To T1 ) ª EPr º
«¬ 2 1»¼
H
(y)
(iii) Checking. Dimensional check: Each term in (3.8), (t), (u), (w) and (x) is dimensionless.
Units of (y) should be W/m 2 :
q cc( H )
k ( W/m o C)(To T1 )( o C)
H (m)
W
m2
Differential equation check: Velocity solution (3.8) satisfies equation (j) and temperature
solution (t) satisfies (p).
Boundary conditions check: Velocity solution (3.8) satisfies boundary conditions (l) and
temperature solution (t) satisfies boundary conditions (r).
Limiting check: (i) If the upper plate is stationary the fluid will also be stationary. Setting U o
in (3.8) gives u ( y ) 0.
0
(ii) If the upper plate is stationary, dissipation will vanish, temperature distribution will be linear
and surface flux at the upper plate will be due to conduction between the two surfaces. Setting
U o 0 in (v) gives E = 0. When this is substituted into (u) and (y) gives the anticipated linear
temperature distribution and a surface flux of
q cc( H )
k (To T1 )
H
(iii) If the fluid is inviscid, dissipation will vanish and temperature should be linear. Setting
P 0 in (v ) gives Pr 0 . When this is substituted into (u) gives a linear temperature
distribution.
PROBLEM 3.1 (continued)
(5) Comments. (i) Treating the plate as infinite is one of the key simplifying assumptions. This
eliminates the x-coordinate as a variable and results in governing equations that are ordinary.
Alternatively, one could state that the streamline are parallel. This means that v wv / wy 0.
substituting this into the continuity equation for two-dimensional incompressible flow gives
wu / wx 0. This is identical equation (b) which is based on assuming infinite plate.
(ii) According to (w), maximum temperature occurs in the upper half of the channel.
T (0) To
PU o2
2k
PROBLEM 3.2
A large plate moves with constant velocity U o
parallel to a stationary plate separated by a
distance H. An incompressible fluid fills the
channel formed by the plates. The upper plate
is maintained at uniform temperature To and
the stationary plate is insulated. A pressure gradient dp / dx is applied to the fluid. Taking into
consideration dissipation, determine the temperature of the insulated plate and the heat flux at
the upper plate. Assume laminar flow and neglect gravity effect.
(1) Observations. (i) Moving plate sets fluid in motion in the x-direction. (ii) Since plates are
infinite the flow field does not vary in the axial direction x. (iii) The effect of pressure gradient
must be included. (iv) The fluid is incompressible. (v) Using Fourier’s law, Temperature
distribution gives surface heat flux of the moving plate. (vi) Use Cartesian coordinates.
(2) Problem Definition. Determine the velocity and temperature distribution.
(3) Solution Plan. Apply continuity and Navier-Stokes equations to determine the flow field.
Apply the energy equation to determine the temperature distribution.
(4) Plan Execution.
(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) constant properties (density, viscosity
and conductivity), (iv) infinite plates, (v) no end effects and (v) negligible gravitational effect.
(ii) Analysis. Since the objective is the determination of temperature distribution and heat
transfer rate, it is logical to begin the analysis with the energy equation. The energy equation for
constant properties is given by (2.19b)
§ wT
wT
wT
wT ·
¸
u
v
w
wx
wy
wz ¸¹
© wt
U c 5 ¨¨
§ w 2 T w 2T w 2 T ·
k ¨¨ 2 2 2 ¸¸ P)
wy
wz ¹
© wx
(2.19b)
where the dissipation function ) is given by (2.17)
)
ª
2 § wv · 2
2 º ª§ wu wv · 2 § wv ww · 2
2º
§ ww wu · »
§ wu ·
§ ww ·
¸¸ ¨
2«¨ ¸ ¨¨ ¸¸ ¨ ¸ » «¨¨ ¸¸ ¨¨ ¸ «© wx ¹
© wx wz ¹ »
© wz ¹ » «© wy wx ¹
© wz wy ¹
© wy ¹
¼
¬
¼ ¬
2
2 § wu wv ww ·
¸¸
¨¨ 3 © wx wy wz ¹
(2.17)
Thus it is clear from (2.19b) and (2.17) that the determination of temperature distribution
requires the determination of the velocity components u , v and w. This is accomplished by
applying continuity and the Navier-Stokes equations. We begin with the continuity equation in
Cartesian coordinates
ª wu w v w w º
wU
wU
wU
wU
(2.2b)
u
v
w
U« » 0
wt
wx
wy
wz
¬ wx wy wz ¼
For constant density
wU wU wU wU
(a)
0
w t wx wy wz
PROBLEM 3.2 (continued)
Since plates are infinite
w
wx
w
wz
w 0
(b)
wv
wy
0
(c)
f (x)
(d)
Substituting (a) and (b) into (2.2b), gives
Integrating (c)
v
To determine the “constant” of integration f (x) , we apply the no-slip boundary condition at the
lower plate
v ( x,0) 0
(e)
Equations (d) and (e) give
f ( x) 0
Substituting into (d)
v 0
(f)
Since the vertical component v vanishes everywhere, it follows that the streamlines are parallel.
To determine the horizontal component u we apply the Navier-Stokes equations (2.10)
§ wu
wu
wu
wu ·
v
w ¸¸
u
wx
wy
wz ¹
© wt
Ug x § w 2u w 2u w 2u ·
wp
P ¨¨ 2 2 2 ¸¸
wx
wz ¹
wy
© wx
(2.10x)
§ wv
wv
wv
wv ·
u
v
w ¸¸
wx
wy
wz ¹
© wt
Ug y § w 2v w 2v w 2v ·
wp
P¨ 2 2 2 ¸
¨ wx
wy
wy
wz ¸¹
©
(2.10y)
U ¨¨
U ¨¨
These equations are simplified as follows: Steady state
wu
wt
(g)
0
Negligible gravity effect
gx
gy
0
(h)
Substituting (b) and (f)-(h) into (2.10x) and (2.10y) gives
wp
wx
P
d 2u
dy 2
(i)
and
wp
(j)
0
wy
Equation (j) shows that pressure does not vary in the y-direction and thus it can either be a
function of x or constant. Integrating (i) twice
u
1 dp 2
y C1 y C2
2 P dx
where C1 and C 2 are constants of integration. The two boundary conditions on u are:
(k)
PROBLEM 3.2 (continued)
(1) u (0)
0
(2) u ( H ) U o
These conditions give
C1
U o H dp
and C 2
H 2 P dx
0
(l)
y ª
H 2 dp §
y ·º
¨1 ¸»
«1 H ¬ 2 PU o dx ©
H ¹¼
(m)
Substituting (l) into (k)
u
Uo
With the velocity distribution determined, we return to the dissipation function and energy
equation. Substituting (b) and (f) into (2.17) gives
)
§ wu ·
¨¨ ¸¸
© wy ¹
2
(n)
Using solution (m) into (n) gives
ªU o H dp 1 dp º
« H 2 P dx P dx y »
¬
¼
)
Noting that for steady state wT / wt
simplifies to
2
(o)
0 and using (b), (f) and (o), the energy equation (2.10b)
ªU
d 2T
H dp 1 dp º
k 2 P« o y»
dy
¬ H 2 P dx P dx ¼
2
(p)
0
In arriving at (p) axial temperature variation was neglected. This is valid for infinite plates at
uniform surface temperature. Equation (p) is solved by direct integration
T
2
P ªU o
2
H dp º 2 P ªU o H dp º 1 dp 3 P ª 1 dp º 4
«
y «
y y B1 y B2
2k ¬ H 2 P dx »¼
3k ¬ H 2 P dx »¼ P dx
12k «¬ P dx »¼
(q)
where B1 and B2 are constants of integration. The two boundary conditions on (q) are
dT (0)
dy
0
(2) T ( H )
To
(1)
These boundary conditions and solution (q) give
B1
B2
P ªU o
(r)
0
2
2
H dp º 2 P ªU o H dp º 1 dp 3 P ª 1 dp º 4
To H «
H H
«
2k ¬ H 2 P dx »¼
3k ¬ H 2 P dx »¼ P dx
12k «¬ P dx »¼
Substituting (r) into (q) and rearranging the result in dimensionless form, give
(s)
PROBLEM 3.2 (continued)
2
T To
H4
Pk
ª dp º
«¬ dx »¼
2
ª
º
ª
º
2
«
»
«
Ȥ
§
·
1 2 PU o
y3 · 1 §
y4 ·
y
1 2 PU o
¨
¸
¨
¸
¨
¸
1
1
1
1
1
«
»
«
»
8 « H 2 dp » ¨© H 2 ¸¹ 8 « H 2 dp »¨© H 3 ¸¹ 12 ¨© H 4 ¸¹
dx ¼
dx ¼
¬
¬
(t)
Surface temperature of the insulated plate, T(0), is obtained by setting y = 0 in (t)
2
ª
º
ª
º
» 1 « 2 PU o
» 1
1 « 2 PU o
1» «
1» «
8 « H 2 dp » 8 « H 2 dp » 12
dx ¼
dx ¼
¬
¬
T (0) To
4
H ª dp º
Pk «¬ dx »¼
2
(u)
Surface heat flux at the upper plate is obtained by applying Fourier’s law at y = H
q cc( H )
k
dT ( H )
dy
Using (t) into the above
2
q cc( H )
H 3 ª dp º
P «¬ dx »¼
2
º
ª
º
ª
» 1
»
1 « 2 PU o
1 « 2 PU o
1» 1» «
«
4 « H 2 dp »
2 « H 2 dp » 3
dx ¼
dx ¼
¬
¬
(v)
(iii) Checking. Dimensional check: Each term in (m), (t), (u) and (v) is dimensionless. Each
term in (q) has units of temperature.
Differential equation check: Velocity solution (m) satisfies equation (i) and temperature solution
(t) satisfies (p).
Boundary conditions check: Velocity solution (m) and temperature solution (t) satisfy their
respective boundary conditions.
Limiting check: (i) If the upper plate is stationary and there is no axial pressure gradient the fluid
will also be stationary. Setting U o dp / dx 0 in (m) gives u ( y ) 0.
(ii) If the upper plate is stationary and there is no axial pressure gradient, there will be no fluid
motion and dissipation will vanish. The temperature distribution will be uniform equal to To .
Setting U o dp / dx 0 in (t) gives T ( y ) To . Similarly, surface heat flux will vanish. Setting
U o dp / dx 0 in (v) gives qcc( H ) 0.
(5) Comments. (i) Treating the plate as infinite is one of the key simplifying assumptions. This
eliminates the x-coordinate as a variable and results in governing equations that are ordinary. (ii)
According to (t), maximum temperature occurs at the insulated plate. (iii) According to the
dimensionless form of solutions (u) and (v), the problems is characterized by the single
dimensionless parameter
2 PU o
dp
H2
dx
PROBLEM 3.3
Incompressible fluid is set in motion between two
large parallel plates by moving the upper plate with
constant velocity U o and holding the lower plate
stationary. The clearance between the plates is H.
The lower plate is insulated while the upper plate
exchanges heat with the ambient by convection. The heat transfer coefficient is h and the
ambient temperature is Tf . Taking into consideration dissipation determine the temperature of
the insulated plate and the heat flux at the moving plate. Assume laminar flow and neglect
gravity effect.
(1) Observations. (i) Moving plate sets fluid in motion in the x-direction. (ii) Since plates are
infinite the flow field does not vary in the axial direction x. (iii) The fluid is incompressible
(constant density). (iv) Use Cartesian coordinates.
(2) Problem Definition. Determine the velocity and temperature distribution.
(3) Solution Plan. Apply continuity and Navier-Stokes equations to determine the flow field.
Apply the energy equation to determine the temperature distribution.
(4) Plan Execution.
(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) constant properties (density, viscosity
and conductivity), (iv) infinite plates, (v) no end effects, (vi) uniform pressure (vi) negligible
gravitational effect.
(ii) Analysis. Since the objective is the determination of temperature distribution and heat
transfer rate, it is logical to begin the analysis with the energy equation. The energy equation for
constant properties is given by (2.19b)
§ wT
wT
wT
wT ·
¸
u
v
w
wx
wy
wz ¸¹
© wt
U c 5 ¨¨
§ w 2 T w 2T w 2 T ·
k ¨¨ 2 2 2 ¸¸ P)
wy
wz ¹
© wx
(2.19b)
where the dissipation function ) is given by (2.17)
)
2
2
ª
2 § wv · 2
2º ª
2º
§ wv ww ·
§ ww · » «§ wu wv ·
§ ww wu · »
§ wu ·
«
¸¸ ¨
¸ 2 ¨ ¸ ¨¨ ¸¸ ¨ ¸ ¨¨ ¸¸ ¨¨ «© wx ¹
© wz ¹ » «© wy wx ¹
© wx wz ¹ »
© wy ¹
© wz wy ¹
¬
¼ ¬
¼
2
2 § wu wv ww ·
¸¸
¨¨ 3 © wx wy wz ¹
(2.17)
Thus it is clear from (2.19b) and (2.17) that the determination of temperature distribution
requires the determination of the velocity components u , v and w. This is accomplished by
applying continuity and the Navier-Stokes equations. We begin with the continuity equation in
Cartesian coordinates
ª wu w v w w º
wU
wU
wU
wU
u
v
w
U« » 0
wt
wx
wy
wz
¬ wx wy wz ¼
For constant density
(2.2b)
PROBLEM 3.3 (continued)
wU
wt
wU
wx
wU
wy
wU
wz
0
(a)
Since plates are infinite
w
wx
w
wz
w 0
(b)
wv
wy
0
(c)
f (x)
(d)
Substituting (a) and (b) into (2.2b), gives
Integrating (c)
v
To determine the “constant” of integration f (x ) we apply the no-slip boundary condition at the
lower plate
v ( x,0) 0
(e)
Equations (d) and (e) give
f ( x) 0
Substituting into (d)
v 0
(f)
Since the vertical component v vanishes everywhere, it follows that the streamlines are parallel.
To determine the horizontal component u we apply the Navier-Stokes equation in the x-direction,
(2.10x)
§ wu
wu ·
wu
wu
w ¸¸
v
u
wz ¹
wy
wx
© wt
U ¨¨
Ug x § w 2u w 2u w 2u ·
wp
P¨ 2 2 2 ¸
¨ wx
wx
wy
wz ¸¹
©
(2.10x)
This equation is simplified as follows: Steady state
wu
wt
0
(g)
gx
0
(h)
wp
wx
0
(i)
0
(j)
Negligible gravity effect
Negligible axial pressure variation
Substituting (b) and (f)-(i) into (2.10x) gives
d 2u
dy 2
The solution to (j) is
u
C1 y C 2
(k)
where C1 and C 2 are constants of integration. The two boundary conditions on u are:
u (0)
These conditions give
0 and u ( H ) U o
(l)
PROBLEM 3.3 (continued)
Uo
and C 2
H
C1
(m)
0
Substituting (m) into (k)
u
Uo
y
H
(3.8)
With the velocity distribution determined, we return to the dissipation function and energy
equation. Substituting (b) and (f) into (2.17) gives
§ wu ·
¨¨ ¸¸
© wy ¹
)
2
(n)
Using solution (3.8) into (n) gives
U o2
)
Noting that for steady state wT / wt
simplifies to
(o)
H2
0 and using (b), (f) and (o), the energy equation (2.10b)
d 2T
k
P
dy 2
U o2
H2
0
(p)
In arriving at (p) axial temperature variation was neglected. This is valid for infinite plates at
uniform surface temperature. Equation (p) is solved by direct integration
T
PU o2
2kH
2
y 2 C3 y C 4
(q)
where C3 and C 4 are constants of integration. The two boundary conditions on (q) are
(1)
dT (0)
dy
(2) k
0
dT ( H )
dy
h>T ( y ) Tf @
These boundary conditions and solution (q) give
C3
C4
Tf 0
PU o2
khH
(r)
PU o2
2k
(s)
Substituting (r) and (s) into (q) and rearranging the result in dimensionless form, give
T Tf
PU o2
k
1 k
1 y2
2 hH 2 H 2
(t)
The dimensionless parameter hH / k is known as the Biot number, Bi . It is associated with
convection boundary conditions.
PROBLEM 3.3 (continued)
The temperature of the insulated surface, T(0), is obtained by evaluating (t) at y = 0.
T (0) Tf 1 k
2 hH
PU o2
k
The heat flux at the moving surface is determined by applying Fourier’s law at y
qcc( H )
k
(u)
H
dT ( H )
dy
Substituting (t) into the above
PU o2
qcc( H )
H
(v)
(iii) Checking. Dimensional check: Each term in (3.8), (t) and (u) is dimensionless. Each term
in solution (v) has units of W/m 2 .
Differential equation check: Velocity solution (3.8) satisfies equation (j) and temperature
solution (t) satisfies (p).
Boundary conditions check: Velocity solution (3.8) satisfies boundary conditions (l) and
temperature solution (t) satisfies boundary conditions (r).
Limiting check: (i) If the upper plate is stationary the fluid will also be stationary. Setting U o
in (3.8) gives u ( y ) 0.
0
(ii) If the upper plate is stationary, dissipation will vanish, temperature distribution will be
uniform equal to the ambient temperature Tf . Setting U o 0 in (u) gives T ( y ) Tf . Similarly
the heat flux qcc( H ) vanishes. Substituting U o 0 (v) gives qcc( H ) 0.
(iii) If the fluid is inviscid, dissipation will vanish and temperature should be uniform equal to
Tf . Setting P 0 in (u) gives T ( y ) Tf . Similarly the heat flux qcc( H ) vanishes. Substituting
P 0 (v) gives qcc( H ) 0.
Global energy balance: Energy leaving the channel must equal to the work done to move the
plate. Consider the work done by the plate on the fluid
W cc W oU o
(w)
where
W cc
Wo
work done per unit surface area by the plate on the fluid
shearing stress at the moving plate
However, shearing stress is given by
Wo
P
wu( H )
wy
(x)
Uo
H
(y)
(3.8) into (x)
Wo
(y) into (w)
P
PROBLEM 3.3 (continued)
W cc
PU o2
(z)
H
This is identical to the heat removed from the upper plate given in equation (v).
(5) Comments. (i) Treating the plate as infinite is one of the key simplifying assumptions. This
eliminates the x-coordinate as a variable and results in governing equations that are ordinary.
Alternatively, one could state that the streamline are parallel. This means that v wv / wy 0. (ii)
The solutions is characterized by a single dimensionless parameter hH / k , which is the Biot
number. (iii) The Nusselt number at moving plate, Nu(H ), is defined as
hH
k
Nu( H )
(w)
The heat transfer coefficient h is based on the overall temperature drop, defined as
h
qcc( H )
T (0) T ( H )
(x)
Centerline temperature and moving plate temperature are obtained by evaluating (t) at y = 0 and
y=H
PU o2 ª 1 k º
T (0) Tf
k «¬ 2 hH »¼
and
1 PU o2
T ( H ) Tf
2 Hh
The above two equations give
T ( 0) T ( H )
1 PU o2
2 k
(y)
(v) and (y) into (x)
h
2
k
H
Substituting into (w) gives the Nusselt number
Nu( H )
2
(z)
PROBLEM 3.4
Two parallel plates are separated by a
distance 2H. The plates are moved in
opposite direction with constant velocity U o .
Each plate is maintained at uniform
temperature To . Taking into consideration
dissipation determine the heat flux at the
plates. Assume laminar flow and neglect
gravity effect
(1) Observations. (i) Moving plates set fluid in motion in the positive and negative x-direction.
(ii) Since plates are infinite the flow field does not vary in the axial direction x. (iii) The fluid is
incompressible (constant density). (iv) The fluid is stationary at the center plane y = 0. (v)
Symmetry dictates that no heat is conducted through the center plane. (vi) Use Cartesian
coordinates.
(2) Problem Definition. Determine the velocity and temperature distribution.
(3) Solution Plan. Apply continuity and Navier-Stokes equations to determine the flow field.
Apply the energy equation to determine the temperature distribution.
(4) Plan Execution.
(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) constant properties (density, viscosity
and conductivity), (iv) infinite plates, (v) no end effects, (vi) uniform pressure (vi) negligible
gravitational effect.
(ii) Analysis. Taking advantage of symmetry only the upper half of the channel is analyzed.
Since the objective is the determination of temperature distribution and heat transfer rate, it is
logical to begin the analysis with the energy equation. The energy equation for constant
properties is given by (2.19b)
§ wT
wT
wT
wT ·
¸
u
v
w
wx
wy
wz ¸¹
© wt
U c 5 ¨¨
§ w 2 T w 2T w 2 T ·
k ¨¨ 2 2 2 ¸¸ P)
wy
wz ¹
© wx
(2.19b)
where the dissipation function ) is given by (2.17)
)
ª
2 § wv · 2
2 º ª§ wu wv · 2 § wv ww · 2
2º
§ ww ·
§ ww wu · »
§ wu ·
¸¸ ¨
¸ 2«¨ ¸ ¨¨ ¸¸ ¨ ¸ » «¨¨ ¸¸ ¨¨ «© wx ¹
© wz ¹ » «© wy wx ¹
© wx wz ¹ »
© wy ¹
© wz wy ¹
¬
¼ ¬
¼
2
2 § wu wv ww ·
¸
¨ 3 ¨© wx wy wz ¸¹
(2.17)
Thus it is clear from (2.19b) and (2.17) that the determination of temperature distribution
requires the determination of the velocity components u , v and w. This is accomplished by
applying continuity and the Navier-Stokes equations. We begin with the continuity equation in
Cartesian coordinates
ª wu w v w w º
wU
wU
wU
wU
u
v
w
U« » 0
wt
wx
wy
wz
¬ wx wy wz ¼
(2.2b)
PROBLEM 3.4 (continued)
For constant density
wU
wt
wU
wx
wU
wy
wU
wz
0
(a)
Since plates are infinite
w
wx
w
wz
w 0
(b)
wv
wy
0
(c)
f (x)
(d)
Substituting (a) and (b) into (2.2b), gives
Integrating (c)
v
To determine the “constant” of integration f (x ) we apply the no-slip boundary condition at the
lower plate
v ( x,0) 0
(e)
Equations (d) and (e) give
f ( x) 0
Substituting into (d)
v 0
(f)
Since the vertical component v vanishes everywhere, it follows that the streamlines are parallel.
To determine the horizontal component u we apply the Navier-Stokes equation in the x-direction,
(2.10x)
§ wu
wu ·
wu
wu
w ¸¸
v
u
wz ¹
wy
wx
© wt
U ¨¨
Ug x § w 2u w 2u w 2u ·
wp
P¨ 2 2 2 ¸
¨ wx
wx
wy
wz ¸¹
©
(2.10x)
This equation is simplified as follows: Steady state
wu
wt
0
(g)
gx
0
(h)
wp
wx
0
(i)
0
(j)
Negligible gravity effect
Negligible axial pressure variation
Substituting (b) and (f)-(i) into (2.10x) gives
d 2u
dy 2
The solution to (j) is
u
C1 y C 2
(k)
where C1 and C 2 are constants of integration. The two boundary conditions on u are:
u (0)
0 and u ( H ) U o
(l)
PROBLEM 3.4 (continued)
These conditions give
Uo
and C 2
H
C1
(m)
0
Substituting (m) into (k)
u
Uo
y
H
(3.8)
With the velocity distribution determined, we return to the dissipation function and energy
equation. Substituting (b) and (f) into (2.17) gives
§ wu ·
¨¨ ¸¸
© wy ¹
)
2
(n)
Using solution (3.8) into (n) gives
U o2
)
Noting that for steady state wT / wt
simplifies to
(o)
H2
0 and using (b), (f) and (o), the energy equation (2.10b)
k
d 2T
dy 2
U o2
P
H2
0
(p)
In arriving at (p) axial temperature variation was neglected. This is valid for infinite plates at
uniform surface temperature. Equation (p) is solved by direct integration
T
PU o2
2kH 2
y 2 C3 y C 4
(q)
where C3 and C 4 are constants of integration. The two boundary conditions on (q) are
dT (0)
dy
0
(2) T ( H )
To
(1)
These boundary conditions and solution (q) give
C3
C4
(r)
0
To PU o2
2k
(s)
Substituting (r) and (s) into (q) and rearranging the result in dimensionless form, give
T To
PU o2
k
1ª
y2 º
1
«
»
2¬ H2¼
The temperature of the insulated surface, T(0), is obtained by evaluating (t) at y = 0.
(t)
PROBLEM 3.4 (continued)
The heat flux at the moving surface is determined by applying Fourier’s law at y
qcc( H )
k
H
dT ( H )
dy
Substituting (t) into the above
qcc( H )
PU o2
H
(u)
Similarly, the heat flux at the lower plate is
qcc( H )
PU o2
H
(v)
(iii) Checking. Dimensional check: Each term in (3.8) and (t) is dimensionless. Each term in
solutions (u) and (v) has units of W/m 2 .
Differential equation check: Velocity solution (3.8) satisfies equation (j) and temperature
solution (t) satisfies (p).
Boundary conditions check: Velocity solution (3.8) and temperature solution (t) satisfy their
respective boundary conditions.
Limiting check: (i) If the two plates are stationary, the fluid will also be stationary. Setting
U o 0 in (3.8) gives u ( y ) 0.
(ii) If the upper and lower plates are stationary, dissipation will vanish, temperature distribution
will be uniform equal to the ambient temperature To . Setting U o 0 in (t) gives T ( y ) To .
Similarly the heat flux qcc(H ) vanishes. Substituting U o 0 (u) gives qcc( H ) 0.
(iii) If the fluid is inviscid, dissipation will vanish and temperature should be uniform equal to
To . Setting P 0 in (t) gives T ( y ) To . Similarly the heat flux qcc(H ) vanishes. Substituting
P 0 (u) gives qcc( H ) 0.
(5) Comments. (i) Treating the plate as infinite is one of the key simplifying assumptions. This
eliminates the x-coordinate as a variable and results in governing equations that are ordinary.
Alternatively, one could state that the streamline are parallel. This means that v wv / wy 0. (ii)
Symmetry provides additional simplification. (iii) The Nusselt number at the upper moving
plate, Nu(H ), is defined as
hH
k
(w)
qcc( H )
T (0) To
(x)
Nu( H )
Defining the heat transfer coefficient is defined as
h
Centerline temperature is obtained by evaluating (t) at y = 0
T (0) To
1 PU o2
2 k
(y)
PROBLEM 3.4 (continued)
(u) and (y) into (x) gives h
h
2
k
H
Substituting into (w) gives the Nusselt number
Nu( H )
2
(z)
A more appropriate definition of the heat transfer coefficient is based on the mean temperature,
Tm , rather than the temperature at the center. That is
h
q cc( H )
Tm To
PROBLEM 3.5
Incompressible fluid flows in a long tube of radius
ro . Fluid motion is driven by an axial pressure
gradient wp / wz. The tube exchanges heat by
convection with an ambient fluid. The heat transfer
coefficient is h and the ambient temperature is Tf .
Taking into consideration dissipation, assuming
laminar incompressible axisymmetric flow, and neglecting gravity, axial temperature variation
and end effects, determine:
[a] Surface temperature.
[b] Surface heat flux.
[c] Nusselt number based on [ T (0) T (ro ) ].
(1) Observations. (i) Fluid motion is driven by axial pressure drop. (ii) For a very long tube the
flow field does not vary in the axial direction z. (iii) The fluid is incompressible (constant
density). (iv) Heat is generated due to viscous dissipation. It is removed from the fluid by
convection at the surface. (v) The Nusselt number is a dimensionless heat transfer coefficient.
(vi) To determine surface heat flux and heat transfer coefficient requires the determination of
temperature distribution. (vii) Temperature distribution depends on the velocity distribution.
(viii) Use cylindrical coordinates.
(2) Problem Definition. Determine the velocity and temperature distribution.
(3) Solution Plan. Apply continuity and Navier-Stokes equations in cylindrical coordinates to
determine the flow field. Apply the energy equation to determine temperature distribution.
Fourier’s law gives surface heat flux. Equation (1.10) gives the heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) axisymmetric flow, (iv) constant
properties (density, viscosity and conductivity), (v) no end effects, (vi) uniform ambient
temperature, (vii) uniform heat transfer coefficient and (viii) negligible gravitational effect.
(ii) Analysis. [a] Since temperature distribution is obtained by solving the energy equation,
we begin the analysis with the energy equation. The energy equation in cylindrical coordinates
for constant properties is given by (2.24)
ª 1 w § wT · 1 w 2T w 2T º
wT v T wT
wT ·
§ wT
2 » P)
vr
vz
¸ k«
¨r
¸ 2
2
r wT
wr
wz ¹
wz ¼
© wt
¬ r wr © wr ¹ r w 0
U cP ¨
(2.24)
where the dissipation function ) is given by (2.25)
2
)
2
2
2
v
1 wv r ·
§ 1 wv T v r ·
§ wv · § wv
§ wv ·
2¨ r ¸ 2¨
¸ 2¨ z ¸ ¨ 0 T ¸ r ¹
r
r w0 ¹
© wz ¹ © wr
© wr ¹
© r wT
2
wv ·
§ 1 wv z wv 0 ·
§ wv
¨
¸ ¨ r z¸
wz ¹
wr ¹
© wz
© r w0
2
(2.25)
Equations (2.24) and (2.25) show that the determination of temperature distribution requires the
determination of the velocity components v r , v T and v z . The flow field is determined by
PROBLEM 3.5 (continued)
solving the continuity and the Navier-Stokes equations. We begin with the continuity equation in
cylindrical coordinates
wU 1 w
1 w
w
U rv r U vT Uv z
r wT
wt r wr
wz
0
(2.4)
For constant density
wU
wt
wU
wr
wU
wT
wU
wz
vT
w
wT
0
0
(a)
For axisymmetric flow
(b)
For a long tube with no end effects axial changes in velocity are negligible
w
wz
0
(c)
Substituting (a)-(c) into (2.4)
d
rv r
dr
0
(d)
f (z )
(e)
Integrating (d)
rv r
To determine the “constant” of integration f (z ) we apply the no-slip boundary condition at the
surface
v (ro , z ) 0
(f)
Equations (e) and (f) give
f ( z) 0
Substituting into (e)
vr 0
(g)
Since the radial component v r vanishes everywhere, it follows that the streamlines are parallel
to the surface. To determine the axial component v z we apply the Navier-Stokes equation in the
z-direction, (2.11z)
wv z v T wv z
wv z wv z ·
vz
¸
wr
wz
wt ¹
r wT
©
ª 1 w § wv z · 1 w 2 v z w 2 v z º
wp
P«
Ug z ¸ 2
¨r
»
2
wz
wz 2 ¼»
¬« r wr © wr ¹ r wT
This equation is simplified as follows:
§
U¨ v r
(2.11z)
Steady state
w
wt
0
(h)
Negligible gravity effect
gr
Substituting (b), (c) and (g)-(i) into (2.11z) gives
gz
0
(i)
PROBLEM 3.5 (continued)
wp
1 d § dv z ·
P
¨r
¸ 0
wz
r dr © dr ¹
(3.11)
Since v z depends on r only, equation (3.11) can be written as
wp
wz
1 d § dv z ·
¨r
¸
r dr © dr ¹
P
g (r )
(j)
Integrating (j) with respect to z
g (r ) z C o
p
(k)
where C o is constant of integration. We turn our attention now to the radial component of
Navier-Stokes equation, (2.11r)
2
§ wv
wv
wv ·
v wv r v T
vz r r ¸
U¨ v r r T
¨
wr
wz
wt ¸¹
r wT
r
©
(2.11r)
2
2
ª w §1 w
w vr º
wp
2 wv
· 1 w vr
P« ¨
2 T (rv r ) ¸ 2
Ug r »
2
wr
r wT
wz 2 ¼»
¹ r wT
¬« wr © r wr
Substituting (b), (g) and (i) into (2.11r), gives
wp
wr
0
(l)
f (z )
(m)
Integrating (l)
p
where f (z ) is “constant” of integration. We now have two solutions for the pressure p: (k) and
(m). Equating the two, gives
p
g (r ) z C o
f ( z)
(n)
One side of (n) shows that the pressure depends on z only while the other side shows that it
depends on r and z. This, of course, is a contradiction. The only possibility for reconciling this is
by requiring that
g(r) = C
(o)
where C is a constant. Substituting (o) into (j)
wp
wz
P
1 d § dv z ·
¸
¨r
r dr © dr ¹
C
Thus the axial pressure gradient in the tube is constant. Equation (p) can now be integrated to
give the axial velocity distribution. Integrating once
r
dv z
dr
Separating variables and integrating again
1 dp 2
r C1
2P d z
(p)
PROBLEM 3.5 (continued)
vz
1 dp 2
r C1 ln r C 2
4P d z
(q)
where C1 and C 2 are constants of integration. The two boundary conditions on v z are
dv z ( 0)
dr
0, v z (ro )
0
(r)
Equations (q) and (r) give C1 and C 2
C1
0, C 2
1 dp 2
ro
4P d z
Substituting into (q)
1 dp 2
(3.12)
(r ro2 )
4P d z
With the velocity distribution determined we return to the energy equation (2.24) and the
dissipation function (2.25). We note that for a long tube at uniform surface temperature with no
end effects, axial temperature variation can be neglected. Thus
vz
w 2T
wz 2
wT
wz
(s)
0
Substituting (b), (c), (g), (h) and (s) into (2.24)
k
1 d § dT ·
¸ P)
¨r
r dr © dr ¹
0
(t)
Using (b), (c) and (g) into (2.25) gives the dissipation function for this flow
)
§ dv z ·
¨
¸
© dr ¹
2
Substituting the velocity solution (3.11) into the above, gives
2
)
§ 1 d p· 2
¸¸ r
¨¨
© 2P d z ¹
(u)
Using (u) to eliminate ) in (t) and rearranging, we obtain
d § dT ·
¨r
¸
dr © dr ¹
2
1 §dp· 3
¸ r
¨
4kP ¨© d z ¸¹
Integrating the above twice
2
T
1 §d p· 4
¨
¸ r C 3 ln r C 4
64kP ¨© d z ¸¹
(v)
Two boundary conditions are needed to evaluate the constants of integration C 3 and C 4 . They
are:
dT (0)
(1)
0
dr
PROBLEM 3.5 (continued)
(2) k
dT (ro )
dr
h[T (ro ) Tf ]
Equations (v) and the two boundary conditions give the two constants
2
r4 § d p ·
k
¸¸ [4
1]
Tf o ¨¨
hro
64kP © d z ¹
0, C 4
C3
Substituting the above into (v)
2
T
r4 § d p ·
k
r4 § d p ·
¸¸ [4
¨
¸
1] Tf o ¨¨
64kP © d z ¹
hro
64kP ¨© d z ¸¹
2
(w)
This solution can be expressed in dimensionless form as
T Tf
ro4
§dp·
¨
¸
64kP ¨© d z ¸¹
The dimensionless parameter
4
2
k
r4
1 4
hro
ro
(x)
hro
in (x) is known as the Biot number.
k
Surface temperature is obtained by setting r
ro in (w)
2
T (ro )
ro3 § d p · 1
¨
¸
Tf 16 P ¨© d z ¸¹ h
(y)
[b] Surface heat flux q cc(ro ) is obtained by applying Fourier’s law
q cc(ro )
dT (ro )
dr
k
Using (w) into the above
ro3 § d p ·
¨
¸
q cc(ro )
16 P ¨© d z ¸¹
2
(z)
[c] The Nusselt number is defined as
hD 2hro
(z-1)
k
k
where D is tube diameter. The heat transfer coefficient h is determined using equation (1.10)
Nu
h
dT (ro )
k
[T (0) T (ro )] dr
(z-2)
4k
ro
(z-3)
Substituting (w) into the above
h
Introducing (z-3) into (z-1)
Nu
8
(z-4)
PROBLEM 3.5 (continued)
(iii) Checking. Dimensional check: Each term in (3.12) has units of velocity. Each term in
(w) has units of temperature. Each term in (x) is dimensionless.
Differential equation check: Velocity solution (3.12) satisfies equation (p) and temperature
solution (w) satisfies (t).
Boundary conditions check: Velocity solution (3.12) satisfies boundary conditions (r) and
temperature solution (w) satisfies boundary conditions (1) and (2).
Limiting check: (i) If pressure is uniform ( dp / dz
dp / dz 0 in (3.12) gives v z 0.
0 ) the fluid will be stationary. Setting
(ii) If pressure is uniform ( dp / dz 0 ) the fluid will be stationary and no dissipation takes place
and thus surface heat transfer should vanish Setting dp / dz 0 in (z) gives q cc(ro ) 0.
(iii) Global conservation of energy. Heat transfer rate leaving the tube must be equal to the work
required to pump the fluid. Pump work for a tube section of length L is
( p1 p 2 )Q
W
(z-1)
Where
p1 = upstream pressure
p 2 = downstream pressure
Q = volumetric flow rate, given by
ro
2S
Q
³ v rdr
z
0
Substituting (3.12) into the above and integrating
Q
S dp 4
ro
8P dz
(z-2)
Combining (z-1) and (z-2))
W
S ro4 dp
( p1 p 2 )
8P dz
(z-3)
Work per unit area W cc is
W cc
W
2S ro L
Substituting (z-3) into the above
W cc
ro3 dp ( p1 p 2 )
L
16P dz
However
( p1 p 2 )
L
dp
dz
Combining this result with (z-4) gives
W cc
ro3 § dp ·
¨ ¸
16 P © dz ¹
2
(z-4)
PROBLEM 3.5 (continued)
This result is identical to surface heat transfer rate given in (z)
(5) Comments. (i) The assumption of a long tube with negligible end effects is a key factor in
simplifying the problem. This assumption eliminates the z-coordinate as a variable and results in
governing equations that are ordinary. Assuming parallel streamlines is equivalent to neglecting
end effects since it leads to the same mathematical simplifications.
(ii) Solution (w) shows that maximum temperature occurs at the center r 0.
(iii) The Nusselt number is constant independent of Reynolds and Prandtl numbers.
PROBLEM 3.6
Fluid flows axially in the annular space
between a cylinder and a concentric rod . The
radius of the rod is ri and that of the cylinder
is ro . Fluid motion in the annular space is
driven by an axial pressure gradient wp / wz.
The cylinder is maintained at uniform
temperature To . Assume incompressible laminar axisymmetric flow and neglect gravity and end
effects. Show that the axial velocity is given by
º
1 (ri / ro ) 2
ro2 dp ª
2
(
/
)
ln(r / ro ) 1»
r
r
«
o
4P dz «¬
ln(ro / ri )
»¼
vz
(1) Observations. (i) Fluid motion is driven by axial pressure drop. (ii) For a very long tube the
flow field does not vary in the axial direction z. (iii) The fluid is incompressible (constant
density). (iv) Use cylindrical coordinates.
(2) Problem Definition. Determine the velocity distribution.
(3) Solution Plan. Apply continuity and Navier-Stokes equations in cylindrical coordinates to
determine the flow field.
(4) Plan Execution.
(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) axisymmetric flow, (iv) constant
properties (density and viscosity), (v) no end effects and (vi) negligible gravitational effect.
(ii) Analysis. The flow field is determined by solving the continuity and the Navier-Stokes
equations. We begin with the continuity equation in cylindrical coordinates
wU 1 w
1 w
w
U rv r U vT Uv z
r wT
wt r wr
wz
0
(2.4)
For constant density
wU
wt
wU
wr
wU
wT
wU
wz
0
(a)
For axisymmetric flow
w
wT
vT
0
(b)
For a long tube with no end effects axial changes in velocity are negligible
w
wz
0
(c)
Substituting (a)-(c) into (2.4)
d
rv r
dr
Integrating (d)
0
(d)
PROBLEM 3.6 (continued)
rv r
(e)
f (z )
To determine the “constant” of integration f (z ) we apply the no-slip boundary condition at the
surface
v (ro , z ) 0
(f)
Equations (e) and (f) give
f ( z) 0
Substituting into (e)
vr 0
(g)
Since the radial component v r vanishes everywhere, it follows that the streamlines are parallel
to the surface. To determine the axial component v z we apply the Navier-Stokes equation in the
wv z v T wv z
wv z wv z ·
vz
¸
wr
wz
wt ¹
r wT
©
z-direction, (2.11z)
ª 1 w § wv z · 1 w 2 v z w 2 v z º
wp
P«
Ug z ¸ 2
¨r
»
2
wz
wz 2 ¼»
¬« r wr © wr ¹ r wT
This equation is simplified as follows:
§
U¨ v r
(2.11z)
Steady state
w
wt
0
(h)
Negligible gravity effect
gr
gz
(i)
0
Substituting (b), (c) and (g)-(i) into (2.11z) gives
wp
1 d § dv z ·
P
¨r
¸ 0
wz
r dr © dr ¹
(3.11)
Since v z depends on r only, equation (3.11) can be written as
wp
wz
P
1 d § dv z ·
¨r
¸
r dr © dr ¹
g (r )
(j)
Integrating (j) with respect to z
p
g (r ) z C o
(k)
where C o is constant of integration. We turn our attention now to the radial component of
Navier-Stokes equation, (2.11r)
§
¨
©
U¨ v r
2
wv r v T wv r v T
wv
wv ·
vz r r ¸
wr
wz
wt ¸¹
r wT
r
2
ª w §1 w
wp
2 wv T w 2 v r º
· 1 w vr
P« ¨
(rv r ) ¸ 2
Ug r »
wr
wz 2 »¼
¹ r wT 2 r 2 wT
«¬ wr © r wr
Substituting (b), (g) and (i) into (2.11r), gives
(2.11r)
PROBLEM 3.6 (continued)
wp
wr
0
(l)
f (z )
(m)
Integrating (l)
p
where f (z ) is “constant” of integration. We now have two solutions for the pressure p: (k) and
(m). Equating the two, gives
(n)
p g (r ) z C o f ( z )
One side of (n) shows that the pressure depends on z only while the other side shows that it
depends on r and z. This, of course, is a contradiction. The only possibility for reconciling this is
by requiring that
g(r) = C
(o)
where C is a constant. Substituting (o) into (j)
wp
wz
P
1 d § dv z ·
¸
¨r
r dr © dr ¹
(p)
C
Thus the axial pressure gradient in the tube is constant. Equation (p) can now be integrated to
give the axial velocity distribution. Integrating once
r
dv z
dr
1 dp 2
r C1
2P d z
Separating variables and integrating again
vz
1 dp 2
r C1 ln r C 2
4P d z
(q)
where C1 and C 2 are constants of integration. The two boundary conditions on v z are
v z (ri )
0, v z (ro )
0
(r)
Equations (q) and (r) give C1 and C 2
C1
1 d p 2 2 ª ro º
( ri ro ) «ln »
4P d z
¬ ri ¼
1
1
C2
ª r º
1 dp 2
1 dp 2
( ri ro2 ) «ln o » ln ri ri
4P d z
4P d z
¬ ri ¼
Substituting into (q) and rearranging
vz
º
ro2 dp ª
1 (ri / ro ) 2
2
(
/
)
r
r
ln(r / ro ) 1»
«
o
4P dz «¬
ln(ro / ri )
»¼
(iii) Checking. Dimensional check: Each term in (3.12) has units of velocity.
Differential equation check: Velocity solution (3.12) satisfies equation (p).
(3.12)
PROBLEM 3.6 (continued)
Boundary conditions check: Velocity solution (3.12) satisfies boundary conditions (r).
Limiting check: (i) If pressure is uniform ( dp / dz
dp / dz 0 in (3.12) gives v z 0.
0 ) the fluid will be stationary. Setting
(5) Comments. The assumption of a long tube with negligible end effects is a key factor in
simplifying the problem. This assumption eliminates the z-coordinate as a variable and results in
governing equations that are ordinary. Assuming parallel streamlines is equivalent to neglecting
end effects since it leads to the same mathematical simplifications.
PROBLEM 3.7
A rod of radius ri is placed concentrically inside a cylinder of radius ro . The rod moves axially
with constant velocity U o and sets the
fluid in the annular space in motion. The
cylinder is maintained at uniform
temperature To . Neglect gravity and end
effects, and assume incompressible
laminar axisymmetric flow
[a] Show that the axial velocity is given by
vz
Uo
ln(r / ro )
ln(ri / ro )
[b] Taking into consideration dissipation, determine the heat flux at the outer surface and the
Nusselt number based on [ T (ri ) To ]. Neglect axial temperature variation.
(1) Observations. (i) Fluid motion is driven by axial motion of the rod. Thus motion is not due
to pressure gradient. (ii) For a very long tube the flow field does not vary in the axial direction z.
(iii) The fluid is incompressible (constant density). (iv) Heat is generated due to viscous
dissipation. It is removed from the fluid by conduction at the surface. (v) The Nusselt number is
a dimensionless heat transfer coefficient. (vi) To determine the heat transfer coefficient require
the determination of temperature distribution. (vii) Temperature distribution depends on the
velocity distribution. (viii) Use cylindrical coordinates.
(2) Problem Definition. Determine the velocity and temperature distribution.
(3) Solution Plan. Apply continuity and Navier-Stokes equations in cylindrical coordinates to
determine the flow field. Apply the energy equation to determine temperature distribution.
Fourier’s law gives surface heat flux. Equation (1.10) gives the heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) axisymmetric flow, (iv) constant
properties (density, viscosity and conductivity), (v) no end effects, (vi) uniform surface
temperature and (vii) constant rod velocity, (viii) negligible axial pressure gradient and (ix)
negligible gravitational effect.
(ii) Analysis. [a] Velocity distribution is governed by the continuity equation and NavierStokes equations of motion. The flow field is determined by solving the continuity and the
Navier-Stokes equations. We begin with the continuity equation in cylindrical coordinates
wU 1 w
1 w
w
U rv r U vT Uv z
wt r wr
wz
r wT
0
(2.4)
For constant density
wU
wt
For axisymmetric flow
wU
wr
wU
wT
wU
wz
0
(a)
PROBLEM 3.7 (continued)
w
wT
vT
(b)
0
For a long tube with no end effects axial changes in velocity are negligible
w
wz
(c)
0
Substituting (a)-(c) into (2.4)
d
rv r
dr
0
(d)
f (z )
(e)
Integrating (d)
rv r
To determine the “constant” of integration f (z ) we apply the no-slip boundary condition at the
surface
v (ro , z ) 0
(f)
Equations (e) and (f) give
f ( z) 0
Substituting into (e)
vr 0
(g)
Since the radial component v r vanishes everywhere, it follows that the streamlines are parallel
to the surface. To determine the axial component v z we apply the Navier-Stokes equation in the
wv z v T wv z
wv z wv z ·
vz
¸
wr
wz
wt ¹
r wT
©
z-direction, (2.11z)
ª 1 w § wv z · 1 w 2 v z w 2 v z º
wp
P«
Ug z ¸ 2
¨r
»
2
wz
wz 2 ¼»
¬« r wr © wr ¹ r wT
This equation is simplified as follows:
§
U¨ v r
(2.11z)
Steady state
w
wt
0
(h)
Negligible gravity effect
gr
gz
(i)
0
Substituting (b), (c) and (g)-(i) into (2.11z) gives
wp
1 d § dv z ·
P
¨r
¸ 0
wz
r dr © dr ¹
(3.11)
Since v z depends on r only, equation (3.11) can be written as
wp
wz
Integrating (j) with respect to z
P
1 d § dv z ·
¨r
¸
r dr © dr ¹
g (r )
(j)
PROBLEM 3.7 (continued)
g (r ) z C o
p
(k)
where C o is constant of integration. We turn our attention now to the radial component of
Navier-Stokes equation, (2.11r)
2
§ wv
wv
wv ·
v wv r v T
vz r r ¸
U¨ v r r T
¨
wr
wz
wt ¸¹
r wT
r
©
(2.11r)
2
ª w §1 w
wp
2 wv T w 2 v r º
· 1 w vr
P« ¨
(rv r ) ¸ 2
Ug r »
wr
wz 2 ¼»
¹ r wT 2 r 2 wT
¬« wr © r wr
Substituting (b), (g) and (i) into (2.11r), gives
wp
wr
0
(l)
f (z )
(m)
Integrating (l)
p
where f (z ) is “constant” of integration. We now have two solutions for the pressure p: (k) and
(m). Equating the two, gives
(n)
p g (r ) z C o f ( z )
One side of (n) shows that the pressure depends on z only while the other side shows that it
depends on r and z. This, of course, is a contradiction. The only possibility for reconciling this is
by requiring that
g(r) = C
(o)
where C is a constant. Substituting (o) into (j)
wp
wz
P
1 d § dv z ·
¸
¨r
r dr © dr ¹
C
(p)
Thus the axial pressure gradient in the tube is constant. Equation (p) can now be integrated to
give the axial velocity distribution. Integrating once
r
dv z
dr
1 dp 2
r C1
2P d z
Separating variables and integrating again
1 dp 2
r C1 ln r C 2
4P d z
vz
(q)
where C1 and C 2 are constants of integration. The two boundary conditions on v z are
v z (ri )
0, v z (ro )
0
Equations (q) and (r) give C1 and C 2
C1
Substituting into (q)
0, C 2
1 dp 2
ro
4P d z
(r)
PROBLEM 3.7 (continued)
vz
1 dp 2
(r ro2 )
4P d z
(3.12)
Since temperature distribution is obtained by solving the energy equation, we begin the analysis
with the energy equation. The energy equation in cylindrical coordinates for constant properties
is given by (2.24)
ª 1 w § wT · 1 w 2T w 2T º
wT v T wT
wT ·
§ wT
2 » P)
vr
vz
¸ k«
¨r
¸ 2
2
r wT
wr
wz ¹
wz ¼
© wt
¬ r wr © wr ¹ r w 0
U cP ¨
(2.24)
where the dissipation function ) is given by (2.25)
2
2
)
2
2
v
1 wv r ·
§ 1 wv T v r ·
§ wv · § wv
§ wv ·
2¨ r ¸ 2¨
¸ 2¨ z ¸ ¨ 0 T ¸ r ¹
r
r w0 ¹
© wz ¹ © wr
© wr ¹
© r wT
(2.25)
2
2
v
w
v
v
v
w
w
w
§1 z
·
§
·
0¸ ¨ r z¸
¨
wz ¹
wr ¹
© wz
© r w0
Equations (2.24) and (2.25) show that the determination of temperature distribution requires the
determination of the velocity components v r , v T and v z .
With the velocity distribution determined we return to the energy equation (2.24) and the
dissipation function (2.25). We note that for a long tube at uniform surface temperature with no
end effects, axial temperature variation can be neglected. Thus
wT
wz
w 2T
wz 2
(s)
0
Substituting (b), (c), (g), (h) and (s) into (2.24)
k
1 d § dT ·
¸ P)
¨r
r dr © dr ¹
(t)
0
Using (b), (c) and (g) into (2.25) gives the dissipation function for this flow
)
§ dv z ·
¨
¸
© dr ¹
2
2
Substituting the velocity solution (3.11) into the above, gives )
§ 1 d p· 2
¨¨
¸¸ r
© 2P d z ¹
(u)
Using (u) to eliminate ) in (t) and rearranging, we obtain
d § dT ·
¨r
¸
dr © dr ¹
2
1 §d p· 3
¸ r
¨
4kP ¨© d z ¸¹
(3.13)
Integrating the above twice
2
T
1 §d p· 4
¨
¸ r C 3 ln r C 4
64kP ¨© d z ¸¹
(v)
PROBLEM 3.7 (continued)
Two boundary conditions are needed to evaluate the constants of integration C 3 and C 4 . They
are:
dT (0)
0 and T (ro ) To
(w)
dr
Equations (v) and (w) give the two constants
2
1 §d p· 4
¸ ro
¨
To 64kP ¨© d z ¸¹
0, C 4
C3
Substituting the above into (v)
T
To ro4 § d p ·
¨
¸
64kP ¨© d z ¸¹
2
§ r4
¨1 ¨ r4
o
©
·
¸
¸
¹
(3.14a)
This solution can be expressed in dimensionless form as
T To
ro4
§dp·
¨
¸
64kP ¨© d z ¸¹
2
§ r4 ·
¨1 ¸
¨ r4 ¸
o ¹
©
(3.14b)
[b] Surface heat flux q cc(ro ) is obtained by applying Fourier’s law
q cc(ro )
k
dT (ro )
dr
Using (3.14) into the above
q cc(ro )
ro3 § d p ·
¸
¨
16 P ¨© d z ¸¹
2
(3.15)
[c] The Nusselt number is defined as
hD 2hro
(x)
k
k
where D is tube diameter. The heat transfer coefficient h is determined using equation (1.10)
Nu
h
Substituting (3.14a) into (y) h
dT (ro )
k
[T (0) To ] dr
4k
ro
(y)
(z)
Substituting (z) into (x)
Nu
8
(3.16)
(iii) Checking. Dimensional check: Each term in (3.12) has units of velocity. Each term in
(3.14a) as units of temperature. Each term in (3.15) has units of W/m 2 .
Differential equation check: Velocity solution (3.12) satisfies equation (p) and temperature
solution (3.14) satisfies (3.13).
PROBLEM 3.7 (continued)
Boundary conditions check: Velocity solution (3.12) satisfies boundary conditions (r) and
temperature solution (3.14) satisfies boundary conditions (w).
Limiting check: (i) If pressure is uniform ( dp / dz
dp / dz 0 in (3.12) gives v z 0.
0 ) the fluid will be stationary. Setting
(ii) If pressure is uniform ( dp / dz 0 ) the fluid will be stationary and no dissipation takes place
and thus surface heat transfer should vanish Setting dp / dz 0 in (3.15) gives q cc(ro ) 0.
(iii) Global conservation of energy. Heat transfer rate leaving the tube must be equal to the rate
of work required to pump the fluid. Pump work for a tube section of length L is
( p1 p 2 )Q
W
(z-1)
Where
p1 = upstream pressure
p 2 = downstream pressure
Q = volumetric flow rate, given by
ro
2S
Q
³ v rdr
z
0
Substituting (3.12) into the above and integrating
Q
S dp 4
ro
8P dz
(z-2)
Combining (z-1) and (z-2))
W
S ro4 dp
( p1 p 2 )
8P dz
(z-3)
Work per unit area W cc is
W cc
Substituting (z-3) into the above W cc
W
2S ro L
ro3 dp ( p1 p 2 )
16P dz
L
However
( p1 p 2 )
L
dp
dz
Combining this result with (z-4) gives
W cc
ro3 § dp ·
¨ ¸
16 P © dz ¹
2
This result is identical to surface heat transfer rate given in (3.15)
(z-4)
PROBLEM 3.7 (continued)
(5) Comments. (i) The assumption of a long tube with negligible end effects is a key factor in
simplifying the problem. This assumption eliminates the z-coordinate as a variable and results in
governing equations that are ordinary. Assuming parallel streamlines is equivalent to neglecting
end effects since it leads to the same mathematical simplifications.
(ii) Solution (3.14) shows that maximum temperature occurs at the center r
0.
(iii) The Nusselt number is constant independent of Reynolds and Prandtl numbers.
PROBLEM 3.8
A liquid film of thickness H flows down an inclined plane due to
gravity. The plane is maintained at uniform temperature To and
the free film surface is insulated. Assume incompressible laminar
flow and neglect axial variation of velocity and temperature and
end effects.
[a] Show that the axial velocity is given by
u
ª y 1 y2 º
U gH 2
sin T « 2»
P
«¬ H 2 H »¼
[b] Taking into consideration dissipation, determine the heat flux at the inclined plane.
(1) Observations. (i) Fluid motion is driven by gravity. (ii) No velocity and temperature
variation in the axial direction. (iii) The fluid is incompressible (constant density). (iv) Heat is
generated due to viscous dissipation. (v) Temperature distribution depends on the velocity
distribution. (vi) Use Cartesian coordinates.
(2) Problem Definition. Determine the velocity and temperature distribution.
(3) Solution Plan. Apply Navier-Stokes and continuity equations in Cartesian coordinates to
determine the flow field. Apply the energy equation to determine temperature distribution.
(4) Plan Execution.
(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) constant properties (density,
viscosity and conductivity), (iv) no end effects, (v) no motion in the z-direction and (vi) velocity
and temperature do not vary in the axial direction. .
(ii) Analysis. [a] The Navier-Stokes equation for constant properties are
§ w 2u w 2u w 2u ·
wp
P¨ 2 2 2 ¸
¨ wx
wx
wy
wz ¸¹
©
(2.10x)
§ w 2v w 2v w 2v ·
§ wv
wp
wv
wv
wv ·
¨
¸
w ¸ Ug y U¨ u v
P¨ 2 2 2 ¸
¨ wx
wx
wy
wz ¹
wy
wy
wz ¸¹
© wt
©
(2.10y)
§ wu
wu
wu
wu ·
w ¸¸
u
v
wx
wy
wz ¹
© wt
U ¨¨
x-direction:
y-direction:
Ug x Note that the sign of the gravity force in (2.10y) is negative since this component of gravity
points in the negative y-direction. That is
gx
g sin T and g y
These equations are simplified as follows:
wu
wt
wv
wt
wu
wx
w
wz
0
g cosT
PROBLEM 3.8 (continued)
Uv
wu
wy
Ug x wp
w 2u
P 2
wx
wy
§ w 2v w 2v ·
§ wv
wp
wv ·
v ¸¸ Ug y P¨ 2 2 ¸
¨ wx
wy
wy ¹
wy ¸¹
© wx
©
U ¨¨ u
(a)
(b)
The continuity equation introduces additional simplifications. For two-dimensional constant
properties, the continuity equations gives
wu wv
wx wy
Since
wu
wx
0
0 , it follows that
wv
wy
0
(c)
v
C
(d)
Integrating
However, the no slip condition at the wall gives
v (0)
0
Applying this condition to (d) gives
C=0
Thus
v
(d)
0
Substituting (d) into (a) and (b)
wp
w 2u
P 2
wx
wy
wp
0 Ug y wy
Ug x 0
(e)
(f)
Integrating (f)
p
U g y y C1
(g)
The pressure at the free surface is atmospheric. Thus
p( H )
pa
(g) gives
pa
U g y H C1
C1
pa U g y H
Substituting into (g)
p
U g y ( H y) pa
This result shows that pressure is independent of x. thus
(h)
PROBLEM 3.8 (continued)
wp
wx
0
(i)
Ug x
(j)
Substituting into (e)
P
w 2u
wy 2
Integrating twice
u
Ug x y 2
C 2 y C3
P 2
(k)
The two boundary conditions are
(1) u (0)
0
du ( H )
0
dy
These boundary conditions give
(2)
C2
U gx
H and C3
P
0
Substituting into (k)
u
Ug x H 2
P
ª y 1 y2 º
« 2»
«¬ H 2 H »¼
(l)
However
gx
g sin T
(m)
Introducing (m) into (l)
u
UH 2 g sin T
P
ª y 1 y2 º
« 2»
«¬ H 2 H »¼
(n)
[b] The energy equation is the starting point for determining the temperature distribution. The
energy equation for constant properties is given by (2.19b)
§ w 2T w 2T w 2T ·
§ wT
wT
wT
wT ·
¸ P)
¸¸ k ¨
u
v
w
¨ wx 2 wy 2 wz 2 ¸
wx
wy
wz ¹
© wt
©
¹
U c p ¨¨
(2.19b)
where the dissipation function ) is given by (2.17)
)
2
2
ª
2 § wv · 2
2º ª
2º
§ wv ww ·
§ ww · » «§ wu wv ·
§ ww wu · »
§ wu ·
«
¸¸ ¨
¸ 2 ¨ ¸ ¨¨ ¸¸ ¨ ¸ ¨¨ ¸¸ ¨¨ «© wx ¹
© wz ¹ » «© wy wx ¹
© wx wz ¹ »
© wy ¹
© wz wy ¹
¬
¼ ¬
¼
2
2 § wu wv ww ·
¨¨ ¸¸
3 © wx wy wz ¹
(2.17)
Equation (2.19b) simplifies to
k
w 2T
wy 2
P) 0
(o)
PROBLEM 3.8 (continued)
The dissipation function simplifies to
)
§ wu ·
¨¨ ¸¸
© wy ¹
2
(p)
Using (n) into (p)
)
( U gH sin T ) 2 ª
yº
1
«
»
¬ H¼
P2
2
( U gH sin T ) 2 ª
yº
1 »
«
P
¬ H¼
2
(q)
Introducing (q) into (o)
k
d 2T
dy 2
0
(r)
Integrating (r) twice
( U gH sin T ) 2 ª y 2 y 3
y4 º
A1 y A2
«
2»
Pk
¬« 2 3H 12 H ¼»
The two boundary conditions are
T
(1) T (0)
(s)
To
dT ( H )
0
dy
The two boundary conditions give the constants A1 and A2
(2)
H3
( U g sin T ) 2
3P k
A2 To
Substituting into (s) and rearranging the result in dimensionless form
A1
T To
H 4 ( U g sin T ) 2
Pk
1 y 1 y2 1 y3
1 y4
3 H 2 H 2 3 H 3 12 H 4
(t)
Surface heat flux is determined using Fourier’s law
q cc(0)
k
dT (0)
dy
H 3 ( U g sin T ) 2
(u)
3P
(iii) Checking. Dimensional check: Equation (n) has units of velocity. Each term in (t) is
dimensionless. Each term in (u) has units of heat flux.
q cc(0) Differential equation check: Velocity solution (n) satisfies equation (j) and temperature solution
(t) satisfies (r).
Boundary conditions check: Velocity solution (n) satisfies the two boundary conditions
following equation (k) and temperature solution (t) satisfies the boundary conditions following
equation (s).
PROBLEM 3.8 (continued)
Limiting check: (i) If gravity or inclination angle vanishes the fluid will be stationary. Setting
g 0 or T 0 in (n) gives u 0.
(ii) If gravity or inclination angle vanishes the fluid will be stationary, dissipation will also
vanish and the temperature will be uniform throughout equal to To . Setting g 0 or T 0 in
(t) gives T To .
(5) Comments. (i) Neglecting axial variation of u and T are key simplifying assumptions in this
problem. (ii) Surface heat flux is negative since all energy generated due to friction must leave
through the inclined plane.
PROBLEM 3.9
A liquid film of thickness H flows down an inclined plane due to gravity. The plane exchanges
heat by convection with an ambient fluid. The heat transfer coefficient is h and the ambient
temperature is Tf . The inclined surface is insulated. Assume incompressible laminar flow and
neglect axial variation of velocity and temperature and end effects.
[a] Show that the axial velocity is given by
u
ª y 1 y2 º
U gH 2
sin T « 2»
P
«¬ H 2 H »¼
[b] Taking into consideration dissipation, determine the heat flux
at the free surface.
(1) Observations. (i) Fluid motion is driven by gravity. (ii) No velocity and temperature
variation in the axial direction. (iii) The fluid is incompressible (constant density). (iv) eHat is
generated due to viscous dissipation. (v) Temperature distribution depends on the velocity
distribution. (vi) the inclined surface is at specified temperature and the free surface exchanges
heat by convection with the ambient. (vii) Use Cartesian coordinates.
(2) Problem Definition. Determine the velocity and temperature distribution.
(3) Solution Plan. Apply Navier-Stokes and continuity equations in Cartesian coordinates to
determine the flow field. Apply the energy equation to determine temperature distribution.
(4) Plan Execution.
(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) constant properties (density,
viscosity and conductivity), (iv) no end effects, (v) no motion in the z-direction, (vi) uniform heat
transfer coefficient and ambient temperature and (vii) velocity and temperature do not vary in the
axial direction.
(ii) Analysis. a[ ]The Navier-Stokes equati on for constant properties are
x-direction:
y-direction:
§ w 2u w 2u w 2u ·
wp
P¨ 2 2 2 ¸
¨ wx
wx
wy
wz ¸¹
©
(2.10x)
§ w 2v w 2v w 2v ·
§ wv
wp
wv
wv
wv ·
w ¸¸ Ug y u
v
P¨ 2 2 2 ¸
¨ wx
wx
wy
wz ¹
wy
wy
wz ¸¹
© wt
©
(2.10y)
§ wu
wu
wu
wu ·
w ¸¸
u
v
wx
wy
wz ¹
© wt
U ¨¨
Ug x U ¨¨
Note that the sign of the gravity force in (2.10y) is negative since this component of gravity
points in the negative y-direction. That is
gx
g cosT
g sin T and g y
These equations are simplified as follows:
wu
wt
wv
wt
wu
wx
w
wz
0
PROBLEM 3.9 (continued)
Uv
wu
wy
Ug x wp
w 2u
P 2
wx
wy
§ w 2v w 2v ·
§ wv
wp
wv ·
v ¸¸ Ug y P¨ 2 2 ¸
¨ wx
wy
wy ¹
wy ¸¹
© wx
©
U ¨¨ u
(a)
(b)
The continuity equation introduces additional simplifications. For two-dimensional constant
properties, the continuity equations gives
wu wv
wx wy
Since
wu
wx
0
0 , it follows that
wv
wy
0
v
C
(c)
Integrating
oHwever, the no slip condition at the wall gives
v (0)
0
Applying this condition gives
C=0
Thus
v
(d)
0
Substituting (d) into (a) and (b)
wp
w 2u
P 2
wx
wy
wp
0 Ug y wy
Ug x 0
(e)
(f)
Integrating (f)
p
U g y y C1
(g)
The pressure at the free surface is atmospheric. Thus
p( H )
pa
(g) gives
pa
U g y H C1
C1
pa U g y H
Substituting into (g)
p
U g y ( H y) pa
This result shows that pressure is independent of x. thus
(h)
PROBLEM 3.9 (continued)
wp
wx
0
(i)
Ug x
(j)
Substituting into (e)
P
w 2u
wy 2
Integrating twice
u
Ug x y 2
C 2 y C3
P 2
(k)
The two boundary conditions are
(1) u (0)
0
du ( H )
0
dy
These boundary conditions give
(2)
C2
U gx
H and C3
P
0
Substituting into (k)
u
Ug x H 2
P
ª y 1 y2 º
« 2»
«¬ H 2 H »¼
(l)
oHwever
gx
g sin T
(m)
Introducing (m) into (l)
u
UH 2 g sin T
P
ª y 1 y2 º
« 2»
«¬ H 2 H »¼
(n)
b[ ] The energy equation is the starting point for determining the temperature distribution. The
energy equation for constant properties is given by (2.19b)
§ w 2T w 2T w 2T ·
§ wT
wT
wT
wT ·
¸ P)
¸¸ k ¨
u
v
w
¨ wx 2 wy 2 wz 2 ¸
wx
wy
wz ¹
© wt
©
¹
U c p ¨¨
(2.19b)
where the dissipation function ) is given by (2.17)
)
2
2
ª
2 § wv · 2
2º ª
2º
§ wv ww ·
§ ww · » «§ wu wv ·
§ ww wu · »
§ wu ·
«
¸¸ ¨
¸ 2 ¨ ¸ ¨¨ ¸¸ ¨ ¸ ¨¨ ¸¸ ¨¨ «© wx ¹
© wz ¹ » «© wy wx ¹
© wx wz ¹ »
© wy ¹
© wz wy ¹
¼
¬
¼ ¬
2
2 § wu wv ww ·
¨¨ ¸¸
3 © wx wy wz ¹
(2.17)
PROBLEM 3.9 (continued)
Equation (2.19b) simplifies to
k
w 2T
wy 2
P) 0
(o)
The dissipation function simplifies to
)
§ wu ·
¨¨ ¸¸
© wy ¹
2
(p)
Using (n) into (p)
)
( U gH sin T ) 2 ª
yº
1 »
«
2
P
¬ H¼
2
(q)
Introducing (q) into (o)
k
d 2T
dy 2
( U gH sin T ) 2 ª
yº
1 »
«
P
¬ H¼
2
0
(r)
Integrating (r) twice
( U gH sin T ) 2 ª y 2 y 3
y4 º
C3 y C2
«
2»
Pk
¬ 2 3H 12 H ¼
The two boundary conditions are
T
(s)
dT (0)
0
dy
dT ( H )
h>T ( H ) Tf @
(2) k
dy
The two boundary conditions give the constants C 3 and C 4
C3 0
(1)
C4
Tf ( U g sin T ) 2 H 3 ª k H º
«¬ 3h 4 »¼
Pk
Substituting into (s) and rearranging
T
Tf H 4 ( U g sin T ) 2
Pk
ª1 k
1 1 y2 1 y3
1 y4 º
«
»
2
3 H 3 12 H 4 ¼
¬ 3 hH 4 2 H
(t)
Rewriting (t) in dimensionless form
T Tf
H 4 ( U g sin T ) 2
Pk
ª1 k
1 1 y2 1 y3
1 y4 º
«
»
2
3 H 3 12 H 4 ¼
¬ 3 hH 4 2 H
Surface heat flux is determined using Fourier’s law
q cc( H )
(t) into the above
k
dT ( H )
dy
(v)
PROBLEM 3.9 (continued)
q cc( H )
H 3 ( U g sin T ) 2
3P
(w)
(iii) Checking. Dimensional check: Equation (n) has units of velocity. Each term in (t) has
units of temperature. Each term in (w) has units of flux. Each term in (v ) and (y) is
dimensionless.
Differential equation check: V
elocity solution (n) satisfies e quation (j) and temperature solution
(t) satisfies (r).
Boundary conditions check: V
elocity solution (n) and temperature solution (t) satisfy their
respective boundary conditions.
Limiting check: (i) If gravity or inclination angle vanishes the fluid will be stationary. Setting
g 0 or T 0 in (n) gives u 0.
(ii) If gravity or inclination angle vanishes the fluid will be stationary, dissipation will also
vanish and the temperature will be uniform throughout equal to Tf . Setting g 0 or T 0 in
(t) gives T Tf .
Qualitative check: All dissipation heat must leave the free surface. Equation (w) shows that
surface heat flux is positive (leaving the fluid).
(5) Comments. (i) Neglecting axial variation in u and T are key simplifying assumptions in this
problem. (ii) Distinction should be made between the ambient heat transfer coefficient and the
liquid film heat transfer coefficient. They are not identical.
PROBLEM 3.10
Lubricating oil fills the clearance space of between a rotating
shaft and its housing. The shaft radius is ri 6 cm and housing
radius is ri 6.1 cm. The angular velocity of the shaft is
Z 3000 RPM and the housing temperature is To 40 o C. Taking
into consideration dissipation, determine the maximum oil
temperature and the heat flux at the housing. Neglect end effects
and assume incompressible laminar flow. Properties of lubricating
oil are: k 0.138 W/m o C and P 0.0356 kg/m s .
Solution
This problem is identical to Example 3.3. The maximum temperature occurs at the shaft surface
r ri . This temperature is given in equation (3.21)
P ª
2Z ri
thermal conductivity = 0 . 138
shaft radius = 0.06 m
housing radius = 0.061 m
W/m
T (ri ) To
2
º
2
«
» (ri / ro ) 2 ln(ro / ri ) 1
4k «¬1 (ri / ro ) 2 ¼»
>
@
(3.21)
where
k
ri
ro
To
P
Z
o
C
housing temperature = 40 o C
viscosity = 0.0356 kg/m s
angular velocity = 3000 RPM
= 100 S rad/s
The heat flux at the housing surface per unit length is given in (3.22)
q c(ro ) 4S P
(Z ri ) 2
(3.22)
1 (ri / ro ) 2
Computation.
Substituting into (3.21)
Tmax
Tmax
T (ri )
40(o C) 2
ª 2(100S )(rad /s )(0.06(m) º ª
0.061(m) º
(0.06 / 0.061)2 (m/m)2 2 ln
1»
«
o
2
2» «
0.06(m)
4(0.138 W/m C) «¬1 (0.06 / 0.061) (m/m) »¼ ¬
¼
0.0356 kg/m s
86.8 o C
Substituting into (3.22)
q c(ro )
2
>
(100 S )(rad / s)(0.06)(m)@
4S (0.0356)(kg/s m)
2
1 (0.06/0.061) (m/m)
2
= 4,888
W
kg m
= 4,888
3
m
s
Comments. (i) The key simplifying assumption is axisymmetry. This resulted in concentric
streamlines with vanishing normal velocity and angular changes.
PROBLEM 3.10 (continued)
(ii) Temperature rise of the lubricating oil and energy dissipation increase as the clearance
between the shaft and the housing is decreased. This is evident from equations (3.22) which
show that in the limit as (ri / ro ) o 1 , q c o f.
(iii) The energy dissipated due to friction is considerable. The heat dissipated for a housing
radius of 1.0 cm is 48.9 W.
PROBLEM 3.11
Consider lubrication oil in the clearance between a shaft and its
housing. The radius of the shaft is ri and that of the housing is ro .
The shaft rotates with an angular velocity Z and its housing
exchanges heat by convection with the ambient fluid. The heat
transfer coefficient is h and the ambient temperature is Tf . Taking
into consideration dissipation, determine the maximum
temperature of the oil and surface heat flux at the housing. Assume
incompressible laminar flow and neglect end effects.
(1) Observations. (i) Fluid motion is driven by shaft rotation (ii)
The housing is stationary. (iii) Axial variation in velocity and temperature are negligible for a
very long shaft. (iv) V
elocity and temperature do not vary with angular position. (v) The fluid is
incompressible (constant density). (vi) H
eat ge nerated by viscous dissipation is removed from
the oil at the housing. (vii) No heat is conducted through the shaft. (viii) The maximum
temperature occurs at the shaft. (ix) eHat fl ux at the housing is determined from temperature
distribution and Fourier’s law of conduction. (x) Use cylindrical coordinates.
(2) Problem Definition. Determine the velocity and temperature distribution in the oil.
(3) Solution Plan. This problem is similar to Example 3.3. The flow field is given by (3.18).
The energy equation is given in (l) of Example (3.18). The only difference between this problem
and Example 3.3 is the boundary condition at the housing. H
ousing heat flux can be determined
using the solution to temperature distribution and Fourier’s law.
(4) Plan Execution
(i) Assumptions. (I) Steady state, (ii) laminar flow, (iii) axisymmetric flow, (iv) constant
properties (density, viscosity and conductivity), (v) no end effects, (vi) uniform ambient
temperature and (vii) negligible gravitational effect.
(ii) Analysis. The velocity distribution is given by
(ro / ri ) 2 (ri / r ) (r / ri )
v T (r )
Z ri
(ro / ri ) 2 1
(3.18)
Following Example 3.3, the energy equation is
k
1 d § dT ·
¨r
¸ P)
r dr © dr ¹
0
(a)
and the solution is
P ª
2Z ri2
2
º 1
T (r ) C 3 ln r C 4
»
«
4k «¬1 (ri / ro ) 2 »¼ r 2
(b)
where C3 and C 4 are the integration constants. Two boundary conditions are needed to
determine C3 and C 4 . They are:
PROBLEM 3.11 (continued)
dT (ri )
0
dr
dT (ro )
(2) k
h >T (ro ) Tf @
dr
(1)
These boundary conditions give the two constants
2Z ri2
P ª
2
º 1
«
»
2k «¬1 (ri / ro ) 2 »¼ ri2
C3
and
C4
Tf ª 2Z ri2 º
«
2»
«¬1 (ri / ro ) »¼
P
4kro2
2
ª k
ro2
2
1
2
«
ri2
«¬ hro
§
k
¨¨ ln ro hro
©
·º
¸¸»
¹»¼
Substituting the above into (b) and rearranging
P ri2 ª
2
º ª
º
k
T (r ) Tf (r / r ) 2 2(ro / ri ) 2 ln(r / ro ) 2
1 (ro / ri ) 2 1»
2 «
2» « o
hro
4k ro «¬1 (ri / ro ) ¼» ¬
¼
2Z ri
(c)
This solution can be expressed in dimensionless form as
· º
k §¨ ro2
1 ri2 ª ro2 ro ro2
¸ 1»
2
ln
2
1
«
¸
r r2
hro ¨© ri2
4 ro2 «¬ ri2
¹ »¼
T (r ) Tf
º
P ª 2Z ri
«
»
k «¬1 (ri / ro ) 2 »¼
2
The maximum temperature is at the shaft’s surface. Setting r
Tmax
ri in (c) gives
2
º ª
º
k
T (ri ) Tf 1 2 ln(ri / ro ) 2
(ri / ro ) 2 1 (ri / ro ) 2 »
«
2» «
hro
4k «¬1 (ri / ro ) »¼ ¬
¼
P ª
2Z ri
(d)
(e)
Energy generated due to dissipation per unit shaft length, q c(ro ), is determined by applying
Fourier’s law at the housing. Thus
q c(ro )
2S ro k
dT (ro )
dr
Using (c), the above gives
q c(ro ) 4S P
(Z ri ) 2
1 (ri / ro ) 2
(f)
(iii) Checking. Dimensional check: each term in solutions (c) has units of temperature. (f) has
the correct units of W/m. and (3.20b) is dimensionless. Equation (3.22) has the correct units of
W/m.
Boundary conditions check: Temperature solution (c) satisfies the two boundary conditions on
temperature.
Limiting check: (i) If the shaft does not rotate no dissipation takes place and thus surface heat
transfer should vanish. Setting Z 0 in (f) gives q c(ro ) 0.
PROBLEM 3.11 (continued)
(ii) If the fluid is inviscid no dissipation takes place and thus surface heat transfer should vanish.
Setting Z 0 in (f) gives q c(ro ) 0.
Global conservation of energy. H
eat transfer rate from the hous ing must equal to work required
to overcome friction at the shaft’s surface. The rate of shaft work per unit length is given by
Wc
2S riW (ri )Z ri
(p)
where
W c = work done on the fluid per unit shaft length
W (ri ) = shearing stress at the shaft’s surface, given by
ª dv
v º
W (ri ) P « 0 T »
r ¼r r
¬ dr
i
(q)
Substituting (3.18) into the above
W (ri ) 2 P
Z
1 (ri / ro ) 2
(r)
Combining (p) and (r) and rearranging, gives
W c 4S P
(Z ri ) 2
1 (ri / ro ) 2
(s)
This result is identical to surface heat transfer rate given in (f)
(5) Comments. (i) The key simplifying assumption is axisymmetry. This resulted in concentric
streamlines with vanishing normal velocity and angular changes.
(ii) Temperature rise of the lubricatig oil and energy dissipation increase as the clearance
between the shaft and the housing is decreased.
(iii) V
elocity distributions are governed by a single parameter (ri / r).o
is governed by two parameters: (ri / ro ) and the B
iot number hro /.k
Temperature distribution
(iv) H
eat transfer rate at the housing, equation (f), is identical to that of Example 3.3 given in
equation (3.22). This is not surprising since dissipation energy for constant property fluids is a
function of flow field. Thus, dissipation energy is the same for problems with identical flow
fields even if they have different temperature boundary conditions.
PROBLEM 3.12
A rod of radius ri is placed concentrically inside a sleeve of radius
ro . Incompressible fluid fills the clearance between the rod and the
sleeve. The sleeve is maintained at uniform temperature To while
rotating with constant angular velocity Z . Taking into consideration
dissipation, determine the maximum fluid temperature and surface
heat flux at the sleeve. Assume incompressible laminar flow and
neglect end effects.
(1) Observations. (i) Fluid motion is driven by sleeve rotation (ii)
The shaft is stationary. (iii) Axial variation in velocity and temperature are negligible for a very
long shaft. (iv) V
elocity and temperature do not vary with angular position. (v) The fluid is
incompressible (constant density). (vi) H
eat ge nerated by viscous dissipation is removed from
the oil at the housing. (vii) No heat is conducted through the shaft. (viii) The maximum
temperature occurs at the shaft. (ix) Use cylindrical coordinates.
(2) Problem Definition. Determine the velocity and temperature distribution in the oil.
(3) Solution Plan. Apply continuity and Navier-Stokes equations in cylindrical coordinates to
determine the flow field. Use the energy equation to determine temperature distribution. Apply
Fourier’s law at the housing to determine the rate of energy generated by dissipation.
(4) Plan Execution
(i) Assumptions. (I) Steady state, (ii) laminar flow, (iii) axisymmetric flow, (iv) constant
properties (density, viscosity and conductivity), (v) no end effects, (vi) uniform surface
temperature and (vii) negligible gravitational effect.
(ii) Analysis. Temperature distribution is obtained by solving the energy equation. Thus we
begin the analysis with the energy equation. The energy equation in cylindrical coordinates for
constant properties is given by (2.24)
ª 1 w § wT · 1 w 2T w 2T º
wT v T wT
wT ·
§ wT
2 » P)
k
vr
vz
¸
¨r
¸ 2
«
2
r wT
r
r
r
w
w
wr
wz ¹
w
wz ¼
r
0
©
¹
© wt
¬
U cP ¨
(2.24)
where the dissipation function ) is given by (2.25)
2
)
2
2
2
v
1 wv r ·
§ 1 wv T v r ·
§ wv · § wv
§ wv ·
2¨ r ¸ 2¨
¸ 2¨ z ¸ ¨ 0 T ¸ r ¹
r
r w0 ¹
© wz ¹ © wr
© wr ¹
© r wT
2
wv ·
§ 1 wv z wv 0 ·
§ wv
¨
¸ ¨ r z¸
wz ¹
wr ¹
© wz
© r w0
2
(2.25)
The solution to (2.24) requires the determination of the velocity components v r , v T and v z .
These are determined by solving the continuity and the Navier-Stokes equations in cylindrical
coordinates. The continuity equation is given by equation (2.4)
wU 1 w
1 w
w
U rv r U vT Uv z
r wT
wt r wr
wz
For constant density
0
(2.4)
PROBLEM 3.12 (continued)
wU
wt
wU
wr
wU
wT
wU
wz
0
(a)
For axisymmetric flow
w
wT
(b)
0
For a long shaft with no end effects axial changes are negligible
w
wz
0
(c)
d
rv r
dr
0
(d)
vz
Substituting (a)-(c) into (2.4)
Integrating (d)
rv r
C
(e)
To determine the constant of integration C we apply the no-slip boundary condition at the
housing surface
v r (ro ) 0
(f)
Equations (e) and (f) give
C 0
Substituting into (e)
vr 0
(g)
Since the radial component v r vanishes everywhere, it follows that the streamlines are
concentric circles. To determine the tangential velocity v T we apply the Navier-Stokes equation
in the T -directions, equation (2.11 T )
§
U¨ v r
©
wv T v T wv T v r v T
wv
wv ·
vz T T ¸
wr
wz
wt ¹
r wT
r
2
ª w §1 w
1 wp
2 wv r w 2 v T º
· 1 w vT
P« ¨
(rv T ) ¸ 2
UgT »
r wT
wz 2 »¼
r 2 wT
¹ r wT 2
¬« wr © r wr
(2.11 T )
For steady state
w
wt
0
(h)
Neglecting gravity and applying (b),(c), (g) and (h), equation (2.11 T ) simplifies to
d §1 d
·
( rv T ) ¸ 0
¨
dr © r dr
¹
(i)
Integrating (i) twice
vT
C1
C
r 2
2
r
where C1 and C 2 are constants of integration. The two boundary conditions on v T are
(j)
PROBLEM 3.12 (continued)
v T (ri )
oundary conditions (j) give
B
v T (ro ) Z ro
0,
(j)
C1 and C 2
2Z ro2
C1
ro2 ri2
, C2
Z ri2 ro2
ro2 ri2
(k)
Substituting (k) into (j) and rearranging in dimensionless form, gives
ª r ri º
« »
1 (ri / ro ) ¬ ri r ¼
v T (r )
Z ro
(ri / ro )
2
(l)
We now return to the energy equation (2.24) and the dissipation function (2.25). Using (b), (c),
(g) and (h), equation (2.24) simplifies to
k
1 d § dT ·
¨r
¸ P)
r dr © dr ¹
(m)
0
The dissipation function (2.25) is simplified using (b), (c) and (g)
§ dv 0 v T ·
¨
¸
r ¹
© dr
)
2
Substituting the velocity solution (l) into the above, gives
2
ª 2Z ri2 º 1
«
2»
4
¬«1 (ri / ro ) ¼» r
)
(n)
Combining (n) and (m) and rearranging, we obtain
P ª
2
2Z ri2
º 1
«
2»
k ¬«1 (ri / ro ) ¼» r 3
d § dT ·
¨r
¸
dr © dr ¹
(o)
Integrating (o) twice
2
2Z ri2
P ª
º 1
T (r ) C 3 ln r C 4
«
»
4k «¬1 (ri / ro ) 2 »¼ r 2
(n)
where C3 and C 4 are the integration constants. Two boundary conditions are needed to
determine C3 and C 4 . They are:
dT (ri )
(o)
0 and T (ro ) To
dr
Equations (n) and (o) give the two constants
C3
P ª
2
2Z ri2
º 1
«
»
2k «¬1 (ri / ro ) 2 »¼ ri2
and
C4
P ª
2Z ri2
2
º ª1
º
2
2 ln ro »
To «
2» « 2
4k ¬«1 (ri / ro ) ¼» «¬ ro
ri
¼»
PROBLEM 3.12 (continued)
Substituting the above into (o)
P ª
2
º
T (r ) To (ri / ro ) 2 (ri / r ) 2 2 ln(ro / r )
«
2»
4k «¬1 (ri / ro ) ¼»
2Z ri
>
@
(p)
This solution can be expressed in dimensionless form as
T (r ) To
P ª
º
2Z ri
«
2»
4k ¬«1 (ri / ro ) ¼»
2
(ri / ro ) 2 (ri / r ) 2 2 ln(ro / r )
The maximum temperature is at the shaft’s surface. Setting r
T (ri ) To
2
ri in (p) gives
º
2
» 1 (ri / ro ) 2 ln(ro / ri )
«
2
4k ¬«1 (ri / ro ) ¼»
P ª
2Z ri
>
(q)
@
(r)
Energy generated due to dissipation per unit shaft length, q c(ro ), is determined by applying
Fourier’s law at the housing. Thus
q c(ro )
2S ro k
dT (ro )
dr
Using (q), the above gives
q c(ro ) 4S P
(Z ri ) 2
(s)
1 (ri / ro ) 2
(iii) Checking. Dimensional check: each term in solutions (l) and (q) is dimensionless.
Equation (s) has the correct units of W/m.
Differential equation check: V
elocity solution (l) satisfies equation (i) and temperature solution
(p) satisfies (o).
Boundary conditions check: eVlocity solution l) satisfies boundary conditions (j) and
temperature solution (p) satisfies boundary conditions (o).
Limiting check: (i) If sleeve does not rotate the fluid will be stationary. Setting Z
gives v T 0.
0 in (l)
(ii) If the sleeve does not rotate no dissipation takes place and thus surface heat transfer should
vanish. Setting Z 0 in (s) gives q c(ro ) 0.
(5) Comments. (i) The key simplifying assumption is axisymmetry. This resulted in concentric
streamlines with vanishing normal velocity and angular changes.
(ii) Temperature rise of the lubricating oil and energy dissipation increase as the clearance
between the shaft and the housing is decreased. This is evident from equation (s) which show
that in the limit as (ri / ro ) o 1 , q c o f.
(iii) V
elocity and temperature distribu tion are governed by a single parameter (ri / ro ).
PROBLEM 3.13
A hollow shaft of outer radius ro rotates with constant angular
velocity Z while immersed in an infinite fluid at uniform
temperature Tf . Taking into consideration dissipation, determine
surface temperature and heat flux. Assume incompressible laminar
flow and neglect end effects.
(1) Observations. (i) Fluid motion is driven by shaft rotation (ii)
Axial variation in velocity and temperature are negligible for a very long shaft. (iii) eVlocity,
pressure and temperature do not vary with angular position. (iv) The fluid is incompressible
(constant density). (v) H
eat ge nerated by viscous dissipation is conducted radially. (vi) The
determination of surface temperature and heat flux requires the determination of temperature
distribution in the rotating fluid. (vii) Use cylindrical coordinates.
(2) Problem Definition. Determine the velocity and temperature distribution in the rotating
fluid.
(3) Solution Plan. Apply continuity and Navier-Stokes equations in cylindrical coordinates to
determine the flow field. Use the energy equation to determine temperature distribution.
(4) Plan Execution.
(i) Assumptions. (I) Steady state, (ii) laminar flow, (iii) axisymmetric flow, (iv) constant
properties (density, viscosity and conductivity), (v) no end effects, (vi) no angular and axial
variation of velocity, pressure and temperature and (vii) negligible gravitational effect.
(ii) Analysis. Temperature distribution is obtained by solving the energy equation. Thus we
begin the analysis with the energy equation. The energy equation in cylindrical coordinates for
constant properties is given by (2.24)
ª 1 w § wT · 1 w 2T w 2T º
wT v T wT
wT ·
§ wT
2 » P)
k
vr
vz
¸
¨r
¸ 2
«
2
r wT
r
r
r
w
w
wr
wz ¹
w
wz ¼
r
0
©
¹
© wt
¬
U cP ¨
(2.24)
where the dissipation function ) is given by (2.25)
2
)
2
2
2
v
1 wv r ·
§ 1 wv T v r ·
§ wv · § wv
§ wv ·
2¨ r ¸ 2¨
¸ 2¨ z ¸ ¨ 0 T ¸ r ¹
r
r w0 ¹
© wz ¹ © wr
© wr ¹
© r wT
2
wv ·
§ 1 wv z wv 0 ·
§ wv
¨
¸ ¨ r z¸
wz ¹
wr ¹
© wz
© r w0
2
(2.25)
The solution to (2.24) requires the determination of the velocity components v r , v T and v z .
These are determined by solving the continuity and the Navier-Stokes equations in cylindrical
coordinates. The continuity equation is given by equation (2.4)
wU 1 w
w
1 w
U rv r U vT Uv z
r wT
wt r wr
wz
For constant density
0
(2.4)
PROBLEM 3.13 (continued)
wU
wt
wU
wr
wU
wT
wU
wz
0
(a)
For axisymmetric flow
w
wT
(b)
0
For a long shaft with no end effects axial changes are negligible
w
wz
0
(c)
d
rv r
dr
0
(d)
vz
Substituting (a)-(c) into (2.4)
Integrating (d)
rv r
C
(e)
To determine the constant of integration C we apply the no-slip boundary condition at the
housing surface
v r (ri ) 0
(f)
Equations (e) and (f) give
C 0
Substituting into (e)
vr 0
(g)
Since the radial component v r vanishes everywhere, it follows that the streamlines are
concentric circles. To determine the tangential velocity v T we apply the Navier-Stokes equation
in the T -direction, equation (2.11 T )
§
U¨ v r
©
wv T v T wv T v r v T
wv
wv ·
vz T T ¸
wr
wz
wt ¹
r wT
r
2
ª w §1 w
1 wp
2 wv r w 2 v T º
· 1 w vT
P« ¨
2
( rv T ) ¸ 2
UgT »
r wT
¹ r wT 2
wz 2 »¼
r wT
¬« wr © r wr
(2.11 T )
For steady state
w
wt
0
(h)
Neglecting gravity and applying (b),(c), (g) and (h), equation (2.11 T ) simplifies to
d §1 d
·
( rv T ) ¸ 0
¨
dr © r dr
¹
(i)
Integrating (3.17) twice
C1
C
r 2
2
r
where C1 and C 2 are constants of integration. The two boundary conditions on v T are
vT
(j)
PROBLEM 3.13 (continued)
v T (ro ) Z ro ,
v T (f )
0
These boundary conditions give C1 and C 2
C1
Z ro2
0 , C2
(k)
Substituting (k) into (j) and rearranging in dimensionless form, gives
v T (r )
Z ro
ro
r
(l)
We now return to the energy equation (2.24) and the dissipation function (2.25). Using (b), (c),
(g) and (h), equation (2.24) simplifies to
k
1 d § dT ·
¨r
¸ P)
r dr © dr ¹
(m)
0
The dissipation function (2.25) is simplified using (b), (c) and (g)
)
§ dv 0 v T ·
¨
¸
r ¹
© dr
2
Substituting the velocity solution (l) into the above, gives
)
>2Z r @
2
o
2
1
(n)
r4
Combining (m) and (n) and rearranging
d § dT ·
¨r
¸
dr © dr ¹
P
>2Z r @
2
o
k
2
1
r3
(o)
Integrating (o) twice
T (r ) P
>2Z r @
4k
2
o
2
1
r2
C 3 ln r C 4
(p)
where C3 and C 4 are the integration constants. Two boundary conditions are needed to
determine C3 and C 4 . They are:
(1) T ( f)
finite
(2) T (f)
Tf
B
oundary condition (1) gives
C3
0
C4
Tf
B
oundary condition (2) gives
Substituting the above into (n)
PROBLEM 3.13 (continued)
T ( r ) Tf P
>2Z r @
4k
2
o
2
1
r2
(q)
This solution can be expressed in dimensionless form as
ro2
r2
(r)
(Z ro ) 2
(s)
T ( r ) Tf
P
(Z ro )
k
Surface temperature obtained by setting r
2
ro in (q)
T ( ro ) Tf P
k
Surface heat flux per unit shaft length, q c(ro ), is determined by applying Fourier’s law at r
q c(ro )
2S ro k
ro
dT (ro )
dr
Using (q) the above gives
q c( ro )
4S P (Z ro ) 2
(t)
(iii) Checking. Dimensional check: each term in solutions (l) and (r) is dimensionless.
Equation (q) has the correct units of o C and equation (t) has units of W/m.
Differential equation check: V
elocity solution (l) satisfies equation (i) and temperature solution
(q) satisfies (o).
Boundary conditions check: V
elocity solution (l) and temperature solution (q) satisfy their
respective boundary conditions.
Limiting check: (i) If shaft does not rotate the fluid will be stationary. Setting Z
v T 0.
0 in (l) gives
(ii) If the shaft does not rotate no dissipation takes place and fluid temperature should be uniform
equal to Tf and surface heat transfer should vanish. Setting Z 0 in (q) gives T ( r ) Tf .
Setting Z 0 in (t) gives q c(ro ) 0.
Global conservation of energy: Surface heat transfer rate must equal to work required to
overcome friction at the shaft’s surface. The rate of shaft work per unit length is given by
Wc
2S roW ( ro )Z ro
(u)
where
W c = work done on the fluid per unit shaft length
W ( ro ) = shearing stress at the shaft’s surface, given by
W ( ro )
ª dv 0 v T º
»
r ¼ r ro
¬ dr
P«
(v)
Substituting (l) into the above
W ( ri ) 2 PZ
(w)
PROBLEM 3.13 (continued)
Combining (u) and (w)
W c 4S P (Z ro ) 2
(y)
This result is identical to surface heat transfer rate given in (t).
(5) Comments. (i) The key simplifying assumption is axisymmetry. This resulted in concentric
streamlines with vanishing normal velocity and angular changes. (ii) Surface temperature is
lowest in the entire region. (iii) eHat flow direct ion is negative. (iii) This problem was solved by
specifying two conditions at infinity. If surface temperature is specified instead of fluid
temperature at infinity, the solution determines T (f).
PROBLEM 3.14
Two large porous plates are separated by a distance H. An incompressible fluid fills the channel
formed by the plates. The lower plate is maintained at temperature T1 and the upper plate at T2 .
An axial pressure gradient dp / dx is applied to the
fluid to set it in motion. A fluid at temperature T1 is
injected through the lower plate with a normal
velocity v o . Fluid is removed along the upper plate
at velocity v o . The injected fluid is identical to the
channel fluid. Neglect gravity, dissipation and
axial variation of temperature.
[a] Show that the axial velocity is given by
u
Hǎ 1 dp ª 1 exp(vo y / ǎ )
yº
»
«
vo P dx ¬1 exp(vo H / ǎ ) H ¼
[b] Determine surface heat flux at each plate.
(1) Observations. (i) Axial pressure gradient sets fluid in motion. (ii) The fluid is
incompressible. (iii) The flow field is determined by solving the continuity and Navier-Stokes
equations. (iv) Energy equation gives the temperature distribution. (v) Fourier’s law and
temperature distribution give surface heat flux. (vi) Axial variation of temperature is neglected.
(viii) Use Cartesian coordinates.
(2) Problem Definition. Determine the velocity and temperature distribution.
(3) Solution Plan. Apply continuity and Navier-Stokes equations to determine the flow field.
Apply the energy equation to determine the temperature distribution.
(4) Plan Execution.
(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) constant properties (density, viscosity
and conductivity), (iv) infinite plates, (v) no axial variation of temperature, (vi) negligible
gravitational and (vii) negligible dissipation.
(ii) Analysis. a[ ]eVlocity distributi on. Applying continuity and the Navier-Stokes equations.
We begin with the continuity equation in Cartesian coordinates
ª wu w v w w º
wU
wU
wU
wU
u
v
w
U« »
wt
wx
wy
wz
¬ wx wy wz ¼
0
(2.2b)
For constant density
wU
wt
wU
wx
wU
wy
wU
wz
0
(a)
Since plates are infinite
w
wx
Substituting (a) and (b) into (2.2b), gives
w
wz
w 0
(b)
PROBLEM 3.14 (continued)
wv
wy
0
(c)
v
C
(d)
Integrating (c)
where C is constant of integration. The boundary condition on v is
v ( 0)
(e)
vo
Equations (d) and (e) give
C
vo
v
vo
Substituting into (d)
(f)
To determine the horizontal component u we apply the Navier-Stokes equations (2.10)
§ wu
wu
wu
wu ·
v
w ¸¸
u
wx
wy
wz ¹
© wt
Ug x § w 2u w 2u w 2u ·
wp
P ¨¨ 2 2 2 ¸¸
wx
wz ¹
wy
© wx
(2.10x)
§ wv
wv
wv
wv ·
u
v
w ¸¸
wx
wy
wz ¹
© wt
Ug y § w 2v w 2v w 2v ·
wp
P¨ 2 2 2 ¸
¨ wx
wy
wy
wz ¸¹
©
(2.10y)
U ¨¨
U ¨¨
These equations are simplified as follows: Steady state
wu
wt
(g)
0
Negligible gravity effect
gx
gy
0
(h)
Substituting (b) and (f)-(h) into (2.10x) and (2.10y) gives
v o du
ǎ dy
1 w p d 2u
P w x dy 2
(i)
and
wp
wy
0
(j)
where ǎ P / U . Equation (j) shows that pressure does not vary in the y-direction and thus it can
be a function of x or constant. Integrating (i) once
vo
ǎ
u C1
1 dp
du
y
P dx
dy
To solve this equation it is rewritten first as
du
P u Q( y )
dy
where
v
1 dp
y C1
P o , Q( y )
ǎ
P dx
(k)
(l)
(m)
PROBLEM 3.14 (continued)
The solution to (l) is
u
Pdy
Pdy
e ³ ª« e ³ Q ( y )dy º» C2
¬
¼
³
(n)
substituting (m) into (n) and evaluating the integrals
u
1 dp ǎ
> y (ǎ / vo )@ ǎ C1 C 2 e (vo / ǎ ) y
vo
P dx v o
(o)
The constants C1 and C 2 are determined form the boundary conditions on u
(1) u (0) 0
(2) u( H ) 0
These conditions give
ǎ º
1
H dp ª
« vo H / ǎ
»
P dx ¬ e
1 vo H ¼
C1
Substituting (l) into (k)
ǎ
(p)
H dp
1
v
H
vo P dx e o / ǎ 1
(q)
1 dp ª e vo y / ǎ 1 y º
»
«
P dx ¬ e vo H / ǎ 1 H ¼
(r)
C2
(p) and (q) into (o)
u
ǎH
vo
b[ ]Temperature distribution and Nusselt number. With the velocity distribution determined, the
energy equation is applied to determine temperature distribution. The energy equation for
constant properties is given by (2.19b)
§ w 2T w 2T w 2 T ·
k ¨¨ 2 2 2 ¸¸ P)
wz ¹
wy
© wx
§ wT
wT
wT
wT ·
¸
u
v
w
wx
wy
wz ¸¹
© wt
U c 5 ¨¨
(2.19b)
Neglecting dissipation ) and using (a) and (b) this equation is simplified
vo
where D
dT
dy
D
d 2T
dy 2
k / U c p is thermal diffusivity. To Integrate (s) it is rewritten as
§ dT ·
d ¨¨
¸¸
© dy ¹
dT
dy
vo
D
dy
Integrating
ln
Rewriting
dT
dy
vo
D
y ln C3
(s)
PROBLEM 3.14 (continued)
1 dT v o
y
C 3 dy D
dT
C3e v o y / D
dy
ln
Integrating again
T
C3
D
vo
e vo y / D C4
(t)
where C3 and C4 are constants of integration. The two boundary conditions on (q) are
(1) T (0) T1
(2) T ( H ) T2
These boundary conditions and solution (t) give
vo
C3
C4
D e
T2 T2 T1
vo H / D
1
(T2 T1 ) e v o H / D
e voH /D 1
(u)
(w)
Substituting (u) into (w) into (t) and rearranging the result in dimensionless form, give
T
T2 (T1 T2 )
exp[v o H / D ] exp[( v o H / D )( y / H )]
exp[v o H / D ] 1
(x)
This result in now expressed in terms of the Prandtl number. Note that
Pr
ǎ
D
Substituting into (x)
T
T2 (T1 T2 )
exp[(vo H / ǎ ) Pr ] exp[(v o H / ǎ ) Pr ( y / H )]
exp[(v o H / ǎ ) Pr ] 1
(y)
This result can be rearranged in dimensionless form as
T T2
T1 T2
exp[(v o H / ǎ ) Pr ] exp[(v o H / ǎ ) Pr ( y / H )]
exp[(v o H / ǎ ) Pr ] 1
(z)
Surface heat flux is determined by applying Fourier’s law at each plate
Substituting (y) into (z-1) and (z-2)
q cc(0)
k
dT (0)
dy
(z-1)
q cc( H )
k
dT ( H )
dy
(z-2)
PROBLEM 3.14 (continued)
vo
q cc(0)
q cc( H )
ǎ
vo
ǎ
kPr
kPr
T1 T2
exp[(vo H / ǎ ) Pr ] 1
(T1 T2 ) exp[(v o H / ǎ ) Pr ]
exp[(v o H / ǎ ) Pr ] 1
(z-3)
(z-4)
Expressed in dimensionless form, (K) and (L
) become
q cc(0)
vo
ǎ
k Pr (T1 T2 )
q cc( H )
vo
ǎ
k Pr (T1 T2 )
1
exp[(v o H / ǎ ) Pr ] 1
(z-5)
exp[(v o H / ǎ ) Pr ]
exp[(vo H / ǎ ) Pr ] 1
(z-6)
(iii) Checking. Dimensional check: Each term in (z), (z-5), and (z-6) is dimensionless. The
exponents of all exponentials are dimensionless.
Differential equation check: eVlocity solution (r) satisfies e quation (i) and temperature solution
(x) satisfies (s).
Boundary conditions check: V
elocity solution (r) and temperature solution (x) satisfy their
respective boundary conditions.
Limiting check: (i) If there is no axial pressure gradient, the fluid will be stationary. Set
dp / dx 0 in (r) gives u ( y ) 0.
(ii) If T1 T2 , surface heat flux will vanish. Set T1
q cc(0) q cc( H ) 0.
T2 in (z-3) and (z-4) gives
(5) Comments. (i) Treating the plate as infinite is one of the key simplifying assumptions. This
eliminates the x-coordinate as a variable and results in governing equations that are ordinary. (ii)
Contrary to expectation, the axial velocity plays no role in the temperature distribution This is
evident from energy equation (s) and temperature solutions (y).
(iii) According to the dimensionless form of solutions (z), (z -5) and (z-6), the problems is
characterized by the following single dimensionless parameter
vo H
ǎ
Pr
Note that this parameter is a combination of Prandtl number, geometry, injection velocity, and
kinematic viscosity ǎ , a property.
(iv) Taking the ratio of (N) to (M
) provides a co mparison of surface heat flux at the two plates
q cc( H )
q cc(0)
exp[(v o H / ǎ ) Pr ]
This result indicates that heat flux at the upper plate is higher than that at the lower plate.
PROBLEM 4.1
Put a check mark in the appropriate column for each of the following statements.
Statement
(a)
(b)
(wu / wx) (wv / wy ) 0 is valid for transient flow.
The y-momentum equation is neglected in
boundary layer flow.
true
false
may be
x
x
(c)
B
oundary layer equatio ns are valid for all
Reynolds numbers.
x
(d)
Pressure gradient is zero outside the boundary
layer.
x
(e)
w 2u
w 2u
2 for a streamlined body.
wx 2
wy
x
(f)
In boundary layer flow fluid velocity upstream
of an object is undisturbed.
(g)
Axial pressure gradient is neglected in boundary layer
flow.
(i)
Axial conduction is neglected in boundary layer
flow.
x
x
x
PROBLEM 4.2
Examine the three governing equations, (2.2), (4.13) and (4.18) for two-dimensional, constant
properties, laminar boundary layer flow.
a[ ] How many dependent variables do these equations have?
b[ ] How is the pressure pf determined?
c[ ] If streamlines are parallel in the boundary layer what terms will vanish?
[d] Can (2.2) and (4.13) be solved for the velocity field u and v independently of the energy
equation (4.18)?
Solution: The three equations are:
wu wv
wx wy
:
u
wu
wu
v
wx
wy
u
[a]There are four dependent variables:
0
(2.2)
1 dp f
w 2u
Q 2
U dx
wy
wT
wT
v
wx
wy
D
(4.13)
w 2T
(4.18)
wy 2
u , v , p f , and T.
b[ ]The pressure p f is determined form the inviscid external flow. The solution to the NavierStokes equations with P 0 (Euler’s equations of motion) for the flow over the same object
gives p f .
c[ ]The following terms will vanish if streamlines are parallel:
If v
0 then
wv
wy
0 . When this is substituted into (2.2) gives
wu
wx
0 . Equations (4.13) and
(4.18) become:
0 w 2u
1 dp f
Q 2
U dx
wy
and
u
wT
wx
D
w 2T
wy 2
d[ ]Equations (2.2) and (4.13) can be solved for u and v independently of energy equation (4.18)
since temperature does not enter in (2.2) and (4.13).
PROBLEM 4.3
Air flows over a semi-infinite plate with a free stream velocity Vf = 0.4 m/s and a free stream
temperature Tf 20 o C. The plate is maintained at Ts 60 o C. Can boundary layer
approximations for the flow and temperature fields be applied at:
a[ ] location x = 1.5 mm?
b[ ] location x = 15 mm?
Note: Evaluate air properties at the average film temperature T f
(1) Observations. (i) This is forced convection flow
over a streamlined body. (ii) V
iscous (velocity) boundary
layer approximations can be made if the Reynolds
number Rex >100. (iii) Thermal (temperature) boundary
layer approximations can be made if the Peclet number
Pex = Rex Pr >100. (iv) The Reynolds number decreases
as the distance along the plate is decreased.
(Ts Tf ) / 2.
Tf ,Vf
y
x
(2) Problem Definition. Determine the local Reynolds
and Peclet numbers at the locations of interest.
G
Gt
Ts
(3) Solution Plan. Write the definitions of Rex and Pex and calculate their values at x = 1.5 mm
and x = 15 mm.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) twodimensional flow and (5) streamlined body.
(ii) Analysis. The local Reynolds and Peclet numbers are defined as number
Rex =
Vf x
Q
(a)
and
Pex = Rex Pr
(b)
where
Pex = local Peclet number
Pr = Prandtl number
Rex = local Reynolds number
Vf = upstream velocity = 0.4 m/s
x = distance from the leading edge of the plate, m
Q = kinematic viscosity, m2 /s
Properties are evaluated at the film temperature Tf defined as
Tf =
where
Tf = film temperature, oC
Ts = surface temperature = 60oC
Tf = free stream temperature = 20oC
Ts Tf
2
(c)
PROBLEM 4.3 (continued)
(iii) Computations. Use (c) to calculate Tf
Tf = (60 +20)( oC)/2 = 40oC
At this temperature Appendix C gives
Pr = 0.71
Q = 16.96u10-6 m2/s
a[ ]At x = 1.5 mm = 0.0015 m, equation (a) gives
Rex =
0.4(m /s) 0.0015(m)
16.96 u 10 6 (m 2 /s )
= 35.4
Since this value is less than 100, it follows that velocity boundary layer approximations can not
be made. Using (b) to calculate Pex
Pex = 35.4 x 0.71 = 25.1
Since this is smaller than 100, it follows that temperature boundary layer approximations can not
be made.
b[ ]At x = 15 mm = 0.015 m the Reynolds number and Peclet number of part a[ ]will increase by
a factor of 10. Thus
Rex = 354
and
Pex = 251
Since both Rex and Pex are larger than 100, it follows that both velocity and temperature
boundary layer approximations can be made at this location.
(iv) Checking. Dimensional check: Computations showed that equation (a) is dimensionally
consistent.
(5) Comments. The Reynolds and Peclet numbers should be calculated to establish if boundary
layer approximations can be made.
PROBLEM 4.4
Water at 25 o C flows with uniform velocity Vf = 2 m/s over a streamlined object. The object is 8
85 o C. Use scaling to: a[ ] show that G / L 1 , [b]
cm long and its surface is maintained at Ts
evaluate the inertia terms u
wu
wu
and v
, and c[ ] evaluate the viscous terms
wx
wy
ǎw
2
u
wx 2
and
ǎw
2
u
wy 2
.
(1) Observations. (i) The surface is streamlined. (ii) The fluid is water. (iii) Inertia and viscous
effects can be estimated using scaling. (iv) If a viscous term is small compared to inertia, it can
be neglected. (v) Properties should be evaluated at the film temperature T f (Ts Tf ) / 2.
(2) Problem Definition. Estimate the magnitudes G / L, u
wu
wu
, v
,
wx
wy
ǎw
2
wx
u
2
and
ǎw
2
u
wy 2
.
(3) Solution Plan. Use the scaling to estimate the magnitudes of the above terms.
(4) Plan Execution.
(i) Assumptions. (1) Continuum and (2) streamlined surface.
(ii) Analysis. Scales:
u a Vf
xaL
y aG
(a)
(b)
(c)
Scaling of continuity equation gives a scale for v
v a Vf
a[ ]A balance between inertia term
G
(4.7d)
L
wu
u
and viscous term
wx
1
G
a
L
Re L
ǎ
w 2u
gives
wy 2
(4.14b)
where Re L is the Reynolds number defined as
Re L
Vf L
ǎ
(d)
where
L 0.08 m
Vf 2 m/s
ǎ
kinematic viscosity, m 2 /s
b[ ]Scales for inertia terms are:
u
and
V
wu
a Vf f
wx
L
(e)
PROBLEM 4.4 (continued)
v
V
wu
av f
wy
G
Using (4.7d) to eliminate v in the above, gives
v
V
wu
a Vf f
L
wy
(f)
Thus the two inertia terms are of the same magnitude.
c[ ]The two viscous terms are scaled as: First term:
2
ǎ w u2 a ǎ Vf2
wx
L
Second term:
w 2u
ǎ 2 a ǎ Vf2
G
wy
(g)
(h)
For G / L 1 , comparing (g) with (h) shows that
w 2u
w 2u
<
wx 2
wy 2
(iii) Computation. Properties are evaluated at the film temperature T f
(85 25)( o C)
2
2
m
0.5116 u 10 6
s
Ts Tf
2
Tf
Q
55 o C
a[ ]Substituting into (d)
2 (m/s)0.08(m)
Re
0.5116 u 10
6
2
(m /s)
3.127 u 10 5
Equation (4.14b) gives
G
L
a
1
3.127 u 10 5
0.00179
G a 0.00179 u 0.08(m) 0.000143 m
Thus G / L 1 .
b[ ]The two inertia terms are of the same order of magnitude, given by (e) or (f)
u
2(m/s)
wu
a 2(m/s)
0.08(m)
wx
50(m/s 2 )
[c]The first viscous term is given by (g)
(4.2)
PROBLEM 4.4 (continued)
2
u
wx
2
ǎw
a 0.5116 u 10 6 (m 2 /s)
2(m/s)
2
2
0.00016 m/s 2
(0.08) (m )
The second viscous term is given by (h)
2
u
wy
2
ǎw
a 0.5116 u 10 6 (m 2 /s)
2(m/s)
2
2
50 m/s 2
(0.000143) (m )
Thus the first viscous term can be neglected since it is much smaller than the second term.
(iv) Checking: Dimensional check: Inertia and viscous terms have the same units.
(5) Comments. Computation showed that the second viscous term is identical to the inertia term.
This is consequence of equating the two terms to derive (4.14b).
PROBLEM 4.5
Water at 25 o C flows with uniform velocity Vf = 2 m/s over a streamlined object. The object is 8
cm long and its surface is maintained at Ts
evaluate the convection terms u
and D
w 2T
wy 2
85 o C. Use scaling to: a[ ] show that G t / L 1 , [b]
wT
wT
w 2T
and v
, and c[ ] evaluate the conduction terms D 2
wx
wy
wx
.
(1) Observations. (i) The surface is streamlined. (ii) The fluid is water. (iii) Convection and
conduction effects can be estimated using scaling. (iv) If a conduction term is small compared to
convection, it can be neglected. (v) The scale for G t / L depends on whether G t ! G or G t G .
(vi) Properties should be evaluated at the film temperature T f (Ts Tf ) / 2.
(2) Problem Definition. Estimate the magnitudes G t / L, u
w 2T
wT
wT
w 2T
,v
, D 2 and D 2 .
wx
wy
wx
wy
(3) Solution Plan. Use the scaling to estimate the magnitudes of the above terms.
(4) Plan Execution.
(i) Assumptions. (1) Continuum and (2) streamlined surface.
(ii) Analysis. Scales: Two scales are used for u depending on whether G t ! G or G t G .
Case (1): G t ! G .
u a Vf
xaL
y a Gt
(a)
(b)
(c)
'T a Ts Tf
(d)
Scaling of continuity equation gives a scale for v
v a Vf
Gt
L
(4.23)
Case (2): G t G .
u a Vf
Gt
G
(4.29)
G t2
LG
(4.30)
Scaling of the continuity equation gives
v a Vf
a[ ]Scaling of
G t / L depends on whether G t ! G or G t G . The two cases are considered.
Case (1): G t ! G . A balance between convection u
wT
w 2T
and normal conduction D 2 gives
wx
wy
PROBLEM 4.5 (continued)
Gt
L
1
a
PrRe L
(4.24)
where Pr is the Prandtl number and Re L is the Reynolds number defined as
Vf L
Re L
(f)
Q
where
L 0.08 m
Vf 2 m/s
Q
kinematic viscosity, m 2 /s
Case (2): G t G . A balance between convection u
Gt
L
wT
w 2T
and normal conduction D 2 gives
wx
wy
1
a
Pr
1/3
(4.31)
Re L
b[ ]Scales for convection terms.
Case (1): G t ! G .
u
T Tf
wT
a Vf s
wx
L
(g)
and
v
T Tf
wT
av s
Gt
wy
Using (4.23) to eliminate v in the above, gives
v
T Tf
wT
a Vf s
wy
L
(h)
Thus the two convection terms are of the same magnitude.
Case (2): G t G . Using (d) and (4.29)
u
G T Tf
wT
a Vf t s
G
L
wx
(i)
where G is scaled as
G a
L
Re L
(4.14b)
Substituting (4.14b) into (i)
u
T T
wT
a VfG t Re L s 2 f
wx
L
(j)
PROBLEM 4.5 (continued)
Similarly, using (c), (d), (4.14b) and (4.30) give the same result of (j).
[c]Scale for conduction terms are: Axial conduction is scaled as
D
w 2T
wx 2
aD
Ts Tf
L2
(i)
normal conduction is scaled as
D
w 2T
wy
2
aD
Ts Tf
G t2
(j)
For G t / L 1 , comparing (i) with (j) shows that
w 2T
wx 2
<
w 2T
wy 2
(iii) Computation. Properties are evaluated at the film temperature T f
(85 25)( o C)
2
Ts Tf
2
Tf
55 o C
Pr = 3.27
m2
s
2
6 m
0.5116 u 10
s
D 1.566 u 10 7
Q
a[ ]Substituting into (f)
2 (m/s)0.08(m)
Re
0.5116 u 10
6
3.127 u 10 5
2
(m /s)
Case (1): G t ! G . Equation (4.24) gives
Gt
L
1
a
(3.27)(3.127 u 10
9.89 u 10 4
5
G t a 9.89 u 10 4 u 0.08(m) 7.91 u 10 5 m
Thus G 1 / L 1 for this case.
Case (2): G t G . Equation (4.31) gives
Gt
L
a
1
1/ 3
(3.27)
3.127 u 10
5
1.205 u 10 3
G t a 1.205 u 10 3 u 0.08(m) 9.64 u 10 5 m
b[ ]The two convection terms are of the same order of magnitude for both cases.
(4.2)
PROBLEM 4.5 (continued)
Case (1): G t ! G . Using (g)
(85 25)( o C)
wT
a 2(m/s)
u
0.08(m)
wx
o
1500
C
s
Case (2): G t G . Noting that G t a 9.64 u 10 5 m , and using (j)
u
wT
(85 20)( o C)
a 2(m/s)9.64 u 10 5 (m) 3.127 u 10 5
wx
(0.08) 2 (m 2 )
1011
c[ ]Axial conduction is given by (i)
D
w 2T
wx
2
a 1.566 u 10 7 (m 2 /s)
(85 25)( o C)
2
2
(0.08) (m )
o
C
s
1.175 u 10 4
Normal conduction is given by(j). Two cases are considered:
Case (1): G t ! G .
G t a 7.91 u 10 5 m
D
w 2T
wy
2
a 1.566 u 10 7 (m 2 /s)
(85 25)( o C)
(7.91 u 10
7 2
2
1502
) (m )
m
s2
Case (2): G t G .
G t a 9.64 u 10 5 m
D
w 2T
wy
2
a 1.566 u 10
7
2
(m /s)
(85 25)( o C)
(9.64 u 10
7 2
2
) (m )
1011
m
s2
Thus axial conduction can be neglected since it is much smaller than normal conduction.
(iv) Checking: Dimensional check: convection and conduction terms have the same units.
(5) Comments. (i) G t / L 1 . (ii) Axial conduction is small compared to normal conduction.
(iii) Computation showed that the normal conduction is identical to the convection term for both
cases. This is a consequence of equating the two terms to derive (4.24) and (4.31).
PROBLEM 4.6
Atmospheric air at 25 o C flows over a surface at 115 o C . The free stream velocity is 10 m/s.
a[ ] Calculate the Eckert number.
b[ ] Use scale analysis to show that the dissipation term P (wu / w y ) 2 is small compared to the
conduction term k (w 2T / w y 2 ).
(1) Observations. (i) The fluid is air. (ii) Dissipation and conduction can be estimated using
scaling. (iii) Dissipation is negligible if the Eckert number is small compared to unity.
(2) Problem Definition. Compute the Eckert number. Estimate the magnitudes of dissipation
and conduction terms.
(3) Solution Plan. Use the definition of Eckert number to compute its value. Apply scaling to
estimate the dissipation and conduction terms.
(4) Plan Execution.
(i) Assumptions. Continuum.
(ii) Analysis. a[ ]The Eckert number is defined as
E
Vf2
c p (Ts Tf )
(a)
where
cp
specific heat,
J
kg- o C
E Eckert number
Ts = surface temperature = 115 o C
Tf = free stream temperature = 25 o C
b[ ]The ratio of dissipation to condu ction is estimated using scaling.
ª wu º
Dissipation = P « »
¬ wy ¼
Conduction = k
where
k
thermal conductivity,
P
viscosity,
Scales:
kg
m-s
W
m- o C
w 2T
wy 2
2
PROBLEM 4.6 (continued)
u a Vf
xaL
y aG
(a)
(b)
(c)
’T a Ts Tf
(d)
The dissipation and conduction terms are estimated using scales (a)-(d).
ªV º
Dissipation a P « f »
¬G ¼
Conduction a k
2
(e)
Ts Tf
(f)
G2
where
Ts
surface temperature = 115 o C
Tf
free stream temperature = 25 o C
Taking the ratio of (e) to (f)
P Vf2
Dissipation
a
Conduction k (Ts Tf )
(g)
(iii) Computation. Properties are evaluated at the film temperature T f
Tf
Ts Tf
2
cp
1008.7
(115 25)( o C)
2
J
70 o C
kg- o C
W
k
0.02922
P
20.47 u 10 6
m- o C
kg
m-s
a[ ]Substituting into (a)
E
b[ ]
(10) 2 (m/s) 2
1008.7(J/kg o C)(115 25)( o C)
0.0011
kg -m 2
J -s 2
Dissipation 20.47 u 10 6 (kg/m s) (10) 2 (m 2 / s 2 )
a
Conduction
0.02922( W/m o C)(115 25)( o C)
0.0011
0.00078
(iv) Checking: Dimensional check: The ratio of dissipation to conduction in (g) must be
dimensionless:
PROBLEM 4.6 (continued)
P (kg/m s) Vf2 (m 2 / s 2 )
kg - m 2
k ( W/m o C)(Ts Tf )( o C)
W - s3
Limiting check: If Vf
1
0 there is no dissipation. Setting Vf
0 in (e) gives the correct result.
(5) Comments. (i) Since the Eckert number is small compared to unity dissipation is negligible.
(ii) Dissipation is negligible compared to conduction.
PROBLEM 4.7
Air at 20 o C flows over a streamlined surface with a free stream velocity of 10 m/s . Use scale
analysis to determine the boundary layer thickness at a distance of 80 cm from the leading edge.
(1) Observations. (i) The surface is streamlined. (ii) The fluid is air.
(2) Problem Definition. Estimate the magnitude of boundary layer thickness G at a specified
distance from the leading edge.
(3) Solution Plan. Use the scaling to estimate the magnitudes of G .
(4) Plan Execution.
(i) Assumptions. (1) Continuum and (2) streamlined surface.
(ii) Analysis. Scale analysis gives G as
G
L
a
1
Re L
(4.14b)
where Re L is the Reynolds number defined as
Re L
Vf L
Q
where
L 0 .8 m
V f 10 m/s
15.09 u 10 6 m 2 /s
Q
(iii) Computation. Substituting into (a)
10 ( m/s)0.8( m)
Re
15.09 u 10
6
2
( m /s)
5.302 u 10 5
Equation (4.14b) gives
G
L
a
1
5.302 u 10
5
0.00137
G a 0.00137 u 0.8( m) 0.0011 m
(iv) Checking: Dimensional check: The Reynolds number is dimesionles.
(5) Comments. G / L 1.
(a)
PROBLEM 4.8
In boundary layer flow, pressure gradient normal to the
flow direction is assumed zero. That is w p / w y | 0. If this is
correct, how do you explain lift on the wing of an airplane
in flight?
Solution
Although w p / w y | 0 in boundary layer flow, w p / w x z 0. Thus, pressure distribution around the
surface is accounted for. L
ift is th e net force acting on a surface in the y-direction.
PROBLEM 4.9
Derive an equation describing the vertical velocity component v at the edge of the boundary
layer for two-dimensional incompressible flow over a semi-infinite flat plate. Assume laminar
flow. Compare your result with scaling estimate.
(1) Observations. (i) This is a forced convection problem over a flat plate. (ii) At the edge of the
lasius solution gives the distribution
thermal boundary layer, the axial velocity is u | Vf . (iii) B
of the velocity components u(x,y) and v(x,y). (iv) Scaling gives an estimate of v(x,y).
(2) Problem Definition. Determine the vertical velocity at the edge of the viscous boundary
layer, v ( x, G ).
v ( x, y ) . Evaluate v ( x, y ) at y G .
(3) Solution Plan. Use B
lasius solution for
(4) Plan Execution.
(i) Assumptions. All assumptions leading to lBasius so lution are applicable. These are: (1)
Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5) laminar flow
(Rex <5 u105), (6) uniform upstream velocity and temperature, (7) flat plate and (8) negligible
changes in kinetic and potential energy.
(ii) Analysis. B
lasius solution for v ( x, y ) is given by
v
Vf
where V f
1 Q
2 Vf x
·
§ df
¨¨K
f ¸¸
¹
© dK
(4.43)
free steam velocity and Q is kinematic viscosity. The variable K and f are defined as
K ( x, y )
y
Vf
Qx
df
dK
u
Vf
(4.41)
(4.42)
At the edge of the thermal boundary layer, y G
u
Vf
According to lBasius solution, at
u
Vf
df
o1
dK
(a)
df
o 1 , Table 4.1,
dK
K ( x, G ) 8
(b)
f (K ) 6.27923
(c)
and
Substituting(b) and (c) into (4.43)
PROBLEM 4.9 (continued)
v(x, G )
Vf
1 Q
(8 6.27923)
2 Vf x
v(x, G )
Vf
0.8604
Re x
Vf x
where the Reynolds number is defined as Re x
Q
(d)
.
Scaling estimate of the velocity component v in the boundary layer is given by
v a Vf
G
x
(4.7d)
where
G
x
a
1
Re x
(4.16)
substituting (4.16) into (4.7d) and rearranging
1
v
a
Vf
Re x
(e)
(iii) Checking. Dimensional check: Since the Reynolds number is dimensionless it follows that
each term in (d) is dimensionless.
Qualitative check: The layer thickness increases with distance x. Solution (c) confirms this
behavior.
(5) Comments. Recalling that B
lasius solution gives
G
5.2
x
Rex
(4.46)
Comparing (d) with (4.46) shows that G t G for Pr 9.8 . Examination of Fig. 4.6 shows that
G t G for all fluids with Pr ! 1.0 and that G t G for Pr 1.0 .
PROBLEM 4.10
Sketch the streamlines in boundary layer flow over a semi-infinite flat plate.
Since the flow within the boundary layer is two-dimensional the vertical velocity component
does not vanish. Thus stream lines are not parallel.
PROBLEM 4.11
Define the thickness of the velocity boundary layer G in Blasius solution as the distance y where
the velocity u = 0.988 Vf . Derive an expression for G /x.
(1) Observations. (i) This is a laminar boundary layer flow problem. (ii) B
lasius solution gives
the velocity distribution for the flow over a semi-infinite flat plate. (iii) A solution for the
boundary layer thickness depends on how the thickness is defined.
(2) Problem Definition. Determine the distance y from the surface of the plate to the location
where the velocity ratio u/Vf = 0.988.
(3) Solution Plan. Use B
lasius solution, Table 4.1, to dete rmine the location of the edge of the
velocity boundary layer where u/Vf = 0.988.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4)
laminar flow (Rex <5 u105), (5) constant properties, (6) uniform upstream velocity, (7) flat plate
and (8) boundary layer flow (Rex >100).
(ii) Analysis. B
lasius solution, Table 4.1, gives the axial velocity ratio
function of the variable K which is defined as
V
K=y f
Qx
where
u/Vf as a
(a)
Vf = free stream velocity, m/s
x = axial coordinate, m
y = normal coordinate, m
K = dimensionless variable
Q = kinematic viscosity, m2/s
The viscous boundary layer thickness G is the value of y where the velocity ratio u/Vf reaches
an arbitrarily selected value. At f c (K) = u/Vf = 0.988, Table 47.1 gives K = 4.8. This value of
K corresponds to the edge of the boundary layer where y = G. Thus (a) gives
4.8 = K = y
Vf
V
= G f
Qx
Qx
(b)
Solving for G
G
4 .8
Qx
(c)
Vf
Dividing both sided of (c) by x, rearranging and using the definition of Reynolds number gives
G
4 .8
4 .8
x
Vf x
Re x
Q
(iii) Checking. Dimensional check: G in equation (c) should have units of length
(d)
PROBLEM 4.11 (continued)
G
Q (m 2/s) x(m)
Vf (m/s)
=m
(5) Comments. (i) The thickness of the viscous boundary layer depends on how it is defined.
Thus, it is not uniquely determined. (ii) Regardless of how G is defined, the solution takes the
form of equation (d). O
nly the cons tant in (d) changes according to how G is defined.
PROBLEM 4.12
Water flows over a semi-infinite plate with an upstream velocity of 0.2
m/s. Blasius solution is used to calculate G at three locations along the
plate. Results are tabulated. Are these results valid? Explain.
x(cm)
300
40
0.01
G (cm)
1.441
0.526
0.0083
(1) Observations. (i) B
lasius solution is valid for laminar boundary
layer flow over a semi-infinite plate. (ii) The transition Reynolds number from laminar to
turbulent flow is 5u 105 . (iii) oBundary layer approximations are valid if the Reynolds number is
greater than 100.
(2) Problem Definition. Determine the Reynolds number at each location.
(3) Solution Plan. Determine the Reynolds number at each location to establish the applicability
of B
lasius solution. Where applicable, use lB asius result to determine the boundary layer
thickness.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to lBasius solution are applicable. These are:(1)
Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5) laminar flow
(Rex <5 u105), (6) viscous boundary layer flow (Rex >100), (7) uniform upstream velocity, (8)
flat plate, (9) negligible changes in kinetic and potential energy and (10) no buoyancy (E = 0 or g
= 0).
(ii) Analysis. The Reynolds number is computed to establish if the flow is laminar and if
boundary layer approximations can be made. The Reynolds number is defined as
Rex =
Vf x
Q
y
(a)
where
Rex = Reynolds number
Vf = upstream velocity = 0.2 m/s
x = distance from the leading edge of the plate, m
Q = kinematic viscosity = 0.5116 u 10 6 m2 /s
G
Vf
0
x
To determine if the flow is laminar or turbulent, compare the Reynolds number with the
transition Reynolds number. For flow over a flat plate the transition Reynolds number Re xt is
Rext
5u 105
(b)
iscous boundary layer approximations are valid for
The flow is laminar if Rex < Re xt . V
Re x ! 100
(c)
B
lasius solution for the boundary layer thickness G is
G
5.2
x
Rex
(d)
PROBLEM 4.12 (continued)
(iii) Computations. Evaluating the Reynolds number at x = 300 cm, equation (a) gives
Rex
0.2(m/s)3(m)
1.173 u 106
6
2
0.5116 u 10 (m /s)
Since this value is larger than the transition Reynolds number given in (b), it follows that the
flow is turbulent and thus B
lasius solu tion (d) does not apply. Using (d) gives
G
5 .2
1.173 u106
3(m)
0.01441 m 1.441cm
Although this is the reported value for G , it is incorrect.
Evaluating the Reynolds number at x = 40 cm, equation (a) gives
Rex
0.2(m/s)0.4(m)
1.5637 u 105
2
6
0.5116 u 10 (m /s)
Since this value is smaller than the transition Reynolds number given in (b), and since it is larger
than 100, it follows that the flow is laminar and thus lBasius solution (d) is applicable. Using (d)
gives
G
5 .2
1.173 u 106
0.4(m) 0.00526 m
0.526 cm
Thus the reported value for G is correct.
Evaluating the Reynolds number at x = 0.01 cm, equation (a) gives
Rex
0.2(m/s)0.0001(m)
0.5116 u 10 6 (m 2 /s)
39
Since this value is smaller than 100, it follows that boundary layer approximations are not valid
and thus lBasius solution (d) doe s not apply. Using (d) gives
G
5.2
0.0001(m) 0.000083 m 0.0083 cm
39
Although this is the reported value for G , it is incorrect
(iv) Checking. Dimensional check: Computations showed that equations (a) and (d) are
dimensionally correct.
(5) Comments. In applying lBasius results it is important to verify that the conditions leading to
B
lasius solution are satisfied.
PROBLEM 4.13
Consider laminar boundary layer flow over a semi-infinite flat plate. Evaluate the wall shearing
stress at the leading edge. Comment on your answer. Is it valid? If not explain why.
(1) Observations. (i) This is an external flow problem over a flat plate. (ii) lBasius’s solution
for the velocity distribution and wall shearing stress is assumed to be applicable. (iii) fOinterest
is the value of the local stress at the leading edge of the plate.
(2) Problem Definition. Determine the local wall shearing stress for laminar boundary layer
flow over a flat plate.
(3) Solution Plan. Use B
lasius solution for the local wall stress. Evaluate wall stress at the
leading edge.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to lBasius so lution are applicable. These are:(1)
Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5) laminar flow
(Rex <5 u105), (6) viscous boundary layer flow (Rex >100, (7) uniform upstream velocity and (8)
flat plate.
(ii) Analysis. B
lasius solution gives
Vf
Qx
Wo = 0.33206 PVf
(4.47)
where
Vf = free stream velocity, m/s
x = axial distance measured from the leading edge, m
P viscosity, kg/s m
Q = kinematic viscosity, m2/s
W o wall shearing stress, N/m 2
Noting that all quantities in equation (4.47) are constant except the variable x, (4.47) can be
rewritten as
Wo
constant
x
(a)
(iii) Computations. To determine the shearing stress at the leading edge, set x = 0 in (a)
W o ( 0)
constant
=f
0
(iv) Checking. Dimensional check: Units of in (4.47) should be N/m2:
Wo
P ( kg/s m)Vf ( m/s)
Vf ( m/s)
Q ( m 2 /s) x ( m)
kg
s m
2
N
m2
PROBLEM 4.13 (continued)
(5) Comments. The shearing stress cannot be infinite. This suggests that lBasius solution is not
valid at the leading edge. O
ne of the assumptions leading to B
lasius solution is
Rex >100.
H
owever, at the leading edge x = 0, the Reynolds number is given by
Rex =
Vf x
Q
=0
Therefore, B
lasius solution ca nnot be used to determine W o at the leading edge. In fact the
solution breaks down at small values of x where the corresponding local Reynolds number and
Peclet number are smaller than 100.
PROBLEM 4.14
Water at 20 o C flows over a 2m u 2 m plate with a free stream velocity of 0.18 m/s. Determine
the force needed to hold the plate in place. Assume laminar boundary layer flow.
(1) Observations. (i) This is an external flow problem over a flat plate. (ii) The force needed to
hold the plate in place is equal to the total shearing force by the fluid on the plate, (iii)
Integration of wall shear over the surface gives the total shearing force. (iv) lBasius’s solution for
the velocity distribution and wall shearing stress is assumed to be applicable.
(2) Problem Definition. Determine the local wall shearing stress for laminar boundary layer
flow over a flat plate.
(3) Solution Plan. Use B
lasius solution for the local wall stress. Integrate shearing force over
the total surface area of plate.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to lBasius so lution are applicable. These are:(1)
Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5) laminar flow
(Rex <5 u105), (6) viscous boundary layer flow (Rex >100, (7) uniform upstream velocity and (8)
flat plate.
(ii) Analysis. Since shearing stress is not uniform,
shearing force must be integrated over the area to obtain the
total restraining force. Thus
L
Vf
x
W
L
F
³W
0
o
(a)
( x )Wdx
where
F
dx
L plate length = 2 m
W plate width = 2 m
x distance along surface, m
W o wall shearing stress, N/m 2
The local shearing stress is given by B
lasius solution
Vf
Qx
Wo = 0.33206 PVf
where
Vf = free stream velocity = 0.18 m/s
P viscosity = 1.003 u 10 3 kg/s m
Q = kinematic viscosity = 1.004 u 10 6 m2/s
Substituting (4.47) into (a)
F
Evaluating the integral
0.33206 PVfW
Vf
Q
³
L
0
x 1 / 2 dx
(4.47)
PROBLEM 4.14 (continued)
F
0.66412 PVfW
Vf L
(b)
Q
(iii) Computations. Equation (b) gives
F
0.66412 (1.003 u 10 3 )( kg/s m)0.18(m/s)2(m)
0.18(m/s)2(m)
1.004 u 10 6 ( m 2 /s)
0.144
kg m
s
F = 0.144 newton
(iv) Checking. Dimensional check: Computations showed that units of force are correct.
(5) Comments. According to (4.47) shearing stress decreases with distance from the leading
edge. Thus doubling the length of the plate increases the total force by a factor of 2 . This is
evident in equation (b) which shows that F v L .
PROBLEM 4.15
Consider Blasius solution for uniform flow over a semi-infinite plate. Put a check mark in the
appropriate column for each of the following statements.
Solution.
Statement
(a)
(b)
true
dpf
0 because the flow is laminar.
dx
Wall shearing stress increases with distance
from the leading edge of plate.
x (2)
Solution is not valid for Re x 100 .
x
(d)
Solution is not valid for Re x ! 5 u 105 .
x
(e)
(i)
The solution is valid for Rex ! 100 .
oBundary layer thickness is uniquely defined.
Solution is not valid for a curved plate.
The solution for the wall shear at the leading
edge (x = 0) is not valid.
The plate does not disturb upstream flow.
(j)
Solution is not valid for Re x 5 u 105 .
(1) In boundary layer flow
dp f
dx
Undetermined
x (1)
(c)
(f)
(g)
(h)
False
x(3)
x
(4)
x
x(5)
x
x(3)
0 for flat plate only.
(2) See equation (7.12).
(3) lBasius solution is valid for 100 Re x 5 u 10 5 .
dp f
z 0.
dx
0 at x = 0. oBundary layer approximation is not valid at
(4) For a curved surface
(5) Re x
Re x 100 .
PROBLEM 4.16
Imagine a cold fluid flowing over a thin hot plate. Using your intuition, would you expect the
fluid just upstream of the plate to experience a temperature rise due to conduction from the hot
plate? How do you explain the assumption in Pohlhausen's solution that fluid temperature is
unaffected by the plate and therefore T(0, y) = Tf ?
(1) Observations. (i) This is an external forced convection problem for flow over a flat plate.
(ii) O
f interest is the region where the upstream fluid reaches the leading edge of the plate. (iii)
The fluid is heated by the plate. (iv) eHat fr om the plate is conducted through the fluid in all
directions. (v) Pohlhausen’s solution assumes that heat is not conducted upstream from the plate
and therefore fluid temperature at the leading edge is the same as upstream temperature.
(2) Problem Definition. Determine the conditions
under which axial conduction in force convection
flow can be neglected.
Vf
T(0,y) = ?
y
Tf
0
G
0
x
Gt
Ts
(3) Solution. In reality heat from the plate is
conducted in all directions and thus one would
expect the temperature of the incoming fluid to be affected by the presence of the plate. This
becomes more obvious if one imagines that fluid velocity is decreased and in the limit the fluid
becomes stationary. Clearly, for a stationary fluid heat is conducted from the plate through the
fluid in all directions. oHweve r, as the velocity of the fluid increases, energy conducted
upstream is carried downstream by convection. This tends to minimize the effect of axial
conduction. and eventually may be ignored when compared with normal conduction. The
condition for this approximation is
Pe = Rex Pr >100
(a)
where
Pe = Peclet number
Pr = Prandtl number
Rex = local Reynolds number
Pohlhausen’s solution is based on the above boundary layer approximation. B
y neglecting axial
conduction the temperature of the incoming fluid at the leading edge is assumed to be the same
as the free stream temperature. That is
T(0,y) = Tf
where
T = temperature distribution in the fluid, oC
Tf = free stream temperature, oC
y = coordinate normal to plate, m
PROBLEM 4.17
Consider laminar boundary layer flow over a semi-infinite flat plat. The plate is maintained at
uniform temperature Ts . Assume constant properties and take into consideration dissipation.
a[ ] Does Blasius solution apply to this case? Explain.
b[ ] Does Pohlhausen’s solution apply to this case? Explain.
Solution
[a]B
lasius solution is based on the assumption
that properties are constant independent of
temperature. Thus, B
lasius solu tion applies to this problem.
[b]Pohlhausen’s solution neglects dissipati on. Thus it does not apply to this problem.
PROBLEM 4.18
A fluid with Prandtl number 9.8 flows over a semi-infinite flat plat. The plate is maintained at
uniform surface temperature. Derive an expression for the variation of the thermal boundary
layer thickness with distance along the plate. Assume steady state laminar boundary layer flow
with constant properties and neglect dissipation. Express your result in dimensionless form.
(1) Observations. (i) This is a forced
convection problem over a flat plate. (ii) At the
edge of the thermal boundary layer, fluid
temperature is T | T f . (iii) Pohlhausen’s
solution gives the temperature distribution in
the boundary layer. (iv) The thermal boundary
layer thickness G t increases with distance from the leading edge. (v) G t depends on the Prandtl
number.
(2) Problem Definition. Determine the variation of the thermal boundary layer thickness with
distance for a fluid with Prandtl number of 9.8.
(3) Solution Plan. Use Pohlhausen’s graphical solution for the temperature distribution of
laminar flow over a flat plate to determine the thermal boundary layer thickness.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are:
(1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5) laminar
flow (Rex <5 u105), (6) viscous and thermal boundary layer flow (Rex >100 and Pex >100), (7)
uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10)
negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no buoyancy
(E = 0 or g = 0) and (13) no energy generation ( q ccc 0 ). In addition, (14) neglect radiation.
(ii) Analysis. Pohlhausen’s solution is shown in graphical form in fig. 4.6
1.0
100 10
1 0.7(air )
0.8
0.1
0.6
Fig. 4.6
T Ts
Tf Ts 0.4
Pr
0.2
0
2
0.01
6
4
K
8
y Vf Q x
10
12
14
PROBLEM 4.18 (continued)
At the edge of the thermal boundary layer, y G t the dimensionless temperature in Fig. 4.6 is
equal to unity
T Ts
|1
(a)
Tf Ts
For a fluid with Pr 9.8 the corresponding value of K ( x, G t ) is obtained from Fig. 4.6 as
Vf
| 2 .8
Qx
Gt
(b)
Solving (b) for G t
G t | 2 .8
Qx
(c)
Vf
Expressing this result in dimensionless form
Gt
x
| 2.8
Q
Vf x
|
2.8
Re x
(d)
(iii) Checking. Dimensional check: Since the Reynolds number is dimensionless it follows
that each term in (d) is dimensionless.
Qualitative check: The thermal boundary layer thickness increases with distance x. Solution (c)
confirms this behavior.
(5) Comments. Recalling that B
lasius solution gives
G
5.2
x
Rex
(4.46)
Comparing (d) with (4.46) shows that G t G for Pr 9.8 . Examination of Fig. 4.6 shows that
G t G for all fluids with Pr ! 1.0 and that G t G for Pr 1.0 .
PROBLEM 4.19
Use Pohlhausen’s solution to determine the heat flux at the leading edge of a plate. Comment on
your answer. Is it valid? If not explain why.
(1) Observations. (i) This is an external forced convection problem for flow over a flat plate.
(ii) Pohlhausen’s solution for the temperature distribution and heat transfer coefficient is
assumed to be applicable. (iii) fOinterest is the va lue of the local heat flux at the leading edge of
the plate. (iv) Knowing the local transfer coefficient V y
f
and using Newton’s law, gives the heat flux
G
(2) Problem Definition. Determine the local heat
transfer coefficient for laminar boundary layer flow
over a flat plate.
Tf
G
0
0
x
t
Ts
(3) Solution Plan. Apply Newton’s law of cooling to determine the local heat flux. Use
Pohlhausen’s solution for the local heat transfer coefficient. Apply the solution at the leading
edge.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These
are:(1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5)
laminar flow (Rex <5 u105), (6) viscous and thermal boundary layer flow (Rex >100 and Pex >
100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat
plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no
buoyancy (E = 0 or g = 0) and (13) no energy generation ( q ccc 0 ).
(ii) Analysis. Application of Newton’s law gives
q csc( x)
h [Ts Tf ]
(a)
where
h = local heat transfer coefficient, W/m2-oC
q csc = surface heat flux, W/m2
Ts = surface temperature, oC
Tf = free stream temperature, oC
B
ased on the above assumptions, Pohlhausen’s soluti on for the local heat transfer coefficient is
h
k
Vf dT (0)
Qx dK
where
k = thermal conductivity, W/m- oC
Vf = free stream velocity, m/s
T = temperature variable, oC
T (T Ts ) /(Tf Ts ) , dimensionless temperature
x = axial distance measured from the leading edge, m
Q = kinematic viscosity, m2/s
(b)
PROBLEM 4.19 (continued)
Vf
, dimensionless variable
Qx
y = vertical coordinate, m
K= y
Noting that all quantities in equation (a) are constant except the variable x, (a) can be rewritten as
h=
constant
x
(c)
(iii) Computations. To determine the heat transfer coefficient at the leading edge, set x = 0
in (c)
h(0) =
cons tan t
=f
0
Substituting into (a)
q csc( x)
f [Ts Tf ]
f
(iv) Checking. Dimensional check: Units of h in (a) should be W/m2-oC:
h = k (W/m-oC)
Vf
Qx
m/s
dT (0)
(1/1) = W/m2-oC
2
m / s m dK
(5) Comments. Physically, the heat transfer coefficient cannot be infinite. This suggests that
Pohlhausen’s solution is not valid at the leading edge. nOe of the assumptions leading to
Pohlhausen’s solution is Rex >100. H
owever, at the leading edge
x = 0, the Reynolds number is
given by
Rex =
Vf x
Q
=0
Therefore, Pohlhausen’s solution cannot be used to determine h at the leading edge. In fact the
solution breaks down at small values of x where the corresponding local Reynolds number and
Peclet number are smaller than 100.
PROBLEM 4.20
Consider laminar boundary layer flow over a semi-infinite flat plate at uniform surface
temperature Ts . The free stream velocity is Vf and the Prandtl number is 0.1. Determine the
temperature gradient at the surface dT ( x,0) / dy.
(1) Observations. (i) This is an external forced convection problem for flow over a flat plate.
(ii) Pohlhausen’s solution for the temperature distribution is assumed to be applicable. (iii) O
f
interest is the value of the normal temperature gradient at the surface.
(2) Problem Definition.
Determine the normal temperature gradient at the surface,
dT ( x,0) / dy, for laminar boundary layer flow over a flat plate.
(3) Solution Plan. Apply Pohlhausen’s solution for the temperature distribution for laminar
flow over a flat plate.
(4) Plan Execution.
(i) Assumptions. All assumptions leading
to Pohlhausen's solution are applicable. These
are:(1) Newtonian fluid, (2) steady state, (3)
constant properties, (4) two-dimensional, (5)
laminar flow (Rex <5 u105), (6) viscous and thermal boundary layer flow (Rex >100 and Pex >
100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat
plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no
buoyancy (E = 0 or g = 0) and (13) no energy generation ( q ccc 0 ).
(ii) Analysis. Surface temperature gradient is given by
dT dT dK
dT dK dy
wT ( x, y )
wy
(a)
where
K ( x, y )
T
y
Vf
Qx
(4.41)
T Ts
Tf Ts
(4.58)
Substituting (4.41) and (4.58) into (a)
wT ( x, y )
wy
Evaluating (b) at y
(Tf Ts )
Vf d T
Q x dK
(b)
0 gives the temperature gradient at the surface
wT ( x,0)
wy
(Tf Ts )
Vf dT (0)
Q x dK
The problem reduces to determining dT (0) / dK. This is given by Pohlhausen’s solution as
(c)
PROBLEM 4.20 (continued)
dT (0)
dK
>0.332@ Pr
f
³
K
2
ªd f º
« 2»
¬ dK ¼
(4.64)
Pr
dK
where f (K ) is given by B
lasius solution an d is listed in Table 4.1. The integral in (6.64) must be
evaluated numerically. An alternate approximate method for determining dT (0) / dK is using
Fig. 4.6 to determine the slope of the Pr 0.098 curve.
(iii) Computations.
Numerical integration of (4.64) gives
dT (0)
dK
0.1568
(d)
Substituting into (c)
wT ( x,0)
V
(e)
0.1568(Tf Ts ) f
wy
Qx
Approximating the Prandtl number of 0.098 by 0.1 and using Fig. 4.6, gives the slope at the wall
as
dT (0)
(f)
| 0.125
dK
(iv) Checking. Dimensional check: Surface temperature gradient in (e) should have units of
C/m:
o
wT ( x,0)
Vf ( m/s)
C
o
(Ts Tf )( C/m)
2
wy
Q ( m /s) x (m) m
Limiting check: Surface temperature gradient should vanish for Ts Tf . Setting Ts Tf in (e)
gives wT ( x,0) / wy 0.
Qualitative check: Surface temperature gradient should increase as the free stream velocity is
increased. This is confirmed by ©
.
Comments. (1) Using numerical integration to evaluate the integral in (4.64) is necessary since
this integral cannot be evaluated analytically. (2) As the thermal boundary layer thickness
increases surface heat flux, and thus surface temperature gradient, should decrease. This follows
from the observation that the thermal boundary layer acts as an insulation layer. Equation (e)
show that surface temperature gradient decreases with distance x. (3) Fig. 4.6 provides a
reasonable estimate surface temperature gradient.
PROBLEM 4.21
Fluid flows between two parallel plates. It enters with uniform velocity Vf and temperature Tf .
The plates are maintained at uniform surface temperature Ts . Assume laminar boundary layer
flow at the entrance. Can Pohlhausen solution be applied to determine the heat transfer
coefficient? Explain.
Solution
The velocity distribution in Pohlhausen’s solution is based on B
lasius solution. B
lasius solution
is limited to the flow over a single plate. For a single plate axial pressure gradient is set equal to
zero. That is, lBasius solution is based on
wp
wx
0
For flow between parallel plates axial pressure gradient does not vanish. That is
wp
z0
wx
In fact, pressure decreases in the flow direction. Thus, lBasius solution does not apply to the flow
between parallel plates. It follows that Pohlhausen’s also does not apply.
PROBLEM 4.22
Two identical rectangles, A and B, of dimensions
L1uL2 are drawn on the surface of a semi-infinite
flat plate as shown. Rectangle A is oriented with
side L1 along the leading edge while rectangle B is
oriented with side L2 along the edge. The plate is
maintained at uniform surface temperature.
a[ ] If the flow over rectangle A is laminar, what is
it for B ?
b[ ] If the heat transfer rate from plate A is
435 W, what is the rate from plate B ?
L2
A
L1
Tf
Vf
Ts
B L2
L1
top view
(1) Observations. (i) This is an external forced convection problem over two flat plates. (ii)
B
oth plates have the same surf ace area. (iii) For flow over a flat plate, the heat transfer
coefficient h decreases with distance from the leading edge. (iv) Since the length in the flow
direction is not the same for the two plates, the average heat transfer coefficient is not the
same. It follows that the total heat transfer rate is not the same. (v) The flow over a flat plate
is laminar if the Reynolds number is less than 5u105.
(2) Problem Definition. Determine the Reynolds number at the trailing end of plate B.
O
btain a solution for the average heat transf er coefficient for laminar forced convection
over a flat plate.
(3) Solution Plan. Examine the Reynolds number at the trailing end of plate B to establish
if the flow is laminar or turbulent. Use Newton’s law of cooling to determine the heat
transfer from each plate. Use Pohlhausen’s solution to obtain a solution for the average heat
transfer coefficient.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable.
These are: (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) twodimensional, (5) laminar flow (Rex <5 u105), (6) viscous and thermal boundary layer flow
(Rex >100 and Pex >100), (7) uniform upstream veloc ity and temperature, (8) uniform
surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy,
(11) negligible dissipation, (12) no buoyancy (E = 0 or g = 0) and (13) no energy generation
( q ccc 0 ). In addition, (14) neglect radiation.
(ii) Analysis. a[ ] To establish if the flow is laminar or turbulent, compare the
Reynolds number with the transition Reynolds number. For the flow over a flat plate the
transition Reynolds number is
Re xt = 5u105
(a)
The flow is considered laminar if Rex < Re xt . The Reynolds number for plate B is
Re L1 =
Vf L1
Q
(b)
PROBLEM 4.22 (continued)
where
Re L1 = Reynolds number at trailing end of plate B
L1 = length of plate B in the flow direction, m
Vf = upstream velocity, m/s
Q = kinematic viscosity, m2 /s
Similarly, for plate A
ReL2 =
Vf L 2
(c)
Q
where
ReL2 = Reynolds number at trailing end of plate A
L2 = length of plate A in the flow direction, m
Taking the ratio of (b) and (c) and rearranging
Re L1 =
L1
ReL2
L2
(d)
Since L1 < L2, equation (d) gives
Re L1 < ReL2
(e)
Thus, Re L1 < ReL2 < Re xt . Since the flow is laminar for plate A, it follows that it is also
laminar for plate B.
b[ ]Application of Newton' s law of cooling gives
q = h A (Ts - Tf)
(f)
where
A = surface area, m2
h = average heat transfer coefficient, W/m2-oC
q = total surface heat transfer rate, W
Ts = surface temperature, oC
Tf = free stream temperature, oC
Applying (f) to the two plates and taking the ratio of the resulting equations
qB
qA
hB
hA
(g)
where the subscripts refer to plates A and B. The average heat transfer coefficient for
laminar flow over a flat plate is given by Pohlhausen's solution, equations (4.67) and (4.71b)
h
where
0.664k
Vf
( Pr )1 / 3
QL
(h)
PROBLEM 4.22 (continued)
k = thermal conductivity, W/m-oC
Pr = Prandtl number
Applying (h) to plates A and B, noting that L = L2 for A and L = L1 for B, and taking the
ratio of the results
hB
hA
L2
L1
(i)
Substituting into (g) and solving for q B
qB
qA
L2
L1
(iii) Computations. With q A = 435 W, equation (j) gives
q B = 435(W)
L2
L1
Since L2 > L1, it follows that the heat transfer rate from B is greater than that from A.
(iv) Checking. Dimensional check: Units of h in equation (h) should be W/m2-oC
h = k(W/m-oC)
Vf
Q L
m/s
2
Pr1/3 = W/m2-oC
(m / s)(m)
Limiting check: For the special case of L1 = L2 (square plate), the heat transfer rate from the
two plates should be the same. Setting L1 = L2 in equation (j) gives q B q A .
(5) Learning and Generalizing. To maximize the rate of heat transfer from a flat
rectangular plate under laminar flow conditions, the long side of the plate should face flow
direction.
(j)
PROBLEM 4.23
A semi-infinite plate is divided into four equal sections of one centimeter long each. Free
stream temperature and velocity are uniform and the flow is laminar. The surface is maintained
at uniform temperature. Determine the ratio of the heat transfer rate from the third section to
that from the second section.
(1) Observations. (i) This is an external forced convection
problem for flow over a flat plate. (ii) Pohlhausen’s solution
for the temperature distribution and heat transfer coefficient is
assumed to be applicable. (iii) O
f interest is the value of the
heat transfer rate from a section of the plate at a specified
location and of a given width. (iv) Newton’s law of cooling
gives the heat transfer rate.
x1
Tf
Vf
2
1
x2
3
x
x3
x4
4
W
Ts
dx
(2) Problem Definition. Determine the local heat transfer coefficient for laminar boundary layer
flow over a flat plate.
(3) Solution Plan. Apply Newton’s law of cooling to determine the local heat flux. Use
Pohlhausen’s solution for the local heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These
are:(1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5)
laminar flow (Rex <5 u105), (6) viscous and thermal boundary layer flow (Rex >100 and Pex >
100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat
plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no
buoyancy (E = 0 or g = 0) and (13) no energy generation ( q ccc 0 ).
(ii) Analysis. Application of Newton’s law to the surface element Wdx gives
dq
h ( x ) [Ts Tf ] Wdx
(a)
where
h = local heat transfer coefficient, W/m2-oC
q = surface heat flux, W
Ts = surface temperature, oC
Tf = free stream temperature, oC
W = plate width, m
x = axial distance measured from the leading edge, m
The local heat transfer coefficient is given by
h( x )
where
k = thermal conductivity, W/m- oC
Vf = free stream velocity, m/s
k
Vf dT (0)
Qx dK
(4.66)
PROBLEM 4.23 (continued)
T (T Ts ) /(Tf Ts ) , dimensionless temperature
Q = kinematic viscosity, m2/s
Vf
, dimensionless variable
Qx
y = vertical coordinate, m
K= y
Substituting (4.66) into (a)
dq [Ts Tf ] Wk
Vf dT (0) dx
Q dK
x
(b)
Integration of (b) between two values of x gives the heat transfer rate for that section.
Application of (b) to sections 3 and 4, give
x
q2
V dT (0) 3 dx
[Ts Tf ] Wk f
Q dK x2 x
q3
V dT (0) 4 dx
[Ts Tf ] Wk f
Q dK x3 x
³
2[Ts Tf ] Wk
Vf dT (0)
( x3 x 2 )
Q dK
(c)
[Ts Tf ] Wk
Vf dT (0)
( x4 x3 )
Q dK
(d)
x
³
where
x2
x3
0.01 m
0.02 m
x4
0.03 m
Taking the ratio of (c) and (d)
x4 x3
q3
q2
(e)
x3 x 2
(iii) Computations. Equation (e) gives
q3
q2
0.03( m) 0.02( m)
0.02( m) 0.01( m)
0.7673
(iv) Checking. Dimensional check: Units of q in (c) should be W:
q
[Ts Tf ]( o C)W ( m)k ( W / m o C)
Vf (m/s) dT (0)
(1 / 1)[ x 4 x3 ] ( m)
Q m 2 /s dK
Limiting check: The heat transfer rate should vanish for Ts
gives q2 q3 0.
Tf . Setting Ts
W
Tf in (c) and (d)
Qualitative check: The heat transfer rate should increase as the free stream velocity, plate width,
or thermal conductivity are increased. This is confirmed by (c) and (d).
Comments. Although each section is rectangular in shape, the same procedure can be followed
to determine the heat transfer rate from any configuration drawn on the plate.
PROBLEM 4.24
A fluid at a uniform velocity and temperature flows over a semi-infinite flat plate. The
surface temperature is uniform. Assume laminar boundary layer flow.
a[ ] What will be the percent change in the local heat transfer coefficient if the free
stream velocity is reduced by a factor of two?
b[ ] What will be the percent change in the local heat transfer coefficient if the distance from
the leading edge is reduced by a factor of two?
(1) Observations. (i) This is an external forced convection problem for flow over a flat
plate. (ii) fOinterest is the variation of the local heat transfer coefficient with free stream
velocity and distance from the leading edge. (iii) Pohlhausen's solution applies to this
problem.
(2) Problem Definition. O
btain a solution for the local heat transfer coefficient and
examine its dependency on the free stream velocity and distance from the leading edge.
(3) Solution Plan. Use Pohlhausen's solution for the loca l heat transfer coefficient over a
semi-infinite flat plate.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable.
These are:(1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional,
(5) laminar flow (Rex <5 u105), (6) viscous and thermal boundary layer flow (Rex >100 and
Pex >100), (7) uniform upstream velocity and te mperature, (8) uniform surface temperature,
(9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible
dissipation, (12) no buoyancy (E = 0 or g = 0) and (13) no energy generation ( q ccc 0 ).
(ii) Analysis. Pohlhausen's solution for the local heat transfer coefficient gives
h
k
Vf dT (0)
Qx dK
where
h = local heat transfer coefficient, W/m2-oC
k = thermal conductivity, W/m-oC
Vf = free stream velocity, m/s
T = temperature variable, oC
Ts = surface temperature, oC
Tf = free stream temperature, oC
T (T Ts ) /(Tf Ts ) , dimensionless temperature
x = axial coordinate, measured from the leading edge, m
y = vertical coordinate, measured from the surface, m
Q = kinematic viscosity, m2/s
(a)
PROBLEM 4.24 (continued)
K= y
Vf
, dimensionless variable
Qx
oHwever, k, Q and
T (0)
are constants. Thus, equation (a) can be rewritten as
dK
h C
Vf
x
(b)
where C is a constant. The percent change in h is given by
Percent change = 100 ( h2 h1 )/h1 = 100 (h2/h1 1 )
(c)
where subscripts 1 and 2 refer to the initial and changed conditions, respectively. Applying
(b) to the two locations and substituting into (c)
ª V x
º
Percent change = 100 « f 2 1 1»
»¼
¬« Vf1 x 2
(d)
(iii) Computations.
a[ ]Percent change in h if the free stream velocity Vf is reduced by a factor of two:
Applying (d) for Vf 2 Vf1 / 2 and x1 = x2, gives
ª V
º
Percent change = 100 « f 2 1» 100 ( 1 / 2 1)
«¬ Vf1
»¼
29.3 %
b[ ]Percent change in h if the distance x from the leading edge is reduced by a factor of two:
Applying (d) for x2 = x1/2 and Vf1 Vf 2 , gives
ª x
º
Percent change = 100 « 1 1» 100( 2 1)
«¬ x 2
»¼
41.4%
(iv) Checking. Dimensional check: Units of h in (a) should be W/m2-oC:
V
h = k (W/m- C) f
Qx
o
m/s
dT * (0)
(1/1) = W/m2-oC
2
m /s m
dK
Qualitative check: As the free stream velocity decreases, the local heat transfer coefficient
should decrease. Computations confirm this as indicated by the negative sign in the percent
change when free stream velocity is reduced by a factor of two.
(5) Comments. (i) The local heat transfer coefficient increases as the distance from the
leading edge is decreased and as the free stream velocity is increased. (ii) Since percent
change involves taking ratios, the problem is solved without knowing the nature of the fluid
and the magnitudes of Vf and x.
PROBLEM 4.25
Use Pohlhausen's solution to derive an expression for the ratio of the thermal boundary layer
thickness for two fluids. The Prandtl number of one fluid is 1.0 and its kinematic viscosity is
012
. u 10 6 m 2 / s . The Prandtl number of the second fluid is 100 and its kinematic viscosity is
6.8 u 10 6 m 2 / s .
0.8
T Ts
Tf Ts
(2) Problem Definition. Derive an expression
for the thermal boundary layer thickness for
laminar flow over a semi-infinite flat plate.
1.0
1.0
100
(1) Observations. (i) This is an external flow
problem. (ii) At the edge of the thermal
boundary layer, y G t , fluid temperature
approaches free stream temperature. That is,
T Tf and T * (Tf Ts ) /(Tf Ts ) 1 . (iii)
According to Pohlhausen's solution, Fig. 4.6,
the thermal boundary layer thickness depends
on the Prandtl number, free stream velocity Vf,
kinematic viscosity Q and location x.
0.71 (air)
1
0.
0.6
0.4
=
Pr
1
0,0
0.2
0
0
2
4
K
6
8
y Vf /Q x
10
12
14
Fig. 7.2
(3) Solution Plan. Use Pohlhausen's solution, Fig. 4.6, to determine G t at Pr = 1 and Pr = 100.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These
are:(1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5)
laminar flow (Rex <5 u105), (6) viscous and thermal boundary layer flow (Rex >100 and Pex >
100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat
plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no
buoyancy (E = 0 or g = 0) and (13) no energy generation ( q ccc 0 ).
(ii) Analysis and Computations. At the edge of the thermal boundary layer, y G t , fluid
temperature is
T ( x, G t ) | Tf
and therefore the dimensionless temperature T * is
T*
(T Ts ) /(Tf Ts ) (Tf Ts ) /(Tf Ts ) 1
(a)
where
T = temperature variable, oC
Ts = surface temperature, oC
Tf = free stream temperature, oC
T * = dimensionless temperature, oC
y = coordinate normal to the plate, m
G t = thermal boundary layer thickness = value of y where T * = 1, m
eLt the subscripts 1 and 2 de note conditions corresponding to Pr = 1 and Pr = 100, respectively.
PROBLEM 4.25 (continued)
Using Fig. 4.6 at Pr = 1, the value of K corresponding to T * = 1 is
K1 G t 1
Vf
| 5 .2
Q1x
(b)
K2
Vf
| 1 .5
Q2x
(c)
Similarly, for Pr = 100
G t2
where
Vf = free stream velocity, m/s
x = coordinate along the plate, m
K = dimensionless variable = y Vf /Q x
Q1 = kinematic viscosity (of fluid with Pr = 1) = 0.12 u 10 6 m2/s
Q2 = kinematic viscosity (of fluid with Pr = 100) = 6.8 u 10 6 m2/s
Taking the ratio of (b) and (c) gives
G t1
G t2
5 .2 Q 1
1 .5 Q 2
5.2 0.12 u 10 6 (m 2 / s)
1.5 6.8 u 10 6 (m 2 / s)
0.46
(iii) Checking. Dimensional check: The right hand side of equation (b) should be
dimensionless:
K1 G t 1 (m)
Vf (m/s)
Q 1 (m 2 /s)x(m)
= dimensionless
(5) Learning and Generalizing. (i) In calculating the ratio G t1 / G t 2 , the location x and the free
stream velocity Vf are assumed to be the same for both fluids. (ii) The ratio G t1 / G t 2 is constant
independent of location.
PROBLEM 4.26
o
Water at 25 C flows over a flat plate with a uniform velocity of 2 m/s. The plate is maintained at
85oC. Determine the following:
a[ ]
b[ ]
c[ ]
d[ ]
The thermal boundary layer thickness at a distance of 8 cm from the leading edge.
The heat flux at this location.
The total heat transfer from the first 8 cm of the plate.
Whether Pohlhausen's solution can be used to find the heat flux at a distance of 80 cm from
the leading edge.
(1) Observations. (i) This is an external forced convection problem for flow over a flat plate.
(ii) The Reynolds number and Peclet number should be checked to determine if the flow is
laminar and if boundary layer approximations are valid. (iii) Pohlhausen's solution is applicable
if 100 < Rex <100 u105 and Pex = Rex Pr >100. (iv) Thermal boundary layer thickness and
heat transfer coefficient vary along the plate. (v) Newton’s law of cooling gives local heat flux.
(vi) The fluid is water.
(2) Problem Definition. The problem is determining the temperature distribution in the fluid.
Knowing the temperature distribution for laminar
Gt
flow (Pohlhausen's solu tion), the thermal boundary Vf y
layer thickness, local and average heat transfer
h(x )
coefficients, local heat flux and total heat transfer Tf 00
rate can be determined.
Ts
qcsc(x ) x
(3) Solution Plan. Check the Reynolds and Peclet
numbers to determine if boundary layer approximations can be made, the flow is laminar and if
Pohlhausen’s solution is applicable. Use Pohlhausen’s solution to determine the local heat
transfer coefficient and Newton’s law of cooling to determine heat flux and heat transfer rate.
(4) Plan Execution.
(i) Assumptions. Assume and verify that Pohlhausen’s solution is applicable. Assumptions
leading to Pohlhausen's solution are: (1) Ne wtonian fluid, (2) steady state, (3) constant
properties, (4) two-dimensional, (5) laminar flow (Rex <5 u105), (6) viscous and thermal
boundary layer flow (Rex >100 and Pex >100), (7) uniform upstream velocity and temperature,
(8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential
energy, (11) negligible dissipation, (12) no buoyancy (E = 0 or g = 0) and (13) no energy
generation ( q ccc 0 ). In addition, (14) no radiation.
(ii) Analysis. The Reynolds and Peclet numbers are defined as
Rex =
Vf x
Q
(a)
and
Pex = Rex Pr
where
Pex = Peclet number
Pr = Prandtl number
Rex = Reynolds number
Vf = upstream velocity = 2 m/s
x = distance from the leading edge of the plate, m
(b)
PROBLEM 4.26 (continued)
Q = kinematic viscosity, m2 /s
Properties are evaluated at the film temperature Tf defined as
Tf =
Ts Tf
2
(c)
where
Tf = film temperature, oC
Ts = surface temperature = 85oC
Tf = free stream temperature = 25oC
To determine if the flow is laminar or turbulent, compare the Reynolds number with the
transition Reynolds number. For the flow over a flat plate, transition Reynolds number Rex is
t
5
Re xt = 5u10
(d)
The flow is laminar if Rex < Re xt . V
iscous boundary layer approximations are valid for
Re x ! 100
Thermal boundary layer approximations are valid for
Pex >100
Substituting into (c)
Tf = (85 +25)( oC)/2 = 55oC
Properties of water at this temperature are
k = 0.6458 W/m-oC
Pr = 3.27
Q = 0.5116u10-6 m2/s
The Reynolds and Peclet numbers are determined at x = 0.08 m
Rex =
2( m / s)0.08( m)
0.5116 u 10 6 m2 / s
312,744
and
Pex = 312,744u3.27 = 1.0227u106
Therefore, boundary layer approximations can be made and the flow is laminar. Pohlhausen’s
solution is applicable.
G t . Pohlhausen’s solution, Fig. 4.6, is used to determine the
a[ ] oBundary layer thickness
thermal boundary layer thickness. At the edge of the thermal boundary layer, y = G t , fluid
temperature is approximately the same as ambient temperature. That is
T(x, G t ) | Tf
O
r, in terms of the dimensionless temperature
T *=
(e)
T*
T ( x, G t ) Ts
|1
Tf Ts
(f)
PROBLEM 4.26 (continued)
where
T = temperature variable, oC
T * = dimensionless temperature
y = distance from the plate, m
G t = thermal boundary layer thickness, m
At T = 1 and Pr = 3.27, Fig. 4.6 gives the thermal boundary layer thickness in terms of the
dimensionless variable K
V
V
(g)
K | 3 = y f = Gt f
Qx
Qx
Solving (g) for G t
Gt = 3
Qx
(h)
Vf
[b]L
ocal surface heat flux
qscc . This is the local heat transfer per unit area. Applying Newton's
law of cooling at location x gives
qscc = h (Ts - Tf )
(i)
where
h = local heat transfer coefficient, W/m2-oC
qscc = local surface heat flux, W/m2
The local heat transfer coefficient is given by Pohlhausen’s solution. For 0.5 < Pr <50, equation
(7.24b) gives
V
h = 0.332 k f Pr1/3
(j)
Qx
Substituting (j) into (i) gives
V
qscc = 0.332 k (Ts - Tf ) f Pr1/3
(k)
Qx
c[ ] Total heat transfer rate qT. Applying Newton’s law of cooling and using the average heat
transfer coefficient gives the total heat transfer from the plate
qT = h LW (Ts Tf )
(l)
where
h = average heat transfer coefficient, W/m2-oC
L = length of plate = 0.08 m
W = width of plate, m
The average heat transfer coefficient is given by equations (4.67) and (4.71b)
h
0.664k
Vf
Pr
vL
1/ 3
Substituting (m) into (l) and rearranging
qT
0.664k (Ts Tf ) Re L ( Pr )1/3
W
(m)
(n)
PROBLEM 4.26 (continued)
where ReL is the Reynolds number at x = L.
(iii) Computations.
a[ ] oBundary layer thickness G t . Substituting into equation (h)
Gt = 3
0.5116 u 10 6 (m 2 / s)0.08(m)
= 0.00043 m = 0.43 mm
2(m / s)
b[ ]Surface heat flux
qscc . Equation (k) gives
qscc = 0.332 (0.6458)(W/m-oC) (85 -25)(oC)
2 ( m / s)
(3.27)1/3
6
2
0.5116 u 10 ( m / s) 0.08( m)
=133,477 W/m2
c[ ]Total heat transfer rate qT . Equation (n) gives
qT
= 0.664 (0.6458)(W/m-oC) (85 - 25)(oC) (312,744)1/2 (3.27)1/3 = 21,356 W/m
W
d[ ]aVlidity of Pohlhausen’s solution at
Rex =
2( m / s)0.8( m)
0.5116 u 10 6 m2 / s
x = 0.8 m. The Reynolds number at x = 0.8 m is
3127
.
u 106
Since the Reynolds number is greater than the transition Reynolds number the flow is turbulent
and therefore Pohlhausen’s solution does not apply.
(iv) Checking. Dimensional check: Computations showed that units for equations (a), (h),
(k) and (n) are dimensionally consistent.
Quantitative check: Equations (4.66) and (4.67) show that for a plate of length L, the average
heat transfer coefficient is twice the local coefficient at x = L. This means that the average heat
flux is twice the flux at x = L. Thus, the average flux for this problem is 2(133,477)(W/m2) =
266,954(W/m2). Therefore
qT LW q csc
(o)
where q csc is the average heat flux. Substituting into (o)
qT
= 0.08(m)266,954(W/m2) = 21,356 W/m
W
Limiting check: If Ts = Tf, the local heat flux and the total heat transfer rate should vanish.
Setting Ts = Tf in equations(k) and (n) gives the expected results.
(5) Comments. (i) It is important to check the conditions for boundary layer flow and for
laminar flow before proceeding with the solution (i.e. check the Reynolds and Peclet numbers).
(ii) Typically, the thickness of the thermal boundary layer is relatively small. In this example, at
a distance of 8 cm from the leading edge is 0.43 mm.
PROBLEM 4.27
The cap of an electronic package is cooled by forced convection. The free stream temperature is
25oC. The Reynolds number at the downstream end of the cap is 110,000. Surface temperature
was found to be 145oC. H
owever, reliability requires that
cap at Ts
Vf , Tf
surface temperature does not exceed 83oC. nOe possible
solution to this design problem is to increase the free stream
velocity by a factor of 3. Y
ou are asked to determine if
surface temperature under this plan will meet design
specification.
(1) Observations. (i) This is an external forced convection problem over a flat plate. (ii)
Increasing the free stream velocity, increases the average heat transfer coefficient. This in turn
causes surface temperature to drop. (iii) B
ased on this observation, it is possible that the
proposed plan will meet design specification. (iv) Since the Reynolds number at the downstream
end of the package is less than 500,000, it follows that the flow is laminar throughout. (v)
Increasing the free stream velocity by a factor of 3, increases the Reynolds number by a factor
of 3 to 330,000. At this Reynolds number the flow is still laminar. (vi) The power supplied to the
package is dissipated into heat and transferred to the surroundings from the surface. (vii)
Pohlhausen's solution can be applied to this problem. (viii) The ambient fluid is unknown.
(2) Problem Definition. Find the relationship between surface temperature, free stream velocity
and heat transfer coefficient.
(3) Solution Plan. Surface temperature is related to heat transfer coefficient by Newton's law of
cooling. H
eat transfer coefficient is related to free stream velocity in Pohlhausen’s solution.
Apply Newton’s law of cooling and Pohlhausen’s solution to the cap.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are:
(1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5) laminar
flow (Rex <5 u105), (6) viscous and thermal boundary layer flow (Rex >100 and Pex >100), (7)
uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10)
negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no buoyancy
(E = 0 or g = 0) and (13) no energy generation ( q ccc 0 ). In addition, (14) neglect radiation and
(15) all power supplied to package leaves the cap as heat.
(ii) Analysis. Application of Newton's law of cooling gives
qT = h A (Ts - Tf)
where
A = surface area, m2
h = average heat transfer coefficient, W/m2-oC
qT = total surface heat transfer rate, W
Ts = surface temperature, oC
Tf = free stream temperature = 25oC
(a)
PROBLEM 4.27 (continued)
The heat transfer rate qT is determined by the power supplied to the package which is assumed
constant. Changing the free stream velocity results in a change in the average heat transfer
coefficient and surface temperature. oHwever, qT , A and Tf remain the same. L
et subscripts 1
and 2 refer to conditions corresponding to the two velocities Vf1 and Vf2, where
Vf2 = 3 Vf1
(b)
qT = h1 A (Ts1 - Tf)
(c)
qT = h2 A (Ts2 - Tf)
(d)
Applying (a) to the two cases gives
and
Taking the ratio of (c) and (d)
h1 ( Ts1 Tf )
h2 ( Ts 2 Tf )
1
Solving the above for Ts2
Ts2 = Tf +
h1
(Ts1 Tf )
h2
(e)
The average heat transfer coefficient for laminar flow over a flat plate is given by Pohlhausen's
solution (7.25)
h
0.664k
Vf
( Pr )1 / 3
QL
(f)
where
k = thermal conductivity, W/m-oC
L = length of cap, m
Pr = Prandtl number
Vf = free stream velocity, m/s
Q = kinematic viscosity, m2/s
Properties in Pohlhausen’s solution are evaluated at the film temperature, T f
(Ts Tf ) / 2 .
Since surface temperature is expected to change, it follows that properties will also change.
oHwever, this effect cannot be considered in th e solution since the cooling fluid is not known. As
a first approximation, changes in properties due to changes in surface temperature will be
neglected. Thus, k, Pr and Q are assumed to remain constant and equation (f) simplifies to
h
C
Vf
(g)
where C is a constant. Applying (g) to the two cases and taking the ratio of the resulting
equations
h1
Vf1
(h)
Vf 2
h2
Substituting (h) into (e) gives
PROBLEM 4.27 (continued)
Ts2 = Tf +
Vf1
(Ts1 - Tf)
Vf 2
(i)
(iii) Computations. Setting Vf2 = 3 Vf1 and Ts1 = 145oC in (i) gives
Ts2 = 25 (oC) + 1 / 3 (145 25)( o C) = 94.3oC
Since this temperature is above the design limit of 83oC, the proposed plan will not work.
(iv) Checking. Dimensional check: Since the ratio of velocities is dimensionless, it follows
that equation (i) is dimensionally consistent.
Qualitative check: As Vf is increased, surface temperature Ts should decrease. This is confirmed
by equation (i).
Limiting check: There should be no change in surface temperature for the special case of Vf2 =
V f1
Vf1. Setting
= 1 in (i) gives the expected result of Ts2 = Ts1.
Vf 2
(5) Comments. (i) This problem is solved without knowing the size of the cap, the magnitude
of the free stream velocity and the nature of the fluid. (ii) Neglecting radiation is a conservative
assumption since it overestimates surface temperature. Taking radiation into consideration
results in a lower temperature than 94.3oC. Thus, the proposed design modification should not
be rejected without first examining the effect of radiation. (iii) If the cooling fluid is known,
changes in properties due to changes in surface temperature can be accounted for using a trial
and error procedure. A value for Ts2 is assumed, properties at Tf2 are determined and h1 / h2 is
calculated and used in equation (e) to calculate Ts2. If the calculated Ts2 is not close to the
assumed value, the procedure is repeated until a satisfactory agreement is obtained between
assumed and calculated values. (iv) For Vf2 = 4.3Vf1, the resulting surface temperature will be
83oC.This meets design specification.
PROBLEM 4.28
The back of the dinosaur Stegosaurus has two rows
of fins. Each row is made up of several fins arranged
in line and separated by a space. One theory
suggests that providing a space between neighboring
fins reduces the weight on the back of the dinosaur
when compared with a single long fin along the back.
On the other hand, having a space between
neighboring fins reduces the total surface area. This
may result in a reduction in the total heat loss.
Model the fins as rectangular plates positioned in line as shown. The length of each plate is L
and its height is H. Consider two fins separated by a distance L. Compare the heat loss from the
two fins with that of a single fin of length 3L and height H. Does your result support the
argument that spaced fins lead to a reduction in heat loss? To simplify the analysis assume
laminar flow.
(1) Observations. (i) This is an external forced convection problem of flow over a flat plate. (ii)
Convection heat transfer from a surface can be determined using Newton’s law of cooling. (iii)
The local heat transfer coefficient changes along the plate. The total heat transfer rate can be
determined using the average heat transfer coefficient. (iv) For laminar flow, Pohlhausen's
solution gives the heat transfer coefficient. (v) For two in-line fins heat transfer from the down
stream fin is influenced by the upstream fin. The further the two fins are apart the less the
interference will be.
(2) Problem Definition. Determine the average heat transfer coefficient for plates of length L
and 3L.
(3) Solution Plan. Use Pohlhausen’s solution to determine the average heat transfer coefficient.
Apply Newton's law of cooling to determine the h eat transfer rate from two plates of length L
each and compare with the heat transfer from a single plate of length 3L.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable.
These are: (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5)
laminar flow (Rex <5 u105), (6) viscous and thermal boundary layer flow (Rex >100 and Pex >
100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat
plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no
buoyancy (E = 0 or g = 0) and (13) no energy generation ( q ccc 0 ). In addition, (14) neglect
radiation and (15) neglect interference between in-line plates.
(ii) Analysis. O
f interest is the ratio of the total heat transfer rate from two single plates
of length L each and a single plate of length 3L. Newton’s law of cooling applied to a single plate
of length L gives
q1 h A(Ts Tf )
(a)
PROBLEM 4.28 (continued)
where
A plate area, m 2
h average heat transfer coefficient, W/m 2 o C
q1 heat transfer rate from a single plate of length L, W
Ts
surface temperature, o C
Tf
free stream temperature, o C
The area of a single plate of length L and height H is
A = HL
(b)
The average Nusselt number is obtained from Pohlhausen's solution
hL
k
Nu L
0.664 Pr 1/ 3 Re L
(c)
where
k = thermal conductivity, W/m-oC
L = plate length, m
Pr Prandtl number
Re L Reynolds number
The Reynolds number is defined as
Re
Vf L
(d)
Q
where
Vf free stream velocity, m/s
Q = kinematic viscosity, m2/s
Substituting (b)-(d) into (a)
q1
Vf
0.664 Pr 1 / 3 kH
Q
(Ts Tf ) L
(e)
The total heat transfer rate from two in-line fins, q 2 , is
q2
2q1
2 u 0.664 Pr 1 / 3 kH
Vf
Q
(Ts Tf ) L
(f)
The heat transfer rate q3 from a single plate of length 3L is obtained from (e) by replacing L
with 3L
V
q3 0.664 Pr 1 / 3 kH f (Ts Tf ) 3L
(g)
Q
Taking the ratio of (f) to (g)
q2
q3
2
3
1.15
(h)
PROBLEM 4.28 (continued)
(iii) Checking. Dimensional check: Units of q1 in equation (e) should be W
q1
0.664 Pr 1/ 3 k ( W/m o C) H (m)
Vf (m / s)
2
Q ( m / s)
(Ts Tf )( o C) L(m) = W
Limiting check: If Ts Tf the heat transfer rate should vanish regardless of plate arrangement.
Setting Ts Tf in (f) and (g) gives
q2
q3
0
(5) Comments. (i) B
ased on the assumption of no inte rference between neighboring in-line
plates, the heat transfer rate from two in-line plates of length L each separated by a distance L
exceeds that of a single plate of length 3L by 15%
. The weight of tw o in-line plates of length L
each is 2/3rd of that of a single plate of length 3L. Thus the two in-line plate arrangement has
advantages in heat transfer rate and weight when compared to a single plate of length 3L. (iii)
The analysis can be generalized to n in-line plates each of length L separated by spacing L as
compared with a single plate of length (2n-1)L. The ratio of the heat transfer rates for this case is
given by
qn
n
(i)
q 2 n 1
2n 1
For example, for n = 10
q10
10
2.29
q19
2 u 10 1
Thus the heat transfer rate from 10 in-line plates of length L each separated by a distance L
exceeds that of a single plate of length 19L by 129%
. The weight is reduced by approximately ½
.
PROBLEM 4.29
A fluid with Prandtl number 0.098 flows over a semiinfinite flat plate. The free stream temperature is Tf and
the free stream velocity is Vf . The surface of the plate is
maintained at uniform temperature Ts . Assume laminar
flow.
Vf
Tf 0
x1
x
x2
W
dx
a[ ] Derive an equation for the local Nusselt number.
[b] Determine the heat transfer rate from a section of the plate between x1 and x 2 . The width of
the plate is W.
c[ ] Derive an equation for the thermal boundary layer thickness G t (x).
(1) Observations. (i) This is an external forced convection problem for flow over a flat plate.
(ii) Pohlhausen’s solution for the temperature distribution and heat transfer coefficient is
assumed to be applicable. (iii) Knowing the heat transfer coefficient, the local Nusselt number
can be determined. (iv) the Newton’s law of cooling gives the heat transfer rate. (iv)
Pohlhausen’s solution gives the thermal boundary layer thickness.
(2) Problem Definition. Determine the local heat transfer coefficient and thermal boundary
layer thickness for laminar boundary layer flow over a flat plate.
(3) Solution Plan. Use Pohlhausen’s solution to determine the local heat transfer coefficient.
Apply Newton’s law of cooling to determine the heat transfer rate. Use Fig. 4.6 to determine the
thermal boundary layer thickness.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These
are:(1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5)
laminar flow (Rex <5 u105), (6) viscous and thermal boundary layer flow (Rex >100 and Pex >
100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat
plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no
buoyancy (E = 0 or g = 0) and (13) no energy generation ( q ccc 0 ).
(ii) Analysis. [a]: The local heat tran sfer coefficient is given by
h( x )
k
Vf dT (0)
Qx dK
where
k = thermal conductivity, W/m- oC
Vf = free stream velocity, m/s
T (T Ts ) /(Tf Ts ) , dimensionless temperature
Q = kinematic viscosity, m2/s
Vf
, dimensionless variable
Qx
y = vertical coordinate, m
K= y
(4.66)
PROBLEM 4.29 (continued)
The local Nusselt number is defined as
hx
k
Nu x
(a)
Substituting (4.66) into (a)
Nu x
Re x
dT (0)
dK
(b)
where Re x is the local Reynolds number defined as
Vf x
Re x
(c)
Q
The constant dT (0) / dK is given in Table 4.2
dT (0)
0.138
dK
Substituting (d) into (b)
Nu x 0.138 Re x
(d)
(e)
[b]Application of Newton’s law to the surface element Wdx gives
h( x ) [Ts Tf ] Wdx
dq
(f)
where
h = local heat transfer coefficient, W/m2-oC
q = surface heat flux, W
Ts = surface temperature, oC
Tf = free stream temperature, oC
W = plate width, m
x = axial distance measured from the leading edge, m
Substituting (4.66) into (f)
dq [Ts Tf ] Wk
Vf dT (0) dx
Q dK
x
(g)
Integration of (g) from x1 to x 2
x
V dT (0) 2 dx
q [Ts Tf ] Wk f
Q dK x
x
1
³
2[Ts Tf ] Wk
Vf dT (0)
( x 2 x1 )
Q dK
Introducing the definition of the local Reynolds number (c) and using (d), the above gives
q 0.276 2[Ts Tf ] Wk ( Re x Re x )
2
1
(h)
[c] At the edge of the thermal boundary layer, y G t the dimensionless temperature in Fig. 4.6
is equal to unity
T Ts
|1
(i)
Tf Ts
For a fluid with Pr 0.098 the corresponding value of K ( x, G t ) is obtained from Fig. 4.6 as
PROBLEM 4.29 (continued)
Kt
Vf
| 12
Qx
Gt
(j)
Solving (b) for G t
Qx
G t | 12
(k)
Vf
Expressing this result in dimensionless form
Gt
x
| 12
Q
Vf x
12
|
(l)
Re x
(iii) Checking. Dimensional check: (1) Each term in (e) and (l) is dimensionless.
(2) Units q in equation (h) are
q
[Ts Tf ]( o C)W (m)k ( W / m o C)
W
Limiting check: The heat transfer rate should vanish for Ts
q 0.
Tf . Setting Ts
Tf in (h) gives
Qualitative check: The thermal boundary layer thickness increases with distance x. Solution (k)
confirms this behavior.
(5) Comments. (i) The value of dT (0) / dK was obtained by interpolation in Table 4.2. A more
accurate value can be obtained using equation (6.46)
dT (0)
dK
>0.332@ Pr
f
³
K
2
ªd f º
« 2»
¬ dK ¼
(4.64)
Pr
dK
H
owever, this requires numerical ev aluation of the integral in (6.46).
(ii)Recalling that lBasius solution gives
G
5.2
x
Rex
(4.46)
Comparing (l) with (4.46) shows that G t ! G for Pr 0.098 . Examination of Fig. 4.6 shows that
G t ! G for all fluids with Pr 1.0 and that G t G for Pr 1.0 .
PROBLEM 4.30
Two identical triangles are drawn on the surface of a flat
plate as shown. The plate, which is maintained at uniform
surface temperature, is cooled by laminar forced
convection. Determine the ratio of the heat transfer rate
from the two triangles, q1/q2.
L
x
Tf
Vf
y1
1
2
y2
H
Ts
top view
(1) Observations. (i) This is an external forced convection
problem of flow over a flat plate. (ii) Convection heat transfer from a surface can be determined
using Newton’s law of cooling. (iii) The local heat transfer coefficient changes along the plate.
(iv) For each triangle the area changes with distance along the plate. (v) The total heat
transfer rate can be determined by integration along the length of each triangle. (vi)
Pohlhausen's solution may be a pplicable to this problem.
(2) Problem Definition. Determine the local heat transfer coefficient along each triangle.
(3) Solution Plan. Apply Newton's law of cooling to an element of each triangle, ydx, determine
the local heat transfer coefficient along the plate and integrate over the area.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable.
These are: (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5)
laminar flow (Rex <5 u105), (6) viscous and thermal boundary layer flow (Rex >100 and Pex >
100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat
plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no
buoyancy (E = 0 or g = 0) and (13) no energy generation ( q ccc 0 ). In addition, (14) neglect
radiation.
(ii) Analysis. O
f interest is the ratio of the total heat transfer rate from triangle 1 to
triangle 2. Since both the heat transfer coefficient and area vary along each triangle, it follows
that Newton's law of cooling s hould be applied to an element dA at a distance x from the leading
edge:
(a)
dq = h(x) (Ts - Tf)dA
where
dA = area of element, m2
dq = rate of heat transfer from element, W
h = local heat transfer coefficient, W/m2-oC
Ts = surface temperature, oC
Tf = free stream temperature, oC
x = distance along plate, m
The local heat transfer coefficient h(x) is obtained from Pohlhausen's solution (4.66)
h
where
k
Vf dT (0)
Qx dK
(b)
PROBLEM 4.30 (continued)
k = thermal conductivity, W/m-oC
Vf = free stream velocity, m/s
T (T Ts ) /(Tf Ts ) , dimensionless temperature
Q = kinematic viscosity, m2/s
Vf
, dimensionless variable
Qx
y = vertical coordinate, m
K= y
Noting that all quantities in equation (b) are constant except the variable x, (b) is rewritten as
h=
constant
(c)
x
The next step is to determine the infinitesimal area dA for each triangle. Using the subscripts 1
and 2 to refer to triangles 1 and 2, respectively,
dA1
y1 ( x)dx
(d)
dA2
y 2 ( x)dx
(e)
and
where
y1 ( x) = side of element in triangle 1, m
y 2 ( x) = side of element in triangle 2, m
Similarity of triangles gives
H
( L x)
L
H
y 2 ( x)
x
L
y1 ( x)
(f)
(g)
Substituting (f) into (d) and (g) into (e) gives
dA1 = (H/L)(L x )dx
(h)
dA2 = (H/L) x dx
(i)
where
H = base of triangle, m
L = length of triangle, m
Substituting (c) and (h) into (a) and integrating from x = 0 to x = L, gives
L
q1 =
³ dq ³
1
H Lx
C (Ts Tf )
dx
L x1 / 2
0
H
C (Ts Tf )
L
L
³
0
Lx
dx
x1 / 2
Carrying out the integration yields
q1
(4 / 3)C (Ts Tf ) HL1 / 2
Similarly, substituting (c) and (i) into (a) and integration from x = 0 to x = L gives
(j)
PROBLEM 4.30 (continued)
L
q2 =
³ dq ³
2
H x
C (Ts Tf )
dx
L x1 / 2
0
H
C (Ts Tf )
L
L
³
x1 / 2 dx
0
Carrying out the integration yields
q2 = (2/3)C(Ts - Tf )H L1/2
(k)
q1
q2
(l)
Taking the ratio of (j) and (k)
2
(iii) Checking. Dimensional check: Units of q1 in equation (j) should be W. First, units
of C are determined
C = k(W/m-oC)[Vf (m/s)/Q(m2/s)]1/2
dT (0)
(1/1) = W/m3/2-oC
dK
Units of q1 are
q1 = C(W/m3/2-oC)(Ts- Tf )(oC)H(m)L1/2 (m1/2) = W
Since q 2 has the same form as q1 , it follows that the units of q 2 in equation (k) are also correct.
Qualitative check: The result shows that the rate of heat transfer from triangle 1 is greater than
that from triangle 2. This is expected since the heat transfer coefficient increases as the distance
from the leading edge is decreased and triangle 1 has its base at x = 0 where h is maximum.
(5) Comments. (i) Although the two triangles have the same area, the rate of heat transfer from
triangle 1 is double that from triangle 2. Thus, orientation and proximity to the leading edge of a
flat plate play an important role in determining the rate of heat transfer. (ii) The same approach
can be used to determine heat transfer for configurations other than rectangles, such as circles
and ellipses.
PROBLEM 4.31
An isosceles triangle is drawn on a semi-infinite flat
plate at a uniform surface temperature Ts . Consider
laminar uniform flow of constant properties fluid
over the plate. Determine the rate of heat transfer
between the triangular area and the fluid.
top view
Tf
Vf
dx
y
H Ts
(1) Observations. (i) This is an external forced
L
convection problem of flow over a flat plate. (ii)
H
eat transfer rate can be determined using Newton’s
law of cooling. (iii) The local heat transfer coefficient changes along the plate. (iv) The area
changes with distance along the plate. (v) The total heat transfer rate can be determined by
integration along the length of the triangle. (vi) Pohlhausen's solu tion may be applicable to this
problem.
(2) Problem Definition. Determine the local heat transfer coefficient along the triangle.
(3) Solution Plan. Apply Newton's law of cooling to an element ydx, determine the local heat
transfer coefficient along the plate and integrate over the area.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable.
These are: (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5)
laminar flow (Rex <5 u105), (6) viscous and thermal boundary layer flow (Rex >100 and Pex >
100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat
plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no
buoyancy (E = 0 or g = 0) and (13) no energy generation ( q ccc 0 ). In addition, (14) neglect
radiation.
(ii) Analysis. Taking advantage of symmetry, we consider the right angle triangle
representing half the isosceles triangle. Newton's law of cooling applied to an element ydx at a
distance x from the leading edge gives
dq = 2h(x) (Ts - Tf)ydx
(a)
where
h = local heat transfer coefficient, W/m2-oC
dq = rate of heat transfer from 2 elements, W
Ts = surface temperature, oC
Tf = free stream temperature, oC
x = distance along plate, m
y = y(x) height of element
Note that the factor 2 is introduced to account for heat transfer from the two right angle triangles
representing the isosceles triangles. The local heat transfer coefficient h(x) is obtained from
Pohlhausen's solution
h
k
Vf dT (0)
Qx dK
(4.66)
PROBLEM 4.31 (continued)
where
k = thermal conductivity, W/m-oC
Vf = free stream velocity, m/s
T (T Ts ) /(Tf Ts ) , dimensionless temperature
Q = kinematic viscosity, m2/s
Vf
, dimensionless variable
Qx
The variable y(x) is given by
K= y
y( x)
H
x
L
(b)
where
H = base of triangle, m
L = length of triangle, m
Substituting (4.66) and (b) into (a) and integrating from x = 0 to x = L, gives
q
V dT ( 0)
H
2(Ts Tf ) k f
Q dK
L
L
³x
1/ 2
dx
0
Carrying out the integration yields
q
V L dT (0)
4
k (Ts Tf ) H f
Q
dK
3
(c)
This result can be expressed in terms of the Reynolds number as
q
dT (0)
4
k (Ts Tf ) H ReL
dK
3
(d)
where
ReL
Vf L
(e)
Q
(iii) Checking. Dimensional check: Units of q in equation (d) should be W
dT (0)
(1/1) = W
dK
Limiting check: The heat transfer rate should vanish for Ts
q 0.
q = k(W/m-oC) (Ts Tf )( o C) H (m)
Tf . Setting Ts
Tf in (d) gives
(5) Comments. Examination of (4.66) shows that the heat transfer coefficient decreases as x is
increased. Thus, the same triangle rotated 180 degrees, with its base at x = 0, will have a higher
heat transfer rate.
PROBLEM 4.32
Determine the total heat transfer rate from a half circle
drawn on a semi-infinite plate as shown. Assume laminar
two-dimensional boundary layer flow over the plate.
(1) Observations. (i) This is an external forced convection
problem of flow over a flat plate. (ii) H
eat transfer rate
can be determined using Newton’s law of cooling. (iii)
The local heat transfer coefficient changes along the
plate. (iv) The area changes with distance along the plate.
(v) The total heat transfer rate can be determined by
integration along over the area of the semi-circle. (vi)
Pohlhausen's solution gives the heat transfer coefficient.
(2) Problem Definition. Determine the local heat transfer coefficient along the semi-circle.
(3) Solution Plan. Apply Newton's law of cooling to an element ydx, determine the local heat
transfer coefficient along the plate and integrate over the area.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable.
These are: (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5)
laminar flow (Rex <5 u105), (6) viscous and thermal boundary layer flow (Rex >100 and Pex >
100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat
plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no
buoyancy (E = 0 or g = 0) and (13) no energy generation ( q ccc 0 ). In addition, (14) neglect
radiation.
(ii) Analysis. Taking advantage of symmetry, we consider the upper half the semi-circle.
Newton's law of cooling applied to an element ydx at a distance x from the leading edge gives
dq = 2h(x) (Ts - Tf)ydx
(a)
where
h = local heat transfer coefficient, W/m2-oC
dq = rate of heat transfer from upper and lower elements, W
Ts = surface temperature, oC
Tf = free stream temperature, oC
x = distance along plate, m
y = y(x) height of element
Note that the factor 2 is introduced to account for heat transfer from the two halves of the semicircle. The local heat transfer coefficient h(x) is obtained from Pohlhausen's solution
h
where
k = thermal conductivity, W/m-oC
k
Vf dT (0)
Qx dK
(4.66)
PROBLEM 4.32 (continued)
Vf = free stream velocity, m/s
T (T Ts ) /(Tf Ts ) , dimensionless temperature
Q = kinematic viscosity, m2/s
V
K = y f , dimensionless variable
Qx
The variable y(x) is given by
y( x)
ro2 ( x ro ) 2
This can be simplified to
y( x)
2 xro x 2
(b)
where
ro
radius of semi-circle, m
Substituting (4.66) and (b into (a) and integrating from x
q
2ro , gives
2 ro
V dT (0)
2(Ts Tf )k f
Q dK ro
³
ro to x
( 2ro x ) dx
Carrying out the integration yields
q
V dT (0)
4
k (Ts Tf ) ro3 / 2 f
Q dK
3
(c)
This result can be expressed in terms of the Reynolds number as
q
dT (0)
4
k (Ts Tf ) ro Rero
dK
3
(d)
where
Rero
Vf ro
(e)
Q
(iii) Checking. Dimensional check: Units of q in equation (d) should be W
dT (0)
(1/1) = W
dK
Limiting check: The heat transfer rate should vanish for Ts
q 0.
q = k(W/m-oC) (Ts Tf )( o C) ro (m)
Tf . Setting Ts
Tf in (d) gives
(5) Comments. Examination of (4.66) shows that the heat transfer coefficient decreases as x is
increased. Thus, the same semi-circle rotated 180 degrees and remaining at distance ro from the
leading edge of the plate, will have a lower heat transfer rate.
PROBLEM 4.33
Consider steady, two-dimensional, laminar boundary
layer flow over a semi-infinite plate. The surface is
maintained at uniform temperature Ts . Determine the
total heat transfer rate from the surface area described
by y ( x) H x / L as shown.
y
y
Tf
Vf
x
H
x
L
H
L
top view
Ts
(1) Observations. (i) This is an external forced
convection problem of flow over a flat plate. (ii) H
eat transfer rate can be determined using
Newton’s law of cooling. (iii) The local heat transfer coefficient changes along the plate. (iv)
The area changes with distance along the plate. (v) The total heat transfer rate can be
determined by integration along the length of the triangle. (vi) Pohlhausen's solution may be
applicable to this problem.
(2) Problem Definition. Determine the local heat transfer coefficient along the triangle.
(3) Solution Plan. Apply Newton's law of cooling to an element ydx, determine the local heat
transfer coefficient along the plate and integrate over the area.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable.
These are: (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5)
laminar flow (Rex <5 u105), (6) viscous and thermal boundary layer flow (Rex >100 and Pex >
100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat
plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no
buoyancy (E = 0 or g = 0) and (13) no energy generation ( q ccc 0 ). In addition, (14) neglect
radiation.
(ii) Analysis. Newton's law of cooling applied to an element ydx at a distance x from the
leading edge gives
dq = h(x) (Ts - Tf)ydx
where
h = local heat transfer coefficient, W/m2-oC
dq = rate of heat transfer from 2 elements, W
Ts = surface temperature, oC
Tf = free stream temperature, oC
x = distance along plate, m
y = y(x) height of element
(a)
The local heat transfer coefficient h(x) is obtained from Pohlhausen's solution
h
where
k = thermal conductivity, W/m-oC
k
Vf dT (0)
Qx dK
(4.66)
PROBLEM 4.33 (continued)
Vf = free stream velocity, m/s
T (T Ts ) /(Tf Ts ) , dimensionless temperature
Q = kinematic viscosity, m2/s
V
K = y f , dimensionless variable
Qx
The variable y(x) is given by
y( x)
x
L
H
(b)
where
H = height at x = L, m
L = length, m
Substituting (4.66) and (b) into (a) and integrating from x = 0 to x = L, gives
V dT (0)
H
k f
(Ts Tf )
Q dK
L
q
L
³ dx
0
Carrying out the integration yields
k (Ts Tf ) H
q
Vf L dT (0)
Q
dK
(c)
This result can be expressed in terms of the Reynolds number as
q
k (Ts Tf ) H ReL
dT (0)
dK
(d)
where
ReL
Vf L
(e)
Q
(iii) Checking. Dimensional check: Units of q in equation (d) should be W
dT (0)
(1/1) = W
dK
Limiting check: The heat transfer rate should vanish for Ts
q 0.
q = k(W/m-oC) (Ts Tf )( o C) H (m)
Tf . Setting Ts
Tf in (d) gives
(5) Comments. Examination of (4.66) shows that the heat transfer coefficient decreases as x is
increased. Thus, the same shape rotated 180 degrees, with its base H at x = 0, will have a higher
heat transfer rate.
PROBLEM 4.34
Fluid flows over a semi-infinite flat plat which is maintained
at uniform surface temperature. It is desired to double the
rate of heat transfer from a circular area of radius R1 by
increasing its radius to R2 . Determine the percent increase
in radius need to accomplish this change. In both cases the
circle is tangent to the leading edge. Assume laminar
boundary layer flow with constant properties.
R1
Tf
R2
Ts
Vf
top view
(1) Observations. (i) This is an external forced convection problem of flow over a flat plate. (ii)
this problem involves determining the heat transfer rate from a circle tangent to the leading edge
of a plate. (iii) H
eat transfer rate can be determined using Newton’s law of cooling. (iv) The
local heat transfer coefficient changes along the plate. (v) The area changes with distance
along the plate. (vi) The total heat transfer rate can be determined by integration along the
length of the triangle. (vii) Pohlhausen's so lution may be applicable to this problem.
(2) Problem Definition. Determine the local heat transfer coefficient along the triangle.
(3) Solution Plan. Determine the heat transfer rate from a circle which is tangent to the leading
edge of a plate. Apply Newton's law of cooling to an element ydx, determine the local heat
transfer coefficient along the plate and integrate over the area.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable.
These are: (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5)
laminar flow (Rex <5 u105), (6) viscous and thermal boundary layer flow (Rex >100 and Pex >
100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat
plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no
buoyancy (E = 0 or g = 0) and (13) no energy generation ( q ccc 0 ). In addition, (14) neglect
radiation.
(ii) Analysis. Consider a circle of radius R
which is tangent to the leading edge of the plate.
Taking advantage of symmetry, we consider the upper
half of the circle. Newton's la w of cooling applied to an
element ydx at a distance x from the leading edge gives
dq = 2h(x) (Ts - Tf)ydx
where
h = local heat transfer coefficient, W/m2-oC
dq = rate of heat transfer from 2 elements, W
Ts = surface temperature, oC
Tf = free stream temperature, oC
x = distance along plate, m
y = y(x) height of element
(a)
y
Tf
Vf
0
x
R
Rx
PROBLEM 4.34 (continued)
Note that the factor 2 is introduced to account for the two haves of the circle. The local heat
transfer coefficient h(x) is obtained from Pohlhausen's solution
h
k
Vf dT (0)
Qx dK
(4.66)
where
k = thermal conductivity, W/m-oC
Vf = free stream velocity, m/s
T (T Ts ) /(Tf Ts ) , dimensionless temperature
Q = kinematic viscosity, m2/s
V
K = y f , dimensionless variable
Qx
The variable y(x) is given by
y( x)
R 2 ( R x)2
2 Rx x 2
(b)
Substituting (4.66) and (b) into (a) and integrating from x = 0 to x = 2R, gives
q
2(Ts Tf )k
Vf dT (0)
Q dK
2R
³
( 2 R x ) dx
0
Carrying out the integration yields
q
V dT (0) 3 / 2
2( 2) 3 / 2
k (Ts Tf ) f
R
3
Q dK
(c)
Applying (c) to two circles of radii R1 and R2 and taking the ratio of the two results, gives
q2
q1
R23 / 2
R13 / 2
(d)
2
(e)
Doubling the heat transfer rate gives
q2
q1
Thus the radius of the circle needed to double the heat transfer rate is obtained by substituting (e)
into (d) and solving for R2
R2
( 2) 2 / 3 R1
1.587 R1
Thus percent increase in R is
%
increase =
R2 R1
100
R1
1.587 R1 R1
R1
58.7
(iii) Checking. Dimensional check: Units of q in equation (c) should be W
(f)
PROBLEM 4.34 (continued)
dT (0)
Vf ( m/s) 3 / 2 3/2
(1/1) = W
R ( m ) (m)
2
dK
Q ( m /s)
Limiting check: (1) The heat transfer rate should vanish for Ts Tf . Setting Ts
q 0.
q = k(W/m-oC) (Ts Tf )( o C)
Tf in (c) gives
(2) The heat transfer rate must vanish for a circle of radius R = 0. Setting R = 0 in (c) gives
q 0.
(5) Comments. Examination of (4.66) shows that the heat transfer coefficient decreases as x is
increased. Thus, moving a circle in the x-direction decreases the rate of heat transfer.
PROBLEM 4.35
Liquid potassium (Pr << 1) flows over a semiinfinite plate. Assume laminar boundary layer
flow, suggest a simplified velocity profile for
solving the energy equation
Solution
For Pr << 1 the viscous boundary layer
thickness G is much thinner than the thermal boundary layer thickness G t . Thus, throughout
much of the thermal boundary layer the axial velocity is u | Vf . Therefore, the term convective
term is approximate as
wT
wT
.
u
| Vf
wx
wx
PROBLEM 4.36
For very low Prandtl numbers the thermal boundary layer is much thinner than the viscous
boundary layer. Thus little error is introduced if the velocity everywhere in the thermal boundary
layer is assumed to be the free stream velocity Vf . Show that for laminar boundary layer flow
over a flat plate at low Prandtl numbers the local Nusselt number is given by
Nu x
0.564 Pr 1 / 2 Re1 / 2
How does this result compare with scaling prediction?
(1) Observations. (i) The flow field for this boundary
layer problem is simplified by assuming that the axial
velocity is uniform throughout the thermal boundary
layer. (ii) Since velocity distribution affects temperature
distribution, the solution for the local Nusselt number can
be expected to differ from Pohlhausen’s solution. (iii)
The Nusselt number depends on the temperature gradient
at the surface.
(2) Problem Definition. Determine the temperature gradient at the surface for laminar flow over
a flat plate using a uniform velocity throughout the thermal boundary layer.
(3) Solution Plan. Follow Pohlhausen’s solution replacing B
l asius solution to the flow field with
uniform axial velocity equal to the free stream velocity.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are:
(1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5) laminar
flow (Rex <5 u105), (6) viscous and thermal boundary layer flow (Rex >100 and Pex >100), (7)
uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10)
negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no buoyancy
(E = 0 or g = 0) and (13) no energy generation ( q ccc 0 ). In addition, (14) neglect radiation and
(15) assume uniform axial velocity equal to the free stream velocity.
(ii) Analysis. The local Nusselt number for laminar flow over a flat plate is given by
Nu x
where
Re x
Vf
Vf x
Q
free stream velocity, m/s
K ( x, y )
T
y
T Ts
Tf Ts
Vf
Qx
dT (0)
Re x
dK
(4.68)
PROBLEM 4.36 (continued)
Ts
surface temperature, o C
Tf
free stream temperature, o C
dT (0)
is given in equation (b) of Appendix B
dK
The temperature gradient at the surface
­°
®
°̄
dT (0)
dK
f
³
0
ª Pr
exp «
¬ 2
K
³
0
½°
º
f (K )dK » dK ¾
°¿
¼
1
(b)
where f (K ) is the solution to the velocity field. In Pohlhausen’s solution f (K ) is obtained from
lBasius solution to the flow fiel d for laminar boundary layer flow over a semi-infinite flat plate.
Rather than use B
lasius solution we now use a simp lified flow field of uniform velocity given by
u
Vf
df
|1
dK
(c)
Integrating (c)
f
K
(d)
Substituting (d) into (b)
­ f
½
ª Pr K
º
exp
KdK » dK ¾
®³
³
«
¬ 2 0
¼
¯ 0
¿
dT (0)
dK
1
Evaluating the integral
dT (0)
dK
­ f
½
ª Pr 2 º
®³ exp « K » dK ¾
¬ 4
¼
¯ 0
¿
1
(e)
The definite integral in (e) is recognized as the error function. Thus
dT (0)
dK
­ f
½
ª Pr 2 º
®³ exp « K » dK ¾
¬ 4
¼
¯ 0
¿
1
Pr
S
0.564 Pr
(f)
Introducing (f) into (4.68)
Nu x
0.564 Pr Re x
(g)
Scaling results for the case of Pr 1 is given by
Nu x a Pr Re x ,
for Pr <
1
(4.55)
This compares favorably with the exact solution (g).
(iii) Checking. Dimensional check;Each term in (g) is dimensionless.
(5) Comments. Scaling prediction of the local Nusselt number gives the same dependency on
the Prandtl and Reynolds numbers as the exact solution. The constant 0.564 in the exact solution
is replaced by unity in the scaling prediction.
PROBLEM 4.37
Consider laminar boundary layer flow over a flat plate at a uniform temperature Ts . When the
Prandtl number is very high the viscous boundary layer is much thicker than the thermal
boundary layer. Assume that the thermal boundary layer is entirely within the part of the
velocity boundary layer in which the velocity profile is approximately linear. Show that for such
approximation the Nusselt number is given by
Nu x
f
³
Note:
exp(cx 3 )dx
0
0.339 Pr 1 / 3 Re1 / 2
c 1 / 3
* (1 / 3) , where ī is the Gamma function.
3
(1) Observations. (i) The flow field for this
boundary layer problem is simplified by assuming
that the axial velocity varies linearly in the ydirection. (ii) Since velocity distribution affects
temperature distribution, the solution for the local
Nusselt number can be expected to differ from
Pohlhausen’s solution. (iii) The Nusselt number
depends on the temperature gradient at the surface.
(2) Problem Definition. Determine the temperature gradient at the surface for laminar flow over
a flat plate using a uniform velocity throughout the thermal boundary layer.
(3) Solution Plan. Follow Pohlhausen’s solution replacing B
l asius solution to the flow field with
linear axial velocity distribution.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are:
(1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5) laminar
flow (Rex <5 u105), (6) viscous and thermal boundary layer flow (Rex >100 and Pex >100), (7)
uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10)
negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no buoyancy
(E = 0 or g = 0) and (13) no energy generation ( q ccc 0 ). In addition, (14) neglect radiation and
(15) assume uniform axial velocity equal to the free stream velocity.
(ii) Analysis. The local Nusselt number for laminar flow over a flat plate is given by
Nu x
where
Re x
Vf
Vf x
Q
free stream velocity, m/s
K ( x, y )
y
Vf
Qx
dT (0)
Re x
dK
(4.68)
PROBLEM 4.37 (continued)
T Ts
Tf Ts
T
Ts
surface temperature, o C
Tf
free stream temperature, o C
dT (0)
is given in equation (b) of Appendix B
dK
The temperature gradient at the surface
­°
®
°̄
dT ( 0)
dK
f
³
0
ª Pr
exp «
¬ 2
K
³
0
½°
º
f (K )dK » dK ¾
°¿
¼
1
(b)
where f (K ) is the solution to the velocity field. In Pohlhausen’s solution f (K ) is obtained from
lBasius solution to the flow fiel d for laminar boundary layer flow over a semi-infinite flat plate.
Rather than use B
lasius solution we now use a si mplified flow field of linear velocity given by
u
Vf
df
| AK
dK
(c)
where A is constant. Integrating (c)
f (K )
A
K2
2
B
(d)
lasius bound ary condition (4.45b) and solution.
The constants A and B are determined from B
oundary condition (4.45b) gives
B
f ( 0)
0
(4.45b)
Table 7.2 of lBasius solution gives
d 2 f ( 0)
dK 2
0.33206
(e)
These two conditions give A and B
A
0.16603 , B
(f)
0
Equation (d) becomes
f (K )
0.16603K 2
(g)
Substituting (g) into (b)
­
®
¯
Evaluating the indefinite integral
dT (0)
dK
dT (0)
dK
f
³
0
­
®
¯
ª
Pr
exp « 0.16603
2
¬
f
³
0
K
³
0
½
º
K dK » dK ¾
¼
¿
½
Pr º
ª
exp « 0.16603 K 3 » dK ¾
6 ¼
¬
¿
The definite integral in (h) is evaluated next. eLt
1
2
1
(h)
PROBLEM 4.37 (continued)
0.16603
Pr K 3
6
z
It follows that
º
ª
6
»
«
¬ 0.16603 Pr ¼
K
1/ 3
z1/ 3
Differentiating
º
1ª
6
»
«
3 ¬ 0.16603 Pr ¼
dK
1/ 3
z 2 / 3 dz
Substituting into (h)
dT ( 0)
dK
1/ 3
­° 1 ª
º
6
® «
»
°̄ 3 ¬ 0.16603 Pr ¼
f
³e
z
z
2 / 3
0
½°
dz ¾
°¿
1
(i)
The definite integral in (i) is recognized as the aGmma function given by
f
³e
z n 1
z
dz
* (n)
n
1
3
n !1
(j)
0
Comparing the integral in (i) with (j) gives
(k)
Using (j) and (k), equation (i) becomes
dT ( 0)
dK
1/ 3
­° 1 ª
½°
º
6
*
(
1
/
3
)
® «
¾
»
°̄ 3 ¬ 0.16603 Pr ¼
°¿
1
(l)
amma function
The value of * (1 / 3) is obtained from tables of G
* (1 / 3)
2.679
(m)
0.339 Pr 1/3
(n)
0.339 Pr 1/ 3 Re1x/ 2
(o)
dT (0)
dK
Substituting (n) into (4.
Nu x
(iii) Checking. Dimensional check: (1) The exponent of the exponential in(h) is
dimensionless. (2) Each term in (o) is dimensionless.
(5) Comments. Scaling prediction of the local Nusselt number gives the same dependency on
the Prandtl and Reynolds numbers as the exact solution (o). The constant 0.339 in the exact
solution is replaced by unity in the scaling prediction. Scaling results for the case of Pr !! 1 is
given by
Nu x a Pr 1/3 Re1x/ 2 ,
for Pr 1>
(4.57)
PROBLEM 4.38
Consider steady, two-dimensional, laminar boundary
y
Tf
layer flow over a porous flat plate at uniform surface
temperature. The plate is subject to a uniform suction
Vf
x
v (x,0) v o . Far away downstream both the axial
vo
velocity and the temperature may be assumed to be
functions of y only. Free stream velocity is Vf and
free stream temperature is Tf . Determine the heat transfer coefficient and Nusselt number in
this region.
(1) Observations. (i) The flow and temperature fields for this boundary layer problem are
simplified by assuming that the axial velocity and temperature do not vary in the x-direction. (ii)
The heat transfer coefficient depends on the temperature gradient at the surface. (iii)
Temperature distribution depends on the flow field. (iv) The effect of wall suction must be taken
into consideration.
(2) Problem Definition. Determine the temperature gradient at the surface for laminar flow over
a flat plate.
(3) Solution Plan. Introduce simplifying assumptions in the energy equation for boundary layer
flow and solve for the temperature distribution for laminar flow over a plate with suction.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) twodimensional, (5) laminar flow (Rex <5 u105), (6) viscous and thermal boundary layer flow (Rex >
100 and Pex >100), (7) uniform upstr eam velocity and temperature, (8) uniform surface
temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible
dissipation, (12) no buoyancy (E = 0 or g = 0) and (13) no energy generation ( q ccc 0 ). In
addition, (14) neglect radiation, (15) uniform suction velocity and (16) negligible axial variation
of velocity and temperature.
(ii) Analysis. The local heat transfer coefficient h is given by
h
where
k
T
Ts
thermal conductivity, W/m o C
fluid temperature, o C
surface temperature, o C
Tf free stream temperature, o C
y normal coordinate, m
The local Nusselt number is defined as
wT ( x,0)
wy
k
Ts Tf
(1.10)
PROBLEM 4.38 (continued)
hx
(a)
k
Temperature distribution in (1.10) is obtained from the solution to boundary layer energy
equation
wT
wT
w 2T
(4.18)
u
v
D 2
wx
wy
wy
where
Nu x
u
v
D
axial velocity, m/s
normal velocity, m/s
thermal diffusivity, m 2 /s
Since temperature variation in the axial direction is neglected, it follows that
wT
wx
0
(b)
(b) into (a)
wT
wy
v
w 2T
D
wy 2
(c)
To solve (c) for the temperature distribution, velocity component v must be determined. The
continuity equation for two-dimensional incompressible flow is given by
wu wv
wx wy
0
(2.2)
Since velocity variation in the axial direction is neglected, it follows that
wu
wx
0
(d)
wv
wy
0
(e)
(d) into (2.2)
Integration of (e) gives
v
(f)
c
where c is constant of integration. Surface boundary condition on v is
v
vo
where
vo
surface suction velocity, m/s
Substituting (g) into (c)
- vo
wT
wy
D
w 2T
wy 2
Since T is independent of x, it follows that the above can be written as
(g)
PROBLEM 4.38 (continued)
dT
dy
- vo
D
d 2T
(h)
dy 2
The solution to (h) requires two boundary conditions. They are:
T ( x,0)
Ts
(i)
T ( x, f )
Tf
(j)
To solve (h), it is rewritten as
wT
wy
dT
dy
d
vo
dy
(k)
y ln C1
(l)
D
Integrate (j)
ln
dT
dy
vo
D
where C1 is constant of integration. To integrate (l) again, it is first rewritten as
ln
1 dT
C1 dy
vo
y
D
or
dT
dy
C1 exp(
vo
D
(m)
y)
Integrate (m)
T
C1
D
vo
exp(
vo
D
(n)
y) C 2
where C 2 is constant of integration. Application of boundary conditions (i) and (j) gives C1 and
C2
vo
(o)
C1
(Ts Tf )
D
C2
(p)
Tf
(o) and (p) into (n)
T
Tf (Ts Tf ) exp(
vo
D
y)
(q)
Substituting (q) into (1.10) gives the heat transfer coefficient h
h k
oHwever,
vo
D
(r)
D is defined as
D
k
Uc p
(s)
PROBLEM 4.38 (continued)
where
cp
U
specific heat, J/kg o C
density, kg/m 3
(s) into (r)
h
U c pv o
(t)
Substituting (t) into (a) gives the local Nusselt number
U c pv ox
Nu x
k
The above can be written in a more revealing way as
c pP U v ox
k
P
Nu x
This is recognized as
Nu x
Pr Re x
(u)
U v ox
P
(v)
where the local Reynolds number is defined as
Re x
(iii) Checking. Dimensional check: (1) The exponent of the exponential in(q) is
dimensionless. (2) Each term in (q) has units of temperature. (3) Equation (t) give the correct
units for h.
Boundary conditions check: Solution (q) satisfies conditions (i) and (j).
Limiting check: For the special case of Ts
Tf . . Setting Ts Tf in (q) gives T Tf .
Tf , fluid temperature should be uniform equal to
(5) Comments. (i) Neglecting axial variation of velocity and temperature are the key
simplifying assumptions in this problem. (ii) Free stream velocity Vf does not enter into the
solution for the temperature distribution and heat transfer coefficient. (iii) The Reynolds number
in solution (u) depends on the suction velocity and not free stream velocity.
PROBLEM 4.39
A semi infinite plate is heated with uniform flux q cc along its length. The free stream temperature
is Tf and free stream velocity is Vf . Since the heat transfer coefficient varies with distance
along the plate, Newton’s law of cooling requires that surface temperature must also vary to
maintain uniform heat flux. Consider the case of laminar boundary layer flow over a plate whose
surface temperature varies according to
Ts ( x) Tf
Cx n
Working with the solution to this case, show that n 1 / 2 corresponds to a plate with uniform
surface flux.
(1) Observations. (i) This is a forced convection flow over a plate with variable surface
temperature. (ii) The local heat flux is determined by Newton’s law of cooling. (iii) The local
heat transfer coefficient and surface temperature vary with distance along the plate. The variation
of surface temperature and heat transfer coefficient must be such that Newton’s law gives
uniform heat flux. (iv) The local heat transfer coefficient is obtained from the local Nusselt
number.
(2) Problem Definition. Determine the value of the exponent n that results in uniform heat flux.
(3) Solution Plan. Apply Newton’s law of cooling to determine the local surface heat flux.
Compute the Reynolds number to establish if the flow is laminar or turbulent. If the flow is
laminar use the result of numerical solution to forced convection over a plate with variable
surface temperature to determine the local Nusselt number and heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) twodimensional, (5) laminar flow (Rex <5 u105, to be verified), (6) viscous and thermal boundary
Pex >100, to be verified), (7) uniform upstream velocity and
layer flow (Rex >100 and
temperature, (8) flat plate, (9) negligible changes in kinetic and potential energy, (10) negligible
dissipation, (11) no buoyancy (E = 0 or g = 0), (12) no energy generation ( q ccc 0 ) and (14)
negligible radiation
(ii) Analysis. Newton’s law of cooling gives
q cc
where
h(x)
h( x)(Ts Tf )
(a)
local heat transfer coefficient, W/.m 2 o C
q cc surface heat transfer rate, W/m 2
Ts (x)
Tf
surface temperature, o C
free stream temperature, o C
Surface temperature variation is given by
Ts ( x) Tf
Substituting (b) into (a)
Cx n
(b)
PROBLEM 4.39 (continued)
Cx n h(x)
(c)
h ( x ) v x1 / n
(d)
q cc
According to (c), surface flux is constant if
The local heat transfer coefficient is obtained from the local Nusselt number defined as
hx
k
Nu x
(e)
Solving for h(x)
h( x )
Nu x
k
x
(f)
The solution to the Nusselt number for a plate with surface temperature variation described in
equation (b) is
dT (0)
Nu x
Re x
(4.80)
dK
where dT (0) / dK is the dimensionless surface temperature gradient. It is a constant which
depends on the Prandtl number and the exponent n. Substituting (4.80) into (f)
h( x )
dT (0)
k
Re x
dK
x
Using the definition of Reynolds number into the above
h( x )
dT (0) Vf x k
Q x
dK
k
dT (0) Vf 1/ 2
x
Q
dK
(g)
Using (g) into (c)
dT (0) Vf 1 / 2
x
(h)
Q
dK
Examination of (h) shows that for q cc to be constant independent of x, the exponent n must be
1
n
(i)
2
q cc
kCx n
(iii) Checking. Dimensional check: (1) Each term in (4.80) is dimensionless. (2)
Equations(g) and (h) are dimensionally correct.
Limiting check: (1) For the special case of C = 0, surface temperature according to (b) will be the
same as ambient temperature. That is Ts Tf . The corresponding heat flux for this case should
vanish. Setting C = 0 in (h) gives q cc 0.
(5) Comments. (i) The key to the solution to this problem is the determination of the variation of
h with distance x.
PROBLEM 4.40
Water flows over a semi-infinite flat
temperature Ts given by
Ts ( x) Tf
plate which is maintained at a variable surface
Vf
Cx 0.75
Tf
where
C = 54.27 oC / (m)0.75
Tf = free stream temperature = 3o C.
x = distance from the leading edge, m
x
L
W
Ts Tf C x0.75
Determine the average heat transfer coefficient for a plate if length L = 0.3 m. Free stream
velocity is 1.2 m/s.
(1) Observations. (i) This is a forced convection flow over a plate with variable surface
temperature. (ii) The Reynolds number should be computed to determine if the flow is laminar or
turbulent. (iii) The local heat transfer coefficient and surface temperature vary with distance
along the plate. (iv) The local heat transfer coefficient is obtained from the solution to the local
Nusselt number. (v) The determination of the Nusselt number requires determining the
temperature gradient at the surface.
(2) Problem Definition. Determine surface temperature gradient for a plate at variable surface
temperature.
(3) Solution Plan. Compute the Reynolds number to establish if the flow is laminar or turbulent.
If the flow is laminar use the result of numerical solution to forced convection over a plate with
variable surface temperature to determine surface temperature gradient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) twodimensional, (5) laminar flow (Rex <5 u105, to be verified), (6) viscous and thermal boundary
Pex >100, to be verified), (7) uniform upstream velocity and
layer flow (Rex >100 and
temperature, (8) flat plate, (9) negligible changes in kinetic and potential energy, (10) negligible
dissipation, (11) no buoyancy (E = 0 or g = 0), (12) no energy generation( q ccc 0 ) and (13)
negligible radiation
(ii) Analysis.
The maximum Reynolds number is given by
Re L
Vf L
Q
where
L = plate length = 0.3 m
Vf free stream velocity = 1.2 m/s
Q kinematic
Properties are evaluated at the film temperature defined as
(a)
PROBLEM 4.40 (continued)
Tf Ts
2
Tf
where
Tf
(b)
free stream temperature = 3 o C
average surface temperature, o C
Ts
The average surface temperature is defined as
Ts
Ts (0) Ts ( L)
2
(c)
Surface temperature is given by
Ts ( x)
Tf Cx 0.75
(d)
where
o
C
54.27
C
m 0.75
Evaluating Ts at x = 0 and x = 0.3 m and substituting into (d) gives
Tf Tf CL0.75
2
Ts
3o C 3o C 54.27( o C/m 0.75 )(0.3) 0.75 (m) 0.75
2
14 o C
Substituting into (b)
(3 14)( o C)
2
Tf
8.5 o C
Properties of water at this temperature are
0.5791
k
Pr
Q
W
m o C
9.942
1.3716 u 10 6
m2
s
Substituting into (c)
Re L
1.2(m/s)0.3(m)
1.3716 u 10 6 (m 2 /s)
= 262,467
Thus this is laminar boundary layer flow. The local Nusselt number for laminar flow over a flat
plate with variable surface temperature is given by
Nu x
where
dT (0)
Re x
dK
(4.80)
PROBLEM 4.40 (continued)
hx
k
2 o
h(x) local heat transfer coefficient, W.m C
x = distance from the leading edge, m
(e)
Nu x
2.0
Substitute (e) into (4.80) and solve for h
h
30
dT (0) k
Re x
dK x
(f)
The dimensionless surface temperature
gradient, dT (0) / dK , is obtained from the
numerical solution for laminar flow over a
flat plate at a surface temperature of the
form
Ts ( x) Tf Cx n
(g)
dT (0)
dK
10
1.0
Pr
0
Fig. 4.8
The solution is presented in Fig. 4.8. The
exponent n in (g) characterizes surface
temperature variation.
0.5
hx
k
1.0
1.5
dT (0)
for plate with varring surface temperature,
dK
Ts - Tf Cx n 4[ ]
(iii) Computations. For the problem under consideration n
gives
dT (0)
| 1.1
dK.
Substituting into (f)
k
Re x
h 1.1
x
The corresponding Nusselt number is
Nu x
n
0.7
0.75 and Pr
1.1 Re x
9.942 , Fig. 4.8
(h)
(i)
(j)
(iv) Checking. Dimensional check: (1) Equation (f) has the correct units for heat transfer
coefficient. (2) The Reynolds number in (a) and Nusselt number in (j) are dimensionless.
(5) Comments. (i) The key to the solution to this problem is the determination of the constant
dT (0) / dK. This constant is determined through the use of Fig. 4.8. (ii) Fig. 4.8 is applicable to a
class of variable surface temperature described by equation (g) only
Ts ( x)
Tf Cx n
(g)
(iii) According to (d) surface temperature varies from 3o C at the leading edge to 25 o C . Clearly,
surface temperature is not uniform. For uniform surface temperature the Nusselt number is given
by (4.72c)
Nu x
0.339 Pr 1 / 3 Re x , for Pr ! 10
(4.72c)
PROBLEM 4.40 (continued)
In this problem Pr
9.942 . Using this value in (4.72c) gives
Nu x
0.729 Re x
This is 33%
smaller than the variable surface temperature solution given in (j).
(k)
PROBLEM 4.41
Air flows over a plate which is heated non-uniformly such that its surface temperature increases
linearly as the distance from the leading edge is increased according to
Ts ( x)
Tf Cx
where
C = 24 oC /cm
Tf = free stream temperature = 20 o C
x = distance from the leading edge, m
Determine the total heat transfer rate from a square
plate 10 cm u 10 cm. Free stream velocity is 3.2 m/s.
Vf
Tf
x
(1) Observations. (i) This is a forced convection flow
L
over a plate with variable surface temperature. (ii) The
Ts Tf Cx
Reynolds number should be computed to determine if
the flow is laminar or turbulent. (iii) Newton’s law of cooling gives the heat transfer rate from
the plate. (iv) The local heat transfer coefficient and surface temperature vary with distance
along the plate. Thus determining the total heat transfer rate requires integration of Newton’s law
along the plate. (v) The local heat transfer coefficient is obtained from the local Nusselt number.
(2) Problem Definition. Determine the average Nusselt number for forced convection over a flat
plate with variable surface temperature.
(3) Solution Plan. Apply Newton’s law of cooling to determine the total heat transfer rate form
the plate. Compute the Reynolds number to establish if the flow is laminar or turbulent. If the
flow is laminar use the result of numerical solution to forced convection over a plate with
variable surface temperature to determine the average Nusselt number.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) twodimensional, (5) laminar flow (Rex <5 u105, to be verified), (6) viscous and thermal boundary
Pex >100, to be verified), (7) uniform upstream velocity and
layer flow (Rex >100 and
temperature, (8) flat plate, (9) negligible changes in kinetic and potential energy, (10) negligible
dissipation, (11) no buoyancy (E = 0 or g = 0), (12) no energy generation ( q ccc ) and (13)
negligible radiation
(ii) Analysis. The total heat transfer from the plate is given by Newton’s law of cooling
L
q
³
h( x)(Ts Tf )Wdx
0
where
h(x) local heat transfer coefficient, W.m 2 o C
L plate length = 0.1 m
q heat transfer rate, W
Ts ( x)
surface temperature, o C
(a)
PROBLEM 4.41 (continued)
Tf free stream temperature = 20 o C
W plate width = 0.1 m
x distance along plate, m
The local heat transfer coefficient is obtained from the local Nusselt number defined as
Nu x
hx
k
(b)
The maximum Reynolds number is given by
Re L
Vf L
(c)
Q
where
Vf
Q
free stream velocity = 3.2 m/s
kinematic
Properties are evaluated at the film temperature defined as
Tf Ts
Tf
2
where
Ts
(d)
average surface temperature, o C
The average surface temperature is defined as
Ts
Ts ( 0 ) Ts ( L )
2
(e)
Surface temperature is given by
Ts ( x)
Tf Cx
(f)
where C = 24 oC /cm. Evaluating Ts at x = 0 and x = 0,1 m and substituting into (e) gives
Ts
Tf Tf CL
2
20 o C 20 o C 24( o C/cm)10(cm)
2
2.0
Substituting into (d)
Tf
(20 140)( o C)
2
30
o
80 C
Properties of air at this temperature are
k
Pr
0.02991
dT (0)
dK
10
1.0
W
Pr
0.7
m o C
0.706
6
Q
140 o C
20.92 u 10
m2
s
0
Fig. 4.8
0.5
n
1.0
1.5
dT (0)
for plate with varring surface temperature,
dK
Ts - Tf Cx n 4[ ]
PROBLEM 4.41 (continued)
Substituting into (c)
Re L
3.2(m/s)0.1(m)
20.92 u 10 6 (m 2 /s)
=15,296
Thus this is laminar boundary layer flow. The local Nusselt number for laminar flow over a flat
plate is given by
dT (0)
Re x
dK
Nu x
(4.80)
where dT (0) / dK. is the dimensionless surface temperature gradient. Numerical solution to
dT (0) / dK for laminar flow over a flat plate at a surface temperature of the form
Ts ( x)
Tf Cx n
(g)
is given in Fig. 4.8. The exponent n in (g) characterizes surface te mperature variation.. Using the
definition of local Nusselt number and Reynolds number, (4.80) is solved for the local heat
transfer coefficient
h
dT (0) k Vf x
dK x Q
k
dT (0) Vf 1
dK
Q
x
(h)
Substituting (f) and (h) into (a)
q
kWC
dT (0) Vf
dK
Q
L
³
x dx
0
Evaluating the integral
dT (0) Vf 3 / 2
2
kWC
L
dK
3
Q
q
Expressing this result in terms of the Reynolds number
q
dT (0) Vf L
2
kWLC
dK
Q
3
q
2
dT (0)
kWLC
Re L
3
dK
(i)
Equation (i) is rewritten in dimensionless form as
q
kWLC
2 dT (0)
Re L
3 dK
(iii) Computations. For the problem under consideration n 1 and Pr
gives
dT (0)
| 0.48
dK.
Substituting into (i)
(j)
0.706 , Fig. 4.8
(h)
PROBLEM 4.41 (continued)
q
2
(0.02991)( W/m o C)(0.1)(m)0.1(m)24( o C/cm)100(cm/m)0.48 15,296
3
28.4 W
(iv) Checking. Dimensional check: (1) Computations showed that (i) has the correct units for
heat transfer rate. (2) Each term in (j) is dimensionless.
Limiting check: (1) If the width of plant is ezro the heat transfer rate will vanish. Setting W = 0 in
(i) gives q = 0. (2) For the special case of C = 0, surface temperature according to (f) will be the
same as ambient temperature. That is Ts Tf . The corresponding heat transfer rate for this case
is q = 0. Setting C = 0 in (i) gives q = 0.
(5) Comments. (i) The key to the solution to this problem is the determination of the constant
dT (0) / dK. This constant is determined through the use of Fig. 4.8. (ii) Fig. 4.8 is applicable to a
class of variable surface temperature described by equation (g) only
Ts ( x)
Tf Cx n
(g)
PROBLEM 4.42
The surface temperature of a plate varies with distance
from the leading edge according to
L
Tf
Ts ( x)
Tf Cx
1
0.8
2
Vf
top view
H
Ts (x )
Two identical triangles are drawn on the surface as
shown. Fluid at uniform upstream temperature Tf and
uniform upstream velocity Vf flows over the plate. Assume laminar boundary layer flow.
Determine the ratio of the heat transfer rate from the two triangles, q1/q2.
(1) Observations. (i) This is an external forced convection problem of flow over a flat plate. (ii)
Convection heat transfer from a surface can be determined using Newton’s law of cooling. (iii)
The local heat transfer coefficient and surface temperature vary along the plate. (iv) For each
triangle the area varies with distance along the plate. (v) The total heat transfer rate can be
determined by integration along the length of each triangle.
(2) Problem Definition. Determine the local heat transfer coefficient along each for laminar
boundary layer flow over a plate with variable surface temperature.
(3) Solution Plan. Apply Newton's law of cooling to an element of each triangle, ydx, determine
the local heat transfer coefficient along the plate and integrate over the area.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) twodimensional, (5) laminar flow (Rex <5 u105), (6) viscous and thermal boundary layer flow (Rex >
100 and Pex >100), (7) uniform upstr eam velocity and temperature, (8) uniform surface
temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible
dissipation, (12) no buoyancy (E = 0 or g = 0) and (13) no energy generation ( q ccc 0 ). In
addition, (14) neglect radiation.
(ii) Analysis. O
f interest is the ratio of the total heat transfer rate from triangle 1 to
triangle 2. Since both the heat transfer coefficient and area vary along each triangle, it follows
that Newton's law of cooling s hould be applied to an element dA = ydx at a distance x from the
leading edge:
(a)
dq = h(x) (Ts - Tf) y(x)dx
where
dq = rate of heat transfer from element, W
h = local heat transfer coefficient, W/m2-oC
Ts = surface temperature, oC
Tf = free stream temperature, oC
x = distance along plate, m
y = y(x) = width of element
L
x
Tf
Vf
y1
1
2
y2
top view
H
Ts
PROBLEM 4.42 (continued)
Integration of (a) gives the total heat transfer rate
L
q
³ h( x)(T T
s
f ) y ( x) dx
(b)
0
To evaluate the integral we must specify ( Ts Tf ), y(x) and h(x). Surface temperature variation
is given as
Ts ( x) Tf Cx 0.8
(c)
Similarity of triangles gives
H
( L x)
L
H
y 2 ( x)
x
L
y1 ( x)
(d)
(e)
The local heat transfer coefficient h(x) for a plate with variable surface temperature described by
(c) is given by (4.78)
h
k
Vf dT (0)
Qx dK
(4.78)
where
k = thermal conductivity, W/m-oC
Vf = free stream velocity, m/s
T (T Ts ) /(Tf Ts ) , dimensionless temperature
Q = kinematic viscosity, m2/s
V
K = y f , dimensionless variable
Qx
y = coordinate normal to plate, m
The dimensionless surface temperature gradient, dT (0) / dK , is obtained from the numerical
solution for laminar flow over a flat plate at a surface temperature of the form
Ts ( x)
Tf Cx n
(f)
Fig. 4.8 gives dT (0) / dK . As shown in Fig. 4.8, dT (0) / dK depends on the Prandtl number and
the exponent n in (f). Since both triangles have the same surface temperature distribution and the
same fluid, it follows that dT (0) / dK is the same for both triangles.
To determine the heat transfer rate, q1 , from triangle 1, equations (c), (d) and (4.78) are
substituted into (b)
L
H Vf d T ( 0 )
( L x ) 0. 8
q1 kC
x dx
L Q dK 0
x
Evaluating the integral
³
PROBLEM 4.42 (continued)
q1
kCH
Vf dT (0) L1.3
Q dK 2.99
(g)
Similarly, for triangle 2, equations (c), (e) and (4.78) are substituted into (b)
q1
H
kC
L
Vf dT (0)
Q dK
L
³
0
x 0.8
x dx
x
Evaluating the integral
q2
kCH
Vf dT (0) L1.3
Q dK 2.3
(h)
Taking the ratio of (g) and (h)
q1
q2
2 .3
2.99
0.769
(i)
(iii) Checking. Dimensional check: (1) Each term in (d) and (e) has units of length. (2)
Noting that units of C are o C/m 0.8 , equations (g) and (h) have the correct units for heat.
Limiting check: For the limiting case of a plate which is maintained at the free stream
temperature, that is Ts Tf , the corresponding heat transfer rate should vanish for both
triangles. According to (f), Ts Tf when C = 0. Setting C = 0 in (g) and (h) gives q1 q 2 0.
(5) Comments. (i) According to (i) the heat transfer ratio q1 / q 2 1. B
y contrast, in Problem
4.30 in which surface temperature is uniform, q1 / q 2 ! 1 . The reason for this reversal is the
increase in surface temperature with x favors triangle 2 where the area also increases with x.
Although the two triangles have the same area, the rate of heat transfer from triangle 1 is double
that from triangle 2. Thus, orientation and proximity to the leading edge of a flat plate play an
important role in determining the rate of heat transfer. (ii) The same approach can be used to
determine heat transfer for configurations other than rectangles, such as circles and ellipses.
PROBLEM 4.43
Construct a plot showing the variation of Nu x / Re x with wedge angle. Where Nu x is the local
Nusselt number and Re x is the local Reynolds number. Assume laminar boundary layer flow of
air.
(1) Observations. (i) This is a forced convection
boundary layer flow over a wedge. (ii) Wedge
surface is maintained at uniform temperature.
(iii) The flow is laminar. (iv) The fluid is air. (v)
Similarity solution for the local Nusselt number
is presented in Section 4.4.3. (vi) The Nusselt
number depends on the Reynolds number and
the dimensionless temperature gradient at the
surface dT (0) / dK. (vii) Surface temperature
gradient depends on wedge angle.
(2) Problem Definition. Determine the variation
of dT (0) / dK with wedge angle.
(3) Solution Plan. Use the wedge solution of Section 4.4.3 to determine the variation of local
Nusselt number with wedge angle.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) twodimensional, (5) laminar flow (Rex <5 u105), (6) viscous and thermal boundary layer flow (Rex >
100 and Pex >100), (7) uniform upstream velocity and temperature, (8) uniform surface
temperature, (9) symmetrical flow, (10) negligible changes in kinetic and potential energy, (11)
negligible dissipation, (12) no buoyancy (E = 0 or g = 0) and (13) no energy generation ( q ccc 0 ).
(ii) Analysis. The local Nusselt number for laminar boundary layer flow over a wedge is
given by
Nu x
dT (0)
Re x
dK
where
Vf x
Re x
Q
Vf (x)
K ( x, y )
T
, local Reynolds number
external flow velocity over the wedge, m/s
y
Vf ( x)
, similarity variable
Qx
T Ts
, dimensionless temperature
Tf Ts
(4.96)
PROBLEM 4.43 (continued)
Rewrite (496)
Nu x
Re x
dT (0)
dK
(a)
Surface temperature gradient depends on wedge angle and Prandtl number. Table 4.3 lists
dT (0) / dK corresponding to four wedge angles and five Prandtl numbers. For air with Prandtl
number Pr 0.7 , Table 4.3 gives the values of dT (0) / dK used to construct a pot of
Nu x / Re x vs. wedge angle S E .
Pr = 0.7
Wedge
dT (0)
angle
dK
SE
0.5
Nu x
Rex
0.4
0
S /5
S /2
S
0.3
0
60
120
0.292
0.331
0.384
0.496
180
wedge angle , deg
(4) Comments. (i) The local heat transfer coefficient decreases with distance along the surface.
(ii) L
ocal Nusselt number and heat transfer coeffi cient increase with wedge angle. The increase
is approximately linear.
PROBLEM 4.44
Consider laminar boundary layer flow over a wedge. Show that the average Nusselt number Nu
for a wedge of length L is given by
2 dT (0)
Re L
m 1 dK
LVf ( L)
where the Reynolds number is defined as Re L
.
Nu
ǎ
(1) Observations. (i) This is a forced convection boundary layer flow over a wedge. (ii) Wedge
surface is maintained at uniform temperature. (iii) The flow is laminar. (iv) The average Nusselt
number depends on the average heat transfer coefficient.. (v) Similarity solution for the local
heat transfer coefficient is presented in Section 4.4.3.
(2) Problem Definition. Determine the average heat transfer coefficient for laminar flow over a
wedge at uniform surface temperature.
(3) Solution Plan. Start with the definitions of the average Nusselt number. Use the wedge
solution of Section 4.4.3 to determine the local heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) twodimensional, (5) laminar flow (Rex <5 u105), (6) viscous and thermal boundary layer flow (Rex >
100 and Pex >100), (7) uniform upstr eam velocity and temperature, (8) uniform surface
temperature, (9) symmetrical flow, (10) negligible changes in kinetic and potential energy, (11)
negligible dissipation, (12) no buoyancy (E = 0 or g = 0) and (13) no energy generation ( q ccc 0 ).
(ii) Analysis. The average Nusselt number for a wedge of length L is defined as
hL
k
Nu L
(a)
where
h = average heat transfer coefficient, W/m 2 o C
k = thermal conductivity, W/m o C
L = wedge length, m
The average heat transfer coefficient is defined as
h
1
L
L
³ h( x)dx
(b)
0
where the local heat transfer coefficient, h(x), is
given by (4.95)
h( x )
k
Vf ( x) dT (0)
ǎ x dK
(4.95)
PROBLEM 4.44 (continued)
where
Vf ( x) external flow velocity over the wedge, m/s
x = distance along wedge surface form the leading end, m
V ( x)
, similarity variable
K ( x, y ) y f
ǎx
T
T Ts
, dimensionless temperature
Tf Ts
ǎ = kinematic viscosity, m 2 s
(4.95) into (b)
k d T ( 0)
ǎ dK
h( x )
L
³
0
Vf ( x )
dx
x
(c)
The external flow velocity, Vf ( x), is given by
Vf ( x )
Cx m
(4.82)
where C is constant and m is a measure of wedge angle ES , defined as
m
E
(4.83)
2E
(4.82) into (c)
k C d T ( 0)
L ǎ dK
h
L
³x
( m 1) / 2
dx
0
Evaluating the integral
h
2 k C dT (0) ( m1) / 2
L
m 1 L ǎ dK
(e)
(e) into (a)
Nu L
C dT (0) ( m1) / 2
2
L
m 1 ǎ dK
To express in terms of the Reynolds number, rewrite the above
Nu L
2 dT (0) (CLm ) L
m 1 dK
Q
Nu L
2 dT (0) Vf ( L) L
Q
m 1 dK
(4.82) into the above
(f)
Introduce the definition of Reynolds number
Nu L
2 dT (0)
Re L
m 1 dK
(g)
PROBLEM 4.44 (continued)
(iii) Checking: Dimensional check: (1) Units of the constant C is determined from (4.82) as
ª m º
C«
» . Using this shows that (e) has the correct units. (2) Equation (f) is dimensionless,
¬ s m m ¼
Limiting check: For a wedge with zero angle ( E m = 0), the solutions should reduce to the flat
plate solution of Pohlhausen. Setting m = 0 in (g) gives
Nu L
2
dT (0)
Re L
dK
(h)
This agrees with Pohlhausen’s result of equation (4.69).
(4) Comments. In determining the average heat transfer coefficient for a wedge, the variation of
the velocity outside the boundary layer with distance x must be taken into consideration. That is,
unlike Pohlhausen’s solution for the flat plate, for the wedge Vf Vf (x).
PROBLEM 4.45
Compare the total heat transfer rate from a 90 o wedge, q w , with that from a flat plate, q p , of
the same length. Construct a plot of q w / q p as a function of Prandtl number.
(1) Observations. (i) Newton’s law of cooling gives the heat transfer rate from a surface. (ii)
Total heat transfer from a surface depends on the average heat transfer coefficient h . (iii) oBth
flat plate and wedge are maintained at uniform surface temperature. (iv) Pohlhausen’s solution
gives h for a flat plate. (v) Similarity solution for the local heat transfer coefficient for a wedge
is presented in Section 4.4.3.
(2) Problem Definition. Determine the average heat transfer
coefficient for laminar flow over a flat plate and a wedge.
(3) Solution Plan. Apply Newton’s law of cooling to flat plate and
wedge. Use Pohlhausen’s and wedge solutions to h .
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3)
constant properties, (4) two-dimensional, (5) laminar flow (Rex <5 u105), (6) viscous and thermal
boundary layer flow (Rex >100 and Pex >100), (7) uniform upstream velocity and temperature,
(8) uniform surface temperature, (9) symmetrical flow over wedge, (10) negligible changes in
kinetic and potential energy, (11) negligible dissipation, (12) no buoyancy (E = 0 or g = 0) and
(13) no energy generation ( q ccc 0 ).
(ii) Analysis. Newton’s law of cooling gives
q
h A(Ts Tf )
(a)
where
A = surface area, m 2
h average heat transfer coefficient, W/m 2 o C
q heat transfer rate, W
Ts
surface temperature, o C
Tf
free stream temperature, o C
eLt the subscript p denote plate and w denote wedge. Apply (a) to the plate and wedge and take
their ratio
q p hp
(b)
q w hw
The problem reduces to determining h p and hw .
The average heat transfer coefficient for boundary layer laminar flow over a flat plate at uniform
surface temperature is given by Pohlhausen’s solution (4.67)
PROBLEM 4.45 (continued)
2
hp
dT p (0)
k
Re L
L
dK p
(4.67)
where
k = thermal conductivity, W/m o C
L = wedge length, m
Vf L
Re L
Vf
Q
free stream velocity, m/s
Vf
, dimensionless similarity variable
Qx
x = axial coordinates, m
y = normal coordinates, m
T p Ts
Tp
, dimensionless temperature
Tf Ts
T p T p ( x, y ) , temperature distribution
K p ( x, y )
y
Q = kinematic viscosity, m 2 s
Using the definition of Reynolds number, (4.67) us rewritten as
2
hp
k Vf L dT p (0)
L Q
dK p
(c)
The average heat transfer coefficient foe a wedge is defined as
hw
1
L
L
³ h ( x)dx
w
(d)
0
The local heat transfer coefficient for a wedge, hw (x ) , is given by (4.95)
hw ( x )
k
Vf ( x ) d T w ( 0 )
Qx
dK w
where
Vf ( x )
K w ( x, y )
Tw
Tw
external flow velocity over the wedge, m/s
y
Vf ( x )
, similarity variable
Qx
Tw Ts
, dimensionless temperature
Tf Ts
Tw ( x, y ) , temperature distribution
Q = kinematic viscosity, m 2 s
(4.95)
PROBLEM 4.45 (continued)
(4.95) into (d)
hw ( x )
L
k dT w (0)
Q dK
Vf ( x)
dx
x
³
0
(e)
The external flow velocity, Vf ( x), is given by
Vf ( x )
Cx m
(4.82)
where C is constant and m is a measure of wedge angle ES , defined as
m
E
(4.83)
2E
(4.82) into (e)
hw
k C dT w (0)
L Q
dK
L
³
x ( m1) / 2 dx
0
Evaluating the integral
hw
2 k C dT w (0) ( m1) / 2
L
m 1 L Q
dK
(f)
Substitute (c) and (f) into (a)
qw
qp
m
1
CL
m 1 Vf
dT w (0)
dK
dT p (0)
(g)
dK p
(iii) Checking: Dimensional check: (1) Equation (4.82) shows that CLm has units of velocity. It
follows that (g) is dimensionless. Similarly, units of (f) are correct.
Limiting check: For a wedge with ezro angle ( E m = 0), the wedge solution should reduce to
the flat plate solution and the heat ratio should be unity. For this case, according to (4.82),
C Vf . Setting m = 0 in (g) gives q w / q p 1.
(4) Comments. (i) The local heat transfer coefficient decreases with distance along the surface.
(ii) L
ocal Nusselt number and heat transfer coeffi cient increase with wedge angle. The increase
is approximately linear.
PROBLEM 4.46
For very low Prandtl numbers the thermal boundary layer is much thinner than the viscous
boundary layer. Thus little error is introduced if the velocity everywhere in the thermal boundary
layer is assumed to be the free stream velocity Vf . Show that for laminar boundary flow layer
flow over a wedge at low Prandtl numbers the local Nusselt number is given by
Nu x
(m 1) Pr
S
Re x
(1) Observations. (i) The flow field for this boundary
layer problem is simplified by assuming that axial
velocity within the thermal boundary layer is the same
as that of the external flow. (ii) Since velocity
distribution affects temperature distribution, the solution
for the local Nusselt number differs from the exact case
of Section 4.4.3. (iii) The local Nusselt number depends
the local heat transfer coefficient which depends on the
temperature gradient at the surface.
(2) Problem Definition. Determine the temperature gradient at the surface for laminar flow over
a wedge for the simplified velocity field described above.
(3) Solution Plan. Follow analysis of section 4.4.3 for determining the local Nusselt number
using a simplified flow field solution.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) twodimensional, (5) laminar flow (Rex <5 u105), (6) viscous and thermal boundary layer flow (Rex >
100 and Pex >100), (7) uniform upstr eam velocity and temperature, (8) uniform surface
temperature, (9) symmetrical flow, (10) negligible changes in kinetic and potential energy, (11)
negligible dissipation, (12) no buoyancy (E = 0 or g = 0), (13) no energy generation ( q ccc 0 ) and
(14) Pr <1. Thus assume that axial velocity with in the thermal boundary layer is the same as
that of the external flow.
(ii) Analysis. The local Nusselt number, Nu x , for a wedge is given by equation (4.96) of
Section 4.4.3
dT (0)
(4.68)
Nu x
Re x
dK
where
Vf ( x) x
Re x
Q
Vf (x)
K ( x, y )
external flow velocity over the wedge, m/s
y
Vf ( x)
, similarity variable
Qx
PROBLEM 4.46 (continued)
T
T Ts
, dimensionless temperature
Tf Ts
Q = kinematic viscosity, m 2 s
The problem reduces to determining the dimensionless temperature gradient at the surface
dT (0) / dK given by equation (4.94)
dT (0)
dK
­°
®
°̄
f
³
0
ª (m 1) Pr
exp «
2
¬
K
½°
º
F (K )dK » dK ¾
°¿
¼
0
1
³
(4.94)
where m is measure of wedge angle SE , defined as
m
E
(4.83)
2E
The function F (K ) is obtained from the solution to the flow field over the wedge. It is defined in
(4.87) as
dF u ( x, y )
(4.87)
dK Vf ( x)
where u(x,y) is the axial velocity within the thermal boundary layer. H
owever, for
assume that
u ( x, y ) Vf ( x)
(a) into (4.87)
dF
1
dK
Integration of (b) gives
F K
(b) into (4.94)
dT ( 0)
dK
­°
®
°̄
f
³
0
ª (m 1) Pr
exp «
2
«¬
K
½°
º
K dK » dK ¾
°¿
»¼
0
Pr <1 we
(a)
(b)
(c)
1
³
(d)
Evaluating the integral in the integrand
dT (0)
dK
­°
®
°̄
f
³
0
½°
ª (m 1) Pr 2 º
K » dK ¾
exp «
2
¬
¼
°¿
1
(e)
The definite integral in (e) is recognize d as the error function. To proceed, let
z
(m 1) Pr
K
4
(f)
Thus
dK
4
dz
(m 1) Pr
(g)
PROBLEM 4.46 (continued)
Substitute (f) and (g) into (e)
dT (0)
dK
­°
4
®
°̄ (m 1) Pr
f
½°
exp( z dz )¾
°¿
0
³
1
2
(h)
owever
H
z
2
erf z
S
³
2
e z dz
(i)
0
and
erf f
2
S
f
³
2
e z dz 1
(j)
0
(i) into (h)
dT ( 0)
dK
­
½
S
®
¾
¯ (m 1) Pr ¿
1
(m 1) Pr
S
(k)
(k) into (4.68)
Nu x
(m 1) Pr
S
Re x
(l)
(iii) Checking. Dimensional check;All equations are dimensionless.
Limiting check: For a wedge with ezro angle ( E m = 0), the solutions should reduce to the flat
plate solution with Pr <1 and u Vf . . Setting m = 0 in (l) gives
Nu x
0.564 PrRe x
(m)
This agrees with the solution to Problem 4.36.
(5) Comments. The assumption that axial velocity within the thermal boundary layer is the same
as that of the external flow provided a major simplification in the solution. It made it possible to
obtain a solution for the Nusselt number without the need for numerical integration.
PROBLEM 4.47
Consider laminar boundary layer flow over a wedge at a uniform temperature Ts . When the
Prandtl number is very high the viscous boundary layer is much thicker than the thermal
boundary layer. Assume that the velocity profile within the thermal boundary layer is
approximately linear. Show that for such approximation the local Nusselt number is given by
Nu x
f
Note:
³
exp(cx 3 )dx
0
0.489>(m 1) F cc(0) Pr @1/3 Re1 / 2
c 1 / 3
* (1 / 3) , where ī is the Gamma function.
3
(1) Observations. (i) The flow field for this boundary
layer problem is simplified by assuming that axial
velocity within the thermal boundary layer varies
linearly with the normal distance. (ii) Since velocity
distribution affects temperature distribution, the solution
for the local Nusselt number differs from the exact case
of Section 4.4.3. (iii) The local Nusselt number depends
on the local heat transfer coefficient which depends on
the temperature gradient at the surface.
(2) Problem Definition. Determine the temperature
gradient at the surface for laminar flow over a wedge
for the simplified velocity field described above.
(3) Solution Plan. Follow the analysis of section 4.4.3 for determining the local Nusselt number
using a simplified flow field solution.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) twodimensional, (5) laminar flow (Rex <5 u105), (6) viscous and thermal boundary layer flow (Rex >
100 and Pex >100), (7) uniform upstr eam velocity and temperature, (8) uniform surface
temperature, (9) symmetrical flow, (10) negligible changes in kinetic and potential energy, (11)
negligible dissipation, (12) no buoyancy (E = 0 or g = 0), (13) no energy generation ( q ccc 0 ) and
(14) Pr >1. Thus assume that axia l velocity within the thermal boundary layer varies linearly
with distance normal to wedge surface.
(ii) Analysis. The local Nusselt number, Nu x , for a wedge is given by equation (4.96) of
Section 4.4.3
dT (0)
(4.68)
Nu x
Re x
dK
where
Vf ( x) x
Re x
Q
Vf (x)
external flow velocity over the wedge, m/s
PROBLEM 4.47 (continued)
K ( x, y )
T
y
Vf ( x)
, similarity variable
ǎx
T Ts
, dimensionless temperature
Tf Ts
ǎ = kinematic viscosity, m 2 s
The problem reduces to determining the dimensionless temperature gradient at the surface
dT (0) / dK given by equation (4.94)
dT ( 0)
dK
­°
®
°̄
f
³
0
ª (m 1) Pr
exp «
2
¬
K
½°
º
F (K )dK » dK ¾
°¿
¼
0
³
1
(4.94)
where m is a measure of wedge angle SE , defined as
E
m
(4.83)
2E
The function F (K ) is obtained from the solution to the flow field over the wedge. It is defined in
(4.87) as
dF u ( x, y )
(4.87)
dK Vf ( x)
where u(x,y) is the axial velocity within the thermal boundary layer. oHwever, for Pr 1> we use
a simplified flow field of linear velocity given by
u
Vf
dF
| AK
dK
(c)
where A is constant. Integrating (c)
F (K )
A
K2
2
B
(d)
The constant B is determined from wedge flow boundary condition. (4.89b) and solution.
B
oundary condition (4.89b) gives
F (0)
0
(4.89b)
Apply (4.89b) to (d)
B=0
(e)
Differentiate (d) twice and apply at the surface
A
d 2F
dK 2
F cc(0)
(f)
The constant F cc(0) depends on wedge angle. It is obtained from the solution to flow field over
the wedge. V
alues of F cc()0 for four angles are listed in Table 4.3. Thus (d) can be written as
PROBLEM 4.47 (continued)
F cc(0) 2
K
2
F (K )
(g)
Substitute (g) into (4.94)
­°
®
°̄
d T ( 0)
dK
f
ª (m 1) F cc(0) Pr
exp «
4
¬«
³
0
K
³
½°
º
K dK » dK ¾
°¿
¼»
1
2
0
\
(h)
Evaluating the integral in the integrand
­°
®
°̄
dT (0)
dK
f
³
0
½°
ª (m 1) F cc(0) Pr 3 º
K » dK ¾
exp «
12
¬
¼
°¿
1
(i)
The definite integral in (i) is evaluated next. eLt
(m 1) F cc(0) Pr 3
K
12
z
It follows that
ª
º
12
« (m 1) F cc(0) Pr »
¬
¼
K
1/ 3
z1/ 3
Differentiating
º
1ª
12
«
3 ¬ (m 1) F cc(0) Pr »¼
dK
1/ 3
z 2 / 3 dz
Substituting into (i)
dT (0)
dK
1/ 3
­° 1 ª
º
12
® «
»
°̄ 3 ¬ (m 1) F cc(0) Pr ¼
f
³e
z
z 2 / 3
0
½°
dz ¾
°¿
1
(j)
The definite integral in (j) is recognized as the aGmma function given by
f
³e
z n 1
z
dz
* (n)
n !1
(k)
0
Comparing the integral in (j) with (k) gives
n
1
3
(l)
Using (k) and (l), equation (j) becomes
dT (0)
dK
1/ 3
­° 1 ª
½°
º
12
® «
» * (1 / 3)¾
°̄ 3 ¬ (m 1) F cc(0) Pr ¼
°¿
1
(m)
PROBLEM 4.47 (continued)
amma function
The value of * (1 / 3) is obtained from tables of G
* (1 / 3)
2.679
(n)
0.489>(m 1) F cc(0)@ 1/ 3 Pr 1/3
(o)
(n) into (m)
dT (0)
dK
Substitute (n) into (o)
Nu x
0.489>(m 1) F cc(0)@ 1/ 3 Pr 1/3 Re1x/ 2
(o)
(iii) Checking. Dimensional check: (1) The exponent of the exponential in (j) is
dimensionless. (2) All equations are dimensionless.
Limiting check: For a wedge with ezro angle ( E m = 0), the solutions should reduce to the flat
plate solution with Pr >1 and linear axial velocity. Setting m = 0 in (o) gives
Nu x
0.489 >F cc(0)@ 1/ 3 Pr 1/3 Re1x/ 2
(p)
oHwever, for m = 0 the flow field reduces to B
lasius solution. Thus, using Table 4.1, gives
F cc(0)
f cc(0)
0.33206
(q)
(q) into (p)
Nu x
0.339 Pr 1/3 Re1x/ 2
This agrees with the solution to Problem 4.37.
(5) Comments. The assumption that axial velocity within the thermal boundary layer varies
linearly in the direction normal to the surface provided a major simplification in the solution. It
made it possible to obtain a solution for the Nusselt number without the need for numerical
integration.
PROBLEM 5.1
For fluids with Pr 1 the thermal boundary
layer thickness is much larger than the viscous
boundary layer. That is G t / G !! 1. It is
reasonable for such cases to assume that fluid
velocity within the thermal layer is uniform
equal to the free stream velocity. That is
u | Vf
Consider uniform laminar boundary layer flow over a flat plate. The surface is maintained at
uniform temperature Ts and has an insulated leading section of length xo . Assume a third degree
polynomial temperature profile, show that the local Nusselt number is given by
Nu x
ª x º
0.53 «1 o »
x¼
¬
where the local Reynolds number is Re x
1/ 2
Pr 1 / 2 Re1x/ 2
Vf x / Q .
(1) Observations. (i) In general, to determine the Nusselt number it is necessary to determine the
velocity and temperature distribution. (ii) Fluid velocity for Pr 1 is assumed to be uniform,
u | Vf . This represents a significant simplification.
(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a
flat plate with insulated leading section.
(3) Solution Plan. Start with equating Newton’s law with Fourier’s law to obtain an equation for
the heat transfer coefficient h. Apply the integral form of the energy equation using a third
degree polynomial temperature profile.
(4) Plan Execution. (i) Assumptions. (1) Steady state, (2) constant properties, (3) twodimensional, (4) laminar flow (Rex <5 u105), (5) thermal boundary layer (Pe >100), (6) uniform
upstream velocity and temperature , (7) flat plate, (8) uniform surface temperature, (9) negligible
changes in kinetic and potential energy, (10) negligible axial conduction, (11) negligible
dissipation, (12) no buoyancy (E = 0 or g = 0) and (13) uniform velocity within the thermal
boundary layer ( Pr 1 ).
(ii) Analysis. The local Nusselt number is defined as
hx
k
Nu x
(a)
where the heat transfer coefficient h is given by equation (1.10)
wT ( x,0)
wy
Ts Tf
k
h
(1.10)
Thus h depends on the temperature distribution T ( x, y ). The integral form of the energy equation
is used to determine the temperature distribution
PROBLEM 5.1 (continued)
wT x,0
D
wy
G t ( x)
d
dx
³
u (T Tf )dy
(5.7)
0
oHwever, For Pr 1 the velocity boundary layer thickness is much smaller than the thermal
boundary layer thickness. Thus we assume
u
(b)
Vf
For the temperature profile we assume a third degree polynomial
T x, y
b0 x b1 x y b2 x y 2 b3 x y 3
(c)
The boundary conditions on the temperature are
(1) T x,0 Ts
(2) T x, G t # Tf
wT x, G t
#0
(3)
wy
(4)
w 2T x,0
wy 2
0
Equation (c) and the four boundary conditions give the coefficients bn (x)
3
1
1
1
(Tf Ts ) , b2 0, b3 (Tf Ts ) 3
b0 Ts , b1
Gt
2
2
Gt
Substituting the above into (c)
ª3 y 1 y3 º
T ( x, y ) Ts (Tf Ts ) «
(d)
»
3
«¬ 2 G t 2 G t »¼
Substituting (d) into (1.10)
3k
h
(e)
2G t
Combining (a) and (e)
3x
Nu x
(f)
2G t
The problem reduces to finding G t which is obtained using the energy equation. Substituting (b)
and (d) into (5.7)
3(Tf Ts )
d
D
Vf (Ts Tf )
2G t
dx
G t ( x)
³
0
ª
3y
y3 º
1
«
» dy
3
«¬ 2G t 2G t »¼
Evaluating the integral in the above
d
3
>G t (3 / 4)G t (1 / 8)G t @ 3 Vf dG t
Vf
D
dx
dx
2G t
8
Separating variables and rearranging
PROBLEM 5.1 (continued)
4D
dx
Vf
G t dG t
(f)
Integrating
8D
xC
Vf
where C is constant of integration. The boundary condition on G t is
G t2
(g)
G t ( xo ) 0
(h)
Applying (h) to (g) gives
C
8D
xo
Vf
Substituting into (g) and solving for G t
Gt
8D ª xo º
1 »
Vf x «¬
x¼
x
(i)
Substituting (i) into (f)
3
Nu x
Vf x
4 2
1
D 1 ( xo / x)
(j)
P/U
k / Uc p
(k)
Noting that
cpP
Pr
k
Q
D
Using (k), equation (j) is expressed in terms of Prandtl and Reynolds numbers
PrRe x
1 ( xo / x)
(iii) Checking. Dimensional check: All terms in equations (i)-(l) are dimensionless.
Nu x
(l)
0.530
Boundary conditions check: Assumed temperature profile satisfies the four boundary conditions
on temperature.
Comparison with scaling results: For the special case of xo
number for Pr <1 is given in equation (4.55)
Nu x a PrRe x ,
0 , scaling estimate of the Nusselt
for Pr 1<
(4.55)
Thus, the dependency on the Prandtl and Reynolds number is the same in (l) and (4.55).
(5) Comments. (i) For the special case of no leading insulated section, equation (l) reduces to
Nu x
0.530 PrRe x
(m
Nu x
0.564 PrRe x
(n)
The exact solution to this case gives
Thus the error in the integral solution is 6%
.
PROBLEM 5.2
For fluids with Pr !! 1 the thermal boundary layer
thickness is much smaller than the viscous boundary
layer. That is G t / G 1. It is reasonable for such
cases to assume that fluid velocity within the thermal
layer is linear given by
y
u Vf
G
Consider uniform laminar boundary layer flow over a flat plate with an insulated leading section
of length xo . The plate is maintained at uniform surface temperature Ts . Assume a third degree
polynomial temperature profile, show that the local Nusselt number is given by
Nu x
>
0.319 1 xo / x
@
3/ 4
1 / 3
Pr 1 / 3 Re1 / 2
(1) Observations. (i) In general, to determine the Nusselt number it is necessary to determine the
velocity and temperature distribution. (ii) Fluid velocity for Pr !! 1 is assumed to be linear,
u Vf ( y / G ) .
(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a
flat plate with insulated leading section.
(3) Solution Plan. Start with equating Newton’s law with Fourier’s law to obtain an equation for
the heat transfer coefficient h. Apply the integral form of the energy equation using a linear
velocity profile and a third degree polynomial temperature profile.
(4) Plan Execution. (i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant
properties, (4) laminar flow (Rex <5 u105), (5) thermal boundary layer (Pe >100), (6) uniform
upstream velocity and temperature , (7) flat plate, (8) uniform surface temperature, (9) negligible
changes in kinetic and potential energy, (10) negligible axial conduction, (11) negligible
dissipation, (12) no buoyancy (E = 0 or g = 0) and (13) linear velocity within the thermal
boundary layer ( Pr !! 1 ).
(ii) Analysis. The local Nusselt number is defined as
hx
k
Nu x
(a)
where the heat transfer coefficient h is given by equation (1.10)
wT ( x,0)
wy
Ts Tf
k
h
(1.10)
Thus h depends on the temperature distribution T ( x, y ). The integral form of the energy equation
is used to determine the temperature distribution
D
wT x,0
wy
d
dx
G t ( x)
³
0
u (T Tf )dy
(5.7)
PROBLEM 5.2 (continued)
oHwever, For Pr !! 1 the velocity boundar (5. (5. (5. (5. (5. (5. (5. (5.r than the thermal
boundary layer thickness. Thus we assume
u
Vf
y
(b)
G
where G is the thickness of the velocity boundary layer. Application of the integral form of the
momentum equation, (5.5) gives G (x). The solution for G (x) for this case is detailed in Example
5.1 and is given by
12
G
(5.26)
x
Re x
This gives
G
12
Q
Vf
x
(c)
For the temperature profile we assume a third degree polynomial
T ( x, y )
b0 ( x) b1 ( x) y b2 ( x) y 2 b3 ( x) y 3
(d)
The boundary conditions on the temperature are
(1) T x,0 Ts
(2) T x, G t # Tf
wT x, G t
#0
(3)
wy
(4)
w 2T x,0
wy 2
0
Equation (d) and the four boundary conditions give the coefficients bn (x )
3
1
1
1
(Tf Ts ) , b2 0, b3 (Tf Ts ) 3
b0 Ts , b1
Gt
2
2
Gt
Substituting the above into (d)
ª3 y 1 y3 º
T ( x, y ) Ts (Tf Ts ) «
(e)
»
3
«¬ 2 G t 2 G t »¼
Substituting (e) into (1.10)
3k
h
(f)
2G t
Combining (a) and (f)
3x
(g)
Nu x
2G t
The problem reduces to finding G t which is obtained using the energy equation. Substituting (b)
and (e) into (5.7)
PROBLEM 5.2 (continued)
G t ( x)
3(Tf Ts )
d
D
Vf (Ts Tf )
2G t
dx
3y
yª
y3 º
3 » dy
«1 G ¬« 2G t 2G t ¼»
³
0
Evaluating the integral in the above
D
uMltiplying and dividing by
Vf d ª G t2 º
« »
10 dx «¬ G »¼
3
2G t
G and rearranging the above
d ª G t2 º
«G
»
dx «¬ G 2 »¼
Gt
G
15D 1
Vf G
(h)
Define
Gt
G
r
(i)
substituting (i) into (h)
15D 1
Vf G
oHwever,
> @
d
G r2
dx
r
(j)
G (x) is given in (c). Substituting (c) into the above and rearranging
15D 1
12Q x
r
d
dx
>
x r2
@
(k)
Differentiating the right hand side and noting that Pr Q / D
1.25 1
Pr x
2 x r2
dr
1 3
r
dx 2 x
(l)
Rearranging and separating variables
4r 2 dr
dx
x
Integrating and using the boundary condition G t ( xo )
x
dx
xo x
³
(m)
(2.5 / Pr ) - r 3
r
³
0
0
4r 2 dr
(2.5 / Pr ) - r 3
(n)
Evaluating the integrals
ln
x
xo
>
ln (2.5 / Pr ) r 3
@
4/3
The above can be written as
x
xo
>(2.5 / Pr ) r @
3 4/3
(o)
PROBLEM 5.2 (continued)
Using the definition of r in (i) and solving (o) for r
r
Gt
G
ª 2 .5
3/ 4 º
« Pr 1 [ xo / x]
»
¬
¼
1/ 3
(p)
Substituting (c) for G
12Q
ª 2.5
º
x «
1 [ x o / x ]3 / 4 »
Vf
¬ Pr
¼
Gt
1/ 3
(q)
Using the definition of Reynolds number
Gt
12 (2.5)1 / 3
x
1
Pr 1/3 Re1/2
>1 ( x
3/ 4
o / x)
@
1/ 3
(r)
Substituting (r) into (g)
Nu x
(iii) Checking.
dimensionless.
>
0.319 1 xo / x
3/ 4
@
1 / 3
Pr 1 / 3 Re1 / 2
(s)
Dimensional check: All terms in equations (g)-(j) and (m)-(p) are
Boundary conditions check: Assumed temperature profile satisfies the four boundary conditions
on temperature.
Comparison with scaling results: For the special case of xo
number for Pr >1 is given in equation (4.57)
Nu x a ( Pr )1/3 Re x ,
0 , scaling estimate of the Nusselt
for Pr 1>
(4.57)
Thus, the dependency on the Prandtl and Reynolds number is the same in (s) and (4.57).
(5) Comments. (i) For the special case of no leading insulated section, equation (s) reduces to
Nu x
1/ 3
0.319 ( Pr )
(t)
Re x
The exact solution to this case gives
Nu x
0.339 ( Pr )1/ 3 Re x ,
Thus the error in the integral solution is 5.9%
.
for Pr >1
(n)
PROBLEM 5.3
A square array of chips is mounted flush on a flat plate. The array measures L cm u L cm . The
forward edge of the array is at a distance xo from the leading edge of the plate. The chips
dissipate uniform surface flux q csc . The plate is cooled by forced convection with uniform
upstream velocity Vf and temperature Tf . Assume laminar boundary layer flow. Assume
further that the axial velocity within the thermal
boundary layer is equal to the free stream velocity.
Use a third degree polynomial temperature profile.
a[ ] Show that the local Nusselt number is given by
Nu x
0.75
Pr Re x
1 ( xo / x)
b[ ] Determine the maximum surface temperature.
(1) Observations. (i) The velocity is assumed to be uniform, u Vf , throughout the thermal
boundary layer. (ii) A leading section of length xo is unheated. (iii) at x ! xo , surface heat flux
is uniform. (iv) The determination of the Nusselt number requires the determination of the
temperature distribution. (v) Surface temperature is unknown. (vi) The maximum surface
temperature for a uniformly heated plate occurs at the trailing end.
(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a
flat plate which heated with uniform surface flux. This reduces to determining the thermal
boundary layer thickness.
(3) Solution Plan. Start with the definition of local Nusselt number. Equate Newton’s law with
Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of
the energy equation using a third degree polynomial temperature profile to determine the
temperature distribution.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties, (4) laminar
flow (Rex <5 u105), (5) thermal boundary layer (Pe >100), (6) flat surfa ce, (7) uniform surface
heat flux, (8) negligible changes in kinetic and potential energy, (9) negligible axial conduction,
(10) negligible dissipation and (11) no buoyancy (E = 0).
(ii) Analysis. The local Nusselt number is defined as
hx
k
where the heat transfer coefficient h is given by equation (1.10)
Nu x
wT ( x,0)
wy
Ts ( x) Tf
(a)
k
h
Substitute (1.10) into (a)
(1.10)
PROBLEM 5.3 (continued)
wT ( x,0)
wy
x
Ts ( x) Tf
Nu x
(b)
Thus the Nusselt number depends on the temperature distribution T ( x, y ). Note that surface
temperature varies with location x and is unknown. The integral form of the energy equation is
used to determine the temperature distribution
wT x,0
D
wy
G t ( x)
d
dx
³
(5.7)
u (T Tf )dy
0
The axial velocity u is assumed to be uniform equal to the free stream velocity. Thus
u
(c)
Vf
(c) into (5.7)
wT x,0
D
wy
d
dx
Vf
G t ( x)
³
(d)
(T Tf )dy
0
O
nce temperature distribution is determine d, surface temperature and maximum surface
temperature will be known.
We assume a third degree polynomial temperature profile
T ( x, y )
b0 ( x) b1 ( x) y b2 ( x) y 2 b3 ( x) y 3
(e)
The boundary conditions on the temperature are
wT ( x,0)
q csc
wy
(2) T ( x, G t ) # Tf
wTx, G t
(3)
#0
wy
(1) k
(4)
w 2T ( x,0)
wy 2
0
Equation (e) and the four boundary conditions give the coefficients bn (x)
b0
Tf 2 q csc
Gt
3 k
q csc
,
k
b1
Tf q csc
k
b2
0,
b3
q csc 1
3k G t2
Substitute the above into (e)
T ( x, y )
ª2
y3 º
y
G
« t
»
3G t2 »¼
«¬ 3
(f)
PROBLEM 5.3 (continued)
Surface temperature is obtained by setting y = 0 in (f)
Ts ( x )
Tf T ( x ,0 )
2q csc
Gt
3k
(g)
The Nusselt number is obtained by substituting (f) and (g) into (b)
3 x
2 Gt
Nu x
(h)
The problem reduces to determining G t . Substitute (f) into (d) and simplify
D
Vf
G t ( x)
ª2
y3 º
d
G
y
« t
» dy
dx
3G t2 »¼
«¬ 3
0
³
Evaluate the integral in the above
D
Vf
d ª2 2 1 2 1 2 º
Gt Gt Gt »
dx «¬ 3
2
12 ¼
1 dG t2
4 dx
D
Vf
(i)
Separate variables and integrate
D
Vf
x
³
1
4
dx
³
Gt
dG t2
Evaluate the integrals
4
D
Vf
G t2 C
x
(j)
where C is constant of integration determined from the boundary condition on G t
Gt
0 at x
xo
(k)
Apply (k) to (j)
C
4
D
Vf
xo
Substitute into (j) and solve for G t
Gt
4
D
Vf
( x xo )
The Nusselt number is obtained by substituting (l) into (h)
Nu x
3
2
x
4
D
Vf
( x xo )
(l)
PROBLEM 5.3 (continued)
This simplifies to
Nu x
Noting that D
Vf x
3
4 D (1 xo / x)
(m)
k / U c p , (m) is rewritten as
Nu x
U c pVf x
3
4 k (1 xo / x)
3 c p P U Vf x
1
4
k
P (1 xo / x)
Introduce the definitions of Prandtl and Reynolds numbers, the above gives
Nu x
Pr Re x
1 ( xo / x)
0.75
(n)
Surface temperature is obtained by substituting (l) in (g)
Ts ( x)
Tf 4q csc D
( x xo )
3k Vf
(o)
According to (o), the maximum surface temperature occurs at the trailing end of the plate.
For the special case where heating begins at the leading edge, xo
0, equations (n) and (o) give
Nu x
0.75
Pr Re x
(p)
Ts ( x)
Tf 4q csc D
3k Vf
(q)
(iii) Checking. Dimensional check: (1) All term in (b), (h), (m) and (n) are dimensionless. (2)
Each term in (f) and (o) has units of temperature.
Boundary conditions check: Assumed temperature profile satisfies the four boundary conditions
on temperature.
Limiting check: If q csc 0 surface temperature will be the same as free stream temperature.
Setting q csc 0 in (0) gives Ts ( x) Tf .
(5) Comments. For the special case of xo
Pr 1 is given in equation (4.55)
0 , scaling estimate of the Nusselt number for
Nu x a Pr Re x ,
for Pr <
1
Thus, the dependency on the Prandtl and Reynolds number is the same in (p) and (4.55).
(4.55)
PROBLEM 5.4
A liquid film of thickness H flows by gravity down
an inclined surface. The axial velocity u is given by
u
ª y
y2 º
u o «2 2 »
«¬ H H »¼
where u o is the free surface velocity. At x ! 0 the
surface is maintained at uniform temperature Ts .
The fluid temperature upstream of this section is
Tf . Assume laminar boundary layer flow and that G t / H 1. Determine the local Nusselt
number and the total surface heat transfer from a section of width W and length L. Neglect heat
loss from the free surface. Use a third degree polynomial temperature profile.
(1) Observations. (i) The velocity distribution is known. (ii) Surface temperature is uniform. (iii)
The determination of the Nusselt number requires the determination of the temperature
distribution. (iv) Newton’s law of cooling gives the heat transfer rate. This requires knowing the
local heat transfer coefficient.
(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a
flat plate at uniform surface temperature. This reduces to determining the thermal boundary layer
thickness.
(3) Solution Plan. Start with the definition of local Nusselt number. Equate Newton’s law with
Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of
the energy equation using a third degree polynomial temperature profile to determine the
temperature distribution. Apply Newton’s law of cooling to determine the heat transfer rate.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties, (4) laminar
flow (Rex <5 u105), (5) thermal boundary layer (Pe >100), (6) flat surfa ce, (7) uniform surface
temperature, (8) negligible changes in kinetic and potential energy, (9) negligible axial
conduction, (10) negligible dissipation and (11) no buoyancy (E = 0).
(ii) Analysis The local Nusselt number is defined as
hx
k
where the heat transfer coefficient h is given by equation (1.10)
Nu x
wT ( x,0)
wy
Ts Tf
(a)
k
h
Substitute (1.10) into (a)
(1.10)
PROBLEM 5.4 (continued)
wT ( x,0)
wy
x
Ts Tf
Nu x
(b)
Thus, the Nusselt number depends on the temperature distribution T ( x, y ). Integration of
Newton’s law of cooling gives the total heat transfer rate
L
q
³ h( x)dx
(Ts Tf )W
(c)
0
The integral form of the energy equation is used to determine the temperature distribution
wT x,0
D
wy
d
dx
G t ( x)
³
u (T Tf )dy
(5.7)
0
The velocity distribution is given by
u
ª y
y2 º
u o «2 2 »
«¬ H H »¼
(d)
We assume a third degree polynomial temperature profile
T ( x, y )
b0 ( x) b1 ( x) y b2 ( x) y 2 b3 ( x) y 3
(e)
The boundary conditions on the temperature are
(1) T ( x,0) Ts
(2) T ( x, G t ) # Tf
wTx, G t
#0
(3)
wy
(4)
w 2T ( x,0)
wy 2
0
Equation (e) and the four boundary conditions give the coefficients bn (x)
b0
Ts ,
b1
3
1
(Tf Ts ) , b2
Gt
2
0,
b3
1
1
(Tf Ts ) 3
2
Gt
Substitute the above into (e)
T ( x, y )
ª3 y 1 y3 º
Ts (Tf Ts ) «
»
3
¬« 2 G t 2 G t ¼»
(f)
The Nusselt number is obtained by substituting (f) into (b)
Nu x
3 x
2 Gt
(g)
PROBLEM 5.4 (continued)
The problem reduces to determining G t . Substitute (d) and (f) into (5.7)
D
3 1
2 Gt
d
dx
uo
G t ( x)
³0
º
ª y
y 2 ºª 3 y 1 y 3
2
1
«
» dy
»
«
«¬ H H 2 »¼ «¬ 2 G t 2 G t3
»¼
Expand the integrand
3 1
D
2 Gt
uo
d
dx
G t ( x)
³0
ª 3 y 2 1 y4 2
3 y3
1 y5 y 2 º
y
«
» dy
3
H
2 H 2 G t 2 H 2 G t3 H 2 »¼
«¬ H G t H G t
Evaluate the integral
D
3 1
2 Gt
uo
d ª 1
1 2º
G t3 Gt »
«
2
dx ¬ 24 H
5H
¼
(h)
To solve this differential equation for G t it is first simplified by noting that
1
24 H
G 3 2 t
1 2
Gt
5H
Thus (h) is approximated by
D
3 1
2 Gt
d
dx
uo
ª 1 2º
«¬ 5 H G t »¼
(i)
Rewrite (i)
D 15 H
uo 2 G t
2G t
dG t
dx
Separate variables and integrate
15 D H
4 uo
x
³ dx ³
Gt
G t2 dG t
Evaluate the integrals
45 D H
x G t3 C
4 uo
(j)
where C is constant of integration determined from the boundary condition on G t
Gt
0 at x
(k)
0
Apply (k) to (j) gives
C
0
Substitute into (j) and solve for G t
Gt
ª 45 D H º
x»
«
¬ 4 uo ¼
1/ 3
The Nusselt number is obtained by substituting (l) into (g)
(l)
PROBLEM 5.4 (continued)
Noting that D
3 ª 4 uo 1 º
Nu x
»
«
2 ¬ 45 D H x ¼
k / U c p , (m) is rewritten as
Nu x
3ª 4 º
4 «¬ 45 »¼
1/ 3
ª U c p uo H x 2 º
«
»
k
H 2 »¼
«¬
1/ 3
1/ 3
x
1
101 / 3
ª U uo H c p P x 2 º
«
»
k H 2 »¼
«¬ Pk
1/ 3
Introduce the definitions of Prandtl and Reynolds numbers, the above gives
1
Nu x
101 / 3
ª
x2 º
Pr
R
e
»
«
H
H 2 »¼
«¬
1/ 3
(m)
where the Reynolds number is defined as
uo H
Re x
(n)
Q
With the Nusselt number determined, the local heat transfer coefficient can be formulated and
the total heat transfer rate computed. Equate (a) and (m)
hx
k
ª
x2 º
Pr
R
e
«
»
H
H 2 »¼
«¬
1
101 / 3
1/ 3
Solve for h
h
k
1/ 3
10
H
2/3
( Pr Re H )1/ 3 ( x) 1/ 3
(o)
Substitute (o) into (c)
(Ts Tf )W
q
L
k
1/ 3
10
H
2/3
( PrRe H )1 / 3
³x
1 / 3
dx
(p)
0
Evaluate the integral
q
3 1
(Ts Tf )Wk ( Pr Re H )1/ 3 ( L / H ) 2 / 3
1
/
3
2 10
(q)
Rewrite (q) in dimensionless form
q
(Ts Tf )Wk
3 1
( Pr Re H )1 / 3 ( L / H ) 2 / 3
1
/
3
2 10
(r)
(iii) Checking. Dimensional check: (1) Each term in (f) has units of temperature. (2) Each
term is (g) has units of heat. (3) each term in (b), (g), (m) and (r) is dimensionless.
PROBLEM 5.4 (continued)
Limiting check: If surface temperature is the same as free stream temperature, Ts
transfer rate will vanish. Setting Ts Tf in (q) gives q = 0.
Tf , the heat
Boundary conditions check: Assumed temperature profile satisfies the four boundary conditions
on temperature.
Qualitative check: The total heat transfer rate is expected to increase with increasing length L.
This is in agreement with result (q).
(5) Comments. (i) The Nusselt number in (m) depends on a single parameter PrRe . This
product is known as the Peclet number Pe. (ii) The dimensionless heat transfer rate depends on
two parameters: the Peclet number, Pe, and the geometric parameter L/H. (iii) The solution is
not valid for G t ! H . Thus there is a maximum length for which the solution is valid. This
maximum length, Lmax , is determined from (l) by setting G t H and letting x Lmax . The
result is
4 uo H 2
(s)
Lmax
45 D
Expressed in dimensionless form, (s) becomes
Lmax
H
4
Pr Re H
45
(t)
PROBLEM 5.5
A thin liquid film flows under gravity down an
inclined surface of width W. The film thickness
is H and the angle of inclination is T . The
solution to the equations of motion gives the
axial velocity u of the film as
u
gH 2 sin T
2Q
ª y
y2 º
2
«
2»
¬ H H ¼
Heat is added to the film along the surface
beginning at x 0 at uniform flux q csc . Determine the total heat added from x 0 to the section
where the thermal boundary layer penetrates half the film thickness. Assume laminar boundary
layer flow and that G t / H 1. Neglect heat loss from the free surface. Use a third degree
polynomial temperature profile.
(1) Observations. (i) The velocity distribution is known. (ii) Total heat transfer is equal to heat
flux times surface area. (iii) eHat fl ux is given. oHwever, the distance x = L at which
G t H / 2 is unknown.
(2) Problem Definition. Determine the thickness of the thermal boundary layer G t (x ).
(3) Solution Plan. (i) Express total heat in terms of heat flux and surface area. (ii) Use the
integral form of the energy equation to determine G t (x ).
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties, (4) laminar
flow (Rex <5 u105), (5) thermal boundary layer (Pe >100), (6) flat surfa ce, (7) uniform surface
heat flux, (8) negligible changes in kinetic and potential energy, (9) negligible axial conduction,
(10) negligible dissipation and (11) no buoyancy (E = 0).
(ii) Analysis. The total heat transfer rate from the surface is
qT
q csc A
q csc LW
(a)
where
A surface area, m 2
L = distance along surface where G t
qT total heat transfer rate, W
W = width of surface, m
H /2, m
To determine G t ( x) , we apply the integral form of the energy equation
D
wT x,0
wy
where the axial velocity u is given by
d
dx
G t ( x)
³
0
u (T Tf )dy
(5.7)
PROBLEM 5.5 (continued)
gH 2 sin T
2Q
u
ª y
y2 º
2
«
2»
¬ H H ¼
(b)
We assume a third degree polynomial temperature profile
b0 ( x) b1 ( x) y b2 ( x) y 2 b3 ( x) y 3
T ( x, y )
(c)
The boundary conditions on the temperature are
wT ( x,0)
q csc
wy
(2) T ( x, G t ) # Tf
wTx, G t
(3)
#0
wy
(1) k
w 2T ( x,0)
(4)
wy 2
0
Equation (c) and the four boundary conditions give the coefficients bn (x)
b0
Tf 2 q csc
Gt
3 k
b1
q csc
,
k
b2
0,
b3
q csc 1
3k G t2
Substituting the above into (c)
T ( x, y )
q cc
Tf s
k
ª2
y3 º
« Gt y 2 »
3G t ¼»
¬« 3
Substituting (b) and (d) into (5.7) and recalling that D
q csc
Uc p
k
Uc p
g sin T
Q
k / Ucp
G t ( x)
gH 2 sin T ª y
y 2 º q csc
d
«2 2 »
2Q
dx
¬ H H ¼ k
0
³
d
dx
ª2
y3 º
« G t y 2 » dy
3G t »¼
«¬ 3
G t ( x)
ª2
º
H 4 Gt 2 1 3
1
2
y y 2 y 5 » dy
« HG t y Hy 2 y 3
2
3G t
36 t
«3
»¼
0 ¬
³
Evaluating the integral in the above
k
Uc p
g sin T d ª H 3 1 4 º
Gt Gt »
Q dx «¬15
72 ¼
Separating variables and rearranging
dx
g sin T
DQ
1
ªH
º
d « G t3 G t4 »
15
72
¬
¼
(d)
PROBLEM 5.5 (continued)
Integrating and noting that G t (0)
0
g sin T ª H 3 1 4 º
(e)
Gt Gt »
72 ¼
DQ «¬15
L , where L is the distance form the leading end to the location
x
Evaluating the above at x
where G t H / 2 , gives
L
g sin T ª H
1
º
( H / 2) 3 ( H / 2) 4 »
«
DQ ¬15
72
¼
43 H 4 g sin T
DQ
5760
(f)
Substituting (f) into (a)
qT
43 H 4 g sin T
Wq csc
DQ
5760
(g)
(iii) Checking. Dimensional check: (i) Each term in (d) has units of temperature. (ii) Each
term is (g) has units of watts.
Boundary conditions check: Assumed temperature profile satisfies the four boundary conditions
on temperature.
Qualitative check: The total heat transfer rate is expected to increase with increasing angle T .
This is in agreement with result (g).
(5) Comments. (i) Equation (g) shows that qT is proportional to H 4 . Thus film thickness has
significant effect on the total heat transfer rate.
PROBLEM 5.6
A plate is cooled by a fluid with Prandtl number Pr 1 . Surface temperature varies with
distance from the leading edge according to
Ts ( x,0)
Tf C x
where C is constant. For such a fluid it is
reasonable to assume that u | Vf . Use a
third degree polynomial temperature
profile to show that the local Nusselt
number is given by
Nu x
0.75 Pr 1 / 2 Re1 / 2
and that surface heat flux is uniform. Assume laminar boundary layer flow.
(1) Observations. (i) The determination of the Nusselt number requires the determination of
the velocity and temperature distributions. (ii). V
elocity is assumed uniform. (iii) Surface
temperature is variable. (iv) Newton’s law of cooling gives surface heat flux. This requires
knowing the local heat transfer coefficient.
(2) Problem Definition. Determine the temperature distribution for boundary layer flow
over a flat plate at variable surface temperature. This reduces to determining the thermal
boundary layer thickness.
(3) Solution Plan. Start with the definition of local Nusselt number. Equate Newton’s law
with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the
integral form of the energy equation using a third degree polynomial temperature profile to
determine the temperature distribution. Apply Newton’s law of cooling to determine surface
heat flux.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties, (3) two-dimensional, (4)
laminar flow (Rex <5 u105), (5) viscous boundary layer flow (Rex >100), (6) thermal
boundary layer (Pe >100), (7) uniform upstream velocity and temperature , (8) flat plate, (9)
(10) negligible changes in kinetic and potential energy, (11) negligible axial conduction,
(12) negligible dissipation, (13) no buoyancy (E = 0 or g = 0) and (14) Pr 1 .
(ii) Analysis. The local Nusselt number is defined as
Nu x
hx
k
(a)
The heat transfer coefficient h given by equation (1.10)
wT ( x,0)
wy
Ts ( x) Tf
k
h
(1.10)
PROBLEM 5.6 (continued)
Thus the temperature distribution T ( x, y ) must be determined. Surface heat flux is obtained
using Newton’s law of cooling
q csc
h(Ts Tf )
(b)
The integral form of the energy equation is used to determine temperature distribution
wT x,0
D
wy
d
dx
G t ( x)
³
u (T Tf )dy
(5.7)
0
Axial velocity distribution u(x,y) for Pr 1 is assume to be the same as free stream
velocity. Thus
u
Vf
(c)
We assume a third degree polynomial
T x, y
b0 x b1 x y b2 x y 2 b3 x y 3
(d)
The temperature boundary conditions are:
(1) T x,0
Ts ( x)
(2) T x, G t # Tf
(3)
wT x, G t
#0
wy
(4)
w 2T x,0
wy 2
0
The four boundary conditions are used to determine the coefficient in (d). The assumed
profile becomes
T ( x, y )
ª3 y 1 y3 º
Ts ( x) >Tf Ts ( x)@ «
»
3
¬« 2 G t 2 G t ¼»
(e)
Substitute (e) into (1.10)
h( x )
3 k
2 Gt
(f)
Nu x
3 x
2 Gt
(g)
Introducing (f) into (a)
Thus the problem reduces to determining the thermal boundary layer thickness G t . This is
accomplished using the integral form of the energy equation (5.7). Substituting (c) and (e)
into (5.7)
PROBLEM 5.6 (continued)
3 D >Ts ( x) Tf @
Gt
2 Vf
G t ( x)
d
dx
ª
y
«¬
2 Gt
>Ts ( x) Tf @ «1 3
³
0
1 y3 º
» dy
2 G t3 »¼
Evaluate the integral
3 D >Ts ( x) Tf @ 3 d
^ >Ts ( x) Tf @ G t `
2 Vf
Gt
8 dx
(h)
H
owever, surface temperature is given by
Ts Tf
(i)
C x
Substitute into (h)
D
4
x
>
d
dx
Vf G t
x Gt
@
(j)
To solve (j) for G t (x) we let
z2
xG t
(k)
Solve (k) for G t
z2
Gt
(l)
x
Substitute (k) and (l) into (j)
4
D
Vf
x z2
dz 2
dx
2z 3
dz
dx
Separate variables
2
D
Vf
x dx
z 3 dz
Integrate
D
Vf
x2
1 4
z Co
4
(m)
where C o is a constant determined from the boundary condition on G t (x) :
G t (0) 0
(n)
Apply (n) to (l) gives
z(0) = 0
(o)
Apply (o) to (m) gives C o = 0. Equation (m) becomes
4
Use (l) to eliminate z in (p)
D
Vf
x2
z4
(p)
PROBLEM 5.6 (continued)
4
D
Vf
x 2 xG t2
Solve the above for G t
Gt
2
D
Vf
x
(q)
Substitute (q) into (g) gives the local Nusselt number
3 Vf x
4 D
Nu x
Noting that D
k / U c p , the above becomes
Nu x
3
4
U c pVf x
k
3 c p P U Vf x
k
P
4
(r)
Expressing this result in terms of the Prandtl and local Reynolds number, gives
Nu x
0.75 Pr 1 / 2 Re1 / 2
(s)
The heat transfer coefficient is determined to examine surface heat flux. Substitute (q) into
(f), gives h(x)
3 Vf
(t)
h( x )
k
4 Dx
Substitute (i) and (t) into (b), simplify
V
3
(u)
q csc
kC f
4
D
This result shows that surface heat flux is uniform.
(5) Checking. Dimensional check: (i) Equations (g), (r) and (s) are dimensionless. (ii)
Equations (e), (f), (h), (q), (t) and (u) are dimensionally correct.
Boundary conditions check: Assumed temperature profile (e) satisfies the four boundary
conditions.
Limiting check: If surface temperature is the same as free stream temperature, the heat flux
will be ezro. According to (i) Ts ( x) Tf when C = 0. Setting C = 0 in (u) gives q csc 0 .
(6) Comments. The solution to the case of a specified uniform surface flux is given in
Section 5.7.3. The corresponding surface temperature is given in (5.35)
q cc
x
(5.35)
Ts ( x) Tf 2.396 s
1/3
k Pr Re1/2
x
Note that the above can be rewritten as
Ts ( x )
Tf C c x
PROBLEM 5.6 (continued)
where C c is constant. This is identical to equation (i) which gives the specified surface
temperature in this problem.
PROBLEM 5.7
A plate is cooled by a fluid with Prandtl number Pr !! 1 . Surface temperature varies with
distance form the leading edge according to
Ts ( x,0)
Tf C x
where C is constant. For such a fluid it
is reasonable to assume that axial
velocity within the thermal boundary
layer is linear given by
u
Vf
y
G
Determine the local Nusselt number and show that surface heat flux is uniform. Use a third
degree polynomial temperature profile and assume laminar boundary layer flow.
(1)Observations. (i) The determination of the Nusselt number requires the determination of
the velocity and temperature distributions. (ii). V
elocity is assumed linear. (iii) Surface
temperature is variable. (iv) Newton’s law of cooling gives surface heat flux. This requires
knowing the local heat transfer coefficient.
(2) Problem Definition. Determine the velocity and temperature distribution for boundary
layer flow over a flat plate at variable surface temperature. This reduces to determining the
viscous and thermal boundary layer thickness.
(3) Solution Plan. Start with the definition of local Nusselt number. Equate Newton’s law
with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the
integral form of the momentum and energy equations to determine the velocity and
temperature distribution. Apply Newton’s law of cooling to determine surface heat flux.
Since
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties, (3) two-dimensional, (4)
laminar flow (Rex <5 u105), (5) viscous boundary layer flow (Rex >100), (6) thermal
boundary layer (Pe >100), (7) uniform upstream velocity and temperature , (8) flat plate, (9)
(10) negligible changes in kinetic and potential energy, (11) negligible axial conduction,
(12) negligible dissipation, (13) no buoyancy (E = 0 or g = 0) and (14) Pr !! 1 .
(ii) Analysis. The local Nusselt number is defined as
Nu x
hx
k
(a)
The heat transfer coefficient h given by equation (1.10)
wT ( x,0)
wy
Ts ( x) Tf
k
h
(1.10)
PROBLEM 5.7 (continued)
Thus the temperature distribution T ( x, y ) must be determined. Surface heat flux is obtained
using Newton’s law of cooling
q csc
h(Ts Tf )
(b)
The integral form of the energy equation is used to determine temperature distribution
wT x,0
D
wy
G t ( x)
d
dx
³
(5.7)
u (T Tf )dy
0
Axial velocity distribution u(x,y) for Pr !! 1 is assume to vary linearly with normal
distance y from the plate. Thus
u
Vf
y
(c)
G
efore proceeding with the determination
where G is the viscous boundary layer thickness. B
of temperature distribution, G must be determined. We apply the integral form of the
momentum equation (5.5)
G x
wu x,0
v
wy
d
Vf
dx
³
0
G x
d
udy dx
³ u dy
2
(5.5)
0
Substitute (c) into (5.5)
G x
v
Vf
G
Vf2
d
dx
1
³
G
0
G x
ydy
Vf2
d
dx
³
0
1
G
2
y 2 dy
Evaluate the integrals and simplify
v
1
G
V f dG
6 dx
Separate variables and integrate
6
Q
Vf
dx
G2
2
C1
(d)
The constant of integration C1 is obtained from the boundary condition on G
G (0) 0
This condition gives C1 = 0. Substituting into (d) and solving for G
G
12Q
x
Vf
(e)
Turning now to the temperature distribution, we assume a third degree polynomial
T x, y
b0 x b1 x y b2 x y 2 b3 x y 3
(f)
PROBLEM 5.7 (continued)
The temperature boundary conditions are:
(1) T x,0
Ts ( x)
(2) T x, G t # Tf
(3)
wT x, G t
#0
wy
(4)
w 2T x,0
wy 2
0
The four boundary conditions are used to determine the coefficient in (f). The assumed
profile becomes
T ( x, y )
ª3 y 1 y3 º
Ts ( x) >Tf Ts ( x)@ «
»
3
¬« 2 G t 2 G t ¼»
(g)
Substitute (g) into (1.10)
h( x )
3 k
2 Gt
(h)
Nu x
3 x
2 Gt
(i)
Introducing (h) into (a)
Thus the problem reduces to determining the thermal boundary layer thickness G t . This is
accomplished using the integral form of the energy equation (5.7). Substituting (c) and (g)
into (5.7)
3 D >Ts ( x) Tf @
Gt
2 Vf
d
dx
G t ( x)
y
³
G
ª
y
¬«
2 Gt
>Ts ( x) Tf @ «1 3
0
1 y3 º
» dy
2 G t3 ¼»
Evaluate the integral
3 D >Ts ( x) Tf @ 1 d ­ ª Ts ( x) Tf º 2 ½
®
» Gt ¾
Gt
G
2 Vf
10 dx ¯ «¬
¼ ¿
(j)
oHwever, surface temperature is given by
Ts Tf
C x
(k)
Substitute into (j)
15
D
x
Vf G t
To solve (l) for G t (x) we first rewrite (l)
d ª x 2 º
Gt »
«
dx ¬ G
¼
(l)
PROBLEM 5.7 (continued)
15
2
x ª G º d ­°
ª G t º ½°
® xG« » ¾
« »
G ¬ G t ¼ dx °̄
¬ G ¼ °¿
D
Vf
(m)
Use (e) to eliminate 1/ G on the left side of (m) and G on the right side
5D
4Q
2
ª G º d ­° ª G t º ½°
®x « » ¾
« »
¬ G t ¼ dx °̄ ¬ G ¼ °¿
(n)
To solve (n) for G t / G , let
ªG º
x« t »
¬G ¼
z2
(o)
substitute into (n)
5D
8Q
z2
x
dz
dx
Separate variables and integrate
10 D 3 / 2
x
24 Q
1 3
z C2
3
(p)
where C 2 is a constant determined from the boundary condition on G t (x)
G t (0) 0
(q)
Apply (q) to (o)
(z0) = 0
(r)
Apply (r) to (p) gives C1 = 0. Equation (p) becomes
5 D 3/ 2
x
4Q
z3
(s)
Use (o) to eliminate z in (s)
5D
4Q
Use (e) to eliminate G and solve for
ªG t º
«G »
¬ ¼
3
Gt
Gt
x
x
5 3 ªD º
2 «¬Q »¼
1/ 3
Q
Vf x
(t)
Note that
Q
D
and
Pr
(u)
PROBLEM 5.7 (continued)
Re x
Vf x
(v)
Q
Substitute (u) and (v) into (t)
Gt
5 3
>Pr @1/ 3 >Re x @1/ 2
2
x
(w)
(w) into (i) gives the local Nusselt number
3
>Pr @1/ 3 >Re x @1/ 2
5
Nu x
(x)
To examine surface heat flux we determine the heat transfer coefficient. Substitute (a) into
(w), and solve for h(x)
3k
>Pr @1/ 3 >Re x @1/ 2
5 x
h
Use (v) to eliminate the Reynolds number in the above
3
ªV º
h
k >Pr @1 / 3 « f »
5
¬Q ¼
Substitute (k) and (y) into (b), simplify
1/ 2
3
ªV º
C k >Pr @1 / 3 « f »
5
¬Q ¼
q csc
x
(y)
1/ 2
(z)
This result shows that surface heat flux is uniform.
(5) Checking. Dimensional check: (i) Equations (i), (t), (w) and (x) are dimensionless. (ii)
Equations (e), (g), (h), (y), and (z) are dimensionally correct.
Boundary conditions check: Assumed temperature profile (g) satisfies the four boundary
conditions.
Limiting check: If surface temperature is the same as free stream temperature, the heat flux
will be ezro. According to (k) Ts ( x) Tf when C = 0. Setting C = 0 in (z) gives q csc 0 .
(6) Comments. The solution to the case of a specified uniform surface flux is given in
Section 5.7.3. The corresponding surface temperature is given in (5.35)
Ts ( x )
Tf 2.396
q csc
x
1/3
k Pr Re1/2
x
(5.35)
Note that the above can be rewritten as
Ts ( x )
Tf C c x
where C c is constant. This is identical to equation (k) which gives the specified surface
temperature in this problem.
PROBLEM 5.8
Surface temperature of a plate increases exponentially with distance from the leading edge
according to
Ts ( x,0)
Tf C exp( E x)
where C and E are constants. The plate is cooled
with a low Prandtl number fluid ( Pr 1 ). Since
for such fluids G G t , it is reasonable to assume
uniform axial velocity within the thermal boundary
layer. That is
u | Vf
Assume laminar boundary layer flow and use a third degree polynomial temperature
profile.
a[ ] Show that the local Nusselt number is given by
Nu x
0.75 E x >1 exp( E x)@
1/ 2
Pr 1 / 2 Re1x/ 2
b[ ] Determine surface flux distribution.
(1) Observations. (i) The determination of the Nusselt number requires the determination of
the velocity and temperature distributions. (ii). V
elocity is assumed uniform. (iii) Surface
temperature is variable. (iv) Newton’s law of cooling gives surface heat flux. This requires
knowing the local heat transfer coefficient.
(2) Problem Definition. Determine the temperature distribution for boundary layer flow
over a flat plate at variable surface temperature. This reduces to determining the thermal
boundary layer thickness.
(3) Solution Plan. Start with the definition of local Nusselt number. Equate Newton’s law
with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the
integral form of the energy equation using a third degree polynomial temperature profile to
determine the temperature distribution. Apply Newton’s law of cooling to determine surface
heat flux.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties, (3) two-dimensional, (4)
laminar flow (Rex <5 u105), (5) viscous boundary layer flow (Rex >100), (6) thermal
boundary layer (Pe >100), (7) uniform upstream velocity and temperature , (8) flat plate, (9)
(10) negligible changes in kinetic and potential energy, (11) negligible axial conduction,
(12) negligible dissipation, (13) no buoyancy (E = 0 or g = 0) and (14) Pr 1 .
(ii) Analysis. The local Nusselt number is defined as
Nu x
hx
k
(a)
PROBLEM 5.8 (continued)
The heat transfer coefficient h given by equation (1.10)
wT ( x,0)
wy
Ts ( x) Tf
k
h
(1.10)
Thus the temperature distribution T ( x, y ) must be determined. Surface heat flux is obtained
using Newton’s law of cooling
q csc
h(Ts Tf )
(b)
The integral form of the energy equation is used to determine temperature distribution
wT x,0
D
wy
d
dx
G t ( x)
³
u (T Tf )dy
(5.7)
0
Axial velocity distribution u(x,y) for Pr 1 is assume to be the same as free stream
velocity. Thus
u
Vf
(c)
We assume a third degree polynomial
T x, y
b0 x b1 x y b2 x y 2 b3 x y 3
(d)
The temperature boundary conditions are:
(1) T x,0
Ts ( x)
(2) T x, G t # Tf
(3)
wT x, G t
#0
wy
(4)
w 2T x,0
wy 2
0
The four boundary conditions are used to determine the coefficient in (d). The assumed
profile becomes
ª3 y 1 y3 º
(e)
T ( x, y ) Ts ( x) >Tf Ts ( x)@ «
»
3
2
2
G
G
»
t ¼
¬« t
Substitute (e) into (1.10)
3 k
(f)
h( x )
2 Gt
Introducing (f) into (a)
3 x
(g)
Nu x
2 Gt
PROBLEM 5.8 (continued)
Thus the problem reduces to determining the thermal boundary layer thickness G t . This is
accomplished using the integral form of the energy equation (5.7). Substituting (c) and (e)
into (5.7)
3 D >Ts ( x) Tf @
Gt
2 Vf
G t ( x)
d
dx
ª
y
¬«
2 Gt
>Ts ( x) Tf @ «1 3
³
0
1 y3 º
» dy
2 G t3 ¼»
Evaluate the integral
3 D >Ts ( x) Tf @ 3 d
^ >Ts ( x) Tf @ G t `
2 Vf
Gt
8 dx
(h)
H
owever, surface temperature is given by
Ts Tf
C exp( E x)
(i)
Substitute into (h)
4
D exp( E x) d
> G t exp( E x)@
Vf
Gt
dx
(j)
Differentiate the right side of (j), the above becomes
4
dG
D exp( E x)
G t E exp( E x) exp( E x) t
Vf
dx
Gt
The above is simplified to
4
D
Vf
E G t2 G t
dG t
dx
Rewrite as
4
D
Vf
E G t2
Gt
dG t
dx
Separate variables
dx
G t dG t
D
4
E G t2
Vf
Integrate
x
º
ª D
1
ln «4
E G t2 » C o
2 E ¬ Vf
¼
(k)
where C o is a constant determined from the boundary condition on G t (x)
G t (0) 0
(l)
Apply (l) to (k) gives
Co
(m) into (k)
D
1
ln 4
2E
Vf
(m)
PROBLEM 5.8 (continued)
4
2E x
ln
D
E G t2
Vf
4
D
(n)
Vf
Rewrite
4
exp(2 E x)
D
Vf
4
E G t2
D
Vf
Solve the above for G t
Gt
4
D
>1 exp(2 E x)@
E Vf
(o)
Substitute (o) into (g) gives the local Nusselt number
3
2
Nu x
E Vf x 2
1
4D >1 exp(2 E x)@
(p)
To express (p) in terms of the Prandtl and local Reynolds numbers, rewrite the above and
note that D k / U c p
Nu x
( E x)
3 c p P UVf x
4
k
P >1 exp(2 E x)@
Nu x
0.75
This is written as
( E x)
PrRe x
>1 exp(2 E x)@
(r)
The heat transfer coefficient is determined to examine surface heat flux. Substitute (o) into
(f), gives h(x)
E Vf
3
1
(s)
h( x )
k
4
D >1 exp(2 E x)@
Substitute (i) and (s) into (b), simplify
q csc
E Vf exp(2 E x)
3
kC
4
D >1 exp(2 E x)@
(t)
This result shows that surface heat flux is varies with distance along the surface.
Expressing the above in dimensionless form, gives
q cc
E Vf
kC
D
3
4
exp(2 E x)
>1 exp(2 E x)@
(u)
PROBLEM 5.8 (continued)
(5) Checking. Dimensional check: (i) Equations (g), (p), (r) and (u) are dimensionless. (ii)
Equations (e), (f), (k), (o), (s) and (t) are dimensionally correct.
Boundary conditions check: Assumed temperature profile (e) satisfies the four boundary
conditions.
Limiting check: (i) If surface temperature is the same as free stream temperature, the heat
flux will be ezro. According to equation (i), Ts ( x) Tf when C = 0. Setting C = 0 in (u)
gives q csc 0 .
(ii) According to equation (i), surface temperature is the same as free stream temperature for
E 0. this case should give ezro heat flux. Setting E 0 in (t) and applying ’LoHspital’s
rule, give q csc 0 .
(6) Comments. Solution (t) shows that surface heat flux increases with x. Recall that for
uniform surface temperature Pohlhausen’s solution shows that the heat transfer coefficient
and surface flux decrease with x. oHwever, in this problem surface temperature and the heat
transfer coefficient increase with x. This results in a flux that increases with x.
PROBLEM 5.9
A square array of chips of side L is mounted flush on a flat
plate. The chips dissipate non-uniform surface flux according to
q cxc
C
x
The plate is cooled by forced convection with uniform upstream
velocity Vf and temperature Tf . Assume laminar boundary
layer flow with G t / G 1. Use third degree polynomials for the
axial velocity and temperature.
a[ ] Show that the local Nusselt number is given by
0.331 Pr 1 / 3 Re1 / 2
Nu x
b[ ] Show that surface temperature is uniform.
(1) Observations. (i) In general, to determine the Nusselt number it is necessary to determine the
velocity and temperature distribution. (ii) Surface heat flux is variable. It decreases with distance
x. (iii) Surface temperature is unknown. (iv) Newton’s law of cooling gives surface temperature.
This requires knowing the local heat transfer coefficient. (v) G t / G 1.
(2) Problem Definition. Determine the velocity and temperature distribution for boundary layer
flow over a flat plate at variable surface temperature. This reduces to determining the viscous
and thermal boundary layer thickness.
(3) Solution Plan. Start with the definition of local Nusselt number. Equate Newton’s law with
Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of
the momentum and energy equations using a third degree polynomial profiles to determine the
velocity and temperature distribution. Apply Newton’s law of cooling to determine surface
temperature.
(4) Plan Execution. (i) Assumptions. (1) Steady state, (2) constant properties, (3) laminar flow
(Rex <5 u105), (4) two-dimensional, (5) viscous boundary layer flow (Rex >100), (6) thermal
boundary layer (Pe >100), (7) uniform upstrea m velocity and temperature , (8) flat plate, (9)
negligible changes in kinetic and potential energy, (10) negligible axial conduction, (11)
negligible dissipation and (12) no buoyancy (E = 0 or g = 0).
(ii) Analysis. [a]The local Nusselt number is defined as
Nu x
hx
k
(a)
where the heat transfer coefficient h is given by equation (1.10)
wT ( x,0)
wy
Ts ( x) Tf
k
h
(1.10)
PROBLEM 5.9 (continued)
Thus h depends on the temperature distribution T ( x, y ). The integral form of the energy equation
is used to determine the temperature distribution
D
wT x,0
wy
d
dx
G t ( x)
³
(5.7)
u (T Tf )dy
0
The velocity solution, u ( x, y ) , for an assumed third degree polynomial is solved in Section 5.7.1
and given by equation (5.9)
u
Vf
3§ y· 1§ y·
¨ ¸ ¨ ¸
2 ©G ¹ 2 ©G ¹
3
(5.9)
where the integral solution to G (x) is
G
4.64
x
Re x
(5.10)
The local Reynolds number is defined as
Vf x
Re x
(b)
Q
For the temperature profile we assume a third degree polynomial
T x, y
b0 x b1 x y b2 x y 2 b3 x y 3
(c)
The boundary conditions on the temperature are
(1) k
wT x,0
wy
C
x
(2) T x, G t # Tf
(3)
wT x, G t
#0
wy
(4)
w 2T x,0
wy 2
0
Equation (c) and the four boundary conditions give the coefficients bn (x)
b0
Tf 2 C
Gt ,
3k x
b1
1 C
,
k x
b2
0,
b3
1 1 C
3k G t2 x
Substituting the above into (c)
T ( x, y )
Tf Surface temperature is obtained by setting y
C ª2
1 3º
« Gt y 2 y »
k x ¬« 3
3G t
»¼
0 in (d)
(d)
PROBLEM 5.9 (continued)
Ts ( x)
T ( x,0)
Tf 2C
3k x
Gt
(e)
Substituting (d) and (e) into (1.10)
3k
2G t
h
(f)
Combining (a) and (f)
3x
(g)
2G t
Thus to determine surface temperature and Nusselt number requires the determination of G t .
Application of the energy equation gives G t . Substitute (5.9) and (d) into (5.7)
Nu x
D
1
Vf
x
G t ( x)
3
º
1 ª3 § y · 1 § y · º ª2
1
d
« ¨ ¸ ¨ ¸ » « G t y 2 y 3 » dy
dx
x «¬ 2 © G ¹ 2 © G ¹ »¼ ¬« 3
3G t
»¼
0
³
Expanding the integrand and evaluating the integral in the above
D
1
Vf
x
d ­° G 2 ª 1
Gt /G
®
dx °̄ x «¬10
3
1
Gt /G
140
°
5 º½
» ¾°
¼¿
(h)
Since G t / G 1 , it follows that
1
Gt /G
140
5
1
Gt /G
10
3
Dropping the last term in (h), gives
10
D
1
Vf
x
d ª 1 G t3 º
«
»
dx ¬« x G »¼
(i)
Separate variables and integrate
10
D
Vf
x
³
1
x
dx
³
Gt
d ª 1 G t3 º
«
»
dx ¬« x G »¼
Evaluate the integrals
D
1 G t3
Co
x G
where C o is constant of integration. The boundary condition on G t is
20
Vf
x
G t (0) 0
Apply (k) to (j) gives C o
(j)
(k)
0 . Substitute into (j) and solve for G t
Gt
Use (5.10) to eliminate G in (l)
º
ª D
G x»
«20
¼
¬ Vf
1/ 3
(l)
PROBLEM 5.9 (continued)
Gt
1/ 3
ª
D
«(20)(4.64)
Vf
«¬
x2 º
»
Re x »¼
ª
D
«(20)(4.64)
Vf x
«¬
º
»
Re x »¼
(m)
Express (m) in dimensionless form
Gt
x
Note that D
1/ 3
1
k / U c p , the above is rewritten as
ª k
P
4.5274«
«¬ c p P UVf x
Gt
x
º
»
Re x »¼
1/ 3
1
Introduce the definition of the Prandtl number, the above gives
Gt
4.5274
x
Pr 1 / 3 Re x1/ 2
(n)
Substitute (n) into (g) gives the local Nusselt number
Nu x
0.331 Pr 1/ 3 Re x1 / 2
(o)
b[ ]Surface temperature is obtained by apply Newton’ s law of cooling or substituting (e) into (e).
Newton’s law gives
q cc
(p)
Ts Tf s
h
Substitute (a) into (o) gives h
k
h 0.331 Pr 1 / 3 Re x1/ 2
(q)
x
Surface flux is given by
C
(r)
q csc
x
(b), (q) and (r) into (p)
xC
Ts Tf V x
0.331 k Pr 1 / 3 f
x
Q
Note that in the above variable x cancels out to give
Ts
Tf C
0.331 k Pr
1/ 3
Vf
Q
(s)
PROBLEM 5.9 (continued)
This result can be expressed in dimensionless form as
Ts Tf
1
C Q
k Vf
0.331 Pr 1 / 3
(t)
This result shows that surface temperature is uniform.
(iii) Checking. Dimensional check: (i) Equations (5.9), (5.10), (b) and (n) are dimensionless.
(ii) Equations (d), (f), (l), (m) and (s) are dimensionally correct.
Boundary conditions check: Assumed temperature profile (d) satisfies the four boundary
conditions.
Limiting check: If surface heat flux vanishes, surface temperature will be the same as free stream
temperature. eZro flux corresponds to C = 0. Setting C = 0 in (s) gives Ts ( x) Tf .
(5) Comments. (i)The solution is valid for G t / G 1. This corresponds to Pr ! 1.
(ii) Pohlhausen’s solution for a plate at uniform surface temperature gives h(x) as
h( x )
k
Vf dT (0)
Q x dK
(4.66)
The corresponding surface heat flux is
q csc
h( x)(Ts Tf )
k (Ts Tf )
Vf dT (0)
Q x dK
This result can be expressed as
q csc
C
(u)
x
In this problem a surface flux of the form (u) results in uniform surface temperature. This is in
agreement with Pohlhausen’s solution. Thus this problem is identical to Pohlhausen’s problem of
flow over a plate at uniform surface temperature.
(iii) Pohlhausen’s solution for the local Nusselt number is
Nu x
0.331 Pr 1/ 3 Re x1/ 2
This is in good agreement with the integral solution(o).
Pr ! 10
(4.72c)
PROBLEM 5.10
A square array of chips of side L is mounted flush on a flat plate. The forward edge of the array
is at a distance xo from the leading edge of the plate. The heat dissipated in each row increases
with successive rows as the distance from the forward edge increases. The distribution of surface
heat flux for this arrangement may be approximated by
q csc
Cx 2
where C is constant. The plate is cooled by forced
convection with uniform upstream velocity Vf and
temperature Tf . Assume laminar boundary layer
flow. Assume further that the axial velocity within
the thermal boundary layer is equal to the free
stream velocity, u | Vf . Use a third degree
polynomial temperature profile.
a[ ] Show that the local Nusselt number is given by
Nu x
>
1 .3 1 ( x o / x ) 3
1 / 2
@
Pr 1 / 2 Re1 / 2
b[ ] Determine the maximum surface temperature
c[ ] How should the rows be rearranged to reduce the maximum surface temperature?
(1) Observations. (i) In general, to determine the Nusselt number it is necessary to determine the
velocity and temperature distribution. (ii) Fluid velocity for Pr 1 is assumed to be uniform,
u | Vf . This represents a significant simplification. (iii) Surface heat flux is variable. It
increases with distance x. (iv) Surface temperature is unknown. Since flux increases with x and
heat transfer coefficient decreases with x, surface temperature is expected to increase with x.
Thus maximum surface temperature is at the trailing end x = L. (v) Newton’s law of cooling
gives surface temperature. This requires knowing the local heat transfer coefficient.
(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a
non-uniformly heated flat plate with insulated leading section.
(3) Solution Plan. Start with equating Newton’s law with Fourier’s law to obtain an equation for
the heat transfer coefficient h. Apply the integral form of the energy equation using a third
degree polynomial temperature profile.
(4) Plan Execution. (i) Assumptions. (1) Steady state, (2) constant properties, (3) laminar flow
(Rex <5 u105), (4) two-dimensional, (5) thermal boundary layer (Pe >100), (6) uniform upstream
velocity and temperature , (7) flat plate, (8) negligible changes in kinetic and potential energy,
(9) negligible axial conduction, (10) negligible dissipation, (11) no buoyancy (E = 0 or g = 0) and
(12) uniform velocity within the thermal boundary layer ( Pr 1 ).
(ii) Analysis. The local Nusselt number is defined as
Nu x
hx
k
(a)
PROBLEM 5.10 (continued)
where the heat transfer coefficient h is given by equation (1.10)
wT ( x,0)
wy
Ts Tf
k
h
(1.10)
Thus h depends on the temperature distribution T ( x, y ). The integral form of the energy equation
is used to determine the temperature distribution
wT x,0
D
wy
G t ( x)
d
dx
³
(5.7)
u (T Tf )dy
0
oHwever, For Pr 1 the velocity boundary layer thickness is much smaller than the thermal
boundary layer thickness. Thus we assume
u
(b)
Vf
For the temperature profile we assume a third degree polynomial
b0 x b1 x y b2 x y 2 b3 x y 3
T x, y
(c)
The boundary conditions on the temperature are
wT x,0
Cx 2
wy
(2) T x, G t # Tf
wT x, G t
#0
(3)
wy
(1) k
(4)
w 2T x,0
wy 2
0
Equation (c) and the four boundary conditions give the coefficients bn (x)
2C 2
x Gt ,
3k
Substituting the above into (c)
b0
Tf T ( x, y )
b1
Tf Surface temperature is obtained by setting y
Ts ( x)
C 2
x ,
k
b2
0,
b3
1 y3 º
C 2 ª2
x « Gt y »
3 G t2 ¼
k ¬3
C x2
3k G t2
(d)
0 in (d)
T ( x,0)
Tf 2C 2
x Gt
3k
(e)
Substituting (d) and (e) into (1.10)
h
3k
2G t
(f)
PROBLEM 5.10 (continued)
Combining (a) and (f)
3x
(g)
2G t
Thus to determine surface temperature and Nusselt number requires the determination of G t .
Application of the energy equation gives G t . Substituting (b) and (d) into (5.7)
Nu x
D x 2 Vf
Gt ( x)
d
dx
ª2
1 y3 º
x2 « Gt y » dy
3 G t2 ¼
¬3
³
0
Evaluating the integral in the above
Dx 2 Vf
d ­ 2 ª 2 2 1 2 1 2 º ½ Vf d 2 2
G t G t G t »¾
x Gt
®x
2
12 ¼ ¿ 4 dx
dx ¯ «¬ 3
>
@
Separating variables and rearranging
>
d x 2G t2
4D 2
x dx
Vf
@
Integrating
4D
Vf
³ d >x G @
2
2
t
³ x dx
2
Performing the integration
4D 3
x Co
3Vf
where C o is constant of integration. The boundary condition on G t is
x 2G t2
G t ( xo ) 0
(h)
(i)
Applying (i) to (h) gives
Co
4D 3
xo
3Vf
Substituting into (h) and solving for G t
4D
1 ( xo / x) 3 x
3Vf
(j)
3 3 Vf x
1
D 1 ( xo / x) 3
4
(k)
>
Gt
@
Substituting (j) into (g)
Nu x
Noting that
cpP
Pr
k
P/U
k / Ucp
Q
D
(l)
a[ ]Using (l), equation (k) is expressed in terms of Prandtl and Reynolds numbers
Nu x
>
1 .3 1 ( x o / x ) 3
1 / 2
@
Pr 1 / 2 Re1 / 2
(m)
PROBLEM 5.10 (continued)
b[ ]Surface temperature is obtained by substituting (j) into (e)
Ts
Tf 4C
D
3 3k Vf
x 5 / 2 1 ( xo / x) 3
(n)
This result shows that surface temperature increases with x. Thus maximum temperature is at the
trailing end x L :
D 5/ 2
4C
(o)
(Ts ) max Tf L
1 ( x o / L) 3
3 3k Vf
c[ ] Since heat transfer coeffici ent decreases with distance from the leading end, rows of high
power density chips should be placed near the leading edge and low density rows towards the
trailing end.
(iii) Checking. Dimensional check: Dimensional check: (1) Equations (g), (k) and (l) are
dimensionless. (2) Equations (d), (f), (j) and (n) are dimensionally correct.
Boundary conditions check: Assumed temperature profile (d) satisfies the four boundary
conditions on temperature.
Limiting check: If surface heat flux vanishes, surface temperature will be the same as free stream
temperature. eZro flux corresponds to C = 0. Setting C = 0 in (n) gives Ts ( x) Tf .
(5) Comments. (i) For the special case of no leading insulated section, equations (m) and (n)
reduce to
Nu x
Ts
Tf 1.3 PrRe x
(p)
D 5/ 2
4C
x
3 3k Vf
(q)
(ii) Application of (m) at x xo gives infinite Nusselt number. This anomaly is due to the fact
that boundary layer approximations (neglecting axial conduction) is not valid near the leading
edge of the thermal boundary layer.
PROBLEM 5.11
Repeat Problem 5.10 using a linear surface flux distribution q csc
Cx.
a[ ] Show that the local Nusselt number is given by
Nu x
>
1.06 1 ( xo / x ) 3
1 / 2
@
Pr 1 / 2 Re1/ 2
b[ ] Determine the maximum surface temperature
c[ ] How should the rows be rearranged to reduce the
maximum surface temperature?
(1) Observations. (i) In general, to determine the Nusselt number it is necessary to determine the
velocity and temperature distribution. (ii) Fluid velocity for Pr 1 is assumed to be uniform,
u | Vf . This represents a significant simplification. (iii) Surface heat flux is variable. It
increases with distance x. (iv) Surface temperature is unknown. Since flux increases with x and
heat transfer coefficient decreases with x, surface temperature is expected to increase with x.
Thus maximum surface temperature is at the trailing end x = L. (v) Newton’s law of cooling
gives surface temperature. This requires knowing the local heat transfer coefficient.
(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a
non-uniformly heated flat plate with insulated leading section.
(3) Solution Plan. Start with equating Newton’s law with Fourier’s law to obtain an equation for
the heat transfer coefficient h. Apply the integral form of the energy equation using a third
degree polynomial temperature profile.
(4) Plan Execution. (i) Assumptions. (1) Steady state, (2) constant properties, (3) laminar flow
(Rex <5 u105), (4) two-dimensional, (5) thermal boundary layer (Pe >100), (6) uniform upstream
velocity and temperature , (7) flat plate, (8) negligible changes in kinetic and potential energy,
(9) negligible axial conduction, (10) negligible dissipation, (11) no buoyancy (E = 0 or g = 0) and
(12) uniform velocity within the thermal boundary layer ( Pr 1 ).
(ii) Analysis. [a] The local Nusselt number is defined as
hx
k
Nu x
(a)
where the heat transfer coefficient h is given by equation (1.10)
wT ( x,0)
wy
Ts Tf
k
h
(1.10)
Thus h depends on the temperature distribution T ( x, y ). The integral form of the energy equation
is used to determine the temperature distribution
wT x,0
D
wy
d
dx
G t ( x)
³
0
u (T Tf )dy
(5.7)
PROBLEM 5.11 (continued)
oHwever, For Pr 1 the velocity boundary layer thickness is much smaller than the thermal
boundary layer thickness. Thus we assume
u
(b)
Vf
For the temperature profile we assume a third degree polynomial
b0 x b1 x y b2 x y 2 b3 x y 3
T x, y
(c)
The boundary conditions on the temperature are
wT x,0
Cx
wy
(2) T x, G t # Tf
wT x, G t
#0
(3)
wy
(1) k
(4)
w 2T x,0
wy 2
0
Equation (c) and the four boundary conditions give the coefficients bn (x)
b0
Tf 2C
xG t ,
3k
C
x,
k
b1
Tf 1 1 3º
C x ª2
y »
« Gt y 3 G t2 ¼
k ¬3
b2
0,
b3
C x
3k G t2
Substituting the above into (c)
T ( x, y )
Surface temperature is obtained by setting y
Ts ( x )
(d)
0 in (d)
T ( x,0)
Tf 2C
xG t
3k
(e)
Substituting (d) and (e) into (1.10)
3k
2G t
h
(f)
Combining (a) and (f)
3x
(g)
2G t
Thus to determine surface temperature and Nusselt number requires the determination of G t .
Application of the energy equation gives G t . Substituting (b) and (d) into (5.7)
Nu x
Gt ( x)
D x Vf
d
dx
³
0
Evaluating the integral in the above
ª2
1 y3 º
x « Gt y » dy
3 G t2 ¼
¬3
PROBLEM 5.11 (continued)
D x Vf
> @
d ª2 2 1 2 1
º
xG t xG t xG t2 »
«
2
12
dx ¬ 3
¼
1
d
xG t2
Vf
4 dx
Separating variables and rearranging
> @
4D
xdx
Vf
> @
4D
Vf
d xG t2
Integrating
³
d xG t2
³ xdx
Evaluating the integral
2D 2
x Co
Vf
where C o is constant of integration. The boundary condition on G t is
xG t2
G t ( xo ) 0
(h)
(i)
Applying (i) to (h) gives
Co
2D 2
xo
Vf
Substituting into (h) and solving for G t
2D
1 ( xo / x ) 2 x
Vf
>
Gt
@
(j)
Substituting (j) into (g)
Nu x
Noting that D
Vf x
3
2 2
1
D 1 ( xo / x ) 2
(k)
k / U c p , the above is expressed as
Nu x
1.06
Pr 1/ 2 Re1/ 2
1 ( xo / x )
(l)
2
[b]Surface temperature is obtaine d by substituting (j) into (e)
Ts
Tf 2 2C
3k
D
Vf
>1 ( x
o
@
/ x)2 x 3
(m)
This result shows that surface temperature increases with x. Thus maximum temperature is at the
trailing end x L :
2 2C D
(n)
Ts ,max Tf 1 ( xo / L) 2 L3
3k
Vf
>
@
[c]According to (l), the heat transfer coefficien t decreases with distance from the leading end.
Thus rows of high power density chips should be placed near the leading edge and low density
rows towards the trailing end.
PROBLEM 5.11 (continued)
(iii) Checking. Dimensional check: Dimensional check: (i) Equations (g), (k) and (l) are
dimensionless. (ii) Equations (d), (f), (j) and (m) are dimensionally correct.
Boundary conditions check: Assumed temperature profile (d) satisfies the four boundary
conditions.
Limiting check: If surface heat flux vanishes, surface temperature will be the same as free stream
temperature. Z
ero flux corresponds to C = 0. Setting C = 0 in (m) gives Ts ( x) Tf .
(5) Comments. (i) For the special case of no leading insulated section, equations (l) and (m)
reduce to
Nu x
Ts
1.06 PrRe x
Tf 2 2C
3k
D
Vf
(o)
x3
(p)
(ii) Application of (l) at x xo gives infinite Nusselt number. This anomaly is due to the fact
that boundary layer approximations (neglecting axial conduction) is not valid near the leading
edge of the thermal boundary layer.
PROBLEM 5.12
A fluid at temperature To and flow rate mo is injected radially between parallel plates. The
spacing between the plates is H. The upper plate is insulated and the lower plate is maintained at
uniform temperature Ts along r t Ro and is insulated along 0 d r d Ro . Consider laminar
boundary layer flow and assume that the radial velocity u does not vary in the direction normal
to the plates (slug flow).
a[ ] Show that for a cylindrical element G t u 2S rdr the external mass flow dme to the thermal
boundary layer is
dme
d
2S U
dr
³
Gt
mo
dG t
H
u r dr
0
b[ ] Show that the integral form of conservation of energy is
wT (r ,0)
kr
wr
mo c p d
2S H dr
³
Gt
(T To )dy
0
c[ ] Assume a linear temperature profile, show that the local Nusselt number is
Nu r
>1 R
2
1
o /r
2
@
1 / 2
Pr 1 / 2 Re1r / 2
where
Rer
U ur
P
mo
2S P H
(1) Observations. (i) This problem is described by cylindrical coordinates. (ii) V
elocity variation
with y is negligible. (iii) Conservation of mass requires that radial velocity decrease with radial
distance r. (iv) Surface temperature is uniform.
(2) Problem Definition. Determine the integral formulation of conservation of mass and energy.
Determine the temperature distribution within the thermal boundary layer. This requires the
determination of G t (x).
(3) Solution Plan. (i) Apply conservation of mass and energy to a cylindrical element
G t u 2S rdr. (ii) Use the integral form of the energy equation to determine G t (x).
PROBLEM 5.12 (continued)
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional temperature field, (3) slug flow
(velocity varies radially only), (4) constant properties, (5) laminar flow, (6) thermal boundary
layer flow, (7) flat surface, (8) uniform surface temperature, (9) negligible changes in kinetic and
potential energy, (10) negligible axial conduction, (11) negligible dissipation and (12) no
buoyancy (E = 0).
(ii) Analysis. a[ ] External mass flow rate. Consider a
volumetric cylindrical element G t u 2S rdr . The diagram
shows the cross section of the element and the external and
radial mass flow rates. Conservation of mass gives
dme mr
mr dmr
dr
dr
dme
mr
mr dmr
dr
dr
Gt
or
dme
dmr
(a)
dr
where
dme external mass flow element into the boundary layer
mr radial mass flow rate entering element
Integral formulation of mr is
Gt
³ 2S U u rdy
mr
0
where u is radial velocity and U is density. For constant density and slug flow the above
simplifies to
Gt
2S Ur u
mr
³ dy
(b)
0
Evaluating the integral
mr
2S U r uG t
(c)
Application of conservation of mass between the inlet and channel section location r, gives
2 U S rHu
mo
Solving for u
u
mo
2 US rH
(d)
mo
Gt
H
(e)
mo
dG t
H
(f)
(d) into (c)
mr
Differentiating (e)
dmr
PROBLEM 5.12 (continued)
(f) into (a)
mo
dG t
H
dme
(g)
b[ ] Conservation of energy fo r the volumetric cylindrical
element G t u 2S rdr shown gives
dq s E r dEe
Er dE r
dr
dr
dEe
Er
Er dEr
dr
dr
Gt
This simplifies to
dq s dE e
dE r
dr
dr
dr
(h)
dqs
where
dE e = energy supplied by the external mass
E r = energy convected radially through the boundary layer
dq s = energy conducted to the element through the surface
We now formulate the three energy components:
dEe
c p To dme
(d) into the above
dEe
Er
c p To
mo
dG t
H
(i)
³ 2S U c ruTdy
p
(f) into the above
Er
U c p mo
H
Gt
³ Tdy
0
Differentiating the above
dE r
dr
U c p mo d
H
dr
Gt
³ Tdy
(j)
0
Fourier’s law gives
dq s k 2S r
wT (r ,0)
dr
wy
(k)
Substituting (i)-(k) into (h)
m
wT (r ,0)
dr c p To o dG t
k 2S r
H
wy
Dividing through by 2S dr and rearrange the above
U c p mo d ª
H
«
dr ¬
³
Gt
0
º
Tdy » dr
¼
PROBLEM 5.12 (continued)
wT (r ,0)
kr
wy
U c p mo ­° d ª
® «
°̄ dr ¬
H
³
Gt
0
º
dG ½°
Tdy » To t ¾
dr °¿
¼
(l)
H
owever, the last term in (l) can be written as
dG
To t
dr
d
dr
Gt
³ T dy
(m)
o
0
(m) into (l)
wT (r ,0)
kr
wr
mo c p d
2S H dr
³
Gt
(T To )dy
(n)
0
[c}Nusselt number. The local Nusselt number is defined as
Nu r
hr
k
(o)
where the heat transfer coefficient h is given by equation (1.10)
wT (r ,0)
wy
Ts To
k
h
(1.10)
Thus h depends on the temperature distribution T (r , y ). The integral form of the energy equation
is used to determine the temperature distribution. We assume a linear temperature profile
T (r , y )
b0 (r ) b1 (r ) y
(p)
The boundary conditions on the temperature are
(1) T (r ,0) Ts
(2) T (r , G t ) # To
Equation (p) and the two boundary conditions give
T (r , y )
Ts (To Ts )
y
Gt
(q)
Substituting (q) into (1.10)
h
k
Gt
When this is substituted into(o), we obtain
Nu r
r
Gt
(s)
Thus the problem becomes one of determining the thermal boundary layer thickness G t . The
integral form of conservation of energy is used to determine G t . Substituting (q) into (n)
PROBLEM 5.12 (continued)
kr
To Ts
Gt
mo c p
d
(To Ts )
2S H
dr
Gt
³
0
ª
yº
«1 » dy
¬ Gt ¼
Evaluating the integral and simplifying
mo c p dG t
4S H dr
kr
Gt
Separating variables
4S kH
rdr
mo c p
G t dG t
Integrating
G t2
2S kH 2
r Co
mo c p
2
(t)
where C o is constant of integration. The boundary condition on G t is
G t ( Ro ) 0
(u)
(t) and (u) give
Co
2S kH 2
Ro
mo c p
Substituting into (t) and solving for G t
>
4S kH 2
r Ro2
mo c p
Gt
@
(v)
(v) into (s)
Nu r
r
mo c p
>r
4S kH
2
Ro2
@
1 / 2
Substituting (d) into the above and rearranging
Nu r
r
ur U c p
>r
2k
2
Ro2
@
1 / 2
This result can be expressed in terms of the Reynolds and Prandtl numbers as
Nu r
>1 R
2
1
o /r
Rer
2
@
U ur
P
1 / 2
Pr 1 / 2 Re1r / 2
(w)
PROBLEM 5.12 (continued)
(iii) Checking. Dimensional check: (1) Units of (d), (e), (i), (k), (n) and (v) are correct. (2)
Equations (s) and (w) are dimensionless.
Boundary conditions check: Assumed temperature profile (q) satisfies the two boundary
conditions on temperature.
(5) Comments. (i) The assumption of bulk flow provided a significant simplification. (ii)
Application of (w) at r Ro gives infinite Nusselt number. This anomaly is due to the fact that
boundary layer approximations (neglecting axial conduction) is not valid near the leading edge
of the thermal boundary layer.
PROBLEM 5.13
The lower plate in Problem 5.12 is heated with uniform flux q csc along r t Ro and insulated
along 0 d r d Ro .
a[ ] Show that for a cylindrical element G t u 2S r dr the external mass flow dme to the thermal
boundary layer is
dme
d
2S U
dr
³
Gt
mo
dG t
H
u r dr
0
b[ ] Show that the integral form of conservation of energy is
mo c p d G t
c
c
(T To )dy
qs
2S H dr 0
³
c[ ] Assume a linear temperature profile show that the local Nusselt number is
Nu r
>1 R
o /r
Rer
U ur
P
2
@
1 / 2
Pr 1 / 2 Re1r / 2
where
mo
2S P H
(1) Observations. (i) This problem is described by cylindrical coordinates. (ii) eVlocity variation
with y is negligible. (iii) Conservation of mass requires that radial velocity decrease with radial
distance r. (iv) Surface heat flux is uniform. (v) Surface temperature is unknown.
(2) Problem Definition. Determine the integral formulation of conservation of mass and energy.
Determine the temperature distribution within the thermal boundary layer. This requires the
determination of G t (x).
(3) Solution Plan. (i) Apply conservation of mass and energy to a cylindrical element
G t u 2S rdr. (ii) Use the integral form of the energy equation to determine G t (x).
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional temperature field, (3) slug flow
(velocity varies radially only), (4) constant properties, (5) laminar flow, (6) thermal boundary
layer, (7) flat surface, (8) uniform surface heat flux, (9) negligible changes in kinetic and
potential energy, (10) negligible axial conduction, (11) negligible dissipation and (12) no
buoyancy (E = 0).
PROBLEM 5.13 (continued)
(ii) Analysis. a[ ] External mass flow rate. Consider a
volumetric cylindrical element G t u 2S rdr . The diagram
shows the cross section of the element and the external and
radial mass flow rates. Conservation of mass gives
dme mr
mr dmr
dr
dr
dme
mr
mr dmr
dr
dr
Gt
or
dme
dmr
(a)
dr
where
dme external mass flow element into the boundary layer
mr radial mass flow rate entering element
Integral formulation of mr is
Gt
³ 2S U u rdy
mr
0
where u is radial velocity and U is density. For constant density and slug flow the above
simplifies to
Gt
mr
2S Ur u
³ dy
(b)
0
Evaluating the integral
2S U r uG t
mr
(c)
Application of conservation of mass between the inlet and channel section location r, gives
2 U S rHu
mo
Solving for u
u
mo
2 US rH
(d)
mo
Gt
H
(e)
mo
dG t
H
(f)
(d) into (c)
mr
differentiating (e)
dmr
(f) into (a)
dme
mo
dG t
H
b[ ]Conservation of energy for th e volumetric cylindrical element G t u 2S rdr shown gives
(g)
PROBLEM 5.13 (continued)
dq s E r dEe
Er dEe
dE r
dr
dr
This simplifies to
dq s dEe
Er
dE r
dr
dr
(h)
Er dEr
dr
dr
Gt
where
dr
dE e = energy supplied by the external mass
E r = energy convected radially through the boundary
layer
dq s = energy conducted to the element through the surface
dqs
We now formulate the three energy components:
dEe
c p To dme
(d) into the above
dEe
Er
c p To
mo
dG t
H
(i)
³ 2S U c ruTdy
p
(f) into the above
U c p mo
Er
H
Gt
³ Tdy
0
Differentiating the above
dE r
dr
U c p mo d
H
dr
Gt
³ Tdy
(j)
0
Fourier’s law gives
2S qscc rdr
dqs
(k)
where
qscc
surface heat flux
Substituting (i)-(k) into (h)
m
2S rqscc dr c p To o dG t
H
U c p mo d ª
H
«
dr ¬
Gt
º
Tdy » dr
¼
0
³
Dividing through by 2S dr and rearrange the above
qscc r
U c p mo ­° d ª
H
H
owever, the last term in (l) can be written as
® «
°̄ dr ¬
Gt
º
dG ½°
Tdy » To t ¾
dr °¿
¼
0
³
(l)
PROBLEM 5.13 (continued)
dG
To t
dr
Gt
d
dr
³ T dy
o
(m)
0
(m) into (l)
qscc r
mo c p d
2S H dr
Gt
³
(T To )dy
(n)
0
[c}Nusselt number. The local Nusselt number is defined as
hr
k
Nu r
(o)
where the heat transfer coefficient h is given by equation (1.10)
wT ( r,0)
wy
Ts ( x ) To
k
h
(1.10)
Thus h depends on the temperature distribution T (r , y ). The integral form of the energy equation
is used to determine the temperature distribution. We assume a linear temperature profile
b0 (r ) b1 (r ) y
T (r , y )
(p)
The boundary conditions on the temperature are
wT ( r,0)
qscc
wy
(2) T (r , G t ) # To
(1) k
Equation (p) and the two boundary conditions give
To T ( r, y )
Surface temperature is obtained by setting y
Ts ( x )
qscc
(G t y )
k
(q)
0 in (q)
To T ( r,0)
qscc
Gt
k
(r)
Substituting (q) and (r) into (1.10)
h
k
Gt
When this is substituted into(o), we obtain
Nu r
r
Gt
(s)
Thus the problem becomes one of determining the thermal boundary layer thickness G t . The
integral form of conservation of energy is used to determine G t . Substituting (q) into (n)
PROBLEM 5.13 (continued)
mo c p d
2S Hk dr
r
Gt
³
(G t y ) dy
0
Evaluating the integral
mo c p dG t2
4S kH dr
r
Separating variables
4S kH
rdr
mo c p
dG t2
Integrating
2S kH 2
r Co
mo c p
G t2
(t)
where C o is constant of integration. The boundary condition on G t is
G t ( Ro ) 0
(u)
(t) and (u) give
Co
2S kH 2
Ro
mo c p
Substituting into (t) and solving for G t
2S kH 2
r Ro2
mo c p
>
Gt
@
(v)
(v) into (s)
Nur
r
mo c p
1
2S kH ( r Ro2 )
2
Substituting (d) into the above and rearranging
Nur
r
ur U c p
>r
k
2
Ro2
@
1 / 2
This result can be expressed in terms of the Reynolds and Prandtl numbers as
Nur
>1 R
o
/r
Rer
2
@
1 / 2
Pr1 / 2 Re1r / 2
U ur
P
Substituting (v) in (r) gives surface temperature distribution
Ts ( x )
T ( r ,0)
This result can be written in dimensionless form as
To qscc
Gt
k
(w)
PROBLEM 5.13 (continued)
Ts ( x )
T ( r,0)
Ts ( x ) To
qscc Ro
k
To qscc
k
>
k
r 2 Ro2
U ruc p
>
@
@
1
( r / Ro ) 2 1
PrRer
(iii) Checking. Dimensional check: (1) Units of (d), (e), (i), (k), (n) and (v) are correct. (2)
Equations (s) and (w) are dimensionless.
Boundary conditions check: Assumed temperature profile (q) satisfies the two boundary
conditions on temperature.
Limiting check: If surface heat flux vanishes, surface temperature will be the same as inlet fluid
temperature To . Setting qscc 0 in (r) gives Ts ( x ) To .
(5) Comments. (i) The assumption of bulk flow provided a significant simplification. (ii) The
solution does not apply to the limiting case of Ro 0 since the flow is three-dimensional and
thus cannot be approximated by bulk conditions. (ii) Application of (w) at r Ro gives infinite
Nusselt number. This anomaly is due to the fact that boundary layer approximations (neglecting
axial conduction) is not valid near the leading edge of the thermal boundary layer.
PROBLEM 5.14
A porous plate with an impermeable and insulated leading section of length xo is maintained at
uniform temperature Ts along x t xo . The plate is cooled by forced convection with a free
stream velocity Vf and temperature Tf .
Fluid at temperature To is injected through
the porous surface with uniform velocity v o .
The injected and free stream fluids are
identical. Assume laminar boundary layer
flow, introduce axial velocity simplification
based on Pr 1 and use a linear
temperature profile to determine the local
Nusselt number.
(1) Observations. (i) In general, to determine the Nusselt number it is necessary to determine the
velocity and temperature distribution. (ii) Fluid velocity for Pr 1 is assumed to be uniform,
u | Vf . This represents a significant simplification. (iii) The plate is porous. (iv) Fluid is
injected through the plate with uniform velocity. (v) The plate is maintained at uniform surface
temperature. (vi) A leading section of the plate is insulated.
(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a
porous flat plate with surface injection and insulated leading section.
(3) Solution Plan. Start with equating Newton’s law with Fourier’s law to obtain an equation for
the heat transfer coefficient h. Apply the integral form of the energy equation using a third
degree polynomial temperature profile.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties, (3) two-dimensional, (4) laminar
flow (Rex <5 u105), (5) thermal boundary layer (Pe >100), (6) uniform upstream velocity and
temperature , (7) flat plate, (8) uniform surface temperature, (9) negligible changes in kinetic and
potential energy, (10) negligible axial conduction, (11) negligible dissipation, (12) no buoyancy
(E = 0 or g = 0), (13) uniform axial velocity within the thermal boundary layer ( Pr 1 ), (14)
uniform porosity and (15) injected fluid is at uniform velocity and temperature and is the same
as the external fluid.
(ii) Analysis. The local Nusselt number is defined as
hx
k
Nu x
(a)
where the heat transfer coefficient h is given by equation (1.10)
wT ( x,0)
wy
Ts Tf
k
h
(1.10)
PROBLEM 5.14 (continued)
Thus h depends on the temperature distribution T ( x, y ). The integral form of the energy equation
is used to determine the temperature distribution
G ( x)
wT x,0
k 1 P
wy
d
dx
t
³
G ( x)
d
U c p uTdy c p Tf
dx
0
t
³
U udy U c p v o P To Tf
(5.6)
0
This equation is simplified for constant properties
G ( x)
wT x,0
D 1 P
wy
d
dx
t
³
u (T Tf )dy v o P To Tf
(b)
0
oHwever, For Pr 1 the velocity boundary layer thickness is much smaller than the thermal
boundary layer thickness. Thus we assume
u
(b)
Vf
Assume a linear temperature profile
T x, y
b0 x b1 x y
(c)
The boundary conditions on the temperature are
(1) T x,0 Ts
(2) T x, G t # Tf
Equation (d) and the two boundary conditions give the coefficients bn (x)
1
b0 Ts , b1 (Tf Ts )
Gt
Substituting the above into (d)
T ( x, y )
Ts (Tf Ts )
y
Gt
(d)
Substituting (e) into (1.10)
h
k
(e)
Gt
Combining (a) and (e)
Nu x
x
Gt
(f)
The problem reduces to finding G t which is obtained using the energy equation. Substituting (c)
and (e) into (b)
D 1 P
(Tf Ts )
Gt
d
Vf(Tf Ts )
dx
Evaluating the integral and rearranging
G ( x)
t
ªy
º
« 1» dy v o P To Tf
¬G t
¼
³
0
PROBLEM 5.14 (continued)
1 dG t v o P To Tf
2 dx
Vf Ts Tf
D 1 P 1
Vf G t
(h)
Rewriting (h)
D 1 P 1
Vf G t
1 dG t
E
2 dx
(i)
where E is constant, defined as
vo P To Tf
Vf Ts Tf
E
(j)
Equation (i) is solved for G t by separating variables
G t dG t
dx
Integrating (k) and noting that G t ( xo )
(k)
2D 1 P
2 EG t
Vf
0
x
³x ³
Gt
dx
o
0
G t dG t
(l)
2D 1 P
2 EG t
Vf
Evaluating the integrals
x xo
Vf (Ts Tf )
D 1 P
Gt ln
2v o P (To Tf )
2 E 2 Vf 1 1
E Vf
Gt
D (1 P)
(m)
This result gives an implicitly solution for G t . The procedure for determining the Nusselt number
at a given x is to select a value for x, use (m) to determine the corresponding G t and substitute into
(g).
(iii) Checking. Dimensional check: Equations (h), (i) and (j) are dimensionally correct.
Boundary conditions check: Assumed temperature profile (d) satisfies the two boundary
conditions on temperature.
(5) Comments. L
imiting checks are on solution (m) do not yield useful results. For example, for
a solid plate, P E 0 , the first term on the right side of (m) becomes infinite. For the limiting
case of Ts Tf corresponds to E f. When this is substituted into (m) gives f / f. These
difficulties arise because solution (m) is not valid for P E 0 or E f. The type of
differential equation (i) changes for these limiting values and consequently solutions different
from (m) must be obtained.
PROBLEM 5.15
A porous plate with an impermeable and
insulated leading section of length xo is
heated with uniform surface flux q csc along
x t xo . The plate is cooled by forced
convection with a free stream velocity Vf
and temperature Tf . Fluid at temperature
To is injected through the porous surface
with uniform velocity v o . The injected and free stream fluids are identical. Assume laminar
boundary layer flow and introduce axial velocity simplification based on Pr 1 . Use a third
degree polynomial temperature profile to determine the local Nusselt number.
(1) Observations. (i) In general, to determine the Nusselt number it is necessary to determine the
velocity and temperature distribution. (ii) Fluid velocity for Pr 1 is assumed to be uniform,
u | Vf . This represents a significant simplification. (iii) The plate is porous. (iv) Fluid is
injected through the plate with uniform velocity. (v) The plate is heated with uniform surface
flux. (vi) Surface temperature is unknown, (vii) A leading section of the plate is insulated.
(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a
uniformly heated porous flat plate with surface injection and insulated leading section.
(3) Solution Plan. Start with equating Newton’s law with Fourier’s law to obtain an equation for
the heat transfer coefficient h. Apply the integral form of the energy equation using a third
degree polynomial temperature profile.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties, (3) two-dimensional, (4) laminar
flow (Rex <5 u105), (5) thermal boundary layer (Pe >100), (6) uniform upstream velocity and
temperature , (7) flat plate, (8) uniform surface heat flux, (9) negligible changes in kinetic and
potential energy, (10) negligible axial conduction, (11) negligible dissipation, (12) no buoyancy
(E = 0 or g = 0), (13) uniform axial velocity within the thermal boundary layer ( Pr 1 ), (14)
uniform porosity and (15) injected fluid is at uniform velocity and temperature and is the same
as the external fluid.
(ii) Analysis. The local Nusselt number is defined as
Nu x
hx
k
(a)
where the heat transfer coefficient h is given by equation (1.10)
wT ( x,0)
wy
Ts Tf
k
h
(1.10)
PROBLEM 5.15 (continued)
Thus h depends on the temperature distribution T ( x, y ). The integral form of the energy equation
is used to determine the temperature distribution
G ( x)
wT x,0
k 1 P
wy
d
dx
t
³
G ( x)
d
U c p uTdy c p Tf
dx
0
t
³
U udy U c p v o P To Tf
(5.6)
0
This equation is simplified for constant properties
G ( x)
wT x,0
D 1 P
wy
d
dx
t
³ u(T T )dy v P T
f
o
o
Tf
(b)
0
oHwever, For Pr 1 the velocity boundary layer thickness is much smaller than the thermal
boundary layer thickness. Thus we assume
u
(c)
Vf
For the temperature profile we assume a third degree polynomial
T x, y
b0 x b1 x y b2 x y 2 b3 x y 3
(d)
The boundary conditions on the temperature are
wT x,0
qscc
wy
(2) T x, G t # Tf
wT x, G t
#0
(3)
wy
(1) k
(4)
w 2T x,0
wy 2
0
Equation (d) and the four boundary conditions give the coefficients bn (x )
b0
Tf 2qscc
Gt,
3k
qscc
,
k
b1
Tf q csc
k
b2
0,
b3
qscc 1
3k G t2
Substituting the above into (d)
T ( x, y )
Surface temperature is obtained by setting y
Ts ( x )
ª2
1 y3 º
G
y
« t
»
3 G t2 »¼
«¬ 3
(e)
0 in (d)
T ( x,0)
Tf 2qscc
Gt
3k
(f)
Substituting (e) and (f) into (1.10)
h
3k
2G t
(g)
PROBLEM 5.15 (continued)
Combining (a) and (g)
Nu x
3x
2G t
(h)
Thus to determine surface temperature and Nusselt number requires the determination of G t .
Substituting (c) and (e) into (b)
D 1 P
d
dx
Vf
G ( x)
t
ª2
kvoP
1 y3 º
To Tf
« Gt y » dy 2
Vf q csc
3 G t ¼»
¬« 3
³
0
Evaluating the integral and rearranging
1 dG t2 k v o P
(To Tf )
4 dx
Vf q csc
(h)
4D 1 P 4k v o P
(To Tf )
Vf
Vf q csc
(i)
D 1 P
Vf
Rewriting (h)
dG t2
dx
Note that the right side of (i) is constant. The boundary condition on (i) is
G t ( xo ) 0
(j)
Integrating (i) and using (j)
G t2
ª 4D 1 P 4k v o P
º
(To Tf )» ( x xo )
«
Vf q csc
¬ Vf
¼
Solving for G t
Gt
ª 4D 1 P 4k v o P
º
(To Tf )»
«
Vf q csc
¬ Vf
¼
1/ 2
( x x o )1 / 2
(k)
(k) into (h) gives the local Nusselt number
º
3 ª 4D 1 P 4k v o P
(To Tf )»
«
2 ¬ Vf
Vf q csc
¼
Nu x
1 / 2
x
( x x o )1 / 2
(l)
This result can be expressed in terms of the Prandtl and local Reynolds number as
Nu x
Pr 1 / 2 Re1x/ 2
3
1/ 2
4ª
Uc p voP
º
(To Tf )» ( x xo )1 / 2
«(1 P) q csc
¬
¼
(m)
Surface temperature is obtained by substituting (k) into (f)
Ts ( x)
º
2q cc ª 4D 1 P 4k v o P
Tf s «
(To Tf )»
3k ¬ Vf
Vf q csc
¼
1/ 2
( x xo )1 / 2
(n)
PROBLEM 5.15 (continued)
(iii) Checking. Dimensional check: (1) Equations (e), (h), (k) and (n) are dimensionally
correct. (2) Equations (l) and (m) are dimensionless.
Boundary conditions check: Assumed temperature profile (e) satisfies the two boundary
conditions on temperature.
Limiting check: For the special case of solid plate, P = 0, (m) and (n) reduce to
Nu x
3 1/ 2 1/ 2
Pr Re x ( x x o ) 1 / 2
4
Ts ( x)
Tf 2q csc
( x xo )1 / 2
3k
Equation (o) is the correct result for this case (see Problem 5.3). For xo
that surface temperature varies with
(o)
(p)
0 , equation (p) shows
x . This result is correct (see Problem 5.6).
(5) Comments. The effect of wall injection on surface temperature is can be evaluated using
solution (n) for Ts (x). If the temperature of the injected fluid is greater than the free stream
temperature, To ! Tf , injection increases surface temperature. nO the other hand, if
injection will lower surface temperature.
To Tf ,
PROBLEM 5.16
Consider steady two-dimensional laminar flow in the inlet region of two parallel plates. The
plates are separated by a distance H. The lower plate is maintained at uniform temperature To
while heat is removed from the upper plate at uniform flux qocc . The inlet temperature is Ti .
Determine the distance from the inlet where the lower and upper thermal boundary layers meet.
Use a linear temperature profile and assume that velocity is uniform equal to Vi . Express your
result in terms of dimensionless quantities.
(1) Observations. (i) There are two thermal boundary layers in this problem. (ii) The upper and
lower plates have different boundary conditions. Thus, temperature distribution is not
symmetrical. (iii) The lower plate is at uniform temperature while heat is removed at uniform
flux along the upper plate. (iv) Fluid velocity is assumed uniform throughout the channel.
qocc
(2) Problem Definition. Determine the
temperature distribution for boundary layer
flow over the lower and upper plates.
G t2
Vi
(3) Solution Plan. Apply the integral form of
the energy equation using linear temperature
profiles for both plates.
Ti
0
x
H
G t1
L
(4) Plan Execution.
To
(i) Assumptions. (1) Steady state, (2) constant properties, (3) two-dimensional, (4) laminar
flow (Rex <5 u105), (5) thermal boundary layer (Pe >100), (6) uniform velocity throughout, (7)
uniform upstream temperature , (7) flat plates, (8) uniform surface temperature at the lower plate,
(9) uniform surface flux at the upper plate, (10) negligible changes in kinetic and potential
energy, (11) negligible axial conduction, (12) negligible dissipation and (13) no buoyancy (E = 0
or g = 0).
(ii) Analysis. At the location where the two thermal boundary layers meets, we have
G t1 ( L) G t2 ( L)
H
(a)
where
H spacing between the two plates
L distance from inlet to location where the two thermal layers meet
G t1 thermal boundary layer for the lower plate
G t2
thermal boundary layer for the upper plate
The integral form of conservation of energy is given in equation (5.7)
G t ( x)
wT x,0
D
wy
d
dx
³
0
H
owever, we assume
u (T Tf )dy
(5.7)
PROBLEM 5.16 (continued)
u
Vi
(b)
where
Vi = fluid axial velocity
Substitute (b) into (a)
G t ( x)
D wT x,0
Vi
d
dx
wy
³
(T Ti )dy
(c)
0
where
Ti
fluid temperature outside the thermal boundary layers
(1) Lower plate. Assume a linear temperature profile
b0 x b1 x y
T x, y
(d)
The boundary conditions on the temperature are
(1) T ( x,0)
To
(2) T ( x, G t1 ) # Ti
Equation (d) and the two boundary conditions give the coefficients bn ( x)
1
b0 To , b1 (Ti To ) 1
Gt
Substitute the above into (d)
To (Ti To )
T ( x, y )
y
G t1
(e)
Substitute (e) into (c)
D (Ti To )
Vi
G t1
G t1
(Ti To )
d
dx
³
0
Evaluate the integral and simplify
D 1
Vi G t1
1 d 1
(G t )
2 dx
Separate variables
2
D
dx G t1
Vi
Integrate and use boundary condition G t1 (0)
4
Solve for G t1
D
Vi
dG t1
dx
0
x
(G t1 ) 2
ª y
º
« 1 1» dy
«¬ G t
»¼
(f)
PROBLEM 5.16 (continued)
D
G t1 2
(g)
x
Vi
(1) Upper plate. Assume a linear temperature profile
b0 x b1 x y
T x, y
(d)
The boundary conditions on the temperature are
(1) k
wT ( x,0)
wy
qocc
(2) T ( x, G t2 ) # Ti
Equation (d) and the two boundary conditions give the coefficients bn ( x)
q cc
q occ
b0 Ti o G t2 b1
k
k
Substitute the above into (d)
q cc
T ( x, y ) Ti o (G t2 y )
k
Substitute (h) into (c)
D q occ
d
dx
Vi k
³
G t2
q occ
k
(h)
(G t2 y )dy
0
Evaluate the integral and simplify
D
1 d 2
(G t )
2 dx
Vi
Separate variables
2
D
Vi
Integrate and use boundary condition G t 2 (0)
2
d (G t2 )
dx
D
Vi
0
x
(G t 2 ) 2
Solve for G t 2
Gt2
L
et
x
2
D
Vi
(i)
x
L in (g) and (i), substitute into (a)
H
Solve for L
2
D
Vi
L 2
D
Vi
L
PROBLEM 5.16 (continued)
L
H2
(2 2 )
Rewrite (j) in dimensionless form and use the D
L
H
Vf
2
D
(j)
k / Ucp
1
U c p HVf
(2 2 ) 2
k
This can be written in terms of the Prandtl number and local Reynolds number
L
H
1
(2 2 ) 2
PrRe H
(k)
where
Re H
U HVf
P
(l)
(iii) Checking. Dimensional check: (1) Equations (e), (g), (h), (i) and (j) are dimensionally
correct. (2) Equations (k) and (l) are dimensionless.
Boundary conditions check: Assumed temperature profiles (e) and (h) satisfy their respective
boundary conditions.
uQalitative check: Increasing the free stream velocity decreases the thermal boundary layer,
resulting in an increase in L. Solution (j) shows that L is directly proportional to Vf .
(5) Comments. (1) To increase L the Reynolds number should be increased. (2) Taking the ratio
of (g) and (i)
G t1
(m)
2
Gt2
Thus the thermal boundary layer for constant wall temperature is thicker than that of uniform
surface flux by a factor of 2 .
PROBLEM 6.1
Use scaling to determine the ratio Lt / Lh . Compare scaling estimates with exact solutions.
(1) Observations. (i) This is an internal forced convection problem. (ii) Scaling gives estimates
of Lh and Lt . (iii) Exact solutions for Lh and Lt are available for laminar flow through
channels. (iv) Exact solutions for Lt depend on channel geometry and surface boundary
conditions.
(2) Problem Definition. Determine the ratio Lt / Lh using scaling and using exact solutions.
(3) Solution Plan. Apply scaling results (6.2) and (6.3) to estimate Lt / Lh . Apply (6.5) and (6.6)
to obtain an exact solution for Lt / Lh .
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) constant properties, (4) uniform
surface heat flux or uniform surface temperature, (5) negligible axial conduction (6) negligible
changes in kinetic and potential energy and (7) negligible dissipation
(ii) Analysis.
Scaling estimates of Lh and Lt are given by equations (6.2) and (6.3)
§ Lh / D ·
¸¸
¨¨
© Re D ¹
§ Lt / D ·
¨¨
¸¸
© Re D Pr ¹
1/ 2
~1
(6.2)
~1
(6.3)
1/ 2
From (6.2) and (6.3) we obtain
Lt / Lh
~1
Pr
(a)
Exact solutions for Lh for laminar flow is given by equation (6.5)
Lh
De
C h Re D e
(6.5)
where De is the equivalent diameter, defined as
De
4Af
P
where A f is channel flow area and P is channel perimeter. The coefficient C h depends on
channel geometry and is given in Table 6.1. Similarly, exact solutions for Lt for laminar flow is
given by equation (6.6)
PROBLEM 6.1 (continued)
Lt
De
(6.6)
C t PrRe D e
where Ct is a constant which depends on channel geometry as well as boundary conditions and
is given in Table 6.1. Taking the ratio of (6.6) to (6.5) and rearranging
Lt / Lh
Pr
(iii) Computations. Using Table 6.1,
the ratio C t / C h is computed for the six
geometries listed in the table for both
uniform surface flux and uniform surface
temperature. Comparisons between scaling
estimate and exact solutions are tabulated.
Tabulation results show that scaling
estimate is close to exact solutions for the
six geometries examined.
(5) Comments. (i) Scaling estimate of Lh
and Lt does not take into consideration
channel geometry. In addition, scaling
does not distinguish between laminar and
turbulent flow. (ii) Examination of the
tabulated results show that scaling
provides reasonable estimates of Lt / Lh
for all Prandtl numbers.
Ct
Ch
(b)
Scaling Estimate
Exact Solution
Lt / Lh
Pr
Lt / Lh
~1
Pr
G
eometry
a
a
b
a/b =1
Ct
Ch
uniform
surface flux
uniform surface
temperature
0.77
0.60
0.73
0.46
0.67
0.57
0.56
0.72
1.09
0.73
a
a/b = 2
b
a
b
a/b = 4
PROBLEM 6.2
Use scaling to estimate the hydrodynamic and thermal entrance lengths for the flow of air in a
3 cm u 3 cm square duct . The mean velocity is 0.8 m/s. Compare scaling estimates with exact
solutions. Evaluate properties at 50 o C.
(1) Observations. (i) This is an internal forced convection problem. (ii) Scaling gives estimates
of Lh and Lt . (iii) Exact solutions for Lh and Lt are available for laminar flow through
channels. (iv) Exact solutions for Lt depend on channel geometry and surface boundary
conditions.
(2) Problem Definition. Determine the ratio Lt / Lh using scaling and using exact solutions.
(3) Solution Plan. Apply scaling results (6.2) and (6.3) to estimate Lt / Lh . Apply (6.5) and (6.6)
to obtain an exact solution for Lt / Lh .
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) constant properties, (4) uniform
surface heat flux or uniform surface temperature, (5) negligible axial conduction (6) negligible
changes in kinetic and potential energy and (7) negligible dissipation
(ii) Analysis.
Scaling estimates of Lh and Lt are given by equations (6.2) and (6.3)
§ Lh / D ·
¸¸
¨¨
© Re D ¹
§ Lt / D ·
¨¨
¸¸
© Re D Pr ¹
1/ 2
~1
(6.2)
~1
(6.3)
1/ 2
Exact solutions for Lh for laminar flow is given by equation (6.5)
Lh
De
C h Re D e
(6.5)
u De
(a)
where the Reynolds number is defined as
Re De
Q
where De is the equivalent diameter, defined as
De
A f = channel flow area, m 2
P = channel perimeter, m
u mean flow velocity = 0.8 m/s
Q
kinematic viscosity, m 2 /s
4Af
P
(b)
PROBLEM 6.2 (continued)
The coefficient C h depends on channel geometry and is given in Table 6.1. Similarly, exact
solutions for Lt for laminar flow is given by equation (6.6)
Lt
(6.6)
C t PrRe D e
De
where Ct is a constant which depends on channel geometry as well as boundary conditions and
is given in Table 6.1.
(iii) Computations. For a square duct of side 0.03 m,
De
4
(0.03) 2 (m 2 )
4(0.03)(m)
0.03 m
Properties of air at 50 o C are
Pr 0.709
m2
s
The Reynolds number is
Q
17.93 u 10 6
Re De
0.08(m / s )(0.03)(m)
17.93 u 10 6 (m 2 / s)
1339.3
Scaling estimates: Substituting into (6.2)
§ Lh / 0.03(m) ·
¨
¸
© 1339.3 ¹
1/ 2
~1
Lh ~ 40.2 m
Equation (6.3) gives
§ Lt / 0.03(m) ·
¨¨
¸¸
© (0.709)(1339.3) ¹
1/ 2
~1
Lt ~ 28.5 m
Exact solution: For a square channel, Table 6.1 gives:
Ch
Ct
Ct
0.09
0.066 , for uniform surface heat flux
0.041 , for uniform surface temperature
Equation (6.5) gives Lh
Lh
0.09 (0.03)(m)1339.3 = 3.61 m
PROBLEM 6.2 (continued)
Equation (6.5) gives Lt
Lt
Lt
0.066 (0.03)(m)(0.709)(1339.3) 1.88 m, for uniform surface heat flux.
0.041 (0.03)(m)(0.709)(1339.3) 1.17 m, for uniform surface temperature.
(5) Comments. (i) Scaling estimate of Lh and Lt does not take into consideration channel
geometry or surface thermal condition.(ii) Scaling overestimates Lh and Lt by and order of
magnitude.
PROBLEM 6.3
Far away from the entrance of a channel the velocity and temperature become fully developed. It
can be shown that under such conditions the Nusselt number becomes constant. Consider air
flowing with a mean velocity of 2 m/s through a long tube of diameter 1.0 cm. The mean
temperature at a section in the fully developed
region is 35oC. The surface of the tube is
L
maintained at a uniform temperature of 130oC.
What is the length of the tube section needed for
the mean temperature to reach 105oC? The Nusselt
number for this case is given by
Nu D
3.657
u
Ts
(1) Observations. (i) This is an internal force convection problem. (ii) The channel is a long tube.
(iii) The surface is maintained at a uniform temperature. (iv) Since the tube section is far away
from the entrance, the velocity and temperature can be assumed fully developed. (v) Tube diameter,
mean velocity and inlet, outlet and surface temperatures are known. The length is unknown. (vi)
The fluid is air.
(2) Problem Definition. Determine the tube length needed to raise the mean temperature to a
specified level.
(3) Solution Plan. Use the analysis of flow in tubes at uniform surface temperature to determine the
required tube length.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) fully developed flow, (3) axisymmetric flow, (4)
constant properties, (5) uniform surface temperature, (6) negligible changes in kinetic and potential
energy, (7) negligible axial conduction, (8) negligible dissipation and (9) no energy generation.
(ii) Analysis. For flow in a tube at uniform surface temperature, conservation of energy and
Newton's law of cooling lead to equation (6.13)
Ph
Tm ( x) Ts (Tmi Ts ) exp[
x]
(a)
mc p
cp = specific heat, J/kg-oC
2 o
h = average heat transfer coefficient for a tube of length L, W/m - C
= mass flow rate, kg/s
m
P = tube perimeter, m
Tm(x) = mean temperature at x, oC
Tmi = mean inlet temperature = 35oC
Ts = surface temperature = 130oC
x = distance from inlet of heated section, m
Applying (a) at the outlet of the heated section (x = L) and solving for L
mc p Ts Tmi
L
ln
Ph Ts Tmo
(b)
PROBLEM 6.3 (continued)
where
Tmo = mean outlet temperature = 105oC
, and h . Air properties are determined
To compute L using (b), it is necessary to determine cp, P, m
at the mean temperature Tm , defined as
Tm =
Tmi Tmo
2
(c)
The perimeter P and flow rate m are given by
P=SD
(d)
And
m S
D2
Uu
4
(e)
where
D = inside tube diameter = 1 cm = 0.01 m
u = mean flow velocity = 2 m/s
U = density, kg/m3
The heat transfer coefficient for this case is determined from the Nusselt number, given by
Nu D
hD
k
3.657
(f)
(iii) Computations. Properties are determined at the mean temperature Tm . Using (c)
(35 105)( o C)
Tm =
70 o C
2
Properties of air at this temperature are:
cp = 1008.7 J/kg-oC
k = 0.02922 W/m-oC
Pr = 0.707
Q = 19.9u10-6 m2/s
U = 1.0287 kg/m3
Substituting into (d), (e) and (f)
P = S 0.01(m) = 0.03142 m
(0.01) 2 (m 2 )
m S
1.0287(kg/m 3 )2(m/s) 0.0001616 kg/s
4
0.02922( W / m o C)
h = 3.657
= 10.69 W/m2-oC
0.01(m)
Substituting into (b)
L
0.0001616(kg / s)1008.7(J / kg o C )
0.03142(m)10.69( W / m 2 o C)
ln
(130 35)( o C)
(130 105)( o C)
= 0.65 m
PROBLEM 6.3 (continued)
(iv) Checking. Dimensional check: Computations showed that equations (b), (d), (e) and
(f) are dimensionally consistent.
Limiting checks: (1) For the special case of Tmo = Tmi , the required length should vanish. Setting
Tmo = Tmi in (b) gives L = 0.
(2) The required length for the outlet temperature to reach surface temperature is infinite. Setting
Tmo = Ts in (b) gives L = f.
Quantitative checks: (1) An approximate check can be made using conservation of energy and
Newton’s law of cooling. Conservation of energy is applied to the air between inlet and outlet
Energy added at the surface = Energy gained by air
(g)
Assuming that air temperature in the tube is uniform equal to Tm , Newton’s law of cooling gives
Energy added at surface = h S D L (Ts Tm )
(h)
Neglecting axial conduction and changes in kinetic and potential energy, energy gained by air is
Energy gained by air = m cp(Tmo Tmi )
(i)
Substituting (h) and (i) into (g) and solving for the resulting equation for L
L
mc p (Tmo Tmi )
(j)
h SD(Ts Tm )
Equation (j) gives
L
0.0001616(kg / s)1008.7(J / kg o C)(105 35)( o C)
10.69( W / m 2 o C)S (0.01)(m)(130 70)( o C)
= 0.57 m
This is in reasonable agreement with the more exact answer obtained above.
(2) The value of h appears to be low compared with typical values listed in Table 1.1 for forced
convection of gases. H
owever, it shoul d be kept in mind that values of h in Table 1.1 are for typical
applications. Exceptions should be expected.
(5) Comments. Equation (f) gives the Nusselt number and heat transfer coefficient for this case.
This equation is valid under certain conditions. Key among the restrictions are: fully developed
laminar flow in tubes at uniform surface temperature.
PROBLEM 6.4
A fluid is heated in a long tube with uniform surface flux. The resulting surface temperature
distribution is found to be higher than design specification. Two suggestions are made for
lowering surface temperature without changing surface flux or flow rate: (1) increasing the
diameter, (2) decreasing the diameter. You are asked to determine which suggestion to follow. The
flow is laminar and fully developed. Under such conditions the Nusselt number is given by
Nu D
4.364
(1) Observations. (i) This is an internal force convection in a tube. (ii) The surface is heated at
uniform flux. (iii) Surface temperature increases along the tube and is unknown. (iv) The flow is
assumed laminar and fully developed. (v) The heat transfer coefficient for fully developed flow
through channels is constant. (vi) According to Newton’s law of cooling, surface temperature is
related to mean fluid temperature, surface heat flux and heat transfer coefficient.
(2) Problem Definition. Derive an equation for surface temperature variation in terms of tube
diameter.
(3) Solution Plan. Apply surface temperature solution for fully developed laminar flow through a
tube with constant surface flux.
(4)Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar, (3) fully developed flow, (4) axisymmetric
flow, (5) constant properties, (6) uniform surface heat flux, (7) negligible changes in kinetic and
potential energy, (8) negligible axial conduction, (9) negligible dissipation and (10) no energy
generation.
(ii) Analysis. Application of conservation of energy and Newton’s law of cooling give the
variation of surface temperature with distance. The solution is given by equation (7.61)
Ts ( x)
§ Px 1 ·
¸
Tmi q csc ¨
¨ m c p h ¸
©
¹
(a)
where
cp = specific heat, J/kg-oC
h = heat transfer coefficient, W/m2-oC
m = mass flow rate, kg/s
x = distance from inlet, m
P = tube perimeter, m
qcsc = surface heat flux, W/m2
Ts (x) = local surface temperature, oC
Tmi = mean inlet temperature, oC
The perimeter is given by
P=SD
(b)
PROBLEM 6.4 (continued)
For fully developed laminar flow through tubes at constant surface flux the Nusselt number is
given by
hD
Nu D
= 4.364
(c)
k
where
D = tube diameter, m
k = thermal conductivity, W/m-oC
NuD = Nusselt number
Solving (c) for h
h = 4.364
k
D
(d)
Substituting (b) and (d) into (a)
§S Dx
D ·
¸
Ts (x) = Tmi + qscc ¨¨
p 4.364 k ¸¹
© mc
(e)
Examination of (e) shows that decreasing the diameter will decrease surface temperature.
(iii) Checking. Dimensional check: The right hand side of (e) should have units of oC.
§
·
§ o C o C· o
S D m x ( m)
D( m)
¨
¸
qscc (W/m ) ¨
o
o
¸ = W ¨© J / s W ¸¹ = C
© m ( kg / s) c p ( J / kg C) 4.364 k (W / m C) ¹
2
Qualitative check: Increasing surface flux, increases Ts. Decreasing the mass flow rate, increases
Ts. This behavior is confirmed by equation (e).
Limiting check: If surface flux qscc = 0, fluid outlet temperature remains constant equal to the inlet
temperature. Setting qscc = 0 in (e) gives Ts = Tmi.
(5) Comments. The effect of diameter, surface flux, mass flow rate, distance along the tube and
fluid properties on surface temperature can be evaluated using the result obtained in (e). oHwever,
attention should be given to the assumptions leading to this result.
PROBLEM 6.5
Two identical tubes are heated with the same uniform flux at their surfaces. Air flows through
one tube while water flows at the same rate through the other. The mean inlet temperature for
both tubes is the same. Which tube will have a higher surface temperature distribution? Assume
laminar flow and neglect entrance effects. For this case the Nusselt number is given by
Nu D
4.364
(1) Observations. (i) This is an internal force convection in a tube. (ii) The surface is heated at
uniform flux. (iii) Surface temperature increases along the tube and is unknown. (iv) The flow is
assumed laminar and fully developed. (v) The heat transfer coefficient for fully developed flow
through channels is constant. (vi) According to Newton’s law of cooling, surface temperature is
related to mean fluid temperature, surface heat flux and heat transfer coefficient.
(2) Problem Definition. Derive an equation for surface temperature variation in terms of tube
diameter.
(3) Solution Plan. Apply surface temperature solution for fully developed laminar flow through
a tube with constant surface flux.
(4)Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar, (3) fully developed flow, (4) axisymmetric
flow, (5) constant properties, (6) uniform surface heat flux, (7) negligible changes in kinetic and
potential energy, (8) negligible axial conduction, (9) negligible dissipation and (10) no energy
generation.
L
Lt
qcsc
Tmi
qcsc
Tmo
u
m
x
D
qcsc
Lh
(ii) Analysis. Application of conservation of energy and Newton’s law of cooling give the
variation of surface temperature with distance. The solution is given by equation (6.10)
§ Px 1 ·
¸
Ts (x) = Tmi + q csc ¨
¨ mc p h ¸
©
¹
where
cp = specific heat, J/kg-oC
h = heat transfer coefficient, W/m2-oC
m = mass flow rate, kg/s
x = distance from inlet, m
(6.10)
PROBLEM 6.5 (continued)
P = tube perimeter, m
qscc = surface heat flux, W/m2
Ts (x) = local surface temperature, oC
Tmi = mean inlet temperature, oC
The perimeter is given by
P=SD
(a)
For fully developed laminar flow through tubes at constant surface flux the Nusselt number is
given by
hD
NuD =
= 4.364
(b)
k
where
D = tube diameter, m
k = thermal conductivity, W/m-oC
NuD = Nusselt number
Solving (b) for h
h = 4.364
k
D
(c)
Substituting (a) and (c) into (6.10)
§S Dx
D ·
¸
Ts (x) = Tmi + qscc ¨¨
p 4.364 k ¸¹
© mc
(d)
Examination of this result shows that decreasing the diameter will decrease surface temperature.
(iii) Checking. Dimensional check: The right hand side of (d) should have units of oC.
§
·
§ o C o C· o
S D m x ( m)
D (m)
¨
¸
qscc (W/m )
=W ¨
¸= C
¨ m(kg/s)c (J/kg o C) 4.364k ( W/m o C) ¸
©J /s W¹
p
©
¹
2
Qualitative check: Increasing surface flux, increases Ts. Decreasing the mass flow rate, increases
Ts. This behavior is confirmed by equation (d).
Limiting check: If surface flux qscc = 0, fluid outlet temperature remains constant equal to the
inlet temperature. Setting qscc = 0 in (d) gives Ts = Tmi.
(5) Comments. The effect of diameter, surface flux, mass flow rate, distance along the tube and
fluid properties on surface temperature can be evaluated using the result obtained in (d).
oHwever, attention should be given to the assumptions leading to this result.
PROBLEM 6.6
Water flows through a tube with a mean velocity of 0.2 m/s. The mean inlet temperature is 20oC
and the inside diameter of the tube is 0.5 cm. The water is heated to 80oC with uniform surface
heat flux of 0.6 W/cm2. Determine surface temperature at the outlet. If entrance effects can be
neglected the Nusselt number for fully developed flow is constant given by
Nu D 4.364
Is it justifiable to neglect entrance effects?
(1) Observations. (i) This is an internal forced convection problem in a tube. (ii) The surface is
heated at uniform flux. (iii) Surface temperature changes along the tube and is unknown. (iv) The
Reynolds number should be checked to determine if the flow is laminar or turbulent. (v) If
hydrodynamic and thermal entrance lengths are small compared to tube length, the flow can be
assumed fully developed throughout. (vi) For fully developed flow, the heat transfer coefficient is
uniform. (vii) The length of the tube is unknown. (viii) The fluid is water.
(2) Problem Definition. (i) Find the required length to heat the water to a given temperature and
(ii) determine the surface temperature at the outlet.
(3) Solution Plan. (i) Since surface flux, mean velocity, diameter, inlet and outlet temperatures
are known, apply conservation of energy between the inlet and outlet to determine the required
tube length. (ii) Check the Reynolds number to determine if the flow is laminar or turbulent. (iii)
Calculate the hydrodynamic and thermal entrance lengths and compare with the tube length. (iv)
Apply surface temperature solution for flow through a tube with constant surface flux.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties, (3) axisymmetric flow, (4) uniform
surface heat flux, (5) negligible changes in kinetic and potential energy, (6) negligible axial
conduction, (7) negligible dissipation and (8) no energy generation.
(ii) Analysis. Determination of tube length. Application of conservation of energy between
the inlet and outlet of the tube gives
S D Lq csc = mcp(Tmo - Tmi)
(a)
where
c p = specific heat, J/kg-oC
D = tube diameter = 0.5 cm = 0.005 m
L = tube length, m
m = mass flow rate, kg/s
Tmi = mean temperature at the inlet = 20oC
Tmo = mean temperature at the outlet = 80oC
qcsc = surface heat flux = 0.6 W/cm2 = 6000 W/m2
Solving (a) for L gives
L =
mc p (Tmo Tmi )
S Dq csc
(b)
PROBLEM 6.6 (continued)
The mass flow rate m is given by
m
(S / 4) D 2 U u
(c)
where
u = mean flow velocity = 0.2 m/s
U = density, kg/m3
To determine surface temperature at the outlet, use the solution for surface temperature
distribution for flow through a tube with uniform surface flux, given by equation (6.10)
ª Px
1 º
Ts (x) = Tmi + q csc «
»
¬« mc p h( x) ¼»
(d)
where
h = local heat transfer coefficient, W/m2-oC
P = tube perimeter, m
Ts (x) = local surface temperature, oC
x = distance from inlet of heated section, m
The perimeter P is given by
P=SD
(e)
Surface temperature at the outlet Ts(L) is obtained by setting x = L in (d). Substituting (e) into (d)
and letting x = L gives
ªS D L
1 º
Ts (L) = Tmi + q csc «
(f)
»
¬« mc p h( L) ¼»
The determination of h(L) requires establishing if the flow is laminar or turbulent and if it is fully
developed at the outlet. Thus, the Reynolds number should be determined. It is defined as
ReD
uD
Q
(g)
where
ReD = Reynolds number
Q = kinematic viscosity, m2/s
Properties of water are determined at the mean temperature T defined as
T =
Substituting into (h)
(20 80)( o C)
T =
2
Tmi Tmo
2
50 o C
Properties of water at this temperature are given in Appendix D
cp = 4182 J/kg-oC
k = 0.6405 W/m-oC
Pr = 3.57
(h)
PROBLEM 6.6 (continued)
-6
Q = 0.5537u10 m2/s
3
U = 988 kg/m
Substituting into (g)
ReD
0.2( m / s)0.005( m)
0.5537 u 10 6 ( m2 / s)
1806
Since the Reynolds number is less than 2300, the flow is laminar. The next step is calculating
the hydrodynamic and thermal entrance lengths Lh and Lt to see if the flow is fully developed at
the outlet. For laminar flow in a tube the hydrodynamic and thermal lengths are given by (7.43)
Lh = Ch D ReD
(i)
Lt = Ct D ReD Pr
(j)
where
Ch = hydrodynamic entrance length constant (Table 6.1) = 0.056
Ct = thermal entrance length constant (Table 6.1) = 0.043
Lh = hydrodynamic entrance length, m
Lt = thermal entrance length, m
Substituting numerical values into (i) and (j)
Lh = 0.056 u 0.005 (m) u 1806 = 0.506 m
and
Lt = 0.043 u 0.005 (m) u 1806 u 3.57 = 1.386 m
If tube length L is larger than Lh and Lt, the flow is fully developed. Thus, it is necessary to
compute L using equation (b). The mass flow rate in equation (b) is given by (c)
m = 988(kg/m3) 0.2(m/s)S (0.005)2(m2)/4 = 0.00388kg/s
Substituting into (b)
L=
0.00388(kg / s)4182(J / kg o C)(80 20)( o C)
S 0.005(m)0.6( W / cm 2 )10 4 (cm 2 / m 2 )
= 10.33 m
Since L > Lt> Lh, the flow is fully developed at the outlet. The heat transfer coefficient for fully
developed laminar flow through a tube with uniform surface flux is given by
NuD =
hD
= 4.364
k
(k)
where
k = thermal conductivity = 0.6405 W/m-oC
NuD = Nusselt number
Solving (k) for h
h
4.364k / D
(l)
PROBLEM 6.6 (continued)
(iii) Computations. To determine surface temperature at the outlet we first use (l) to compute
h(L)
h(L) = 4.364
0.6405( W / m o C )
= 559 W/m2-oC
0.005( m)
With L, m and h(L) determined, equation (f) gives the surface temperature at the outlet
ª
º
S 0.005(m)10.43(m)
1
o
Ts (L) = 20oC + 6000( W / m 2 ) «
» = 91.3 C
o
2 o
«¬ 0.00388(kg / s)4182(J / kg C) 559( W / m C) »¼
(iv) Checking. Dimensional check: Computations showed that equations (b), (c), (e), (f), (g),
(i), (j) and (l) are dimensionally correct.
Quantitative checks: (1) Alternate approach to determining Ts(L): Application of Newton’s law of
cooling at the outlet gives
qcsc = h [Ts(L) - Tmo ]
(m)
solving for Ts(L)
Ts(L) = Tmo +
qscc
0.6(W / cm2 ) u 104 (cm2 / m2 )
= 80 (oC) +
= 90.7oC
h
559(W / m2 o C)
(2) The value of h is within the range reported in Table 1.1 for forced convection of liquids.
Limiting check: If Tmi = Tmo, the required length should vanish. Setting Tmi = Tmo into (b) gives L
= 0.
(5) Comments. (i) As long as the outlet is in the fully developed region, surface temperature at
the outlet is determined entirely by the local heat transfer coefficient. Therefore, it is not necessary
to justify neglecting entrance length to solve the problem. (ii) In solving internal forced convection
problems, it is important to establish if the flow is laminar or turbulent and if it is developing or
fully developed.
PROBLEM 6.7
Fluid flows with a mean axial velocity u in a tube of
diameter D. The mean inlet temperature is Tmi . The
surface is maintained at uniform temperature Ts .
Show that the average Nusselt number for a tube of
length L is given by
Nu L
T Ts
Re D Pr
ln mi
4
Tm ( L) Ts
L
u
Tmi
Ts
hL L
uD
, Re D
and hL is the average
k
Q
heat transfer coefficient over the length L.
where Nu L
(1) Observations. (i) This is an internal force convection problem. (ii) The channel is a tube. (iii)
The surface is maintained at a uniform temperature. (iv) Entrance effect is important in this
problem. (v) The average Nusselt number for a tube of length L depends on the average heat
transfer coefficient over the length.
(2) Problem Definition. Determine the average heat transfer coefficient for a tube of length L
which is maintained at uniform surface temperature.
(3) Solution Plan. Start with the definition of the average Nusselt number. Use the analysis of flow
in tubes at uniform surface temperature to determine the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) axisymmetric flow, (3) constant properties, (4) uniform
surface temperature, (5) negligible changes in kinetic and potential energy, (6) negligible axial
conduction, (7) negligible dissipation and (8) no energy generation.
(ii) Analysis. The average Nusselt number for a tube of length L is given by
Nu L
hL L
k
(a)
where hL is the average heat transfer coefficient over the length L, defined as
hL
1
L
L
³ h( x)dx
(6.12)
0
For flow in a tube at uniform surface temperature, conservation of energy and Newton's law of
cooling lead to equation (6.13)
Ph
Tm ( x) Ts (Tmi Ts ) exp[
x]
(6.13)
mc p
c p = specific heat, J/kg-oC
hL = average heat transfer coefficient for a tube of length x, W/m2-oC
= mass flow rate, kg/s
m
h
PROBLEM 6.7 (continued)
P = tube perimeter, m
Tm(x) = mean temperature at x, oC
Tmi = mean inlet temperature, oC
Ts = surface temperature, oC
x = distance from inlet of heated section, m
Applying (a) at the outlet of the heated section (x = L) and solving for hL
mc p
hL
PL
ln
Ts Tmi
Ts Tm ( L)
(b)
where
Tm (L) = mean outlet temperature
Substitute (b) into (a)
mc p
Nu L
Pk
ln
Ts Tmi
Ts Tm ( L)
(c)
oHwever, the mass flow rate m is given by
D2
m S
Uu
4
(d)
where
D = inside tube diameter, m
u = mean flow velocity, m/s
U = density, kg/m3
The perimeter of a tube is
P
SD
(e)
Substitute (d) and (e) into (c)
Nu L
U Du c p
4k
ln
Ts Tmi
Ts Tm ( L)
(f)
The coefficient in (f) can be expressed in terms of Prandtl and Reynolds number as
U Du c p
k
c p P U Du
k
P
PrRe D
(g)
(g) into(f)
Nu L
T Tmi
PrRe D
ln s
4
Ts Tm ( L)
(h)
(iii) Checking. Dimensional check: (1) Equations (6.12), (6.13), (b) and (d) are dimensionally
consistent. (2) Equations (c) and (f) are dimensionless.
(5) Comments. Equation (f) or (h) can be used to experimentally determine the average heat
transfer coefficient and average Nusselt number.
PROBLEM 6.8
Water flows through a 0.75 cm u 0.75 cm square duct with
a mean velocity of 0.12 m/s. The duct is heated with a
uniform surface flux of 0.25 W/cm2. The mean inlet
temperature is 25oC. The maximum allowable surface
temperature is 95oC. Justify neglecting entrance effects.
And determine maximum outlet mean temperature.
L
Tmo
u
Ts
Tmi
(1) Observations. (i) This is an internal forced convection problem. (ii) The fluid is heated at
uniform wall flux. (iii) Surface temperature changes with distance along the channel. It reaches a
maximum value at the outlet. (iv) The Reynolds and Peclet numbers should be checked to
establish if the flow is laminar or turbulent and if this is an entrance or fully developed problem.
(v) The channel has a square cross-section. (vi) Application of Newton’s law of cooling at the
outlet relates outlet temperature to surface temperature, surface flux and heat transfer coefficient.
(vii) Application of conservation of energy gives a relationship between heat added, inlet
temperature, outlet temperature, specific heat and mass flow rate.
(2) Problem Definition. [a] Determine the outlet temper ature corresponding to a specified
surface temperature and flux. [b]Determine the require d channel length to heat the water to outlet
temperature and compare with entrance lengths.
(3) Solution Plan. Apply Newton’s law of cooling at the outlet to determine the mean outlet
temperature. This requires determining the heat transfer coefficient. Check the Reynolds and
Peclet numbers to establish if the flow is laminar or turbulent and if this is an entrance or fully
developed problem. O
btain a solution to the heat transfer coefficient (Nusselt number).
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties, (3) uniform surface heat flux, (4)
negligible changes in kinetic and potential energy, (5) negligible axial conduction, (6) negligible
dissipation and (7) no energy generation.
Lt
L
qcsc
Tmi
Tmo
u
m
Lh
qcsc
Ts (L)
S
S
qcsc
(ii) Analysis. [a]Determination of Tmo. Applying Newton’s law of cooling at the outlet
q scc = h(L) [Ts (L) Tmo ]
(a)
Solving (a) for Tmo
Tmo = Ts(L) q csc
hL
(b)
PROBLEM 6.8 (continued)
where
h(L) = heat transfer coefficient at the outlet, W/m2-oC
L = channel length, m
2
2
q scc = surface heat flux = 0.25 W/cm = 2500 W/m
Tmo = mean outlet temperature, oC
Ts ( L) = surface temperature at the outlet = 95oC
Equation (b) gives Tmo in terms of the heat transfer coefficient at the outlet, h(L). The value of
h(L) depends on whether the flow is laminar or turbulent and if the flow is developing or fully
developed at the outlet. To establish these conditions, the Reynolds and Prandtl numbers are
determined. The Reynolds number for flow through a square channel is defined as
ReDe
uDe
Q
(c)
where
De = equivalent diameter, m
ReDe = Reynolds number
u = mean velocity = 0.12 m/s
Q = kinematic viscosity, m2/s
The equivalent diameter for a square channel is defined as
De = 4
A
S2
= 4
=S
P
4S
(d)
where
A = channel flow area = S2, m2
P = channel perimeter in contact with the fluid = 4S, m
S = side dimension of the square channel = 0.75 cm = 0.0075 m
Water properties are evaluated at the mean temperature, Tm , defined as
Tm = (Tmi + Tmo)/2
(e)
where
Tm = mean fluid temperature in channel, oC
Tmi = mean inlet temperature = 25oC
oHwever, since Tmo is unknown, a solution is obtained using a trial and error procedure. A value
for Tmo is assumed, (e) is used to calculate Tm and (b) is used to calculate Tmo. The calculated Tmo
is compared with the assumed value. The procedure is repeated until a satisfactory agreement is
obtained between assumed and calculated values of Tmo.
L
et Tmo = 85oC
Equation (e) gives
o
o
Tm = (25 +85)( C)/2 = 55 C
Properties of water at this temperature are
cp = specific heat = 4184 J/kg-oC
k = thermal conductivity = 0.6458 W/m-oC
PROBLEM 6.8 (continued)
Pr = Prandtl number = 3.27
Q = kinematic viscosity = 0.5116u10-6 m2/s
U = density =985.7 kg/m3
Using (d) to calculate De
De = 0.0075 m
Substituting into (c)
Re De =
0.12(m / s)0.0075(m)
0.5116 u 10 6 (m 2 /s)
= 1759
Since the Reynolds number is smaller than 2300, the flow is laminar. To establish if the flow is
developing or fully developed at the outlet, the hydrodynamic length, thermal entrance length and
tube length must be determined. Equations (7.43a) and (7.43b) give
Lh = Ch De Re De
(f)
Lt
(g)
and
C t De Re De Pr
where
Ch = velocity entrance length constant (Table 7.2) = 0.09
Ct = temperature entrance length constant (Table 7.2) = 0.066
Lh = hydrodynamic entrance length, m
Lt = thermal entrance length, m
Substituting numerical values into (f) and (g)
Lh = 0.09 (0.0075)(m)(1759) = 1.19 m
and
Lt = 0.066(0.0075)(m)(1759) (3.27) = 2.85 m
These two lengths should be compared with the tube length L. To determine L, conservation of
energy is applied to the fluid between inlet and outlet
Energy added at the surface = Energy gained by the fluid
(h)
Neglecting axial conduction and changes in kinetic and potential energy, (h) gives
cp(Tmo Tmi )
q scc (4 S L) = m
or
L=
m c p Tmo Tmi
4 S qscc
(i)
where the mass flow rate m is given by
= U u A = U u S2
m
(j)
Since Tmo is unknown, L cannot be computed. To proceed, assume that the flow is fully developed
at the outlet, determine h(L), use (b) to compute Tmo and (i) to compute L. If the computed length
is larger than Lh and Lt, the assumption of fully developed flow is verified. The Nusselt number for
fully developed laminar flow in a square channel with uniform surface heat flux is given by
equation (7.58) and Table 7.3
PROBLEM 6.8 (continued)
hDe
Nu De
= 3.608
k
(k)
or
h = 3.608 k / De
(l)
(iii) Computations. a[ ]Determination of Tmo. Equation (l) gives h
h = 3.608(0.6458)(W/m-oC)/0.0075(m) = 310.7 W/m2-oC
Substituting into (b)
Tmo = 95(oC) 2500( W / m 2 )
2
o
310.7( W / m C)
= 87oC
This is close to the assumed value of 85oC used to obtain approximate water properties.
b[ ]Determination of channel length L. Equation (j) gives m
= 985.7(kg/m3)0.12(m/s)(0.0075)2(m2) = 0.00665 kg/s
m
Substituting into (i)
L=
0.00665(kg / s)4184(J / kg o C) 87 25 ( o C)
4(0.0075)(m)2500( W / m 2 )
= 23 m
Since L > Lt > Lh, the flow is fully developed at the outlet.
(iv) Checking. Dimensional check: Computations showed that equations (b), (c), (f), (g), (i),
(j) and (l) are dimensionally consistent.
Quantitative check. The value of h is within the range listed in Table 1.1.
Qualitative check: As surface heat flux is decreased, channel length should increase. Equation (i)
shows that L is inversely proportional to qscc .
Limiting check. If Tmo = Tmi, channel length should be zero. Setting Tmo = Tmi in (i) gives L = 0.
(5) Comments. (i) This problem illustrates how analysis cannot always be completed without
carrying out some computations. This can occur if it is necessary to establish if the flow is laminar
or turbulent or if entrance effects can be neglected or not. (ii) A solution is obtained without the
need to neglect entrance effects. As long as the outlet is in the fully developed region, water outlet
temperature is determined entirely by the local heat transfer coefficient.
PROBLEM 6.9
Two experiments were conducted on fully developed laminar flow through a tube. In both
experiments surface temperature is 180 o C and the mean inlet temperature is 20 o C . The mean
outlet temperature for the first experiment is found to be 120 o C . In the second experiment the
flow rate is reduced by a factor of 2. All other conditions remained the same. Determine:
a[ ] The outlet temperature of the second experiment.
b[ ] The ratio of heat transfer rate for the two experiments.
(1) Observations. (i) This is an internal forced convection problem in tubes. (ii) The flow is
laminar and fully developed. (iii) The surface is maintained at uniform temperature. (iv) All
conditions are identical for two experiments except the flow rate through one is half that of the
other. (v) The total heat transfer rate depends on the outlet temperature.
(2) Problem Definition. Determine the outlet temperature for fully developed laminar flow
through a tube at uniform surface temperature.
(3) Solution Plan. Use the analysis of fully developed laminar flow in tubes at uniform surface
temperature to determine the outlet temperature. Apply conservation of energy to obtain an
equation for the heat transfer.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) fully developed laminar flow, (3) axisymmetric flow,
(4) constant properties, (5) uniform surface temperature, (6) negligible changes in kinetic and
potential energy, (7) no axial conduction, (8) no dissipation and (9) no energy generation.
(ii) Analysis. For flow through tubes with uniform surface temperature, conservation of
energy and Newton’s law of cooling lead to equation (6.13)
Tm ( x)
Ts (Tmi Ts ) exp[
Ph
x]
m cp
(a)
c p = specific heat, J/kg o C
h = average heat transfer coefficient for a tube of length x, W/m 2 o C
m = mass flow rate, kg/s
P = tube perimeter, m
Tm ( x) = mean temperature at x, o C
Tmi = mean inlet temperature = 20 o C
Ts = surface temperature = 180 o C
x = distance from inlet of heated section, m
Applying (a) at the outlet ( x
L)
Tmo ( x)
where
L = tube length, m
Tmo = mean outlet temperature, o C
Ts (Tmi Ts ) exp[
Ph
L]
m cp
(b)
PROBLEM 6.9 (continued)
The quantities L, P and c p are the same for both experiments. Furthermore, the flow remains
laminar and fully developed when the flow rate is reduced in the second experiment. Thus the
heat transfer coefficient is the same for both experiments. Equation (b) is rewritten in
dimensionless form as
Tm o Ts
exp(C / m)
(c)
Tmi Ts
where
PLh
(d)
C
cp
Rewriting (c)
ln
Tmi Ts
Tm o Ts
C
(e)
m
L
et the subscripts 1 and 2 refer to the first an d second experiments. Applying (e) to the two
experiments gives
T Ts
C
ln mi
(f)
Tm o1 Ts m1
ln
Tmi Ts
Tm o 2 Ts
C
(g)
m2
Taking the ratio of (g) to (f) and rearranging
ln
Tmi Ts
Tm o 2 Ts
m1
m2
ln
Tmi Ts
Tm o1 Ts
rO
Tmi Ts
Tm o 2 Ts
ª Tmi Ts º
«
»
«¬ Tm o1 Ts »¼
m1
m2
Solving for Tmo 2
Tm o 2
Ts (Tmi
ª Tm Ts º
Ts ) « o1
»
¬ Tmi Ts ¼
m1
m2
(h)
This result gives the outlet temperature when the flow rate is reduced. Application of
conservation of energy to the fluid between the inlet and outlet gives the heat transfer rate q
q
mc p Tmo Tmi
(i)
Applying (i) to the flow in the two tubes and taking the ratio of the two results
q2
q1
m2 Tmo 2 Tmi
m1 Tmo1 Tmi
(iii) Computations. Substituting numerical values into (h) and noting that m 1 / m 2
(j)
2
PROBLEM 6.9 (continued)
2
Tm o 2
ª (120 180)( o C) º
o
180( C) (20 180)( C) «
» = 157.5 C
o
¬ (20 180)( C) ¼
o
o
Equation (j) gives
q2
q1
1 157.5 20
2 120 20
0.6875
(iv) Checking. Dimensional check: Each term in (h) has units of temperature.
Limiting check: If m1
(h) gives Tmo 2 Tmo1 .
m2 , the two outlet temperatures must be the same. Setting m1
m2 in
(5) Comments. (i) Although tube size, fluid nature a nd flow rate are not known, it was possible
to obtain a solution to the problem. Taking the ratio of two operating conditions results in the
cancellation of the unknown factors. (ii) Although the outlet temperature increases as the flow
rate is decreased, the rate of heat transfer decreases. (iii) Decreasing the flow rate without
changing inlet and surface temperatures and heat transfer coefficient is expected to increase the
outlet temperature. In the limit as the flow rate approaches ezro (m2 0), the corresponding
outlet temperature becomes equal to surface temperature. This follows from (h).
PROBLEM 6.10
A long rectangular duct with a 4cm u 8 cm cross section is
used to heat air from –19.6oC to 339.6oC. The mean
velocity in the duct is 0.2 m/s and surface temperature is
340 oC. Determine the required duct length. Is neglecting
entrance effects justified?
L
Tmo
Ts
u
Tmi
(1) Observations. (i) This is an internal forced convection
problem. (ii) The channel has a rectangular cross section. (iii) Surface temperature is uniform.
(iv) The Reynolds and Peclet numbers should be checked to establish if the flow is laminar or
turbulent and if entrance effects can be neglected. (v) Channel length is unknown. (vi) The fluid
is air.
(2) Problem Definition. Determine the required channel length to heat air to a specified outlet
temperature. Determine the hydrodynamic and thermal entrance lengths and compare with
channel length.
(3) Solution Plan. Use the analysis of flow through tubes with uniform surface temperature to
determine channel length. Check the Reynolds and Peclet numbers to establish if the flow is
laminar or turbulent. Compute the hydrodynamic and thermal entrance lengths and compare
with channel length to determine if entrance effects can be neglected. O
btain a solution to the
heat transfer coefficient (Nusselt number).
(4) Plan Execution.
(i) Assumptions. (1) Steady
state, (2) constant properties, (3)
uniform surface temperature, (4)
negligible changes in kinetic and
potential energy, (5) negligible
axial conduction, (6) negligible
dissipation and (7) no energy
generation.
Lh
Ts
L
u
u
Tmi
m
Tmo
a
b
Lt
(ii) Analysis. For flow through tubes with uniform surface temperature, conservation of
energy and Newton’s law of cooling lead to equation (6.13)
Tm ( x)
Ts (Tmi Ts ) exp[
Ph
x]
m cp
where
cp = specific heat, J/kg-oC
2 o
h = average heat transfer coefficient for a channel of length x, W/m - C
m = mass flow rate, kg/s
P = channel perimeter, m
Tm(x) = mean temperature at x, oC
Tmi = mean inlet temperature, -19.6oC
Ts = surface temperature = 340oC
x = distance from inlet of heated section, m
(a)
PROBLEM 6.10 (continued)
Applying (a) at the outlet, x = L, and solving for L
L
mc p
Ph
ln
Ts Tmi
Ts Tmo
(b)
where
L = channel length, m
Tmo = mean outlet temperature = 339.6oC
Equation (b) gives L in terms of cp, P, m, and h . The average heat transfer coefficient depends
on whether the flow is laminar or turbulent and if the flow is developing or fully developed. To
establish these conditions, the Reynolds and Prandtl numbers are determined. The Reynolds
number for flow through a rectangular channel is defined as
ReDe
uDe
Q
(c)
where
De = equivalent diameter, m
ReDe = Reynolds number
u = mean velocity = 0.2 m/s
Q = kinematic viscosity, m2/s
The equivalent diameter for a rectangular channel is defined as
De = 4
ab
A
= 2
P
( a b)
(d)
where
A = channel flow area = ab, m2
P = channel perimeter in contact with the fluid = 2(a + b), m
a = width of rectangular channel = 8cm = 0.08 m
b = height of rectangular channel = 4 cm = 0.04 m
The perimeter P and flow rate m are given by
P = 2(a + b)
(e)
and
m
U Au
U (ab)u
(f)
where
U = density, kg/m3
Air properties are evaluated at the mean temperature, Tm , defined as
Tm = (Tmi + Tmo)/2
(g)
where
Tm = mean fluid temperature in channel, oC
The mean temperature is calculated in order that properties are determined. Substituting into (g)
Tm =
(19.6 339.6)( o C)
2
160 o C
PROBLEM 6.10 (continued)
Properties of air at this temperature are given in Appendix C
cp = 1018.5 J/kg-oC
k = 0.3525 W/m-oC
Pr = 0.701
Q = 29.75u10-6, m2/s
U = 0.8342 kg/m3
Substituting into (d)
De = 2
0.04(m)0.08(m)
= 0.0533 m
(0.04 0.08)(m)
Substituting into (c)
ReDe
0.2(m / s )0.0533(m)
358.3
29.75 u 10 6 (m 2/s)
Since the Reynolds number is smaller than 2300, the flow is laminar. To determine if entrance
effects can be neglected, the hydrodynamic and thermal entrance length must be compared with
channel length. For laminar flow, equations (6.5) and (6.6) give
Lh = Ch De ReDe
(h)
Lt
(i)
and
Ct De ReDe Pr
where
Ch = hydrodynamic entrance length constant (Table 6.1, for a/b = 0.08m/0.04m = 2) = 0.085
Ct = thermal entrance length constant (Table 6.1, for a/b = 0.08m/0.04m = 2) = 0.049
Lh = hydrodynamic entrance length, m
Lt = thermal entrance length, m
Substituting numerical values into (h) and (i)
Lh = 0.085 (0.0533)(m)(358.3) = 1.623 m
and
Lt = 0.049(0.0533)(m) (358.3) (0.706) = 0.656 m
These two lengths should be compared with channel length L. oHwever, L can be determined
only after h is computed. To compute h , L must be known!To proceed, assume that entrance
length effects are negligible (fully developed flow throughout channel), determine h , use (b) to
compute L and compare it with Lh and Lt. For fully developed laminar flow through a rectangular
channel at constant surface temperature, the Nusselt number is given in Table 6.2
Nu De
h De
= 3.391
k
(j)
or
h = 3.391 k / De
(iii) Computations. Equations (e), (f) and (k) give
P = 2(0.08 +0.04)(m) = 0.24 m
(k)
PROBLEM 6.10 (continued)
m = 0.0.8342(kg/m3)0.08(m)0.04(m)0.2(m/s) = 0.0005339 kg/s
h = 3.391(0.03525)(W/m-oC)/0.0533(m) = 2.242 W/m2-oC
Substituting into (b)
L
0.0005339(kg/s)1018.5(J/kg o C)
0.24(m)2.242(W/m 2 o C)
ln
(340 19.6)( o C)
(340 339.6)( o C)
= 6.873 m
Comparing Lh and Lt with L
Lh/L = 1.623(m)/6.873(m) = 0.236
and
Lt/L = 0.656(m)/6.87(m) = 0.095
(iv) Checking. Dimensional check: Computations showed that equations (b) (f ) , (h), (i)
and (l) are dimensionally consistent.
Limiting checks: (1) For the special case of Tmo = Tmi , the required length should vanish.
Setting Tmo = Tmi in (b) gives L = 0. (2) The required length for the outlet temperature to reach
surface temperature is infinite. Setting Tmo = Ts in (b) gives L = f.
Quantitative check: The value of h appears to be low compared with typical values listed in
Table 1.1 for forced convection of gases. oHwe ver, equation (l) shows that for laminar flow
through channels, h is inversely proportional to D. A large D can result in a small h . In
addition, values of h in Table 1.1 are for typical applications. Exceptions should be expected.
(5) Comments. (i) This problem illustrates the importance of establishing if entrance effects can
be neglected or not. (ii) Neglecting thermal entrance length is justified since Lt is less than 10%
of L. oHwever, neglecting the viscous entrance length requires careful judgment. This
assumption underestimates h and consequently, according to (b), it overestimates L. Thus it is a
conservative assumption.
PROBLEM 6.11
A rectangular duct with inside dimensions of
2 cm u 4 cm is used to heat water from 25 o C to
115 o C . The mean water velocity is 0.018 m/s. The
surface of the duct is maintained at 145 o C . Determine
the required duct length. Assume fully developed flow
conditions throughout.
L
Tmo
Ts
u
Tmi
(1) Observations. (i) This is an internal force convection problem. (ii) The channel is a
rectangular duct. (iii) The surface is maintained at a uniform temperature. (iv) The velocity and
temperature are fully developed. (v) The Reynolds number should be checked to determine if the
flow is laminar or turbulent. (vi) Duct size, mean velocity and inlet, outlet and surface
temperatures are known. The length is unknown. (vii) Duct length depends on the heat transfer
coefficient. (vii) The fluid is water.
(2) Problem Definition. Determine the duct length needed to raise the mean temperature to a
specified level. This requires determining the heat transfer coefficient.
(3) Solution Plan. Use the analysis of flow in channels at uniform surface temperature to
determine the required duct length.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) fully developed flow, (3) constant properties, (4)
uniform surface temperature, (5) negligible changes in kinetic and potential energy, (6)
negligible axial conduction, (7) negligible dissipation and (8) no energy generation.
(ii) Analysis. For flow in a channel at uniform surface temperature, conservation of energy
and Newton's law of cooling lead to equation (6.13)
Ts (Tmi Ts ) exp[
Tm ( x)
Ph
x]
m cp
(a)
where
c p = specific heat, J/kg o C
h = average heat transfer coefficient for a channel of length L, W/m 2 o C
m = mass flow rate, kg/s
P = cross section perimeter, m
Tm(x) = mean temperature at x, o C
Tmi = mean inlet temperature = 25 o C
Ts = surface temperature = 145 o C
x = distance from inlet, m
Applying (a) at the outlet (x = L) and solving for L
L
mc p
Ph
ln
Ts Tmi
Ts Tmo
(b)
PROBLEM 6.11 (continued)
where
Tmo = mean outlet temperature = 115 o C
To compute L using (b), it is necessary to determine cp, P, m , and h . Water properties are
determined at the mean temperature Tm , defined as
Tm =
Tmi Tmo
2
(c)
The perimeter P is given by
P = 2(a + b)
(d)
U (ab)u
(e)
Where
a = duct width = 4 cm = 0.04 m
b = duct height = 2 cm = 0.02 m
The mass flow rate m is given by
m
where
U = density, kg/m 3
The heat transfer coefficient for fully developed flow is uniform along a channel. Its value
depends on whether the flow is laminar or turbulent. To proceed, it is necessary to calculate the
Reynolds number to determine if the flow is laminar or turbulent. For flow in a rectangular duct
the Reynolds number is defined as
ReDe
u De
Q
(f)
where
De = equivalent diameter, m
ReDe = Reynolds number
u = mean velocity = 0.018 m/s
Q = kinematic viscosity, m 2 /s
The equivalent diameter for a rectangular channel is defined as
De = 4
ab
A
= 2
P
( a b)
where
A = duct flow area = ab, m2
The mean temperature is calculated to determine water properties. Substituting into (c)
(25 115)( o C)
70o C
2
Properties of water at this temperature are:
Tm =
c p = 4191 J/kg o C
(g)
PROBLEM 6.11 (continued)
k = 0.6594 W/m o C
Pr = 2.57
Q = 0.4137 u 10 6 m 2 /s
U = 977.7 kg/m3
Substituting into (h)
De = 2
0.04(m)0.02(m)
(0.04 0.02)(m)
0.02667 m
Equation (h) gives
ReDe
0.018(m/s)0.02667(m)
1160
0.4137 u 10 6 (m 2 /s)
Since the Reynolds number is smaller than 2300, the flow is laminar. The Nusselt number for
fully developed laminar flow through rectangular channels at uniform surface temperature is
given by equation (7.57) and Table 7.3. Thus
Nu De
h De
= 3.391
k
(h)
Solving for h
h
3.391
k
De
(i)
(iii) Computations. Substituting into (d), (e) and (i)
P
2(0.04 0.02)(m) = 0.12 m
m
977.7(kg/m 3 ) 0.04(m) 0.02(m) 0.018(m/s)
0.01408 kg/s
0.6594( W/m o C)
= 83.84 W/m 2 o C
0.02667(m)
Substituting into (b)
h = 3.391
L
4191(J/kg o C) 0.01408 (kg / s)
(145 25)( o C)
ln
0.12(m) 83.84 ( W/m 2 o C)
(145 115)( o C)
8.13 m
(iv) Checking. Dimensional check: Computations showed that equations (b), (d), (e), (f), (g)
and (i) are dimensionally consistent.
Limiting checks: (1) For the special case of Tmo = Tmi , the required length should be zero.
Setting Tmo = Tmi in (b) gives L = 0.
(2) The required length for the outlet temperature to reach surface temperature is infinite. Setting
Tmo = Ts in (b) gives L = f.
Quantitative checks: (1) An approximate check can be made using conservation of energy and
Newton’s law of cooling. Conservation of energy is applied to the water between inlet and outlet
Energy added at the surface = Energy gained by water
(j)
PROBLEM 6.11 (continued)
Assuming that water temperature in the tube is uniform equal to Tm , Newton’s law of cooling
gives
Energy added at surface = h P L (Ts Tm )
(k)
Neglecting axial conduction and changes in kinetic and potential energy, energy gained by the
water is
Energy gained by air = m cp(Tmo Tmi )
(l)
Substituting (k) and (l) into (j) and solving for the resulting equation for L
L
m c p (Tmo Tmi )
h P(Ts Tm )
(m)
Equation (m) gives
L
0.01408 (kg/s)4191(J/kg o C)(115 25)( o C)
83.84( W/m 2 o C) 0.12)(m)(145 70)( o C)
= 7.04 m
This is in reasonable agreement with the more exact answer obtained above.
(2) The value of h within the range listed in Table 1.1 for forced convection of liquids.
(5) Comments. This problem is simplified by two conditions: fully developed and laminar flow.
PROBLEM 6.12
Air is heated in a 4cm u 4 cm square duct at
uniform surface flux of 590 W/m 2 . The mean air
velocity is 0.32 m/s. At a section far away from
the inlet the mean temperature is 40 o C . The
mean temperature is 120 o C . Determine the
maximum surface temperature.
Tmo
L
u
T
qcc
s
mi
(1) Observations. (i) This is an internal forced
convection problem in a channel. (ii) The surface is heated at uniform flux. (iii) Surface
temperature changes along the channel. It reaches a maximum value at the outlet. (iv) The
Reynolds number should be checked to determine if the flow is laminar or turbulent. (v) eVlocity
and temperature profiles become fully developed far away from the inlet. (vi) The heat transfer
coefficient is uniform for fully developed flow. (vii) The channel has a square cross section.
(viii) tube length is unknown. (ix) The fluid is air.
(2) Problem Definition. (i) Find the required length to heat the air to a given temperature and
(ii) determine surface temperature at the outlet.
(3) Solution Plan. (i) Since surface flux, mean velocity, duct size, inlet and outlet temperatures
are known, application of conservation of energy between the inlet and outlet gives the required
duct length. (ii) Check the Reynolds number to determine if the flow is laminar or turbulent. (iii)
Apply surface temperature solution for flow through a channel with constant surface flux.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties, (3) uniform surface heat flux, (4)
negligible changes in kinetic and potential energy, (5) negligible axial conduction, (6) negligible
dissipation and (7) no energy generation.
(ii) Analysis. Application of conservation of energy between the inlet and outlet gives the
required channel length
P L q csc m c p (Tmo Tmi )
(a)
where
c p = specific heat, J/kg o C
L = channel length, m
m = mass flow rate, kg/s
P = perimeter, m
q csc = surface heat flux = 590 W/m 2
Tmi
40 o C
Tmo
120 o C
Solving (a) for L
L
m c p (Tmo Tmi )
P q cc
(b)
PROBLEM 6.12 (continued)
The mass flow rate and perimeter are given by
m
U S2u
(c)
4S
(d)
P
where
S = duct side = 0.04 m
u = mean flow velocity = 0.32 m/s
U = density, kg/m 3
Substituting (c) and (d) into (b)
L
U S u c p (Tmo Tmi )
4 q cc
(e)
To determine surface temperature at the outlet, use the solution for surface temperature
distribution for channel flow with uniform surface flux, given by equation (6.10)
§ Px
1 ·¸
Ts (x) = Tmi + q csc ¨
¨ mc p h( x) ¸
©
¹
(f)
where
h(x) = local heat transfer coefficient, W/m 2 o C
Ts ( x) = local surface temperature, o C
x = distance from inlet of heated section, m
Surface temperature at the outlet, Ts(L), is obtained by setting x = L in (f). Substituting (c) and
(d) into (f)
§ 4L
1 ·¸
Ts (L) = Tmi + qcsc ¨
(g)
¨ U S u c p h( L) ¸
©
¹
Finally, it remains to determine the heat transfer coefficient at the outlet, h(L). This requires
establishing whether the flow is laminar or turbulent. Thus, the Reynolds number should be
determined. The Reynolds number for flow through a square channel is defined as
ReDe
u De
Q
(h)
where
De = equivalent diameter, m
Q = kinematic viscosity, m 2 /s
The equivalent diameter for a square channel is defined as
De = 4
Substituting (i) into (h)
A
S2
= 4
=S
4S
P
(i)
PROBLEM 6.12 (continued)
ReDe
uS
(j)
Q
Properties of air are determined at the mean temperature Tm defined as
Tm =
Tmi Tmo
2
(k)
Substituting into (k)
Tm =
(40 120)( o C)
2
80o C
Properties of air at this temperature are:
c p = 1009.5 J/kg o C
k = 0.02991 W/m o C
Pr = 0.706
Q = 20.92 u 10 6 m2/s
3
U = 0.9996 kg/m
Substituting into (j)
ReDe
0.32(m/s)0.04(m)
20.92 u 10 6 (m 2 /s)
611.9
Since the Reynolds number is smaller than 2300, the flow is laminar. The heat transfer
coefficient for fully developed laminar flow through a square channel with uniform surface flux
is constant. It is given by equation (6.55) and Table 6.2
Nu De
where h
h De
= 3.608
k
(l)
h . Solving (k) for h
h
3.608
k
De
(m)
(iii) Computations. Substituting numerical values in (e) gives required channel length
L
0.9996(kg/m 3 ) 0.04(m)0.32(m/s) 1009.5(J/kg-o C)(120 40)( o C)
(4) 590( W/m 2 )
0.4378 m
To determine surface temperature at the outlet, the heat transfer coefficient is computed using
(m)
0.02991( W/m o C)
h(L) = h = 3.608
= 2.7 W/m 2 o C
0.04(m)
Equation (g) gives the surface temperature at the outlet
PROBLEM 6.12 (continued)
Ts ( L)
·
§
4(0.4378)( m )
1
¸
¨ 0.9996( kg/m 3 ) 0.04( m)0.32 ( m/s) 1009.5( J / kg o C) 2.7( W/m 2 o C) ¸
¹
©
o
2
40( C) 590( W/m ) ¨
Ts (L) = 338.5 o C
(iv) Checking. Dimensional check: Computations showed that equations (e), (g), (j), and
(m) are dimensionally correct.
Quantitative checks: (1) Alternate approach to determining Ts(L): Application of Newton’s law
of cooling at the outlet gives
qscc = h [Ts(L) - Tmo ]
(n)
solving for Ts(L)
q csc
590( W/m 2 )
o
= 120( C) Ts(L) = Tmo +
h
2.7( W/m 2 o C)
338.5o C
(2) The value of h is within the range reported in Table 1.1 for forced convection of liquids.
Limiting check: If Tmi = Tmo, the required length should be zero. Setting Tmi = Tmo into (e) gives
L = 0.
(5) Comments. (i) As long as the outlet is in the fully developed region, surface temperature at
the outlet is determined entirely by the local heat transfer coefficient. (ii) In solving internal
forced convection problems, it is important to establish if the flow is laminar or turbulent and if it
is developing or fully developed.
PROBLEM 6.13
Consider fully developed laminar flow in two tubes having the same length. The flow rate, fluid,
inlet temperature and surface temperature are the same for both tubes. However, the diameter
of one tube is double that of the other. Determine the ratio of the heat transfer rate from the two
tubes.
(1) Observations. (i) This is an internal forced convection problem in tubes. (ii) The flow is
laminar and fully developed. (iii) The surface is maintained at uniform temperature. (iv) All
conditions are identical for two tubes except the diameter of one is twice that of the other. (v)
The total heat transfer in each tube depends on the outlet temperature.
(2) Problem Definition. Compare the outlet temperatures of the two tubes.
(3) Solution Plan. Apply conservation of energy to obtain an equation for the heat transfer in
each tube. Use the analysis of fully developed laminar flow in tubes at uniform surface
temperature to determine the outlet temperatures.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) fully developed laminar flow, (3) axisymmetric flow,
(4) constant properties, (5) uniform surface temperature, (6) negligible changes in kinetic and
potential energy, (7) no axial conduction, (8) no dissipation and (9) no energy generation.
(ii) Analysis. Application of conservation of energy to the fluid between the inlet and
outlet of tube, gives
q = m c p Tmo Tmi
(a)
where
cp = specific heat, J/kg-oC
= mass flow rate, kg/s
m
q = rate of heat transfer, W
Tmi = inlet mean temperature, oC
Tmo = outlet mean temperature, oC
Applying (a) to the flow in the two tubes, noting that cp, m and Tmi are the same for both tubes,
and taking the ratio of the two results
Tmo 2 Tmi
q2
(b)
q1
Tmo1 Tmi
where the subscripts 1 and 2 refer to the small tube and large tube, respectively. Thus, the
problem becomes one of determining the outlet temperatures. For flow through tubes with
uniform surface temperature, conservation of energy and Newton’s law of cooling lead to
equation (7.13)
Ph
Tm ( x) Ts (Tmi Ts ) exp[
x]
(c)
m cp
2 o
h = average heat transfer coefficient for a tube of length x, W/m - C
= mass flow rate, kg/s
m
P = tube perimeter, m
Tm(x) = mean temperature at x, oC
Tmi = mean inlet temperature, 35oC
PROBLEM 6.13 (continued)
Ts = surface temperature, oC
x = distance from inlet of heated section, m
Applying (c) at the outlet where x = L and Tm(x) = Tmo and rearranging
Tm o Tmi
(Ts Tmi )[1 exp( Ph L / m c p )]
(d)
where
L = tube length, m
Tmo = mean outlet temperature, oC
Examination of (d) shows that all quantities are identical for both tubes except P and h .
Applying (d) to the two tubes and taking the ratio of the resulting equations
Tmo 2 Tmi
Tmo1 Tmi
1 exp( P2 h2 L / m c p )
1 exp( P1h1 L / m c p )
(e)
Thus, the two outlet temperatures will differ according to how the product of P h differs for the
two tubes. The perimeter P is given by
P=SD
(f)
where D is tube diameter. For fully developed laminar flow with uniform surface temperature,
the heat transfer coefficient is uniform along the tube, given by (7.57)
NuD =
hD
= 3.66
k
(g)
where
NuD = Nusselt number
k = thermal conductivity of fluid, W/m-oC
The product P h can now be constructed from (f ) and (g)
P h = 3.66 S D
k
= 3.66 S k
D
(h)
Thus, the product P h is independent of tube size. It follows from (e) that the two outlet
temperatures are identical
Tmo 2 Tmi 1 exp(3.66SkL / m c p )
1
(i)
Tmo1 Tmi 1 exp(3.66SkL / m c p )
Substituting (i) into (b) gives
q 2 / q1 1
(j)
(iii) Checking. Dimensional check: The exponent in equation (d) should be dimensionless.
Ph L
m c p
( m)( W / m2 o C )( m)
( kg / s)( J / kg o C )
W
J/s
W
=1
W
Limiting check: In the limit as L o f , the outlet temperature becomes equal to the surface
temperature regardless of tube size. Setting L = f in (d) gives Tmo = Ts.
(5) Comments. The result is somewhat surprising. n
Oe would expect that increasing the
diameter, increases the heat transfer rate. oHwever, according to (g), the heat transfer coefficient
is inversely proportional to diameter. O
n the other hand, the perimeter is
PROBLEM 6.13 (continued)
(6) directly proportional to diameter. These two effects cancel each other resulting in identical
outlet temperatures regardless of tube size. This is true for the assumptions listed above and as
long as all conditions are the same for both tubes.
PROBLEM 6.14
To evaluate the accuracy of scaling prediction of the thermal entrance length and Nusselt
number, compare scaling estimates with the exact results of Graetz solution for flow through
tubes.
(1) Observations. (i) This is an internal forced convection problem. (ii) Equation (6.3) gives
scaling estimate of the thermal entrance length. (iii) Equation (6.20b) gives scaling estimate of
the local Nusselt number. (iv) The G
raetzproblem d eals with laminar flow in the entrance of a
tube at uniform surface temperature. (v) G
rae tz solutions gives the thermal entrance length
(distance to reach fully developed temperature) and local Nusselt number.
(2) Problem Definition. Determine the thermal entrance length and Nusselt number using
scaling and compare with rGaetzresults.
(3) Solution Plan. (i) Use G
raetzsolution (Table 6.4 or Fig. 6.9) to determine the distance from
the entrance to the section where the Nusselt number is constant (fully developed temperature).
Compare with scaling estimate, equation (6.3) (ii) Use rGaetzso lution (Table 6.4) to determine
the Nusselt number at various distances from the entrance. Compare with scaling estimate,
equation (6.20b).
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties, (3) uniform surface temperature,
(4) negligible axial conduction (5) negligible changes in kinetic and potential energy and (6)
negligible dissipation
(ii) Analysis. Equation (6.3) gives scaling estimate of the thermal entrance length Lt
§ Lt / D ·
¨¨
¸¸
© Re D Pr ¹
1/ 2
~1
(a)
where
D = diameter
Pr = Prandtl number
Re D = Reynolds number
Scaling estimate of the Nusselt number is given by equation (6.20b)
Nu D
1/ 2
~1
(b)
§ PrRe ·
¸
¨
© x/D ¹
G
raetzsolution for the variation of the local Nusse lt number with distance from the entrance is
presented in Table 6.4. Since the Nusselt is constant in the fully developed region, G
raetz
solution can also be used to determine the entrance length. The fully developed Nusselt number
is
Nu D | 3.66
Table 6.4 gives the dimensionless distance [ corresponding to Nu D | 3.66 as
PROBLEM 6.14 (continued)
[
x/D
Re D Pr
Lt / D
Re D Pr
(c)
0 .1
To compare with scaling result (a), equation (c) is rewritten in the same form as (a)
§ Lt / D ·
¨¨
¸¸
© Re D Pr ¹
Thus the rGaetzsolution constant
1/ 2
0.1
(d)
0.316
0.316 is replaced by unity in scaling.
G
raetzsolution, Table 6.4, shows that the Nusse lt number depends on the dimensionless axial
distance [ , defined in (c). Rewriting scaling result (b) in terms of [ , gives
[ Nu D ~ 1
(e)
To facilitate comparison of scaling estimate (e) with rGaetz
solution, Table 6.4 is modified to include
[ Nu D .
Examination the result shown in Table 6.4a shows that
exact values of [ Nu D . range from 0.286 to 1.157. Scaling
predicts these constants to be unity, as shown in (e).
(iv) Checking.
dimensionless.
Dimensional check: All equations are
(5) Comments. (i) To compare scaling estimate with exact
solution for the Nusselt number, it is necessary to cast both
results in the same form. (ii) Scaling estimate of the Nusselt
number is surprisingly good.
Table 6.4a
Local Nusselt number for tube at
uniform surface temperature
[=
x/D
Re D Pr
0.0005
0.002
0.005
0.02
0.04
0.05
0.1
Nu ([ )
12.8
8.03
6.00
4.17
3.77
3.71
3.66
[ Nu ([ )
0.286
0.359
0.424
0.590
0.754
0.830
1.157
PROBLEM 6.15
Use scaling to estimate the heat transfer coefficient for plasma at a distance of 9 cm from the
entrance of a vessel. The mean plasma velocity is 0.042 m/s and the diameter is 2.2 mm.
Properties of plasma are:
3900 J/kg o C , k
cp
0.94 u 10 6 m 2 /s , U
0.5 W/m o C , Q
1040 kg/m 3
(1) Observations. (i) This is an internal forced convection problem. (ii) Equation (6.20b) gives
scaling estimate of the local Nusselt number. (iii) The rGaetzproblem deals with laminar flow in
the entrance of a tube at uniform surface temperature.
(2) Problem Definition. Determine the Nusselt number using scaling.
(3) Solution Plan. Use (6.20b) to estimate the Nusselt number in the entrance region of a tube.
(4) Plan Execution.
(i) Assumptions. (1)
temperature.
Steady state, (2) constant properties and (3) uniform surface
(ii) Analysis. Scaling estimate of the Nusselt number is given by equation (6.20b)
§ PrRe D ·
Nu D ~ ¨
¸
© x/D ¹
1/ 2
(a)
where
D = diameter = 2 mm = 0.002 m
Pr = Prandtl number
Re D = Reynolds number
The Reynolds number is defined as
uD
Re D
(b)
Q
where
u
mean velocity = 0.042 m/s
The Prandtl number is given by
Pr
cpP
c pQU
k
k
where
c p = specific heat = 3900 J/kg o C
k
thermal conductivity = 0.5 W/m o C
Q
kinematic viscosity = 0.94 u 10 6 m 2 /s
U = density = 1040 kg/m 3
The Nusselt number is defined as
(c)
PROBLEM 6.15 (continued)
Nu D
h
hD
k
(d)
heat transfer coefficient
Substitute (d) into (a) and solve for h
1/ 2
k § PrRe D ·
h~ ¨
¸
D © x/D ¹
(iii) Computations. Use (b) and (c) to compute the Reynolds and Prandtl numbers
0.024(m/s) 0.0022 (m)
Re D
0.94 u 10 6 m 2 /s
98.3
3900 (J/kg o C)1040 ( kg/m 3 )0.94 u 10 6 ( m 2 /s)
Pr
(e)
0.5 W/m o C
= 7.63
Substitute into (e)
1/ 2
·
0.5(W/m- o C) §
7.63 u 98.3
¨¨
¸
h~
0.0022(m) © 0.09(m)/0.0022(m) ¸¹
973 W/m 2 - o C
The exact solution to this problem is given in Table 6.4 and Fig. 6.9. The local Nusselt is given
in terms of the dimensionless distance [ , defined as
x/D
[
(f)
Re D Pr
Computing [
[
0.09(m)/0.0022(m)
= 0.05454
7.63 u 98.3
At this value of [ , Table 6.4 gives
Nu D
3 .7
Substituting into (d)
h
3.7
0.5(W/m-o C)
0.0022(m)
841 W/m 2 - o C
(iv) Checking. Dimensional check: (1) Each term in (e) has units of heat transfer coefficient. (2)
Equations (a)-(d) are dimensionless.
(5) Comments. Scaling estimate of the heat transfer coefficient is surprisingly good.
PROBLEM 6.16
Air flows with fully developed velocity through a tube of inside diameter 2.0 cm. The flow is fully
developed with a mean velocity of 1.2 m/s. The surface is maintained at a uniform temperature
of 90oC. Inlet temperature is uniform equal 30 oC. Determine the length of tube needed to
increase the mean temperature to 70oC.
(1) Observations. (i) This is an internal forced convection problem in a tube. (ii) The velocity is
fully developed. (iii) The temperature is developing. (iv) Surface is maintained at uniform
temperature. (v) The Reynolds number should be computed to establish if flow is laminar or
turbulent. (vi) Tube length is unknown.
Lh
(vii) The determination of tube length
Ts
L
requires determining the heat transfer
Tmo
u
coefficient.
u
(2) Problem Definition. Find the
required tube length to increase the air
temperature to a specified level. This
reduces to determining the heat transfer
coefficient.
Tmi
m
Lt
(3) Solution Plan. Compute the Reynolds number to establish that the flow is laminar. Use the
results of Section 6.5 on flow through tubes at uniform surface temperature. Use G
raetzsolution
for fully developed laminar flow and developing temperature in tubes to determine the heat
transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) fully developed velocity, (3) axisymmetric flow, (4)
constant properties, (5) uniform surface temperature, (6) negligible changes in kinetic and
potential energy, (7) negligible axial conduction, (8) negligible dissipation and (9) no energy
generation.
(ii) Analysis.. For flow in a tube at uniform surface temperature, conservation of energy and
Newton's law of cooling lead to equation (6.13)
Tm ( x )
Ts (Tmi Ts ) exp[ Ph
x]
m cp
cp = specific heat, J/kg-oC
2 o
h = average heat transfer coefficient for a tube of length L, W/m - C
L = length of tube, m
m = mass flow rate, kg/s
P = tube perimeter, m
Tm(x) = mean temperature at x, oC
Tmi = mean inlet temperature = 30oC
Ts = surface temperature = 90oC
x = distance from inlet of heated section, m
Applying (6.13) at the outlet (x = L) and solving for L
(6.13)
PROBLEM 6.16 (continued)
L
mc p
Ph
ln
Ts Tmi
Ts Tmo
(a)
where
Tmo = mean outlet temperature = 70oC
To compute L using (b), it is necessary to determine cp, P, m , and h . All properties are
determined at the mean temperature Tm defined as
Tm =
Tmi Tmo
2
(b)
The perimeter P and flow rate m are given by
P=SD
(c)
and
m
S
D2
Uu
4
(d)
where
D = inside tube diameter = 2 cm = 0.02 m
u = mean flow velocity = 1.2 m/s
U = density, kg/m3
The average heat transfer coefficient, h , for fully developed velocity and developing
temperature is given in rGaetzsolution, Section 6. 8 (Table 6.4 and Fig. 6.9). To proceed, it is
necessary to calculate the Reynolds number to determine if the flow is laminar or turbulent. For
flow in a tube the Reynolds number is defined as
uD
ReD
(e)
Q
where
ReD = Reynolds number
Q = kinematic viscosity, m2/s
The mean temperature is calculated in order that properties are determined. Substituting into (b)
Tm =
(30 70)( o C)
2
50 o C
Properties of air at this temperature are
cp = 1007.4 J/kg-oC
k = 0.02781 W/m-oC
Pr = 0.709
Q = 17.92u10-6, m2/s
U = 1.0924 kg/m3
Substituting into (f)
ReD
12
. ( m / s)0.02( m)
17.92 u 10 6 ( m2 / s)
1339.3
PROBLEM 6.16 (continued)
Since the Reynolds number is smaller than 2300, the flow is laminar. Attention is focus on the
determination of h using either Table 6.4 or Fig. 6.9. H
owever, h depends on the length L
which is unknown. Thus the problem is solved by trial and error. Assume L, use Table 6.4 or Fig.
6.9 to determine h and substitute into (a) to calculate L. If the calculated L is not the same as the
assumed value, the procedure is repeated until a satisfactory agreement between assume and
calculated L is obtained.
Table 6.4 gives the average Nusselt number, Nu([ ) , as a function of dimensionless axial distance
[ . These are defied as
hx
k
(f)
[
x/D
PrReD
(g)
h
Nu([ )
Nu([ )
The variable [ is defined in (6.21) as
Solving (f) for h
k
D
(h)
(iii) Computations. Substituting into (d) and (e)
P = S 0.02(m) = 0.06283 m
(0.02) 2 (m 2 )
m ʌ
1.0924(kg/m 3 )1.2(m/s) 0.0004118kg/s
4
The result of the trial and error procedure described above is:
Assume: x = L = 1.15 m. Substitute into (g)
[
1.14( m)/0.02(m)
0.709 u 1339.3
0.0606
At this value of [ Table 6.4 gives
Nu([ )
4.536
Substitute into (h)
h
0.02781(W/m o C)
4.536
0.02(m)
6.307 W/m 2 o C)
Substitute into (a)
L
0.0004118(kg/s)1007.4(J/kg o C ) (90 30)( o C)
ln
0.06283(m)6.307(W/m 2 o C)
(90 70)( o C)
1.15 m
Thus the calculated value of Lis the same as the assumed value.
(iv) Checking. Dimensional check: Computations showed that equations (a), (b), (c), (e),
(g) and (h) are dimensionally consistent.
PROBLEM 6.16 (continued)
Limiting check: For the special case of Tmo
Tmo Tmi in (a) gives x = L = 0.
Tmi , the required length should vanish. Setting
Quantitative checks: The value of h appears to be low compared with typical values listed in
Table 1.1 for forced convection of gases. H
owe ver, it should be remembered that values of h in
Table 1.1 are for typical applications. Exceptions should be expected.
(5) Comments. (i) The thermal entrance length Lt is determined using equation (6.6)
Lt
Ct PrReD
(6.6)
D
Table 6.1 gives Ct 0.033 . Substituting into (6.6) gives Lt 0.627 m. Since this is not small
compared to L , entrance length must be taken into consideration in solving this problem.
(ii) If entrance effects are neglected and temperature is assumed full developed, the
corresponding Nusselt number will be 3.66. Substituting this value in (a) gives L = 1.425 m.
This is 24%
larger than the more accu rate result of entrance length analysis.
PROBLEM 6.17
Air flows with a mean velocity of 2 m/s through a tube of diameter 1.0 cm and length 14 cm. The
velocity is fully developed throughout. The mean temperature at the inlet is 35oC. The surface of
the tube is maintained at a uniform temperature of 130oC. Determine the outlet temperature.
(1) Observations. (i) This is an internal forced convection problem in a tube. (ii) The velocity is
fully developed. (iii) The temperature is developing. (iv) Surface is maintained at uniform
temperature. (v) The Reynolds number should be computed to establish if flow is laminar or
turbulent. (vi) uOtlet mean temperature
is unknown. (vii) The determination of
Lh
Ts
L
outlet temperature requires determining
Tmo
the heat transfer coefficient. (viii) Since
u
u
outlet temperature is unknown, air
m
properties can not be determined. Thus a Tmi
trial and error procedure is needed to
Lt
solve the problem.
(2) Problem Definition. Find the outlet temperature of air heated in tube at uniform surface
temperature. This reduces to determining the heat transfer coefficient.
(3) Solution Plan. Compute the Reynolds number to establish that the flow is laminar. Use the
results of Section 6.5 on flow through tubes at uniform surface temperature. Use G
raetzsolution
for fully developed laminar flow and developing temperature in tubes to determine the heat
transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) fully developed velocity, (3) axisymmetric flow, (4)
constant properties, (5) uniform surface temperature, (6) negligible changes in kinetic and
potential energy, (7) negligible axial conduction, (8) negligible dissipation and (9) no energy
generation.
(ii) Analysis. For flow in a tube at uniform surface temperature, conservation of energy and
Newton's law of cooling lead to equation (6.13)
Tm ( x )
Ts (Tmi Ts ) exp[ Ph
x]
m cp
cp = specific heat, J/kg-oC
2 o
h = average heat transfer coefficient for a tube of length L, W/m - C
m = mass flow rate, kg/s
P = tube perimeter, m
Tm(x) = mean temperature at x, oC
Tmi = mean inlet temperature = 35oC
Ts = surface temperature = 130oC
x = distance from inlet of heated section, m
Applying (6.13) at the outlet (x = L) and solving for Tmo
(6.13)
PROBLEM 6.17 (continued)
Tmo
ª LPh º
Ts (Ts Tmi ) exp « »
¬« mc p ¼»
(a)
where
L = length of tube = 0.14 m
o
Tmo mean outlet temperature, C
To compute Tmo using (b), it is necessary to determine cp, P, m , and h . All properties are
determined at the mean temperature Tm defined as
Tm =
Tmi Tmo
2
(b)
The perimeter P and flow rate m are given by
P=SD
(c)
and
m
S
D2
Uu
4
(d)
where
D = inside tube diameter = 1 cm = 0.01 m
u = mean flow velocity = 2 m/s
U = density, kg/m3
The average heat transfer coefficient, h , for fully developed velocity and developing
temperature is given in rGaetzsolution, Section
6.8 (Table 6.4 and Fig. 6.9). H
owever, this
solution is valid for laminar flow. Thus, to proceed with the analysis, it is necessary to calculate
the Reynolds number to determine if the flow is laminar or turbulent. For flow in a tube the
Reynolds number is defined as
uD
(e)
ReD
Q
where
ReD = Reynolds number
Q = kinematic viscosity, m2/s
Since mean outlet temperature Tmo is unknown, properties can not be determined using (b). A
trial an error procedure is required in which a value for Tmo is assumed, properties determined
using the assumed value and (a) is used to calculate Tmo . If the calculated Tmo is equal to the
assumed value, the procedure is repeated until a satisfactory agreement is obtained.
Assume Tmo
Tm =
65o C . (b) gives
(35 65)( o C)
2
50o C
Properties of air at this temperature are
cp = 1007.4 J/kg-oC
k = 0.02781 W/m-oC
PROBLEM 6.17 (continued)
Pr = 0.709
Q = 17.92u10-6, m2/s
U = 1.0924 kg/m3
Substituting into (f)
2(m/s)0.01(m)
ReD
17.92 u 10 6 (m 2/s)
1116.1
Since the Reynolds number is smaller than 2300, the flow is laminar. Attention is now focus on
the determination of h using either Table 6.4 or Fig. 6.9. Table 6.4 gives the average Nusselt
number, Nu([ ) , as a function of dimensionless axial distance [ . These are defied as
hx
k
(f)
[
x/D
PrReD
(g)
h
Nu([ )
Nu([ )
The variable [ is defined in (6.21) as
Solving (f) for h
k
D
(h)
(iii) Computations. Substituting into (d) and (e)
P = S 0.01(m) = 0.03141 m
m
ʌ
(0.01) 2 (m 2 )
1.0924(kg/m 3 )2(m/s) 0.0001716 kg/s
4
To determine h , use (g) and Table 6.4
[
0.14( m)/0.01(m)
0.709 u 1116.07
0.01769
At this value of [ Table 6.4 gives
Nu([ ) 5.95
Substitute into (h)
h
0.02781(W/m o C)
5.95 16.55 W/m 2 o C)
0.01(m)
Substitute into (a)
Tmo
ª 0.14( m)0.03141( m)16.55( W/m 2 o C) º
130( o C) (130 35) exp « »
0.0001716( kg/s)1007.4( J/kg o C) ¼»
¬«
This is close to the assume value. Thus Tmo
67.6 o C .
67.6 o C
PROBLEM 6.17 (continued)
(iv) Checking. Dimensional check: Computations showed that equations (a), (b), (c), (e),
(g) and (h) are dimensionally consistent.
Limiting check: For the special case of Ts Tmi , the outlet temperature should be the same as
the inlet. Setting Ts Tmi in (a) gives To Tmi .
Quantitative checks: The value of h appears to be low compared with typical values listed in
Table 1.1 for forced convection of gases. H
owe ver, it should be remembered that values of h in
Table 1.1 are for typical applications. Exceptions should be expected.
(5) Comments. (i) The thermal entrance length Lt is determined using equation (6.6)
Lt
D
Ct PrReD
(6.6)
Table 6.1 gives Ct 0.033 . Substituting into (6.6) gives Lt 0.261 m. Since this is not small
compared to L , entrance length must be taken into consideration in solving this problem. In fact
the thermal boundary layer is developing throughout the tube.
(ii) B
ecause outlet temperature is unknown propertie s can not be determined a priori. Thus trial
and error procedure is needed to solve the problem.
PROBLEM 6.18
A research apparatus for a pharmaceutical laboratory requires heating plasma in a tube 0.5 cm
in diameter. The tube is heated by uniformly wrapping an electric element over its surface. This
arrangement provides uniform surface heat flux. The plasma is monitored in a 15 cm long
section. The mean inlet temperature to this
L
section is 18 o C and the mean velocity is 0.025
test
section
D
m/s. The maximum allowable temperature is
u
42 o C. You are asked to provide the designer of
the apparatus with the outlet temperature and
required power corresponding to the maximum
Tmi
Tmo
+
temperature. Properties of plasma are:
cp
3900 J/kg o C , k
0.5 W/m o C , Q
0.94 u 10 6 m 2 /s , U
1040 kg/m 3
(1) Observations. (i) This is an internal forced convection problem in a tube. (ii) The velocity is
fully developed and the temperature is developing. (iii) The surface is heated with uniform flux.
(iv) The Reynolds number should be computed to establish if the flow is laminar or turbulent. (v)
Compute thermal entrance length to determine if it can be neglected. (vi) Surface temperature
varies with distance from entrance. It is maximum at the outlet. Thus surface temperature at the
outlet is known. (vii) Analysis of uniformly heated channels gives a relationship between local
surface temperature, heat flux and heat transfer coefficient. (viii) The local heat transfer
coefficient varies with distance form the inlet. (ix) Knowing surface heat flux, the required
power can be determined. (x) Newton’s law of cooling applied at the outlet gives outlet
temperature.
(2) Problem Definition. Determine the local heat transfer coefficient at the outlet h(L).
(3) Solution Plan. Apply channel flow heat transfer analysis for uniform surface flux to
determine surface heat flux. Compute the Reynolds number to establish if the flow is laminar or
turbulent. Compute entrance length to determine if it can be neglected. Select an applicable
equation for determining the Nusselt number at the outlet. Apply Newton’s law of cooling at the
outlet to determine outlet temperature.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties, (e) uniform surface heat flux, (4)
negligible changes in kinetic and potential energy, (5) negligible axial conduction and (6)
negligible dissipation.
(ii) Analysis. Equation (6.10) gives the result of heat transfer analysis for channel flow with
uniform surface heat
§ Px
1 ·¸
Ts ( x ) Tmi q scc ¨
¨ m c p h( x ) ¸
©
¹
where
c p = specific heat, 3900 J/kg o C
h(x) = local heat transfer coefficient, W/m 2 o C
(6.10)
PROBLEM 6.18 (continued)
m = mass flow rate, kg/s
P = tube perimeter, m
Tmi
mean inlet temperature = 18 o C
qscc = surface heat flux, W/m 2
Ts (x) = local surface temperature, o C
x = distance from inlet of heated section, m
Apply (6.10) at the outlet, x = L, and solve for qscc
q scc
§
·
>Ts ( L) Tmi @ ¨¨ PL 1 ¸¸
© m c p h ( L) ¹
1
(a)
where
L = 0.15 m
Ts ( L )
42 o C
The perimeter P is
P SD
(b)
where
D = tube diameter = 0.005 cm
The flow rate is given by
SD 2
m
4
Uu
(c)
where
mean velocity = 0.025m/s
U density = 1040 kg/m3
u
The heat transfer coefficient at the outlet, h(L). This requires establishing whether the flow is
laminar or turbulent. Thus, the Reynolds number should be computed
uD
Re D
Q = kinematic viscosity
(d)
Q
0.94 u 10 6 m 2 /s
The Peclet number, Pe, is computed to determine if axial conduction can be neglected
Pe
(e)
PrRe D
The Prandtl number, Pr, is given by
Pr
cpP
c p UQ
k
k
(f)
where P is viscosity.
The thermal entrance length Lt is determined using equation (6.6)
Lt
D
Ct PrReD
(6.6)
PROBLEM 6.18 (continued)
where Ct is a constant given in Table 6.1. For tubes at uniform surface heat flux, Ct
0.033 .
To determine the required power, the surface flux is multiplied by the total surface are
Power SDLqscc
(g)
To determine outlet temperature, apply Newton’s law of cooling at the outlet
h( L)>Ts Tmo @
q scc
Solve for Tmo
Tmo
Ts qscc
h(L)
(h)
Substitute into (c)
Re D
0.025(m/s)0.005(m)
0.94 u 10 6 (m 2 /s)
133
Thus the flow is laminar. Substitute into (f)
Pr
3900(J/kg-o C)1040(kg/m 3 )0.94 u 10 6 (m 2 /s)
0.5(W/m 2 - o C)
7.625
Substitute into (e)
Pe 7.625 u 133 1014
Thus axial conduction can be neglected.
Equation (6.6) is used to compute Lt
Lt
0.033 u 0.005( m)7.625 u 132 0.166 m
Therefore, the thermal boundary layer is still developing at the outlet. It follows that entrance
effects are important and that the heat transfer coefficient at the outlet should be obtained from
Fig. 6.11. For laminar flow in the entrance region of a tube at fully developed velocity profile
and uniform heat flux Fig. 6.11 gives the local Nusselt numbers Nu(x) as a function of
dimensionless axial distance [ , defined as
[
x/D
Re D Pr
(i)
The local heat transfer coefficient is given by
h( x )
k
Nu ( x)
D
(iii) Computation. Compute [ at x = L
[
At [
0.15( m ) / 0.005( m )
133 u 7.625
0.0296
0.0296 , Fig. 6.11 gives Nu D ( L) | 5 . Equation (j) gives
(j)
PROBLEM 6.18 (continued)
0.5(W/m o C)
5 500 W/m 2 o C
0.005(m)
h ( L)
Equation (b) is used to compute P
P S (0.005)( m)
0.01571 m
Equation (c) gives the flow rate
S (0.005) ( m 2 )
m
4
1040( kg/m3 )0.025( m/s) 0.0005105 kg/s
Substitute into (a)
qscc
·
§
0.0157( m)0.15( m)
1
¸
42( C) 18( C) ¨
¨ 0.0005105( kg/s) 3900( J/kg-o C) 500 W/m 2 o C ¸
¹
©
>
o
o
@
1
7540 W/m 2
Substitute into (g)
Power S 0.005( m)0.15( m)7540( W/m 2 ) 17.77 W
Substitute into (h)
Tmo
42( o C) 7540( W/m 2 )
500( W/m 2 o C)
26.9 o C
(iv) Checking. Dimensional check: (i) Computations showed that equations (a), (b), (c), (e), (g),
(h) and (j) are dimensionally consistent. (ii) Equations (d), (f) and (i) are dimensionless.
Limiting checks: For the special case of Ts Tmi , the required surface flux should vanish and
Tmo Tmi . Setting Ts Tmi in (a) gives q csc 0. When this result is substituted into (g) gives
Tmo
Tmi .
Quantitative checks: (i) The value of h within the range listed in Table 1.1 for forced convection
of liquids.
Global energy balance: energy added at the surface (power) should be equal to energy change of
mass flow rate
SDLqscc mc p (Tmo Tmi )
or
SDLq scc
mc p (Tmo Tmi )
1
The above gives
S 0.005( m)0.15( m)7540( W/m 2 )
0.0005105( kg/s)3900( J/kg-o C)( 26.9 18)( o C)
1.0026
(5) Comments. (i) Using Fig. 6.1 to determine h introduces a small error. (ii) If entrance effects
are neglected and the temperature is assumed fully developed at the outlet, the corresponding
PROBLEM 6.18 (continued)
Nusselt number is 4.364. Using this value gives h 436.4 W/m2 o C, qscc 6978 W/m 2 , power =
16.45 W and Tmo
28 o C.
(iii) In solving internal forced convection problems, it is important to establish if the flow is
laminar or turbulent and if it is developing or fully developed.
PROBLEM 6.19
An experiment is designed to investigate heat transfer in rectangular ducts at uniform surface
temperature. One method for providing heating at uniform surface temperature is based on
wrapping a set of electric elements around the surface. Power supply to each element is
individually adjusted to provide uniform surface temperature. This experiment uses air flowing
in a 4 cm u 8 cm rectangular duct 32 cm
L
long. The air is to be heated from 22 o C
- - - -- to 98 o C. The velocity is fully developed
a
with a mean value of 0.15 m/s. Your task
u
b
is to provide the designer of the
experiment with the heat flux distribution
+ + + + +
along the surface. This data is needed to
Tmo
Tmi
determine the power supplied to the
individual elements.
(1) Observations. (i) This is an internal forced convection problem in a rectangular channel. (ii)
The velocity is fully developed and the temperature is developing. (iii) The surface is maintained
at uniform temperature. (iv) The Reynolds number should be computed to establish if the flow is
laminar or turbulent. (v) Compute entrance lengths to determine if they can be neglected. (vi)
Surface flux varies with distance from entrance. It is minimum at outlet. (vii) Newton’s law
gives surface flux in terms of the local heat transfer coefficient h(x) and the local mean
temperature Tm (x) . (viii) The local and average heat transfer coefficient decrease with distance
form the inlet. (ix) The local mean temperature depends on the local average heat transfer
coefficient h (x). (x) Surface temperature is unknown.
(2) Problem Definition. Determine the local and average heat transfer coefficient temperature to
a specified level. This requires determining the local heat transfer coefficients h(x) and h (x) and
surface temperature.
(3) Solution Plan. Apply Newton’s law of cooling. Use the analysis of flow in channels at
uniform surface temperature to determine the local mean temperature Tm (x). Compute the
Reynolds number to establish if the flow is laminar or turbulent. Compute entrance lengths to
determine if entrance or fully developed analysis is required. If the thermal entrance can be
neglected, use fully developed Nusselt number results. If entrance region is significant, use
rGaetz solution to determine the local a nd average heat transfer coefficients.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties, (e) uniform surface temperature,
(4) negligible changes in kinetic and potential energy, (5) negligible axial conduction and (6)
negligible dissipation.
(ii) Analysis. Newton’s law of cooling gives
q csc ( x)
h( x)>Ts Tm ( x)@
where
h(x) = local heat transfer coefficient, W/m 2 o C
(a)
PROBLEM 6.19 (continued)
Ts
q csc (x) = local surface heat flux, W/m 2
o
Tm (x) = local mean temperature, C
Tmi
u
o
Ts = surface temperature, C
0
Tm ( x)
a
x
For flow through a channels at uniform
surface temperature, conservation of
energy and Newton's law of cooling lead
to equation (6.13)
b
Gt
qcsc
Ts (Tmi Ts ) exp[
Ph
x]
mcp
(6.13)
where
c p = specific heat, J/kg o C
h = average heat transfer coefficient for a channel of length x, W/m 2 o C
m = mass flow rate, kg/s
P = duct perimeter, m
Tmi mean inlet temperature 22 o C
x = distance from inlet, m
Substituting (6.13) into (a)
q csc ( x)
h( x)(Ts Tmi ) exp[
Ph ( x )
x]
mcp
(b)
Thus we need to determine: c p , P, m, Ts , h(x) and h (x). Properties are determined at the mean
temperature Tm , defined as
Tmi Tmo
2
Surface temperature is determined by applying (6.13) at the outlet (x = L) where Tm ( L)
Tm =
(c)
Tmo ,
and solving for Ts
Ts
1
Tmi Tmo exp( Ph L / mc p )
1 exp( Ph L / mc p )
>
@
(d)
where
L = channel length = 32 cm = 0.32 m
Tmo = outlet temperature = 98 o C
The perimeter P is
P
where
a = channel height = 8 cm = 0.08 m
b = channel width = 4 cm = 0.04 m
2(a b)
(e)
PROBLEM 6.19 (continued)
aMss flow rate is given by
U (ab)u
m
(f)
where
U = density, kg/m 3
u = mean velocity = 0.15 m/s
The determination of h(x) and h (x) requires computing the Reynolds number to establish if the
flow is laminar or turbulent and computing the thermal entrance lengths to determine if it is
important. The Reynolds number is
u De
Re D e
(g)
Q
where
De = equivalent diameter, m
Q = kinematic viscosity, m2/s
The equivalent diameter is defined as
Af
ab
2ab
(h)
P
2(a b) (a b)
is flow area. To proceed with the analysis the Reynolds number must be computed
De
where A f
4
4
first. Properties are determined at Tm
(22 98)( o C)
2
Tm =
60 o C
Properties of air at this temperature are
c p = 1008 /Jkg- oC
k = 0.02852 W/m-oC
Pr = 0.708
-6
Q = 18.9u10 m2/s
3
U = 1.0596 kg/m
Equation (h) gives
De
2(0.08)(m)0.04)(m)
(0.08 0.04)(m)
0.05333 m 2
Substituting into (g)
Re D 2
0.15(m/s)0.053333(m)
18.9 u 10 6 (m 2 /s)
423.28
Since the Reynolds number is less than 2300, the flow is laminar. The next step is to compute
the thermal entrance length Lt . For laminar flow through channels equation (6.6) gives
Lt
C t De PrRe De
(6.6)
PROBLEM 6.19 (continued)
where
Ct = thermal entrance length coefficient, uniform surface temperature (Table 6.1) = 0.049
Substituting numerical values into (6.6)
Lt = 0.049 u 0.053333 (m) u 423.28 u 0.708 = 0.783 m
Since Lt is larger than channel length L, it follows that entrance effects must be taken into
consideration in determining h(x) and h ( x). For laminar flow in the entrance region of a tube at
fully developed velocity profile and uniform surface temperature, rGaetz solution gives h()x and
h ( x). Fig. 6.9 and Table 6.4 give the average the local and average Nusselt numbers Nu(x) and
Nu as a function of dimensionless axial distance [ , defined as
[
x / De
Re De Pr
(i)
The average heat transfer coefficient h (x) is given by
k
Nu ( x)
D
h ( x)
(j)
Similarly, the local heat transfer coefficient is given by
h( x )
k
Nu ( x)
D
(k)
(iii) Computation. Surface temperature is determined using (d). This requires computing
h (L). Thus we compute [ at x = L
[
At [
0.32(m) / 0.053333(m)
423.28 u 0.708
0.02 , Table 6.4 gives Nu
0.02002
5.81
Substituting into (j)
h ( L)
0.02852(W/m o C)
5.81 3.11 W/m 2 o C
0.053333(m)
Equation (c) is used to compute P
P
2>(0.08)(m) (0.04)(m)@ 0.24 m
Equation (d) gives the flow rate
m 1.0596(kg/m 3 )(0.08)(m)(0.04)(m)0.15(m/s)
0.0005086 kg/s
Before substituting into (d), the exponent of the exponential is calculated
Ph L
mc p
0.24(m)( 3.11)(W/m 2 o C)0.32(m)
0.0005086(kg/s)1008(J/kg o C)
0.4659
PROBLEM 6.19 (continued)
Substituting into (d)
>
1
22( o C) 98( o C) exp(0.4659
1 exp(0.4659)
Ts
@
226.07 o C
Equation (b) is used to determine the heat flux variation with x. The procedure is to select a value
of x, use (i) to compute the corresponding [ , use Table 6.4 or Fig. 6.9 to determine h(x) and
h (x). Substituting in (b) gives the local heat flux. For example, to determine the heat flux at
x L, we apply (b) at x L
q csc ( L)
Table 6.4 gives the local Nusselt number Nu(L) at x = L ( [
0.02852(W/m o C)
4.17
0.053333(m)
Substituting into (l)
h( L )
q csc ( L)
Ph ( L)
L]
mcp
h( L)(Ts Tmi ) exp[
(l)
0.02 ) as Nu
4.17 . Thus
2.23 W/m 2 o C
2.23( W/m 2 o C)(226.07 22)( o C) exp(0.4659)
285.6 W/m 2
The flux at other locations x along the channel is given in the following table.
x
(m)
[
0
0.008
0.016
0.032
0.08
0.12
0.16
0.32
0
0.0005
0.001
0.002
0.005
0.0075
0.01
0.02
Nu(x)
h(x)
( W/m 2 o C )
Nu (x)
h (x)
q cscc(x)
W/m 2
f
1344.9
1024
794.6
547.9
471.9
411.8
285.6
f
f
f
( W/m 2 o C )
f
12.8
10
8.03
6.0
5.5
5
4.17
6.85
5.35
4.29
3.21
2.94
2.67
2.23
19.29
16
12.09
8.92
8
7
5.81
10.32
8.56
6.47
4.77
4.28
3.74
3.11
(iv) Checking. Dimensional check: (i) Computations showed that equations (a)-(k) are
dimensionally consistent. (ii) The Reynolds number and the exponent of the exponential are
dimensionless.
Limiting checks: For the special case of Tmi
vanish. Setting Tm ( x) Ts in (a) gives q csc 0.
Tm ( x)
Ts , the required surface flux should
Qualitative check: As anticipated, the local and average heat transfer coefficients and surface
heat flux decrease with distance from the inlet.
Quantitative checks: (i) The value of h is outside the range listed in Table 1.1 for forced
convection of gases. Examination of equation (j) of (k) shows that the heat transfer coefficient is
inversely proportional to the diameter. Thus, as diameter increases the heat transfer coefficient
decreases.
PROBLEM 6.19 (continued)
(5) Comments. (i) Using Fig. 6.9 to determine h and h introduces a small error. (ii) If entrance
effects are neglected and the flow is assumed fully developed throughout, the corresponding
Nusselt number is 3.66. Using this value gives h 1.96 W/m 2 o C.
PROBLEM 7.1
Explain why
(a) G t can not be larger than G .
(b) G can be larger than G t .
Solution
[a] The driving force in free convection is
buoyancy. Thus, wherever the local temperature
is different from the ambient temperature, the
fluid will move. This means that G t can not be
larger than G .
[b] Fluid inside the thermal boundary layer moves due to buoyancy force. Because of viscous
forces this moving fluid drags fluid layers outside the thermal boundary layer and cause it to
move. That is, fluid motion outside G t is due to viscous force and not buoyancy. Consequently,
G can be larger than G t .
PROBLEM 7.2
A vertical plate 6.5 cm high and 30 cm wide is maintained at
82 o C . The plate is immersed in water at 18 o C . Determine:
(a) The viscous boundary layer thickness.
(b) The thermal boundary layer thickness at the trailing end of Tf
g
the plate.
(c) The average heat transfer coefficient.
(d) Total heat added to water.
L
Ts
W
(1) Observations. (i) This is an external free convection problem over a vertical plate. (ii) The
Rayleigh number should be computed to determine if the flow is laminar or turbulent. (iii) The
solution for laminar flow is given in Section 7.4. (iv) For laminar flow, Fig.7.2 gives the viscous
boundary layer thickness G and Fig. 7.3 gives the thermal boundary layer thickness G t . (v)
Newton’s law of cooling gives the heat transfer rate. (vi) Equation (7.23) gives the average heat
transfer coefficient h . (vii) The fluid is water.
(2) Problem Definition. Determine flow and heat transfer characteristics for free convection
over a vertical plate at uniform surface temperature.
(3) Solution Plan. Compute the Rayleigh number to determine if the flow is laminar. For
laminar flow use the result of Section 7.4 to determine G , G t and h. Apply Newton's law of
cooling to determine the total heat transfer qT .
(4) Plan Execution.
(i) Assumptions. (1) Continuum, (2) Newtonian, (3) steady state, (4) two-dimensional, (5)
constant properties (except in buoyancy), (6) boundary layer flow, (7) laminar flow, (8) uniform
surface temperature, (9) negligible radiation and (10) quiescent fluid.
(ii) Analysis. The Rayleigh number RaL is calculated first to determine the appropriate
correlation equation for h . The Rayleigh number is defined as
RaL=
E g Ts Tf L3
ǎ2
Pr
(a)
where
g = gravitational acceleration = 9.81 m/s2
L = plate dimension in the direction of gravity = 6.5 cm = 0.065 m
Pr = Prandtl number
RaL = Rayleigh number
E = coefficient of thermal expansion, 1/K
v = kinematic viscosity, m2/s
Properties of air are determined at the film temperature Tf defined as
Tf =
Ts Tf
2
(b)
PROBLEM 7.2 (continued)
Ts = surface temperature = 82oC
Tf = ambient air temperature = 18oC
Viscous boundary layer thickness G is determined using Fig. 7.2. This requires computing the
following:
1/ 4
y
§ Gr ·
(c)
Ș ¨ x¸
x
© 4 ¹
and
E g Ts Tf x 3
(d)
Grx
2
ǎ
Thermal boundary layer thickness G t is determined using Fig. 7.3. The average heat transfer
coefficient is given by equation (7.27)
1/ 4
h
4 k § GrL ·
¨
¸
3 L© 4 ¹
dT (0)
dK
(7.27)
h = average heat transfer coefficient, W/m2-oC
k = thermal conductivity, W/m-oC
Temperature gradient dT (0) / dK is given in Table 7.1. Newton’s law of cooling gives the total
heat transfer rate
qT = h A (Ts - Tf)
(e)
where
A = surface area of the two vertical sides, m2
qT = heat transfer from the surface to the ambient air, W
Surface area of the two vertical sides is given by
A = 2 LW
W = plate width = 30 cm = 0.3 m
(iii) Computations. Properties of air are determined at the film temperature Tf defined as
Tf =
Ts Tf
(82 18)( o C)
= 50oC
=
2
2
Water properties at this temperature
k = 0.64056 W/m-oC
Pr = 3.57
E 0.462 u 10 3 1/K
ǎ
0.5537 u 10 6 m2/s
Substituting into (a)
RaL =
0.462 u 10 3 (1/K)9.81(m/s 2 )(82 18)( o C)(0.065) 3 (m 3 )
3.57 = 0.927577 u 10 9
6 2
4 2
(0.5537 u 10 ) (m /s )
(f)
PROBLEM 7.2 (continued)
At Pr = 3.57, Fig. 7.2 gives
§ Gr ·
Ș ¨ L¸
© 4 ¹
1/ 4
G
L
(g)
|5
Use (d) to compute GrL at the trailing end x = L
GrL
0.462 u 10 3 (1/K)9.81(m/s 2 )(82 18)( o C)(0.065) 3 (m 3 )
0.259826 u 10 9
4 2
6 2
(0.5537 u 10 ) (m /s )
Substituting into (g)
5
§ 0.259826 u 10 9 ·
¨
¸
¨
¸
4
©
¹
1/ 4
G
0.065(m)
Solving the above for G
G
3.62 u 10 3 m
3.62 mm
Fig. 7.3 gives the thermal boundary layer thickness G t . At Pr = 3.57, Fig. 7.3 gives
§ 0.259826 u 10 9 ·
¸
K t | 2.5 ¨¨
¸
4
©
¹
1/ 4
Gt
0.065(m)
Solving for G t 1.81 u 10 3 m 1.81 mm
At Pr = 3.57, Table 7.1 gives the gradient dT (0) / dK
dT (0)
| 0.86
dK
Substituting into (7.27)
h
4 0.6405(W/m o C) § 0.25982 u 10 9 ·
¨
¸
¨
¸
3
0.065(m)
4
©
¹
1/4
0.86 1014
W
2
m o C
Substituting into (e)
qT = 1014(W/m 2 o C)0.065(m)0.3(m)(82 18)( o C) 1265 W
(iv) Checking.
Dimensional check: Computations showed that equations (a)-(g) are
dimensionally consistent.
Quantitative check: The magnitude of h is within the range given in Table 1.1 for free
convection of gases.
Qualitative check: Increasing surface temperature Ts should increase the heat transfer rate.
According to equation (e), qT is directly proportional to h and Ts. According to (7.27), an
increase in Ts brings about an increase in h .
(5) Comments. The computed values of G and G t are approximate since the corresponding
values of K cannot be read accurately from Fig. 7.2 and Fig. 7.3.
PROBLEM 7.3
Use Fig. 7.3 to determine dT (0) / dK for Pr = 0.01 and 100. Compare your result with the value
given in table 7.1.
(1) Observations. (i) This is an external free convection problem for flow over a vertical plate.
(ii) Laminar flow solution for temperature distribution for a plate at uniform surface temperature
is given in Fig. 7.3 . (iii) The dimensionless temperature gradient at the surface is given in Table
7.1. (iv) The solution depends on the Prandtl number.
(2) Problem Definition. Determine the normal temperature gradient at the surface, dT (0) / dK ,
for laminar boundary layer flow over a vertical plate.
(3) Solution Plan. Use the temperature solution presented graphically in Fig. 7.3 and compare
with the exact value listed in Table 7.1.
(4) Plan Execution.
(i) Assumptions (a)
(i) Assumptions. (1) Newtonian fluid, (2)
steady state, (3) Boussinesq approximations, (4)
two-dimensional, (5) laminar flow ( Rax 10 9 ),
(6) vertical flat plate (7) uniform surface
temperature, (8) no dissipation and (9) no
radiation.
(ii) Analysis. Fig. 7.3 is a plot of the
dimensionless temperature T vs. the similarity
variable K . Temperature gradient dT (0) / dK
can be determined from this figure by
graphically evaluating the slope is the slope at
K 0. This slope is expressed as
dT (0)
dK
'T
'K K 0
(a)
(iii) Computations. For Pr = 0.01, equation (a) and Fig. 7.3 give
dT (0)
dK
'T
1 0.545
|
'K K 0
6
Table 7.1 gives
dT (0)
dK
0.076
0.0806 .
For Pr = 100, equation (a) and Fig. 7.3 give
PROBLEM 7.3 (continued)
dT (0)
dK
'T
1
|
'K K 0
0.49
Table 7.1 gives
dT (0)
dK
2.04
2.191 .
(iv) Checking. u
Qantitative check: Numerical results
agreement with the exact solutions of Table 7.1.
obtained using Fig. 7.3 are in good
Comments. Using Fig. 7.3 to determine dT (0) / dK has an inherent error associated with reading
its scale. Nevertheless, for Prandtl numbers 0.01 and 100 the error is less than 7%.
PROBLEM 7.4
In designing an air conditioning system for a pizza restaurant an estimate of the heat added to
the kitchen from the door of the pizza oven is needed.
The rectangular door is
50 cm u 120 cm with its short side along the vertical direction. Door surface temperature is
110oC. Estimate the heat loss from the door if the ambient air temperature is 20oC.
(1) Observations. (i) This is a free convection problem. (ii) Heat is lost from the door to the
surroundings by free convection and radiation. (iii) To determine the rate of heat loss, the door
can by modeled as a vertical plate losing heat by free convection to an ambient air. (iv) As a
first approximation, radiation can be neglected. (v) Newton’s law of cooling gives the rate of
heat transfer. (vi) The Rayleigh number should be computed to determine if the flow is laminar
or turbulent. (vii) For laminar flow the solution of Section 7.4 is applicable.
(2) Problem Definition. Determine the average heat transfer coefficient for free convection from
a vertical plate.
(3) Solution Plan. Apply Newton's law of cooling to the door. For laminar flow use results of
Section 7.4.
(4) Plan Execution.
(i) Assumptions. (1) steady state, (2) two-dimensional, (3) constant properties (except in
buoyancy), (4) uniform surface temperature, (5) negligible radiation, (6) quiescent ambient fluid
and (7) door is in the closed position at all times.
(ii) Analysis. Application of Newton's law of cooling to the surface gives
q = h A ( Ts - Tf)
(a)
where
A = surface area, m2
h = average heat transfer coefficient, W/m2-oC
q = heat transfer rate, W
Ts = surface temperature = 110oC
Tf = ambient air temperature = 20oC
L
Tf
g
Surface area is given by
Ts
W
A = LW
(b)
where
L = door height = 50 cm = 0.5 m
W = door width = 120 cm = 1.2 m
To determine the average heat transfer coefficient h , the Rayleigh number RaL is calculated first
to determine if the flow is laminar or turbulent.
PROBLEM 7.4 (continued)
RaL =
E g Ts Tf L3
Q2
(c)
Pr
where
g = gravitational acceleration = 9.81 m/s2
L = door side in the direction of gravity = 50 cm = 0.5 m
Pr = Prandtl number of air
RaL = Rayleigh number
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
For ideal gases, the coefficient of thermal expansion E is given by
E=
1
Tf (K )
1
o
20( C) 273.15
0.003411 (1/K)
Properties are determined at the film temperature Tf defined as
Tf =
Ts Tf
(110 20)( o C)
= 65oC
=
2
2
Properties of air at this temperature are
k = 0.02887 W/m-oC
Pr = 0.7075
Q = 19.4u10-6 m2/s
Substituting into (c) gives
RaL =
0.003411(1/ o C)9.81(m / s 2 )(110 20)( o C)(0.5) 3 (m 3 )
6 2
4
2
(19.4 u 10 ) (m / s )
0.7075 = 0.70766u109
Since RaL < 109, the flow is laminar and h is given by (7.23)
1/ 4
h
4 k § GrL ·
¨
¸
3 L© 4 ¹
dT (0)
dK
(7.23)
where GrL is the G
rashof number given by
GrL =
E g Ts Tf L3
Q
2
Ra L
Pr
(d)
and dT (0) / dK is a dimensionless temperature gradient which depends on the Prandtl number. It
is listed in Table 7.1.
(iii) Computations. Substitution into (d) gives
GrL
0.70766 u 10 9
0.7075
1.00226 u 10 9
PROBLEM 7.4 (continued)
At Pr = 0.7075, Table 7.1 gives
dT (0)
dK
0.501
Substitute into (7.23)
h
4 0.02887(W/m o C) §¨ 1.00226 u 10 9
¨
3
0.5(m)
4
©
1/ 4
·
¸
¸
¹
0.501
4.853 W/m 2 o C
Substitute into (a) and use (b) give the heat transfer rate from door
q = 4.853 (W/m2-oC) (0.5)(m)(1.2)(m) (110 20) (oC) = 261.9 W
(iii) Checking. Dimensional check: Computations showed that equations (a), (c) and (7.23)
are dimensionally consistent.
Quantitative check: The magnitude of h is within the range given in Table 1.1.
(5) Comments. (i) The model used to solve this problem is not conservative due to neglecting
radiation. If the door is treated as a small object enclosed by a much larger surface at the
ambient temperature, an estimate can be made of the radiation heat loss. Assuming that the door
is made of stainless steel with an emissivity of 0.25, radiation heat loss will be 120 W. This is
significant when compared with free convection heat loss. (ii) Opening and closing the door
results in transient effects not accounted for in the above model. In addition, when the door is
open radiation from the interior of oven may be significant.
PROBLEM 7.5
To compare the rate of heat transfer by radiation with that by free convection, consider the
following test case. A vertical plate measuring 12 cm u 12 cm is maintained at a uniform
surface temperature of 125oC. The ambient air and the surroundings are at 25oC. Compare the
two modes of heat transfer for surface emissivities of 0.2 and 0.9. A simplified model for heat
loss by radiation q r is given by
qr
4
)
H V A(Ts4 Tsur
where A is surface area, H is emissivity and V 5.67 u 10 8 W/m 2 K 4 . Surface and
Surroundings temperatures are measured in degrees kelvin
(1) Observations. (i) This is a free convection and radiation problem. (ii) The geometry is a
vertical plate. (iii) Surface temperature is uniform. (iv) Newton’s law of cooling gives
convection heat transfer rate while Stefan-Boltzmann law gives radiation heat transfer rate. (v)
The Rayleigh number should be computed to determine if the flow is laminar or turbulent. (vi)
For laminar flow the solution of Section 7.4 is applicable. (vii) Since radiation heat transfer is
considered in this problem, all temperatures should be expressed in degrees kelvin. (viii) The
fluid is air.
(2) Problem Definition. Determine heat transfer rate by free convection and by radiation from
a vertical plate in air. Convection requires the determination of the heat transfer coefficient.
(3) Solution Plan. Apply Newton's law of cooling to determine the rate of heat loss by
convection. Apply Stefan-Boltzmann radiation law to determine the rate of heat loss by
radiation. Compute the Rayleigh number and select an appropriate correlation equations to
obtain the average heat transfer coefficient. For laminar flow use results of Section 7.4.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in
buoyancy), (4) uniform surface temperature, (5) the plate is a small surface enclosed by a much
larger surface at a uniform temperature and (6) quiescent ambient.
(ii) Analysis. Application of Newton's law of cooling to the vertical plate gives
qc
h A(Ts Tf )
where
A = surface area of vertical side, m2
h = average heat transfer coefficient, W/m2-oC or W/m2-K
qc = convection heat transfer rate, W
Ts = surface temperature = 125(oC) + 273.15 = 398.15 K
Tf = ambient temperature = 25(oC) + 273.13 = 298.15 K
Surface area is
surroundin gs
(a)
Tsur
L
Tf
g
Ts
L
PROBLEM 7.5 (continued)
A = L2
(b)
where
L = side of square plate = 12 cm = 0.12 m
To determine the average heat transfer coefficient h , the Rayleigh number RaL is calculated first
to determine if the flow is laminar or turbulent.
RaL =
E g Ts Tf L3
Q2
(c)
Pr
where
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number of air
RaL = Rayleigh number
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
Properties are determined at the film temperature Tf defined as
Tf =
(29815
. 39815
. )( K )
Ts Tf
=
= 348.15K
2
2
Properties of air at this temperature are
k = 0.02957 W/m-oC
Pr = 0.7065
Q = 20.41u10-6 m2/s
For ideal gases the coefficient of thermal expansion E is given by
E=
1
T f (K )
1
348.15
0.002872 (1/K)
Substituting into (c) gives
RaL =
0.002872(1/ o C)9.81(m/ s 2 )(125 25)( o C)(0.12) 3 (m 3 )
6 2
4
2
(20.41 u 10 ) (m / s )
0.7065 = 8.257 u 10 6
Since RaL < 109, the flow is laminar and h is given by (7.23)
1/ 4
h
4 k § GrL ·
¨
¸
3 L© 4 ¹
dT (0)
dK
(7.23)
rashof number given by
where GrL is the G
GrL =
E g Ts Tf L3
Q
2
Ra L
Pr
(d)
and dT (0) / dK is a dimensionless temperature gradient which depends on the Prandtl number. It
is listed in Table 7.1.
PROBLEM 7.5 (continued)
Radiation heat loss q r is given by the Stefan-Boltzmann law. Assuming that the plate is a small
surface which is surrounded by a much larger surface, q r is given by
4
q r = H V A ( Ts4 Tsur
)
(e)
where
q r = radiation heat loss, W
Tsur = surroundings temperature = 25(oC) + 273.13 = 298.15 K
H = emissivity
V = Stefan-Boltzmann constant = 5.67u10-8 W/m2-K4
(iii) Computations. Convection heat loss: substitution into (d) gives
8.257 u 10 6
0.7065
GrL
11.687 u 10 6
At Pr = 0.7065, Table 7.1 gives
dT (0)
dK
0.5009
Substitute into (7.23)
4 0.02957(W/m o C) §¨ 11.687 u 10 6
¨
3
0.12(m)
4
©
1/ 4
·
¸ 0.5009 6.804 W/m 2 o C
h
¸
¹
Substitute into (a) and use (b) give the heat transfer rate from door
qc = 6.804 (W/m2-oC) (0.12)(m)(0.12)(m) (125 -25)(oC) = 9.798 W
Radiation heat loss: equation (e) for H = 0.2, equation (e) gives
q r = 0.2u5.67u10-8 (W/m2-K4) 0.12(m) 0.12(m) ( 398.15 4 298.15 4 )(K4) = 2.81 W
For H = 0.9:
q r = 0.9u5.67u10-8 (W/m2-K4) 0.12(m) 0.12(m) ( 398.15 4 298.15 4 )(K4) = 12.66 W
(iv) Checking. Dimensional check: Computations showed that equations (a), (c), (7.23)
and (e) are dimensionally consistent.
Quantitative check: The magnitude of h is within the range given in Table 1.1 for free
convection of gases.
Limiting check: For Ts Tf Tsur , heat transfer by convection and radiation vanish. Setting
Ts Tf Tsur in (a) and (e) gives qc q r 0.
(5) Comments. (i) When compared with free convection, radiation heat loss can be significant
PROBLEM 7.5 (continued)
and in general should not be neglected. (ii) The magnitude of E is the same whether it is
expressed in units of degree Celsius or kelvin. The reason is that E is measured in terms of
degree change. One degree change on the Celsius scale is equal to one degree change on the
kelvin scale. This is also true of units of heat transfer coefficient and specific heat. (iii) Because
temperature in the Stefan-Boltzmann radiation law must be expressed in degrees kelvin, care
should be exercised in using the correct units when carrying radiation computations.
PROBLEM 7.6
A sealed electronic package is designed to be cooled by free convection. The
package consists of components which are mounted on the inside surfaces of
two cover plates measuring 7.5 cm u 7.5 cm cm each. Because the plates are air
made of high conductivity material, surface temperature may be assumed
uniform. The maximum allowable surface temperature is 70oC. Determine Tf
the maximum power that can be dissipated in the package without violating g
design constraints. Ambient air temperature is 20oC.
(1) Observations. (i) This is a free convection problem. (ii) The power
components
dissipated in the electronic package is transferred to the ambient fluid
by free convection. (iii) As the power is increased, surface temperature increases. (iv) The
maximum power dissipated corresponds to the maximum allowable surface temperature. (v)
Surface temperature is related to surface heat transfer by Newton’s law of cooling. (vi) The
problem can be modeled as free convection over a vertical plate. (vii) The Rayleigh number
should be computed to determine if the flow is laminar or turbulent. (viii) For laminar flow the
solution of Section 7.4 is applicable. (ix) The fluid is air.
(2) Problem Definition. Since dissipated power is related to heat loss from the surface to the
surroundings, Newton's law of cooling should be applied. Thus, the problem becomes finding
the average heat transfer coefficient.
(3) Solution Plan. Apply Newton's law of cooling to the surface of the package. Check the
Rayleigh number to determine if the flow is laminar or turbulent. Use appropriate correlation
equations to determine the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in
buoyancy), (4) uniform surface temperature, (5) negligible heat loss from the top and bottom
surfaces, (6) the sides are vertical flat plates, (7) negligible radiation and (8) quiescent fluid.
(ii) Analysis. Application of Newton's law of cooling to the surface gives
P = q = h A ( Ts Tf )
(a)
where
A = surface area of the two vertical sides, m2
h = average heat transfer coefficient, W/m2-oC
P = power dissipated in package, W
q = heat transfer from the surface to the ambient air, W
Ts = surface temperature = 70oC
Tf = ambient air temperature = 20oC
Surface area of the two vertical sides is given by
A = 2 LW
L
Tf
g
Ts
W
(b)
PROBLEM 7.6 (continued)
where
L = package height = 7.5 cm = 0.75 m
W = package width = 7.5 cm = 0.75 m
To determine the average heat transfer coefficient h , the Rayleigh number RaL is calculated first
to determine if the flow is laminar or turbulent.
RaL =
E g Ts Tf L3
Q2
Pr
(c)
where
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number of air
RaL = Rayleigh number
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
Properties are determined at the film temperature Tf defined as
Tf =
(70 20)( o C)
Ts Tf
=
= 45oC
2
2
Air properties at this temperature are
k = 0.02746 W/m-oC
Pr = 0.7095
Q = 17.44u10-6 m2/s
For ideal gases the coefficient of thermal expansion E is given by
E=
1
T f (K )
1
o
45( C) 273.15
Substituting into (c) gives
RaL =
0.003143 (1/K)
0.003143(1/ o C)9.81(m/s 2 )(70 20)( o C)(0.075) 3 (m 3 )
(17.44 u 10
6 2
4
2
) (m /s )
0.7095 = 1.5171 u 10 6
Since RaL < 109, the flow is laminar and h is given by (7.23)
1/ 4
h
4 k § GrL ·
¨
¸
3 L© 4 ¹
dT (0)
dK
(7.23)
where GrL is the G
rashof number given by
GrL =
E g Ts Tf L3
Q
2
Ra L
Pr
(d)
and dT (0) / dK is a dimensionless temperature gradient which depends on the Prandtl number. It
is listed in Table 7.1.
(iii) Computations. Substitution into (d) gives
PROBLEM 7.6 (continued)
8.257 u 10 6
0.7065
GrL
1.5171 u 10 6
At Pr = 0.7095, Table 7.1 gives
dT (0)
dK
0.5017
Substitute into (7.23)
h
4 0.02746(W/m o C) §¨ 1.5171 u 10 6
¨
3
0.075(m)
4
©
1/ 4
·
¸
¸
¹
0.5017
6.08 W/m 2 o C
Equations (a) and (b) give the maximum power dissipated from the two sides of the package
P = q = 2 u 6.08 (W/m2-oC) (0.075)(m)(0.075)(m) (70-20)(oC) = 3.42 W
(iv) Checking. Dimensional check: Computations showed that equations (a), (c) and (7.23)
dimensionally consistent.
Quantitative check: The magnitude of h is within the range given in Table 1.1 for free
convection of gases.
Qualitative check: Increasing the allowable surface temperature Ts should increase the maximum
power P. According to equation (a), q is directly proportional to Ts . Furthermore, h increases
when Ts is increased.
Limiting check: The dissipated power should vanish (P = 0) if heat cannot be removed by free
convection (q = 0). Setting h = 0 in (a) gives P = q = 0.
(5) Comments. (i) The model used to solve this problem is conservative due to both neglecting
radiation and heat loss from the side surfaces. (ii) The maximum power dissipated is relatively
small, indicating the limitation of free convection in air as a cooling mode for such applications.
The maximum dissipated power in water is 494.7 W (Problem 7.7). (iii) The magnitude of E is
the same whether it is expressed in units of degree Celsius or kelvin. The reason is that E is
measured in terms of degree change. One degree change on the Celsius scale is equal to one
degree change on the kelvin scale. This is also true of units of heat transfer coefficient and
specific heat.
PROBLEM 7.7
Assume that the electronic package of Problem 7.6 is to be used in an underwater application.
Determine the maximum power that can be dissipated if the ambient water temperature is 20oC.
(1) Observations. (i) This is a free convection problem. (ii) The power dissipated in the
electronic package is transferred to the ambient fluid by free convection. (iii) As the power
is increased, surface temperature increases. (iv) The maximum power dissipated corresponds to
the maximum allowable surface temperature. (v) Surface temperature is related to surface heat
transfer by Newton’s law of cooling. (vi) The problem can be modeled as free convection over a
vertical plate. (vii) The Rayleigh number should be computed to determine if the flow is laminar
or turbulent. (viii) For laminar flow the solution of Section 7.4 is applicable. (ix) The fluid is
water.
(2) Problem Definition. Since dissipated power is related to heat loss from the surface to the
surroundings, Newton's law of cooling should be applied. Thus, the problem becomes finding
the average heat transfer coefficient.
(3) Solution Plan. Apply Newton's law of cooling to the surface of the package. Check the
Rayleigh number to determine if the flow is laminar or turbulent. Use appropriate correlation
equations to determine the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in
buoyancy), (4) uniform surface temperature, (5) negligible heat loss from the top and bottom
surfaces, (6) the sides are vertical flat plates, (7) negligible radiation and (8) quiescent fluid.
(ii) Analysis. Application of Newton's law of cooling to the surface gives
P = q = h A ( Ts Tf )
(a)
where
A = surface area of the two vertical sides, m2
h = average heat transfer coefficient, W/m2-oC
P = power dissipated in package, W
q = heat transfer from the surface to the ambient air, W
Ts = surface temperature = 70oC
Tf = ambient air temperature = 20oC
Surface area of the two vertical sides is given by
A = 2 LW
L
Tf
g
Ts
W
(b)
where
L = package height = 7.5 cm = 0.75 m
W = package width = 7.5 cm = 0.75 m
To determine the average heat transfer coefficient h , the Rayleigh number RaL is calculated first
to determine if the flow is laminar or turbulent.
PROBLEM 7.7 (continued)
RaL =
E g Ts Tf L3
Q2
Pr
(c)
where
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number of water
RaL = Rayleigh number
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
Properties are determined at the film temperature Tf defined as
Tf =
(70 20)( o C)
Ts Tf
=
= 45oC
2
2
Water properties at this temperature are
k = 0.6286 W/m-oC
Pr = 4.34
E = 0.000389 1/K
Q = 0.6582u10-6 m2/s
Substituting into (c) gives
RaL =
0.000389(1/ o C)9.81(m/s 2 )(70 10)( o C)(0.075) 3 (m 3 )
(0.6582 u 10
6 2
4
2
) (m /s )
4.34 = 0.96778 u 10 9
Since RaL < 109, the flow is laminar and h is given by (7.23)
1/ 4
h
4 k § GrL ·
¨
¸
3 L© 4 ¹
dT (0)
dK
(7.23)
where GrL is the G
rashof number given by
GrL =
E g Ts Tf L3
Q
2
Ra L
Pr
(d)
and dT (0) / dK is a dimensionless temperature gradient which depends on the Prandtl number. It
is listed in Table 7.1.
(iii) Computations. Substitution into (d) gives
GrL
0.96778 u 10 9
4.34
0.22299 u 10 9
At Pr = 4.34, Table 7.1 gives
dT (0)
dK
0.9108
Substitute into (7.23)
PROBLEM 7.7 (continued)
h
4 0.6286(W/m o C) §¨ 0.22299 u 10 9
¨
3
0.075(m)
4
©
·
¸
¸
¹
1/ 4
0.9108 879.5 W/m 2 o C
Equations (a) and (b) give the maximum power dissipated from the two sides of the package
P = q = 2 u 879.5 (W/m2-oC) (0.075)(m)(0.075)(m) (70-20)(oC) = 494.7 W
(iv) Checking. Dimensional check: Computations showed that equations (a), (c) and (7.23)
dimensionally consistent.
Quantitative check: The magnitude of h is within the range given in Table 1.1 for free
convection of liquids
Qualitative check: Increasing the allowable surface temperature Ts should increase the maximum
power P. According to equation (a), q is directly proportional to Ts . Furthermore, h increases
when Ts is increased.
Limiting check: The dissipated power should vanish (P = 0) if heat cannot be removed by free
convection (q = 0). Setting h = 0 in (a) gives P = q = 0.
(5) Comments. (i) The model used to solve this problem is conservative due to neglecting heat
loss from the side surfaces. (ii) The maximum power dissipated is relatively large, indicating the
effectiveness of water as a free convection medium. The maximum power in air (Problem 7.6) is
3.42 W. (iii) The magnitude of E is the same whether it is expressed in units of degree Celsius or
kelvin. The reason is that E is measured in terms of degree change. One degree change on the
Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat
transfer coefficient and specific heat.
PROBLEM 7.8
Consider laminar free convection from a vertical plate at uniform
surface temperature. Two 45q triangles are drawn on the plate as
shown. Determine the ratio of the heat transfer rates from two
triangles.
g
Tf
2
1
(1) Observations. (i) This is a free convection problem. (ii) The
surface is maintained at uniform temperature. (iii) Newton’s law of
cooling determines the heat transfer rate. (iv) Heat transfer rate
depends on the heat transfer coefficient. (v) The heat transfer coefficient decreases with distance
from the leading edge of the plate. (vi) The width of each triangle changes with distance from the
leading edge. (vii) For laminar flow the solution of Section 7.4 is applicable.
(2) Problem Definition. Examine the variation of local heat transfer coefficient with distance
and determine the heat transfer rate from each triangle.
(3) Solution Plan.
Apply Newton’s law of cooling to an element of each triangle. Formulate an equation for
element area and heat transfer coefficient h(x) for laminar free convection over a flat plate.
(4) Plan Execution.
(i) Assumptions. (1) Laminar flow, (2) steady state, (3) two-dimensional, (4) constant
properties (except in buoyancy), (5) uniform surface temperature, (6) quiescent fluid and (7) no
radiation.
(ii) Analysis. Consider triangle 1 first. Application of Newton’s law of cooling to the element
b1 ( x )dx
dq1 h( x )(Ts Tf )b1 ( x )dx
Integrate
H
q1
³
(Ts Tf ) h ( x )b1 ( x )dx
b 2(x)
(a)
0
where
b1 ( x )
2
1
H
width of element 1, m
2
o
h(x ) local heat transfer coefficient, W/m C
q1 = heat transfer rate, W
Ts surface temperature, oC
Tf = ambient temperature, oC
x = distance along plate, m
dx
x
Tf
g
)
b 1(x
dx
B
Similarly, for the second triangle
H
q2
³
(Ts Tf ) h ( x )b2 ( x )dx
0
where
(b)
PROBLEM 7.8 (continued)
b2 ( x ) width of element 2, m
To evaluate the integrals in (a) and (b) it is necessary to determine the variation with x of h(x),
b1 ( x ) and b2 ( x ) . The local heat transfer coefficient for free convection laminar flow over a
vertical plate is given by (7.21)
k
x
h
ª Grx º
« 4 »
¬
¼
1/ 4
dT (0)
dK
(7.21)
where
k
thermal conductivity, W/m-oC
dT (0) / dx dimensionless temperature gradient at the surface
Grx = rGashof number, defined as
Grx
E g (Ts Tf ) x 3
Q2
(c)
g = gravitational acceleration, m/s2
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
(c) into (7.21)
h
k
x
ª E g (Ts Tf ) x 3 º
«
»
4Q 2
«¬
»¼
1/ 4
dT ( 0 )
dK
(d)
According to (d), h(x) decreases varies with x as
h( x )
C x 1 / 4
(e)
The geometric functions b1 ( x ) and b2 ( x ) are determined using similarity of triangles. Thus
x
b1(x) = B(1 - )
H
and
x
b2(x) = B
H
where
B = base of triangle, m
H = height of triangle, m
(f)
(g)
Substitute (e) and (f) into (a)
H
q1
BC Ts Tf
³ 1 x / H
x 1 / 4 dx
0
Evaluate the integral
q1
(16 / 21)(Ts Tf ) BCH 3 / 4
Similarly, substitute (d) and (g) into (b)
(h)
PROBLEM 7.8 (continued)
q2
BC Ts Tf
³
H
( x / H ) x 1 / 4 dx
0
Evaluate the integral
q2
( 4 / 7)(Ts Tf ) BCH 3 / 4
(i)
Taking the ratio of (i) and (i)
q1
4
=
q2
3
(j)
(iii) Checking. Dimensional check: Units of C are determined using (d):
C = W/m7/4-oC
Thus units of h(x) in (d) are
h(x) = C (W/m7/4-oC) x-1/4(m)-1/4 = W/m2-oC
Examining units of q1 in (h)
q1 = B(m) C(W/m7/4-oC) (Ts -Tf) (oC) H
3/ 4
(m3/4) = W
Limiting check: If E = 0 or g = 0 or Ts = Tf, no free convection takes place and consequently q1
= q2 = 0. Any of these limiting cases give C = 0. Thus, according to (h) and (i), q1 = q2 = 0.
(5) Comments. (i) o
Mre heat is transferred from triangle 1 than triangle 2. This follows from the
fact that h decrease with distance x. This favors triangle 1 since its large base is at x = 0 where h
is maximum.
(ii) The result applies to any right angle triangles and is not limited to 45o
triangles. (iii) Heat transfer from a surface of fixed area depends on its orientation relative to the
leading edge. (iv) This problem illustrates how integration is used to account for variations in
element area and heat transfer coefficient. The same approach can be applied if surface
temperature and/or ambient temperature vary over a surface area.
PROBLEM 7.9
A vertical plate measuring 21 cm u 21 cm is at a uniform surface temperature of 80oC. The
ambient air temperature is 25oC. Determine the heat flux at 1 cm, 10 cm and 20 cm from the
lower edge.
(1) Observations. (i) This is a free convection problem over a vertical plate. (ii) The surface is
maintained at uniform temperature. (iii) Local heat flux is determined by Newton’s law of
cooling. (iv) Heat flux depends on the local heat transfer coefficient. (v) Free convection heat
transfer coefficient for a vertical plate decreases with distance from the leading edge. Thus, the
flux also decreases. (vi) The Rayleigh number should be computed to determine if the flow is
laminar or turbulent. For (vii) Laminar flow the solution of Section 7.4 is applicable.
(viii) The fluid is air.
(2) Problem Definition. Determine the local heat transfer coefficient for free convection over a
vertical plate at uniform surface temperature.
(3) Solution Plan. Apply Newton’s law of cooling, compute the Rayleigh number and select an
appropriate Nusselt number correlation equation.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in
buoyancy), (4) uniform surface temperature, (5) quiescent fluid and (6) negligible radiation.
(ii) Analysis. Application of Newton's law gives
q cc = h(x) (Ts - Tf)
(a)
where
h(x) = local heat transfer coefficient, W/m2-oC
q cc = local heat flux, W
Ts = surface temperature = 80oC
Tf = ambient temperature = 25oC
The Rayleigh is computed to determine if the flow is laminar or turbulent.
Rax =
E g Ts Tf L3
Pr
Q2
(b)
where
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number
Ra L = Rayleigh number at L
L = plate height = 0.21 m
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
Properties of air are determined at the film temperature T f defined as
Tf = (Ts Tf ) / 2 = (80 25)( o C) / 2 = 52.5oC
L
Tf
g
Ts
L
PROBLEM 7.9 (continued)
Properties of air at this temperature are
k = 0.02799 W/m-oC
Pr = 0.709
Q = 18.165u10-6 m2/s
For ideal gases the coefficient of thermal expansion E is given by
E=
1
Tf ( K )
1
0.003071 (1/K)
o
52.5( C) 273.15
Substituting into (b)
RaL =
0.003071(1/ o C)9.81(m /s 2 )(80 25)( o C)(0.2) 3 (m 3 )
6 2
4
2
(18.165 u 10 ) (m /s )
0.709 = 2.8482u107
Since RaL < 109, the flow is laminar over the region of interest. The local heat transfer
coefficient for free convection laminar flow over a vertical plate is given by (7.21)
k ª Grx º
x «¬ 4 »¼
h
1/ 4
dT (0)
dK
(7.21)
where
dT (0) / dx dimensionless temperature gradient at the surface
Grx = rGashof number, defined as
Grx
E g (Ts Tf ) x 3
Q2
(c)
Note that dT (0) / dx depends on the Prandtl number and is listed in Table T.1.
(iii) Computations. At Pr = 0.709, Table 7.1 gives
dT (0)
dK
0.5015 =
At x = 0.01 m, (c) gives
0.003071(1/ o C)9.81(m /s 2 )(80 25)( o C)(0.01) 3 (m 3 )
Grx
6 2
4
2
(18.165 u 10 ) (m /s )
= 0.5022 u 10 4
Substitute into (7.21)
h
0.02799( W/m o C) ª 5022 º
«¬ 4 »¼
0.01( m)
1/ 4
( 0.5015) = 8.356 W/m2-oC
Substitute into (a)
qcc 8.355( W/m 2 o C)(80 25)( o C)
459.6 W/m 2
PROBLEM 7.9 (continued)
The same procedure is followed to determine the flux at x = 10 and x = 20 cm. Results are
tabulated below.
x (cm)
1
10
20
Grx
h(x)(W/m2-oC)
q cc (W/m2)
0.5022 u 104
0.5022 u 107
0.4018 u 108
8.355
4.699
459.6
258.4
3.951
217.3
(iv) Checking. Dimensional check: Computations showed that equations (a), (b) and (7.21)
are dimensionally consistent.
Quantitative check: The values of h are approximately within the range given in Table 1.1 for
free convection of gases.
Limiting check: The flux should vanish for Ts = Tf. Setting Ts = Tf in (a) gives q cc 0 .
(5) Comments. According to (7.21) and (a), surface heat flux decreases with distance from the
leading edge as
C
qcc
1/ 4
x
Thus, high heat flux components should be place close to the leading edge.
PROBLEM 7.10
200 square chips measuring 1 cm u 1 cm each are mounted on both
sides of a thin vertical board 10 cm u 10 cm. The chips dissipate 0.035
W each. Assume uniform surface heat flux. Determine the maximum
surface temperature in air at 22oC.
Tf g
(1) Observations. (i) This is a free convection problem over a vertical
plate. (ii) The power dissipated in the chips is transferred to the air by
free convection. (iii) This problem can be modeled as free convection over a vertical plate with
constant surface heat flux. (iv) Surface temperature increases as the distance from the leading
edge is increased. Thus, the maximum surface temperature occurs at the top end of the plate
(trailing end). (v) The Rayleigh number should be computed to determine if the flow is laminar
or turbulent. (vii) For laminar flow the analysis of Section 7.5 gives surface temperature
distribution. (vii) The fluid is air. (viii) Properties depend on the average surface temperature Ts .
Since Ts is unknown, the problem must be solved by trail and error.
(2) Problem Definition. Determine surface temperature distribution for a vertical plate with
uniform surface heat flux under free convection conditions.
(3) Solution Plan. Apply the analysis of Section 7.5 for surface temperature distribution of a
vertical plate with uniform surface heat flux in laminar free convection. Compute the Rayleigh
number to confirm that the flow is laminar.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in
buoyancy), (4) uniform surface heat flux, (5) all dissipated power leaves surface as heat, (6)
negligible radiation and (7) quiescent ambient fluid.
(ii) Analysis. Surface temperature distribution for a vertical plate with uniform surface flux is
given by equation (7.27)
Ts ( x) Tf
ª ǎ 2 (q csc ) 4 º
x»
«5
4
»¼
¬« E g k
1/ 5
T ( 0)
(7.27)
where
g = gravitational acceleration = 9.81 m/s2
k = thermal conductivity, W/m-oC
Pr = Prandtl number
q csc = surface flux, W/m2
RaL = Rayleigh number at x = L
Ts = surface temperature, oC
Tf = ambient temperature = 22oC
x = distance from leading edge, m
E = coefficient of thermal expansion, 1/K
T (0) = dimensionless surface temperature
ǎ = kinematic viscosity, m2/s
The dimensionless surface temperature, T (0) , depends the Prandtl number. Values corresponding
to four Prandtl numbers are listed in Table 7.2.
PROBLEM 7.10 (continued)
For laminar flow
RaL =
E g Ts Tf L3
ǎ
2
Pr < 109
(a)
where
L = vertical side of plate = 10 cm = 0.1 m
If all dissipated power in the chip leaves the surface, conservation of energy gives
qscc
P
A
(b)
where
A = chip surface area = 1 cm2 = 0.0001 m2
P = power dissipated in chip = 0.035 W
Properties are evaluate at the film temperature defined as
Ts ( L / 2) Tf
(c)
2
where Ts(L/2) is surface temperature at the mid-point. However, since Ts(x) is unknown, an
iterative procedure is required to obtain a solution. An assumed value for Ts(L/2) is used to
calculate the film temperature at which properties are determined. Equation (a) is then used to
calculate Ts(L/2). If the calculated value does not agree with the assumed temperature, the
procedure is repeated until a satisfactory agreement is obtained.
Tf
(iii) Computations. Equation (b) gives surface flux
q csc = 0.035(W)/0.0001(m2) = 350 W/m2
Assume Ts (L/2) = 58oC. Equation (c) gives
Tf = (58 + 22)(oC)/2 = 40oC
Properties of air at this temperature are
cp = 1006.8 J/kg- oC
k = 0.0271 W/m-oC
Pr = 0.71
ǎ = 16.96u10-6 m2/s
Coefficient of thermal expansion for an ideal gas is given by
E =
1
T f 273.15
1
o
40( C) 273.15
0.003193 1/K
At Pr = 0.71, Table 7.2 gives T (0) 1.806.
Substituting into (7.27) and letting x = L/2 = 0.1(m)/2 = 0.05 m
1/ 5
Ts L/2
4
ª
º
(16.96 u 10 6 ) 2 (m/s)2 §¨ 350(W/m 2 ) ·¸
o
«
22( C) 5
0.05( m)»
« 0.003193(1/ o C)9.81(m/s2 ) ¨ 0.0271(W/m o C) ¸
»
©
¹
¬
¼
1.806 = 87.7 o C
PROBLEM 7.10 (continued)
o
Ts(L/2) = 87.7 C
Since this is higher than the assumed value of 58oC, the procedure is repeated with a new
assumed temperature at mid-point. Assume Ts(L/2) = 78oC. The following results are obtained
Tf = 50oC
cp = 1007.4 J/kg- oC
k = 0.02781 W/m-oC
Pr = 0.709
D = 25.27u10-6 m2/s
E = 0.0030945 1/K
ǎ = 17.92u10-6 m2/s
3
U = 1.0924 kg/m
Substituting into (a) gives Ts(L/2) = 76.8oC. This is close to the assumed value of 78oC.
Surface temperature at the trailing end is now computed by evaluating (a) at x = L = 0.1 m
1/ 5
Ts L/2
4
ª
º
·
4 9(0.709)1 / 2 10(0.709) §¨ 25.27 u 10 6 (m 2 / s)17.92 u 10 6 (m 2 / s) ·¸§¨
350( W / m 2 )
¸ 0.1(m)»
22( C) «
o
2
o
¨
¸¨ 0.02781( W / m C) ¸
«
»
0.709
0.0030945(1/ C)9.81( m / s )
©
¹©
¹
¬
¼
o
Ts(L) = 84.9oC
The condition on the Rayleigh number in equation (b) is verified next. Substituting into (c)
RaL =
0.0030945(1/ o C)9.81(m / s 2 )(84.9 22)( o C)(0.1) 3 (m 3 )
6 2
4
2
(17.92 u 10 ) (m / s )
0.709 = 4.22u106
This satisfies the condition on RaL given in equation (b).
(iv) Checking. Dimensional check: Equations (a), (c) and (d) are dimensionally consistent..
Quantitative check: The heat transfer coefficient at the mid-point, h(L/2), can be calculated using
Newton's law of cooling:
h(L/2) = q csc /[Ts ( L / 2) Tf ] = 350(W/m2)/(76.8 - 22)(oC) = 6.39 W/m2-oC
This is within the range given in Table 1.1 for free convection of gases.
Validity of correlation equation (a): The conditions listed in (b) are met.
(5) Comments. (i) Surface temperature is determined without calculating the heat transfer
coefficient. This is possible because equation (a) combines the correlation equation for the heat
transfer coefficient and Newton's law of cooling to eliminate h and obtain an equation for surface
temperature in terms of surface heat flux. (ii) The magnitude of E is the same whether it is
expressed in units of degree Celsius or kelvin. The reason is that E is measured in terms of
degree change. One degree change on the Celsius scale is equal to one degree change on the
kelvin scale. This is also true of units of heat transfer coefficient and specific heat.
PROBLEM 7.11
12 cm u 12 cm power board dissipates 15 watts uniformly. Assume that all energy leaves the
board from one side. The maximum allowable surface temperature is 82 o C. The ambient fluid is
air at 24 o C. Would you recommend cooling the board by free convection?
(1) Observations. (i) This is a free convection problem over a vertical plate. (ii) The power
dissipated in the chips is transferred to the air by free convection. (iii) This problem can be
modeled as free convection over a vertical plate with constant surface heat flux. (iv) Surface
temperature increases as the distance from the leading edge is increased. Thus, the maximum
surface temperature occurs at the top end of the plate (trailing end). (v) The Rayleigh number
should be computed to determine if the flow is laminar or turbulent. (vii) For laminar flow the
analysis of Section 7.5 gives surface temperature distribution. (vii) The fluid is air. (viii)
Properties depend on the average surface temperature Ts . Since Ts is unknown, the problem
must be solved by trail and error.
(2) Problem Definition. Determine surface temperature distribution for a vertical plate with
uniform surface heat flux under free convection conditions.
(3) Solution Plan. Apply the analysis of Section 7.5 for surface temperature distribution of a
vertical plate with uniform surface heat flux in laminar free convection. Compute the Rayleigh
number to confirm that the flow is laminar.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in
buoyancy), (4) uniform surface heat flux, (5) all dissipated power leaves surface as heat, (6)
negligible radiation and (7) quiescent ambient fluid.
(ii) Analysis. Surface temperature distribution for a vertical plate with uniform surface flux is
given by equation (7.27)
Ts ( x) Tf
ª Q 2 (q csc ) 4
«5
4
«¬ E g k
where
g = gravitational acceleration = 9.81 m/s2
k = thermal conductivity, W/m-oC
q csc = surface flux, W/m2
Ts = surface temperature, oC
Tf = ambient temperature = 22oC
x = distance from leading edge, m
E = coefficient of thermal expansion, 1/K
T (0) = dimensionless surface temperature
Q = kinematic viscosity, m2/s
º
x»
»¼
1/ 5
T (0)
(7.27)
L
Tf
g
x
0
L
qsc
The dimensionless surface temperature, T (0) , depends the Prandtl number. It can be determined
from Table 7.2 or using correlation equation (7.29):
PROBLEM 7.11 (continued)
ª 4 9 Pr 1 / 2 10 Pr º
«
»
5Pr 2
¬«
¼»
T ( 0)
1/ 5
, 0.001 Pr 1000
(7.29)
The heat flux is defined as
q scc
P
A
(a)
A
L2
(b)
where
A = surface area, m 2
P = dissipated power = 15 W
Surface area is
where
L = side of power board = 0.12 m
The Rayleigh number is used to determine if the flow is laminar. The criterion is
RaL =
E g Ts Tf L3
Pr < 109
2
Q
(c)
Equation (7.27) is used to determine maximum surface temperature, Ts (L) , corresponding to the
specified surface flux. Properties are evaluate at the film temperature defined as
Tf
Ts ( L / 2) Tf
2
(d)
where Ts(L/2) is surface temperature at the mid-point. However, since Ts(x) is unknown, an
iterative procedure is required to obtain a solution. An assumed value for Ts(L/2) is used to
calculate the film temperature at which properties are determined. Equation (7.27) is then used
to calculate Ts(L/2). If the calculated value does not agree with the assumed temperature, the
procedure is repeated until a satisfactory agreement is obtained.
(iii) Computations. Assume Ts (L/2) = 76oC. Equation (d) gives
Tf = (76 + 24)(oC)/2 = 50oC
Properties of air at this temperature are
k = 0.02781 W/m-oC
Pr = 0.709
Q = 17.92u10-6 m2/s
Coefficient of thermal expansion for an ideal gas is given by
E =
1
T f 273.15
1
o
50( C) 273.15
0.0030945 1/K
At Pr = 0.709, equation (7.29) gives T (0)
PROBLEM 7.11 (continued)
1/ 5
ª 4 9 0.709 10(0.709) º
«
»
5(0.709) 2
»¼
«¬
T ( 0)
1.49337
Equations (a) and (b) give surface heat flux
q scc
15(W)
1041.67
2
0.12 u 0.12(m )
W
m2
Substituting into (7.27) and letting x = L/2 = 0.12(m)/2 = 0.06 m
1/ 5
Ts L/2
4
ª
º
2
6 2
§ 1041.67(W/m 2 ) ·
u
(17.92
10
)
(m/s)
o
¨
¸ 0.06( m)»
24( C) «5
« 0.00309451/ o C)9.81(m/s 2 ) ¨ 0.02781(W/m o C) ¸
»
©
¹
¬
¼
1.49337 159.9 o C
Since this is higher than the assumed value of 58oC, the procedure is repeated with a new
assumed temperature at mid-point. Assume Ts(L/2) = 156oC. The following results are obtained
Tf = 90oC
k = 0.03059 W/m-oC
Pr = 0.705
E = 0.0027537 1/K
Q = 21.35u10-6 m2/s
T (0) 1.4958
Substituting into (7.27) gives Ts(L/2) = 162.4oC. Further iteration will bring Ts (L / 2) between
156 oC and 162 oC. Surface temperature at the trailing end will be even higher. Therefore, board
temperature will exceed the maximum allowable of 82 o C. It follows that cooling by free
convection is not recommended.
Surface temperature at the trailing end is now computed by evaluating (7.27) at x = L = 0.12 m
1/ 5
Ts L/2
Ts (L)
4
ª
º
(21.35 u 10 6 ) 2 (m 2 /s) 2 §¨ 1041.67(W/m 2 ) ·¸
o
«
0.12(m)»
24( C) 5
« 0.0027537(1/ o C)9.81(m/s 2 ) ¨ 0.03059(W/m o C) ¸
»
©
¹
¬
¼
1.4958
183.1oC
The Rayleigh number is computed to confirm that the flow is laminar. Substitute into (a)
RaL =
0.0027537(1/ o C)9.81(m/s 2 )(183.1 24)( o C)(0.12) 3 (m 3 )
(21.35 u 10
6 2
4 2
0.705 0.1149 u 10 7
) (m /s )
Since this is less than 10 9 , the flow is laminar.
(iv) Checking.
consistent..
Dimensional check: Equations (7.27) and (a)-(d) are dimensionally
Quantitative check: The heat transfer coefficient at the mid-point, h(L/2), can be calculated using
Newton's law of cooling:
PROBLEM 7.11 (continued)
h(L/2) = q csc /[Ts ( L / 2) Tf ] = 1041.67(W/m2)/(159.0 - 24)(oC) = 7.66 W/m2-oC
This is within the range given in Table 1.1 for free convection of gases.
(5) Comments. (i) High surface temperature is due to high surface heat flux. Forced convection
cooling is required to meet design conditions on maximum temperature. (ii) A trial and error
procedure was required to solve this problem because properties depend on surface temperature
which is unknown a priori.
PROBLEM 7.12
Use the integral method to obtain a solution to the local Nusselt number for laminar flow
over a vertical plate at uniform surface temperature Ts . Assume G G t and a velocity and
temperature profiles given by
u x, y
a0 ( x) a1 ( x) y a 2 ( x) y 2 a3 ( x) y 3
and
T ( x, y )
b0 ( x) b1 ( x) y b2 ( x) y 2 b3 ( x) y 3
Since there is a single unknown G (x), either the momentum or energy equation may be
used. Select the energy equation to determine G t (x ).
(1) Observations. (i) This is a free convection problem over a vertical plate at uniform
surface temperature. (ii) In general, to determine the Nusselt number it is necessary to
determine the velocity and temperature distribution. (iii) The integral method can be used to
determine the velocity and temperature distribution. (iv) Application of the integral method
reduces to determining the velocity and temperature boundary layer thickness.
(2) Problem Definition. Determine the temperature distribution for boundary layer flow
over a flat plate at uniform surface temperature.
(3) Solution Plan. Start with the definition of local Nusselt number. Equate Newton’s law
with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the
integral form of the energy equation using a third degree polynomial temperature profile.
(4) Plan Execution. (i) Assumptions. (1) Steady state, (2) constant properties except for
buoyancy, (3) Boussinesq approximations are valid, (4) two-dimensional, (5) laminar flow
( Ra L 10 9 ), (6) flat plate, (7) uniform surface temperature, (8) negligible changes in kinetic
and potential energy, (9) negligible axial conduction, (10) negligible dissipation, and (11)
G Gt .
(ii) Analysis. The local Nusselt number is defined as
hx
k
Nu x
(a)
where
h local heat transfer coefficient, W/m 2 o C
k thermal conductivity, W/m o C
Nu x local Nusselt number
x distance along plate measured from the leading edge, m
where the heat transfer coefficient h is given by equation (1.10)
wT ( x,0)
wy
Ts Tf
k
h
where
(1.10)
PROBLEM 7.12 (continued)
Ts
surface temperature, o C
Tf ambient temperature, o C
y coordinate normal to plate, m
Thus h depends on the temperature distribution T ( x, y ). The integral form of the energy
equation, (7.36), is used to determine the temperature distribution
wT x,0
D
wy
d
dx
Gt
³ u(T T
(b)
f )dy
0
where
u axial velocity, m/s
D
thermal diffusivity, m 2 /s
Assume a third degree polynomial for the axial velocity u(x,y)
u x, y
a0 ( x) a1 ( x) y a 2 ( x) y 2 a3 ( x) y 3
(c)
The coefficients a n ( x) are determined using the following known exact and approximate
boundary conditions on the velocity
(1) u ( x,0)
0
(2) u ( x, G t ) # 0
(3)
(4)
wu ( x, G t )
#0
wy
w 2 u ( x ,0 )
wy 2
Eg
(Ts Tf )
Q
Condition (4) is obtained by setting y 0 in the x-component of the equations of motion,
(7.5). Equation (c) and the four boundary conditions give the coefficients a n (x )
a0
0,
a1
E g (Ts Tf )
E g (Ts Tf )
,
G t , a2 2Q
4Q
a3
E g (Ts Tf ) 1
4Q
Gt
Substituting the above into (c) and rearranging
u
ª
E g (Ts Tf )
y y2 º
G t y «1 2 2 »
4Q
G t G t ¼»
¬«
Note that velocity profile (d) is based on the assumption that G
For the temperature profile we assume a third degree polynomial
(d)
Gt.
PROBLEM 7.12 (continued)
T ( x, y )
b0 ( x ) b1 ( x ) y b2 ( x ) y 2 b3 ( x ) y 3
(e)
The boundary conditions on the temperature are
(1) T x,0 Ts
(2) T x, G t # Tf
wT x, G t
(3)
#0
wy
(4)
w 2T x,0
wy 2
0
Equation (c) and the four boundary conditions give the coefficients bn (x)
3
1
1
1
b0 Ts , b1
(Tf Ts ) , b2 0, b3 (Tf Ts ) 3
Gt
2
2
Gt
Substituting the above into (e)
ª3 y 1 y3 º
(f)
T ( x, y ) Ts (Tf Ts ) «
»
3
«¬ 2 G t 2 G t »¼
Substituting (f) into (1.10)
3k
(g)
h
2G t
Combining (a) and (g)
3x
(h)
Nu x
2G t
The problem reduces to finding G t which is obtained using the energy equation. Substituting
(d) and (f) into (b)
D
3(Ts Tf ) 1
2
Gt
Eg (Ts Tf ) d
dx
4Q
G t ( x)
³
0
ª
y
¬«
Gt
G t y «1 ª 3 y 1 y3 º
y2 º
T
T
dy
(
)
»
f «1 s
3»
G t2 ¼»
¬« 2 G t 2 G t ¼»
Expand the integrand and simplify
Gt
3 1
D
2 Gt
ª
Eg (Ts Tf ) d
7 y2 y4
y3 y5 1 y6 º
3 4 2 4 Gt «y » dy
4Q
2 Gt Gt
dx
G t G t 2 G t5 ¼»
«¬
0
³
Evaluate the integral
D
3 1
2 Gt
Simplify and rearrange
Eg (Ts Tf ) d ª 1 3 7 3 1 3
1
1
º
G t G t G t G t3 G t3 G t3 »
«
4Q
dx ¬ 2
6
5
6
14 ¼
PROBLEM 7.12 (continued)
Gt
dG t3
dx
315
DQ
2 Eg (Ts Tf )
G t3
dG t
dx
105
DQ
2 Eg (Ts Tf )
Rewrite the above
Separating variables and integrating
³
Gt
105
DQ
2 Eg (Ts Tf )
G t3 dG t
0
³
Gt
dx
0
Evaluate the integrals and rearrange
G t4
105
DQ
x
2 Eg (Ts Tf )
4
Solve for G t / x
ª
º
«
»
210
«
»
« Eg (Ts Tf ) x 3 »
«
»
DQ
¬
¼
Gt
x
1/ 4
(i)
This result can be expressed in terms of the Rayleigh number as
Gt
3.806
x
Ra1x/ 4
(j)
Substitute (j) into (h)
0.394 Ra1x/ 4
(k)
0.394>Grx Pr @1 / 4
(l)
Nu x
An alternate form is
Nu x
(iii) Checking. Dimensional check: Equations (a), (h), (i), (j) and (k) are dimensionless.
Units of (b), (d) and (f) are correct.
Boundary conditions check: Velocity profile (d) and temperature profile (f) satisfy their
respective boundary conditions.
(5) Comments. (i) The integral form of the momentum equation was not used in the
solution. Therefore the result does not satisfy momentum and thus it is not expected to be
accurate. (ii) To examine the accuracy of this model, equation (l) is rewritten as
ª Grx º
« 4 »
¬
¼
1 / 4
Nu x
0.394( Pr)1/ 4
(m)
This result is compared with similarity solution (7.49) and integral solution (7.50) which
satisfies both momentum and energy. As expected, the accuracy of the integral method
PROBLEM 7.12 (continued)
deteriorates when the momentum equation is neglected. Only at Prandtl numbers of order
unity good accuracy is obtained.
ª Grx º
« 4 »
¬
¼
1 / 4
Nu x
Exact
Pr
0.01
0.09
0.5
0.72
1.0
2.0
10
100
Integral
omentum &
Eq. (7.49) M
energy, Eq (7.50)
0.0806
0.176
0.219
0.305
0.442
0.469.
0.5045
0.513
0.5671
0.557
0.7165
0.663
1.1649
0.991
2.191
1.762
Energy
Eq. (m)
0.0725
0.2166
0.4627
0.5361
0.6078
0.7751
1.2488
2.2665
PROBLEM 7.13
Consider laminar free convection over a vertical plate at uniform surface flux q csc . Assume
G G t and a third degree polynomial velocity profile given by
y ª yº
u o ( x) «1 »
G ¬ G¼
u x, y
2
Show that:
[a] An assumed second degree polynomial for the temperature profile gives
T ( x, y )
Tf 1 q csc ª
y2 º
«G 2 y »
2 k ¬«
G »¼
[b] The local Nusselt number is given by
Nu x
ª 4( Pr ) 2 Egq csc 4 º
x »
«
2
»¼
¬« 36 45 Pr kQ
1/ 5
(1) Observations. (i) This is a free convection problem
over a vertical plate at uniform surface heat flux. (ii) In
general, to determine the Nusselt number it is necessary
to determine the velocity and temperature distribution.
(iii) The integral method can be used to determine the
velocity and temperature distribution. (iv) Application of
the integral method reduces to determining the velocity
and temperature boundary layer thickness.
(2) Problem Definition. Determine the temperature
distribution for boundary layer flow over a flat plate
which is heated with uniform surface flux.
(3) Solution Plan. Start with the definition of local
Nusselt number. Equate Newton’s law with Fourier’s law to obtain an equation for the heat
transfer coefficient h. Apply the integral form of the energy equation using a second degree
polynomial temperature profile.
(4) Plan Execution. (i) Assumptions. (1) Steady state, (2) constant properties except for
buoyancy, (3) Boussinesq approximations are valid, (4) two-dimensional, (5) laminar flow
( Ra L 10 9 ), (6) flat plate, (7) uniform surface temperature, (8) negligible changes in kinetic
and potential energy, (9) negligible axial conduction, (10) negligible dissipation, and (11)
G Gt .
(ii) Analysis. The local Nusselt number is defined as
Nu x
where
h
local heat transfer coefficient, W/m 2 o C
hx
k
(a)
PROBLEM 7.13 (continued)
k thermal conductivity, W/m o C
Nu x local Nusselt number
x distance along plate measured from the leading edge, m
where the heat transfer coefficient h is given by equation (1.10)
wT ( x,0)
wy
Ts Tf
k
h
(1.10)
where
Ts
surface temperature, o C
Tf ambient temperature, o C
y coordinate normal to plate, m
Thus h depends on the temperature distribution T ( x, y ). The integral form of the energy
equation, (7.36), is used to determine the temperature distribution
wT x,0
D
wy
d
dx
Gt
³ u(T T
(b)
f )dy
0
where
u axial velocity, m/s
D
thermal diffusivity, m 2 /s
A third degree polynomial for the axial velocity u(x,y) gives
u x, y
u o ( x)
y ª yº
1
G «¬ G »¼
2
(c)
Assume a second degree polynomial temperature profile
T ( x, y )
b0 ( x) b1 ( x) y b2 ( x) y 2
The boundary conditions on the temperature are
wT ( x,0)
q csc
wy
(2) T x, G t # Tf
wT x, G t
#0
(3)
wy
(1) k
Equation (d) and the three boundary conditions give the coefficients bn (x )
b0
Tf Substituting the above into (d)
q csc
G,
2k
b1
q csc
, b2
k
q csc 1
2k G
(d)
PROBLEM 7.13 (continued)
ª
y2 º
2
y
G
«
»
G »¼
«¬
Surface temperature is determined by setting y = 0 in (e)
T ( x, y )
q csc
2k
Tf Tf T ( x, y )
(e)
q csc
G
2k
(f)
Substituting (f) into (1.10)
2
h
k
(g)
G
Combining (a) and (g)
2
Nu x
x
(h)
G
The problem reduces to finding G . The two unknown functions u o ( x) and G (x) are
determined using momentum equation (7.35) and energy equation (7.36):
G
wu ( x,0)
Q
E g (T Tf ) dy
wy
0
d
dx
³
wT x,0
D
wy
d
dx
³
G
u 2 dy
(7.35)
0
G ( x)
³
(7.36)
u (T Tf )dy
0
Substitute (c) and (e) into (7.35)
q cc
Q
Eg s
G
2k
uo
G
³
0
d ­° u o2
®
dx °̄G 2
ª
y2 º
«G 2 y » dy
G »¼
¬«
G
³
0
4
½°
yº
ª
y 2 «1 » dy ¾
¬ G¼
°¿
(i)
Evaluate the integrals in (i)
Q
uo
G
E gq csc
6k
G2
> @
1 d 2
uo G
105 dx
(j)
Similarly, substitute (c) and (e) into (7.36)
­
q csc d ° u o
®
k
2k dx ° G
¯
Evaluate the integrals and rearrange
D q csc
G ( x)
³
0
60D
yº
ª
y «1 »
¬ G¼
>
2
d
u oG 2
dx
½
ª
y2 º °
«G 2 y » dy ¾
G »¼ °
«¬
¿
(k)
@
(l)
Equations (j) and (l) are two simultaneous firs t order ordinary differential equations. The
two dependent variables are G (x) and u o (x). We assume a solution of the form
PROBLEM 7.13 (continued)
u o ( x)
Ax m
(m)
G ( x)
Bx n
(n)
where A, B, m and n are constants. To determine these constants we substitute (m) and (n)
into (j) and (l) to obtain
A m n E g q csc 2 2 n
vx
B x
B
6k
2m n 2 2 m n 1
A Bx
105
(o)
and
60D
(m 2n) AB 2 x m 2n1
(p)
To satisfy (o) and (p) at all values of x, the exponents of x in each term must be identical.
Thus, (o) requires that
mn
2n
2m n 1
(q)
Similarly, (p) requires that
m 2n 1 0
(r)
Solving (q) and (r) for m and n gives
m
1
5
3
, n
5
(s)
Introducing (s) into (o) and (p) gives two simultaneous algebraic equations for A and B
A E g q csc 2
B
B
6k
Q
2m n 2
A B
105
(t)
and
60D
(m 2n) AB 2
(u)
Solving equations (t) and (u) for A and B, gives
ª 360D k § 4
·º
A 60D «
¨ D Q ¸»
¹¼
¬ E g q csc © 5
2 / 5
(v-1)
and
ª (3600 / 75)D 2 60Q D º
«
»
E g q csc /(6k )
«¬
»¼
B
1/ 5
(w-1)
Note that
Pr
Q
D
(x)
Substitute into (v-1) and (w-1)
A
ª 360D 2 k § 4
·º
60D «
¨ Pr ¸»
¹»¼
«¬ E g q csc © 5
2 / 5
(v-2)
PROBLEM 7.13 (continued)
B
ª 360 a 2 k ^(4 / 5) Pr`º
«
»
E g q csc
¬«
¼»
1/ 5
(w-2)
Substitute (s) and (w-2) into (n), rearranging and introducing the definition of Rayleigh
number, gives the solution to G ( x) / x
G
x
ª 360 a 2 k ^(4 / 5) Pr`º
«
»
E g q csc
¬«
¼»
1/ 5
x (1 / 5)1
(y)
Introduce (y) into (h), use (x) and rearrange gives the local Nusselt number
Nu x
ª 4 Pr 2 E g q csc 4 º
x »
«
2
«¬ 36 45Pr kQ
»¼
1/ 5
(z)
(iii) Checking. Dimensional check: Equations (c), (e), (j), (l), (t), (u), (v-1), (v-2), (w-1)
and are dimensionally correct. Equations (a), (h), (y) and (z) are dimensionless.
Boundary conditions check: Temperature profile (e) satisfy the three listed boundary
conditions.
Limiting check: If q csc
Nu x 0.
0 , the Nusselt number should vanish. Setting q csc
0 in (z) gives
(5) Comments. (i) The same approach can be used to solve the corresponding problem of
variable surface flux, q csc (x).
(ii) The accuracy of the integral solution can be evaluated by comparing (z) with the exact
solution. Equation (7.32) gives the exact solution to the local Nusselt number for free
convection over a vertical plate at uniform surface flux
Nu x
ª E g q cc º
« 2 s x4 »
»¼
¬« 5Q k
1/ 5
1
T ( 0)
(7.32)
where the parameter T (0) is determined using correlation equation (7.33)
T ( 0)
ª 4 9 Pr 1 / 2 10 Pr º
«
»
5 Pr 2
¬«
¼»
1/ 5
, 0.001 Pr 1000
(7.33)
To facilitate the comparison, the two solutions are rearranged. Integral solution (z) is
rewritten as
ª E g q csc 4 º
Nu x «
x »
2
»¼
¬« kQ
1 / 5
ª 4 Pr 2 º
«
»
«¬ 36 45Pr »¼
1/ 5
Similarly, exact solution (7.32) is rewritten using (7.33) to eliminate T (0)
(A-1)
PROBLEM 7.13 (continued)
ª E g q cc º
Nu x « 2 s x 4 »
»¼
¬« Q k
15
ª
º
Pr 2
«
»
1/ 2
¬« 4 9 Pr 10 Pr ¼»
Thus the accuracy of the integral
solution can be evaluated by comparing
(A-1) and (A-2). The following table
compares the two solutions.
The agreement between the two solution
is excellent. At Pr = 0.01 the error is
8.3%. At all other Prandtl numbers from
0.1 to 100 the error ranges from 1% to
3.3%.
1/ 5
(A-2)
ª E g q cc º
Nu x « 2 s x 4 »
»¼
«¬ Q k
Pr
0.01
0.1
0.5
1.0
5
10
100
Integral (A-1)
0.1019
0.2505
0.4432
0.5479
0.8254
0.9618
1.5455
15
Exact (A-2
0.1111
0.2637
0.4388
0.5340
0.8046
0.9453
1.5567
PROBLEM 8.1
Water at 120oC boils inside a channel with a flat
surface measuring 45 cm u 45 cm. Air at 62 m/s
and 20oC flows over the channel parallel to the
surface. Determine the heat transfer rate to the air.
Neglect wall resistance.
Vf
Tf
air
water
water
(1) Observations. (i) This is an external forced
convection problem. (ii) The geometry can be modeled as a flat plate. (iii) Surface temperature
is uniform. (iv) Newton’s law of cooling gives heat transfer rate from the surface to the air. (v)
The average heat transfer coefficient must be determined. (vi) The Reynolds number should be
evaluated to establish if the flow is laminar, turbulent or mixed. (vii) Analytic or correlation
equations give the heat transfer coefficient.
(2) Problem Definition. Determine the average heat transfer coefficients for forced convection
over a flat plate.
(3) Solution Plan. Compute the Reynolds number at the trailing edge to establish if it is laminar
or turbulent. Apply Newton’s law of cooling. Use analytic or correlation equations to determine
the local heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4)
uniform upstream velocity and temperature, (5) flat plate, (6) uniform surface temperature, (7)
constant properties, (8) no buoyancy (E = 0 or g = 0) and (9) no radiation
(ii) Analysis. Newton’s law of cooling gives the total heat transfer rate
qT
h L2 (Ts Tf )
(a)
h = average heat transfer coefficient, W/m 2 o C
L = length and width of the flat surface = 0.45 m
qT = total heat transfer rate, W
Ts = surface temperature = 120 o C
Tf = free stream temperature = 20 o C
To determine the average heat transfer coefficient, the Reynolds number is computed to establish
if the flow is laminar or turbulent. The Reynolds number is defined as
Re x =
Vf x
Q
where
Re x = Reynolds number
Vf = upstream velocity = 62 m/s
x = distance from the leading edge of the plate, m
Q = kinematic viscosity, m 2 /s
Properties are evaluated at the film temperature T f defined as
(b)
PROBLEM 8.1 (continued)
Tf
Ts Tf
2
(c)
For the flow over a flat plate, transition Reynolds number Re x t is
Re x t = 5u105
(d)
The flow is laminar if Re x Re x t . Substituting into (c)
T f = (120 + 20)( o C )/2 = 70oC
Properties of air at this temperature are given in Appendix D
k = 0.02922 W/m o C
Pr = 0.707
Q = 19.9u10-6 m 2 /s
Evaluating the Reynolds number in (b) at x
ReL =
62(m/s)0.45(m)
19.9 u 10 6 (m 2 /s)
L
1.402 u 10 6
Therefore, the flow is mixed over the plate. The average Nusselt, Nu L , number for a plate with
laminar and turbulent flow is given by equation (8.7b)
Nu L
^0.664( Rex )
hL
k
t
1/ 2
>
0.037 ( Re L ) 4 / 5 ( Re x t ) 4 / 5
@` Pr
1/ 3
(e)
Equation (e) is subject to the limitations on th e Pohlhausen’s solution and following conditions
flat plate, constant Ts
5 u10 5 < Re x < 10 7
0.6 < Pr < 60
properties at T f
All conditions are satisfied.
(iii) Computations. Solving (e) for h
h=
0.02922( W/m o C)
0.664(5 u 10 5 )1 / 2 0.037[(1.4 u 10 6 ) 4 / 5 (5 u 10 5 ) 4 / 5 ] (0.707)1 / 3
0.45(m)
^
`
h = 126.6 W/m 2 o C
Substituting into (a)
qT = 126.6( W/m 2 o C ) (0.45) 2 (m 2 ) (120 - 20)( o C ) = 2,562.8 W
(iv) Checking. Dimensional check: Computations showed that equations (a), (b) and (e) are
dimensionally consistent.
Quantitative check: The magnitudes of the heat transfer coefficient is within the range given in
Table 1.1 for force convection of gases.
PROBLEM 8.1 (continued)
Limiting check: If the flow is laminar over the entire plate, ReL = Re xt , equation (e) reduces to
Pohlhausen’s solution.
(5) Comments. (i) It is important to compute the Reynolds number to determine if the flow is
laminar or turbulent. (ii) For laminar forced convection over a flat plate, the heat transfer
coefficient decreases as the distance from the leading edge is increased. However, if transition
takes place, the heat transfer coefficient increases at the transition location and drops as the
distance from the leading edge is increased.
PROBLEM 8.2
Steam at 105oC flows inside a specially designed narrow channel.
Water at 25oC flows over the channel with a velocity of 0.52 m/s.
Assume uniform outside surface temperature Ts = 105oC.
water
L
W
[a] Determine surface heat flux at 20 cm and 70 cm down- stream
from the leading edge of the channel.
[b] Determine the total heat removed by the water if the length is L
= 80 cm and the width is W = 100 cm.
steam
(1) Observations. (i) This is an external forced convection problem. (ii) The geometry can be
modeled as a flat plate. (iii) Surface temperature is uniform. (iv) To determine the heat flux at
a given location, the local heat transfer coefficient must be determined. (v) The average heat
transfer coefficient is needed to determine the total heat transfer rate. (vi) Newton’s law of
cooling gives surface flux and total heat transfer rate. (vii) The Reynolds number should be
checked to establish if the flow is laminar, turbulent or mixed. (viii) Analytic or correlation
equations give the heat transfer coefficient.
(2) Problem Definition. Determine the local and average heat transfer coefficients for forced
convection over a flat plate.
(3) Solution Plan.
[a] Apply Newton's law of cooling to determine the local heat flux. Check the Reynolds number
at 0.2 m and 0.7 m from the leading edge to see if it is laminar or turbulent. Use analytic or
correlation equations for the local heat transfer coefficient.
[b] Apply Newton's law of cooling to the entire plate to determine the total heat loss. Check the
Reynolds number at 0.8 m from the leading edge. Select an appropriate equation to determine the
average heat transfer coefficient for the plate.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4)
uniform upstream velocity and temperature, (5) flat plate, (6) uniform surface temperature, (7)
constant properties, (8) no buoyancy (E = 0 or g = 0) and (9) no radiation.
(ii) Analysis.
qscc = h (Ts - Tf )
(a)
where
water
Vf
Tf
h = local heat transfer coefficient, W/m2-oC
qscc = local surface heat flux, W/m2
Ts = surface temperature = 105oC
Tf = free stream temperature = 25oC
qcsc
L
[a] Newton’s law of cooling gives the local heat flux
0
W
x
Ts
steam
To determine the local heat transfer coefficient, the Reynolds number is computed to establish if
the flow is laminar or turbulent. The Reynolds number is defined as
Rex =
Where
Vf x
Q
(b)
PROBLEM 8.2 (continued)
Rex = Reynolds number
Vf = upstream velocity = 0.52 m/s
x = distance from the leading edge of the plate, m
Q = kinematic viscosity, m2 /s
Properties are evaluated at the film temperature Tf defined as
Ts Tf
2
For the flow over a flat plate, transition Reynolds number Rex is
Tf =
(c)
t
Re xt = 5u105
(d)
The flow is laminar if Rex < Re xt .Substituting into (c)
Tf = (105 + 25)(oC)/2 = 65oC
Properties of water at this temperature are given in Appendix D
k = 0.6553 W/m-oC
Pr = 2.77
Q = 0.4424u10-6 m2/s
The Reynolds number at x = 20 cm = 0.2 m is
Rex =
0.52(m/s)0.2(m)
0.4424 u 10 6 (m 2 /s)
235,081
Comparing this with the transition Reynolds number in (d) shows that the flow is laminar at x =
20 cm. Similarly, the Reynolds number at x = 70 cm = 0.7 m is
Rex =
0.52(m / s)0.7(m)
0.4424 u 10 6 (m 2 / s )
822,785
Thus, the flow is turbulent at x = 70 cm. The local heat transfer coefficient for laminar flow is
given by Pohlhausen’s solution. For 0.5 < Pr < 50, equation (7.24b) gives
h = 0.332 k
Vf
Pr1/3
Qx
(e)
The local Nusselt number for turbulent flow over a flat plate is given by equation (8.4a)
Nu x
hx
k
0.0296( Re x ) 4 / 5 ( Pr )1 / 3
(f)
This correlation equation applies to
flat plate, constant Ts
5u105 < Rex < 107
0.6 < Pr < 60
properties at Tf
(g)
Since all conditions in (g) are satisfied, equation (f) is applicable to this case.
[b] Newton’s law of cooling gives the total heat transfer rate
qT = h LW (Ts Tf )
(h)
PROBLEM 8.2 (continued)
where
h = average heat transfer coefficient, W/m2-oC
L = length of plate = 80 cm = 0.8 m
qT = total heat transfer rate, W
W = width of plate = 100 cm = 1.0 m
The Reynolds number at x = L = 0.8 m is
0.52(m / s)0.8(m)
ReL =
940,325
0.4424 u 10 6 (m 2 / s)
Therefore, the flow is mixed over the plate. The average Nusselt number for a plate with laminar
and turbulent flow is given by equation (8.7b)
Nu L
hL
k
^0.664( Re t )
x
1/ 2
>
@`
0.037 ( Re L ) 4 / 5 ( Re xt ) 4 / 5 Pr
1/ 3
(i)
where
h = average heat transfer coefficient, W/m2-oC
Equation (i) is subject to the condi tions listed in (g) and the limitations on Pohlhausen’s solution.
Both are satisfied by this case.
(iii) Computations.
[a] Heat flux at x = 20 cm. The local heat transfer coefficient for laminar flow at x = 20 cm is
determined using (e). Solving (e) for h
0.6553( W / m o C)
0.332(235,081)1 / 2 (2.77)1 / 3 = 740.7 W/m2-oC
0.2(m)
Substituting this result in (a)
h=
q csc = 740.7(W/m2-oC)(105 - 25)(oC) = 59,256 W/m2
Heat flux at x = 70 cm. The local heat transfer coefficient is given by (f). Solving (f) for h
h = 0.0296
0.6553( W / m o C)
(822,785)4/5 (2.77)1/3 = 2,101 W/m2-oC
0.7(m)
Substituting into (a)
q csc = 2101(W/m2-oC)(105 - 25)(oC) = 168,080 W/m2
[b] Total heat transfer from plate. Solving (i) for h
>
^
@`
h
k
0.664( Re xt )1 / 2 0.037 ( Re L ) 4 / 5 ( Re xt ) 4 / 5 Pr
L
h=
0.6553( W / m o C)
0.664(500,000)1 / 2 0.037[(940,325) 4 / 5 (500,000) 4 / 5 ] (2.77)1 / 3 = 1554.3 W/m2-oC
0.8(m)
1/ 3
^
Substituting into (h)
qT = 1554.3(W/m2-oC) 0.8(m)1.0(m) (105 - 25)(oC) = 99,475 W
`
PROBLEM 8.2 (continued)
(iv) Checking. Dimensional check: Computations showed that units of equations (a), (b),
(e), (f), (h) and (i) are dimensionally consistent.
Quantitative check: The magnitudes of heat transfer coefficients are within the range given in
Table 1.1 for force convection of liquids.
Limiting check: If the flow is laminar over the entire plate, ReL = Re xt , equation (i) reduces to
Pohlhausen’s solution.
(5) Comments. (i) It is important to check the Reynolds number to determine if the flow is
laminar or turbulent. (ii) For laminar forced convection over a flat plate, the heat transfer
coefficient decreases as the distance from the leading edge is increased. However, if transition
takes place, the heat transfer coefficient increases at the transition location and drops as the
distance from the leading edge is increased. (iii) The fact that the average heat transfer
coefficient for the entire plate (L = 0.8 m) is larger than the local heat transfer coefficient at x =
0.2 m is due to transition from laminar to turbulent flow.
PROBLEM 8.3
Electronic components are mounted on one side of a circuit board. The board is cooled on the
other side by air at 23oC flowing with a velocity of 10 m/s. The length of the board is L = 20 cm
and its width is W = 25 cm. Assume uniform board temperature.
[a] Determine the maximum power that can be dissipated in the package if surface temperature
is not to exceed 77oC. Assume that all dissipated power is conducted through the plate to the air.
[b] To increase the maximum power without increasing surface temperature, it is recommended
that the boundary layer be tripped to turbulent flow very close to the leading edge. Is this a valid
recommendation? Substantiate your view.
(1) Observations. (i) This is an external forced convection problem of flow over a flat plate.
(ii) Surface temperature is assumed uniform. (iii) The heat transfer coefficient in turbulent flow
is greater than that in laminar flow. Thus higher heat transfer rates
can be sustained in turbulent flow than laminar flow. (iv) The
L
Tf
Reynolds number should be checked to establish if the flow is
laminar, turbulent or mixed. (v) Heat loss from the surface is V
f
W
approximately equal to the power dissipated in the package. (vi)
Newton’s law of cooling gives a relationship between heat transfer
components
rate, surface area, heat transfer coefficient, surface temperature and
ambient temperature. (vii) The fluid is air.
(2) Problem Definition. Determine the average heat transfer coefficient.
(3) Solution Plan. Apply Newton’s law of cooling to determine the maximum heat removed,
check the Reynolds number to establish if the flow is laminar, turbulent or mixed and use
appropriate average Nusselt number solutions or correlation equations to determine the average
heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4)
uniform upstream velocity and temperature, (5) flat plate, (6) uniform surface temperature, (7)
constant properties, (8) energy leaves the top surface only, (9) no buoyancy ( E = 0 or g = 0) and
(9) negligible radiation.
(ii) Analysis.
[a] Applying Conservation of energy to the package gives
P=q
(a)
where
P = power dissipated in package, W
q = heat removed from top surface, W
Application of Newton’s law of cooling to the top surface
q = h A (Ts - Tf)
where
A = plate area, m2
(b)
PROBLEM 8.3 (continued)
h = average heat transfer coefficient, W/m2-oC
Ts = surface temperature = 77oC
Tf = ambient temperature = 23oC
Surface area is
A = LW
(c)
where
L = plate length = 20 cm = 0.2 m
W = plate width = 25 cm = 0.25m
To determine h it is necessary to first establish if the flow is laminar or turbulent. If the
Reynolds number at the trailing end is smaller than the transition number, the flow is laminar
throughout. Define
V L
(d)
ReL = f
Q
and
Re xt = 5u105
(e)
where
Re L = Reynolds number at the trailing end of plate
Re xt = transition Reynolds number for flow over a flat plate
Vf = free stream velocity = 10 m/s
Q = kinematic viscosity, m2/s
Properties are evaluated at the film temperature Tf
Tf = (Ts + Tf)/2 = (77 + 23)(oC)/2 = 50oC
Properties of air at this temperature are given in Appendix C
k = thermal conductivity = 0.02781 W/m-oC
Pr = Prandtl number = 0.709
Q = kinematic viscosity = 17.92 u10-6 m2/s
Substituting into (d)
Re L =
10( m / s)0.2( m)
= 111,607
17.92 u 10 6 ( m 2 / s)
Since this is smaller than Re xt , it follows that the flow is laminar throughout. Therefore,
Pohlhausen's solution (7.26)for the average Nusselt number is applicable.
Nu L
hL
k
0.664 Re L
1/ 2
Pr
1/ 3
(f)
where Nu L is the average Nusselt number. Solving (f) for h
h
0.664
k
Re L
L
1/ 2
Pr
1/ 3
[b] The average Nusselt number for mixed flow over a flat plate of length L is given by (8.7b)
(g)
PROBLEM 8.3 (continued)
Nu L
hL
k
^0.664( Re t )
>
1/ 2
@`
0.037 ( Re L ) 4 / 5 ( Re xt ) 4 / 5 Pr
x
1/ 3
(h)
If the boundary layer is tripped at the leading edge, the flow will be turbulent throughout. Since
transition is assumed to take place at x = 0, it follows that
Re xt = 0
(i)
Substituting (i) into (h)
Nu L
hL
k
0.037 ( Re L )4/5 (Pr)1/3
(j)
k
( Re L )4/5 (Pr)1/3
L
(k)
Solving (j) for h
h = 0.037
(iii) Computations.
[a] Laminar flow. Equation (c) gives surface area
A = 0.2(m) 0.25(m) = 0.05 m2
Use (g) to calculate h
h
0.664
0.02781( W / m o C)
111,607
0.2( m)
1/ 2
0.709
1/ 3
= 27.5 W/m2-oC
Substituting into (b) and using (a) gives
P = q = 27.5 (W/m2-oC) 0.05(m2) (77 - 23)(oC) = 74.25 W
[b] Turbulent flow. Use (k) to obtain h
h = 0.037
0.02781( W / m o C)
(111,607)4/5 (0.709)1/3 = 50.09 W/m2-oC
0.2 ( m)
Substituting into (b) and using (a)
P = q = 50.09 (W/m2-oC) 0.05(m2) (77 - 23)(oC) = 135.2 W
(iv) Checking. Dimensional check: Computations showed that equations (b), (c), (d), (g)
and (k) are dimensionally consistent.
Quantitative check: The magnitude of the heat transfer coefficient for both laminar and turbulent
flow is within the range of values listed in Table 1.1. Furthermore, as expected, the average heat
transfer coefficient for turbulent flow is higher than laminar flow.
(5) Comments. (i) Tripping the boundary layer increases the maximum power by 82 % without
increasing surface temperature. (ii) The disadvantage of tripping the boundary layer is the
corresponding increase in pressure drop.
PROBLEM 8.4
Water at 15oC flows with a velocity of 0.18 m/s over a plate of length L = 20 cm and width W =
25 cm. Surface temperature is 95oC. Determine the heat transfer rate from the leading and
trailing halves of the plate.
(1) Observations. (i) This is an external forced convection problem. (ii) The geometry is a flat
plate. (iii) Surface temperature is uniform. (iv) Newton’s law of cooling gives the heat transfer
rate. (v) The Reynolds number should be checked to establish if the flow is laminar, turbulent or
mixed. (vi) Analytic or correlation equations give the heat transfer coefficient. (vii) If the flow is
laminar throughout, heat transfer from the first half should be greater than that from the second
half. (viii) Second half heat transfer can be obtained by subtracting first half heat rate from the
heat transfer from the entire plate. (ix) The fluid is water.
(2) Problem Definition. Determine the average heat transfer coefficient for the first half and for
the entire plate.
(3) Solution Plan. Apply Newton’s law of cooling to the first half and to the entire plate. Check
the Reynolds number at end of the first half and second half to establish if the flow is laminar,
turbulent or mixed. Use analytic or correlation equations for the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4)
uniform upstream velocity and temperature, (5) flat plate, (6) uniform surface temperature, (7)
constant properties, (8) no buoyancy ( E = 0 or g = 0 ) and (9) no radiation.
(ii) Analysis. Application of Newton’s law of cooling to the first half gives
q1 = h1 A1(Ts - Tf )
(a)
where
A1 = surface area of half plate, m2
h1 = average heat transfer coefficient for the first half, W/m2-oC
q1 = heat transfer rate from the first half, W
Ts = surface temperature = 95oC
Tf = free stream temperature = 15oC
Tf
0
Heat transfer from the second half is given by
q2
qT q1
(b)
where
q 2 = heat transfer rate from the second half, W
qT = heat transfer rate from entire plate, W
Heat transfer from the entire plate is given by
Vf
L/2
q1
q2
x
L/2
qT = hT AT(Ts - Tf )
where
AT = surface area of entire plate of length L, m2
hT = average heat transfer coefficient for entire plate of length L, W/m2-oC
(c)
PROBLEM 8.4 (continued)
The areas A1 and AT are
A1 = WL/2
(d)
AT = WL
(e)
and
where
L = plate length = 20 cm = 0.2 m
W = plate width = 25 cm = 0.25 m
To determine the average heat transfer coefficient, the Reynolds number is computed to establish
if the flow is laminar or turbulent. The Reynolds number is defined as
V x
(f)
Rex = f
Q
where
Rex = Reynolds number
Vf = upstream velocity = 0.18 m/s
x = distance from the leading edge of the plate, m
Q = kinematic viscosity, m2 /s
Properties of water are evaluated at the film temperature Tf defined as
Ts Tf
2
For the flow over a flat plate, transition Reynolds number Re xt is
Tf =
(g)
Re xt = 5u105
(h)
The flow is laminar if Rex < Re xt .Substituting into (g)
Tf = (95 + 15)(oC)/2 = 55oC
Properties of water at this temperature are given in Appendix D
k = 0.6458 W/m-oC
Pr = 3.27
Q = 0.5116u10-6 m2/s
The Reynolds number for the first half is evaluated at x = L/2 = 20 cm/2 = 10 cm = 0.1 m
ReL/2 =
0.18(m / s)0.1(m)
0.5116 u 10 6 m 2 / s
35,184
Therefore, the flow is laminar over the first half. The average Nusselt number for laminar flow is
given by Pohlhausen’s solution. For 0.5 < Pr < 50, equation (7.26) gives
h1 ( L / 2)
Nu L / 2
0.664 Pr 1/3 ( Re L / 2 )1 / 2
(i)
k
where
Nu L / 2 = average Nusselt number for the first half
The Reynolds number for the entire plate is evaluated at x = L = 20 cm = 0.2 m
ReL =
0.18(m / s)0.2(m)
0.5116 u 10 6 m 2 / s
70,367
PROBLEM 8.4 (continued)
Therefore, the flow is laminar over the entire plate. The average Nusselt number for laminar is
hT L
Nu L
0.664 Pr 1/3 Re1L/ 2
(j)
k
(iii) Computations. Substituting into equations (d) and (e)
A1 = 0.2(m)0.25(m)/2 = 0.025 m2
and
AT = 0.2(m)0.25(m) = 0.05 m2
Solving (i) for h1
h1
2k
o
0.664 Pr 1/3 Re1L//22 = 2(0.6458)( W / m C) 0.664(3.27)1 / 3 (35,184)1 / 2 = 1193.9 W/m2-oC
L
0.2(m)
Similarly, equation (j) gives
k
o
hT
0.664 Pr 1/3 ( Re L )1 / 2 = (0.6458)( W / m C) 0.664(3.27)1 / 3 (70,367)1 / 2 = 844 W/m2-oC
L
0.2(m)
Substituting into (a)
q1 = 0.025(m2)1193.9(W/m2-oC)(95 15 )(oC) = 2388 W
Equation (c) gives
qT = 0.05(m2)844(W/m2-oC)(95 15 )(oC) = 3376 W
Substituting into (b) gives the heat transfer rate from the second half
q 2 = 3376 (W) 2388 (W) = 988 W
(iv) Checking. Dimensional check: Computations showed that units of equations (a), (c)-(f),
(i) and (j) are dimens ionally consistent.
Quantitative check: The magnitudes of heat transfer coefficients are within the range given in
Table 1.1 for forced convection of liquids.
Qualitative check: As anticipated, heat transfer from the first half is greater than that from the
second half.
(5) Comments (i) For laminar flow, two half plates oriented in parallel (side by side) transfer
more heat than two placed in series. (ii) An alternate method for determining q 2 is to determine
the average heat transfer coefficient for the second half. This requires integration of the local
heat transfer coefficient from x = 0.1 m to x = 0.2 m.
PROBLEM 8.5
A chip measuring 5 mm u 5 mm is placed flush on a flat plate 18 cm from the leading edge. The
chip is cooled by air at 17oC flowing with a velocity of 56 m/s. Determine the maximum power
that can be dissipated in the chip if its surface temperature is not to exceed 63oC. Assume no
heat loss from the back side of the chip.
(1) Observations. (i) The chip is cooled by forced convection. (ii) This problem can be
modeled as a flat plate with an unheated leading section. (iii) Newton's law of cooling can be
applied to determine the rate of heat transfer between the chip and the air. (iv) Check the
Reynolds number to establish if the flow is laminar or turbulent.
(2) Problem Definition. Find the average heat transfer coefficient over the chip.
(3) Solution Plan. Apply Newton’s law of cooling to determine the maximum heat removed
from the chip, check the Reynolds number to establish if the flow is laminar or turbulent, model
the chip as flat heated surface with an insulated
S
leading section and use appropriate Nusselt number T
f
0
solutions or correlation equations to determine the
x
chip
average heat transfer coefficient.
Vf
xo
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4)
uniform upstream velocity and temperature, (5) flat plate, (6) chip surface temperature is
uniform, (7) constant properties, (8) all power dissipated in the chip is transferred to the air from
its surface, (9) the chip is mounted on an insulated plate, (10) no buoyancy ( E = 0 or g = 0 ) and
(10) no radiation.
(ii) Analysis. Applying Conservation of energy to the chip
P=q
(a)
where
P = power dissipated in chip, W
q = heat removed from top surface, W
Application of Newton’s law of cooling to the top surface
q = h A (Ts - Tf)
(b)
where
A = surface area of chip, m2
h = average heat transfer coefficient, W/m2-oC
Ts = surface temperature = 63oC
Tf = ambient temperature = 17oC
The surface area is
A = LW
(c)
where
L = chip length = 5 mm = 0.005 m
W = chip width = 5 mm = 0.005 m
To determine h it is necessary to first establish if the flow over the chip is laminar or turbulent.
PROBLEM 8.5 (continued)
If the Reynolds number at the leading end of the chip is larger than the transition number, the
flow is turbulent over the chip. Define
V x
Rex = f
(d)
Q
and
Re xt = 5u105
(e)
where
Re x = local Reynolds number
Re xt = transition Reynolds number for flow over a flat plate
Vf = free stream velocity = 56 m/s
x = variable, measured from the leading edge of plate, m
Q = kinematic viscosity, m2/s
Properties are evaluated at the film temperature Tf
Tf = (Ts + Tf)/2 = (63 + 17)(oC)/2 = 40oC
Properties of air at this temperature are given in Appendix C
k = thermal conductivity = 0.0271 W/m-oC
Pr = Prandtl number = 0.71
Q = kinematic viscosity = 16.96 u10-6 (m2/s)
Evaluating the Reynolds number in (d) at the leading edge of the chip x = xo = 0.18 m
Re xo =
56( m / s) 018
. ( m)
16.96 u 10 6 ( m2 / s)
= 594,340
Since this is larger than Re xt , it follows that the flow is turbulent over the chip. Equation (8.8)
gives the local Nusselt number for flow over a plate with an insulated leading section
Nu x
hx
k
0.0296 ( Re x ) 4 /5 ( Pr ) 1/3
> 1 ( xo / x ) 9/10 @1/9
(f)
where
Nux = local Nusselt number
xo = distance from the leading edge of the plate to the chip = 0.18 m
However, what is needed in equation (b) is the average heat transfer coefficient. Since the chip
is small compared to the distance xo, it is reasonable to assume that the average Nusselt number
is approximately equal to the local value at the center of the chip, s
Nu s
h s 0.0296( Re s ) 4 / 5 ( Pr )1/3
|
1/ 9
k
1 ( x / s) 9 / 10
>
o
@
where
Nus = average Nusselt number over the chip
s = distance from leading edge of plate to center of chip, m
(g)
PROBLEM 8.5 (continued)
Reynolds number Res and distance s are given by
Res =
Vf s
Q
(h)
and
s = xo + L/2
(i)
(iii) Computations. Substituting into (i) and (h)
s = 0.18(m) + 0.005(m)/2 = 0.1825 m
Res =
56( m / s) 01825
.
( m)
= 602,594
6
2
16.96 u 10 ( m / s)
Solving (g) for h and substituting numerical values
h
4/5
1/3
k 0.0296( Re s ) ( Pr )
1
/
9
s
1 ( x o / s) 9 / 10
>
@
0.0271( W / m o C) 0.0296(602,594) 4 / 5 (0.71)1 / 3
0.1825(m)
[1 (0.18 m / 0.1825 m) 9 / 10 ]1 / 9
h = 268.8 W/m2-oC
Substituting into (b) and using (a) and (c)
P = q = 268.8 (W/m2-oC)0.005(m)0.005(m) ( 63 17 )(oC) = 0.309 W
(iv) Checking. Dimensional check: Computations showed that equations (b)-(d) and (g)-(i)
are dimensionally consistent.
Quantitative check: The magnitude of the heat transfer coefficient is within the range of values
listed in Table 1.1 for forced convection of gases.
(5) Comments. The assumption that the average heat transfer coefficient over the chip is
approximately equal to the local value at the center of the chip was made to avoid the need to
integrate the local value, equation (f), over the surface of the chip. This approximation becomes
less reasonable as the dimension of the chip in the x direction becomes large.
PROBLEM 8.6
A 1.2 m u 1.2 m solar collector is mounted flush on
the roof of a house. The leading edge of the collector
is located 5 m from the leading edge of the roof.
Estimate the heat loss to the ambient air on a typical
winter day when wind speed parallel to the roof is 12
m/s and air temperature is 5oC. Outside collector
surface temperature is estimated to be 35oC.
5 .6 m
Vf
Tf
solar
collector
(1) Observations. (i) Heat transfer from the collector to the air is by forced convection. (ii) This
problem can be modeled as a flat plate with an unheated leading section. (iii) Newton's law of
cooling can be applied to determine the rate of heat transfer between the collector and air. (iv)
The heat transfer coefficient varies along the collector. (v) The Reynolds number should be
computed to establish if the flow is laminar or turbulent.
(2) Problem Definition. Determine the heat transfer coefficient over the collector.
(3) Solution Plan. Apply Newton’s law of cooling to the collector, compute the Reynolds
number to establish if the flow is laminar or turbulent, model the collector surface as flat heated
surface with an insulated leading section and use appropriate Nusselt number solutions or
correlation equations to determine the heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4)
uniform upstream velocity and temperature, (5) flat surface, (6) collector surface temperature is
uniform, (7) constant properties, (8) roof is insulated, (9) no buoyancy ( E = 0 or g = 0 ) and (9)
no radiation.
(ii) Analysis. Since the heat transfer coefficient varies with distance along the collector,
integration of Newton’s law of cooling gives the total heat transfer to the air
L
qT
³
(Ts Tf ) w h( x) dx
(a)
0
where
h ( x ) = local heat transfer coefficient, W/m 2 o C
L = length of collector = 1.2 m
qT = total heat transfer rate, W
Ts = surface temperature = 35 o C
Tf = ambient temperature = 5 o C
w = width of collector = 1.2 m
To determine the local heat transfer coefficient, the Reynolds number is computed to establish if
the flow is laminar or turbulent. If the Reynolds number at the leading end of the collector is
larger than the transition number, the flow is turbulent over the collector. Define
Rex =
Vf x
Q
(b)
and
Re xt = 5u105
(c)
PROBLEM 8.6 (continued)
where
Re x = local Reynolds number
Re xt = transition Reynolds number for flow over a flat plate
Vf = wind speed = 12 m/s
x = distance from the leading edge of the roof, m
Q = kinematic viscosity, m 2 /s
Properties are evaluated at the film temperature T f , defined as
Ts Tf
2
Tf
(d)
Substituting into (d)
T f = (35 + 5)( o C )/2 = 20oC
Properties of air at this temperature are
k = 0.02564 W/m o C
Pr = 0.713
Q = 15.09u10-6 m 2 /s
Evaluating the Reynolds number in (b) at x
ReL =
12(m/s)5(m)
15.09 u10 6 (m 2 /s)
5m
xo
3.9761u10 6
Since the Reynolds number is greater than the transition number it follows that the flow is
turbulent over the collector. Equation (8.8) gives the local Nusselt number for flow over a plate
with an insulated leading section
Nu x
hx
k
0.0296 ( Re x ) 4 / 5 ( Pr) 1/3
> 1 (x
o /x)
@
9 /10 1/ 9
(e)
where
Nux = local Nusselt number
xo = distance from the leading edge of the roof to the collector = 5 m
Solving (e) for h
h
k 0.0296( Re x ) 4 / 5 ( Pr )1/3
1/ 9
x
1 ( x /x ) 9 / 10
>
o
@
(f)
However, when (f) is substituted in (a), the resulting integral can not be evaluated analytically.
Thus, numerical integration is required. An approximate approach is to assume that the heat
transfer coefficient over the collector is uniform equal to the local value at the center of the
collector. Thus
h ( x ) | h ( xc )
(g)
where
xc = distance from the leading edge of the roof to the center of collector = 5.6 m
Substituting (g) into (a) and evaluating the integral
PROBLEM 8.6 (continued)
qT
Evaluating (f) at x
(Ts Tf ) hc wL
(h)
xc
hc
4/5
1/3
k 0.0296( Rex c ) ( Pr )
1/ 9
xc
1 ( xo / x c) 9 /10
>
(i)
@
(iii) Computations. The Reynolds number at collector center x
Re xc =
12(m/s)5.6(m)
15.09 u10 6 (m 2 /s)
xc
5.6 m is
4.4533 u10 6
Substituting into (i)
hc
0.02564( W/m o C) 0.0296(4.4533 u10 6 ) 4 / 5 (0.713)1/ 3
5.6(m)
[1 (5 m / 5.6 m) 9 / 10 ]1/ 9
32.7 W/m 2 o C
Substituting (g)
qT
(35 5)( o C) 32.7( W/m 2 o C)1.2 (m)1.2 (m) 1412.6 W
(iv) Checking. Dimensional check: Computations showed that equations (b), (h) and (i) are
dimensionally consistent.
Quantitative check: The magnitude of the heat transfer coefficient is within the range of values
listed in Table 1.1 for forced convection of gases.
(5) Comments. The assumption that the average heat transfer coefficient over the collector is
approximately equal to the local value at the center of the collector was made to avoid the need
to numerically evaluate the integral in equation (a). This approximation becomes less reasonable
as the dimension of the collector in the x direction becomes large.
PROBLEM 8.7
Water at 20oC flows over a rectangular plate of length L =
1.8 m and width W = 0.3 m. The upstream velocity is 0.8
m/s and surface temperature is 80oC. Two orientations are
considered. In the first orientation the width W faces the
flow and in the second the length L faces the flow. Which
orientation should be selected to minimize heat loss from the
plate? Determine the heat loss ratio of the two orientations.
L1
x
0
1
Vf
W1
W2
0
x
Tf
2
(1) Observations. (i) This is an external forced convection
problem. (ii) The flow is over a flat plate. (iii) Surface
L2
temperature is uniform. (iv) Plate orientation is important.
(v) Variation of the heat transfer coefficient along the plate affects the total heat transfer. (vi)
The heat transfer coefficient for laminar flow decreases as the distance from the leading edge is
increased. However, at the transition point it increases and then decreases again. (vii) Higher rate
of heat transfer may be obtained if the wide side of a plate faces the flow. On the other hand,
higher rate may be obtained if the long side of the plate is in line with the flow direction when
transition takes place. (viii) The fluid is water.
(2) Problem Definition. Determine the average heat transfer coefficient for the flow over a
rectangular plate for two orientations: [a] wide side facing the flow and [b] short side facing the
flow.
(3) Solution Plan. Apply Newton's law of cooling for flow over a flat plate. Check the
Reynolds number for the two orientations and select appropriate equations for the average
Nusselt number to obtain the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4)
uniform upstream velocity and temperature, (5) flat plate, (6) uniform surface temperature, (7)
constant properties, (8) no buoyancy (E = 0 or g = 0 ) and (9) negligible radiation.
(ii) Analysis. Newton's law of cooling gives the total heat transfer rate
q = h A(Ts - Tf)
(a)
where
A = plate area, m2
h = average heat transfer coefficient, W/m2-oC
q = heat transfer rate, W
Ts = surface temperature = 80oC
Tf = ambient temperature = 20oC
Surface area is
A = LW
(b)
where
L = plate length = 1.8 m
W = plate width = 0.3 m
To determine h it is necessary to first establish if the flow is laminar or turbulent. If the Reynolds
PROBLEM 8.7 (continued)
number at the trailing end is smaller than the transition number, the flow is laminar throughout.
Define
V L
ReL = f
(c)
Q
and
Re xt = 5u105
(d)
where
ReL = Reynolds number at the trailing end of plate
Re xt = transition Reynolds number for flow over plate
Vf = free stream velocity = 0.8 m/s
Q = kinematic viscosity, m2/s
Properties are evaluated at the film temperature Tf
Tf = (Ts + Tf)/2 = (80 + 20)(oC)/2 = 50oC
Properties of water at this temperature are given in Appendix D
k = thermal conductivity = 0.6405 W/m-oC
Pr = Prandtl number = 3.57
Q = kinematic viscosity = 0.5537 u10-6 m2/s
Consider two orientations, 1 and 2 as shown. In orientation 1, the short side faces the flow. In
orientation 2, the long side faces the flow. The Reynolds numbers corresponding to the two
orientations are computed using (c)
0.8(m / s)1.8(m)
= 2,600,686, turbulent
0.5537 u 10 6 (m 2 / s)
Similarly, for orientation 2
0.8(m / s)0.3(m)
= 433,448, laminar
Re L2 =
0.5537 u 10 6 (m 2 / s)
Re L1 =
where
L1 = L = 1.8 m
L2 = W = 0.3 m
Comparing Re L and Re L with Re xt shows that the flow is mixed (laminar and turbulent) for
1
2
orientation 1, and laminar for orientation 2. The average Nusselt number for mixed flow over a
flat plate of length L1 is given by (8.7b)
Nu L 1
h1 L1
k
^0.664 Re t
x
1/ 2
>
0.037 Re L1
4/5
Re xt
4/5
@` Pr
1/ 3
(e)
Valid for
flat plate, constant Ts
5u105 < Rex < 107
0.6 < Pr < 60
properties at Tf
(f)
Since all conditions in (f) are satisfied, equation (e) is applicable to this case. For orientation 2,
Pohlhausen's solution (7.26) gives the average Nusselt number
PROBLEM 8.7 (continued)
h2 L 2
k
Nu L 2
0.664( Re L2 )1 / 2 Pr
1/ 3
(g)
(iii) Computations. Solving (E) for h1 and substituting numerical values
>
^
@`
h1
k
0.664( Re xt )1 / 2 0.037 ( Re L1 ) 4 / 5 ( Re xt ) 4 / 5
L1
h1
0.6405( W / m o C)
0.664(500,000)1 / 2 0.037 (2,600,686) 4 / 5 (500,000) 4 / 5 (3.57)1 / 3
1.8(m)
Pr
1/ 3
>
^
@`
h1 = 2253.5 W/m2-oC
Similarly, solving (g) for h2 and substituting numerical values
h2 = 0.664
k
( Re L2 )1 / 2 Pr
L2
1/ 3
0.6405( W / m o C)
(433,448)1 / 2 (3.57)1 / 3 = 1426.5W/m2-oC
0.3(m)
Substituting into equation (a) and using (b) gives the heat transfer rate for each orientation
h2 = 0.664
q1 = 2253.5(W/m2-oC)1.8(m)0.3(m)( 80 20 )(oC) = 73,013 W
and
q 2 = 1426.5(W/m2-oC)1.8(m)0.3(m)( 80 20 )(oC) = 46,219 W
The ratio of the two heat transfer rates is
q1
q2
73,013( W )
= 1.58
46,219( W )
(iv) Checking. Dimensional check: Computations showed that equations (a), (b), (c), (e)
and (g) are dimensionally consistent.
Quantitative check: Values of the heat transfer coefficients are within the range listed in Table
1.1 for forced convection of liquids.
Limiting check: For the special case of a square plate L1 = L2, the ratio of the two heat transfer
rates should be unity. Either equation (e) or (g) gives h1 = h2 . Equation (a) gives q1 = q 2 .
(5) Comments. If the flow is laminar for both orientations, orientation 1 will have a lower heat
transfer rate than orientation 2. However, if transition takes place in one or both orientations,
numerical calculations must be carried out to determine which orientation has the higher heat
transfer rate.
PROBLEM 8.8
100 flat chips are placed on a 10 cm u 10 cm circuit board and cooled by forced convection of
air at 27oC. Each chip measures 1 cm u 1 cm and dissipates 0.13 W. The maximum allowable
chip temperature is 83oC. Free stream air velocity is 5 m/s. Tests
showed that several chips near the trailing end of the board
Tf
exceeded the allowable temperature. Would you recommend
tripping the boundary layer to turbulent flow at the leading edge to
Vf
solve the overheating problem? Substantiate your recommendation.
(1) Observations. (i) This is an external forced convection problem. (ii) The flow is over a flat
plate. (iii) The problem can be modeled as flow over a flat plate with uniform surface heat flux.
(iv) Surface temperature varies with distance along plate. The highest surface temperature is at
the trailing end. (v) Tripping the boundary layer at the leading edge changes the flow from
laminar to turbulent. This increases the heat transfer coefficient and lowers surface temperature.
(vi) Newton’s law of cooling gives surface temperature.
(2) Problem Definition. Determine the local heat transfer coefficient at the trailing end for
turbulent flow.
(3) Solution Plan. Apply Newton's law of cooling at the trailing end and use a correlation
equation for turbulent flow over a flat plate at constant surface heat flux to determine the local
heat transfer coefficient. Solve Newton's law for the surface temperature.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4)
uniform upstream velocity and temperature, (5) flat plate, (6) uniform surface heat flux, (7)
constant properties, (8) no buoyancy ( E = 0 or g = 0 ), (9) negligible radiation and (10) turbulent
flow.
(ii) Analysis. Apply Newton's law of cooling
qscc = h(x) [Ts(x) - Tf ]
(a)
where
h(x) = local heat transfer coefficient, W/m2-oC
qscc = surface heat flux = 0.13 W/cm2 = 1300 W/m2
Ts(x) = local surface temperature, oC
Tf = free stream temperature = 27oC
x = distance from the leading edge, m
qcsc
L
Vf
Tf
0
x
Ts (x )
Solving equation (a) for Ts(x)
Ts ( x) Tf q csc
h( x )
(b)
The local Nusselt number for turbulent flow over a flat plate at constant surface heat flux is
given by equation (8.9)
hx
Nu x
0.030Re x4 / 5 Pr 1/ 3
(c)
k
PROBLEM 8.8 (continued)
where
k = thermal conductivity, W/m-oC
Nux = local Nusselt number
Rex = local Reynolds number = Vf x / Q
Pr = Prandtl number
Vf = free stream velocity = 5 m/s
Q = kinematic viscosity, m2/s
Equation (c) gives the local heat transfer coefficient. Properties are evaluated at the film
temperature defined as
(d)
T f (Ts Tf ) / 2
where
T f = film temperature, oC
Ts = average surface temperature, oC
Since surface temperature varies along the plate, an average value, Ts , is used to determine film
temperature. Ts is approximated by
[Tf Ts ( L)] / 2
Ts
(e)
(iii) Computations. To determine air properties, surface temperature at the trailing end,
Ts(L), is needed to compute Tf. However, Ts(L) is unknown. In fact, the objective of the problem
is determining Ts(L). To proceed, a solution is obtained by trial and error procedure. A value for
Ts(L)is assumed, (d) and (e) are used to determine Tf, (c) is used to calculate h(L) and (b) is used
to calculated Ts(L). If the calculated Ts(L) is not equal to the assumed value, the process is
repeated until a satisfactory agreement is obtained between assumed and calculated values.
Assume Ts(L)= 79oC. Equations (e) and (d) give
Ts = (27 + 79)(oC)/2 = 53oC
and
Tf = (53 + 27)(oC)/2 = 40oC
Air properties at this temperature are
k = 0.0271 W/m-oC
Pr = 0.71
Q = 16.96u10-6 m2/s
Thus, the Reynolds number at x = L = 0.1 m is
ReL =
Vf L
5(m / s)0.1(m)
Q
16.96 u 10 6 (m 2 / s)
29,481
Solving (c) for h and evaluating the resulting equation at x = L = 0.1 m
h( L) 0.03
k
( Re L ) 4 / 5 ( Pr )1 / 3
L
Substituting into (b)
0.03
0.0271( W / m o C)
(29,481) 4 / 5 (0.71)1 / 3 = 27.3 W/m2-oC
0.1(m)
PROBLEM 8.8 (continued)
Ts(L) = 27oC +
1300( W / m 2 )
= 74.6oC
27.3( W / m 2 o C)
This is reasonably close to the assumed value of 79oC. Repeating the calculation with a new
assumed value of 75oC will result in a minor change in the result.
(iv) Checking. Dimensional check: Computations showed that units for equations (b) and
(c) are consistent.
Quantitative check: the value of h(L) is within the range given in Table 1.1 for forced convection
of gases.
(5) Comments. (i) By tripping the boundary layer to cause transition to turbulent flow, surface
temperature at the trailing end will not exceed the maximum allowable level. (ii) Since the
Reynolds number at the trailing end is less than the transition value of 5u105, the flow will be
laminar if it is not tripped. To determine surface temperature under laminar flow conditions, the
corresponding heat transfer coefficient must be computed. The Nusselt number for laminar flow
over a flat plate with uniform surface flux is given by equation (7.31)
Nu x
hx
k
0.453Pr 1 / 3 Re1x / 2
(f)
Solving this equation for h(L)
h( L) 0.453
k
( Pr )1 / 3 ( Re L )1 / 2
L
0.453
0.0271( W / m o C)
(0.71)1 / 3 (29,481)1 / 2 = 18.8 W/m2-oC
0.1(m)
Substituting into (b)
Ts(L) = 27oC +
1300( W / m 2 )
18.8( W / m 2 o C)
= 96.1oC
Since this is considerably higher than the assumed value of 79oC, the procedure is repeated with
a new assumed value of Ts(L) = 99oC. This yields a calculated value of Ts(L) = 96.2oC. Thus for
laminar flow, surface temperature exceeds the allowable maximum level of 83oC.
PROBLEM 8.9
Water at 27oC flows normally over a tube with a velocity of 4.5 m/s. The outside diameter of the
tube is 2 cm. Condensation of steam inside the tube results in a uniform outside surface
temperature of 98oC. Determine the length of tube needed to transfer 250,000 W of energy to the
water.
(1) Observations. (i) This is an external forced convection problem. (ii) The flow is normal to a
tube. (iii) Surface temperature is uniform. (iv) Tube length is unknown. (v) Newton’s law of
cooling can be used to determine surface area. Tube length is related to surface area. (vi) The
fluid is water.
(2) Problem Definition. The required tube length can be determined from Newton's law of
cooling. Thus, the problem is finding the average heat transfer coefficient for flow normal to a
cylinder.
(3) Solution Plan. Apply Newton's law of cooling to the flow over tube. Use forced convection
correlation equation to determine the average Nusselt number.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4)
uniform upstream velocity and temperature, (5) uniform surface heat flux, (6) constant
properties, (7) no buoyancy ( E = 0 or g = 0 ) and (8) no radiation.
(ii) Analysis. Newton's law of cooling gives
q = h A(Ts Tf )
where
A = plate area, m2
h = average heat transfer coefficient,
W/m2-oC
q = heat transfer rate = 250,000 W
Ts = surface temperature = 98oC
Tf = free stream temperature = 27oC
(a)
L
Ts
D
Vf Tf
Vf Tf
Surface area is
A = SDL
where
D = outside tube diameter = 2 cm = 0.02 m
L = tube length, m
(b)
Substituting (b) into (a) and solving the resulting equation for L
q
L
SDh (Ts Tf )
(c)
Equation (8.10a) gives a correlation for the average Nusselt number for forced convection
normal to a cylinder
Nu D
hD
k
0.3 0.62 Re1D/ 2 Pr 1 / 3
>1 4 / Pr @
2 / 3 1/ 4
>1 Re
D
/ 282,000
@
5/8 4/5
(d)
PROBLEM 8.9 (continued)
Valid for:
flow normal to cylinder
Pe = Re D Pr > 0.2
properties at Tf
(e)
where
k = thermal conductivity, W/m-oC
Nu D = average Nusselt number
Pe = Peclet number = PrRe D
Pr = Prandtl number
Re D = Reynolds number
T f = film temperature, oC
Reynolds number is defined as
Re D
Vf D
(f )
Q
where
Vf = free stream velocity = 4.5 m/s
Q = kinematic viscosity, m2/s
Properties are determined at the film temperature T f defined as
Tf = (Ts Tf ) / 2
(g)
The Reynolds number is computed to establish if (e) is satisfied. Substituting into (g)
Tf = (98 + 27)(oC)/2 = 62.5oC
Properties of water at this temperature are
k = thermal conductivity = 0.653 W/m-oC
Pr = Prandtl number = 2.885
Q = kinematic viscosity = 0.4586u10-6 m2/s
Thus
Re D
4.5(m / s)0.02(m)
0.4586 u 10 6 (m 2 / s)
= 196,249
and
Pe = Re D Pr = 196,249 (2.885) = 5.66u105
Therefore, (e) is satisfied and correlation equation (d) is applicable.
(iii) Computations. Equation (d) gives the average heat transfer coefficient
Nu D
hD
k
Solving for h
0.3 0.62(196,249)1 / 2 (2.885)1/3
>1 0.4 / 2.885 @
2 / 3 1/ 4
>1 196,249 / 282,000 @
5/8 4/5
= 589
PROBLEM 8.9 (continued)
h = 589
0.653( W / m o C)
k
= 19,231 W/m2-oC
= 589
D
0.02(m)
Substituting into (c) gives the required length
L=
250,000( W )
= 2.91 m
S (0.02)(m)19,231( W/m 2 o C)(98 27)( o C)
(iv) Checking. Dimensional check: Computations showed that units for equations (c), (d)
and (f) are dimensionally consistent.
Quantitative check: The value of h is within the range given in Table 1.1 for forced convection
of liquids.
Limiting check: An infinitely long tube is needed if Ts
gives L = f.
Tf . Setting Ts
Tf in equation (c)
(5) Comments. (i) The required length to transfer 250 kw of heat is only 2.43 m. This appears
unreasonably short. A review of the analysis and calculations uncovered no errors. The relatively
short length needed is due to the high heat transfer coefficient associated with forced convection
of water. Note that the calculated heat transfer coefficient is at the high end of values given in
Table 1.1. (ii) It was not necessary to consider the thermal interaction between the surface and
the fluid inside the tube because outside surface temperature was specified.
PROBLEM 8.10
A proposed steam condenser design for marine applications is based on the concept of rejecting
heat to the surrounding water while a boat is in motion. The idea is to submerge a steamcarrying tube in the water such that its axis is normal to boat velocity. Estimate the rate of
steam condensation for a 75 cm long tube with an outside diameter of 2.5 cm. Assume a
condensation temperature of 90oC and a uniform surface temperature of 88oC. Ambient water
temperature is 15oC and boat speed is 8 m/s.
(1) Observations. (i) Heat is removed by the water from the steam causing it to condense. (ii)
The rate at which steam condenses inside the tube depends on the rate at which heat is removed
from the outside surface. (iii) Heat is removed from the outside surface by forced convection.
(iv) This is an external forced convection problem of flow normal to a tube. (v) Newton’s law of
cooling gives the rate of heat loss from the surface.
(2) Problem Definition. Determine the rate of heat transfer from the outside surface to the
ambient water.
(3) Solution Plan. Apply conservation of energy to the condensing steam inside the tube. Use
Newton’s law of cooling to determine the heat removed from the tube. Use correlation equation
for forced convection flow normal to a tube to determine the average heat transfer coefficient
(Nusselt number).
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4)
uniform upstream velocity and temperature, (5) uniform surface temperature, (6) constant
properties, (7) no radiation, (8) negligible changes in kinetic and potential energy of steam, (9)
negligible axial conduction in tube and steam, constant properties, (10) no buoyancy ( E = 0 or g
= 0 ) and (11) no radiation.
. (ii) Analysis. Applying conservation of energy to the steam between the inlet and outlet of the
tube gives
q = m hi ho = m hˆ fg
(a)
or, solving (a) for m
q
=
(b)
m
ˆh
fg
where
h
= steam enthalpy at inlet of tube, J/kg
i
ho = steam enthalpy at outlet of tube, J/kg
ĥ fg = latent heat of condensation = 2283.2 kJ/kg (at 90 oC)
= rate of steam condensation, kg/s
m
q = rate of heat removed from steam, W
The latent heat, ĥ fg , is defined by the temperature of the condensing steam. Thus to determine
steam condensation rate from (b), the rate of heat removal from steam must be obtained.
Applying conservation of energy to the tube gives
PROBLEM 8.10 (continued)
q = Energy removed from steam and added to tube
= Energy removed from tube and added to water
(c)
Newton’s law of cooling gives the energy removed from tube surface by convection and added
to water
q = h SDL (Ts - Tf)
(d)
where
D = diameter of tube = 2.5 cm = 0.025 m
2 o
h = average heat transfer coefficient at the outer surface of tube, W/m - C
L = length of tube = 75 cm = 0.75 m
Ts = surface temperature = 88oC
Tf = ambient temperature = 15oC
Equation (8.10a) gives a correlation for the average Nusselt number for forced convection
normal to a cylinder
0.62 Re1D/ 2 Pr 1 / 3
hD
5/8 4/5
Nu D
0.3 1 Re D / 282,000
1/ 4
2
/
3
k
1 0.4 / Pr
Valid for:
flow normal to cylinder
Pe = Re D Pr > 0.2
properties at Tf
where
>
@
>
k = thermal conductivity, W/m-oC
Nu D = average Nusselt number
Pe = Peclet number = PrRe D
Pr = Prandtl number
Re D = Reynolds number
T f = film temperature, oC
@
(e)
(f)
Tf
Vf
Ts
Reynolds number is defined as
Re D
Vf D
Q
(g)
where
Vf = free stream velocity = 8 m/s
Q = kinematic viscosity, m2/s
Properties are determined at the film temperature T f defined as
Tf = (Ts Tf ) / 2
The Reynolds number is computed to establish if (f) is satisfied. Substituting into (h)
Tf = (88 + 15)(oC)/2 = 51.5oC
Properties of water at this temperature are
k = thermal conductivity = 0.6421 W/m-oC
(h)
PROBLEM 8.10 (continued)
Pr = Prandtl number = 3.48
Q = kinematic viscosity = 0.5411u10-6 m2/s
Thus
Re D
8(m / s)0.025(m)
0.5411 u 10 6 (m 2 / s)
= 369,617
and
Pe = Re D Pr = 369,617(3.48) = 1.286u106
Therefore, (f) is satisfied and correlation equation (e) is applicable.
(iii) Computations. Equation (e) gives the average Nusselt number and average heat
transfer coefficient
Nu D
hD
k
0.3 0.62(369,617)1 / 2 (3.48)1/3
>1 0.4 / 3.48 @
2 / 3 1/ 4
>1 369,617 / 282,000 @
5/8 4/5
= 1012.3
Solving for h
h = 1012.3
0.6421( W / m o C)
k
= 1012.3
= 26,000 W/m2-oC
D
0.025(m)
Substituting into (d)
q = 26,000(W/m2-oC) S 0.025(m) 0.75(m) (88 - 15)(oC) = 111,801 W = 111.8 kW
Latent heat of condensation at 90oC is
ĥ fg = 2283.2 kJ/kg
Substituting into (b) gives the condensation rate m
=
m
111.8(kW )
(kJ / s)
= 0.04897
= 0.04897 kg/s
2283.2(kJ / kg )
(kJ / kg)
(iv) Checking. Dimensional check: Computations showed that equations (b), (d), (e) and
(g) are dimensionally consistent.
Quantitative check: The value of h is outside the range given in Table 1.1 for forced convection
of liquids. A review of the analysis and calculations uncovered no errors. It should be kept in
mind that Table 1.1 gives rough estimates of h for typical applications. Exceptions are expected.
Qualitative check: Increasing the tube’s length and/or boat speed, increases condensation rate.
Equation (b) shows that condensation rate is directly proportional to q. According to equation
(d), q increases as L is increased. Similarly, according to (e), an increase in Vf results in an
increase in h which in turn increases q.
(5) Comments. Although condensation rate may be adequate when the boat is in motion, it
decreases when the boat is stationary.
PROBLEM 8.11
An inventive student wanted to verify the speed of a boat
using heat transfer analysis. She used a 10 cm long
electrically heated tube with inside and outside diameters
of 1.1 cm and 1.2 cm, respectively. She immersed the tube
in the water such that its axis is normal to boat velocity.
She recorded the following measurements:
Vf
Tf
Water temperature = 16.5oC
Outside surface temperature of tube = 23.5oC
Electric energy dissipated in tube = 480 W
Ts
+
Determine the speed of the boat.
(1) Observations. (i) Electric power is dissipated into heat and is removed by the water. (ii) This
velocity measuring concept is based on the fact that forced convection heat transfer is affected by
fluid velocity. (iii) velocity affects the heat transfer coefficient which in term affects surface
temperature. (iv) Newton’s law of cooling relates surface heat loss to the heat transfer
coefficient, surface area and surface temperature. (v) This problem can be modeled as external
flow normal to a cylinder. (vi) The fluid is water.
(2) Problem Definition. Formulate a relationship between fluid velocity, heat transfer rate and
surface temperature for flow normal to a cylinder.
(3) Solution Plan. Apply conservation of energy and Newton's law of cooling to the tube. Use a
correlation equation to relate heat transfer coefficient to fluid velocity.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) no axial
conduction through tube, (5) constant boat velocity, (6) uniform surface flux, (7) uniform surface
temperature (8) uniform water temperature, (9) no buoyancy (E = 0 or g = 0) and (10) negligible
radiation.
(ii) Analysis. Conservation of energy applied to the tube gives
P
q
(a)
where
P = electric power supplied to tube = 480 W
q = heat transfer rate from tube surface to water, W
Application of Newton’s law of cooling to the tube and gives
q
h S Do L (Ts Tf )
where
Do = outside tube diameter = 1.2 cm = 0.012 m
h = average heat transfer coefficient, W/m 2 o C
L = tube length = 10 cm = 0.1 m
Ts = surface temperature = 23.5 o C
Tf = water temperature = 16.5 o C
(b)
PROBLEM 8.11 (continued)
Substituting (a) into (b)
hS Do L (Ts Tf )
P
Since h is expected to depend on velocity, equation (a) is solved for h
P
S Do L (Ts Tf )
h
(c)
Equation (8.10a) gives a correlation for the average Nusselt number for forced convection
normal to a cylinder
NuD
h Do
k
0.3 0.62 Re1D/ 2 Pr 1/ 3
>1 0.4 / Pr @
2 / 3 1/ 4
>1 Re
D
/ 282,000
@
5/8 4/5
(d)
Valid for:
flow normal to cylinder
Pe = Re D Pr > 0.2
properties at Tf
where
k = thermal conductivity, W/m-oC
Nu D = average Nusselt number
Pe = Peclet number = PrRe D
Pr = Prandtl number
Re D = Reynolds number
T f = film temperature, oC
Reynolds number is defined as
Vf Do
ReD
(e)
Q
where
Vf = boat speed, m/s
Q = kinematic viscosity, m2/s
Substituting (e) into (d) and solving for h
h
k
Do
­°
0.62(Vf Do /Q )1/ 2 Pr1/ 3
1 (Vf Do /Q ) / 282,000
®0.3 2 / 3 1/ 4
°̄
1 0.4 / Pr
>
@
>
½
5 / 8 4 / 5°
@
¾
°¿
(f)
Properties of water are determined at the film temperature T f defined as
Tf
Ts Tf
2
(g)
Equation (f) gives a relationship between h and the boat speed Vf. Substituting (f) into (c) gives
the desired relationship between P , Vf and Ts. However, the resulting equation cannot be solved
explicitly for Vf. The solution is obtained by trial and error. Equation (c) is used to calculate h ,
a value for Vf is selected and substituted into (f) to calculate h . If the calculated h using (f) is
PROBLEM 8.11 (continued)
not the same as that obtained from (c), the procedure is repeated until a satisfactory agreement is
obtained between the two values.
(iii) Computations. Equation (c) is used to calculate h
h
480( W )
S 0.012(m)0.1(m) (23.5 16.5)( c C)
18,189 W/m 2 o C
Equation (f) is used to calculate T f
Tf = (23.5 + 16.5)( o C )/2 = 20 o C
Properties of water at this temperature are:
k = 0.5996 W/m o C
Pr = 6.99
Q = 1.004u10-6 m 2 /s
Assume Vf = 10 m/s. Substituting into (f)
h
0.3
0.5996( W/m o C)
0.012(m)
ª 10(m/s)0.012(m) º
0.62«
»
2
6
«¬1.004 u 10 (m /s) »¼
1/2
>1 0.4 / 6.99 @
2/3
4/5
(6.99)1 / 3 ª
5/8
º º
ª
10(m/s)0.012(m)
0.5996( W/m o C)
«1 «
»
»
1/ 4
« ¬«1.004 u 10 6 (m 2 /s)282,000 ¼» »
0.012(m)
¬
¼
h = 28,606 W/m 2 o C
Since this is larger than h = 18,189 W/m 2 o C obtained from (c), the procedure is repeated
using a lower value for Vf. Assume Vf 5 m/s and substituting into (f) gives h = 18,089
W/m 2 o C . This agrees within 0.5% of the value obtained from (c). Thus the speed of the boat is
5 m/s.
(iv) Checking. Dimensional check: Computations showed that equations (c) and (f) are
dimensionally consistent.
Quantitative check: The calculated value of h = 18,189 W/m 2 o C is within the range given in
Table 1.1 for forced convection in liquids.
(5) Comments. (i) The velocity measuring method suggested by the student is indeed sound. It
is based on the observation that fluid velocity affects heat transfer coefficient. Thus heat transfer
coefficient may be used as a measure of velocity. (ii) Since correlation equations for heat transfer
coefficients are not exact, a velocity measuring instrument which is based on this concept must
be calibrated to obtain accurate velocity measurements.
PROBLEM 8.12
A thin electric heater is wrapped around a rod of diameter 3 cm. The heater dissipates energy
uniformly at a rate of 1300 W/m. Air at 20oC flows normal to the rod with a velocity of 15.6 m/s.
Determine the steady state surface temperature of the heater.
(1) Observations. (i) This is an external forced convection problem. (ii) The flow is normal to a
rod. (iii) Surface heat transfer rate per unit length is known. However, surface temperature is
unknown. (iv) In general, surface temperature varies along the circumference. However, the rod
can be assumed to have a uniform surface temperature. (v) This problem can be modeled as
forced convection normal to a rod with uniform surface flux or temperature. (vi) Newton’s law
of cooling gives surface temperature. (vii) The fluid is air.
(2) Problem Definition. Surface temperature can be determined from Newton's law of cooling
if the heat transfer coefficient is known. Thus, the problem is finding the average heat transfer
coefficient for flow normal to a cylinder.
(3) Solution Plan. Apply Newton's law of cooling to the flow over a rod. Use forced convection
correlation equation to determine the average Nusselt number.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4)
uniform upstream velocity and temperature, (5) uniform surface flux, (6) constant properties, (7)
all energy dissipated in electric heater leaves surface (no axial conduction), (8) no buoyancy (E =
0 or g = 0) and (9) negligible radiation.
(ii) Analysis. Newton's law of cooling gives
q = h SDL ( Ts Tf )
where
D = diameter of rod = 3 cm = 0.03 m
h = average heat transfer coefficient,
W/m2-oC
L = tube length, m
q = heat transfer rate, W
Ts = average surface temperature, oC
Tf = free stream temperature = 20oC
L
D
_
Ts
Vf Tf
Solving (a) for Ts
Ts
(a)
Tf q/L
SDh
Vf Tf
(b)
where
q/L = energy dissipated per unit length = 1300 W/m
Equation (8.10a) gives a correlation for the average Nusselt number for forced convection
normal to a cylinder at uniform surface temperature or surface flux
PROBLEM 8.12 (continued)
Nu D
hD
k
0.3 0.62 Re1D/ 2 Pr 1 / 3
>1 0.4 / Pr @
2 / 3 1/ 4
>1 Re
D
/ 282,000
@
5/8 4/5
(c)
Valid for:
flow normal to cylinder
Pe = Re D Pr > 0.2
properties at Tf
(d)
where
k = thermal conductivity, W/m-oC
Nu D = average Nusselt number
Pe = Peclet number = PrRe D
Pr = Prandtl number
Re D = Reynolds number
T f = film temperature, oC
Reynolds number is defined as
Re D
Vf D
Q
(e)
where
Vf = free stream velocity = 15.6 m/s
Q = kinematic viscosity, m2/s
Properties are determined at the film temperature T f defined as
Tf = (Ts Tf ) / 2
(f)
Since surface temperature is unknown, a trial and error procedure is needed to solve the problem.
A value for Ts is assumed, (f) is used to determine Tf and (b) is used to calculate Ts . If the
calculated Ts is not equal to the assumed value, the process is repeated until a satisfactory
agreement is obtained between assumed and calculated Ts .
Assume Ts = 100oC. Equation (f) gives
Tf = (100 + 20)(oC)/2 = 60oC
At this temperature air properties are
k = thermal conductivity = 0.02852 W/m-oC
Pr = Prandtl number = 0.708
Q = kinematic viscosity = 18.9u10-6 m2/s
The Reynolds number is computed to establish if (d) is satisfied. Substituting into (e)
Re D
and
15.6(m / s)0.03(m)
18.9 u 10 6 (m 2 / s)
= 24,762
PROBLEM 8.12 (continued)
Pe = Re D Pr = 24,762(0.708) = 1.753u104
Therefore, (d) is satisfied and correlation equation (c) is applicable.
(iii) Computations. Equation (c) gives the average Nusselt number and average heat
transfer coefficient
Nu D
hD
k
0.3 0.62(24,762)1 / 2 (0.708)1/3
>1 0.4 / 0.708 @
2 / 3 1/ 4
>1 24,762 / 282,000 @
5/8 4/5
= 89.42
Solving for h
h = 89.42
0.02852( W / m o C)
k
= 85 W/m2-oC
= 89.42
D
0.03(m)
Substituting into (b) gives surface temperature
Ts = 20oC +
1,300( W / m)
S (0.03)(m)85( W / m 2 o C)
= 182.3oC
This value is considerably higher than the assumed one of 100oC. The procedure is repeated with
another assumed temperature. Assume Ts = 180oC. This gives Tf = 100oC, h = 83.2 W/m2- oC
and a calculated surface temperature Ts = 185.8oC. This is close to the assumed value. Thus the
resulting surface temperature is 185.8oC.
(iv) Checking. Dimensional check: Computations showed that units for equations (b), (c)
and (e) are dimensionally consistent.
Quantitative check: The value of h is within the range given in Table 1.1 for forced convection
of gases.
Limiting check: If the heater is turned off, surface temperature should be the same as free stream
temperature. Setting q = 0 in (b) gives Ts Tf .
(5) Comments. (i) The trial and error procedure converges rapidly towards a satisfactory
solution. This is because heat transfer coefficient is not very sensitive to the temperature at which
properties are determined. Changing film temperature from 60oC to 100oC in the above example
resulted in a 2% change in h . (ii) The assumption that radiation loss is negligible should be
examined in view of the high surface temperature. Since surface emissivity is unknown one can
only make a rough approximation of radiation. Using Stefan-Boltzmann law, assuming that the
rod is a small surface enclosed by a much larger surface and assuming an emissivity of 1.0,
radiation loss is found to be 296 W/m. However, if the surface emissivity is 0.1, then the
radiation loss is 2.2%, which is negligible.
PROBLEM 8.13
A fluid velocity measuring instrument consists of a wire which is heated electrically. By
positioning the axis of the wire normal to flow direction and measuring surface temperature and
dissipated electric power, fluid velocity can be estimated. Determine the velocity of air at 25oC
for a wire diameter of 0.5 mm, dissipated power 35 W/m and surface temperature 40oC.
(1) Observations. (i) Electric power is dissipated into heat and is removed by the fluid. (ii) This
velocity measuring instrument is based on the fact that forced convection heat transfer is affected
by fluid velocity. (iii) velocity affects the heat transfer coefficient which in term affects surface
temperature and heat flux. (iv) Newton’s law of cooling relates surface heat loss to the heat
transfer coefficient, surface area and surface temperature. (v) This problem can be modeled as
external flow normal to a cylinder. (vi) The fluid is air.
(2) Problem Definition. Formulate a relationship between fluid velocity, heat transfer and
surface temperature for flow normal to a cylinder.
(3) Solution Plan. Apply Newton's law of cooling to the wire. Use a correlation equation to
relate heat transfer coefficient to the fluid velocity.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) no axial
conduction through wire, (5) uniform upstream velocity and temperature, (6) constant properties,
(7) uniform surface flux and surface temperature, (8) no buoyancy (E = 0 or g = 0) and (9) no
radiation.
(ii) Analysis. Application of Newton’s law of cooling to the wire gives
q = h SDL (Ts - Tf)
(a)
where
D = wire diameter = 0.5 mm = 0.0005 m
2 o
h = average heat transfer coefficient, W/m - C
L = wire length, m
Ts = surface temperature = 40oC
Tf = free stream temperature = 25oC
Since h is expected to depend on velocity, equation (a) is
solved for h
h=
Ts
_
Vf
Tf
q/L
SD(Ts Tf )
q/L
(b)
where
q
= power or heat dissipated in wire per unit length = 35 W/m
L
Equation (8.10a) gives a correlation for the average Nusselt number for forced convection
normal to a cylinder at uniform surface temperature or surface flux
PROBLEM 8.13 (continued)
Nu D
hD
k
0.3 0.62 Re1D/ 2 Pr 1 / 3
>1 0.4 / Pr @
2 / 3 1/ 4
>1 Re
D
/ 282,000
@
5/8 4/5
(c)
Valid for:
flow normal to cylinder
Pe = Re D Pr > 0.2
properties at Tf
(d)
where
k = thermal conductivity, W/m-oC
NuD = average Nusselt number
Pe = Peclet number = PrRe D
Pr = Prandtl number
Re D = Reynolds number
T f = film temperature, oC
The Reynolds number is defined as
ReD =
Vf D
(e)
Q
where
Vf = free stream velocity, m/s
Q = kinematic viscosity = m2/s
Properties of air are evaluated at the film temperature Tf defined as
Tf = (Ts + Tf)/2
(f)
The objective is to express h in terms of free stream velocity Vf , substitute into (b) and obtain
an equation relating q/L, Vf and Ts. Substituting (e) into (c) and solving the resulting equation
for h gives
h
0 .3
k 0.62 Vf D / Q
D
1 0.4 / Pr
>
5/8
Pr 1 / 3 ª § Vf D / Q · º
1
«
»
¨
¸
2 / 3 1/ 4 «
282,000 ¹ »
¬ ©
¼
1/2
@
4/5
k
D
(g)
Equation (g) gives a relationship between h and the free stream velocity Vf. Substituting this
result into (b) gives the desired relationship between q/L, Vf and Ts. However, the resulting
equation cannot be solved explicitly for Vf. The solution is obtained by trial and error. Equation
(b) is used to calculate h , a value for Vf is selected and substituted into (g) to calculate h . If the
calculated h using (g) is not the same as that obtained from (b), the procedure is repeated until a
satisfactory agreement is obtained between the two values.
(iii) Computations. Equation (f) is used to calculate Tf
Tf = (40 + 25)(oC)/2 = 32.5oC
Properties of air at this temperature are given in Appendix C
k = 0.02656 W/m-oC
Pr = 0.7115
PROBLEM 8.13 (continued)
Q = 16.2475u10-6 m2/s
Substituting into (b)
h=
35( W / m)
S 0.0005(m)(40 25)( o C)
= 1485.4 W/m2-oC
Assume Vf = 50 m/s. Substituting into (g)
h
0.3
0.02656( W / m o C)
0.0005(m)
1/2
ª 50(m / s )0.0005(m) º
0.62 «
»
6
2
¬«16.24 u 10 (m / s ) ¼»
4/5
(0.7115)1 / 3 ª
5/8
ª
º º
50(m / s )0.0005(m)
0.02656(W / moC )
«1 «
»
»
1
/
4
6
2
«
»
0.0005(m)
«16.24 u 10 (m / s )282,000 ¼»
1 0.4 / 0.7115 2 / 3
¬ ¬
¼
>
@
2 o
h = 1060.1 W/m - C
Since this is less than h = 1485.4 W/m2-oC, the procedure is repeated using a higher value for
Vf. The results of five trials are tabulated below.
Assumed Vf
m/s
50
70
90
100
96
Calculated h
W/m2-oC
1059.9
1259.7
1434.9
1515.6
1484.2
The result shows that the free stream velocity is Vf = 96 m/s. With Vf determined, it remains to
verify that condition (d) is satisfied by calculating the Peclet number.
Pe = ReD Pr = (Vf D / Q ) Pr = [96(m / s)0.0005(m) / 16.2475 u 10 6 (m 2 / s)]0.7115 = 2102
Therefore, condition (d) is satisfied.
(iv) Checking. Dimensional check: Computations showed that equations (b), (e) and (g) are
dimensionally consistent.
Qualitative check: If q/L is held constant and Vf is increased, surface temperature should
decrease. According to (g), increasing Vf increases h . An increase in h results in a decrease
in Ts, as indicated by equation (a).
Quantitative check: The calculated value of h = 1484.2 W/m2-oC is outside the range suggested
in Table 1.1. This is due to the fact that the diameter of the wire is very small (0.0005 m). This
is not among the typical application considered in Table 1.1.
(5) Comments. This velocity measuring instrument is based on the observation that fluid
velocity affects heat transfer coefficient. Since correlation equations are not exact, it is necessary
to calibrate such an instrument to obtain accurate velocity measurements.
PROBLEM 8.14
Students were asked to devise unusual methods for determining the height of a building. One
student designed and tested the following system. A thin walled copper balloon was heated to
133oC and parachuted from the roof of the building. Based on aerodynamic consideration, the
student reasoned that the balloon dropped at approximately constant speed. The following
measurements were made:
D = balloon diameter = 13 cm
M = mass of balloon = 150 grams
To = balloon temperature at landing = 47oC
Tf = ambient air temperature = 20oC
U = balloon velocity = 4.8 m/s
Determine the height of the building.
(1) Observations. (i) The sphere cools off as it drops. Heat loss from the sphere is by forced
convection. (ii) The height of the building can be determined if the time it takes the sphere to
land is known. (iii) Time to land is the same as cooling time. (iv) Transient conduction
determines cooling time. (v) If the Biot number is less than 0.1, lumped capacity method can be
used to determine transient temperature. (vi) Cooling rate depends on the heat transfer
coefficient.
(2) Problem Definition. Determine transient temperature of sphere. This requires determining
the drop time and the heat transfer coefficient.
(3) Solution Plan. Apply Newton’s law of motion to the falling sphere and use lumped capacity
method to determine drop time. Use correlation equation to determine the heat transfer
coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) Constant heat transfer coefficient, (3) Biot
number < 0.1 (to be verified ), (4) constant properties, (5) constant sphere velocity, (6) uniform
ambient conditions, (7) negligible wind speed, (8) no radiation and (9) no heat loss to the sphere
interior.
(ii) Analysis. Application of Newton’s law of motion to the falling sphere gives
H
U to
(a)
where
H = building height, m
t o = drop time, s
U = balloon velocity, m/s
The dorp time is determined from transient temperature analysis. For Bi << 1, the lumped
capacity model gives the transient temperature solution for the sphere. Equation (5.7) gives
T (t ) Tf (Ti Tf ) exp[(h As /Uc pV ) t]
where
As = surface area, m2
(b)
PROBLEM 8.14 (continued)
c p = specific heat of copper = 385 J/kg o C
h = average heat transfer coefficient, W/m 2 o C
t = time, s
T (t ) = temperature variable, o C
Tf = ambient temperature = 20oC
Ti = initial temperature of sphere = 133 o C
V = volume, m3
U = density of copper = 8933 kg/m3
The product of U V in equation (b) is equal to the mass of the sphere. That is
UV=M
(c)
where
M = mass of sphere = 150 g = 0.15 kg
Surface area of sphere is
S D2
As
(d)
where
D = diameter = 13 cm = 0.13 m
Substituting (c) and (d) into (b) and solving the resulting equation for t
cpM
t
Applying this result at the drop time, t
S D 2h
ln
Ti Tf
T (t ) Tf
(e)
Ti Tf
To Tf
(f)
to
to
cpM
S D2h
ln
where
To = sphere temperature at landing = 47 o C
Thus the only unknown in (f) is the average heat transfer coefficient. Equation (8.12a) gives a
correlation for the average Nusselt number for forced convection over a sphere
NuD
hD
k
P
2 0.4 Re1D/ 2 0.06 ReD2 / 3 Pr 0.4 P
s
>
subject to the following limitations:
3.5 < ReD < 7.6 u104
0.71 < Pr < 380
1.0 < (P /P s) <3.2
properties at Tf , Ps at Ts
where
k = thermal conductivity of air = 0.02564 W/m-oC
NuD = average Nusselt number
Pr = Prandtl number of air = 0.713
Re D = Reynolds number
@
1/ 4
(g)
PROBLEM 8.14 (continued)
P = viscosity of air at the free stream temperature = 18.17u10-6 kg/m-s
P s = viscosity of air at surface temperature, kg/m-s
The Reynolds number is defined as
ReD
UD
(h)
Q
where
Q = kinematic viscosity of air = 15.09u10-6 m2/s
Since surface temperature changes with time, Ps should be evaluated at the average surface
temperature Ts defined as
Ti To
Ts
(i)
2
(iii) Computations. The average heat transfer coefficient h is computed using (g).
Substituting numerical values into (i)
(133 47)( o C)
90 o C
2
The viscosity at this temperature is
Ts
21.35 u 10 6 kg/m s
Ps
The Reynolds number is
4.8(m/s)0.13(m)
15.09 u 10 6 (m 2 /s)
ReD
41,352
Substituting into (g)
Nu D
hD
k
hD
k
Solving for h
Nu D
h
130.4
ª 18.17 u 10 6 (kg/m s) º
2 0.4(41,352)1 / 2 0.06(41,352) 2 / 3 (0.713) 0.4 «
»
6
¬ 21.35 u 10 (kg/m s) ¼
^
`
1/ 4
130.4
02564( W/m o C)
k
130.4
D
0.13(m)
25.7 W/m 2 o C
Substituting into (f)
to
(133 20)
385(J/kg o C)0.15(kg)
ln
2
2
2 o
S (0.13) (m )25.7( W / m - C) (47 20)
60.6 s
Substituting into (a) gives the building height
H
4.8(m/s)60.6(s) 290.9 m
The Biot number can now be computed to establish the validity of the lumped capacity method.
It is defined as
PROBLEM 8.14 (continued)
Bi
hG
kc
(j)
where
k c = thermal conductivity of copper = 397 W/m o C (at 90 o C )
G = wall thickness of sphere, m
U c = density of copper = 8933 kg/m 3
Sphere thickness is determined from its mass, density and volume
G |
M
U cS D 2
0.15(kg )
8933(kg/m 3 )S (0.13) 2 (m 2 )
3.162 u 10 4 m
Substituting into (j)
Bi
25.7( W/m 2 o C)3.162 u 10 4 (m)
397( W/m o C)
2.047 u 10 5
Since the Biot number is much smaller than unity it follows that the lumped capacity method is
applicable.
(iv) Checking. Dimensional check: Computations showed that equations (a), (f), (g), (h)
and (j) are dimensionally consistent.
Limiting check: If building height is infinite, the final temperature should be the same as ambient
temperature. Setting To Tf in (f) gives t o f . When this is substituted into (a) gives H f .
Quantitative check: The value of h is within the range listed in Table 1.1 for forced convection
of gases.
(5) Comments. Neglecting wind speed is a key assumption in the method used. Wind speed can
introduce significant error in estimating the height of the building.
PROBLEM 8.15
A 6 cm diameter sphere is used to study skin friction characteristics at elevated temperatures.
The sphere is heated internally with an electric heater and placed in a wind tunnel. To obtain a
nearly uniform surface temperature the sphere is made of copper. Specify the required heater
capacity to maintain surface temperature at 140oC. Air velocity in the wind tunnel is 18 m/s and
its temperature is 20oC.
(1) Observations. (i) The electric energy dissipated inside the sphere is removed from the
surface as heat by forced convection. (ii) This problem can be modeled as external flow over
a sphere. (iii) Newton’s law of cooling relates heat loss from the surface to heat transfer
coefficient, surface area and surface temperature. (iv) The fluid is air.
(2) Problem Definition.
forced convection.
Determine the rate of heat transfer from the surface of a sphere by
(3) Solution Plan. Apply conservation of energy to the sphere. Apply Newton's law of cooling
to obtain a relationship between heat removed from the sphere and its surface temperature. Use
a correlation equation to determine the average heat transfer coefficient for the flow over a
sphere.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) uniform
surface flux, (5) uniform surface temperature, (6) uniform wind tunnel conditions, (7) constant
properties, (8) no buoyancy (E = 0 or g = 0) and (9) no radiation.
(ii) Analysis. Applying Conservation of energy to the sphere
P=q
where
P = electric power dissipated in sphere, W
q = heat removed from surface, W
(a)
Application of Newton’s law of cooling to the surface of sphere
q = h A (Ts - Tf)
where
A = surface area, m2
h = average heat transfer coefficient, W/m2-oC
Ts = surface temperature = 140oC
Tf = ambient temperature = 20oC
(b)
Ts q
P
Tf
Vf
Surface area of sphere is
A = S D2
+
(c)
-
where
D = diameter = 0.06 m
The only remaining unknown in (b) is the heat transfer coefficient h . Equation (8.12a) gives a
correlation for the average Nusselt number for forced convection over a sphere
Nu D
hD
k
>
@
P
2 0.4 Re1D/ 2 0.06( Re D ) 2 / 3 Pr 0.4 P
s
1/ 4
(d)
PROBLEM 8.15 (continued)
Valid for:
3.5 < ReD < 7.6 u104
0.71 < Pr < 380
1.0 < (P /P s) <3.2
properties at Tf , Ps at Ts
(e)
where
k = thermal conductivity of air = 0.02564 W/m-oC
NuD = average Nusselt number
Pr = Prandtl number of air = 0.713
ReD = Reynolds number
P = viscosity of air at the free stream temperature = 18.17u10-6 kg/m-s
Ps = viscosity of air at surface temperature = 23.44 u10-6 kg/m-s
The Reynolds number is given by
ReD =
Vf D
(f)
Q
where
Vf = free stream velocity = 18 m/s
Q = kinematic viscosity of air = 15.09u10-6 m2/s
(iii) Computations. Appendix C gives air properties at Tf = 20oC. Substituting into (f)
ReD =
18( m / s) 0.06( m)
15.09 u 10 6 ( m2 / s)
= 71,571
Thus, the Reynolds and Prandtl numbers are within the limitations in (e). Next compute P /P s
P /P s = 18.17u10-6 (kg/m-s)/ 23.44 u10-6 (kg/m-s) = 0.775
Although this is outside the range given in (e), it represents a small deviation particularly since
this ratio is raised to the 1/4 power. Substituting into (d)
NuD
§ 18.17 u 10 6 ( kg / m s) ·
hD
¸
= 2 0.4(71,571)1 / 2 0.06(71,571) 2 / 3 (0.713) 0.4 ¨¨
6
¸
k
© 23.44 u 10 (kg / m s) ¹
>
@
1/ 4
= 174.5
Solving the above for h
h = 174.5
k
= 174.5 (0.02564) (W/m-oC)/0.06(m) = 74.57 W/m2-oC
D
Substituting into (b) and using (c) gives the required heater capacity P
P = q = 74.57(W/m2-oC)S (0.06)2(m2) (140 -20)(oC) = 101.2 W
(iv) Checking. Dimensional check: Computations showed that equations (b), (d) and (f) are
dimensionally consistent.
Quantitative check: The value of heat transfer coefficient is within the range shown in Table 1.1.
PROBLEM 8.15 (continued)
Limiting check: For the special case of Ts = Tf the required power should be zero. This is
confirmed by equations (a) and (b).
(5) Comments. (i) Since not all conditions listed in (e) on correlation equation (d) have been
met, the result should be viewed as an approximation. An alternate approach is to search the
literature for another correlation equation that will meet the conditions of this problem. (ii) If
radiation is included in the analysis, the required heater capacity will be greater than that
determined by the above model.
PROBLEM 8.16
A hollow aluminum sphere weighing 0.2 kg is initially at 200oC. The sphere is parachuted from
a building window 100 m above street level. You are challenged to catch the sphere with your
bare hands as it reaches the street. The sphere drops with an average velocity of 4.1 m/s. Its
diameter is 40 cm and the ambient air temperature is 20oC. Will you accept the challenge?
Support your decision.
(1) Observations. (i) The sphere cools off as it drops. Heat loss from the sphere is by forced
convection. (ii) This is an external flow problem with a free stream velocity that changes with
time. (iii) This is a transient conduction problem. The cooling time is equal to the time it takes
the sphere to drop to street level. (iv) If the Biot number is less than 0.1, lumped capacity method
can be used to determine transient temperature. (v) Cooling rate depends on the heat transfer
coefficient.
(2) Problem Definition. Determine transient temperature of sphere. This requires determining
the drop time and the heat transfer coefficient.
(3) Solution Plan. Use the lumped capacity method to determine transient temperature. Use
correlation equation to determine the heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) constant heat transfer coefficient, (3) Biot
number < 0.1 (to be verified ), (4) constant properties, (5) constant sphere velocity, (6) uniform
ambient conditions, (7) negligible wind speed, (8) no heat loss to the air inside the sphere, (9) no
buoyancy (E = 0 or g = 0) and. (10) no radiation and
(ii) Analysis. Equation (5.7) gives the transient temperature for the lumped capacity model
T (t )
Tf (Ti Tf ) exp[
h As
t]
U c pV
(a)
where
As = surface area, m2
c p = specific heat of aluminum = 902 J/kg- oC
h = average heat transfer coefficient, W/m2-oC
t = time, s
T (t ) = temperature variable, oC
Tf = ambient temperature = 20oC
Ti = initial temperature of sphere = 200oC
V = volume, m3
U = density of aluminum = 2702 kg/m3
The product of U V in equation (a) is equal to the mass of the sphere. That is
UV=M
(b)
As = SD2
(c)
where
M = mass of sphere = 0.2 kg
Surface area of sphere is
PROBLEM 8.16 (continued)
where
D = diameter = 0.4 m
There are two unknowns in (a): drop time to and the average heat transfer coefficient h . For
constant sphere velocity, drop time is
s
to
(d)
Vf
where
s = drop distance = 100 m
Vf = sphere velocity = 4.1 m/s
Equation (8.12a) gives a correlation for the average Nusselt number for forced convection over a
sphere
NuD
hD
k
P
2 0.4 Re1D/ 2 0.06 ReD2 / 3 Pr 0.4 P
s
>
@
1/ 4
(e)
Valid for:
3.5 < ReD < 7.6 u104
0.71 < Pr < 380
1.0 < (P /P s) <3.2
properties at Tf , Ps at Ts
(f)
where
k = thermal conductivity of air = 0.02564 W/m-oC
NuD = average Nusselt number
Pr = Prandtl number of air = 0.713
ReD = Reynolds number
P = viscosity of air at the free stream temperature = 18.17u10-6 kg/m-s
Ps = viscosity of air at surface temperature, kg/m-s
The Reynolds number is defined as
ReD =
Vf D
Q
(g)
where
Q = kinematic viscosity of air = 15.09u10-6 m2/s
Since surface temperature changes with time, Ps should be evaluated at the average surface
temperature Ts defined as
Ts = (Ti + To)/2
(h)
where To is the final sphere temperature obtained from (a). Since this temperature is unknown,
the procedure becomes one of trial and error. A value for To is assumed and air viscosity
P s determined at Ts . Equation (a) is used to calculate To. The calculated To is compared with the
assumed value and the process is repeated until a satisfactory agreement is obtained between
assumed and calculated values.
(iii) Computations. To calculate the final sphere temperature from (a), the drop time to and
the average heat transfer coefficient h need to be determined. Equation (d) gives to
to =100(m)/4.1(m/s) = 24.39 s
PROBLEM 8.16 (continued)
To determine Ps at the mean surface temperature Ts , assume a final sphere temperature To =
80oC. Equation (h) gives
Ts = (200 + 80)(oC)/2 = 140oC
Thus
Ps = 23.44u10-6 kg/m-s
The Reynolds number is
ReD =
4.1(m / s)0.4(m)
= 108,681
15.09 u 10 6 (m 2 / s)
This is somewhat outside the range of applicability of correlation equation (e). Therefore, results
based on using this equation are approximate. Substituting into (e)
Nu D
1/ 4
§ 18.17 u 106 (kg / m s) ·
hD
¸
= 2 0.4(108681)1 / 2 0.06(108681)2 / 3 (0.713)0.4 ¨¨
6
¸
k
© 23.44 u 10 (kg / m s) ¹
>
@
o
= 222.1
2 o
h = 222.1(k/D) = 222.1 (0.02564) (W/m- C)/0.4(m) = 14.2 W/m - C
Substituting into (a) and using (c)
ª 14.2( W / m 2 o C)S (0.4) 2 (m 2 )24.39(s) º
o
» = 88.6 C
0.2(kg)902(J / kg o C)
«¬
»¼
To = 20(oC) + [(200 20 )(oC)] exp «
Although this is slightly different from the assumed value of To = 80oC, repeating the procedure
with a new value of To = 88.6oC will have a minor effect on the result.
The Biot number can now be checked to establish the validity of the lumped capacity method.
The Biot number is defined as
Bi = h G /kal
(i)
where
Bi = Biot number
kal = thermal conductivity of aluminum = 236 W/m-oC
G = wall thickness of sphere, m
The thickness G is determined from its mass and volume
G = M/ UalS D2 = 0.2(kg)/2702(kg/m3)S (0.4) 2(m2) = 0.000147 m
When this is substituted into (i) gives Bi = 8.8u10-6. Thus, the use of the lumped capacity
method is justified.
(iv) Checking. Dimensional check: Computations showed that equations (a), (d), (e) and
(g) are dimensionally consistent.
Limiting check: If the sphere is dropped from an infinite height, its temperature at landing should
be equal to the ambient temperature. Setting s = f in (d) gives to = f. When this is substituted
into (a) gives To = Tf.
Quantitative check: The value of h is slightly outside the range listed in Table 1.1 for forced
convection of gases.
PROBLEM 8.16 (continued)
(5) Comments. (i) The analysis shows that it is not safe to catch the sphere since its
temperature at landing is 88.6oC. However, assumptions (8) and (10) are conservative since they
result in an overestimate of surface temperature at landing. (ii) Both the Reynolds number and
the viscosity correction factor in (e) are outside the range of applicability of correlation equation
(e). However, the effect on the accuracy of the result should be minor.
PROBLEM 8.17
Steam condenses on the outside surface of a 1.6 cm diameter tube. Water enters the tube at
12.5oC and leaves at 27.5oC. The mean water velocity is 0.405 m/s. Outside surface temperature
is 34 oC. Neglecting wall thickness, determine tube length.
(1) Observations. (i) This is an internal forced convection problem. (ii) The channel is a tube.
(iii) The outside surface is maintained at a uniform temperature. (iv) Neglecting tube thickness
resistance means that the inside and outside surface temperatures are identical. (v) Fluid
temperature is developing. (vi) Inlet and outlet temperatures are known. (vii) The Reynolds
number should be determined to establish if the flow is laminar or turbulent. (viii) The required
tube length depends on the heat transfer coefficient. (ix) The fluid is water.
(2) Problem Definition. Determine the required tube length to heat water at a specified flow
rate to a specified temperature. This requires the determination of the average heat transfer
coefficient in a tube.
(3) Solution Plan. Apply the analysis of internal flow through a tube at uniform surface
temperature. Compute the Reynolds number to establish if the flow is laminar or turbulent. Use
analytic solutions or correlation equations to determine the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) axisymmetric flow, (4) constant
properties, (5) uniform surface temperature, (6) negligible changes in kinetic and potential
energy, (7) negligible axial conduction, (8) negligible dissipation, (9) negligible wall thickness,
(10) no energy generation and (11)
smooth tube.
L
D
(ii) Analysis. For flow through a
tube at uniform surface temperature, Tmi
conservation of energy and Newton's
law of cooling lead to equation (7.64)
Tm ( x)
u
0
x
m
Tmo
Ts
Ts (Tmi Ts ) exp [
Ph
x]
m c p
(a)
where
c p = specific heat, J/kg o C
h = average heat transfer coefficient for a tube of length L, W/m 2 o C
= mass flow rate, kg/s
m
P = tube perimeter, m
Tm ( x) = mean temperature at x, o C
Tmi = mean inlet temperature = 12.5 o C
Ts = surface temperature = 34 o C
x = distance from inlet of heated section, m
The perimeter P is given by
P SD
where
(b)
PROBLEM 8.17 (continued)
D = inside tube diameter = 1.6 cm = 0.016 m
Substituting (b) into (a) and applying the resulting equation at the outlet, x = L
Tmo
Ts (Tmi Ts ) exp [
S DPh
m c p
L]
(c)
where
Tmo = outlet temperature = 27.5 o C
Solving (c) for the length L
m c p
L
S Dh
ln
Tmi Ts
Tmo Ts
(d)
Conservation of mass gives the flow rate m
m
Uu A
UuS D2
4
(e)
where
u = mean velocity = 0.405 m/s
U = density, kg/m 3
To compute L using (d), it is necessary to determine h . The Reynolds number is computed to
establish if the flow is laminar or turbulent. Reynolds number is defined as
ReD
uD
Q
(f)
where
Re D = Reynolds number
Q = kinematic viscosity, m 2 /s
Water properties are determined at the mean temperature Tm , defined as
Tm =
Substituting numerical values into (g)
Tm =
(12.5 27.5)( o C)
2
20 o C
Properties of water at this temperature are
c p = 4182 J/kg o C
k = 0.5996 W/m o C
Pr = 6.99
Q = 1004 u10 6 m 2 /s
U = 998.3 kg/m 3
Substituting into (e)
Tmi Tmo
2
(g)
PROBLEM 8.17 (continued)
m
998.34(kg/m 3 )0.405(m/s)S (0.016) 2 (m 2 )
= 0.08129 kg/s
4
Finally, the heat transfer coefficient is needed to determine L from equation (d). The Reynolds
number is computed to determine if the flow is laminar or turbulent. Equation (f) gives
0.405(m/s)0.016(m)
1.004 u 10 6 (m 2 /s)
Re D
6454
Since this is greater than the transition number of 2300, it follows that the flow is turbulent. The
appropriate correlation for the Nusselt number is given by the nGielin ski equation (8.17a)
Nu D
hD
k
( f / 8) Re D 1000 Pr
>1 12.7( f / 8)
1/ 2
Pr 2 / 3 1
>1 ( D / L ) @
2/3
@
(h)
Valid for
valid for 0 < D/L <1
developing or fully developed turbulent flow through tubes
2300 < ReD < 5 u 106
0.5<Pr < 2000
properties at Tm
where
N u D = average Nusselt number in turbulent flow
f = friction factor
For a smooth pipe f is given by
(0.79 ln ReD 1.64) 2
f
(i)
(iii) Computations. The mass flow rate is calculated using (e)
998.3(kg/m 3 )0.405(m/s) S (0.016) 2 (m 2 )
4
m
0.08129 kg/s
The Nusselt number is calculated using (h). The friction factor f is determined using (i)
f
(0.79 ln 6374.5 1.64) 2
0.0357
To determine h from equation (h) the length L must be known. Since L is unknown, a trial and
error procedure is used. A value for L is assumed and equation (h) is used to obtain a first
approximation for h . This approximate value is used in (d) to compute L. An improved value for
h is then obtained by substituting the calculated L into (h). Equation (d) is used again to obtain
an new value for L. This procedure is repeated until a satisfactory agreement is obtained between
assumed and calculated L. As a first approximation, let L = 0. Equation (f) gives
NuD
hD
k
(0.0357 / 8) 6454 1000 6.99
1 12.7(0.0401 / 8)1 / 2 (6.99) 2 / 3 1
>
@
52.3
Solving for h
h
k
0.5996( W/m o C)
Nu D = 52.3
= 1960 W/m 2 o C
0.016(m)
D
PROBLEM 8.17 (continued)
Substituting into (d)
L
0.08129(kg/s)4182(J/kg- o C) (12.5 34)( o C)
ln
S (0.016)(m)1960( W/m 2 - o C) (27.5 34)( o C)
Substituting this value into (h) gives h
gives L
4.13 m
1984.2 W/m 2 o C . With this value of h equation (d)
4.077 m. Further iteration gives h
1985 W/m 2 o C and L
4.075 m.
(iv) Checking. Dimensional check: Computations showed that equations (d)-(i) are
dimensionally consistent.
Quantitative check: (1) The value of h is within the range given in Table 1.1 for forced
convection of liquids.
(2) An approximate value for L can be obtained based on the assumption that the fluid is at an
average temperature Tm 20 o C . Conservation of energy gives
m c p (Tmi Tmo ) S DL h (Ts Tm )
Solving for L
L
m c p (Tmi Tmo )
S D h (Ts Tm )
Substituting numerical values into the above
L
0.08129(kg/s)4182(J/kg- o C) (27.5 12.5)( o C)
S (0.016)(m)1985( W/m 2 -o C) (34 20)( o C)
3.65 m
This result differs from the exact solution by 10%.
(5) Comments. The fact that assuming L = 0 in calculating the heat transfer coefficient from
equation (h) introduces a small error in h implies that entrance effects are negligible.
PROBLEM 8.18
A 150 cm long tube with 8 mm inside diameter passes through a laboratory chamber. Air enters
the tube at 12oC with fully developed velocity and a flow rate 0.0005 kg/s. Assume uniform
surface temperature of 25oC, determine outlet air temperature.
(1) Observations. (i) This is an internal force convection problem. (ii) The channel is a tube. (iii)
The surface is maintained at a uniform temperature. (iv) The velocity is fully developed. (v) The
temperature is developing. (vi) The outlet temperature is unknown..(vii) The Reynolds number
should be checked to establish if the flow is laminar or turbulent. (viii) The fluid is air.
(2) Problem Definition. Determine the outlet air temperature. This requires the determination
of the average heat transfer coefficient.
(3) Solution Plan. Apply the analysis of internal flow through a tube at uniform surface
temperature. Check the Reynolds number to determine if the flow is laminar or turbulent. Use
analytic solutions or correlation equations to determine the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) fully developed velocity, (4)
axisymmetric flow, (5) constant properties, (6) uniform surface temperature, (7) negligible
changes in kinetic and potential energy, (8) negligible axial conduction, (9) no dissipation, (10)
no energy generation and (11) smooth tube.
(ii) Analysis. For flow in a tube
at uniform surface temperature,
conservation of energy and Newton's
law of cooling lead to equation Tmi
(7.64)
L
D
u
0
x
m
Tmo
Ts
Tm ( x)
Ts (Tmi Ts ) exp [
Ph
x]
m c p
where
cp = specific heat, /Jkg- oC
2 o
h = average heat transfer coefficient for a tube of length L, W/m - C
= mass flow rate = 0.0005 kg/s
m
P = tube perimeter, m
Tm(x) = mean temperature at x, oC
Tmi = mean inlet temperature = 12oC
Ts = surface temperature = 25oC
x = distance from inlet of heated section, m
The perimeter P is given by
P=SD
where
(a)
(b)
D = inside tube diameter = 8 mm = 0.008 m
Substituting (b) into (a) and setting x = L gives the outlet temperature
Tmo ( x)
Ts (Tmi Ts ) exp [
S Dh
m c p
L]
(c)
PROBLEM 8.18 (continued)
where
L = tube length = 150 cm = 1.5 m
Tmo = mean outlet temperature, oC
To compute Tmo using (c), it is necessary to determine h . The Reynolds number is calculated to
establish if the flow is laminar or turbulent. Reynolds number is defined as
ReD u D /Q
where
ReD = Reynolds number
u = mean flow velocity, m/s
Q = kinematic viscosity, m2/s
The mean velocity is determined from mass flow rate
(d)
Uu A Uu SD 2 / 4
m
or
u
4m / SUD 2
(e)
where
A = cross section area, m2
U = density, kg/m3
Air properties are determined at the mean temperature Tm , defined as
Tmi Tmo
(f)
2
Since outlet temperature Tmo is unknown, the solution is obtained by a trial and error procedure.
A value for Tmo is assumed, (f) is used to calculate Tm , a first approximation of properties is
determined at this temperature and (c) is used to calculate Tmo. If the calculated value is not the
same as the assumed value, the procedure is repeated until a satisfactory agreement is obtained.
Tm =
Assume Tmo = 18oC. Equation (f) gives
(12 18)( o C)
15 o C
2
Properties of air at this temperature are
Tm =
cp = 1005.95 /Jkg- oC
k = 0.02526 W/m-oC
Pr = 0.7145
Q = 14.64u10-6 m2/s
U = 1.22545 kg/m3
The mean velocity is obtained from (e)
u=
4(0.0005)(kg / s)
S (1.22545)(kg / m 3 )(0.008) 2 (m 2 )
The Reynolds number is obtained from (d)
ReD
8.117(m / s)0.008(m)
14.64 u 10 6 (m 2 / s)
4435.5
= 8.117 m/s
PROBLEM 8.18 (continued)
Since this is greater than the transition number of 2300, it follows that the flow is turbulent. The
appropriate correlation for the Nusselt number is given by the G
nielin ski equation (8.17a)
Nu D
hD
k
( f / 8) Re D 1000 Pr
>1 12.7( f / 8)
1/ 2
Pr 2 / 3 1
>1 ( D / L ) @
2/3
@
(g)
Valid for
valid for 0< D/L <1
developing or fully developed turbulent flow through tubes
2300 < ReD < 5 u 106
0.5<Pr < 2000
properties at Tm
(h)
where
N u D = average Nusselt number in turbulent flow
f = friction factor
For a smooth pipe f is given by
f = (0.79lnReD 1.64)–2
(i)
(iii) Computations. The Nusselt number is calculated using properties at the assumed
outlet temperature Tmo = 18oC. The friction factor f is determined first using (i)
f = (0.79 ln 4435.5 - 1.64)-2 = 0.0401
Substituting into (g)
NuD
hD
k
(0.0401 / 8) 4435.5 1000 0.7145
1 12.7(0.0401 / 8)
1/ 2
>(0.7145)
2/3
>1 [0.008(m) / 1.5(m)] @= 15.47
2/3
@
1
Solving for h
h
0.02526( W / m o C)
k
Nu D = 15.47
= 48.8 W/m2-oC
D
0.008(m)
Substituting into (c) gives the outlet temperature
Tmo
25( o C) (12 25)( o C) exp [
S (0.008)(m)48.8( W/m 2 o C)
0.0005(kg/s)1005.95(J/kg o C)
1.5(m)
24.7 o C
Since this is greater than the assumed value of 18oC, the procedure is repeated with a new
assumed value of 23oC. Based on this value, the calculated outlet temperature is found to be
24.7oC. Thus, the outlet temperature is 24.7oC.
(iv) Checking. Dimensional check: Computations showed that equations (c), (d), (e) and (g)
are dimensionally consistent.
Quantitative check: The value of h is within the range given in Table 1.1 for forced convection
of gases.
(5) Comments. (i) The outlet temperature is within 0.3oC of the maximum value that it can
reach. (ii) If one incorrectly assumes that the flow is laminar, the corresponding Nusselt number
is 3.66. This is considerably smaller than the turbulent Nusselt number calculated above.
PROBLEM 8.19
Water enters a tube with a fully developed velocity and uniform temperature Tmi = 18oC. The
inside diameter of the tube is 1.5 cm and its surface temperature is uniform at Ts = 125oC.
Neglecting wall thickness, determine the length of the tube needed to heat the water to 82oC at a
flow rate of 0.002 kg/s.
(1) Observations. (i) This is an internal forced convection problem. (ii) Tube surface is
maintained at uniform temperature. (iii) The velocity is fully developed. (iv) The length of tube
is unknown. (v) The temperature is developing. However, depending on tube length relative to
the thermal entrance length, temperature may be considered fully developed throughout. (vi) The
Reynolds number should be checked to determine if the flow is laminar or turbulent. (vii) The
fluid is water.
(2) Problem Definition. Determine the length of tube needed to increase the temperature of
water to a specified level at a given flow rate.
(3) Solution Plan. Apply the analysis of internal flow through a tube at uniform surface
temperature. Use analytic solutions or correlation equations to determine the average heat
transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4)
axisymmetric flow, (5) uniform surface temperature, (6) fully developed velocity, (7) no axial
conduction, (8) negligible changes in potential and kinetic energy, (8) no dissipation, (9) no
energy generation and (10) smooth tube.
(ii) Analysis. For flow in a tube at uniform surface temperature, conservation of energy and
Newton's law of cooling lead to equation (7.64)
Tm ( x)
Ts (Tmi Ts ) exp [
where
Ph
x]
m c p
L
cp = specific heat, /Jkg- oC
h = average heat transfer coefficient, u
x
Tmi
W/m2-oC
= mass flow rate = 0.002 kg/s
m
P = perimeter, m
Ts = surface temperature = 125oC
Tmi = inlet temperature = 18oC
Tm (x) = mean temperature at any location x along the tube, oC
x = distance along the tube measured from the inlet, m
(a)
Ts
D
Tmo
Ts
To determine the required tube length, set x = L and Tm (L) = Tmo and solve (a) for L
L
where
§ m c p
¨
¨ hP
©
· Ts Tmi
¸ ln
¸ T T
s
mo
¹
(b)
PROBLEM 8.19 (continued)
Tmo = mean outlet temperature = 82oC
The perimeter P is given by
P =S D
where
D = inside diameter = 1.5 cm = 0.015 m
(c)
Thus, all quantities in (b) are known except h . To determine h , the Reynolds number is
calculated to establish if the flow is laminar or turbulent. The Reynolds number is defined as
uD
(d)
ReD =
Q
where
u = mean velocity, m/s
Q = kinematic viscosity, m2/s
The mean velocity is determined from mass flow rate equation
D2
m U u S
4
where
U = density, kg/m3
(e)
Solving (e) for u
u =
4m
US D2
(f)
Properties are evaluated at the mean temperature Tm , the average of inlet and outlet temperatures
Tmi Tmo (18 82)( o C)
50 oC
2
2
Properties of water at this temperature are given in Appendix D
Tm
cp = specific heat = 4182 J/kg- oC
k = thermal conductivity = 0.6405 W/m-oC
Pr = Prandtl number = 3.57
Q = kinematic viscosity = 0.5537u10-6 m2/s
U = density = 988 kg/m3
Equations (f) and ( d) give
u =
4( 0.002)( kg / s)
= 0.01146 m/s
988( kg / m 3 ) S (0.015) 2 ( m 2 )
and
ReD =
0.01146( m / s) 0.015( m)
= 310
0.5537 u 10 6 ( m 2 / s)
Since ReD < 2300, the flow is laminar. The next step is establishing if thermal entrance length Lt
is small compared to tube length L. If this is the case, the flow can be considered thermally fully
developed throughout. The thermal entrance length for laminar flow in a constant temperature
tube is given by equation (7.43b)
Lt = 0.033 D ReD Pr
(g)
PROBLEM 8.19 (continued)
Substituting into (h)
Lt = 0.033 (0.015)(m) (310) (3.57) = 0.548 m
Since L is unknown, comparison can not be made with Lt. Assuming that Lt is not small
compared to L, the flow must be treated as thermally developing and correlation equation (8.14a)
should be used
0.0668( D / L) Re D Pr
hD
Nu D
3.66 (h)
k
1 0.04>( D/L) Re D Pr )@1 / 3
Valid for
entrance region of tube
uniform surface temperature Ts
fully developed laminar flow (ReD < 2300)
developing temperature
properties at Tm (Tmi Tmo ) / 2
(i)
To calculate h from this equation, the length L must be known. Thus, a trial and error procedure
is required to solve the problem. A value for L is assumed, equation (h) is used to calculate h
and the result substituted into (b) to determine L. The procedure is repeated until a satisfactory
agreement is obtained between assumed and calculated values.
(iii) Computations. Assume L = 0.5 m. Substituting into (h)
Nu D
hD
k
3.66 0.0668 [0.015(m) / 0.5(m) ]310(3.57)
1 0.04>^0.015(m)/0.5(m)`310(3.57)@2 / 3
5.229
Solving for h
0.6405( W / m o C)
k
h
Nu D =
5.229 223.29 W / m 2 o C
D
0.015(m)
Substituting into (b)
§ 0.002(kg / s)4182(J / kg o C) · (125 18)( o C)
¸ ln
L ¨¨
2 o
o
¸
© 223.29( W / m C)S (0.015)(m) ¹ (125 82)( C)
0.725 m
Since calculated L is not equal to the assumed value of 0.5 m, the process is repeated with
another assumed value of L = 0.73 m. Results of four trials are tabulated below.
Assumed L Calculated h
m
W/m2-oC
0.5
223.29
0.73
205.38
0.79
202.23
0.80
201.74
Calculated L
m
0.725
0.788
0.8
0.802
Therefore, L = 0.802 m
(iv) Checking. Dimensional check: Computations showed that equations (b), (d), (f), (g) (h)
PROBLEM 8.19 (continued)
and (i) are dimensionally consistent.
Quantitative check: The value of the heat transfer coefficient is within the range shown in Table
1.1 for forced convection of liquids.
Qualitative check: The value of h for developing temperature should be larger than h for fully
developed temperature. Equation (7.57) gives the Nusselt number for the fully developed case
hD
Nu D
3.66
(j)
k
0.6405( W / m o C)
k
h 3.66
156.3 W / m 2 o C
3.66
D
0.015(m)
This is smaller than h = 201.74 W/m2-oC for the developing case.
Limiting check: If Tmo = Tmi, the required length should be zero. Setting Tmo = Tmi in (b) gives the
L = 0.
(5) Comments. (i) Since Lh is not small compared to L, temperature entrance effects can not be
neglected. (ii) Equation (h) converges to the limiting case of fully developed temperature when L
o f. Setting L = f in (h) gives the fully developed solution (j).
PROBLEM 8.20
Cold air is supplied to a research apparatus at a rate of 0.14 g/s. The air enters a 20 cm long
tube with uniform velocity and uniform temperature of 20oC. The inside diameter of the tube is
5 mm. The inside surface is maintained at 30oC. Determine the outlet air temperature.
(1) Observations. (i) This is an internal forced convection problem. (ii) Tube surface is
maintained at a uniform temperature. (iii) The velocity and temperature are developing. Thus,
entrance effects may be important. (iv) The outlet temperature is unknown. (v) The fluid is air.
(2) Problem Definition. Determine air outlet temperature. This requires determining the
average heat transfer coefficient.
(3) Solution Plan. Apply the analysis of internal flow through a tube at uniform surface
temperature. Check the Reynolds number to determine if the flow is laminar or turbulent.
Compute the hydrodynamic and thermal entrance lengths to establish if entrance effects can be
neglected. Use an appropriate correlation equation to compute the average heat transfer
coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4)
axisymmetric flow, (5) uniform surface temperature, (6) uniform inlet velocity and temperature,
(7) no axial conduction, (8) negligible changes in potential and kinetic energy, (9) no dissipation
and (10) no energy generation.
.
(ii) Analysis. For flow in a tube at uniform surface temperature, conservation of energy and
Newton's law of cooling lead to equation (7.64)
Tm ( x)
Ts (Tmi Ts ) exp[
Ph
x]
m c p
(a)
where
cp = specific heat, /Jkg- oC
2 o
h = average heat transfer coefficient for a tube of length L, W/m - C
= mass flow rate = 0.14 g/s = 0.00014 kg/s
m
P = tube perimeter, m
Tm(x) = mean temperature at x, oC
Tmi = mean inlet temperature = 20 oC
Ts = surface temperature = 30oC
x = distance from inlet of heated section, m
The perimeter P is given by
P=SD
L
(b)
where
D = inside tube diameter
= 5 mm = 0.005 m
Substituting (b) into (a) and setting
x = L gives the outlet temperature
u
Tmi
x
D
Tmo
Ts
PROBLEM 8.20 (continued)
Ts (Tmi Ts ) exp[
Tmo ( x)
S Dh
m c p
L]
(c)
where
L = tube length = 20 cm = 0.2 m
Tmo = mean outlet temperature, oC
To compute Tmo using (c), it is necessary to determine h . The Reynolds number is calculated to
establish if the flow is laminar or turbulent. Reynolds number is defined as
ReD
uD
Q
(d)
where
ReD = Reynolds number
u = mean flow velocity, m/s
Q = kinematic viscosity, m2/s
The mean velocity is determined from mass flow rate
Uu A Uu SD 2 / 4
m
or
u
4m / SUD 2
(e)
where
A = cross section area, m2
U = density, kg/m3
Air properties are determined at the mean temperature Tm , defined as
Tmi Tmo
(f)
2
Since outlet temperature Tmo is unknown, the solution is obtained by a trial and error procedure.
A value for Tmo is assumed, (f) is used to calculate Tm , a first approximation of properties is
determined at this temperature and (c) is used to calculate Tmo. If the calculated value of Tmo is
not the same as the assumed value, the procedure is repeated until a satisfactory agreement is
obtained.
Tm =
Assume Tmo = 0oC. Equation (f) gives
(20 0)( o C)
10 o C
2
Properties of air at this temperature are
Tm =
cp = 1005.6 J/kg- oC
k = 0.02329 W/m-oC
Pr = 0.721
Q = 12.46u0-6 m2/s
U = 1.3414 kg/m3
The mean velocity is obtained from (e)
u=
4(0.00014)(kg / s)
S (1.3414)(kg / m 3 )(0.005) 2 (m 2 )
= 5.315 m/s
PROBLEM 8.20 (continued)
The Reynolds number is obtained from (d)
ReD
5.315(m / s)0.005(m)
12.46 u 10 6 (m 2 / s)
2133
Since this is less than the transition number of 2300, it follows that the flow is laminar. The next
step is to determine if entrance effects can be neglected. For laminar flow, the hydrodynamic
entrance length Lh for a constant surface temperature tube is given by equation (7.43a)
Lh/D = 0.056 ReD
(g)
The thermal entrance length Lt is given by equation (7.43b)
Lt/D = 0.033 ReD Pr
(h)
Equations (g) and (h) give
Lh = 0.056(2,133)0.005(m) = 0.597m
and
Lt = 0.033(2,133)(0.721) 0.005(m) = 0.254 m
Thus, both velocity and temperature are developing. The appropriate correlation for the average
Nusselt number is given by equation (8.15a)
Nu D
hD
§P ·
1.86> ( D/L) Re D Pr @1 / 3 ¨ P ¸
k
© s¹
0.14
(i)
Valid for:
entrance region of tube
uniform surface temperature Ts
laminar flow (ReD < 2300)
developing velocity and temperature
0.48 < Pr < 16700
P
0.0044 < Ps < 9.75
properties at Tm , P s at Ts
(j)
where
P = viscosity at mean temperature = 16.71u10-6 kg/s-m
P s = viscosity at surface temperature = 18.65u10-6 kg/s-m
Conditions (j) are satisfied.
(iii) Computations. Equation (i) is used to calculate h
Nu D
hD
k
ª 0.005(m)
º
2133(0.721)»
1.86 «
¬ 0.2(m)
¼
1/ 3
ª16.71 u 10 6 (kg / s m) º
«
»
6
¬18.65 u 10 (kg / s m) ¼
or
0.02329( W / m o C)
k
Nu D
6.18 = 28.8 W/m2-oC
D
0.005(m)
Substituting into (c)
h
0.14
= 6.18
PROBLEM 8.20 (continued)
ª S (0.005)(m)28.8( W / m 2 o C)0.2(m) º
o
Tmo = 30(oC) (30 20) (oC) exp «
» 3.7 C
o
0
.
00014
(
kg
/
s
)
1005
.
6
(
J
/
kg
C
)
¬
¼
Since this is higher than the assumed temperature, the above procedure is repeated with assumed
Tmo = 4oC. The corresponding heat transfer coefficient and calculated outlet temperature are h =
28.94 W/m2-oC and Tmo = 3.8oC.
(iv) Checking. Dimensional check: Computations showed that units of equations (b)-(i) are
dimensionally consistent.
Qualitative check: Since entrance effects are important, it follows that the heat transfer
coefficient is greater than that of fully developed flow. For laminar fully developed flow, the
Nusselt number is given by Equation (7.57)
hD
3.66
(k)
Nu D
k
Using this equation to compute h
h
3.66
k
D
3.66
0.02329( W / m o C)
17 W / m 2 o C
0.005(m)
This is smaller than h = 28.8 W/m2-oC for the developing case.
Limiting check: In the limit as L o 0, the outlet temperature approaches inlet temperature.
Setting L = 0 in equation (c) gives Tmo = Tmi.
(5) Comments. (i) If entrance effects are neglected, h will be underestimated ( h = 17 W/m2o
C) and the corresponding outlet temperature will be Tmo = 4 oC. (ii) The Reynolds number is
very close to the transition Reynolds number of 2300. Since this value of transition Reynolds
number is not exact, depending on surface roughness and other factors, it is uncertain if the flow
is laminar or turbulent for this case.
PROBLEM 8.21
Water flows through a tube of inside diameter 2.5 cm. The inside surface temperature is 230oC
and the mean velocity is 3 cm/s. At a section far away from the inlet the mean temperature is
70oC.
[a] Calculate the heat flux at this section
[b] What will the flux be if the mean velocity is increased by a factor of ten?
(1) Observations. (i) This is an internal forced convection problem. (ii) Tube surface is
maintained at uniform temperature. (iii) The section of interest is far away from the inlet. This
means that flow and temperature can be assumed fully developed and the heat transfer
coefficient uniform. (iv) It is desired to determine the surface flux at this section. Newton’s law
of cooling gives a relationship between local flux, surface temperature and heat transfer
coefficient. (v) The Reynolds number should be checked to determine if the flow is laminar or
turbulent. (vi) The fluid is water.
(2) Problem Definition. Determine surface heat flux corresponding to two mean flow
velocities. Since the flux can be obtained from Newton's law of cooling, the problem is one of
finding the heat transfer coefficient or fully developed flow corresponding to the two velocities.
(3) Solution Plan. Apply Newton's law of cooling at the specified section of the tube. Use
analytic solutions or correlation equations to determine the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4)
axisymmetric flow, (4) uniform surface temperature, (5) fully developed velocity and
temperature, (6) no axial conduction, (7) negligible changes in potential and kinetic energy, (8)
no dissipation, (9) no energy generation and (10) smooth tube.
(ii) Analysis. Apply Newton’s law of cooling
q cc = h (Ts Tf )
(a)
where
h = local heat transfer coefficient, W/m2-oC
q cc = surface flux, W/m2
Ts = surface temperature = 230oC
Tm = mean temperature of water = 70oC
Since flow and temperature are fully developed, the heat transfer coefficient is uniform (local
and average coefficients are identical). To determine h, the Reynolds number is calculated to
establish if the flow is laminar or turbulent. The Reynolds number is defined as
ReD =
uD
Q
fully developed
(b)
where
D = inside tube diameter = 2.5 cm = 0.025 m
ReD = Reynolds number
u = mean velocity = 3 cm/s = 0.03 m/s
Q = kinematic viscosity = 0.4137u10-6 m2/s
Tm
x
Ts
D
u
Ts
PROBLEM 8.21 (continued)
Properties are evaluated at the mean temperature Tm = 70oC. Appendix D gives
k = thermal conductivity = 0.6594 W/m-oC
Pr = Prandtl number =2.57
Equation (b) gives
ReD =
0.03( m / s) 0.025( m)
= 1,813
0.4137 u 10 6 ( m2 / s)
Since ReD < 2300 it follows that the flow is laminar. The Nusselt number for fully developed
laminar flow in tubes at uniform surface temperature is given by equation (7.57)
Nu D = NuD =
hD
= 3.66
k
(c)
When the velocity is increased by a factor of 10 the Reynolds number increases to
ReD = 18,130
Thus, the flow becomes turbulent. For this case the average Nusselt number is given by the
G
nielinski equation (8.17a)
NuD
hD
k
f
8
ReD 1000 Pr
ª1 12.7 f
8
¬«
1/ 2
Pr 2 / 3 1 º»
¼
>1 ( D / L ) @
2/3
(d)
Valid for
valid for 0< D/L <1
developing or fully developed turbulent flow through tubes
2300 < ReD < 5 u 106
0.5 <Pr < 2000
properties at Tm
(e)
where
NuD = average Nusselt number in turbulent flow
f = friction factor
For a smooth pipe f is given by
f = (0.79lnReD 1.64)–2
For fully developed flow the factor D/L in equation (d) is set to zero.
(iii) Computations.
[a] For the case where u = 0.03 m/s and the flow is laminar. Equation (c) gives
k
0.6594( W / m o C)
h = 3.66 = 3.66
= 96.5 W/m2-oC
D
0.025( m)
Substituting into (a)
q cc = 96.5(W/m2-oC)(230 - 70)(oC) = 15,440 W/m2
(f)
PROBLEM 8.21 (continued)
[b] For the case where the flow is turbulent, equations (d) and (f) give h . The Reynolds number
for this case is 18,130. Equation (f) gives f
f = [0.79ln (18,130) 1.64]–2 = 0.02682
Substituting into (d) and setting D/L = 0
NuD
hD
k
0.02682 / 8 18,130 1000 2.57
>1 12.7 0.02682 / 8
1/ 2
2.57 2 / 3 1
@
= 89.76
Solving the above for h
k
0.6594( W / m o C)
= 89.76
= 2367.5 W/m2-oC
D
0.025( m)
Equation (a) gives the flux
h = 89.76
q cc = 2367.5 (W/m2-oC)(230 - 70)(oC) = 378,800 W/m2
(iv) Checking. Dimensional check: Computations showed that equations (a)-(d) are
dimensionally consistent.
Qualitative check: As expected turbulent heat flux is greater than that of laminar flow.
Limitations on correlation equation (d): The conditions listed in (e) are met.
(5) Comments. Although the velocity is increased by a factor of 10, the flux is increased by a
factor of 25. This large increase is a result of transition from laminar to turbulent. If the flow
remains laminar, then according to (c), there will be no change in h even though the Reynolds
number is increased.
PROBLEM 8.22
Air flows through a tube of inside diameter 5 cm. At a section far away from the inlet the mean
temperature is 30oC. At another section further downstream the mean temperature is 70oC.
Inside surface temperature is 90oC and the mean velocity is 4.2 m/s. Determine the length of this
section.
(1) Observations. (i) This is an internal forced convection problem in a tube. (ii) Both velocity
and temperature are fully developed. (iii) Tube surface is maintained at uniform temperature. (iv)
The Reynolds number should be computed to establish if flow is laminar or turbulent. (v) eMan
velocity, mean inlet and outlet temperatures and tube diameter are known. (vi) The fluid is air.
(2) Problem Definition. Find the required tube length to increase the air temperature by a given
amount.
(3) Solution Plan. Use the analysis of fully developed laminar flow in tubes at uniform surface
temperature to determine the required tube length.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) fully developed flow, (4)
axisymmetric flow, (5) constant properties, (6) uniform surface temperature, (7) negligible
changes in kinetic and potential energy, (8) negligible axial conduction, (9) no dissipation, (10)
no energy generation and (11) smooth tube.
L
Lt
Gt
Ti
0
Vi
Tmi
G
x
D
u
m
Tmo
Ts
Lh
(ii) Analysis. For flow in a tube at uniform surface temperature, conservation of energy and
Newton's law of cooling lead to equation (7.64)
Tm ( x)
Ts (Tmi Ts ) exp[
Ph
x]
m c p
cp = specific heat, /Jkg- oC
h = average heat transfer coefficient for a tube of length L, W/m2-oC
L = length of tube, m
= mass flow rate, kg/s
m
P = tube perimeter, m
Tm(x) = mean temperature at x, oC
Tmi = mean inlet temperature = 30oC
Ts = surface temperature = 90oC
x = distance from inlet of heated section, m
Applying (a) at the outlet of the heated section (x = L) and solving for L
(a)
PROBLEM 8.22 (continued)
L
p
mc
Ph
ln
Ts Tmi
Ts Tmo
(b)
where
Tmo = mean outlet temperature = 70oC
, and h . All properties are
To compute L using (b), it is necessary to determine cp, P, m
determined at the mean temperature Tm defined as
Tm =
Tmi Tmo
2
(c)
The perimeter P and flow rate m are given by
P=SD
(d)
m S D 2 U u / 4
(e)
and
where
D = inside tube diameter = 5 cm = 0.05 m
u = mean flow velocity = 4.2 m/s
U = density, kg/m3
The heat transfer coefficient for fully developed flow is uniform along a channel. Its value
depends on whether the flow is laminar or turbulent. To proceed, it is necessary to calculate the
Reynolds number to determine if the flow is laminar or turbulent. For flow in a tube the
Reynolds number is defined as
uD
ReD
(f)
Q
where
ReD = Reynolds number
Q = kinematic viscosity, m2/s
The mean temperature is calculated in order that properties are determined. Substituting into (c)
Tm =
(30 70)( o C)
2
50 o C
Properties of air at this temperature are given in Appendix C
cp = 1007.4 /Jkg- oC
k = 0.02781 W/m-oC
Pr = 0.709
Q = 17.92u10-6, m2/s
U = 1.0924 kg/m3
Substituting into (f)
ReD
4.2(m / s)0.05(m)
17.92 u 10 6 (m 2 / s)
11,719
Since the Reynolds number is greater than 2300, the flow is turbulent. The Nusselt number for
turbulent flow through a tube is given by the nGielinski equation (8.17a)
PROBLEM 8.22 (continued)
Nu D
hD
k
f
8 Re D 1000 Pr
ª1 12.7 f
8
«¬
1/ 2
Pr
2/3
>1 ( D / L ) @
2/3
1 º
»¼
(g)
Valid for
valid for 0< D/L <1
developing or fully developed turbulent flow through tubes
2300 < ReD < 5 u 106
0.5<Pr < 2000
properties at Tm
(h)
where
N u D = average Nusselt number in turbulent flow
f = friction factor
For a smooth pipe f is given by
f = (0.79lnReD 1.64)–2
(i)
For fully developed flow, set D/L = 0 in (g)
Nu D
hD
k
( f / 8) Re D 1000 Pr
>1 12.7( f / 8)
1/ 2
Pr 2 / 3 1
@
(j)
where the average and local heat transfer coefficients, h and h are identical in the fully
developed region.
(iii) Computations. Substituting into (d) and (e)
P = S 0.05(m) = 0.1571 m
m S
(0.05) 2 (m 2 )
1.0924(kg / m 3 )4.2(m / s) 0.009009 kg / s
4
The Nusselt number is calculated using equation (j). The friction factor f is determined using (i)
f = (0.79 ln11,719 1.64 )-2 = 0.0301
Substituting into (j)
Nu D
hD
k
(0.0301 / 8) 11,719 1000 0.709
>
@
1 12.7(0.0301 / 8)1 / 2 (0.709) 2 / 3 1
= 34.2
Solving for h
h
0.02781( W / m o C)
k
Nu D = 34.2
= 19.02 W/m2-oC
D
0.05(m)
Substituting into (b) gives the required tube length
L
0.009009(kg / s)1007.4(J / kg o C )
0.1571(m)17.3( W / m 2 o C)
ln
(90 30)( o C)
(90 70)( o C)
= 3.34 m
(iv) Checking. Dimensional check: Computations showed that equations (b), (d), (e), (f),
and (j) are dimensionally consistent.
PROBLEM 8.22 (continued)
Limiting check: For the special case of Tmo = Tmi , the required length should vanish. Setting Tmo
= Tmi in (b) gives L = 0.
Quantitative check: The value of h is on the low end of values listed in Table 1.1 for forced
convection of gases. It should be kept in mind that values of h in Table 1.1 are for typical
applications. Exceptions should be expected.
(5) Comments. For fully developed laminar flow in tubes at uniform surface temperature, the
Nusselt number is given by equation (7.57)
Nu D
hD
= 3.66
k
If one incorrectly uses this equation, the heat transfer coefficient and required length become
h = 1.85 W/m2-oC
and
L = 34.3 m
Thus, the error in using (k) is significant.
(k)
PROBLEM 8.23
Two identical tubes have inside diameters of 6 mm. Air flows through one tube at a rate of 0.03
kg/hr and through the other at a rate of 0.4 kg/hr. Far away from the inlets of the tubes the
mean temperature is 120oC for both tubes. The air is heated at a uniform surface temperature
which is identical for both tubes. Determine the ratio of the heat flux of the two tubes at this
section.
(1) Observations. (i) This is an internal forced convection problem. (ii) The surface of each
tube is maintained at uniform temperature which is the same for both. (iii) The velocity and
temperature are fully developed. Thus, the heat transfer coefficient is uniform. (iv) Air flows
through each tube at different rates. (v) The Reynolds number should be computed to establish if
the flow is laminar or turbulent. (vi) Surface heat flux depends on the heat transfer coefficient.
(2) Problem Definition. Determine the ratio of surface heat flux for the two tubes at sections
here the mean temperature is the same for both.
(3) Solution Plan. Apply Newton’s law of cooling to each tube. Compute the Reynolds number
to establish if the flow is laminar or turbulent. Use analytic solutions or correlation equations to
determine the heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4)
axisymmetric flow, (5) uniform surface temperature, (6) no axial conduction, (7) fully developed
velocity and temperature, (8) no changes in potential and kinetic energy and (0) no dissipation.
(ii) Analysis. Local heat flux is given by Newton’s law of cooling
q cc = h (Ts - Tm)
(a)
where
h = heat transfer coefficient, W/m2-oC
q cc = surface heat flux, W/m2
Tm = mean temperature = 120oC
Ts = surface temperature, oC
Applying (a) to the two tubes, noting that Ts and Tm are the same for both tubes, and taking the
ratio of the two equations
q1cc h1
(b)
q2cc h2
where the subscripts 1 and 2 refer to tubes 1 and 2. To determine the ratio of the heat transfer
coefficients, the Reynolds number for each tube should be computed to establish if the flow is
laminar or turbulent. The Reynolds number is defined as
ReD =
where
D = tube diameter = 0.006 m
u = mean velocity, m/s
ReD = Reynolds number
Q = kinematic viscosity = 25.19u10-6 m2/s
uD
Q
(c)
PROBLEM 8.23 (continued)
Properties of air are determined at the mean temperature Tm = 120oC. The mean velocity is
determined from the flow rate
US u D2 / 4
(d)
m
or
4m
u
(e)
SUD2
where
= mass flow rate, kg/s
m
U = density = 0.8979 kg/m3
The mass flow rates for the two tubes are
1 = 0.03 kg/hr
m
2 = 0.4 kg/hr
m
Substituting into (e)
u1
4(0.03)(kg / hr )
S 0.8979(kg / m 3 )(0.006) 2 (m 2 )3600( s / hr )
= 0.328 m/s
and
u2
4(0.4)(kg / hr )
S 0.8979(kg / m 3 )(0.006) 2 (m 2 )3600( s / hr )
= 4.38 m/s
Substituting into (c)
ReD1 = = 78.1
and
ReD2 = = 1,043.3
Thus, the flow is laminar in both tubes. For laminar fully developed flow through tubes with
uniform surface temperature, the Nusselt number is given by
NuD = hD/k = 3.66
or
h = 3.66 k/D
(f)
where
NuD = Nusselt number
k = thermal conductivity of air = 0.03261 W/m-oC
(iii) Computations. Applying (f) to the two tubes and noting that D and k are the same for
both tubes
h1 = h2 = 3.66 (0.03261)(W/m-oC)/ 0.006(m) = 19.9 W/m2-oC
Substituting into equation (b)
q1cc
q2cc
19.9(W / m2 o C)
19.9(W / m2 o C)
1
PROBLEM 8.23 (continued)
(iv) Checking. Dimensional check: Computations showed that equations (c), (e) and (f) are
dimensionally correct.
Quantitative check: The value of h is within the range given in Table 1.1 for force convection of
gases.
(5) Comments. It is surprising that although the velocity in tube 2 is over 13 times greater than
that in tube 1, the heat flux is the same for both tubes. This is due to the fact that for laminar
fully developed flow, the Nusselt number is independent of the Reynolds number.
PROBLEM 8.24
Two concentric tubes of diameters 2.5 cm and 6.0 cm are used as a heat exchanger. Air flows
through the inner tube with a mean velocity of 2 m/s and mean temperature of 190oC. Water
flows in the annular space between the two tubes with a mean velocity of 0.5 m/s and a mean
temperature of 30oC. Determine the inside and outside heat transfer coefficients.
(1) Observations. (i) This is an internal forced convection problem. (ii) The geometry consists
of two concentric tubes. (iii) Air flows in the inner tube while water flows in the annular space
between the two tubes. (iv) The Reynolds number should be computed for both fluids to
establish if the flow is laminar or turbulent. (v) Convection resistance depends on the heat
transfer coefficient.
(2) Problem Definition. Determine the air side and water side heat transfer coefficients.
(3) Solution Plan. Compute the Reynolds numbers for air in the tube and for water in the
annular space to establish if the flow is laminar or turbulent. Use analytic solutions or correlation
equations to determine the heat transfer coefficients.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4)
axisymmetric flow, (5) uniform surface temperature, (6) fully developed velocity and
temperature, (7) negligible wall conduction resistance, (8) negligible wall thickness and (9)
smooth tubes.
(ii) Analysis. To determine hi and ho, the two Reynolds numbers are calculated first to establish
if the flow is laminar or turbulent. For air flow inside the inner tube the Reynolds number is
Re Di u a Di / Q a
(a)
where
Di = diameter of inner tube = 2.5 cm = 0.025 m
Re Di = Reynolds number for air flow through inner tube
u a = mean air velocity in inner tube = 2 m/s
Qa = kinematic viscosity of air = 33.34u10-6 m2/s
Substituting into (b)
Re Di =
water
air
air
Do
Di
hi
ho
2(m / s)0.025(m)
= 1499.7
33.34 u 10 6 (m 2 / s)
Thus, the flow is laminar in the inner tube. The Nusselt number for fully developed laminar flow
in tubes depends on surface boundary condition. Assuming uniform surface temperature, the
Nusselt number is given by equation (7.57)
Nu Di =
hi D i
= 3.66
ka
where
ka = thermal conductivity of air = 0.03718 W/m-oC
Nu Di = average Nusselt number
The Reynolds number for water flow in the annular space between the two tubes is defined as
(b)
PR0BLEM 8.24 (continued)
u w De
ReDe
(c)
Qw
where
De = equivalent diameter of the annular space, m
Re De = Reynolds number for water flow through the annular space
u w = mean water velocity in the annular space = 0.5 m/s
Q w = kinematic viscosity of water = 0.8012u10-6 m2/s
The equivalent diameter is defined as
4A
P
De
(d)
where
A = flow area, m2
P = wet perimeter, m
The flow area and wet perimeter for the annular space are given by
A = S ( Do2 Di2 ) / 4
(e)
P S ( Di D o )
(f)
and
where
Do = diameter of outer tube = 6 cm = 0.06 m
Substituting (e) and (g) into (d)
De
D o Di
(g)
Using (h), equation (d) becomes
Re De =
0.5(m / s)(0.06 0.025)(m)
0.8012 u 10 6 (m 2 / s)
= 21,842
Therefore, the flow in the annular space is turbulent. The Nusselt number for turbulent flow
through channels is given by the nGielinski equation (8.17a)
Nu De
ho De
kw
( f / 8)( Re D e 1000) Prw
>1 12.7( f / 8)
1/ 2
( Prw2/3
>1 ( D
1)@
e
/ L) 2 / 3
@
(h)
Valid for
valid for 0< De/L <1
developing or fully developed turbulent flow through channels
2300 < ReDe < 5 u 106
0.5 <Pr < 2000
properties at Tm
where
f = friction factor
kw = thermal conductivity of water = 0.615 W/m-oC
Prw = Prandtl number of water = 5.42
(i)
PR0BLEM 8.24 (continued)
Nu De = Nusselt number in turbulent flow
For a smooth tube f is given by
f = (0.79lnReDe 1.64)–2
(j)
For fully developed flow, set De/L = 0 in (h)
Nu D e
hDe
kw
( f / 8)( Re D e 1000) Pr
>1 12.7( f / 8)
1/ 2
( Pr 2 / 3 1)
@
(k)
where the average and local heat transfer coefficients, ho and ho are identical in the fully
developed region.
(iii) Computations. Equation (b) gives hi
hi = 3.66 (0.03718)(W/m-oC)/0.025(m) = 5.44 W/m2-oC
Equation (k) gives f
f = (0.79 ln 21,842 1.64 )-2 = 0.02557
Substituting into (k)
Nu De
ho D e
kw
(0.02557 / 8)(21,842 1000)5.42
1 12.7 0.02557 / 8
1/ 2
(5.42 2 / 3 1)
= 144.6
and
ho = 144.6 (0.615)(W/m-oC)/(0.06-0.025)(m) = 2541 W/m2-oC
(iv) Checking. Dimensional check: Computations showed that the units of equations (a)- (c)
and (k) are dimensionally correct.
Quantitative check: The values hi for air and ho for water are within the approximate range
shown in Table 1.1 for forced convection.
(5) Comments. Resistance to heat transfer is inversely proportional to the heat transfer
coefficient. Thus, the total resistance in this example is dominated by air side convection
resistance. The contribution to the total resistance of the much larger water side heat transfer
coefficient is insignificant and therefore can be neglected.
PROBLEM 8.25
A heat exchanger consists of a tube and square duct. The tube is
placed co-axially inside the duct. Hot water flows through the tube
while cold water passes through the duct. The inside and outside
diameters are 5 cm and 5.2 cm, respectively. The side of the duct is
10 cm. At a section far away from the inlet the mean hot water
temperature is 90oC and the mean cold water temperature is 30oC.
The mean hot water velocity is 1.32 m/s and the mean cold water
velocity is 0.077 m/s. Determine the inside and outside heat transfer
coefficients.
Do
Di
hot
water
S
cold water
(1) Observations. (i) This is an internal forced convection problem. (ii) The geometry consists
of a tube concentrically placed inside a square duct,. (iii) Water flows in the tube and the duct.
(iv) The Reynolds number should be computed for the two fluids to establish if the flow is
laminar or turbulent. (v) Far away from the inlet the velocity and temperature may be assumed
fully developed.
(2) Problem Definition. Determine the heat transfer coefficient for the flow inside the tube and
for the flow in the duct.
(3) Solution Plan. Compute the Reynolds number for the flow in the tube and in the duct to
establish if the flow is laminar or turbulent. Use analytic solutions or correlation equations to
determine the heat transfer coefficients.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4)
axisymmetric flow, (5) uniform surface temperature, (6) fully developed velocity and
temperature and (7) smooth channels.
(ii) Analysis. To determine hi and ho , the two Reynolds numbers are calculated first to establish
if the flow is laminar or turbulent. For water flow inside the tube the Reynolds number is
ReDi
ui Di
Qi
(a)
where
Di = inside tube diameter = 5 cm = 0.05 m
Re Di = Reynolds number for flow in tube
ui = mean water velocity in tube = 1.32 m/s
Qi = kinematic viscosity of water in tube = 0.3264 u 10 6 m 2 /s
Substituting into (a)
Re Di =
1.32(m/s)0.05(m)
0.3264 u 10 6 (m 2 /s)
2.022 u 105
Thus the flow is turbulent in the tube. The Nusselt number for fully developed turbulent flow is
given by (set D/L = 0 in eq. 8.17a)
PROBLEM 8.25 (continued)
Nu Di
hi Di
ki
( f / 8)( ReD i 1000) Pri
>1 12.7( f / 8)
1/ 2
( Pri2/3 1)
@
(b)
Valid for
fully developed turbulent flow through channels
0 Di / L 1
2300 < < 5 u 106
2300 ReDi 5 u 105
0.5 < Pri < 2000
properties at Tmi
prop
where
f = friction factor
ki = conductivity of water in tube = 0.6727 W/m o C
Pri = Prandtl number of water in tube = 1.97
Tmi = mean temperature of water in tube = 90 o C
For a smooth tube f is given by
f = (0.79ln ReDi 1.64)–2
(c)
The Reynolds number for fluid flow in the annular duct space is defined as
ReDe
uo De
Qo
(d)
where
De = equivalent diameter of the duct annular space, m
Re De = Reynolds number for water flow through the duct
uo = mean water velocity through the duct = 0.077 m/s
Q o = kinematic viscosity of water in duct = 0.8012 u 10 6 m 2 /s
The equivalent diameter is defined as
De
4A
P
(e)
where
A = flow area, m2
P = wet perimeter, m
The flow area and wet perimeter for the annular duct space are given by
A = S 2 S Do2 / 4
(f)
P SDo 4S
(g)
and
where
S
10 cm = 0.1 m
Substituting (f) and (g) into (e)
PROBLEM 8.25 (continued)
4S 2 S Do2
De
S Do 4S
(h)
Using (h)
4(0.1) 2 (m 2 ) S (0.052) 2 (m 2 )
S (0.052)(m) 4(0.1)(m)
De
Re De =
0.05592 m
0.077(m/s)0.05592(m)
= 5,374
0.8012 u 10 6 (m 2 /s)
Therefore, the flow in the duct space is turbulent. The Nusselt number for fully developed
turbulent flow through channels is given by equation (b) with Di replaced by De
( f / 8)( ReD e 1000) Pro
ho De
ko
Nu De
>1 12.7( f / 8)
1/ 2
( Pro2/3 1)
@
properties at Tmo
where
ko = conductivity of water in duct = 0.615 W/m o C
Pro = Prandtl number of water in duct = 5.42
Tmo = mean temperature of water in duct = 30 o C
(iii) Computations. For the flow through the tube, equation (c) gives f
f
>0.79 ln 2.022 u10
5
1.64
@
2
0.01558
Substituting into (b)
hi Di
ki
Nu Di
(0.01558 / 8)(2.022 u 105 1000)1.97
= 584.6
1 12.7 0.01558 / 8 1 / 2 (1.97 2 / 3 1)
Thus
hi
584.6
ki
Di
584.6
0.6727( W/m o C)
0.05(m)
7865 W/m 2 o C
For the flow in the annular duct space equation (c) gives f
f
>0.79 ln 5374 1.64@2
0.03777
Applying (b) to the annular duct space and setting Di
Nu De
ho Do
ko
De
(0.03777 / 8)(5374 1000)5.42
1 12.7 0.03777 / 8
1/ 2
(5.42 2 / 3 1)
39.74
Thus
ho
39.74
ko
De
39.74
0.615( W/m o C)
0.05592(m)
437.1 W/m 2 o C
(i)
PROBLEM 8.25 (continued)
(iv) Checking. Dimensional check: Computations showed that the units of equations (a), (b),
and (h) are dimensionally correct.
Quantitative check: The values hi for and ho for water are within the range shown in Table 1.1 for
forced convection of liquids.
(5) Comments. Since convection resistance to heat transfer is inversely proportional to the heat
transfer coefficient, The total resistance in this example is dominated by the hot water convection
resistance in the tube.
PROBLEM 8.26
In designing an air conditioning system for a pizza restaurant an estimate of the heat added to
the kitchen from the door of the pizza oven is needed. The rectangular door is 50 cm u 120 cm
with its short side along the vertical direction. Door surface temperature is 110oC. Ambient air
and surroundings temperatures are 20oC and 24oC, respectively. Door surface emissivity is 0.08.
Estimate the heat loss from the door.
(1) Observations. (i) Heat is lost from the door to the surroundings by free convection and
radiation. (ii) To determine the rate of heat loss, the door can by modeled as a vertical plate
losing heat by free convection to an ambient air and by radiation to a large surroundings. (iii)
Newton’s law of cooling gives the rate of heat transfer by convection and Stefan-Boltzmann
relation gives the heat loss by radiation.
(2) Problem Definition. Determine the average heat transfer coefficient for free convection from
a vertical plate.
(3) Solution Plan. Apply Newton's law of cooling to the door. Use correlation equations to
determine the average heat transfer coefficient. Apply Stefan-Boltzmann relation, equation
(1.12), to determine the heat transfer by radiation
(4) Plan Execution.
(i) Assumptions. (1) steady state, (2) two-dimensional, (3) constant properties (except in
buoyancy), (4) uniform surface temperature, (5) quiescent ambient fluid, (6) door is in the closed
position at all times, (7) surroundings is at uniform temperature and (8) the door is small
compared to the surroundings.
(ii) Analysis. Total heat transfer q is given by
q
qc q r
(a)
where
qc
qr
heat transfer by convection, W
heat transfer by radiation, W
Application of Newton's law of cooling to the surface gives
qc = h A ( Ts - Tf)
(b)
where
A = surface area, m2
h = average heat transfer coefficient, W/m2-oC
Ts = surface temperature = 110oC = 283.15 K
Tf = ambient air temperature = 20oC
A = LW
L = door height = 50 cm = 0.5 m
Tf
g
Ts
W
Surface area is given by
where
L
(c)
PROBLEM 8.26 (continued)
W = door width = 120 cm = 1.2 m
The average heat transfer coefficient h is determined from correlation equations for free
convection over a vertical plate. The Rayleigh number RaL is calculated first to determine the
appropriate correlation equation for h . The Rayleigh number is defined as
RaL =
E g Ts Tf L3
Q2
(d)
Pr
where
g = gravitational acceleration = 9.81 m/s2
L = door side in the direction of gravity = 50 cm = 0.5 m
Pr = Prandtl number
RaL = Rayleigh number
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
Properties of air are determined at the film temperature Tf defined as
Ts Tf
(110 20)( o C)
=
= 65oC
2
2
Air properties at this temperature are
Tf =
k = 0.02887 W/m-oC
Pr = 0.7075
Q = 19.4u10-6 m2/s
For ideal gases, the coefficient of thermal expansion E is given by
E=
1
T f (K )
1
0.002957 (1/K)
o
65( C) 273.15
Substituting into (d) gives
RaL =
0.002957(1/ o C)9.81(m /s 2 )(110 20)( o C)(0.5) 3 (m 3 )
(19.4 u 10 6 ) 2 (m 4 /s 2 )
9
0.7075 = 0.61349u10
Since RaL < 109, the flow is laminar. Thus, the appropriate equation for the average Nusselt
number is given by (8.25b)
Nu L
hL
k
Pr
ª
º
« 2.435 4.884 Pr 1 / 2 4.953Pr »
¬
¼
1/ 4
Ra L
1/ 4
(e)
Valid for
vertical plate
constant surface temperature Ts
laminar free convection
104 < RaL < 109
0 < Pr < f
properties at Tf
where
(f)
PROBLEM 8.26 (continued)
N u L = average Nusselt number
Radiation heat transfer is given by equation (1.12)
qr
4
)
H V A(Ts4 Tsur
(g)
where
Tsur surroundings temperature = 24 o C = 24 + 273.15 = 297.15 K
H surface emissivity = 0.08
V Stefan-Boltzmann constant = 5.67 u 10 8 W/m 2 K 4
(iii) Computations. Substitution into (e) gives
Nu L
hL
k
ª
º
0.7075
«
»
1/ 2
4.953(0.7075) ¼»
¬« 2.435 4.884(0.7075)
1/ 4
(0.61349 u 10 9 )1 / 4 = 81.07
Solving the above for h
k
= 81.07 (0.02887)(W/m-oC)/0.5(m) = 4.68 W/m2-oC
L
Equation (b) gives
h = 84.02
qc = 4.68 (W/m2-oC) (0.5)(m)(1.2)(m) (110 20) (oC) = 252.7 W
Equation (g) gives
qr
>
(0.08) 5.67 u 10 8 ( W/m 2 K 4 ) (0.5)(m)(1.2)(m) (383.15) 4 (K 4 ) (297.15) 4 (K 4 )
@
37.4 W
Substituting into (a)
q = 252.7 + 37.4 =290.1 W
(iii) Checking. Dimensional check: Computations showed that equations (b), (d), (e) and
(g) are dimensionally consistent.
Quantitative check: The magnitude of h is within the range given in Table 1.1.
Validity of correlation equation (d): The conditions listed in (f) are met.
(5) Comments. (i) Equation (8.26a) applies to both laminar and turbulent flow. However, it is
less accurate than (8.25b) or (e) above. Equation (8.26a) gives h = 6.37 W/m2-oC and q = 344.2
W. This is 31% higher than that found using the laminar free convection equation (e). (ii)
Although radiation accounts for only 13% of the total heat added to the room, the contribution of
radiation is minimized due to the low surface emissivity of the door. (iii) Opening and closing
the door results in transient effects not accounted for in the above model. In addition, when the
door is open radiation from the interior of oven may be significant.
PROBLEM 8.27
To compare the rate of heat transfer by radiation with that by free convection, consider the
following test case. A vertical plate measuring 12 cm u 12 cm is maintained at a uniform
surface temperature of 125oC. The ambient air and the surroundings are at 25oC. Compare the
two modes of heat transfer for surface emissivities of 0.2 and 0.9.
(1) Observations. (i) This is a free convection and radiation problem. (ii) The geometry is a
vertical plate. (iii) Surface temperature is uniform. (iv) Newton’s law of cooling gives
convection heat transfer rate while Stefan-Boltzmann law gives radiation heat transfer rate. (v)
The Rayleigh number should be computed to determine if the flow is laminar or turbulent. (vi)
Since radiation heat transfer is considered in this problem, all temperatures should be expressed
in degrees kelvin. (vii) The fluid is air.
(2) Problem Definition. Determine heat transfer rate by free convection and by radiation from
a vertical plate in air.
(3) Solution Plan. Apply Newton's law of cooling to determine the rate of heat loss by
convection. Apply Stefan-Boltzmann radiation law to determine the rate of heat loss by
radiation. Compute the Rayleigh number and select an appropriate correlation equations to
obtain the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in
buoyancy), (4) uniform surface temperature, (5) the plate is a small surface enclosed by a much
larger surface at a uniform temperature and (6) quiescent ambient.
(ii) Analysis. Application of Newton's law of cooling to the vertical plate gives
qc = h A (Ts - Tf)
(a)
where
A = surface area of vertical side, m2
h = average heat transfer coefficient, W/m2-oC or W/m2-K
g
qc = convection heat transfer rate, W
o
T
f
Ts = surface temperature = 125( C) + 273.15 = 398.15 K
Tf = ambient temperature = 25(oC) + 273.13 = 298.15 K
surface area is
2
A=L
(b)
where
L
Ts
L
qc
qr
g
din
un
o
r
sur
sa
t T su
L = side of square plate = 12 cm = 0.12 m
The average heat transfer coefficient h is determined from correlation equations for free
convection over vertical plates. The Rayleigh number RaL is calculated first to determine the
appropriate correlation equation for h . The Rayleigh number is defined as
RaL =
where
E g Ts Tf L3
Pr
Q2
(c)
r
PROBLEM 8.27 (continued)
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number
RaL = Rayleigh number
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
Properties of air are determined at the film temperature Tf defined as
Tf =
(29815
. 39815
. )( K )
Ts Tf
= 348.15K
=
2
2
Appendix C gives air properties at Tf = 348.15(K) - 273.15 = 75oC
k = 0.02957 W/m-oC
Pr = 0.7065
Q = 20.41u10-6 m2/s
For ideal gases the coefficient of thermal expansion E is given by
E=
1
T f (K )
1
348.15
0.002872 (1/K)
Substituting into (c) gives
RaL =
0.002872(1/ o C)9.81(m/ s 2 )(125 25)( o C)(0.12) 3 (m 3 )
(20.41 u 10 6 ) 2 (m 4 / s 2 )
0.7065 = 8.257 u 10 6
Since RaL < 109, the flow is laminar. Thus the appropriate equation for the average Nusselt
number is given by equation (8.25b)
Nu L
hL
k
Pr
ª
º
« 2.435 4.884 Pr 1 / 2 4.953Pr »
¬
¼
1/ 4
Ra L
1/ 4
(d)
Valid for
vertical plate
constant surface temperature Ts
laminar free convection
104 < RaL < 109
0 < Pr < f
properties at Tf
(e)
where
NuL = average Nusselt number
Radiation heat loss qr is given by the Stefan-Boltzmann law. Assuming that the plate is a small
surface which is surrounded by a much larger surface, qr is given by equation (1.15)
4
qr = H V A ( Ts4 Tsur
)
where
qr = radiation heat loss, W
Tsur = surroundings temperature = 25(oC) + 273.13 = 298.15 K
(f)
PROBLEM 8.27 (continued)
H = emissivity
V = Stefan-Boltzmann constant = 5.67u10-8 W/m2-K4
(iii) Computations. Convection heat loss. Substitution into (d) gives
Nu L
hL
k
ª
º
0.7065
«
»
1/ 2
¬ 2.435 4.884(0.7065) 4.953(0.7065) ¼
1/ 4
(8.257 u 10 6 )1 / 4 = 27.6
Solving the above for h
h = 27.6
k
= 27.6(0.02957)(W/m-oC) / 0.12(m) = 6.8 W/m2-oC
L
Substituting into (a) and using (b) gives the rate of heat loss by convection
qc = 6.8 (W/m2-oC) (0.12)(m)(0.12)(m) (125 -25)(oC) = 9.79 W
Radiation heat loss. Equation (f) is used to determine qr. For H = 0.2:
qr = 0.2u5.67u10-8 (W/m2-K4) 0.12(m) 0.12(m) ( 39815
. 4 29815
. 4 )(K4) = 2.81 W
For H = 0.9:
qr = 0.9u5.67u10-8 (W/m2-K4) 0.12(m) 0.12(m) ( 39815
. 4 29815
. 4 )(K4) = 12.66 W
(iv) Checking. Dimensional check: Computations showed that equations (a)-(d) and (f) are
dimensionally consistent.
Quantitative check: The magnitude of h is within the range given in Table 1.1 for free
convection of gases.
Limiting check: For Ts = Tf = Tsur, both convection and radiation vanish. Setting Ts = Tf = Tsur
in (a) and (f) gives qc = qr = 0.
Limitations on correlation equation (d): The conditions listed in (e) are met.
(5) Comments. (i) When compared with free convection, radiation heat loss can be significant
and in general should not be neglected. (ii) The magnitude of E is the same whether it is
expressed in units of degree Celsius or kelvin. The reason is that E is measured in terms of
degree change. One degree change on the Celsius scale is equal to one degree change on the
kelvin scale. This is also true of units of heat transfer coefficient and specific heat. (iii) Because
temperature in the Stefan-Boltzmann radiation law must be expressed in degrees kelvin, care
should be exercised in using the correct units when carrying radiation computations.
PROBLEM 8.28
A sealed electronic package is designed to be cooled by free convection.
The package consists of components which are mounted on the inside
surfaces of two cover plates measuring 10 cm u 10 cm each. Because the air
plates are made of high conductivity material, surface temperature may
be assumed uniform. The maximum allowable surface temperature is Tf
70oC. Determine the maximum power that can be dissipated in the g
package without violating design constraints. Ambient air temperature is
20oC. Neglect radiation.
components
(1) Observations. (i) This is a free convection problem. (ii) The power
dissipated in the electronic package is transferred to the ambient fluid by free convection.
(iii) As the power is increased, surface temperature increases. (iv) The maximum power
dissipated corresponds to the maximum allowable surface temperature. (v) Surface temperature
is related to surface heat transfer by Newton’s law of cooling. (vi) The problem can be modeled
as free convection over a vertical plate. (vii) The Rayleigh number should be computed to
determine if the flow is laminar or turbulent. (viii) The fluid is air.
(2) Problem Definition. Since dissipated power is related to heat loss from the surface to the
surroundings, Newton's law of cooling should be applied. Thus, the problem becomes finding
the average heat transfer coefficient.
(3) Solution Plan. Apply Newton's law of cooling to the surface of the package. Check the
Rayleigh number to determine if the flow is laminar or turbulent. Use appropriate correlation
equations to determine the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in
buoyancy), (4) uniform surface temperature, (5) negligible heat loss from the top and bottom
surfaces, (6) the sides are vertical flat plates, (7) negligible radiation and (8) quiescent fluid.
(ii) Analysis. Application of Newton's law of cooling to the surface gives
P = q = h A (Ts - Tf)
(a)
where
A = surface area of the two vertical sides, m2
h = average heat transfer coefficient, W/m2-oC
P = power dissipated in package, W
q = heat transfer from the surface to the ambient air, W
Ts = surface temperature = 70oC
Tf = ambient air temperature = 20oC
L
Tf
g
Ts
W
Surface area of the two vertical sides is given by
A = 2 LW
where
L = package height = 10 cm = 0.1 m
(b)
PROBLEM 8.28 (continued)
W = package width = 10 cm = 0.1 m
The average heat transfer coefficient h is determined from correlation equations for free
convection over vertical plates. The Rayleigh number RaL is calculated first to determine the
appropriate correlation equation for h . The Rayleigh number is defined as
RaL=
E g Ts Tf L3
Pr
Q2
(c)
where
g = gravitational acceleration = 9.81 m/s2
L = plate dimension in the direction of gravity = 10 cm = 0.1 m
Pr = Prandtl number
RaL = Rayleigh number
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
Properties of air are determined at the film temperature Tf defined as
Tf =
(70 20)( o C)
Ts Tf
= 45oC
=
2
2
Appendix C gives air properties at this temperature
k = 0.02746 W/m-oC
Pr = 0.7095
Q = 17.44u10-6 m2/s
For ideal gases the coefficient of thermal expansion E is given by
E=
1
T f (K )
1
45( o C) 273.15
Substituting into (c) gives
0.003143 (1/K)
0.003143(1/ o C)9.81(m / s 2 )(70 20)( o C)(0.1) 3 (m 3 )
0.7095 = 3.5962 u 10 6
(17.44 u 10 6 ) 2 (m 4 / s 2 )
Since RaL < 109, the flow is laminar. Thus, the appropriate equation for the average Nusselt
number is (8.25b)
RaL =
Nu L
hL
k
Pr
ª
º
« 2.435 4.884 Pr 1 / 2 4.953Pr »
¬
¼
1/ 4
Ra L
1/ 4
(d)
Valid for
vertical plate
constant surface temperature Ts
laminar free convection
104 < RaL < 109
0 < Pr < f
properties at Tf
where
(e)
PROBLEM 8.28 (continued)
Nu L = average Nusselt number
(iii) Computations. Substitution into (d) gives
N uL
hL
k
ª
º
0.7095
«
»
1/ 2
¬ 2.435 4.884(0.7095) 4.953(0.7095) ¼
1/ 4
3.5962 u 10 6
1/ 4
= 22.44
Solving the above for h
h = 22.44
k
= 22.44(0.02746)(W/m-oC)/0.1(m) = 6.16 W/m2-oC
L
Equations (a) and (b) give the maximum power dissipated from the two sides of the package
P = q = 6.16 (W/m2-oC)2(0.1)(m)(0.1)(m) (70 - 20)(oC) = 6.16W
(iv) Checking.
Dimensional check: Computations showed that equations (a)-(d) are
dimensionally consistent.
Quantitative check: The magnitude of h is within the range given in Table 1.1 for free
convection of gases.
Qualitative check: Increasing the allowable surface temperature Ts should increase the maximum
power P. According to equation (a), q is directly proportional to Ts. Furthermore, h increases
when Ts is increased.
Limiting check: The dissipated power should vanish (P = 0) if heat cannot be removed by free
convection (q = 0). Setting h = 0 in (a) gives P = q = 0.
Limitations on correlation equation (d): The conditions listed in (e) are met.
(5) Comments. (i) The model used to solve this problem is conservative due to both neglecting
radiation and heat loss from the top and bottom surfaces. (ii) The maximum power dissipated is
relatively small, indicating the limitation of free convection in air as a cooling mode for such
applications. (iii) The magnitude of E is the same whether it is expressed in units of degree
Celsius or kelvin. The reason is that E is measured in terms of degree change. One degree
change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of
units of heat transfer coefficient and specific heat.
PROBLEM 8.29
Assume that the electronic package of Problem 8.28 is to be used in an undersea application.
Determine the maximum power that can be dissipated if the ambient water temperature is 10oC.
(1) Observations. (i) This is a free convection problem. (ii) The power
dissipated in the electronic package is transferred to the ambient
water
fluid by free convection. (iii) As the power is increased, surface
temperature increases. (iv) The maximum power dissipated corresponds
Tf
to the maximum allowable surface temperature. (v) Surface temperature
g
is related to surface heat transfer by Newton’s law of cooling. (vi) The
problem can be modeled as free convection over a vertical plate. (vii)
The Rayleigh number should be computed to determine if the flow is
laminar or turbulent. (viii) The fluid is water.
components
(2) Problem Definition. Since dissipated power is related to heat loss from the surface to the
surroundings, Newton's law of cooling should be applied. Thus, the problem becomes finding
the average heat transfer coefficient.
(3) Solution Plan. Apply Newton's law of cooling to the surface of the package. Check the
Rayleigh number to determine if the flow is laminar or turbulent. Use appropriate correlation
equations to determine the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in
buoyancy), (4) uniform surface temperature, (5) negligible heat loss from the top and bottom
surfaces, (6) the sides are vertical flat plates, (7) negligible radiation and (8) quiescent fluid.
(ii) Analysis. Application of Newton's law of cooling to the surface gives
P = q = h A (Ts - Tf)
(a)
where
A = surface area of the two vertical sides, m2
h = average heat transfer coefficient, W/m2-oC
P = power dissipated in package, W
q = heat transfer from the surface to the water, W
Ts = surface temperature = 70oC
Tf = ambient water temperature = 10oC
L
g
Tf
Ts
W
Surface area of the two vertical sides is given by
A = 2 LW
where
L = package height = 10 cm = 0.1 m
W = package width = 10 cm = 0.1 m
(b)
PROBLEM 8.29 (continued)
The average heat transfer coefficient h is determined from correlation equations for free
convection over vertical plates. The Rayleigh number RaL is calculated first to determine the
appropriate correlation equation for h . The Rayleigh number is defined as
RaL=
E g Ts Tf L3
Pr
Q2
(c)
where
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number
RaL = Rayleigh number
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
Properties of air are determined at the film temperature Tf defined as
Tf =
Ts Tf
(70 10)( o C)
=
= 40oC
2
2
Appendix C gives air properties at this temperature
k = 0.6286 W/m-oC
Pr = 4.34
E = 0.000389 1/K
Q = 0.6582u10-6 m2/s
Substituting into (c) gives
0.000389(1/ o C)9.81(m / s 2 )(70 10)( o C)(0.1) 3 (m 3 )
4.34 = 2.294 u 10 9
(0.6582 u 10 6 ) 2 (m 4 / s 2 )
Since RaL > 109, the flow is turbulent. Thus, the appropriate equation for the average Nusselt
number is (8.26a)
RaL =
Nu L
hL
k
­°
0.387 Ra L 1 / 6
0.825
®
°̄
1 0.492 / Pr 9/16
>
@
½°
8/27 ¾
°¿
2
(d)
Valid for
vertical plate
uniform surface temperature Ts
laminar, transition, and turbulent
10 1 < Ra L <1012
0 < Pr < f
properties at Tf
where
Nu L = average Nusselt number
(e)
PROBLEM 8.29 (continued)
(iii) Computations. Substitution into (d) gives
2
Nu L
hL
k
­°
0.387[2.294 u 10 9 ]1 / 6 ½°
= 191.7
®0.825 8/27 ¾
°¿
°̄
1 0.492 / 4.34 9/16
>
@
Solving the above for h
h = 191.7
k
= 191.7(0.6286)(W/m-oC)/0.1(m) = 1205 W/m2-oC
L
Equations (a) and (b) give the maximum power dissipated from the two sides of the package
P = q = 1205(W/m2-oC)2(0.1)(m)(0.1)(m) (70 - 10)(oC) = 1446 W
(iv) Checking. Dimensional check: Computations showed that equations (a), (b), (c) and (d)
are dimensionally consistent.
Quantitative check: The magnitude of h is within the approximate range given in Table 1.1 for
free convection of liquids.
Qualitative check: Increasing the allowable surface temperature Ts should increase the maximum
power P. According to equation (a), q is directly proportional to Ts. Furthermore, h increases
when Ts is increased.
Limiting check: The dissipated power should vanish (P = 0) if heat cannot be removed by free
convection (q = 0). Setting h = 0 in (a) gives P = q = 0.
Limitations on correlation equation (d): The conditions listed in (e) are met.
(5) Comments. (i) The model used to solve this problem is conservative due to both neglecting
heat loss from the top and bottom surfaces. (ii) The maximum power dissipated is relatively
large, indicating the effectiveness of water as a free convection medium. (iii) The magnitude of E
is the same whether it is expressed in units of degree Celsius or kelvin. The reason is that E is
measured in terms of degree change. One degree change on the Celsius scale is equal to one
degree change on the kelvin scale. This is also true of units of heat transfer coefficient and
specific heat.
PROBLEM 8.30
A plate 20 cm high and 25 cm wide is placed vertically in water at 29.4oC. The plate is
maintained at 70.6oC. Determine the free convection heat transfer rate from each half.
(1) Observations. (i) This is a free convection problem. (ii) The
surface is maintained at uniform temperature. (iii) The heat transfer
coefficient decreases with distance from the leading edge of the plate.
(iv) The heat transfer rate from the lower half 1 is greater than that
from the upper half 2. (v) Total heat transfer from each half can be
determined using the average heat transfer coefficient. (vi) Heat
transfer from the upper half is equal to the heat transfer from the
entire plate minus heat transfer from the lower half.
Tf
g
w
2
x
1
H
0
(2) Problem Definition. Determine the heat transfer rate from each vertical rectangle.
(3) Solution Plan. Since the heat transfer coefficient changes with distance, the average heat
transfer coefficient should be used in Newton's law of cooling. The Rayleigh number should be
computed to determine if the flow is laminar or turbulent.
(4) Plan Execution.
(i) Assumptions. (1) steady state, (2) two-dimensional, (3) constant properties (except in
buoyancy), (4) uniform surface temperature, (5) negligible radiation and (6) quiescent fluid.
(ii) Analysis. Application of Newton's law to rectangle 1
q1
h1 (Ts Tf ) w H / 2
(a)
where
h1 = average heat transfer coefficient for rectangle 1, W/m 2 o C
H = plate height = 20 cm = 0.2 m
q1 = heat transfer rate from rectangle 1, W
Ts = surface temperature = 70.6 o C
Tf = ambient temperature = 29.4oC
w = plate width = 25 cm = 0.25 m
The heat transfer rate from rectangle 2 is given by
q2
q1 2 q1
(b)
where
q2 = heat transfer rate from rectangle 2, W
q1 2 = total heat transfer rate from rectangles 1 and 2, W
Application of Newton’s law of cooling to the entire plate gives q1 2
q1 2
h1 2 (Ts Tf ) w H
where
h1 2 = average heat transfer coefficient for rectangles 1 and 2, W/m 2 o C
(c)
PROBLEM 8.30 (continued)
To determine the average heat transfer coefficient for free convection over a vertical plate, the
Rayleigh number is computed to establish if the flow is laminar or turbulent. The Rayleigh
number for rectangle 1 is defined as
E g (Ts Tf )( H / 2)3
Ra1
Pr
(d)
2
Q
where
g = gravitational acceleration = 9.81 m/s 2
Pr = Prandtl number
Ra1 = Rayleigh number for rectangle 1
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
Properties are determined at the film temperature T f defined as
Tf
(Ts Tf ) / 2
(e)
Substituting into (e)
(70.6 29.4) / 2 50 o C
Tf
Properties of water at this temperature are
k = 0.6405 W/m o C
Pr = 3.57
E 0.462 u 10 3 1/K
Q
0.5537 u 10 6 m 2 /s
Substituting into (d)
Ra1
0.462 u 10 3 (1 / K )9.81(m/s 2 )(70.6 29.4)(o C)(0.2 / 2)3 (m 3 )
3.57
(0.5537 u 10 6 ) 2 (m 4 /s 2 )
2.174 u 109
Since the Rayleigh number is greater than 109 , it follows that the flow turbulent. Similarly, the
Rayleigh number for the entire plate, Ra1 2 is
Ra1 2
E g (Ts Tf ) H 3
Pr
Q2
(f)
Substituting into (f)
Ra1 2
0.462 u 10 3 (1 / K )9.81(m/s 2 )(70.6 29.4)( o C)(0.2)3 (m 3 )
3.57 17.3947 u 109
(0.5537 u 10 6 ) 2 (m 4 /s 2 )
Thus the flow is turbulent over the entire upper half of the plate. The applicable correlation
equation for the average Nusselt number for a vertical plate of height H/2 is
Nu H / 2
h1 ( H / 2)
k
1/ 6
­°
0.387 Ra1
®0.825 °̄
1 (0.492 / Pr )9 / 16
>
½°
8 / 27 ¾
°¿
@
2
(g)
PROBLEM 8.30 (continued)
Valid for
vertical plate
constant surface temperature Ts
laminar, transition, and turbulent
10 1 RaH / 2 1012
0 < Pr < f
properties at Tf
Similarly, the average Nusselt number for the entire plate is
h1 2 H
k
Nu H
1/ 6
½°
­°
0.387 Ra1 2
¾
®0.825 8
/
27
°¿
°̄
1 (0.492 / Pr )9 / 16
>
2
(h)
@
(iii) Computations. Substituting into (g)
h1 ( H / 2)
k
Nu H / 2
1/ 6
­°
½°
0.387 2.174 u 109
®0.825 8 / 27 ¾
°̄
°¿
1 (0.492 / 3.57)9 / 16
>
2
@
185.81
Solving for h1
h1 185.81
k
0.6405( W/m o C)
185.81
1190 W/m 2 o C
H /2
(0.2 / 2)(m)
Substituting into (a)
q1 1190( W/m 2 o C)(70.6 29.4) ( o C)0.25(m)(0.2)(m) / 2 = 1225.7 W
Similarly, (h) gives the average heat transfer coefficient for the entire plate
Nu H
1/ 6 ½
­°
0.387 17.3947 u 109
°
®0.825 ¾
9 / 16 8 / 27
°̄
°¿
1 (0.492 / 3.57)
h1 2 H
k
>
@
2
358.58
Solving for h1 2
h1 2
358.58
k
H
358.58
0.6405( W/m o C)
1148.3 W/m 2 o C
0.2(m)
Substituting into (c)
q1 2 1148.3( W/m 2 o C)(70.6 29.4)( o C)0.25(m)0.2(m) 2365.5 W
Substituting into (b) gives the heat transfer rate from rectangle 2
q2
2365.5( W ) 1225.7( W ) 1139.8 W
(iii) Checking. Dimensional check: Computations showed that units of equations (a)-(d), (g)
and (h) are dimensionally consistent.
Limiting check: If Ts
Tf , no free convection takes place and consequently q1
q1 2
0.
PROBLEM 8.30 (continued)
Setting Ts
Tf in (a) and (c) gives the anticipate result.
Qualitative check: The heat transfer rate from the leading rectangle 1 should be higher than that
from the trailing rectangle 2. This is in agreement with the results obtained.
(5) Comments. (i) o
Mre heat is transferred from rectangle 1 than rectangle 2. This was
predicted prior to solving the problem analytically. However, the difference between the two
heat transfer rates is not very significant. (ii) The average heat transfer coefficients are slightly
outside the range given in Table 1.1 for free convection in liquids. It should be remembered that
values listed in Table 1.1 are for typical applications. Exceptions should be expected.
PROBLEM 8.31
Consider laminar free convection from a vertical plate at uniform surface temperature. Two 45q
triangles are drawn on the plate as shown.
[a] Explain why free convection heat transfer from triangle 1 is greater than that from the
triangle 2.
[b] Determine the ratio of the heat transfer from two triangles.
(1) Observations. (i) This is a free convection problem. (ii) The
surface is maintained at uniform temperature. (iii) The heat transfer
coefficient decreases with distance from the leading edge of the plate.
(iv) The width of each triangle changes with distance from the
leading edge.
Tf
g
2
1
(2) Problem Definition. Examine the variation of local heat transfer
coefficient with distance and determine the heat transfer rate from each triangle.
(3) Solution Plan.
[a] Formulate an equation for h(x) for laminar free convection over a flat plate.
[b] Since the heat transfer coefficient and area change with distance, Newton's law of cooling
should be applied to an infinitesimal area of each triangle. Integration over the area gives the
total heat transfer rate.
(4) Plan Execution.
(i) Assumptions. (1) Laminar flow, (2) steady state, (3) two-dimensional, (4) constant
properties (except in buoyancy), (5) uniform surface temperature, (6) quiescent fluid and (7) no
radiation.
(ii) Analysis.
[a] The heat transfer coefficient for laminar free convection over a vertical plate is given by
equation (8.25a)
Nu x
hx
k
Pr
3ª
º
4 «¬ 2.435 4.884 Pr 1 / 2 4.953Pr »¼
1/ 4
Ra x
1/ 4
(a)
and
Rax =
E g (Ts Tf ) x 3
Pr
Q2
where
g = gravitational acceleration, m/s2
h = local heat transfer coefficient, W/m2-oC
k = thermal conductivity, W/m-oC
Pr = Prandtl number
Rax = local Rayleigh number
x = vertical distance from the leading edge of plate, m
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
Valid for
(b)
PROBLEM 8.31 (continued)
vertical plate
constant surface temperature Ts
laminar free convection
104 < Rax < 109
0 < Pr < f
properties at Tf
(c)
Substituting (b) into (a) and solving for h
h(x) = C x-1/4
(d)
where the constant C is defined as
C=
º
Pr 2 E g Ts Tf
3 ª
k«
1/ 2
2»
4 «¬ 2.435 4.884 Pr 4.953Pr Q »¼
1/ 4
(e)
According to (d), h(x) decreases with distance along the plate x.
It is infinite at x 0. Triangle 1 has its wide base starting at
x 0 while triangle 2 has its apex at x 0 . Since heat transfer
rate is proportional to the product of heat transfer coefficient and
area, it follows that triangle 1 transfers more heat than triangle 2.
b 2(x)
[b] Application of Newton's law to an infinitesimal area gives
dq = (Ts - Tf)h(x) dA
(f)
where
2
1
H
dx
x
Tf
g
)
b 1(x
dx
A = surface area, m2
q = heat transfer rate, W
Ts = surface temperature, oC
Tf = ambient temperature, oC
B
The area dA for each triangle is
dA1 = b1(x) dx
(g)
dA2 = b2 (x) dx
(h)
and
Similarity of triangles give b1(x) and b2(x)
b1(x) = B(1 -
x
)
H
(i)
and
b2(x) = B
x
H
(j)
x
)dx
H
(k)
where
B = base of triangle, m
H = height of triangle, m
Substituting (i) into (g) and (j) into (h)
dA1 = B(1 and
PROBLEM 8.31 (continued)
dA2 = B
x
dx
H
(m)
Applying (f) to triangle 1, substituting (d) and (k) into (f) and integrating from x = 0 to x = H
gives the total heat transfer from triangle 1
H
q1 =
³
H
dq1
BC Ts Tf
0
x 1 / 4 dx = (16/21) B C (Ts -Tf) H 3 / 4
³ 1 x / H
(n)
0
Similarly, applying (f) to triangle 2, substituting (d) and (m) into (f) and integrating from x = 0 to
x = H gives the total heat transfer, q2, from triangle 2
H
q2 =
³
dq 2
H
BC Ts Tf
0
³ x/H
x 1 / 4 dx = (4/7) B C (Ts - Tf ) H 3 / 4
(o)
0
Taking the ratio of (n) and (o)
q1
4
=
3
q2
(iii) Checking. Dimensional check: To check the units of (d), units of C in equation (e) is
determined first
2
o
2
o
§ W ·ª Pr E (1/ C)g(m / s )(Ts Tf )( C) º
C = k¨
¸
«
»
© m o C ¹«¬ (2.5 5 Pr 1 / 2 5 Pr) Q 2 (m 4 / s 2 ) »¼
1/ 4
= W/m7/4-oC
Thus units of h(x) in (d) are
h(x) = C (W/m7/4-oC) x-1/4(m)-1/4 = W/m2-oC
Examining units of q1 in (n)
q1 = B(m) C(W/m7/4-oC) (Ts -Tf) (oC) H
3/ 4
(m3/4) = W
Limiting check: If E = 0 or g = 0 or Ts = Tf, no free convection takes place and consequently q1
= q2 = 0. Any of these limiting cases give C = 0. Thus, according to (n) and (o), q1 = q2 = 0.
(5) Comments. (i) M
ore heat is transferred from triangle 1 than triangle 2. This was predicted
prior to solving the problem analytically. (ii) The result applies to any right angle triangles and
is not limited to 45o triangles. (iii) Heat transfer from a surface of fixed area depends on its
orientation relative to the leading edge. (iv) This problem illustrates how integration is used to
account for variations in element area and heat transfer coefficient. The same approach can be
applied if surface temperature and/or ambient temperature vary over a surface area.
PROBLEM 8.32
A vertical plate measuring 21 cm u 21 cm is at a uniform surface temperature of 80oC. The
ambient air temperature is 25oC. Determine the free convection heat flux at 1 cm, 10 cm and 20
cm from the lower edge.
(1) Observations. (i) This is a free convection problem over a vertical plate. (ii) The surface is
maintained at uniform temperature. (iii) Local heat flux is determined by Newton’s law of
cooling. (iv) Heat flux depends on the local heat transfer coefficient. (v) Free convection heat
transfer coefficient for a vertical plate decreases with distance from the leading edge. Thus, the
flux also decreases. (vi) The Rayleigh number should be computed to select an appropriate
Nusselt number correlation equation. (vii) The fluid is air.
(2) Problem Definition. Determine the local heat transfer coefficient for free convection over a
vertical plate at uniform surface temperature.
(3) Solution Plan. Apply Newton’s law of cooling, compute the Rayleigh number and select an
appropriate Nusselt number correlation equation.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in
buoyancy), (4) uniform surface temperature, (5) quiescent fluid and (6) negligible radiation.
(ii) Analysis. Application of Newton's law gives
q cc = h(x) (Ts - Tf)
where
h(x) = local heat transfer coefficient, W/m2-oC
q cc = local heat flux, W
Ts = surface temperature = 80oC
Tf = ambient temperature = 25oC
(a)
The local heat transfer coefficient is determined from correlation equations for free convection
over a vertical plate. The Rayleigh number Rax is calculated to select an appropriate correlation
equation for h. The Rayleigh number is defined as
Rax =
E g Ts Tf x 3
Pr
Q2
(b)
where
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number
Rax = local Rayleigh number
x = distance from leading edge, m
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
L
Tf
g
Properties of air are determined at the film temperature Tf defined as
Tf = (Ts Tf ) / 2 = (80 25)( o C) / 2 = 52.5oC
Appendix C gives air properties at this temperature
x
Ts
W
PROBLEM 8.32 (continued)
k = 0.02799 W/m-oC
Pr = 0.709
Q = 18.165u10-6 m2/s
For ideal gases the coefficient of thermal expansion E is given by
E=
1
T f (K )
1
0.003071 (1/K)
o
52.5( C) 273.15
Substituting into (b) to evaluate the Rayleigh number at the trailing end x = L = 20 cm = 0.2 m
RaL =
0.003071(1/ o C )9.81(m / s 2 )(80 25)( o C )(0.2) 3 (m 3 )
(18.165 u 10 6 ) 2 (m 4 / s 2 )
0.709 = 2.8482u107
Since RaL < 109, the flow is laminar over the region of interest. Thus, the appropriate equation
for the local Nusselt number Nu x is given by (8.25a)
Nu x
hx
k
Pr
3ª
º
«
4 ¬ 2.435 4.884 Pr 1 / 2 4.953Pr »¼
1/ 4
Ra x
1/ 4
(c)
Valid for
vertical plate
constant surface temperature Ts
laminar free convection
104 < Rax < 109
0 < Pr < f
properties at Tf
(d)
(iii) Computations. To evaluate the flux at x = L = 20 cm = 0.2 m, the heat transfer
coefficient at this location is computed using (c)
Nu L
hL
k
º
3ª
0.709
«
»
1/ 2
4 ¬ 2.435 4.884(0.709) 4.953(0.709) ¼
1/ 4
2.8482 u 10 7
1/ 4
= 28.2
Solving the above for h(L)
k
h(L)= 28.2
= 28.2(0.02799)(W/m-oC)/0.2(m) = 3.95 W/m2-oC
L
Substituting into (a)
q cc = 3.95(W/m2-oC)(80 25 )(oC) = 217.3 W/m2
The same procedure is followed to determine the flux at x = 1 cm and x = 10 cm. Results for the
three locations are tabulated.
x (cm)
Rax
Nux
1
10
20
0.356u104
0.356u107
2.8482u107
2.98
16.78
28.2
h(x)(W/m2-oC)
8.34
4.7
3.95
q cc (W/m2)
458.7
258.3
217.3
PROBLEM 8.32 (continued)
(iv) Checking. Dimensional check: Computations showed that equations (a), (b) and (c) are
dimensionally consistent.
Quantitative check: The magnitudes of h are approximately within the range given in Table 1.1
for free convection of gases.
Limiting check: The flux should vanish for Ts = Tf. Setting Ts = Tf in (a) gives q cc 0 .
Limitations on correlation equation (c): The conditions listed in (d) are met except at x = 1 cm
where the Rayleigh number is below the lower limit. The accuracy of the computed flux at the
location is in doubt.
Validity of correlation equation (c): The conditions listed in (d) are met.
(5) Comments. (i) This problem illustrates the importance of verifying the applicability of
correlation equations to specific cases. It should be noted that correlation equations usually do
not suddenly break down outside the limits of their applicability. Instead, their accuracy begins
to deteriorate. (ii) The heat transfer literature should be consulted for applicable correlation
equations whenever the need arises. (iii) The magnitude of E is the same whether it is expressed
in units of degree Celsius or kelvin. The reason is that E is measured in terms of degree change.
One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This
is also true of units of heat transfer coefficient and specific heat.
PROBLEM 8.33
200 square chips measuring 1 cm u 1 cm each are mounted on both sides of
a thin vertical board measuring 10 cm u 10 cm. The chips dissipate 0.035
W each. Assume uniform surface heat flux. Determine the maximum
surface temperature in air at 22oC. Neglect heat exchange by radiation.
Tf
g
(1) Observations. (i) This is a free convection problem over a vertical
plate. (ii) The power dissipated in the chips is transferred to the air by free
convection. (iii) This problem can be modeled as free convection over a
vertical plate with constant surface heat flux. (iv) Surface temperature
increases as the distance from the leading edge is increased. Thus, the maximum surface
temperature occurs at the top end of the plate (trailing end). (v) Newton’s law of cooling relates
surface temperature to heat flux and heat transfer coefficient. (vi) The fluid is air.
(2) Problem Definition. Determine surface temperature distribution for a vertical plate with
uniform surface heat flux under free convection conditions.
(3) Solution Plan. Apply the analysis of surface temperature distribution of a vertical plate with
uniform surface heat flux in free convection.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in
buoyancy), (4) uniform surface heat flux, (5) all dissipated power leaves surface as heat, (6)
negligible radiation and (7) quiescent ambient fluid.
(ii) Analysis. Surface temperature distribution for a vertical plate with uniform surface flux
is given by equation (8.29a)
Ts x Tf
ª 4 9 Pr 1/ 2 10 Pr § DQ · § q scc· 4 º
¨
¸ ¨ ¸ x»
«
Pr
© E g ¹ © k ¹ »¼
«¬
1/ 5
(a)
Valid for:
vertical plate
uniform surface flux q scc
laminar, 104 < RaL < 109
0 < Pr <f b where
g = gravitational acceleration = 9.81 m/s2
k = thermal conductivity, W/m-oC
Pr = Prandtl number
q csc = surface flux, W/m2
RaL = Rayleigh number at x = L
Ts = surface temperature, oC
Tf = ambient temperature = 22oC
x = distance from leading edge, m
D = thermal diffusivity, W/m2-s
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
PROBLEM 8.33 (continued)
The Rayleigh number is defined as
E g Ts Tf L3
Pr
Q2
RaL =
(c)
where
L = vertical side of plate = 10 cm = 0.1 m
If all dissipated power in the chip leaves the surface, conservation of energy gives
q csc = P/A
(d)
where
A = chip surface area = 1 cm2 = 0.0001 m2
P = power dissipated in chip = 0.035 W
Properties are evaluate at the film temperature defined as
Ts ( L / 2) Tf
Tf
(e)
2
where Ts(L/2) is surface temperature at the mid-point. However, since Ts(x)is unknown, an
iterative procedure is required to obtain a solution. An assumed value for Ts(L/2) is used to
calculate the film temperature at which properties are determined. Equation (a) is then used to
calculate Ts(L/2). If the calculated value does not agree with the assumed temperature, the
procedure is repeated until a satisfactory agreement is obtained.
(iii) Computations. Equation (d) gives surface flux
q csc = 0.035(W)/0.0001(m2) = 350 W/m2
Assume Ts (L/2) = 58oC. Equation (e) gives
Tf = (58 + 22)(oC)/2 = 40oC
Properties of air at this temperature are
cp = 1006.8 J/kg- oC
k = 0.0271 W/m-oC
Pr = 0.71
Q = 16.96u10-6 m2/s
3
U = 1.1273 kg/m
Coefficient of thermal expansion for an ideal gas is given by
E =
1
T f 273.15
1
o
40( C) 273.15
0.003193 1/K
Thermal diffusivity is defined as
D
k
Uc p
0.0271( W / m o C)
1.1273(kg / m 3 )1006.8(J / kg o C)
23.877 u 10 6 m 2 / s
Substituting into (a) and letting x = L/2 = 0.1(m)/2 = 0.05 m
1/ 5
Ts L/2
4
ª
º
4 9(0.71)1 / 2 10(0.71) §¨ 23.877 u 10 6 (m 2 / s)16.96 u 10 6 (m 2 / s) ·¸§¨ 350( W / m 2 ) ·¸
22( C) «
0.05(m) »
o
2
o
¨
¸
¨
¸
«
»
0.71
0.003193(1/ C)9.81(m / s )
©
¹© 0.0271( W / m C) ¹
¬
¼
o
PROBLEM 8.33 (continued)
o
Ts(L/2) = 76.3 C
Since this is higher than the assumed value of 58oC, the procedure is repeated with a new
assumed temperature at mid-point. Assume Ts(L/2) = 78oC. The following results are obtained
Tf = 50oC
cp = 1007.4 J/kg- oC
k = 0.02781 W/m-oC
Pr = 0.709
D = 25.27u10-6 m2/s
E = 0.0030945 1/K
Q = 17.92u10-6 m2/s
3
U = 1.0924 kg/m
Substituting into (a) gives Ts(L/2) = 76.8oC. This is close to the assumed value of 78oC.
Surface temperature at the trailing end is now computed by evaluating (a) at x = L = 0.1 m
1/ 5
Ts L/2
4
ª
º
·
4 9(0.709)1 / 2 10(0.709) §¨ 25.27 u 10 6 (m 2 / s)17.92 u 10 6 (m 2 / s) ·¸§¨
350( W / m 2 )
¸ 0.1(m)»
22( C) «
o
2
o
¨
¸
¨
¸
«
»
0.709
0.0030945(1/ C)9.81( m / s )
©
¹© 0.02781( W / m C) ¹
¬
¼
o
Ts(L) = 84.9oC
The condition on the Rayleigh number in equation (b) is verified next. Substituting into (c)
RaL =
0.0030945(1/ o C)9.81(m / s 2 )(84.9 22)( o C)(0.1) 3 (m 3 )
(17.92 u 10 6 ) 2 (m 4 / s 2 )
0.709 = 4.22u106
This satisfies the condition on RaL given in equation (b).
(iv) Checking. Dimensional check: Equations (a), (c) and (d) are dimensionally consistent..
Quantitative check: The heat transfer coefficient at the mid-point, h(L/2), can be calculated using
Newton's law of cooling:
h(L/2) = q csc /[Ts ( L / 2) Tf ] = 350(W/m2)/(76.8 - 22)(oC) = 6.39 W/m2-oC
This is within the range given in Table 1.1 for free convection of gases.
Validity of correlation equation (a): The conditions listed in (b) are met.
(5) Comments. (i) Surface temperature is determined without calculating the heat transfer
coefficient. This is possible because equation (a) combines the correlation equation for the heat
transfer coefficient and Newton's law of cooling to eliminate h and obtain an equation for surface
temperature in terms of surface heat flux. (ii) The magnitude of E is the same whether it is
expressed in units of degree Celsius or kelvin. The reason is that E is measured in terms of
degree change. One degree change on the Celsius scale is equal to one degree change on the
kelvin scale. This is also true of units of heat transfer coefficient and specific heat.
PROBLEM 8.34
An apparatus is designed to determine surface emissivity of
materials. The apparatus consists of an electrically heated
cylindrical sample (disk) of diameter D and thickness G . the
disk is insulated along its heated side and rim. It is placed
horizontally with its heated surface facing down in a large
chamber whose surface is maintained at uniform temperature
Tsur . The sample is cooled by free convection and radiation
from its upper surface. To determine the emissivity of a
sample, measurements are made of the diameter D, electric
power input P, surface temperature Ts , surroundings
temperature Tsur and ambient temperature Tf . Determine
the emissivity of a sample using the following data:
D = 12 cm, G
0.5 cm, P
13.2 W, Ts
98 o C , Tsur
Tsur
g
27 o C , Tf
Tf
D
Ts H
-
+
22 o C
(1) Observations. (i) Power supply to the disk is lost from the surface to the surroundings by
free convection and radiation. (ii) To determine the rate of heat loss, the disk can by modeled as
a horizontal plate losing heat by free convection to an ambient air and by radiation to a large
surroundings. (iii) Newton’s law of cooling gives the rate of heat transfer by convection and
Stefan-Boltzmann relation gives the heat loss by radiation. (iv) Free convection correlations give
the heat transfer coefficient. (v) Conservation of energy at the surface gives the emissivity, if it is
the only unknown.
(2) Problem Definition. Determine the average heat transfer coefficient for free convection from
a vertical plate.
(3) Solution Plan. Apply conservation of energy at the exposed surface. Use Newton's law of
cooling Stefan-Boltzmann relation, equation. Use correlation equations to determine the average
heat transfer coefficient for free convection from a horizontal surface.
(4) Plan Execution.
(i) Assumptions. (1) steady state, (2) constant properties (except in buoyancy), (3) uniform
surface temperature, (4) quiescent ambient fluid, (5) the heated surface and the rim are perfectly
insulated, (6) surroundings is at uniform temperature and (7) the disk is small compared to the
surroundings.
(ii) Analysis. Conservation of energy at the surface gives
P
qc qr
(a)
where
P = electric energy power supplied to the disk = 13.2 W
qc heat transfer by convection, W
q r heat transfer by radiation, W
Application of Newton's law of cooling to the surface gives
qc = h A ( Ts - Tf)
(b)
PROBLEM 8.34 (continued)
where
A = surface area, m2
h = average heat transfer coefficient, W/m2-oC
Ts = surface temperature = 98oC = 371.15 K
Tf = ambient air temperature = 22oC = 295.15 K
Surface area is given by
A
S
4
D2
(c)
where
D = disk diameter = 12 cm = 0.12 m
Radiation heat transfer is given by equation (1.12)
4
)
H V A(Ts4 Tsur
qr
(d)
where
Tsur surroundings temperature = 27 + 273.15 = 300.15 K
H surface emissivity
V
Stefan-Boltzmann constant = 5.67 u 10 8 W/m 2 K 4
(c) and (d) into (a)
4
P = h A ( Ts - Tf) + H V A(Ts4 Tsur
)
(e)
Solving (e) for H
H
( P / A) h (Ts Tf )
(f)
4
V (Ts4 Tsur
)
The average heat transfer coefficient h for a heated plate facing up is determined from (8.29)
Nu L
Nu L
hL
k
hL
k
0.54 Ra L 1 / 4 for 10 5 Ra L 2 u 10 7
(8.29a)
0.14 Ra L 1 / 3 for 2 u 10 7 Ra L 3 u 1010
(8.29b)
horizontal plate
hot surface up or cold surface down
properties, except E , at T f
(8.29c)
E at T f for liquids, Tf for gases
The Rayleigh number RaL is defined as
RaL =
E g Ts Tf L3
Q2
where
g = gravitational acceleration = 9.81 m/s2
L = characteristic length for the horizontal disk, m
Pr
(g)
PROBLEM 8.34 (continued)
Pr = Prandtl number
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
The length L is defined as
A
p
L
S D2 / 4
SD
D
4
(i)
Properties of air are determined at the film temperature Tf defined as
Tf =
Ts Tf
(98 22)( o C)
=
= 60oC
2
2
at this temperature air properties are
k = 0.02852 W/m-oC
Pr = 0.708
Q = 18.9u10-6 m2/s
For ideal gases, the coefficient of thermal expansion E for a horizontal surface is given by
1
Tf ( K )
E=
1
o
0.0033881 (1/K)
22( C) 273.15
Equation (i) gives L
L = 0.12 (m)/4 =0.03 m
Substituting into (g)
RaL =
0.0033881(1/ o C)9.81(m /s 2 )(98 22)( o C)(0.03) 3 (m 3 )
(18.9 u 10 6 ) 2 (m 4 /s 2 )
0.708 = 1.3518 u 10 5
Thus equation (8.29a) is applicable. Radiation heat transfer is given by equation (1.12)
qr
4
H V A(Ts4 Tsur
)
where
surroundings temperature = 27 o C = 27 + 273.15 = 300.15 K
Tsur
V
Stefan-Boltzmann constant = 5.67 u 10 8 W/m 2 K 4
(iii) Computations. Substitution into (8.29a) gives
Nu L
hL
k
0.54(1.3518 u 10 5 )1/ 4 = 10.35
Solving the above for h
k
= 10.35 (0.02852)(W/m-oC)/0.03(m) = 9.84 W/m2-oC
L
Surface area is given by (c)
h = 10.35
(j)
PROBLEM 8.34 (continued)
S
(0.12) 2 (m 2 ) = 0.01131 m 2
4
Equation (f) gives H
A
H
[13.2(W)/0.01131(m 2 ) 9.84(W/m 2 o C)(371.15 295.15)
5.67 u 10 8 [(371.15) 4 (300.15) 4 ](K 4 )
0.681
(iii) Checking. Dimensional check: Computations showed that equations (c), (f) and (g) are
dimensionally consistent.
Quantitative check: The magnitude of h is within the range given in Table 1.1.
Validity of correlation equation (8.29c): The conditions listed in (8.29c) are met.
(5) Comments. (i) Disk thickness G is not needed for the determination of H . (ii) It is important
to provide good insulation of the heated surface and the rim. (iii) According to equation (f),
increasing the operating surface temperature minimizes the error in the measurement of the
surroundings temperature.
PROBLEM 8.35
It is desired to increase the heat by free convection from a wide vertical plate without increasing
its surface temperature. Increasing the height of the plate is ruled out because of the limited
vertical space available. It is suggested that a taller plate can be accommodated in the same
vertical space by tilting it 45o. Explore this suggestion and make appropriate recommendations.
Assume laminar flow.
(1) Observations. (i) This is a free convection problem. (ii) The geometry is a flat plate. (iii)
Heat transfer from two plates is to be compared. One plate is vertical and the other is inclined.
Both plates fit in the same vertical space. Thus, the inclined plate is longer than the vertical
plate. (iv) Both plates are maintained at uniform surface temperature. (v) Heat transfer depends
on surface area and average heat transfer coefficient.
(2) Problem Definition. Determine the average heat transfer coefficient for a vertical plate and
an inclined plate under free convection conditions.
(3) Solution Plan. Apply Newton’s law of cooling and use appropriate correlation equations for
the average Nusselt number to determine the average heat transfer coefficient for each
orientation.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in
buoyancy), (4) uniform surface temperature, (5) negligible radiation and (6) quiescent fluid.
(ii) Analysis. Newton’s law of cooling gives
q = h A (Ts - Tf) = h LS(Ts - Tf)
(a)
where
A = surface area, m2
2 o
h = average heat transfer coefficient, W/m - C
L = plate length, m
q = heat transfer rate, W
S = plate width, m
Ts = surface temperature, oC
Tf = ambient temperature, oC
Applying (a) to the two plates and using the subscripts i and vȱto refer to the inclined and vertical
plates, respectively
qi = hi Li S (Ts - Tf)
(b)
qvȱ= hv Lv S (Ts - Tf)
(c)
hL
qi
= i i
qv
hv L v
(d)
Taking the ratio of (e) and (f)
The average Nusselt number for laminar free convection flow over a vertical plate of height Lv
is
PROBLEM 8.35 (continued)
hv Lv
k
Nu L v
Pr
ª
º
«¬ 2.435 4.884 Pr 1 / 2 4.953Pr »¼
1/ 4
( Ra Lv )1 / 4
(e)
where
E g (Ts Tf ) L3v
Ra Lv
Q2
(f)
Pr
where
g = gravitational acceleration, m/s2
hv = local heat transfer coefficient for vertical plate, W/m2-oC
k = thermal conductivity, W/m-oC
Pr = Prandtl number
Ra Lv = Rayleigh number for vertical plate
Lv = length of vertical plate, m
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
Valid for
vertical plate
constant surface temperature Ts
laminar free convection
104 < Rax < 109
0 < Pr < f
properties at Tf
(g)
Solving (a) and (b) for hv
hv
ªg º
C« »
¬ Lv ¼
1/ 4
(h)
where C is defined as
C
º
ª
Pr 2 E (Ts Tf )
k«
1/ 2
2»
¬« (2.435 4.884 Pr 4.953Pr )Q ¼»
1/ 4
(i)
Applying (h) to the inclined plate
hi
ª g cosT º
C«
»
¬ Li ¼
1/ 4
(j)
Substitute (h) and (j) into (d)
ªL º
qi
= « i »
qv
¬ Lv ¼
3/ 4
(cosT )1 / 4
(k)
However, the length of the inclined plate is given by
Li =
Combining (k) and (m)
Lv
cosT
(m)
PROBLEM 8.35 (continued)
qi
1
=
qv
(cosT )1 / 2
(n)
(iii) Computations. For a plate inclined at 45o equation (m) gives
qi
1
= 1.19
=
qv
(cos45 o ) 1 / 2
(iv) Checking. Dimensional check: Equation (n) is dimensionless.
Limiting check: For the special case of T = 0, the heat transfer from the two plates should be
identical. Setting T = 0 in (n) gives
qi
=1
qv
(5) Comments. (i) The inclined plate loses more heat than the vertical plate. The increase in
heat transfer rate is due to an increase in area rather than heat transfer coefficient. In fact if the
two plates have the same length, equation (k) shows that the inclined plate loses less heat than
the vertical plate. (ii) According to equation (n), the greater the angle of inclination, the more
heat will transfer from the plate. (iii) The above analysis assumes that conditions (g) are satisfied
by both plates. (iv) The magnitude of E is the same whether it is expressed in units of degree
Celsius or kelvin. The reason is that E is measured in terms of degree change. One degree
change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of
units of heat transfer coefficient and specific heat.
PROBLEM 8.36
Estimate the free convection heat transfer rate from five sides of a cubical
ceramic kiln. Surface temperature of each side is assumed uniform at
70oC and the ambient air temperature is 20oC. Each side measures 48
cm.
(1) Observations. (i) This is a free convection problem. (ii) The kiln has
four vertical sides and a horizontal top. (iii) All surfaces are at the same
uniform temperature. (iv) Newton’s law of cooling gives the heat transfer
rate. (v) The sides can be modeled as vertical plates and the top as a
horizontal plate. (vi) The fluid is air.
Tf
g
(2) Problem Definition. Determine the average heat transfer coefficient for a vertical plate and
for a horizontal plate.
(3) Solution Plan. Apply Newton’s law of cooling and use appropriate correlation equations for
the average Nusselt number to determine the average heat transfer coefficients.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties (except in buoyancy), (3) uniform
surface temperature, (4) quiescent fluid, (5) no heat loss from the bottom surface and (6)
negligible radiation.
(ii) Analysis. Heat transfer from the five sides is given by
q
4q s qt
(a)
where
q = heat transfer rate from five surfaces, W
qs = heat transfer rate from each of four sides, W
qt = heat transfer rate from the top, W
Application of Newton's law of cooling gives
qs
hs A(Ts Tf )
(b)
qt
ht A(Ts Tf )
(c)
and
where
A = surface area of one side of cube, m2
hs = average heat transfer coefficient for a vertical side, W/m2-oC
ht = average heat transfer coefficient for top surface, W/m2-oC
Ts = surface temperature = 70oC
Tf = ambient air temperature = 20oC
Surface area of each side is given by
A = L2
where
L = length of each square side = 48 cm = 0.48 m
(d)
PROBLEM 8.36 (continued)
The average heat transfer coefficient is determined from correlation equations for free
convection over a vertical plate. The Rayleigh number RaL is calculated first to select an
appropriate correlation equation for h . The Rayleigh number is defined as
RaL =
E g (Ts Tf ) L3
Pr
Q2
(e)
where
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number
RaL = Rayleigh number
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
Properties of air are determined at the film temperature Tf defined as
Ts Tf
(70 20)( o C)
= 45oC
=
2
2
At this temperature air properties are
Tf =
k = 0.02746 W/m-oC
Pr = 0.7095
Q = 17.44u10-6 m2/s
For ideal gases the coefficient of thermal expansion E for the sides is given by
Es =
1
T f (K )
1
o
45( C) 273.15
0.003143 (1/K)
Substituting into (e) gives
RaL =
0.003143(1/ o C)9.81(m / s 2 )(70 20)( o C)(0.48) 3 (m 3 )
(17.44 u 10 6 ) 2 (m 4 / s 2 )
0.7095 = 0.3977u109
Since RaL < 109, the flow is laminar. Thus, the appropriate equation for the average Nusselt
number for a vertical plate is given by (8.25b)
Nu L
hL
k
Pr
º
ª
« 2.435 4.884 Pr 1 / 2 4.953Pr »
¼
¬
1/ 4
Ra L
1/ 4
(f)
This equation is valid for
vertical plate
constant surface temperature Ts
laminar free convection
104 < RaL < 109
0 < Pr < f
properties at Tf
where
N u L = average Nusselt number
(g)
PROBLEM 8.36 (continued)
Correlation equations for a heated horizontal plate is given by equation (8.29)
Nu L
0.54 Ra L
Nu L
0.14 Ra L
1/ 4
1/ 3
for 10 5 Ra L 2 u 10 7
(8.29a)
for 2 u 10 7 Ra L 3 u 1010
(8.29b)
horizontal plate
hot surface up or cold surface down
properties, except E , at T f
(8.29c)
E at T f for liquids, Tf for gases
Thus the Rayleigh number for the top surface must be determined. It is based on the
characteristic length L defined in equation (8.31) as
Lt
A
p
(h)
where p is perimeter. Using (h)
Lt
L2
4L
L
4
0.48(m)
4
0.12 m
For a horizontal plat in air E t is evaluated at Tf
Et =
1
Tf ( K )
1
20( o C) 273.15
0.003411 (1/K)
(e) gives
RaL =
0.003411(1/ o C)9.81(m /s 2 )(70 20)( o C)(0.12) 3 (m 3 )
(17.44 u 10 6 ) 2 (m 4 /s 2 )
0.7095
6.744 u 10 6
Therefore, (8.29a) is applicable.
(iii) Computations. Substituting into (f) gives hs
Nu L
hs L
k
ª
º
0.7095
«
»
1/ 2
¬ 2.435 4.884(0.7095) 4.953(0.7095) ¼
1/ 4
0.3977 u 10 9
Solving the above for hs
k
hs = 72.77 = 72.77 (0.02746)(W/m-oC)/0.48(m) = 4.16 W/m2-oC
L
Equations (b) and (d) give the heat transfer rate from each side
q s = 4.16 (W/m2-oC) (0.48)2(m2)(70 20) (oC) = 47.9 W
Heat transfer coefficient for the top surface is computed from equation (8.29a)
1/ 4
= 72.77
PROBLEM 8.36 (continued)
Nu L
ht L
k
0.54(6.744 u 10 6 )1/ 4 = 27.52
Solving for ht
k
0.02746(W/m o C)
27.52
= 6.3 W/m2-oC
Lc
0.12(m)
Equations (c) and (d) give the heat transfer rate from the top surface
ht = 27.52
qt = 6.3(W/m2-oC) (0.48)2(m2)(70 20) (oC) = 72.6 W
Equation (a) gives the total heat loss from the five surfaces
q = 4(47.9)(W) + 72.6(W) = 264.2 W
(iv) Checking. Dimensional check: Computations showed that equations (a)-(f) and (h) are
dimensionally consistent.
Quantitative check: Heat transfer coefficients are within the approximate range of Table 1.1.
Validity of correlation equations (f) and (h): The conditions listed in (g) and (i) are met.
(5) Comments. (i) M
odeling each side as a free standing vertical plate is an approximation if
the kiln rests on a flat surface. (ii) Heat loss from the top surface is considerably more than form
a side surface. (iii) Depending on surface emissivity, radiation loss may be appreciable. (iv) The
magnitude of E is the same whether it is expressed in units of degree Celsius or kelvin. The
reason is that E is measured in terms of degree change. One degree change on the Celsius scale
is equal to one degree change on the kelvin scale. This is also true of units of heat transfer
coefficient and specific heat.
POBLEM 8.37
Determine the surface temperature of a single burner electric stove when its power supply is 70
W. The diameter of the burner is 18 cm and its emissivity is 0.32. The ambient air temperature is
30 o C and the surroundings temperature is 25 o C .
(1) Observations. (i) Heat transfer from the surface is by free convection and radiation. (ii)
The burner can be modeled as a horizontal disk with its heated side facing down. (iii) Newton’s
law of cooling gives heat transfer by convection and Stefan-Boltzmann relations gives heat
transfer by radiation. (iv) Both convection and radiation depend on surface temperature. (v) If the
burner is well insulated at the bottom heated surface and its rim, then the electric power supply is
equal to surface heat transfer.
(2) Problem Definition. Determine the average heat transfer coefficient for a for a horizontal
plate.
(3) Solution Plan. Apply Newton’s law of cooling and use appropriate correlation equations for
the average Nusselt number to determine the average heat transfer coefficients. Use StefanBoltzmann relation to determine heat loss by radiation.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties (except in buoyancy), (3) uniform
surface temperature, (4) quiescent fluid, (5) no heat loss from the bottom surface and rim, (6)
uniform surroundings temperature and (7) surface of burner is small compared to surroundings
area.
D
(ii) Analysis. Based on assumption (5), conservation of
energy at the surface gives
P
Tf
Ts
(a)
qc q r
q r qc
g
where
P = power supply to burner = 70 W
qc convection heat transfer rate, W
q r radiation heat transfer rate, W
+
-
Application of Newton's law of cooling gives
qc
h A(Ts Tf )
(b)
where
A = surface area of one side of cube, m2
h = average heat transfer coefficient, W/m2-oC
Ts = surface temperature, oC
Tf = ambient air temperature = 30oC
Surface area is given by
A=
where
D = diameter = 18 cm = 0.18 m
S
4
D2
(c)
PROBLEM 8.37 (continued)
The average heat transfer coefficient is determined from correlation equations for free
convection over a horizontal plate given by (8.29)
Nu L
Nu L
0.54 Ra L
0.14 Ra L
1/ 4
1/ 3
for 10 5 Ra L 2 u 10 7
for 2 u 10 7 Ra L 3 u 1010
horizontal plate
hot surface up or cold surface down
properties, except E , at T f
(8.29a)
(8.29b)
(8.29c)
E at T f for liquids, Tf for gases
The Rayleigh number RaL should be calculated to select an appropriate correlation equation from
the above. The Rayleigh number is defined as
RaL =
E g (Ts Tf ) L3
Pr
Q2
(d)
where
g = gravitational acceleration = 9.81 m/s2
L = characteristic length
Pr = Prandtl number
RaL = Rayleigh number
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
The characteristic length L is defined in equation (8.31) as
A
(e)
p
where p is perimeter. Examination of (d) shows that the Rayleigh number depends on surface
temperature Ts . However, Ts is unknown. This suggest that the an iterative procedure must be
used to solve the problem.
L
Radiation heat transfer is given by equation (1.12)
qr
4
)
H V A(Ts4 Tsur
(f)
where
Tsur surroundings temperature = 25 + 273.15 = 298.15 K
H surface emissivity = 0.3
V Stefan-Boltzmann constant = 5.67 u 10 8 W/m 2 K 4
(b) and (f) into (a)
4
P = h A ( Ts - Tf) + H V A(Ts4 Tsur
)
(g)
Equation (g) is used to determine Ts using an iterative procedure. A value for Ts is assumed,
properties of air are determined, (d) is used to compute the Rayleigh number and an equation is
PROBLEM 8.37 (continued)
selected for the free convection Nusselt number, and finally (g) is used to calculate the dissipated
power P. If the calculated P is not equal to 70 W, the procedure is repeated until a satisfactory
agreement is obtained.
(iii) Computations. Assume Ts
SD 2 / 4
SD
L
0.18(m)
4
D
4
190 o C . Using (e)
0.045 m
Properties of air are determined at the film temperature Tf defined as
Tf =
Ts Tf
(190 30)( o C)
=
= 110oC
2
2
At this temperature air properties are
k = 0.03194 W/m-oC
Pr = 0.704
Q = 24.1u10-6 m2/s
For a horizontal plate in air E is evaluated at Tf
E=
1
Tf ( K )
1
o
30( C) 273.15
0.0032987 (1/K)
(d) gives
RaL =
0.0032987(1/ o C)9.81(m /s 2 )(190 30)( o C)(0.045) 3 (m 3 )
(24.1 u 10 6 ) 2 (m 4 /s 2 )
0.704
5.7188 u 10 5
Therefore, (8.29a) is applicable.
Nu L
hL
k
0.54(5.7188 u 10 5 )1/ 4 = 14.85
Solving for h
h = 14.85
k
= 14.85(0.03194)(W/m-oC)/0.045(m) = 10.54 W/m2-oC
L
Surface area is given by
A SD 2 / 4 S (0.18m) 2 / 4
0.025447 m 2
Substituting into (g)
P 10.54(W/m 2 o C)0.025447(m 2 )(190 30)( o C)
0.3(5.67 u 10 8 )(W/m 2 K 4 )[(190 273.15) 4 (25 273.15) 4 ](K 4 )
P = 59.4 W
This calculated value is not close to the given power P
assume value Ts
70 W. The procedure is repeated for an
210 o C. The following result is obtained at this assume surface temperature:
PROBLEM 8.37 (continued)
Tf
120 o C
k = 0.03261 W/m-oC
Pr = 0.703
Q = 25.19u10-6 m2/s
E 0.0032987
Ra L
5.8806 u 10 5
hL
14.95
k
h =10.83 W/m2-oC
qc 49.64 W
q r 20.17
P = (49.64 + 20.17) = 69.8 W
Nu L
This is close to the given value of P = 70 W.
(iv) Checking. Dimensional check: Computations showed that equations (d,) (f) and (g) are
dimensionally consistent.
Quantitative check: Heat transfer coefficient is within the approximate range of Table 1.1.
Validity of correlation equations (8.29a): The conditions listed in (8.29c) are met.
(5) Comments. (i) Radiation loss is appreciable and thus can not be neglected. (ii) In practice
some of the electric power supplied to the burner is lost through the bottom side and rim. This
has the effect of lowering surface temperature. Thus the model used to solve the problem
overestimates surface temperature.
PROBLEM 8.38
A test apparatus is designed to determine surface emissivity of material. Samples are machined
into disks of diameter D. A sample disc is heated electrically on one side and allowed to cool off
on the opposite side. The heated side and rim are well insulated. The disk is first placed
horizontally with its exposed surface facing up in a large chamber. At steady state the exposed
surface temperature is measured. The procedure is repeated, without changing the power
supplied to the disk, with the exposed surface facing down. Ambient air temperature in the
chamber is recorded.
[a] Show that surface emissivity is given by
h1 (Ts1 Tf ) h2 (Ts 2 Tf )
Tsur
V (Ts42 Ts41 )
Ts
Tf
V
average heat transfer coefficient,
W/m 2 o C
surface temperature,= K
ambient temperature, K
Stefan-Boltzmann constant, W/m 2 K 4
Tf
Ts H
g
+
h
-
D
where subscripts 1 and 2 refer to the exposed
surface facing up and down, respectively, and
+
H
H Ts Tf
-
D
1
2
[b] Calculate the emissivity for the following case:
D
12 cm, Ts1
260 o C
533.15 K , Ts 2
300 o C
573.15 K, Tf
20 o C
293.15 K
(1) Observations. (i) Heat transfer from the surface is by free convection and radiation. (ii)
The sample can be modeled as a horizontal disk with its heated side facing down or up. (iii)
Newton’s law of cooling gives heat transfer by convection and Stefan-Boltzmann relation gives
heat transfer by radiation. (iv) Radiation depends on surface emissivity. (v) If the disk is well
insulated at the heated surface and its rim, then the electric power supply is equal to surface heat
transfer. (vi) Since the electric power is the same for both orientations, it follows that surface
heat transfer rate is also the same. (vii) Each orientation has its own Nusselt number correlation
equation.
(2) Problem Definition. Determine surface heat transfer by convection and radiation for two
horizontal orientations of a plate: heated side facing up and heated side facing down.
(3) Solution Plan. Apply Newton’s law of cooling and use appropriate correlation equations for
the average Nusselt number to determine the average heat transfer coefficients. Use StefanBoltzmann relation to determine heat loss by radiation. Equate total surface heat transfer rate for
both orientations.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties (except in buoyancy), (3) uniform
surface temperature, (4) quiescent fluid, (5) no heat loss from the heated power supply surface
and rim, (6) uniform surroundings temperature and (7) surface of disk is small compared to
surroundings area.
PROBLEM 8.38 (continued)
(ii) Analysis. Based on assumption (5), conservation of energy at the surface gives
qc q r
(a)
h A(Ts Tf )
(b)
P
where
P
qc
qr
power supply to disk, W
convection heat transfer rate, W
radiation heat transfer rate, W
Application of Newton's law of cooling gives
qc
where
A = surface area, m2
h = average heat transfer coefficient, W/m2-oC
Ts = surface temperature, oC
Tf = ambient air temperature = 20oC
Radiation heat transfer is given by equation (1.12)
qr
4
)
H V A(Ts4 Tsur
(c)
where
Tsur surroundings temperature
H surface emissivity
V
Stefan-Boltzmann constant = 5.67 u 10 8 W/m 2 K 4
(b) and (c) into (a)
4
P = h A ( Ts - Tf) + H V A(Ts4 Tsur
)
(d)
This result is applied to orientation 1 (heated surface facing up) and orientation 2 (heated surface
facing down), noting that P is the same for both
4
)
P = h1 A ( Ts1 - Tf) + H V A(Ts41 Tsur
(e)
4
)
P = h2 A ( Ts 2 - Tf) + H V A(Ts42 Tsur
(f)
Equating (e) and (f) and solving for H gives
H
h1 (Ts1 Tf ) h2 (Ts 2 Tf )
V (Ts42 Ts41 )
(g)
[b] The average heat transfer coefficient is determined from correlation equations for free
convection over a horizontal plate. For a plate with the heated side facing up (orientation 1), use
(8.29)
N u L1
N u L1
0.54 Ra L1 1 / 4 for 10 5 Ra L1 2 u 10 7
(8.29a)
0.14 Ra L1 1 / 3 for 2 u 10 7 Ra L1 3 u 1010
(8.29b)
PROBLEM 8.38(continued)
horizontal plate
hot surface up or cold surface down
properties, except E , at T f
(8.29c)
E at T f for liquids, Tf for gases
The Rayleigh Ra L1 number is defined as
Ra L1
E g (Ts1 Tf ) L3
ǎ 12
Pr1
(h)
where
g = gravitational acceleration = 9.81 m/s2
L = characteristic length
Pr1 = Prandtl number
RaL1 = Rayleigh number
E = coefficient of thermal expansion, 1/K
v 1 = kinematic viscosity, m2/s
For the average heat transfer coefficient for horizontal plate with the heated side facing down
(orientation 2), use (8.30)
Nu L2
0.27 Ra L 2 1 / 4 for 3 u 10 5 Ra L 2 3 u 1010
(8.30a)
horizontal plate
hot surface down or cold surface up
properties, except E , at T f
(8.30b)
E at T f for liquids, Tf for gases
The Rayleigh number Ra L 2 is given by
Ra L 2
E g (Ts 2 Tf ) L3
v 22
Pr2
(i)
The characteristic length L is defined as
L
surface area
perimeter
SD 2
4SD
D
4
where
D = diameter = 14 cm = 0.14 m
(iii) Computations. Consider orientation 1 first. Equation (j) gives
L
0.14(m)
4
0.035 m
Properties of air are determined at the film temperature T f 1 defined as
(j)
PROBLEM 8.38(continued)
Ts1 Tf
(260 20)( o C)
=
= 140oC
2
2
Tf1
At this temperature air properties are
k1 = 0.03394 W/m-oC
Pr1 = 0.702
v 1 = 27.44u10-6 m2/s
For a horizontal plate in air E is evaluated at Tf
E=
1
Tf ( K )
1
o
0.0034112 (1/K)
20( C) 273.15
Equation (h) gives
RaL1 =
0.0034112(1/ o C)9.81(m /s 2 )(260 20)( o C)(0.035) 3 (m 3 )
(27.44 u 10 6 ) 2 (m 4 /s 2 )
0.702 3.21 u 10 5
Substituting into (8.29a)
h1 L
k1
N u L1
0.54(3.21 u 10 5 )1 / 4
12.85
Solving for h1
h1 = 14.85
k1
L
0.03394(W/m- o C)
12.46 W/m2-oC
0.035(m)
Consider orientation 2. Properties of air are determined at the film temperature T f 2 defined as
Tf 2
(300 20)( o C)
Ts 2 Tf
=
= 160oC
2
2
At this temperature air properties are
k 2 = 0.03525 W/m-oC
Pr2 = 0.701
v 2 = 29.75u10-6 m2/s
Equation (I) gives
RaL2 =
0.0034112(1/ o C)9.81(m /s 2 )(300 20)( o C)(0.035) 3 (m 3 )
(29.75 u 10 6 ) 2 (m 4 /s 2 )
Substituting into (8.29a)
Nu L2
h2 L
k2
Solving for h2
0.27(3.182 u 10 5 )1 / 4
6.41
0.701 3.182 u 10 5
PROBLEM 8.38(continued)
h2 = 6.41
k2
L
0.03525(W/m- o C)
0.035(m)
6.46 W/m2-oC
Substituting into (g)
H
12.46(W/m 2 o C)(260 20)( o C) 6.46(W/m 2 o C)(300 20)( o C)
5.67 u 10 8 (W/m 2 K 4 ) [(300 273.15) 4 (K) 4 (260 273.15) 4 (K) 4 ]
0.769
(iv) Checking. Dimensional check: Computations showed that equations (g)-(j) are
dimensionally consistent.
Quantitative check: Heat transfer coefficients are within the approximate range of Table 1.1.
Validity of correlation equations (8.29a) and (8.30a): The conditions listed in (8.29c) and (8.30b)
are met.
(5) Comments. (i) The determination of emissivity using this method does not require
measuring the surroundings temperature. However, surroundings temperature must be uniform
and the same for both orientations. (ii) Emissivity can be determined without knowing the power
supply as long as it is the same for both orientations. (iii) The disk need not be perfectly
insulated as long as heat loss from the insulated surfaces is the same for both orientations. (iv).
Based on the calculated emissivity, heat transfer from the surface q is given by (b), (c) and (d).
However, to compute heat transfer by radiation the surroundings temperature must be known.
Assume Tsur 20 o C . Thus, for orientation 1:
qc
12.46(W/m 2 o C)S 0.14/2) 2 (m 2 )(260 20)( o C)
8
2
4
2
46.03 W
2
qr
0.769(5.67 u 10 )(W/m K )S (0.14 / 2) (m )[(260 273.15) 4 (20 273.15) 4 ](K 4 ) = 49.27W
q
46.03 W + 49.27 W = 95.3 W
Similarly, for orientation 2
q c 27.84 W
q r 67.46 W
q 95.3 W
This confirms that the electric power supply to both positions is the same.
POBLEM 8.39
A hot water tank of diameter 65 cm and height 160 cm loses heat by free convection. Estimate
the free convection heat loss from its cylindrical and top surfaces. Assume a surface temperature
of 50oC and an ambient air temperature of 20oC.
(1) Observations. (i) This is a free convection problem. (ii) Heat is transferred from the
cylindrical surface and top surface of tank to the ambient air. (iii) Under certain conditions a
vertical cylindrical surface can be modeled as a vertical plate. (iv) Newton’s law of cooling
gives the heat transfer rate from tank. (v) The fluid is air.
(2) Problem Definition. Determine the average heat transfer coefficient for a vertical
cylindrical surface and a heated horizontal surface facing up.
(3) Solution Plan. Apply Newton’s law of cooling and use appropriate correlation equations for
the average Nusselt number to determine the average heat transfer coefficient for a vertical
cylinder and a heated horizontal plate.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties (except in buoyancy), (3) uniform
surface temperature, (4) quiescent fluid and (6) negligible radiation.
(ii) Analysis. The total heat transfer from tank is
q = qc + qt
(a)
qc
D
where
qt
q = total rate of heat transfer from tank, W
qc = rate of heat transfer from cylindrical surface, W
qt = rate of heat transfer from top surface, W
qc
Applying Newton’s law of cooling to the two surfaces
qc = hc Ac (Ts - Tf)
(b)
qt = ht At (Ts - Tf)
(c)
L
and
Tf
g
where
Ac = area of cylindrical surface, m2
At = area of top surface, m2
hc = average heat transfer coefficient for the cylindrical surface, W/m2-oC
2 o
ht = average heat transfer coefficient for the top surface, W/m - C
Ts = surface temperature = 50oC
Tf = ambient temperature = 20oC
The two areas Ac and At are
Ac = S D L
(d)
At = S D2/4
(e)
and
where
PROBLEM 8.39 (continued)
D = tank diameter = 0.65 m
L = tank height = 1.6 m
The heat transfer coefficients are determined form correlation equations for the Nusselt number.
To determine if the cylindrical surface can be modeled as vertical plate the following criterion is
checked
D
35
!
(f)
L (GrL ) 1 / 4
where the rGashof number GrL is defined as
GrL =
E g Ts Tf L3
(g)
Q2
where
g = gravitational acceleration = 9.81 m/s2
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
Properties are determined at the film temperature Tf defined as
Tf = (Ts Tf ) / 2 = (50 + 20)(oC)/2 = 35oC
Properties of air at this temperature are given in Appendix C
k = 0.02674 W/m-oC
Pr = 0.711
Q = 16.485u10-6 m2/s
For an ideal gas E is given by
E =
1
1
= 0.003245 1/K
=
o
T f (K )
35 C 273.15
Substituting into (g)
GrL =
0.003245(1/ o C)9.81(m / s 2 ) 50 20 ( o C)(1.6) 3 (m 3 )
6 2
2
(16.485 u 10 ) (m / s)
2
= 14.394u109
The right hand side of (f) is
(35/GrL)1/4 = (35/14.394u109)1/4 = 0.00702
The left hand side of (f) is
D/L = 0.65(m)/1.6(m) = 0.406
Equation (f) is satisfied and thus, the cylindrical surface can be treated as a vertical plate.
The Rayleigh number RaL is computed next to select an appropriate correlation equation for the
average Nusselt number for the cylindrical surface
RaL = GrL Pr = 14.394u109 u 0.711 = 10.234u109
For this Rayleigh number the correlation equation is given by (8.26a)
PROBLEM 8.39 (continued)
­°
0.387 Ra 1L/ 6
®0.825 °̄
1 (0.492 / Pr ) 9 / 16
hc L
k
Nu L
>
@
½°
8 / 27 ¾
°¿
2
(h)
valid for
vertical plate
uniform surface temperature Ts
laminar, transition, and turbulent
10 1 < Ra L < 1012
0 < Pr < f
properties at Tf
(i)
where
Nu L = average Nusselt number
Heat loss from the top surface is modeled as a horizontal heated plate facing up. The Rayleigh
umber RaD for the top is computed to select an appropriate correlation equation for the Nusselt
number. The Rayleigh number is defined as
RaD =
RaD =
E g (Ts Tf ) D 3
Pr
Q2
0.003245(1/ o C)9.81(m / s 2 ) 50 20 ( o C)(0.65) 3 (m 3 )
(16.485 u 10 6 ) 2 (m 2 / s) 2
0.711 = 0.6862u109
Thus, the appropriate equation for this Raleigh number is given by (8.31b)
Nu D =
ht D
= 0.15 (RaD)1/3
k
(j)
valid for
8 u 10 6 Ra L 1.6 u 10 9
(iii) Computations. For the cylindrical surface, equation (h) gives
2
Nu L
hc L
k
­°
0.387(10.234 u 10 9 )1 / 6 ½°
= 254.2
®0.825 8 / 27 ¾
°¿
°̄
1 0.492 / 0.711 9 / 16
>
@
Solving the above for hc
hc = 254.2 k/L = 254.2(0.02674)(W/m-oC)/1.6(m) = 4.25 W/m2-oC
Equation (i) gives the heat transfer coefficient for the top surface
ht D
= 0.15 (0.6862u109)1/3 = 132.3
k
Solving for ht
Nu D =
ht = 132.3 k/D = 132.3 (0.02674)(W/m-oC)/0.65(m) = 5.44 W/m2-oC
The surface areas are calculated from (d) and (e)
(k)
PROBLEM 8.39 (continued)
Ac = S 0.65(m) 1.6(m) = 3.267 m2
and
At = S (0.65)2(m2)/4 = 0.3318 m2
Equations (b) and (c) give the rate of heat transfer from the two surfaces
qc = 4.25(W/m2-oC) 3.267(m2)(50 - 20)(oC) = 416.5 W
and
qt = 5.44 (W/m2-oC) 0.3318(m2) (50 - 20)(oC) = 54.1 W
The total heat loss from the tank is
q = 416.5 (W) + 54.1 (W) = 470.6 W
(iv) Checking. Dimensional check: Computations showed that equations (b)-(h) and (j) are
dimensionally consistent.
Quantitative check. The computed values of the heat transfer coefficient are representative of
values listed in Table 1.1 for free convection of gases.
Validity of correlation equations (h) and (j): The conditions lis ted in (i) and (k) are met.
(5) Comments. (i) It is important to check the Rayleigh and G
rashof number s to determine the
applicable correlation equations for the Nusselt number. (ii) The heat loss from the tank is
significant. To conserve energy the tank should be insulated. (iii) The magnitude of E is the
same whether it is expressed in units of degree Celsius or kelvin. The reason is that E is
measured in terms of degree change. One degree change on the Celsius scale is equal to one
degree change on the kelvin scale. This is also true of units of heat transfer coefficient and
specific heat.
PROBLEM 8.40
Hot gases from a furnace are discharged through a round horizontal duct 30 cm in diameter.
The average surface temperature of a 3 m duct section is 180oC. Estimate the free convection
heat loss from the duct to air at 25oC.
(1) Observations. (i) This is a free convection problem. (ii) The geometry is a horizontal round
duct. (iii) Heat is transferred from duct surface to the ambient air. (iv) According to Newton’s
law of cooling, the rate of heat transfer depends on the heat transfer coefficient, surface area and
surface and ambient temperatures.
(2) Problem Definition. Determine the free convection heat transfer coefficient for a horizontal
round duct (cylinder).
(3) Solution Plan. Apply Newton’s law of cooling. Use Nusselt number correlation equations
to determine the average heat transfer coefficient for a horizontal cylinder in free convection.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional (neglect end effects), (3) constant
properties (except in buoyancy), (4) uniform surface temperature, (5) quiescent fluid and (6)
negligible radiation.
(ii) Analysis. Applying Newton’s law of cooling to the duct
q = h A (Ts - Tf) = h S DL(Ts - Tf)
where
A = surface area, m2
L
D = duct diameter = 3 cm = 0.3 m
2 o
h = average heat transfer coefficient,W/m - C
L = duct length = 3 m
q = heat transfer rate, W
Tf
Ts = surface temperature = 180oC
o
Tf = ambient temperature = 25 C
(a)
D
g
Ts
Tf
The heat transfer coefficient is obtained from correlation equations for the Nusselt number for a
horizontal cylinder in free convection given by equation (8.35a)
Nu D
valid for:
hD
k
½°
­°
0.387( Ra D )1 / 6
®0.60 8 / 27 ¾
°¿
°̄
1 0.559 / Pr 9 / 16
>
2
@
horizontal cylinder
uniform surface temperature or flux
105 < RaD < 1012
properties at Tf
where
k = thermal conductivity, W/m-oC
Nu D = average Nusselt number
Pr = Prandtl number
RaD = Rayleigh number
The Rayleigh number is defined as
(b)
(c)
PROBLEM 8.40 (continued)
RaD =
E g (Ts Tf ) D 3
Pr
Q2
(d)
where
g = gravitational acceleration = 9.81 m/s2
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
For an ideal gas E is given by
E=
1
T f 273.15 K
(e)
Properties are determined at the film temperature Tf, defined as
Tf = (Ts + Tf)/2
(f)
(iii) Computations. The film temperature is calculated using (f)
Tf = (180 +25)(oC)/2 = 102.5oC
Properties of air at this temperature are
k = 0.03144W/m-oC
Pr = 0.704
E = 1/(102.5 + 273.15)(K) = 0.0026621(1/K)
Q = 23.29u10-6 m2/s
Substituting into (d)
RaD =
0.0026621(1/ o C)9.81(m / s 2 )(180 25)( o C)(0.3) 3 (m 3 )
(23.29 u 10 6 ) 2 (m 2 / s) 2
0.704 = 1.418u108
Equation (b) is applicable for this Rayleigh number. Substituting into (b)
2
Nu D
hD
k
­°
0.387(1.418 u 10 8 )1 / 6 ½°
¾ =66.86
®0.60 8 / 27
°¿
°̄
1 (0.559 / 0.704) 9 / 16
>
@
Solving for h
o
2 o
h = 66..86 k/D = 66.86(0.03144)(W/m- C)/0.3(m) = 7.01 W/m - C
Equation (a) gives
q = 7.01(W/m2-oC) S 0.3(m)2.5(m) ( 180 25 )(oC) = 2560 W
(iv) Checking. Dimensionless check: Computations showed that equations (a), (b), (d) and
(e) are dimensionally consistent.
Quantitative check: The value of the heat transfer coefficient is within the approximate range
given in Table 1.1 for free convection of gases.
Validity of correlation equation (b): The conditions listed in (c) are met.
(5) Comments. Since surface temperature is relatively high, radiation heat loss may be
significant.
PROBLEM 8.41
A 6 m long horizontal steam pipe has a surface temperature of 120oC. The diameter of the pipe
is 8 cm. It is estimated that if the pipe is covered with a 2.5 cm thick insulation material its
surface temperature will drop to 40oC. Determine the free convection heat loss from the pipe
with and without insulation. The ambient air temperature is 20oC.
(1) Observations. (i) This is a free convection problem. (ii) The geometry is a horizontal pipe.
(iii) Heat is transferred from pipe surface to the ambient air. (iv) Adding insulation material
reduces heat loss from pipe. (v) According to Newton’s law of cooling, the rate of heat transfer
depends on the heat transfer coefficient, surface area and surface and ambient temperatures. (vi)
Heat transfer coefficient and surface area change when insulation is added. (vii) The fluid is air.
(2) Problem Definition. Determine the free convection heat transfer coefficient for a horizontal
pipe (cylinder) with and without insulation.
(3) Solution Plan. Apply Newton’s law of
cooling. Use correlation equations for the
Nusselt number to determine the average heat
transfer coefficient for a horizontal cylinder in
free convection.
L
D
g
Ts
Tf
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional (neglect end effects), (3) constant
properties (except in buoyancy), (4) uniform surface temperature, (5) quiescent fluid and (6)
negligible radiation.
(ii) Analysis. Applying Newton’s law of cooling to the bare and insulated pipe
qo = ho Ao (Tso - Tf) = ho S DoL(Tso - Tf )
(a)
and
qi = hi Ai(Tsi - f) = hi S DiL(Tsi - Tf)
(b)
where the subscripts i and o refer to insulated and bare pipes, respectively, and
A = surface area, m2
Di = insulation diameter = 8 cm + 2(2.5)(cm) = 13 cm = 0.13 m
Do = pipe diameter = 8 cm = 0.08 m
hi = average heat transfer coefficient for insulated pipe, W/m2-oC
ho = average heat transfer coefficient for bare pipe, W/m2-oC
L = pipe length = 6 m
Tsi = surface temperature of insulated pipe = 40oC
Tso = surface temperature of bare pipe = 120oC
Tf = ambient temperature = 20oC
The heat transfer coefficient is obtained from correlation equations for the Nusselt number for a
horizontal cylinder in free convection given by equation (8.35a)
Nu D
hD
k
­°
0.387 Ra D 1 / 6
®0.60 °̄
1 0.559 / Pr 9 / 16
>
@
½°
¾
8 / 27
°¿
2
(c)
PROBLEM 8.41 (continued)
Valid for
horizontal cylinder
uniform surface temperature or flux
105 < RaD < 1012
properties at Tf
(d)
where
D = diameter, m
k = thermal conductivity, W/m-oC
Nu D = average Nusselt number
Pr = Prandtl number
RaD = Rayleigh number
The Rayleigh number is defined as
RaD =
E g Ts Tf D 3
Pr
Q2
(e)
where
g = gravitational acceleration = 9.81 m/s2
E = coefficient of thermal expansion, 1/K (or 1/oC)
Q = kinematic viscosity, m2/s
For an ideal gas E is given by
E=
1
(T f 273.15)(K )
(f)
Properties are determined at the film temperature Tf
Tf = (Ts + Tf)/2
(g)
(iii) Computations. Consider the bare pipe first. Film temperature is given by (g)
Tf = (120 +20)(oC)/2 = 70oC
Properties at this temperature are
k = 0.02922 W/m-oC
Pr = 0.707
E = 1/(70 + 273.15) K = 0.002914 1/K
Q = 19.9u10-6 m2/s
Substituting into (e)
Ra Do =
0.002914(1/ o C)9.81(m / s 2 )(120 20)( o C)(0.08) 3 (m 3 )
(19.9 u 10 6 ) 2 (m 2 / s) 2
0.707 = 0.261u107
Equation (c) is applicable for this Rayleigh number. Substituting into (c)
2
Nu D o
ho Do
k
Solving for ho
­°
0.387(0.261 u 10 7 )1 / 6 ½°
®0.60 ¾ = 19.08
8 / 27
°̄
°¿
1 (0.559 / 0.707) 9 / 16
>
@
PROBLEM 8.41(continued)
ho = 19.08k/Do = 19.08(0.02922)(W/m-oC)/0.08(m) = 6.97 W/m2-oC
Equation (a) gives
qo = 6.97(W/m2-oC)S 0.08(m)6(m) (120-20)(oC) = 1051.1 W
Consider next the insulated pipe. Film temperature is calculated using (g)
Tf = (40 +20)(oC)/2 = 30oC
Properties at this temperature are
k = 0.02638 W/m-oC
Pr = 0.712
E = 1/(30 + 273.15) K = 0.003299 1/K (ideal gas)
Q = 16.01u10-6 m2/s
Substituting into (e)
3
Ra Di =
0.003299(1/ o C)9.81(1/ o C) 40 20 ( o C) 0.13 (m 3 )
6 2
2
(16.01 u 10 ) (m / s)
2
0.712 = 0.395u107
Equation (c) is applicable for this Rayleigh number. Substituting into (c)
2
N u Di
hi Di
k
­°
0.387(0.395 u 10 7 )1 / 6 ½°
®0.60 ¾ = 21.53
8 / 27
°̄
°¿
1 (0.559 / 0.712) 9 / 16
>
@
Solving for hi
hi = 21.53k/Di = 21.53(0.02638)(W/m-oC)/0.13(m) = 4.37 W/m2-oC
Equation (b) gives
qi = 4.37(W/m2-oC)S 0.13(m)6(m) (40-20)(oC) = 214.2 W
(iv) Checking. Dimensionless check: Computations showed that equations (a)-(c) and (e)
are dimensionally consistent.
Quantitative check: The value of the heat transfer coefficient is within the approximate range
given in Table 1.1 for free convection of gases.
Validity of correlation equation (c): The conditions listed in (d) are met.
(5) Comments. (i) Heat loss from the pipe is significantly reduced when insulation is added.
The reduction is due to a decrease in surface temperature and heat transfer coefficient. (ii) The
magnitude of E is the same whether it is expressed in units of degree Celsius or kelvin. The
reason is that E is measured in terms of degree change. One degree change on the Celsius scale
is equal to one degree change on the kelvin scale. This is also true of units of heat transfer
coefficient and specific heat.
PROBLEM 8.42
An electric wire dissipates 0.6 W/m while suspended horizontally in air at 20oC. Determine its
surface temperature if the diameter is 0.1 mm. Neglect radiation.
(1) Observations. (i) This is a free convection problem. (ii) The geometry is a horizontal wire
(cylinder). (iii) Under steady state conditions the power dissipated in the wire is transferred to
the surrounding air. (iv) According to Newton’s law of cooling, surface temperature is
determined by the heat transfer rate, heat transfer coefficient, surface area and ambient
temperature. (v) The fluid is air.
(2) Problem Definition. Determine the free convection heat transfer coefficient for a horizontal
wire.
(3) Solution Plan. Apply Newton’s law of cooling. Use Nusselt number correlation equations
to determine the average heat transfer coefficient for a horizontal cylinder in free convection.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional (neglect end effects), (3) constant
properties (except in buoyancy), (4) uniform surface flux and temperature, (5) no axial
conduction, (6) quiescent fluid and (7) negligible radiation.
(ii) Analysis. Applying Newton’s law of cooling to the wire
q = h A (Ts - Tf) = h S DL(Ts - Tf)
Solving the above for Ts
Ts = Tf +
q/L
(a)
S Dh
where
A = surface area, m2
D = wire diameter = 0.1 mm = 0.0001 m
h = average heat transfer coefficient, W/m2-oC
L = wire length, m
q/L = power dissipated per unit length = 0.6 W/m
Ts = surface temperature, oC
Tf = ambient temperature = 20oC
g
Ts
q/ L
Tf
The heat transfer coefficient is obtained from correlation equations for the Nusselt number for a
horizontal cylinder in free convection, given by equation (8.35a)
Nu D
hD
k
­°
0.387 Ra D 1 / 6
®0.60 °̄
1 (0.559 / Pr ) 9 / 16
>
@
½°
¾
8 / 27
°¿
2
(b)
Valid for
horizontal cylinder
uniform surface temperature or flux
105 < RaD < 1012
properties at Tf
where
k = thermal conductivity, W/m-oC
(c)
PROBLEM 8.42 (continued)
Nu D = average Nusselt number
Pr = Prandtl number
RaD = Rayleigh number
The Rayleigh number is defined as
RaD =
E g Ts Tf D 3
Pr
Q2
(d)
where
g = gravitational acceleration = 9.81 m/s2
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
Properties are determined at the film temperature Tf
Tf = (Ts + Tf)/2
(e)
(iii) Computations. Examining (a), (b) and (c) shows that h depends on Ts through the
Rayleigh number. Thus, equation (a) must be solved for Ts by trial and error. A value for Ts is
assumed, properties are determined based on the assumed temperature, equation (d) is used to
calculate RaD, (b) is used to determine h and (a) is used to calculated Ts. If the calculated
temperature is not close to the assumed value, the procedure is repeated until a satisfactory
agreement is obtained.
Assume Ts = 200oC. Using (e)
Tf = (200 + 20)(oC)/2 = 110oC
Properties at this temperature are
k = 0.03194 W/m-oC
Pr = 0.704
E = 1/(Tf + 273.15) K = 1/(110 + 273.15) K = 0.00261 (1/K) (ideal gas)
Q = 24.1u10-6 (m2/s)
Substituting into (d)
3
RaD =
0.00261(1/ o C)9.81(m / s 2 ) 200 20 ( o C) 0.0001 (m 3 )
(24.1 u 10 6 ) 2 (m 2 / s) 2
0.704 = 5.5863u10-3
Equation (b) can be used for this Rayleigh number. Substituting into (b)
2
Nu D
­°
0.387 0.0055863 1 / 6 ½°
®0.60 ¾ = 0.5406
8 / 27
°̄
°¿
1 (0.559 / 0.704) 9 / 16
hD
k
>
@
Solving for h
o
2 o
h = 0.5406 k/D = 0.5406(0.03194)(W/m- C)/0.0001(m) = 172.7 W/m - C
Equation (a) gives
Ts = 20(oC) +
0 .6 W / m
S 0.0001 m 172.7 W / m 2 o C
= 31.1oC
PROBLEM 8.42 (continued)
This calculated value of Ts is significantly different from the assumed value. Repeating the
procedure with a new assumed value of Ts = 30oC gives h = 127.84 W/m2-oC and a calculated Ts
= 34.9oC. Since this is close to the assumed value, it follows that the estimated surface
temperature is 34.9oC.
(iv) Checking. Dimensionless check: Computations showed that equations (a), (b) and (d)
are dimensionally consistent.
Limiting check: If no energy is dissipated in the wire, its temperature should be the same as the
ambient temperature. Setting q = 0 in (a) gives Ts = Tf.
Quantitative check: The value of the heat transfer coefficient departs significantly from typical
free convection values listed in Table 1.1. A review of the computations revealed no errors in the
analysis or calculations. The departure from the range of h in Table 1.1 is due to the unusual
nature of this application in which the diameter is very small.
Validity of correlation equation (b): The conditions listed in (c) are met.
(5) Comments. (i) Solution by trial and error was necessary because properties depend on the
unknown surface temperature and equations (a), (b) and (d) can not be solved explicitly for Ts.
(ii) Table 1.1 should be used as a guide only, keeping in mind that exceptions to values listed
should be expected. (iii) Taking radiation into consideration has the effect of reducing surface
temperature. (iv) The magnitude of E is the same whether it is expressed in units of degree
Celsius or kelvin. The reason is that E is measured in terms of degree change. One degree
change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of
units of heat transfer coefficient and specific heat.
PROBLEM 8.43
The diameter of a 120 cm long horizontal section of a neon sign is 1.5 cm. Estimate the surface
temperature in air at 25oC if 12 watts are dissipated in the section. Neglect radiation heat loss.
(1) Observations. (i) This is a free convection problem. (ii) The geometry is a horizontal tube.
(iii) Under steady state conditions the power dissipated in the neon tube is transferred to the
surrounding air. (iv) According to Newton’s law of cooling, surface temperature is determined
by the heat transfer rate, heat transfer coefficient, surface area and ambient temperature. (v) The
fluid is air.
(2) Problem Definition. Determine the free convection heat transfer coefficient for a horizontal
tube.
(3) Solution Plan. Apply Newton’s law of cooling. Use Nusselt number correlation equations
to determine the average heat transfer coefficient for a horizontal cylinder in free convection.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional (neglect end effects), (3) constant
properties (except in buoyancy), (4) uniform surface temperature, (5) no axial conduction, (6)
quiescent fluid and (7) negligible radiation.
(ii) Analysis. Applying Newton’s law of cooling to the wire
q = h A (Ts - Tf) = h S DL(Ts - Tf)
Solving the above for Ts
Ts = Tf +
q
(a)
S DLh
where
A = surface area, m2
D = tube diameter = 1.5 cm = 0.015 m
h = average heat transfer coefficient, W/m2-oC
L = wire length = 120 cm = 1.2 m
q = power dissipated in wire per unit length = heat transfer rate from wire =12 W
Ts = surface temperature, oC
Tf = ambient temperature = 25oC
The heat transfer coefficient is obtained from correlation equations for the Nusselt number for a
horizontal cylinder in free convection, given by equation (8.35a)
Nu D
hD
k
­°
0.387 Ra D 1 / 6
®0.60 °̄
1 (0.559 / Pr ) 9 / 16
>
@
½°
¾
8 / 27
°¿
2
(b)
valid for
horizontal cylinder
uniform surface temperature or flux
105 < RaD < 1012
properties at Tf
where
k = thermal conductivity, W/m-oC
(c)
PROBLEM 8.43 (continued)
Nu D = average Nusselt number
Pr = Prandtl number
RaD = Rayleigh number
The Rayleigh number is defined as
RaD =
E g Ts Tf D 3
Pr
Q2
(d)
where
g = gravitational acceleration = 9.81 m/s2
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
Properties are determined at the film temperature Tf
Tf = (Ts + Tf)/2
(e)
(iii) Computations. Examining (a), (b) and (c) shows that h depends on Ts through the
Rayleigh number. Thus, equation (a) must be solved for T- by trial and error. A value for Ts is
assumed, properties are determined based on the assumed temperature, equation (d) is used to
calculate RaD, (b) is used to determine h and (a) is used to calculated Ts. If the calculated
temperature is not close to the assumed value, the procedure is repeated until a satisfactory
agreement is obtained.
Assume Ts = 75oC. Using (e)
Tf = (25 + 75)(oC)/2 = 50oC
Air properties at this temperature are given in Appendix C
k = 0.02781 W/m-oC
Pr = 0.709
E = 1/(50oC + 273.15) = 0.0030945 1/K (ideal gas)
Q = 17.92 u 10-6 m2/s
Substituting into (d)
RaD =
0.0030945(1/ o C)9.81( m / s2 )(75 25)( o C)(0.015) 3 ( m3 )
0.709 = 1.131u104
(17.92 u 10 6 ) 2 ( m4 / s2 )
NuD
hD
k
and
2
­°
0.387 (1.131 u 10 4 )1 / 6 ½°
= 4.503
®0.60 8 / 27 ¾
°̄
°¿
1 0.559 / 0.709 9 / 16
>
@
Solving the above for h
h = 4.503
0.02781( W/m o C )
= 8.35 W/m2-oC
0.015(m)
Substituting into (a) gives the surface temperature
PROBLEM 8.43 (continued)
Ts = 25oC +
12( W )
S 0.015(m)1.2(m)8.35( W / m 2 o C)
= 50.4oC
This is not close to the assumed value of 75oC. The procedure is repeated for Ts = 50oC and 55
o
C. The tabulated results below shows that Ts | 53.4oC.
Assumed Ts (oC)
h (W/m2-oC)
Calculated Ts (oC)
75
50
55
8.348
7.16
7.46
50.4
54.6
53.4
(iv) Checking. Dimensionless check: Computations showed that equations (a), (b) and (d)
are dimensionally consistent.
Limiting check: If no power is dissipated (neon sign is off), surface temperature should be the
same as the ambient temperature. Setting q = 0 in (a) gives Ts = Tf.
Quantitative check: The value of the heat transfer coefficient is within the range of values listed
in Table 1.1 for free convection of gases.
Validity of correlation equation (b): The conditions listed in (c) are met.
(5) Comments. (i) Solution by trial and error was necessary because properties depend on the
unknown surface temperature and equations (a), (b) and (d) can not be solved explicitly for Ts.
(ii) Taking radiation into consideration has the effect of reducing surface temperature. (iii) The
magnitude of E is the same whether it is expressed in units of degree Celsius or kelvin. The
reason is that E is measured in terms of degree change. One degree change on the Celsius scale
is equal to one degree change on the kelvin scale. This is also true of units of heat transfer
coefficient and specific heat.
PROBLEM 8.44
An air conditioning duct passes horizontally a distance of 2.5 m through the attic of a house. The
diameter is 30 cm and the average surface temperature is 10oC. The average ambient air
temperature in the attic during the summer is 42oC. Duct surface emissivity is 0.1. Estimate the
rate of heat transfer to the cold air in the duct.
(1) Observations. (i) This is a free convection problem. (ii) The geometry is a round horizontal
round duct. (iii) Heat is transferred from the ambient air to the duct. (iv) According to Newton’s
law of cooling, the rate of heat transfer to the surface depends on the heat transfer coefficient,
surface area and surface and ambient temperatures. (v) The fluid is air.
(2) Problem Definition. Determine the average free convection heat transfer coefficient for a
horizontal round duct.
(3) Solution Plan. Apply Newton’s law of cooling. Use Nusselt number correlation equations
to determine the average heat transfer coefficient for a horizontal cylinder in free convection.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional (neglect end effects), (3) constant
properties (except in buoyancy), (4) uniform surface temperature, (5) quiescent fluid and (6)
negligible radiation.
(ii) Analysis. Applying Newton’s law of cooling to the round duct
q = h A(Tf - Ts) = h S DL(Tf - Ts)
(a)
where
A = surface area, m2
D = duct diameter = 0.3 m
h = average heat transfer coefficient, W/m2-oC
L = duct length = 2.5 m
q = rate of heat transfer, oC
Ts = surface temperature = 10oC
Tf = ambient temperature = 42oC
L
D
g Tf T
s
q
The heat transfer coefficient is obtained from correlation equations for the Nusselt number for a
horizontal cylinder in free convection, given by equation (8.35a)
Nu D
hD
k
­°
0.387 Ra D 1 / 6
0
.
60
®
°̄
1 0.559 / Pr 9 / 16
>
@
½°
8 / 27 ¾
°¿
2
(b)
valid for:
horizontal cylinder
uniform surface temperature or flux
105 < RaD < 1012
properties at Tf
where
k = thermal conductivity, W/m-oC
Nu D = average Nusselt number
Pr = Prandtl number
(c)
PROBLEM 8.44 (continued)
RaD = Rayleigh number
The Rayleigh number is defined as
RaD =
E g (Tf Ts ) D 3
Pr
Q2
(d)
where
g = gravitational acceleration = 9.81 m/s2
E = coefficient of thermal expansion, 1/K (or 1/oC)
Q = kinematic viscosity, m2/s
Properties are determined at the film temperature Tf
Tf = (Ts + Tf)/2
(e)
For an ideal gas, E is given by
E=
1
(T f 273.15)(K)
(f)
(iii) Computations. The film temperature is calculated first using (e)
Tf = (10 +42)(oC)/2 = 26oC
Properties of air at this temperature are
k = 0.02608W/m-oC
Pr = 0.7124
E = 1/(26 + 273.15) K = 0.003343 (1/K) = 0.003343(1/oC)
Q = 15.64u10-6 (m2/s)
Substituting into (d)
3
RaD =
0.003343(1/ o C)9.81(m / s 2 ) 42 10 ( o C) 0.3 (m 3 )
6 2
2
(15.64 u 10 ) (m / s)
2
0.7124 = 8.2522u107
Equation (b) is applicable for this Rayleigh number. Substituting into (b)
2
Nu D
hD
k
1/ 6
­°
½°
0.387 8.2522 u 10 7
®0.60 ¾ = 53.37
8
/
27
°̄
°¿
1 0.559 / 0.7124 9 / 16
>
@
Solving for h
o
2 o
h = 53.37 k/D = 53.37(0.02608)(W/m- C)/0.3(m) = 4.64 W/m - C
Equation (a) gives
q = 4.64 (W/m2-oC) S 0.3(m) 2.5(m) (42 - 10)(oC) = 349.8 W
(iv) Checking. Dimensionless check: Computations showed that equations (a), (b) and (d)
are dimensionally consistent.
Limiting check: No heat will transfer to the duct if the ambient temperature is the same as
surface temperature. Setting Tf = Ts in (a) gives q = 0.
PROBLEM 8.44 (continued)
Quantitative check: The value of the heat transfer coefficient within the approximate range
given in Table 1.1 for free convection of gases.
Validity of correlation equation (b): The conditions listed in (c) are met.
(5) Comments. (i) Heat added to the cold air in the duct is significant. At $0.15/Kw-Hr it
represents a cost of $38/month. (ii) In practice, air conditioning and heating duct passing through
unoccupied areas are insulated. (iii) The magnitude of E is the same whether it is expressed in
units of degree Celsius or kelvin. The reason is that E is measured in terms of degree change.
One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This
is also true of units of heat transfer coefficient and specific heat.
PROBLEM 8.45
Estimate the surface temperature of a light bulb if its capacity is 150 W and the ambient air is at
23oC. Model the bulb as a sphere of diameter 9 cm. Neglect radiation.
(1) Observations. (i) This is a free convection and radiation problem. (ii) The geometry is a
sphere. (iii) Under steady state conditions the power dissipated in the bulb is transferred to the
surroundings by free convection and radiation and through the base by conduction. (iv)
According to Newton’s law of cooling and Stefan-Boltzmann radiation law, heat loss from the
surface depends on surface temperature. (v) The ambient fluid is air.
(2) Problem Definition. Determine the free convection heat transfer coefficient and radiation
heat loss for the sphere.
(3) Solution Plan. Apply Newton’s law of cooling and Stefan-Boltzmann law. Use Nusselt
number correlation equations to determine the average heat transfer coefficient for a sphere in
free convection.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties (except in buoyancy), (3)
uniform surface temperature, (4) no conduction through the bulb base, (5) all bulb power is
transmitted to the surroundings by surface convection and radiation, (6) filament does not
radiate to the surroundings directly. It radiates to the bulb surface, (7) bulb is a small surface
surrounded by a much larger surface, (8) bulb is spherical, (9) surroundings is at the ambient air
temperature and (10) quiescent fluid.
(ii) Analysis. Application of conservation of energy to the bulb gives
P = qc + qr
where
P = bulb power capacity = 150 W
qc = heat transfer by convection, W
qr = heat transfer by radiation, W
Applying Newton’s law of cooling to the sphere
(a)
qc = h A (Ts - Tf) = h S D2 (Ts Tf )
(b)
Application of Stefan-Boltzmann law
qr
4
H V SD 2 (Ts4 Tsur
)
where
A = surface area, m2
D = bulb diameter = 9 cm = 0.09 m
2 o
h = average heat transfer coefficient, W/m - C
P = bulb power = 150 W
qc = convection heat transfer rate, W
qr = radiation heat transfer rate, W
Ts = surface temperature, K
Tsur = surroundings temperature = 23oC + 273.15 = 296.15 K
Tf = ambient air temperature = 23oC + 273.15 = 296.15 K
H = emissivity of glass = 0.94
V = Stefan-Boltzmann constant = 5.67u10-8 W/m2-K4
(c)
qc
qr
D
g Tf
Ts
PROBLEM 8.45 (continued)
Substituting (b) and (c) into (a)
4
P = h S D2 (Ts Tf ) + H V S D 2 (Ts4 Tsur
)
(d)
The heat transfer coefficient is obtained from correlation equations for the Nusselt number for a
sphere in free convection, given by equation (8.36a)
hD
k
Nu D
2
0.589( Ra D )1 / 4
ª § 0.469 · 9 / 16 º
«1 ¨© Pr ¸¹
»
¬
¼
4/9
(e)
Valid for:
sphere
uniform surface temperature or flux
RaD < 1011
Pr > 0.7
properties at Tf
(f)
where
k = thermal conductivity, W/m-oC
Nu D = average Nusselt number
Pr = Prandtl number
RaD = Rayleigh number
The Rayleigh number is defined as
RaD =
E g Ts Tf D 3
Pr
Q2
(g)
where
g = gravitational acceleration = 9.81 m/s2
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity of air, m2/s
Properties are determined at the film temperature Tf
Tf = (Ts + Tf)/2
(h)
Equation (d) can not be solved explicitly for Ts because air properties depend Ts, the heat transfer
coefficient depends on Ts through the Rayleigh number, and the non-linear Ts4 term in radiation.
Thus, equation (d) must be solved for Ts by trial and error. A value for Ts is assumed, properties
are determined based on the assumed temperature, equation (g) is used to calculate RaD, (e) is
used to determine h and (d) is used to calculate P. If the calculated power is not close to the
given value of 150 W, the procedure is repeated until a satisfactory agreement is obtained.
(iii) Computations.
Assume Ts = 177oC + 273.15 = 450.15 K. Using (h)
Tf = (296.15 + 450.15)(K)/2 = 373.15 K = 100oC
Properties of air at this temperature are given in Appendix C
k = 0.03127 W/m-oC
Pr = 0.704
E = 1/(100oC + 273.15) = 0.00268 1/K (ideal gas)
PROBLEM 8.45 (continued)
-6
2
Q = 23.02u10 m /s
Substituting into (g) and (e)
RaD =
0.00268(1 / K )9.81(m / s 2 )(450.15 296.15)(K )(0.09) 3 (m 3 )
6 2
4
2
(23.02 u 10 ) (m / s )
0.704 = 3.921u106
and
NuD
hD
k
2
0.589 (3.921 u 10 6 )1 / 4
>1 0.469 / 0.704 @
9 / 16 4 / 9
= 22.21
Solving the above for h
0.03127( W / m o C )
= 7.72 W/m2-oC
h = 22.21
0.09(m)
Substituting into (d)
P = 7.72(W/m2-oC) S (0.09)2(m2)(177 23) (oC) +
0.94(5.67u10-8)(W/m2-K4) S (0.09)2(m2)[(450.15)4 (296.15) 4 ](K4)
P = 30.25 (W) + 45.26 (W) = 75.51 W
This is not close to the given power of 150 W. The procedure is repeated for other assumed
values of Ts and the results are tabulated below. At Ts = 535.15 K the calculated power is 151.46
W. This is close enough to the given power.
Assume Ts
K
Tf
K
RaD
450.15
530.15
535.15
373.13
413.15
415.65
3.921u106
3.776u106
3.757u106
h
W/m2-oC
7.72
8.3
8.33
qc
W
30.25
49.42
50.66
qr
W
45.26
96.7
100.8
qc +qr
W
75.51
146.12
151.46
(iv) Checking. Dimensionless check: Computations showed that equations (b)-(e) and (g)
are dimensionally consistent.
Limiting check: If no power is dissipated (bulb is off), surface temperature should be the same as
ambient temperature. Examination of (d) shows that the sum of the two terms will vanish only if
each term vanishes. Setting each term equal to zero gives Ts = Tsur = Tf.
Quantitative check: The value of the heat transfer coefficient is within the range of values listed
in Table 1.1 for free convection of gases.
Validity of correlation equation (e): The conditions listed in (f) are met.
(5) Comments. (i) Solution by trial and error was necessary because equation (d) can not be
solved explicitly for Ts. (ii) Radiation accounts for 66% of the total heat loss from the bulb. (iii)
The estimated surface temperature of 535 K or 262oC appears to be high. Direct radiation
between the filament and surroundings acts to lower surface temperature.
PROBLEM 8.46
A sphere of radius 2.0 cm is suspended in a very large water bath at
25oC. The sphere is heated internally using an electric coil. Determine
the rate of electric power that must be supplied to the sphere so that its
average surface temperature is 85oC. Neglect radiation.
(1) Observations. (I) At steady state, power supply to the sphere must
be equal to the heat loss from the surface. (ii) Heat loss from the surface
is by free convection. (iii) The surface is maintained at uniform temperature.
water
(2) Problem Definition. Determine the free convection heat transfer rate from the surface of a
sphere.
(3) Solution Plan. Apply Newton’s law of cooling to the sphere. Use an appropriate correlation
equation to determine the heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties (except in buoyancy), (3) uniform
surface temperature, (4) negligible radiation and (5) quiescent fluid.
(ii) Analysis. Conservation of energy for the sphere
P
q
(a)
where
P = power supply to the sphere, W
q = surface heat transfer rate from the sphere, W
Application of Newton's law of cooling
h A(Ts Tf )
q
(b)
where
A = surface area of sphere
h = average heat transfer coefficient, W/m 2 o C
Ts = surface temperature = 85 o C
Tf = ambient fluid temperature = 25 o C
Surface area of a sphere is
A S D2
(c)
where
D = Diameter of sphere = 4 cm = 0.04 m
Substituting (c) into (b) and (a)
P S h D(Ts Tf )
The average heat transfer coefficient is determined from correlation equations. For free
convection over a sphere the appropriate equation is given in equation (8.36)
(d)
PROBLEM 8.46 (continued)
Nu D
hD
k
2
0.589 RaD
1/ 4
(e)
>1 (0.469 / Pr ) @
9 / 16 4 / 9
valid for
uniform surface temperature
ReD 1011
Pr ! 0.7
properties at T f
The Rayleigh number is defined as
Ra D
E g (Ts Tf ) D 3
Pr
Q2
(f)
where
g = gravitational acceleration = 9.81 m/s 2
k = thermal conductivity, W/m o C
Pr = Prandtl number
RaD = Rayleigh number
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
Properties are determined at the film temperature T f defined as
(Ts Tf ) / 2
Tf
(g)
(iii) Computations. Substituting into (g)
(85 25) / 2 55 o C
Tf
Properties of water at this temperature are
k = 0.6458 W/m o C
Pr = 3.27
E 0.496 u 10 3 1/K
Q
0.5116 u 10 6 m 2 /s
Substituting into (f)
0.496 u 10 3 (1 / K )9.81(m/s 2 )(85 25)(o C)(0.04)3 (m 3 )
3.27
(0.5116 u 10 6 ) 2 (m 4 /s 2 )
Ra1
Substituting into (e)
Nu D
hD
k
2
0.589 2.33436 u 108
1/ 4
>1 (0.469 / 3.27) @
9 / 16 4 / 9
66.02
Solving for h
h
66.02
k
D
66.02
0.6458( W/m o C)
1065.9 W/m 2 o C
0.04(m)
2.33436 u 108
PROBLEM 8.46 (continued)
Substituting into (d)
P S (1065.9) ( W/m 2 o C)(0.04) 2 (m 2 )(85 25 ( o C) = 321.5 W
(iii) Checking. Dimensional check: Computations showed that units of equations (d)-(f) are
dimensionally consistent.
Limiting check: If Ts Tf , no free convection takes place and consequently P
Ts Tf in (d) gives the anticipate result.
0.
Setting
(5) Comments. The average heat transfer coefficient is slightly outside the range given in Table
1.1 for free convection in liquids. It should be remembered that values listed in Table 1.1 are for
typical applications. Exceptions should be expected.
PROBLEM 8.47
A fish tank at a zoo is designed to maintain water
temperature at 4 o C . Fish are viewed from outdoors
through a glass window L = 1.8 m high and w = 3 m wide.
The average ambient temperature during summer months
is 26 o C . To reduce water cooling load it is proposed to
create an air enclosure over the entire window using a
pexiglass plate. Estimate the reduction in the rate of heat
transfer to the water if the air gap thickness is G 6 cm.
Neglect radiation. Assume that the cold side of the
enclosure is at the same temperature as the water and the
warm side is at ambient temperature.
Tf
L
g
G
(1) Observations. (i) Heat is transferred from the ambient air to the water in the fish tank. (ii)
Adding an air enclosure reduces the rate of heat transfer. (iii) To estimate the reduction in
cooling load, heat transfer from the ambient air to the water with and without the enclosure must
be determined. (iv) Neglecting the thermal resistance of glass, the resistance to heat transfer form
the air to the water is primarily due to the air side free convection heat transfer coefficient. (v)
Installing an air cavity introduces an added thermal resistance. (vi) The problem can be modeled
as a vertical plate and as a vertical rectangular enclosure. (vii) The outside surface temperature
of the enclosure is unknown. (viii) Newton’s law of cooling gives the heat transfer rate. (ix) The
Rayleigh number should be determined for both vertical plate and rectangular enclosure so that
appropriate correlation equations for the Nusselt number are selected. However, since the outside
surface temperature of the enclosure is unknown, the Rayleigh number can not be determined.
The problem must be solved using an iterative procedure.
(2) Problem Definition. Determine the average free convection heat transfer coefficient for (1)
vertical plate at uniform surface temperature, and (2) for a rectangular cavity at uniform surface
temperatures.
(3) Solution Plan. (i) Apply Newton's law of cooling to the window with and without the
enclosure. (ii) Assume an outside enclosure surface temperature and compute the Rayleigh
number. (iii) Select appropriate Nusselt number correlations equations for the two cases. (iv)
Check the assumed temperature using conservation of energy from the tank to the ambient air.
(4) Plan Execution.
(i) Assumptions. (1) Outside surface temperature of the window is the same as water
temperature, (2) insulated top and bottom surfaces of the enclosure, (3) negligible radiation, (4)
uniform surface temperatures and (5) quiescent ambient air.
(ii) Analysis and Computations. Consider first the window without the added cavity.
Newton's law of cooling gives
q1 h1 A(Tw Tf )
(a)
where
A = surface area of window and enclosure = 1.8(m)u3.0(m) = 5.4 m2
h1 = average heat transfer coefficient, W/m2-oC
q1 heat transfer rate to water, W
Tf = ambient air temperature = 26oC
Tw = cold surface temperature = outside surface temperature of window = 4oC
PROBLEM 8.47 (continued)
The problem reduces to determining the heat transfer coefficients h1 for the
window without the enclosure. The window is modeled as a vertical plate
at uniform temperature. The Rayleigh number is calculated to determine the
appropriate correlation equation for the average heat transfer coefficient h1.
The Rayleigh number is defined as
Ra L
Eg Tf Tw L3
Pr
Q2
g
L
Tf
Tw
(b)
where
g = gravitational acceleration = 9.81 m/s2
L = window height = 1.8 m
Pr = Prandtl number
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
Air properties are evaluated at the film temperature T f defined as
Tw Tf (4 26)( o C)
15 o C
2
2
At this temperature air properties are
Tf
k = thermal conductivity = 0.02526 W/m-oC
Pr = 0.7145
Q = 14.64 u 106 m2/s
For an ideal gas E is given by
E
1
Tf
(c)
where Tf in this equation is in degrees kelvin. Thus
E = 1/(15 + 273.15)K = 0.0034704 1/K
Substituting into (b)
Ra L
0.0034704 1/K 9.81(m/s 2 )(26 4)( $ C) 1.8 3 (m 3 )
(14.64 u 10 6 ) 2 (m 4 /s 2 )
0.7145 1.4562 u 1010
Thus the flow is turbulent and the appropriate correlation equation is (8.22a)
Nu L
hL
k
1/ 6
­
0.387 Ra L
°
0
.
825
®
°̄
1 0.492 / Pr 9 / 16
>
@
½
°
8 / 27 ¾
°¿
2
(d)
Valid for:
vertical plate
uniform surface temperature
laminar, transition, and turbulent
10 1 < Ra L < 1012
0 < Pr < f
properties at T f
(e)
PROBLEM 8.47 (continued)
Applying (d) to the window gives h1
h1 L
k
Nu L
h1
1/ 6
­
½
0.387 1.4562 u 1010
°
°
0
.
825
®
¾
8
/
27
9
/
16
°̄
°¿
1 0.492 / 0.7145
>
2
@
284.4 u 0.02526( W/m - $ C )
1.8(m)
284.4
3.99 W/m 2 - $ C
Substituting into (a)
q1 = 3.99(W/m2-oC) 1.8(m)(3)(m) (26–4)(oC) = 474.1 W
Consider next the window with the added enclosure. To determine the appropriate correlation
equation for a vertical rectangular cavity the aspect ratio and Rayleigh number are calculated.
The aspect ration is defined as
L
aspect ratio =
(f)
G
where
L
G
G
length of rectangle = 1.8 m
width of rectangle = 0.06 m
g
Equation (b) gives
L
G
1.8(m)
0.06(m)
L
Tf
30
Tw
Ts
The Rayleigh number for the enclosure is defined as
RaG
Eg (Ts Tw )G 3
Q2
Pr
(g)
where Ts is outside surface temperature of enclosure. This temperature is needed to determine
both the enclosure heat transfer coefficient h and the outside surface coefficient h2 . An iterative
procedure is required to determine Ts , h and h2 . A value for Ts is assumed and h and h2 are
determined from applicable correlation equations. To check the assumed Ts , conservation of
energy is applied to heat transfer from the water to the air. This gives
h2 (Tf Ts )
Assume Ts
h (Ts Tw )
(h)
14 o C . First determine the outside heat transfer coefficient h2 using (b) and (d). Air
properties are evaluated at the film temperature at T which is the average temperature of the two
vertical surfaces of the enclosure given by
T
Ts Tf
2
(14 26)( o C)
2
20 o C
At this temperature air properties are
PROBLEM 8.47 (continued)
k = thermal conductivity = 0.02564 W/m-oC
Pr = 0.713
Q = 15.09 u 106 m2/s
E = 1/(20 + 273.15)K = 0.003411 1/K
Substituting into (b)
Ra L
0.003411 1/K 9.81(m/s 2 )(26 14)( $ C) 1.8 3 (m 3 )
6 2
4
2
(15.09 u 10 ) (m /s )
0.713
7.3326 u 10 9
Thus the flow is turbulent and the appropriate correlation equation is (d)
Nu L
h2 L
k
1/ 6
­°
½°
0.387 7.3326 u 10 9
®0.825 8 / 27 ¾
°̄
°¿
1 0.492 / 0.713 9 / 16
>
@
228.89 u 0.02564( W/m - $ C )
1.8(m)
h2
2
228.89
3.26 W/m 2 - $ C
To determine cavity heat transfer coefficient h2 Rayleigh number RaG is computed. Air
properties are determined at T , given by
T
Ts Tw
2
(14 4)( o C)
2
9o C
At this temperature air properties are
k = 0.02479 W/m-oC
Pr = 0.716
Q = 14.102 u 106 m2/s
E = 1/(9 + 273.15)K = 0.0035442 1/K
RaG
0.0035442 1/K 9.81(m/s 2 )(14 4)( $ C) 0.06 3 (m 3 )
(14.102 u 10
6 2
4
2
0.716
) (m /s )
2.7039 u 10 5
Thus correlation equation (8.39a) is applicable
Nu G
hG
k
ªLº
0.42 >Pr @ 0.012 >RaG @ 0.25 « »
¬G ¼
0.3
(8.39a)
Valid for
vertical rectangular enclosure
L
10 40
G
1 Pr 2 u 10 4
10 4 RaG 10 7
properties at T
Substituting into (8.39a)
(8.39b)
(Tc Th ) / 2
PROBLEM 8.47 (continued)
Nu G
h
hG
k
>
0.42 >0.716@ 0.012 2.7039 u 10 5
3.438 u 0.02479( W/m - $ C )
0.06(m)
@ 0.25 >30@
0.3
3.438
1.42 W/m 2 - $ C
Use (h) to check the assume temperature
3.26( W/m 2 o C)(26 14)
? 1.42( W/m 2 o C)(14 4)( o C)
39.1( W/m 2 ) z 14.2( W/m 2 )
Since (h) is not satisfied the procedure is repeated until a satisfactory agreement is obtained.
Assume Ts 18 o C . At this temperature the following result is obtained:
Ra L
4.7391 u 10 9
Nu L 1.9953
h2 2.86 ( W/m 2 o C)
RaG
3.7173 u 10 5
Nu G 3.723
h 1.548
Substituting into (h)
2.86( W/m 2 o C)(26 18)( o C)
? 1.548( W/m 2 o C)(18 4)( o C)
22.88( W/m 2 ) | 21.67( W/m 2 )
The heat transfer rate is given by applying Newton’s law of cooling between the ambient air and
the outside surface of the enclosure
q 2 = 2.86( W/m 2 o C)1.8(m)3(m)(26 18)( o C) 123.6 W
Thus the reduction in the cooling load due to the addition of the cavity is
Load reduction = 474.1 W - 123.6 W
350.5 W
(iv) Checking. Dimensional check: Computations showed that equations (a), (b), (d) and
(8.39a) are dimensionally consistent.
Quantitative check: The magnitude of h , h1 and h2 are in line with typical free convection
values for air given in Table 1.1.
Validity of correlation equations (d) and (8.39a): Conditions listed in equations (e) (8.39b) are
satisfied.
(5) Comments. The addition of a rectangular cavity reduces the heat transfer to the water by
74%. This is a significant saving in energy.
PROBLEM 8.48
It is proposed to replace a single pane observation window with double pane. On a typical
winter day the inside and outside air temperatures are Ti 20 o C and To 10 o C . The inside
and outside heat transfer coefficients are hi 9.4 W/m 2 o C and ho 37 W/m 2 o C . The height
of the window is L 0.28 m and its width is w = 3 m. The thickness of glass is t = 0.3 cm and its
conductivity is k g 0.7 W/m o C. Estimate the savings in energy if the single pane window is
replaced. Note that for the single pane window there are three resistances in series and the heat
transfer rate q1 is given by
A(Ti To )
q1
t
1
1
hi k g ho
For the double pane window, two additional resistances are added. The width of the air space in
the double pane is G 3 cm. In determining the heat transfer coefficient in the cavity, assume
that enclosure surface temperatures are the same as the inside and outside air temperatures.
(1) Observations. (i) Heat is transferred from the inside to the outside. (ii) Adding an air
enclosure reduces the rate of heat transfer. (iii) To estimate the savings in energy, heat transfer
through the single and double pane windows must be determined. (iv) The double pane window
introduces an added glass conduction resistance and a cavity convection resistance. (v) the
problem can be modeled as a vertical rectangular enclosure. (vi) Newton’s law of cooling gives
the heat transfer rate. (vii) The aspect ratio and Rayleigh number should be determined for the
rectangular enclosure so that an appropriate correlation equation for the Nusselt number can be
selected.
(2) Problem Definition. Determine the average free convection heat transfer coefficient for a
rectangular cavity.
(3) Solution Plan. (i) Apply Newton's law of cooling to the single pane and double pane
windows taking into consideration the multiple resistances in series. (ii) Select an appropriate
Nusselt number correlation equation for the rectangular cavity.
(4) Plan Execution.
(i) Assumptions. (1) Insulated top and bottom surfaces of the enclosure, (2) negligible
radiation and (3) uniform surface temperatures.
(ii) Analysis. Consider first the single pane window
q1
A(Ti To )
t
1
1
hi k g ho
where
A = surface area of window and enclosure = 0.28(m)u3.0(m) = 0.84 m2
hi = inside heat transfer coefficient = 9.4 W/m2-oC
ho = outside heat transfer coefficient = 37 W/m2-oC
k g glass conductivity = 0.7 W/m-oC
q1
heat transfer rate form single pane window, W
(a)
PROBLEM 8.48 (continued)
t glass thickness = 0.3 cm = 0.003 m
Ti inside air temperature = 20oC
To outside air temperature = -10oC
For the double pane window, two additional resistances are added in series: a glass conduction
resistance and a rectangular cavity convection resistance. Equation (a) is modified to
q2
A(Ti To )
1 2t
1 1
hi k g ho h
(b)
where
h
q2
rectangular cavity convection heat transfer coefficient, W/m2-oC
heat transfer rate from the double pane window, W
To determine cavity heat transfer coefficient h the aspect ratio and Rayleigh number RaG are
computed. The aspect ratio is given by
Aspect ratio =
L
G
0.28(m)
0.03(m)
9.33
The Rayleigh number for the enclosure is defined as
RaG
where
Eg (Th Tc )G 3
Pr
Q2
(c)
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number
Tc enclosure cold surface temperature, o C
Th enclosure hot surface temperature, o C
E = coefficient of thermal expansion, 1/K
G enclosure air thickness = 3 cm = 0.03 m
Q = kinematic viscosity, m2/s
Air properties are determined at T which is the average temperature of the two vertical surfaces
of the enclosure, Tc and Th . Both temperatures are unknown . They can be determined by an
iterative procedure. A simpler approach is to assume that the two surfaces are at the inside and
outside temperatures. That is, Th Ti and Tc To . This assumption is conservative in that it
will overestimate the heat loss from the double pane window. Using this approximation gives
T
Ti To
2
(20 10)( o C)
2
5o C
At this temperature air properties are
k = 0.02448 W/m-oC
Pr = 0.717
Q = 13.75 u 106 m2/s
PROBLEM 8.48 (continued)
E = 1/(5 + 273.15)K = 0.003595 1/K
0.003595 1/K 9.81(m/s 2 )(20 10)( $ C) 0.03 3 (m 3 )
RaG
(13.75 u 10 6 ) 2 (m 4 /s 2 )
0.717 1.0834 u 10 5
Thus correlation equation (8.38a) is applicable
hG
k
Nu G
ª Pr
º
RaG »
0.22 «
¬ 0.2 Pr
¼
0.28
ªLº
«G »
¬ ¼
0.25
(8.38a)
Valid for
vertical rectangular enclosure
L
2 10
G
Pr 10
(8.38b)
5
10 3 RaG 1010
properties at T
(Tc Th ) / 2
(iii) Computations. Equation (a) gives
q1
q1
0.84(m)(20 10)( o C)
1
0.003(m)
1
2 o
2 o
9.4( W/m C) 0.7( W/m C) 37( W/m 2 o C)
25.2
W
0.10638 0.00429 0.02703
183 W
Substituting into (8.38a)
Nu G
h
hG
k
3.018
ª 0.717
º
0.22 «
1.0834»
¬ 0.2 0.717
¼
0.02448( W/m o C)
0.03(m)
0.28
>9.333@0.25
3.018
2.463W/m 2 o C
Equation (b) gives
q2
0.84(m)(20 10)( o C)
1
2 u 0.003(m)
1
1
9.4( W/m 2 o C) 0.7( W/m 2 o C) 37( W/m 2 o C)
2.463( W/m 2 o C)
25.2
W 46.3 W
0.10638 0.00429 0.02703 0.406
Thus the percent savings in energy are
%Savings =
(183 46.3) W
186 W
75%
PROBLEM 8.48 (continued)
(iv) Checking. Dimensional check: Computations showed that equations (a), (b), (c) and
(8.38a) are dimensionally consistent.
Quantitative check: The magnitude of h is in line with typical free convection values for air
given in Table 1.1.
Validity of correlation equations (8.38a): Conditions listed in equation (8.38b) are satisfied.
(5) Comments. (i) Using double pane window reduces the heat loss by 75%. This is a
significant saving in energy. (ii) The assumption that Th Ti and Tc To overestimates h . Thus
in fact heat loss form the double pane window is less than 46.3 W and the savings are more than
75%.
.
PROBLEM 8.49
To reduce heat loss form an oven, a glass door with a rectangular air
cavity is used. The cavity has a baffle at its center. The height of the
door is L 65 cm and its width w 70 cm . The air space thickness is
G 1.5 cm . Estimate the heat transfer rate through the door if the
L/2
L
inside and outside surface temperatures of the cavity are 198 o C and
42 o C .
(1) Observations. (i) Heat is transferred through the door from the inside to the outside. (ii)
Newton’s law of cooling gives the heat transfer rate. (iii) The aspect ratio and Rayleigh number
should be determined for the rectangular enclosure so that an appropriate correlation equation for
the Nusselt number can be selected. (iv) The baffle divides the vertical cavity into two equal
parts. This has the effect of decreasing the aspect ratio by a factor of two.
(2) Problem Definition. Determine the average free convection heat transfer coefficient for a
rectangular cavity.
(3) Solution Plan. (i) Apply Newton's law of cooling (ii) Select an appropriate Nusselt number
correlation equation for the rectangular cavity.
(4) Plan Execution.
(i) Assumptions. (1) Insulated top and bottom surfaces of the enclosure, (2) negligible
radiation and (3) uniform surface temperatures.
(ii) Analysis. Newton’s law of cooling gives
q h A(Th Tc )
where
A = surface area of door = L u w 0.065(m)u0.7(m) = 0.455 m2
h = enclosure heat transfer coefficient, W/m2-oC
q heat transfer rate, W
Tc cavity cold side surface temperature = 42oC
Th cavity hot side temperature = 198oC
(a)
To determine cavity heat transfer coefficient h the aspect ratio and Rayleigh number RaG are
computed. The aspect ratio is given by
Aspect ratio =
L
G
(0.65 / 2)(m)
0.015(m)
21.667
The Rayleigh number for the enclosure is defined as
RaG
where
Eg (Th Tc )G 3
Pr
Q2
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number
E = coefficient of thermal expansion, 1/K
G enclosure air thickness = 1.5 cm = 0.015 m
Q = kinematic viscosity, m2/s
(b)
PROBLEM 8.49 (continued)
Air properties are determined at T
Tc Th (42 198)( o C)
120 o C
2
2
At this temperature air properties are
T
k = 0.03261 W/m-oC
Pr = 0.703
Q = 25.19 u 106 m2/s
E = 1/(120 + 273.15)K = 0.002544 1/K
0.002544 1/K 9.81(m/s 2 )(198 42)( $ C) 0.015 3 (m 3 )
RaG
6 2
4
2
(25.19 u 10 ) (m /s )
0.703 1.4557 u 10 4
Thus correlation equation (8.39a) is applicable
hG
k
Nu G
ªLº
0.42 >Pr @ 0.012 >RaG @ 0.25 « »
¬G ¼
0.3
(8.39a)
Valid for
vertical rectangular enclosure
L
10 40
G
(8.39b)
1 Pr 2 u 10 4
10 4 RaG 10 7
properties at T
(Tc Th ) / 2
(iii) Computations. Equation (8.39a)) gives
0.25
hG
0.42 >0.703@ 0.012 1.4557 u 10 4
>21.667@0.3
k
0.03261( W/m o C)
1.826
3.969 W/m 2 o C
0.015(m)
Nu G
h
>
@
1.826
Substituting into (b)
q
3.969( W/m 2 o C)0.455(m 2 )(198 42)( o C)
281.7 W
(iv) Checking. Dimensional check: Computations showed that equations (a), (b) and (8.39a)
are dimensionally consistent.
Quantitative check: The magnitude of h is in line with typical free convection values for air
given in Table 1.1.
Validity of correlation equations (8.39a): Conditions listed in equation (8.39b) are satisfied.
(5) Comments. (i) Even with a double pane door the heat loss is significant. This added energy
to the surroundings raises the temperature in the cooking area. (ii) Without the baffle the aspect
PROBLEM 8.49 (continued)
ratio is 43.334. Although this is slightly over the limit of (8.39a), this equation can still be used
without introducing significant error. With no baffle (8.39a)
h
q
3.224 W/m 2 o C
228.8 W
Therefore the cavity acts to increase the heat loss and thus should be eliminated.
.
PROBLEM 8.50
The ceiling of an exhibit room is designed to provide
natural light by using an array of horizontal skylights.
Each unit is rectangular with an air gap G 6.5 cm thick.
The length and width of each unit are L 54 cm and
w 120 cm . On a typical day the inside and outside glass
G
w
g
L
surface temperatures are 15 o C and 15 o C . Estimate the rate of heat loss from each unit.
(1) Observations. (i) Heat is transferred through the skylight from the inside to the outside. (ii)
Newton’s law of cooling gives the heat transfer rate. (iii) The aspect ratio and Rayleigh number
should be determined for the rectangular enclosure so that an appropriate correlation equation for
the Nusselt number can be selected.
(2) Problem Definition. Determine the average free convection heat transfer coefficient for a
rectangular cavity.
(3) Solution Plan. (i) Apply Newton's law of cooling (ii) Select an appropriate Nusselt number
correlation equation for the rectangular cavity.
(4) Plan Execution.
(i) Assumptions. (1) Insulated end surfaces of the enclosure, (2) negligible radiation, (3)
uniform surface temperatures and (4) negligible glass resistance no temperature drop across the
glass).
(ii) Analysis. Newton’s law of cooling gives
q h A(Th Tc )
where
A = surface area of skylight = L u w 0.54(m)u1.2(m) = 0.648 m2
h = enclosure heat transfer coefficient, W/m2-oC
q heat transfer rate, W
Tc cavity cold side surface temperature = -15oC
Th cavity hot side temperature = 15oC
(a)
To determine cavity heat transfer coefficient h the aspect ratio and Rayleigh number RaG are
computed. The aspect ratio is given by
Aspect ratio =
L
G
(0.54)(m)
0.065(m)
8.3077
The Rayleigh number for the enclosure is defined as
RaG
Eg (Th Tc )G 3
Pr
Q2
where
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number
E = coefficient of thermal expansion, 1/K
G enclosure air thickness = 6.5 cm = 0.065 m
Q = kinematic viscosity, m2/s
(b)
PROBLEM 8.50 (continued)
Air properties are determined at T
Tc Th (15 15)( o C)
0o C
2
2
At this temperature air properties are
T
k = 0.02408 W/m-oC
Pr = 0.718
Q = 25.19 u 106 m2/s
E = 1/(0 + 273.15)K = 0.003661 1/K
0.003661 1/K 9.81(m/s 2 )(15 15)( $ C) 0.065 3 (m 3 )
RaG
6 2
4
2
(13.31 u 10 ) (m /s )
0.718 1.1922 u 10 6
Thus correlation equation (8.39a) is applicable
hG
k
Nu G
0.069>RaG @1/ 3 >Pr @ 0.074
(8.41a)
Valid for
horizontal rectangular enclosure
heated from below
3 u 10 5 RaG 7 u 10 9
properties at T
(8.41b)
(Tc Th ) / 2
(iii) Computations. Equation (8.39a)) gives
Nu G
h
hG
k
7.1389
>
0.069 1.1922 u 10 6
0.02408( W/m o C)
0.065(m)
@
1/ 3
>0.718@ 0.074
7.1389
2.645W/m 2 o C
Substituting into (b)
q
2.645( W/m 2 o C)0.0.648(m 2 )(15 15)( o C)
51.4 W
(iv) Checking. Dimensional check: Computations showed that equations (a), (b) and (8.41a)
are dimensionally consistent.
Quantitative check: The magnitude of h is in line with typical free convection values for air
given in Table 1.1.
Validity of correlation equations (8.41a): Conditions listed in equation (8.41b) are satisfied.
(5) Comments. Using double pane skylight reduces heat loss to the surroundings.
PROBLEM 8.51
Repeat Example 8.4 using inclination angles of 0 o , 60 o , 90 o , 120 o , 150 o and 175 o . Plot heat
transfer rate q vs. inclination angle T .
(2) Problem Definition. Determine the average free convection
heat transfer coefficient h for a rectangular enclosure at various
inclination angles.
L
pl
a
Th
Tc
te
G
ab
so
rb
er
(1) Observations. (i) Power requirement is equal to the heat
transfer rate through the enclosure. (ii) The problem can be
modeled as a rectangular cavity at specified hot and cold surface
temperatures. (iii) The inclination angle varies from 0 o to 175 o . (iv)
Newton’s law of cooling gives the heat transfer rate. (v) The aspect
ratio and critical inclination angle should be computed to determine
the applicable correlation equation for the Nusselt number.
g
T
(3) Solution Plan. (i) Apply Newton's law of cooling. (ii) Compute the aspect ratio and critical
inclination angle. (iii) Select an appropriate Nusselt number correlation equation for convection
for a horizontal, vertical and inclined rectangular cavity.
(4) Plan Execution.
(i) Assumptions. (1) Uniform hot and cold surface temperatures and (2) insulated end
surfaces, (3) negligible radiation.
(ii) Analysis. Newton's law of cooling gives
P q h A(Th Tc )
where
A = surface area of rectangle = 0.7(m)u0.7(m) = 0.49 m2
h = average heat transfer coefficient, W/m2-oC
P power requirement, W
q heat transfer rate through cavity, W
Th = hot surface temperature = 27oC
Tc = cold surface temperature = 23oC
(a)
The aspect ratio is defined as
aspect ratio =
where
L
G
L
G
(b)
length of rectangle = 0.7 m
width of rectangle = 0.05 m
Equation (b) gives
L
G
0.7(m)
0.05(m)
14
According to Table 8.1, the critical angle is T c 70 o . For L / G ! 12 and 0 T 70 o , the
applicable correlation equation for the Nusselt number is (8.42a)
PROBLEM 8.51 (continued)
Nu G
hG
k
ª
1708 º
1 1.44«1 »
¬ RaG cosT ¼
*
*
ª 1708(1.8 sin T )1.6 º ª ( RaG cosT )1/ 3 º
1» (8.42a)
«1 »«
18
RaG cosT
«¬
»¼ «¬
»¼
Valid for
inclined rectangular enclosure
L / G t 12
0 T Tc
(8.42b)
set > @ * 0 when negative
properties at T (Tc Th ) / 2
The Rayleigh number is defined as
RaG
Eg Th Tc G 3
Pr
Q2
(c)
where
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
Water properties are evaluated at the film temperature T defined as
(Th Tc )
(d)
2
To determine the applicable correlation equations for the horizontal, vertical and inclination
angles 175$ t T ! 90 $ , the Rayleigh number is computed. Equation (d) gives
T
(27 23)( o C)
25 o C
2
Properties of water at this temperature are:
T
k = thermal conductivity = 0.6076 W/m o C
Pr = 6.13
E 0.259 u 10 3 1/K
0.8933 u 10 6 m2/s
Substituting into (c)
Q
RaG
0.259 u 10 3 1/K 9.81(m/s 2 )(27 23)( $ C) 0.05 3 (m 3 )
(0.8933 u 10 6 ) 2 (m 4 /s 2 )
6.13 9.75898 u 10 6
Thus the applicable correlation equation for the horizontal position is (8.41a)
Nu G
Valid for
hG
k
0.069>RaG @1/ 3 >Pr @ 0.074
(8.41a)
PROBLEM 8.51 (continued)
horizontal rectangular enclosure
heated from below
3 u 10 5 RaG 7 u 10 9
(8.41b)
(Tc Th ) / 2
properties at T
For the vertical orientation the applicable correlation equation is (8.39a)
Nu G
hG
k
0.046 >RaG @1 / 3
(8.39a)
Valid for
vertical rectangular enclosure
L
1 40
G
(8.39b)
1 Pr 20
10 6 RaG 10 9
properties at T
(Tc Th ) / 2
For 175$ t T ! 90 $ the applicable correlation equation is
Nu G
hG
k
>
@
1 Nu G (90 o ) 1 sin T
(8.45a)
Valid for
inclined rectangular enclosure
all L / G
90 o T 180 o
properties at T
(Tc Th ) / 2
(iii) Computations.
(1) Horizontal position: T
Nu G (0 o )
h (0 o )G
k
h (0 o ) 16.862
k
G
0 $ . Substituting into (8.41a)
>
0.069 9.75898 u 10 6
16.862
@
1/ 3
0.6076( W/m o C)
0.05(m)
>6.13@ 0.074 = 16.862
204.9 W/m 2 o C
The corresponding power is given by (a)
P (0 o )
204.9( W/m 2 o C)(0.49)(m 2 )(27 23)( o C)
401.6 W
(8.45b)
PROBLEM 8.51 (continued)
60 $ . Use (8.42a)
(2) Inclination angle T
Nu G (60 o )
h (60 o )G
k
1708
ª
º
1 1.44«1 ¬ 9.75898 u 10 6 cos 60 o »¼
ª (9.75898 u 10 6 cos 60)1 / 3 º
1»
«
18
¼
¬
k
10.86
132 W/m 2 o C
10.86
P (60 o ) 132( W/m 2 o C)(0.49)(m 2 )(27 23)( o C)
(3) Vertical orientation, T
258.7 W
90 $ . Use (8.39a)
1/ 3
h (90 o )G
0.046 9.75898 u 10 6
9.825
k
0.6076( W/m o C)
k
9.825
9.825
119.4 W/m 2 o C
G
0.05(m)
>
Nu G (90 o )
h (90 o )
ª
1708(1.8 sin 60 o )1.6 º
«1 6
o»
¬ 9.75898 u 10 cos 60 ¼
*
0.6076( W/m o C)
G
0.05(m)
Equation (a) gives the required power
h (60 o ) 10.86
*
@
P (90 o ) 119.4( W/m 2 o C)(0.49)(m 2 )(27 23)( o C)
(4) Inclination angle T
234 W
$
120 . Use (8.45a)
o
h (120 )G
1 >9.825 1@ sin 120 o 8.643
k
0.6076( W/m o C)
k
8.643
8.643
105.03 W/m 2 o C
G
0.05(m)
Nu G (120 o )
h (120 o )
P ( 120 o ) 105.03( W/m 2 o C)(0.49)(m 2 )(27 23)( o C)
(5) Inclination angle T
206 W
150 $ . Use (8.45a)
h (150 o )G
1 >9.825 1@ sin 150 o 5.413
k
0.6076( W/m o C)
k
5.413
5.413
65.8 W/m 2 o C
G
0.05(m)
Nu G (150 $ )
h (150 o )
P ( 150 o )
65.8( W/m 2 o C)(0.49)(m 2 )(27 23)( o C)
(6) Inclination angle T
Nu G (175 o )
h (175 o )
129 W
175$ . Use (8.45a)
h (175 o )G
1 >9.825 1@ sin 175 o 1.77
k
0.6076( W/m o C)
k
8.647
1.77
21.5 W/m 2 o C
G
0.05(m)
PROBLEM 8.51 (continued)
P ( 175 o )
21.5( W/m 2 o C)(0.49)(m 2 )(27 23)( o C)
Using the result of Example
8.4 and the above data, the
required power at various
angles is tabulated and
plotted.
42.1 W
400
300
P(W) 200
100
0
0
30
90
60
T
120
150
180
T (o )
P(W)
0
30
60
90
120
150
180
401.6
303.8
258.7
234
206
129
42.1
o
(iv)
Checking.
Dimensional
check: Computations showed
that
equations
(a), (c), (8.39a), (8.41a), (8.42a) and (8.45a) are dimensionally correct.
Quantitative check: The magnitude of h is in line with typical free convection values for liquids
given in Table 1.1.
Validity of correlation equations (8.39a), (8.41a), (8.42a) and (8.45a): Conditions listed in
equations (8.39b), (8.41b), (8.42b) and (8.45b)are satisfied.
(5) Comments. (i) If the device is to be used continuously, the estimate power requirement is
relatively high. Decreasing the temperature difference between the hot and cold surfaces will
reduce the power requirement. (ii) The ambient temperature plays a role in the operation of the
proposed device. The design must take into consideration changing ambient temperature. (iii)
Specification of the driving motor should be based on highest power which corresponds to the
horizontal orientation.
PROBLEM 8.52
L
be
rp
Th
Tc
la
te
G
ab
so
r
A rectangular solar collector has an absorber plate of length
L 2.5 m and width w 4.0 m. A protection cover is used to
form a rectangular air enclosure of thickness G 4 cm to
provide insulation. Estimate the heat loss by convection from
the plate when the enclosure inclination angle is 45 o and its
surfaces are at 28 o C and 72 o C.
g
(1) Observations. (i) The absorber plate is at a higher
T
temperature than the ambient air. Thus heat is lost through the
rectangular cavity to the atmosphere. (ii) The problem can be
modeled as an inclined rectangular cavity at specified hot and cold surface temperatures. (iii)
Newton’s law of cooling gives the heat transfer rate. (iv) The aspect ratio and critical inclination
angle should be computed to determine the applicable correlation equation for the Nusselt
number.
(2) Problem Definition. Determine the average free convection heat transfer coefficient h for
an inclined rectangular enclosure.
(3) Solution Plan. (i) Apply Newton's law of cooling. (ii) Compute the aspect ratio and critical
inclination angle. Select an appropriate Nusselt number correlation equation for convection in an
inclined rectangular cavity.
(4) Plan Execution.
(i) Assumptions. (1) Uniform hot and cold surface temperatures, (2) insulated end surfaces
and (3) negligible radiation.
(ii) Analysis. Newton's law of cooling gives
q h A(Th Tc )
where
A = surface area of rectangle = 2.5(m)u4(m) = 10 m2
h = average heat transfer coefficient, W/m2-oC
q heat transfer rate through cavity, W
Th = hot surface temperature = 72oC
Tc = cold surface temperature = 28oC
(a)
The aspect ratio is defined as
aspect ratio =
where
L
G
length of rectangle = 2.5 m
width of rectangle = 4 cm = 0.04 m
Equation (b) gives
L
G
2.5(m)
0.04(m)
62.5
L
G
(b)
PROBLEM 8.52 (continued)
According to Table 8.1, the critical angle is T c 70 o . Since L / G ! 12 and 0 T T c , it
follows that the applicable correlation equation for the Nusselt number is
Nu G
hG
k
ª
1708 º
1 1.44«1 »
Ra
G cos T ¼
¬
*
*
ª 1708(1.8 sin T )1.6 º ª ( RaG cosT )1/ 3 º
1» (8.42a)
«1 »«
RaG cosT
18
»¼
¬«
¼» ¬«
Valid for
inclined rectangular enclosure
L / G t 12
0 T Tc
(8.42b)
> @*
0 when negative
properties at T (Tc Th ) / 2
set
The Rayleigh number is defined as
Eg Th Tc G 3
Pr
Q2
RaG
(c)
where
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
Water properties are evaluated at the film temperatureT defined as
T
(Th Tc )
2
(d)
(iii) Computations. Equation (d) gives
(72 28)( o C)
50 o C
2
Properties of water at this temperature are:
T
k = thermal conductivity = 0.0.02781 W/m o C
Pr = 0709
E 1 /(50 273.15)(K) 0.0030946 1/K
Q
17.92 u 10 6 m2/s
Substituting into (c)
RaG
0.0030946 1/K 9.81(m/s 2 )(72 28)( $ C) 0.04 3 (m 3 )
Substituting into (8.42a)
(17.92 u 10 6 ) 2 (m 4 /s 2 )
0.709 1.88745 u 10 5
PROBLEM 8.52 (continued)
Nu G
hG
k
1708
ª
º
1 1.44«1 5
o»
¬ 1.88742 u 10 cos 45 ¼
ª (1.88742 u 10 5 cos 45)1 / 3 º
1»
«
18
¬
¼
*
ª
1708(1.8 sin 45 o )1.6 º
«1 5
o»
¬ 1.88742 u 10 cos 45 ¼
*
4.2338
0.02781( W/m o C)
2.9435 W/m 2 o C
0.04(m)
G
Equation (a) gives the heat transfer rate
q 2.9435( W/m 2 o C)(10)(m 2 )(72 28)( o C) 1295 W
h
4.2338
k
4.2338
(iv) Checking. Dimensional check: Computations showed that equations (a), (8.42a) and (c)
are dimensionally consistent.
Quantitative check: The magnitude of h is in line with typical free convection values for air
given in Table 1.1.
Validity of correlation equation (8.42a): Conditions listed in equation (8.42b) are satisfied.
(5) Comments. The rate of energy loss from the collector is significant. Increasing the thickness
of the air gap will reduce the heat lost to the atmosphere.
PROBLEM 8.53
A liquid-vapor mixture at Ti 20 o C flows inside a tube of diameter
Di 4 cm and length L 3 m.
The tube is placed concentrically
inside another tube of diameter Do 6 cm. Surface temperature of the
outer tube is at To 10 o C. Air fills the annular space. Determine the
heat transfer rate from the mixture.
Do
To
Ti
Di
(1) Observations. (i) Heat is transferred through the annular space
from the outer cylinder to the inner. (ii) Newton’s law of cooling gives
the heat transfer rate. (iii) The Rayleigh number should be determined
for the enclosure formed by the concentric cylinders so that an
appropriate correlation equation can be selected. (iv) The cylinders are
horizontally oriented.
(2) Problem Definition. Determine the average free convection heat transfer coefficient annular
cavity between two concentric horizontal cylinders.
(3) Solution Plan. (i) Use equation (8.46) describing heat transfer between two concentric
cylinders. (ii) Select an appropriate correlation equation for this geometry.
(4) Plan Execution.
(i) Assumptions. (1) negligible radiation and (2) uniform surface temperatures.
(ii) Analysis. Heat transfer between two concentric cylinders is given by equation (8.46)
qc
2S k eff
ln( Do / Di )
(To Ti )
(8.46)
where
Di
Do
k eff
4 cm = 0.04 m
6 cm =0.06 m
effective conductivity, W/m o C
q c heat transfer rate per unit length of tube, W/m
Ti = inner surface temperature (cold) = -20oC
To = outer surface temperature (hot) = 10oC
Using (8.46), the total heat transfer rate from a tube of length L is
q
2S k eff
ln( Do / Di )
(To Ti ) L
(a)
where
L tube length = 3 m
q total heat transfer rate, W
Correlation equation for the effective conductivity k eff is
k eff
k
where
ª
º
Pr
0.386 «
Ra * »
P
r
0
.
861
¬
¼
1/ 4
(8.47a)
PROBLEM 8.53 (continued)
>ln( Do / Di )@4
Ra*
>
G 3 ( Di ) 3 / 5 ( Do ) 3 / 5
@
5
(8.47b)
RaG
Do Di
2
G
(8.47c)
Valid for
concentric cylinders
(8.47d)
10 2 Ra * 10 7
properties at T (Ti To ) / 2
The Rayleigh number RaG in (8.47b) is defined as
RaG
Eg (Th Tc )G 3
Pr
Q2
(b)
where
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number
E = coefficient of thermal expansion, 1/K
Q = kinematic viscosity, m2/s
(iii) Computations. Equation (8.47) gives G
(0.06 0.04)(m)
0.01 m
2
Air properties are determined at T
G
Ti To (20 10)( o C)
5 o C
2
2
At this temperature air properties are
T
k = 0.023698 W/m-oC
Pr = 0.7195
Q = 12.885 u 106 m2/s
E = 1/(-5 + 273.15)K = 0.0037293 1/K
Substituting into (b)
RaG
0.0037293 1/K 9.81(m/s 2 )(15 15)( $ C) 0.01 3 (m 3 )
(12.885 u 10 6 ) 2 (m 4 /s 2 )
0.7195
Use (8.47b)
Ra *
>ln(0.06 / 0.04)@4
>
(0.01) 3 (0.04 m) 3 / 5 (0.06 m) 3 / 5
@
5
4756.4
455.2
4756.4
PROBLEM 8.53 (continued)
Thus condition (8.47d) is satisfied. Substituting into (8.47a)
k eff
k
k eff
0.7195
ª
º
0.386 «
455.2»
¬ 0.861 0.7195
¼
1.465 k
1/ 4
1.465
1.465(0.02369)(W/m o C)
Equation (a) gives q
q
2S (0.0347)( W/m o C)
(10 20)( o C)3(m)
ln(0.06m / 0.04m)
48.4 m
(iv) Checking. Dimensional check: Computations showed that equations (a), (b), (8.47a)
and (8.47b) are dimensionally consistent.
Quantitative check: The magnitude of h is in line with typical free convection values for air
given in Table 1.1.
Validity of correlation equations (8.47a): Conditions listed in equation (8.47d) are satisfied.
(5) Comments. The concentric annular space provides good insulation. This is indicated by the
low value of k eff / k = 1.465. A ratio of unity corresponds to pure conduction with no fluid
circulation.
PROBLEM 9.1
The speed of sound, c, in an ideal gas is given by
J RT
c
where J is the specific heat ratio R is gas constant and T is temperature. Show that
S
Kn
2
J
M
Re
where M is Mach number defined as
V
c
(1) Observations. (i) Definitions of Knudsen number, Reynolds number, and Mach number
are needed. (ii) Fluid velocity appears in the definition of Reynolds number and Mach
number.
M
(2) Problem Definition. Show that the Knudsen number can be expressed in terms of
Reynolds and Mach numbers.
(3) Solution Plan. Star with the definition of Knudsen number and multiply and divide by
variables to form the Reynolds and Mach numbers
(4) Plan Execution.
(i) Analysis. The Knudsen number is defined as
O
Kn
(1.2)
De
The mean free path for an ideal gas is given by
O
P S
p
2
RT
(9.2)
Since the Reynolds number is expressed in terms of density U , use the ideal gas law to
eliminate p in (9. 2)
p URT
(9.31)
Substitute (9.31) into (9.2)
O
P
S
U RT
2
RT
P
S
U RT
2
(a)
(a) into (1.2)
Kn
Multiply and divide (b) by V J
P
S
U De RT
2
(b)
PROBLEM 9.1 (continued)
S
Kn
2
J
P
V
U DeV
J RT
(c)
Introduce the definition of the Reynolds and Mach number
U V De
P
Re
M
V
c
V
J RT
(d)
(e)
Substitute (d) and (e) into (c)
Kn
S
2
J
M
Re
(f)
(ii) Checking: Dimensional check: Both sides of (f) are dimensionless.
(5) Comments. In determining the number of governing parameters in flow through
microchannels, it should be noted that the three parameters, Kn, Re and M are not
independent.
PROBLEM 9.2
Reported discrepancies in experimental data on the fiction factor f are partially attributed to
errors in measurements. One of the key quantities needed to calculate f is channel diameter D.
Show that
f v D5
(1) Observations. (i) The definition of friction factor shows that it depends on pressure drop,
diameter, length and mean velocity. (ii) Mean velocity is determined from flow rate
measurements and channel flow area.
(2) Problem Definition. Determine the dependency of friction factor on diameter.
(3) Solution Plan. Starting with the definition of friction factor f, express it in terms of diameter.
(4) Plan Execution.
(i) Assumptions. Continuum.
(ii) Analysis. Friction factor f is defined as
f
1 D 'p
2 L U u m2
(9.6b)
Pressure drop is determined by measuring the pressure at the inlet and outlet chambers. If
pressure drop at the inlet and outlet can be neglected, then ' p is independent of diameter. Mean
velocity is determine from flow rate measurements:
D2
m U um A U um S
(a)
4
where
A = flow area
m = mass flow rate
Solve (a) for u m
um
4m 1
S U D2
(b)
Substitute (b) into (9.6b)
f
1D
S2
'p
D4
2
2 L
16m
S 2 U' p
32 L m 2
D5
This result shows that f is proportional to the fifth power of diameter.
(iii) Checking.
Dimensional check: (f) should be dimensionless:
(f)
PROBLEM 9.2 (continued)
f
S 2 U (kg/m 3 ) ' p (kg/s 2 m)
32
2
L (m) m (kg/s)
2
D 5 (m 5 )
unity
(5) Comments. Accurate measurements of diameter or channel spacing in microchannels is
critical in obtaining accurate data on friction factor.
PROBLEM 9.3
Consider shear driven Couette flow between parallel plates separated by a distance H. The
lower plate is stationary while the upper plate moves with a velocity u s . Assume that no heat is
conducted through the lower plate and that the upper plate is maintained at uniform temperature
Ts . Taking into consideration dissipation, velocity slip and temperature jump, determine the
Nusselt number. Assume steady state ideal gas flow.
(1) Observations. (i) The determination of the Nusselt number requires the determination of the
temperature distribution. (ii) Temperature field depends on the velocity field. (iii) The velocity
field for Couette flow with a moving upper plate is give in Section 9.6.2. (iv) The solution to the
energy equation gives the temperature distribution.
(2) Problem Definition. Determine the temperature distribution for Couette flow with insulated
stationary plate and uniform temperature moving plate.
(3) Solution Plan. Start with the definition of the Nusselt number, use the velocity solution for
Couette flow of Section 9.6.2, formulate the energy equation and boundary conditions, and solve
for the temperature distribution.
(4) Plan Execution.
(i) Assumptions. (1) Constant
Ts
viscosity, conductivity and specific y
us
heat, (2) infinite plates, (3) uniform
boundary conditions, (4) uniform
H
spacing between plates, (5) no variation
u
x
of density along y, (6) no gravity, (7)
V u V T 1.0, (8) velocity field is
independent of temperature, (9) ideal gas, and (10) continuum, slip flow regime conditions
apply.
(ii) Analysis. The Nusselt number for flow between parallel plates is defined as
Nu
2 Hh
k
(a)
The heat transfer coefficient h for channel flow is defined as
wT ( H )
wy
Tm Ts
k
h
(b)
(b) into (a)
Nu
where
k
thermal conductivity of fluid
wT ( H )
wy
2 H
Tm Ts
(9.19)
PROBLEM 9.3 (continued)
T fluid temperature function (variable)
Tm fluid mean temperature
Ts plate temperature
The mean temperature Tm , as defined in Section 6.6.2, is
H
mc p Tm
W
³ U c uT dy
(9.21)
p
0
where
c p = specific heat
m = mass flow rate
T = temperature distribution
u = velocity distribution
W = plate width
U = density
The velocity distribution is given in Section 9
u
us
1
( y Kn)
1 2 Kn H
u
m U WH s
2
(9.14)
(9.16)
(9.16) into (9.21)
2
us H
Tm
H
³ uT dy
(9.22)
0
Temperature distribution is governed by the energy equation. Based pm the above assumptions,
energy equation (2.15) simplifies to
w 2T
(9.23)
k 2 P) 0
wy
where
§ wu ·
¨¨ ¸¸
© wy ¹
)
2
(9.24)
(9.24) into (9.23)
d 2T
dy 2
P § du ·
2
¨ ¸
k ¨© dy ¸¹
(9.25)
Note that T is independent of x. Substitute (9.14) into (9.25)
d 2T
dy 2
Defining the constant M as
Pª
us
º
«
k ¬ H (1 2 Kn) »¼
2
(c)
PROBLEM 9.3 (continued)
Pª
us
º
«
k ¬ H (1 2 Kn) »¼
M
2
(d)
Substituting (d) into (c)
d 2T
M
dy 2
(e)
This energy equation requires two boundary conditions. They are:
dT (0)
dy
0
(f)
The second boundary condition is at y = H. Plate temperature is specified at this boundary.
However, the boundary condition must be associated with the fluid at y = H and not the plate.
Knowing plate temperature, temperature jump condition (9.11) gives fluid temperature T(H). For
n H y and V T 1 , (9.11) gives
Ts
T (H ) 2J O dT ( H )
1 J Pr dy
(9.20)
Ts 2J O dT ( H )
1 J Pr dy
(g)
y 2 Cy D
(h)
Solve for T (H )
T (H )
Integration of (e) gives
T
M
2
where C and D are constants of integration. Application of boundary conditions (f) and (g) gives
the two constants:
C 0
(i)
and
H 2M
2J Kn 2
D Ts H M
(j)
2
J 1 Pr
Substituting into (h)
T
H 2M
2J Kn 2
y H M Ts
2
2
J 1 Pr
M
2
To determine the Nusselt number using (9.19), equation (k) is used to formulate
(k)
dT ( H )
and
dy
Tm . Differentiating (k)
dT ( H )
dy
HM
Tm is determined by substituting (9.14) and (k) into (9.22)
(l)
PROBLEM 9.3 (continued)
2
H (1 2 Kn)
Tm
H
³
0
(
M
y
Kn)( y 2 D) dy
H
2
(m)
where D is defined in (j). Evaluating the integral, gives
Tm
2 ª 1 2
1
º
H M KnH 2M » D
«
1 2 Kn ¬ 8
6
¼
Substituting (j) into the above
Tm
2
2 ª 1 2
1
2J Kn 2
2 º H M
H
M
KnH
M
H M Ts
«
»
1 2 Kn ¬ 8
6
2
J 1 Pr
¼
or
Tm
1 ª1 2
2
º 2J Kn 2
H M KnH 2M » H M Ts
«
1 2 Kn ¬ 4
3
¼ J 1 Pr
(n)
Using (l) and (n) into (9.19) gives the Nusselt number
Nu
2 H 2M
1 ª1 2
2
º 2J Kn 2
H M KnH 2M » H M
«
1 2 Kn ¬ 4
3
¼ J 1 Pr
This simplifies to
Nu
8(1 2 Kn)
8
8J (1 2 Kn) Kn
1 Kn J 1
Pr
3
(o)
(iii) Checking.
Dimensional check: (i) Noting that units of M are o C/m 2 , each term in (n) has units of
temperature. (ii) The Nusselt number in (o) is dimensionless.
Limiting check: Setting Kn = 0 in (o) gives
Nu o
8
(p)
This is the correct value of Nusselt number for macrochannel flow
(5) Comments.
(i) The Nusselt number is independent of the Reynolds number. This is also the case with
macrochannel flows.
(ii) Unlike macrochannels, the Nusselt number depends on the fluid.
(iii) The Knudsen number in (o) represents the effect of rarefaction while the third term in the
denominator represents the effect of temperature jump. Both act to reduce the Nusselt number.
PROBLEM 9.3 (continued)
(iv) If dissipation is neglected ( M
as
0) , equation (k) gives the corresponding temperature solution
T
Ts
Thus, the temperature is uniform and no heat transfer takes place.
PROBLEM 9.4
A large plate moves with constant velocity u s
parallel to a stationary plate separated by a
distance H. An ideal gas fills the channel
formed by the plates. The stationary plate is
at temperature To and the moving plate is at
temperature Ts . Assume laminar flow and
take into consideration dissipation and
velocity slip and temperature jump:
Ts
y
x
us
H
u
To
(a) Show that temperature distribution is given by
P u s2
ª 2J Kn y
Ts To
y2 º
»
2 « J 1 Pr
2
2J Kn
H H ¼
2k (1 2 Kn) ¬
1 2
J 1 Pr
(b) Determine the heat flux at the plates.
T
To ª 2J Kn y º
« J 1 Pr H »
¼
¬
(1) Observations. (i) Temperature distribution depends on the velocity field. (ii) The velocity
field for Couette flow with a moving upper plate is give in Section 9.6.2. (iii) The solution to the
energy equation gives the temperature distribution. (iv) Two temperature boundary conditions
must be specified. (v) Temperature distribution and Fourier’s law give surface heat flux.
(2) Problem Definition. Determine the temperature distribution for Couette flow with specified
surface temperature on both plates.
(3) Solution Plan. Use the velocity solution for Couette flow of Section 9.6.2, formulate the
energy equation and boundary conditions, and solve for the temperature distribution.
(4) Plan Execution.
(i) Assumptions. (1) Constant viscosity, conductivity and specific heat, (2) infinite plates, (3)
uniform boundary conditions, (4) uniform spacing between plates, (5) no variation of density
along y, (6) no gravity, (7) V u V T 1.0, (8) the velocity field is independent of temperature,
and (9), ideal gas, and (10) continuum, slip flow regime conditions apply.
(ii) Analysis.
(a) Temperature distribution is governed by the energy equation.
assumptions, energy equation (2.15) simplifies to
k
w 2T
wy 2
)
where
k = thermal conductivity of fluid
u = axial velocity
P = fluid viscosity
P)
§ wu ·
¨¨ ¸¸
© wy ¹
0
Based pm the above
(9.23)
2
(9.24)
PROBLEM 9.4 (continued)
(9.24) into (9.23)
d 2T
dy 2
P § du ·
2
¨ ¸
k ¨© dy ¸¹
(9.25)
The velocity distribution is given in Section 9
u
us
1
( y Kn)
1 2 Kn H
d 2T
Pª
us
º
«
k ¬ H (1 2 Kn) »¼
dy 2
(9.14)
2
(a)
Defining the constant M as
M
Pª
us
º
«
k ¬ H (1 2 Kn) »¼
2
(b)
Substituting (b) into (a)
d 2T
dy 2
M
(c)
This energy equation requires two boundary conditions. The temperature of each plate is
specified. However, the boundary conditions must be associated with the fluid at y = H and not
the plates. Knowing plate temperature, temperature jump condition (9.11) gives fluid
temperature. At y 0, fluid temperature T(0) is obtained from (9.11). At n = y = 0 and for
V T 1 , equation (9.11) gives
T (0)
Similarly, at n = H – y and V T
To 2J O dT (0)
1 J Pr dy
(d)
1 , (9.11) gives
T (H )
Ts 2J O dT ( H )
1 J Pr dy
(e)
y 2 Cy D
(h)
Integration of (c) gives
T
M
2
where C and D are constants of integration. Application of boundary conditions (d) and (e) gives
the two constants:
MH 2
C
and
2
2J Kn
M H 2 Ts To
J 1 Pr
2J Kn
H 2H
J 1 Pr
(i)
PROBLEM 9.4 (continued)
To D
2J Kn
HC
J 1 Pr
(j)
Substitute (i) and (j) into (h)
T
MH 2 ª
2
Ts To
2J Kn º
y2
y
« 2 »
4J Kn
H J 1 Pr »¼
«¬ H
1
J 1 Pr
ªy
2J Kn º
« H J 1 Pr » To
¬
¼
(k)
Using the definition of M in (b) into (k) and rearranging, gives
P u s2
ª 2J Kn y
Ts To ª 2J Kn y º
y2 º
(l)
»
»
«
«
2k (1 2 Kn) 2 ¬ J 1 Pr H H 2 ¼ 1 4J Kn ¬ J 1 Pr H ¼
J 1 Pr
This result can be written in dimensionless form as
ª 2J Kn y
P u s2
T To
ª 2J Kn y º
1
y2 º
(m)
«
2
2»
4J Kn «¬ J 1 Pr H »¼
Ts To 2k (1 2 Kn) (Ts To ) ¬ J 1 Pr H H ¼
1
J 1 Pr
T
To (b) Heat flux q cc. Application of Fourier’s law at y = 0
q cc(0)
dT (0)
dy
k
(n)
(l) into (m)
q cc(0)
P u s2
2 H (1 2 Kn)
2
k (Ts To )
4J Kn
H
H
J 1 Pr
(o)
(o) is written in dimensionless form as
q cc(0)
k (Ts To )
H
P u s2
2
2(1 2 Kn) (Ts To )
1
4J Kn
1
J 1 Pr
(p)
Similarly, at y = H
q cc( H )
k
dT ( H )
dy
(q)
(l) into (q)
q cc( H )
P u s2
2 H (1 2 Kn)
Written in dimensionless form, (r) becomes
2
k (Ts To )
4J Kn
H
H
J 1 Pr
(r)
PROBLEM 9.4 (continued)
q cc( H )
k (Ts To )
H
P u s2
2
2(1 2 Kn) (Ts To )
1
4J Kn
1
J 1 Pr
(s)
(iii) Checking.
Dimensional check: (i) Noting that units of M are o C/m 2 , each term in (k) has units of
temperature. (ii) Each term in (o) and (r) has units of heat flux. (iii) each term in (m), (p) and (s)
is dimensionless.
Limiting check: (i) Since the velocity profile is linear, dissipation is uniform. Thus, if Ts To ,
heat flux at each plate should be equal in magnitude and opposite in direction. Setting Ts To in
(o) and (r) gives
q cc(0)
q cc( H )
P u s2
(t)
2 H (1 2 Kn) 2
P u s2
(u)
2 H (1 2 Kn) 2
(ii) If dissipation is neglected, temperature distribution should be linear. Setting P
T
To Ts To ª 2J Kn y º
4J Kn «¬ J 1 Pr H »¼
1
J 1 Pr
0 in (l) gives
(s)
(iii) If dissipation and rarefaction are neglected, temperature distribution should be the same as
one dimensional conduction. Setting Kn = 0 in (l) gives
T
To (Ts To )
y
H
(v)
(5) Comments.
(i) The Knudsen number in (l), (o) and (r) represents the effect of rarefaction while the Prandtl
number terms represents the effect of temperature jump.
(ii) The solution is governed by two parameters:
Dissipation parameter =
P u s2
k (1 2 Kn) 2 (Ts To )
Temperature jump parameter =
J
Kn
J 1 Pr
(w)
(x)
PROBLEM 9.5
Consider Couette flow between two parallel plates separated by a distance H. The lower
plate moves with velocity u s1 and the upper plate moves in the opposite direction with
velocity u s 2 . The channel is filled with ideal gas. Assume velocity slip conditions,
determine the mass flow rate. Under what condition will the net flow rate be zero?
(1) Observations. (i) To determine mass flow rate it is necessary to determine the
velocity distribution. (ii) Velocity slip takes place at both boundaries of the flow channel.
(iii) Because plates move in opposite directions, the fluid moves in both directions. This
makes it possible for the net flow rate to be zero.
(2) Problem Definition. Determine the velocity distribution in the channel.
(3) Solution Plan. Apply the Navier-Stokes equations and formulate the velocity slip
boundary conditions. Follow the analysis of Section 9.6.2 and Example 9.1.
us2 y
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar
H
u
flow, (3) one-dimensional (no variation with
us1
axial distance x and normal distance z), (4) slip
x
flow regime (0.001 < Kn < 0.1), (5) ideal gas,
(6) constant viscosity (7) negligible lateral variation of density, (8) the velocity
accommodation coefficient is equal to unity, V u 1.0, (9) continuum, slip flow regime
conditions apply, (10) ideal gas, and (11) no gravity.
(ii) Analysis. Mass flow rate is given The flow rate, m , for a channel of width W is
given by
H
m W
³ U u dy
(9.15)
0
where
u axial velocity
U density
To determine u we follow the analysis of Section 9.6.2. The axial component of the
Navier-Stokes equations for Couette flow between parallel plates is given by (9.12)
d 2u
dy 2
(9.12)
0
Boundary conditions for (9.12) are given by (9.10)
2 Vu
wu ( x,0)
Vu
wn
Applying (9.10) to the lower surface, n y 0 , and setting V u
du (0)
u (0) u s1 O
dy
u ( x ,0 ) u s
O
(9.10)
1
(a)
PROBLEM 9.5 (continued)
For the upper surface, n = H – y, (9.10) gives
u(H ) u s2
O
du ( H )
dy
(b)
The solution to (9.12) is
u
Ay B
(c)
Boundary conditions (a) and (b) give the two constants of integration A and B
A
u s1 u s 2
, B
H (1 2 Kn)
u s1 (u s1 u s 2 ) Kn
(1 2 Kn)
(d)
where Kn is the Knudsen number, defined as
Kn
O
(9.13)
H
Substituting (d) into (c)
u
u s1 (u s1 u s 2 )
y
( Kn )
(1 2 Kn)
H
(e)
Substituting (e) into (9.15) and noting that U is assumed constant along y, gives
H
m
UW
³
0
u s1 u s 2 y
ª
º
«u s1 1 2 Kn ( H Kn)» dy
¬
¼
(f)
Evaluating the integral
m
UWH
2
>u s1 u s 2 @
(g)
Examination of this result shows that the net mass flow rate is zero when the two
velocities are the same. That is m = 0 for
u s1
us2
(h)
(iii) Checking.
Dimensional check: Equation (g) has the correct units for mass flow rate.
Limiting Check: For the special case of stationary lower plate, u s 2
reduces to
UWH
m
u s1
2
This agrees with (9.17) of Section 9.6.2.
0, equation (g)
(i)
Boundary conditions check: Solution (d) satisfies boundary conditions (a) and (b).
(5) Comments. (i) The effect of slip is to decrease fluid velocity at the upper and lower
surfaces. (ii) Because the velocity distribution is linear, slip velocity is the same for both
plates. (iii) The mass flow rate is independent of Knudsen number.
PROBLEM 9.6
Determine the frictional heat generated by the fluid in Example 9.1.
(1) Observations. (i) In Example 9.1, Couette
T2
us2 y
flow between parallel plates is used to model the
flow in the channel between the shaft and
H
housing. (ii) At steady state, the heat generated
u
us1
due to friction (dissipation) is equal to the net
heat conducted from the channel. (iii) Since no
x
heat is transferred to the shaft, the net heat
leaving the channel is equal to the heat
conducted through the housing surface. (iv) Velocity slip takes place at both boundaries of the
flow channel. (v) To determine heat transfer rate it is necessary to determine fluid temperature
distribution. This requires the determination of the velocity field.
(2) Problem Definition. Determine the velocity and temperature distribution in the channel.
(3) Solution Plan. (i) Model channel flow as Couette flow between parallel plates. (ii) Apply
Fourier’s law at the housing surface to determine heat leaving the channel. (iii) Apply the
Navier-Stokes equations and formulate the velocity slip boundary conditions. Follow the analysis
of Section 9.6.2 and Example 9.1. (iv) Use the energy equation to determine the temperature
distribution
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) one-dimensional (no variation with axial
distance x and normal distance z), (4) slip flow regime (0.001 < Kn < 0.1), (5) ideal gas, (6)
constant viscosity (7) negligible lateral variation of density, (8) the velocity accommodation
coefficients are equal to unity, V u V T 1.0, (9) continuum, slip flow regime conditions apply,
(10) ideal gas, and (11) no gravity.
(ii) Analysis. Apply Fourier’s law at the housing surface (upper plate)
q( H )
kA
dT ( H )
dy
(a)
where
A
k
q
T
surface area
fluid thermal conductivity
heat transfer rate
fluid temperature distribution
Surface area A for a shaft of length L is
A
2S ( R H ) L
(b)
(b) into (a)
q( H )
2k S ( R H ) L
dT ( H )
dy
(c)
PROBLEM 9.6 (continued)
Flow Field. To determine u we follow the analysis of Section 9.6.2. The axial component of the
Navier-Stokes equations for Couette flow between parallel plates is given by (9.12)
d 2u
(9.12)
0
dy 2
Boundary conditions for (9.12) are given by (9.10)
2 Vu
u ( x ,0 ) u s
Applying (9.10) to the lower surface, n
O
Vu
wu ( x,0)
wn
0 , and setting V u
y
u (0) u s1
(9.10)
1
du (0)
dy
O
(d)
For the upper surface, n = H – y, (9.10) gives
O
u(H ) u s2
du ( H )
dy
(e)
The solution to (9.12) is
Ay B
u
(f)
Boundary conditions (a) and (b) give the two constants of integration A and B
A
u s1 u s 2
, B
H (1 2 Kn)
u s1 (u s1 u s 2 ) Kn
(1 2 Kn)
(g)
where Kn is the Knudsen number, defined as
Kn
O
(9.13)
H
Substituting (g) into (f)
u
u s1 (u s1 u s 2 )
y
( Kn )
(1 2 Kn)
H
(h)
Temperature Field. Temperature distribution is governed by the energy equation. Based pm
the above assumptions, energy equation (2.15) simplifies to
k
w 2T
P)
wy 2
0
(9.23)
where
)
§ wu ·
¨¨ ¸¸
© wy ¹
2
(9.24)
(9.24) into (9.23)
d 2T
dy 2
P § du ·
¨ ¸
k ¨© dy ¸¹
2
(9.25)
PROBLEM 9.6 (continued)
Note that T is independent of x. Substitute (h) into (9.25)
d 2T
dy 2
P ª u s1 u s 2 º
2
k «¬ H (1 2 Kn) »¼
(i)
Defining the constant M as
M
P ª us us2 º
2
k «¬ H (1 2 Kn) »¼
(j)
Substituting (d) into (c)
d 2T
dy 2
M
(k)
This energy equation requires two boundary conditions. They are:
dT (0)
dy
0
(l)
The second boundary condition is at y = H. Plate temperature is specified as T2 at this boundary.
However, the boundary condition must be associated with the fluid at y = H and not the plate.
Knowing plate temperature, temperature jump condition (9.11) gives fluid temperature T(H). For
n H y and V T 1 , (9.11) gives
2J O dT ( H )
T2 T ( H ) (9.20)
1 J Pr dy
Solve for T (H )
2J O dT ( H )
T ( H ) T2 (m)
1 J Pr dy
Integration of (k) gives
T
E
2
y 2 Cy D
(n)
where C and D are constants of integration. Application of boundary conditions (l) and (m) gives
the two constants:
C 0
(o)
and
H 2M
2J Kn 2
D T2 H M
(p)
J 1 Pr
2
Substituting into (n)
H 2M
2J Kn 2
T y H M T2
J 1 Pr
2
2
Rewriting (q) in dimensionless form
M
2
T T2
H 2E
1ª
y 2 º 2J Kn
«1 »
2 ¬« H 2 ¼» J 1 Pr
(q)
PROBLEM 9.6 (continued)
Using the definition of M in the above
T T2
P ª u s u s2 º
k «¬ (1 2 Kn) »¼
2
1ª
y 2 º 2J Kn
1
«
»
2 «¬ H 2 »¼ J 1 Pr
Frictional Heat. Using (r) to form the temperature gradient at y
gives the heat generated by fluid friction
q( H )
2S ( R H ) P
L ª us us2 º
H «¬ (1 2 Kn) »¼
(r)
H and substituting into (c)
2
(s)
(iii) Checking.
Dimensional check: Each term in (h) has units of velocity. Each term in (q) has units of
temperature. Each term in (r) is dimensionless. q(H) in (s) is expressed in watts.
Governing equations check: Velocity solution (h) satisfies (9.12). Temperature solution (r)
satisfies (9.25).
Boundary conditions check: Velocity solution (h) satisfies boundary conditions (d) and (e).
Temperature solution (r) satisfies boundary conditions (l) and (m).
Limiting checks:
(i) If the two plates are stationary (u s1 u s 2 M 0) , there is no fluid motion, gas temperature is
uniform, and there is no frictional energy. Setting u s1 u s 2 0 in (h) gives u = 0. Setting M 0
in (q) gives T T2 . Setting u s1 u s 2 0 in (s) gives q(H) = 0.
(5) Comments. (i) The velocity field is governed by a single parameter, Kn. The temperature
2J Kn
.
field is governed by the parameter
J 1 Pr
(ii) Equation (s) shows that the narrower the gap H between the rotor and housing, the greater is
the frictional energy. In addition, frictional energy decreases as the Knudsen number is
increased.
PROBLEM 9.7
Consider shear driven Couette flow between
parallel plates. The upper plate moves with velocity
u s and is maintained at uniform temperature Ts .
The lower plate is heated with uniform flux q occ . The
fluid between the two plates is an ideal gas. Taking
into consideration velocity slip, temperature jump,
and dissipation, determine the temperature of the
lower plate.
Ts
y
u
x
us
H
qocc
(1) Observations. (i) To determine the temperature of the lower plate, fluid temperature
distribution must be known. (ii) Temperature distribution depends on the velocity field. (iii) The
velocity field for Couette flow with a moving upper plate is given in Section 9.6.2. (iv) The
solution to the energy equation gives the temperature distribution. (v) Two temperature boundary
conditions must be specified.
(2) Problem Definition. Determine the temperature distribution for Couette flow with uniform
flux at the lower plate and specified temperature at the upper plate.
(3) Solution Plan. Use the velocity solution for Couette flow of Section 9.6.2, formulate the
energy equation and boundary conditions, and solve for the temperature distribution.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) Constant viscosity, conductivity and
specific heat, (4) infinite plates, (5) uniform boundary conditions, (6) uniform spacing between
plates, (7) no variation of density and pressure along y, (8) no gravity, (7) V u V T 1.0, (9) the
velocity field is independent of temperature, (10), ideal gas, and (11) continuum, slip flow
regime conditions apply.
(ii) Analysis. Surface temperature is related to fluid temperature through temperature jump
condition (9.11)
2 V T 2J O wT ( x,0)
V T 1 J Pr w n
T ( x , 0 ) Ts
Solving the above for plate temperature at y = 0, To , and assuming V T
To
T ( x,0) 2J O wT ( x,0)
1 J Pr w n
(9.11)
1 , gives
(a)
where T (x,0) fluid temperature at the lower plate n = y . Thus fluid temperature distribution is
needed to determine To . Temperature distribution is governed by the energy equation. Based on
the above assumptions, energy equation (2.15) simplifies to (see Section 9.6.2)
k
where
w 2T
wy 2
P)
0
(9.23)
PROBLEM 9.7 (continued)
§ wu ·
¨¨ ¸¸
© wy ¹
)
2
(9.24)
where
k = thermal conductivity of fluid
u = axial velocity
P = fluid viscosity
Noting that velocity and temperature are independent of axial distance, (9.24) into (9.23)
d 2T
dy 2
P § du ·
2
¨ ¸
k ¨© dy ¸¹
(9.25)
The velocity distribution is given in Section 9.6.2
u
us
1
( y Kn)
1 2 Kn H
d 2T
Pª
us
º
«
k ¬ H (1 2 Kn) »¼
dy 2
(9.14)
2
(b)
Defining the constant M as
M
Pª
us
º
«
k ¬ H (1 2 Kn) »¼
2
(c)
Substituting (c) into (b)
d 2T
dy 2
M
(d)
This energy equation requires two boundary conditions. Heat flux is specified at the lower plate.
Fourier’s law gives
dT (0)
k
q occ
(e)
dy
Surface temperature is specified at the upper plate. However, this boundary condition must be
associated with the fluid at y = H and not the plate. Knowing plate temperature, temperature
jump condition (9.11) gives fluid temperature. Thus at n = H – y and V T 1 , (9.11) gives
T (H )
Ts 2J O dT ( H )
1 J Pr dy
(f)
y 2 Cy D
(g)
Integration of (d) gives
T
M
2
where C and D are constants of integration. Application of boundary conditions (e) and (f) gives
the two constants:
PROBLEM 9.7 (continued)
C
q occ
,
k
qocc H
k
D
ª
2J Kn º M H 2
1
« J 1 Pr » 2
¬
¼
ª
4J Kn º
«1 J 1 Pr » Ts
¬
¼
(h)
ª
M 2
4J Kn º qocc
«1 J 1 Pr » k y 2 y
¬
¼
(i)
Substitute (h) into (g)
T
Ts qocc H
k
ª
2J Kn º M H 2
1
« J 1 Pr » 2
¬
¼
Using the definition of M in (c), the above becomes
T
q cc H
Ts o
k
º
ª
2J Kn º P ª u s
«1 J 1 Pr » 2k « (1 2 Kn) »
¬
¼
¬
¼
2
2
º y2
ª
4J Kn º qocc
1 P ª us
1
y
(j)
« J 1 Pr » k
2 k «¬ (1 2 Kn) »¼ H 2
¬
¼
Surface temperature of the lower plate is determined by substituting (j) into (a) and setting y = 0
To
ª q cc H P
ºª
u s2
4J Kn º
Ts « o 1
»
«
2k (1 2 Kn) 2 »¼ ¬ J 1 Pr »¼
«¬ k
(k)
(iii) Checking.
Dimensional check: (i) Noting that units of M are
temperature.
o
C/m 2 , each term in (j) has units of
Limiting check: (i) If dissipation is neglected, temperature distribution should be linear. Setting
P 0 in (j) gives
T
Ts qocc H
k
ª
2J Kn º qocc
«1 J 1 Pr » k y
¬
¼
(l)
(ii) If dissipation and rarefaction are neglected, the process reduces to pure conduction. Setting
P Kn 0 in (j) gives
q cc ª
yº
(m)
T Ts o «1 »
k ¬ H¼
This is the one dimensional conduction solution to the problem.
(5) Comments. (i) The Knudsen number in (j) and (l) represents the effect of rarefaction while
the Prandtl number term represents the effect of temperature jump.
(ii) Fluid temperature adjacent to the lower surface is obtained by setting y = 0 in (j)
T (0)
Ts qocc H
k
ª
2J Kn º P
«1 J 1 Pr » 2k
¬
¼
ª us
º
« (1 2 Kn) »
¬
¼
2
ª
4J Kn º
«1 J 1 Pr »
¬
¼
(n)
PROBLEM 9.7 (continued)
To examine the difference between plate and fluid temperature at y = 0, (n) is subtracted from (k)
2J Kn qocc H
J 1 Pr k
To T (0)
(o)
This result shows that departure of plate temperature from fluid temperature at y = 0 increases
with increasing heat flux and rarefaction.
(iii) To identify the governing parameters, solution (j) is expressed in dimensionless form
T Ts
qocc H
k
ª
P
2J Kn º
«1 J 1 Pr » 2q cc H
¬
¼
o
ª us
º
« (1 2 Kn) »
¬
¼
2
ª
P
4J Kn º y
«1 J 1 Pr » h 2q cc H
¬
¼
o
2
ª us
º y2
« (1 2 Kn) »
2
¬
¼ H
(p)
This result shows that temperature distribution is governed by two parameters:
º
2J Kn
P ª us
and
«
J 1 Pr
2q occ H ¬ (1 2 Kn) »¼
2
(q)
PROBLEM 9.8
Pressure distribution in Poiseuille flow between parallel plates is given by
p( x)
po
2
ª
ª
pi º
pi2
pi º x
Kn
6 Kno «6 Kno (
1
)
12
(
1
)»
«
o
»
po ¼
p o ¼» L
p o2
«¬
¬
(9.35)
This equation was derived in Section 9.6.3 using the continuity equation to determine the y
velocity component v. An alternate approach to derive (9.35) is based on the condition that for
steady state the flow rate is invariant with axial distance x. That is
dm
dx
d ª
«2W
dx ¬«
³
H /2
º
U udy »
0
¼»
0
where W is channel width. Derive (9.35) using this approach.
(1) Observations. (i) To use the proposed approach, the solution to the axial velocity distribution
must be know. (ii) The velocity distribution for Poiseuille flow between parallel plates is given
by equation (9.30) of Section 9.6.3.
(2) Problem Definition. Determine the mass flow rate.
(3) Solution Plan. Use the solution to the axial velocity for Poiseuille flow, equation (9.30), to
determine the mass flow rate.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) constant viscosity (4) infinite plates,
(5) uniform spacing between plates, (6) no variation of density and pressure along y, (7) no
gravity, (8) V u 1.0, (9) the velocity field is independent of temperature, (10) ideal gas, and
(11) continuum, slip flow regime conditions apply.
(ii) Analysis. To derive pressure solution (9.35), it is proposed to use the following
conservation of mass equation
dm
dx
d ª
«2W
dx «
¬
³
H /2
0
º
U udy »
»¼
0
(a)
where W = plate width. The solution to the axial velocity u is given by (9.30)
u
H 2 dp ª
y2 º
«1 4 Kn( p) 4 2 »
8P dx ¬
H ¼
(9.30)
Substituting (9.30) into (a) and noting that U varies along x and is assumed constant along y, we
obtain
H /2
dm d ª
H 2 dp ª
y2 º º
«2WU
1
4
(
)
4
dy » 0
Kn
p
«
2»
8
P
dx dx «
dx
H
¬
¼ »¼
0
¬
This result simplifies to
³
PROBLEM 9.8 (continued)
dm
dx
d ª dp
«U
dx « dx
¬
H /2
³
0
ª
y2 º º
1
4
Kn
(
p
)
4
«
» dy »
H 2 ¼ »¼
¬
0
(b)
Evaluating the integral in (b)
d ª dp § H
·º
U ¨ 2 H Kn( p) ¸»
«
dx ¬ dx © 3
¹¼
0
(c)
To proceed, the density and Knudsen number in (c) must be expressed in terms of pressure. Ideal
gas law (9.31) gives
p
U
(d)
RT
The Knudsen number is expressed in terms of pressure in (9.33)
Kn
O
P
S
H
H
2
RT
1
p
(9.33)
(d) and (9.33) into (c)
d ª p dp § H
S
1 ·º
¨ 2P
¸»
RT
«
dx «¬ RT dx ¨© 3
p ¸¹»¼
2
0
Assuming isothermal flow, the above simplifies to
d ª dp § 1 2P S
1 ·º
RT ¸¸»
« p ¨¨ dx «¬ dx © 3 H 2
p ¹»¼
0
(e)
Integrating (e) once
p
dp § 1 2P S
1 ·
¨ RT ) ¸¸
¨
dx © 3 H 2
p ¹
C
Rewriting the above
dp
1 pdp 2 P S
RT
dx
H 2
3 dx
C
(f)
Integrating again
1 2 P
p H
6
2SRT p
Cx D
(g)
po
(h)
The boundary conditions on p are
p (0)
pi ,
p ( L)
Here L is channel length. Equation (g) and (h) are identical to (o) and (q) of Section 9.6.3. Thus
the solution to p is the same for both, given in (9.35)
2
ª
ª
pi º
pi2
p ºx
(9.35)
6 Kno «6 Kno (
1
) 12 Kno (1 i )»
«
»
2
po ¼
p o ¼» L
po
¬
¬«
(4) Comments. This approach for determining p(x) is simpler than that used in Section 9.6.3
where it was necessary to first determine the normal velocity component v.
p( x)
po
PROBLEM 9.9
One of the factors affecting mass flow rate through a microchannels is channel height H. To
examine this effect, consider air flow through two microchannels. Both channels have the same
length, inlet pressure and temperature and outlet pressure. The height of one channel is double
that of the other. Compute the mass flow ratio for the following case:
H 1 5 P m, H 2 10 P m, pi 420 kPa, p o 105 kPa, Ti 30 o C
(1) Observations. (1) This is a pressure driven microchannel Poiseuille flow between parallel
plates. (ii) The solution to mass flow rate through microchannels is given in Section 9.6.3. (iii)
Channel height affects the Knudsen number.
(2) Problem Definition. Determine the mass flow rate for microchannel Poiseuille flow between
parallel plates.
(3) Solution Plan. Apply the mass flow solution, equation (9.39), to two channels having
different heights and take their ratio.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) constant viscosity (4) infinite plates,
(5) uniform spacing between plates, (6) no variation of density and pressure along y, (7) no
gravity, (8) V u 1.0, (9) the velocity field is independent of temperature, (10) ideal gas, (11)
continuum, slip flow regime conditions apply, and (12) fully developed flow.
(ii) Analysis. The mass flow rate through Poiseuille flow microchannels is given by equation
(9.39):
º
pi
1 W H 3 po2 ª pi2
1)»
« 2 1 12 Kno (
24 P LRT «¬ po
po
»¼
m
(9.39)
Apply (9.39) to two channels having heights H 1 and H 2 , and outlets Knudsen numbers Kno1
and Kno 2
m1
º
pi
1 W H 13 p o2 ª pi2
1)»
« 2 1 12 Kno1 (
24 P LRT «¬ po
po
»¼
(a)
m2
º
pi
1 W H 23 p o2 ª pi2
1)»
« 2 1 12 Kno 2 (
24 P LRT «¬ po
po
»¼
(b)
Take the ratio of (b) and (a)
2
m2
m1
ª pi º
pi
1)
» 1 12 Kno 2 (
3 «
po
ª H 2 º ¬ po ¼
«
»
¬ H1 ¼ ª p º 2
pi
i
1)
« » 1 12 Kno1 (
po
¬ po ¼
Equation (9.34) gives the outlet Knudsen number
(c)
PROBLEM 9.9 (continued)
Kno
P
S
H po
2
RT
(9.34)
Apply (9.34) to the two channels
Kno1
Kno 2
P
S
H po1
2
P
S
H po 2
2
RT
(d)
RT
(e)
(iii) Computations. The following date is given
5 u 10 -6 m
H1
5Pm
H2
10 P m 10 u 10 -6 m
po
105 kPa
pi
420 kPa
Ti # To
kg
s -m
kg
420,000 2
s -m
105,000
2
30 o C
Properties of air at this temperature are
Pr
0.712
m2
R
287 2
s -K
6
kg
P 18.65 u 10
s-m
J
287
kg - K
Substituting into (d) and (e)
kg
)
S
m2
s-m
(287)( 2
)(303)(K )
kg
2
-6
s
K
5 u 10 (m) (105,000)( 2
)
s -m
18.65 u 10 6 (
Kno1
18.65 u 10 6 (
Kno 2
kg
)
s-m
kg
10 u 10 (m) (105,000)( 2
)
s -m
-6
Substituting into (c)
S
2
(287)(
m2
)(303)(K )
s2 - K
0.01313
0.006566
PROBLEM 9.9 (continued)
2
m2
m1
ª 420 º
420 1)
3 «
» 1 12 (0.006566)(
105
ª10 º ¬ 105 ¼
«¬ 5 »¼
2
420
ª 420 º
1
12
(
0
.
01313
)
1)
(
« 105 »
105
¼
¬
7.878
(4) Checking.
Dimensional check: Computations showed that units of Kn in (d) and (e) are dimensionless.
Limiting check: If H 1
H 2 , the mass ratio should be unity. Setting H 1
Kno1
Setting H 1
H 2 and Kno1
H 2 in (d) and (e) gives
Kno 2
Kno 2 in (c) gives
m1
m2
(5) Comments. (i) The effect of channel size on mass flow rate is significant. (ii) Setting
Kno1 Kno 2 0 in (c) gives the mass ratio for macrochannels
m2
m1
8
This indicates that the effect of rarefaction on the mass ratio is 1.5%.
PROBLEMT 9.10
A micro heat exchanger consists of rectangular channels of height H
25 P m, width
W 600 P m, and length L 10 mm. Air enters the channels at temperature Ti 20 o C and
pressure pi 420 kPa. The outlet pressure is po 105 kPa. The air is heated with uniform
surface heat flux q csc 1100 W/m 2 . Taking into
consideration velocity slip and temperature jump,
assume fully developed conditions, compute the
following:
(a) Mass flow rate, m.
(b) Mean outlet temperature, Tmo .
(c) Heat transfer coefficient at the outlet, h( L).
(d) Surface temperature at the outlet, Ts ( L).
q cc
H
q cc
L
W
(1) Observations. (i) This is a pressure driven microchannel Poiseuille flow. (ii) Since channel
height is much smaller than channel width, the rectangular channel can be modeled as Poiseuille
flow between parallel plates. (iii) Channel surface is heated with uniform flux. (iv) The solution
to mass flow rate, temperature distribution, and Nusselt number for fully developed Poiseuille
channel flow with uniform surface flux is presented in Section 9.6.3.
(2) Problem Definition. Determine flow and heat transfer characteristics of fully developed
Poiseuille channel flow with uniform surface heat flux.
(3) Solution Plan. Apply the analysis and results of Section 9.6.3.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) constant viscosity (4) infinite plates,
(5) uniform spacing between plates, (6) no variation of density and pressure along y, (7)
V u V T 1.0, (8) no gravity, (9) the velocity field is independent of temperature, (10) ideal
gas, (11) continuum, slip flow regime conditions apply, and (12) fully developed flow.
(ii) Analysis.
(a) Mass flow rate m. Equation (9.39) gives the mass flow rate through the channel:
m
º
pi
1 W H 3 po2 ª pi2
1)»
« 2 1 12 Kno (
24 P LRT «¬ po
po
»¼
where
H channel height 25 P m 25 u 10 -6 m
Kno outlet Knudsen number = Kn( po)
L
channel length = 0.01 m
(9.39)
PROBLEMT 9.10 (continued)
pi
inlet pressure
po
outlet pressure 105 kPa
q csc
1100 W/m 2
W
channel width
R
gas constant
420 kPa
420,000 kg/s 2 - m
105,000 kg/s 2 - m
600 u 10 6 m
287 J/kg - K
287 m 2 / s 2 - K
P 18.17 u 10 6 kg/s - m
Equation (9.34) gives the outlet Knudsen number
Kno
where To
Kn( po )
O
H
P
H
1
S
RTo
po
2
(9.34)
Tmo is the mean outlet temperature.
(b) Mean outlet temperature, Tmo . The local mean temperature Tm (x ) is given by equation
(9.60):
2q csc
x Tmi
(9.60)
Tm ( x)
Uc p u m H
where
cp
specific heat 998.3 J/kg o C
q csc
surface heat flux 1100 W/m 2
Tmi
mean inlet temperature Ti
m
mean velocity,
s
um
U
20 o C
density, kg/m 3
The mean outlet temperature is obtained by setting x = L in (9.60)
Tmo
Tm ( L)
2q csc
L Tmi
U c pum H
The product U u m is determined from the mass flow rate using the continuity equation:
m
U um
HW
(b) into (a)
2 q csc
Tmo Tm ( L)
LW Tmi
c pm
k
0.02564 W/m o C
Ti # To
20 o C
(a)
(b)
(c)
PROBLEMT 9.10 (continued)
(c) Heat transfer coefficient at the outlet, h(L). The Nusselt number is used to determine the
heat transfer coefficient. Nusselt number for channel flow is defined as
Nu
2hH
k
(d)
where
k
thermal concuctivity
0.02564
W
m o C
Applying (d) at the outlet, x = L and solving for h(L)
k
Nu ( L)
h( L)
2H
The Nusselt number is given by (9.64)
Nu
(e)
2
3
(1 6 Kn)
­1
5
1
13
13 º ½ 2J 1
ª
( Kn) 2 Kn Kn
® Kn ¾
«
48 (1 6 Kn) ¬
40
560 »¼ ¿ J 1 Pr
¯2
(9.64)
where
Kn local Knudsen number
Pr Prandtl number 0.713
J = specific heat ratio = 1.4
Evaluation (9.64) at x = L where Kn
Nu
Kno
2
­1
3
5
1
13
13 º ½ 2 J 1
ª
Kno Kno zzz
( Kno ) 2 ® Kno ¾
«
(1 6 Kno ) ¯ 2
48 (1 6 Kno ) ¬
40
560 »¼ ¿ J 1 Pr
(f)
(d) Surface temperature at the outlet, Ts (L). Surface temperature distribution is given by
(9.63):
3q csc H ª 1
5 º 2J q csc H
(9.63)
Ts ( x )
Kn
Kn g ( x)
k (1 6 Kn) «¬ 2
48 »¼ J 1 kPr
where g(x) is given by (9.62):
g ( x)
Tmi 2q csc
3q csc H
x
U c pum H
k (1 6 Kn) 2
Substituting (b) into (9.62), setting x = L and Kn
g ( L) Tmi 13
13 º
ª
2
«¬( Kn) 40 Kn 560 »¼
(9.62)
Kno , gives
2q cscW
3q csc H
L
c pm
k (1 6 Kn) 2
13 º
ª
2 13
«¬( Kno ) 40 Kno 560 »¼
(g)
PROBLEMT 9.10 (continued)
To determine surface temperature at the outlet, Ts (L), the Knudsen number in (9.62) and (9.63) is
evaluated at outlet pressure and g(x) is evaluated at x = L.
(iii) Computations.
(a) Mass flow rate m. Equation (9.39) for m is based on the assumption that the flow is
isothermal. Since the outlet temperature To is not yet determined, as a first approximation we
assume To { Ti 293 K in (9.34)
18.17 u 10 6 ( kg/s - m)
Kno
-6
2
S
25 u 10 (m)(105,000)(kg/s - m)
2
(287)(m 2 / s 2 - K )(293)(K)
0.002516
Substituting into (9.39)
>
@
2
m
1 600 u 10 6 (m)(25 u 10 6 ) 3 (m) 3 (105,000) 2 kg/s 2 - m
24 18.17 u 10 6 (kg/s - m) (0.01)(m)287(m 2 / s 2 - K )(293)(K)
m
4.25126 u 10 6 kg/s
ª§ 420000 · 2
420000
Ǭ
1
¸ 1 12 (0.002516
105000
«© 105000 ¹
¬
(
(b) Mean outlet temperature, Tmo . Equation (c) gives
2 qcsc
Tmo Tm ( L)
LW Tmi
c pm
2(1100 ) ( W/m 2 )
Tmo
Tm ( L)
Tmo
23.11o C
o
-6
998.3(J/kg C)(4.25126 u 10 )(kg/s)
3
1 6(0.002516)
(d)
Kno , and substituting into (9.64),
2
­1
5
1
13
13 º ½ 2(1.4) (0.002516)
ª
(0.002516) 2 (0.002516) ® (0.002516) ¾
«
48 1 6(0.002516) ¬
40
560 »¼ ¿ 1.4 1 (0.713)
¯2
Nu ( L) 8.14
Equation (e) gives h(L)
h( L )
»
¼
(0.01)(m)(600 u 10 - 6 )(m) 20( o C)
(c) Heat transfer coefficient at the outlet, h(L). Setting Kn
gives the Nusselt number at the outlet
Nu ( L)
º
)»
0.02564( W/m o C)
8.14
2(25 u 10- 6 )(m)
4174 W/m 2 o C
(d) Surface temperature at the outlet, Ts (L). Use (f) to compute g(L)
PROBLEMT 9.10 (continued)
20( o C) g ( L)
2(1100) ( W/m 2 )(600 u 10 6 )(m)
0.01(m) 998.3(J/kg o C)(4.25126 u 10 -6 )(kg/s)
3(1100) ( W/m 2 )(25 u 10 -6 )(m)
13 º
ª
2 13
«¬(0.002516) 40 (0.002516) 560 »¼
0.02564( W/m C)(1 6 u 0.002516)
o
2
23.04 o C
g ( L)
Substitute into (9.63)
Ts ( L )
3(1100) ( W/m 2 )(25 u 10 -6 )(m)
5 º 2(1.4) (1100) ( W/m 2 )(25 u 10 -6 )(m)
ª1
(
0
.
002516
)
(0.002516)
«
48 »¼ 1.4 1 0.02564( W/m o C)(0.713)
0.02564( W/m o C)(1 6 u 0.002516) ¬ 2
Ts ( L)
23.373 o C
(iv) Checking.
Dimensional check: computations showed that equations (9.34), (9.39), (9.60), (9.62) and (9.63)
are dimensionally correct.
Surface temperature check: Application of Newton’s law at the outlet gives
q csc
h( L)>Ts ( L) Tmo @
Solving for Ts (L)
Ts ( L)
q csc
Tmo
h( L )
Using this equation to compute Ts (L) , gives
Ts ( L)
(1100) ( W/m 2 )
2
o
4174( W/m C)
23.11o C
23.374 o C
This is close to the value determined above.
(5) Comments. (i) The flow field solution (velocity, mass flow rate and pressure) is based on
the assumption that the flow is isothermal. Computation showed that the outlet temperature is
Tmo 23.11o C . Since the outlet is 3.11o C above the inlet temperature, it follows that the
assumption of isothermal flow is reasonable. (ii) The heat transfer coefficient at the outlet is high
compared to values for air encountered in typical macrochannels applications. (iii) The Nusselt
number for slip theory for fully developed macrochannel flow is obtained by setting Kno 0 in
(f). This gives
Nu o 8.235
Thus macrochannel theory overestimates the Nusselt number if applied to microchannels.
PROBLEM 9.11
Rectangular microchannels are used to remove heat
from a device at uniform surface heat flux. The height,
width, and length of each channel are H 6.29 P m,
W 90 P m, and L 10 mm, respectively. Using air
qcs
y
W
L
H
x
at Ti 20 o C as the coolant fluid, determine the mass
flow rate and the variation of Nusselt number along
the channel. Inlet and outlet pressure are pi 410
kPa, po 105 kPa. Assume steady state fully
developed slip flow and temperature jump conditions.
m
qcs
(1) Observations. (i) The problem can be modeled as pressure driven Poiseuille flow between
two parallel plates with uniform surface flux. (ii) Assuming fully developed velocity and
temperature, the analysis of Section 9.6.3 gives the mass flow rate and Nusselt number. (iii) The
Nusselt number depends on the Knudsen number, Kn. Since Kn varies along the channel due to
pressure variation, it follows that pressure distribution along the channel must be determined.
(2) Problem Definition. Determine the flow and temperature fields for fully developed
Poiseuille flow with uniform surface flux.
(3) Solution Plan. Apply the results of Section 9.6.3 for the mass flow rate, pressure distribution,
and Nusselt number.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) two-dimensional (no variation along the
width W), (4) slip flow regime (0.001 < Kn < 0.1), (5) ideal gas, (6) constant viscosity,
conductivity and specific heats, (7) negligible lateral variation of density and pressure, (8) the
accommodation coefficients are equal to unity, V u V T 1.0, (9) negligible dissipation, (10)
uniform surface flux, (11) negligible axial conduction, and (12) no gravity.
(ii) Analysis. Assuming isothermal flow, the results of Section 9.6.3 give the mass flow
rate as
m
3 2
2
º
pi
1 W H p o ª pi
1)»
« 2 1 12 Kn o (
24 P LRTo ¬« p o
po
»¼
(9.39)
The Knudsen number at the exit, Kno is
Kno
O ( po )
P
S
H
H po
2
RTo
(9.34)
where the temperature To at the outlet is assumed to be the same as inlet temperature and the
viscosity P is based on inlet temperature.
The Nusselt number, Nu , is given by
PROBLEM 9.11 (continued)
Nu
2
­1
3
5
1
13
13 º ½ 2J 1
ª
( Kn) 2 Kn Kn
® Kn ¾
«
(1 6 Kn) ¯ 2
48 (1 6 Kn) ¬
40
560 »¼ ¿ J 1 Pr
(9.64)
The local Knudsen number, Kn, depends on the local pressure p(x) according to
Kn
O
P
S
H
Hp
2
RT
(9.33)
Equation (9.35) gives p(x)
2
ª
ª
pi º
pi2
pi º x
Kn
6 Kno «6 Kno (
1
)
12
(
1
)»
«
o
»
po ¼
p o ¼» L
p o2
«¬
¬
p( x)
po
(9.35)
Thus, (9.35) is used to determine p(x), (9.33) to determine Kn(x), and (9.64) to determine the
variation of the Nusselt number along the channel.
(iii) Computations. Air properties are determined at 20 o C. To compute p(x), Kn(x),
and Nu , the following data is used
H
6.29 P m
pi
420 u 10 3 kg /s 2 m
po 105 u 10 3 kg /s 2 m
Pr 0.713
R 287 J / kg K 287 m 2 / s 2 K
T # Ti # To
W 90 P m
J 1.4
20 o C
P 18.17 u 10 6 kg /s m
Substituting into (9.34)
Kno
Kno
18.17 u 10 6 (kg /s m)
6.29 u 10
6
S
287(m 2 / s 2 - K)(293.15)(K)
2
( m)105 u 10 (kg /s m)
3
0.01
Using (9.39) and noting that pi / po
m
2
4
1 90 u 10 6 ( m)( 90 u 10 6 ) 3 (m 3 )(105 u 10 3 ) 2 (kg 2 /s 4 m 2 )
(4) 2 1 12 u 0.01(2 1)
24 18.17 u 10 6 (kg /s m)0.01(m)287(m 2 / s 2 K)293.15(K)
m 1.03776 u 10 12 kg/s
>
@
PROBLEM 9.11 (continued)
Axial pressure variation is obtain from (9.35)
p( x)
po
>
6 u 0.01 (6 u 0.01 4) 2 1 (4) 2 12 u 0.01(1 4)
p ( x)
po
@ Lx
0.06 16.4836 15.36
Equation (a) is used to tabulate pressure variation with
x/L. Equations (9.33) and (9.64) are used to compute the
corresponding Knudsen and Nusselt numbers.
(iii) Checking. Dimensional check: Units for equations
(9.33), (9.35), (9.39) and (9.64) are consistent.
Limiting check: No-slip macrochannel Nusselt number is
obtained by setting Kn 0 in (9.64). This gives Nu =
8.235. This agrees with the value given in Table 6.2.
x
L
(a)
x/L p/ po
0
0.2
0.4
0.6
0.8
1.0
4.0
3.602
3.155
2.636
1.988
1.000
Kn
Nu
0.0250
0.00278
0.00317
0.00379
0.00529
0.0100
8.141
8.130
8.115
8.092
8.036
7.862
(5) Comments. (i) To examine the effect of rarefaction and compressibility on the mass flow
rate, equation (9.41) is used to calculate m / mo :
m
mo
1
2
ª pi
º
1 12 Kno »
«
¬ po
¼
1
(4 1 12 u 0.01)
2
2.56
This shows that incompressible no-slip theory will significantly underestimate the mass flow
rate. If rarefaction is neglected ( Kno 0) , the above gives
m
mo
1
2
ª pi
º
1»
«
¬ po ¼
1
(4 1)
2
2 .5
Thus, compressibility plays a dominant role in the mass flow rate.
(ii) No-slip Nusselt number for fully developed Poiseuille flow between parallel plates with
uniform surface heat flux is Nu = 8.235. Thus, no-slip theory overestimates the Nusselt number
if applied to microchannels.
(iii) It should be noted that the equations used to compute m, p(x), and Nu are based on the
assumptions of isothermal conditions in the determination of the flow field. This is a reasonable
approximation for typical applications.
PROBLEM 9.12
A micro heat exchanger consists of rectangular
channels of height H 6.7 P m, width W 400 P m,
and length L 8 mm. Air enters the channels at
temperature Ti 30 o C and pressure pi 510 kPa.
The outlet pressure is p o 102 kPa. Channel surface
is at uniform temperature Ts 50 o C. Assume fully
developed flow and temperature, compute:
Ts
H
Ts
m
L
W
(a) Mass flow rate, m.
(b) Heat transfer coefficient at the inlet, h(0), and outlet, h(L).
(c) Mean outlet temperature, Tmo .
(d) Surface heat flux at the outlet, q csc (L).
(1) Observations. (i) This is a pressure driven microchannel Poiseuille flow. (ii) Since channel
height is much smaller than channel width, the rectangular channel can be modeled as Poiseuille
flow between parallel plates. (iii) Channel surface is maintained at uniform temperature. (iv) The
solution to velocity, pressure, and mass flow rate is presented in Section 9.63. (v) The solution to
the temperature distribution and Nusselt number for fully developed Poiseuille channel flow with
uniform surface temperature is presented in Section 9.6.4. (vi) Surface heat flux is determined
using Newton’s law.
(2) Problem Definition. Determine flow and heat transfer characteristics of fully developed
Poiseuille channel flow with uniform surface temperature.
(3) Solution Plan. To determine velocity, mass flow rate and pressure, apply the analysis and
results of Section 9.6.3. To determine temperature and Nusselt number, apply the analysis and
results of Section 9.6.4.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) constant viscosity (4) infinite plates,
(5) uniform spacing between plates, (6) no variation of density and pressure along y, (7)
V u V T 1.0, (8) no gravity, (9) the velocity field is independent of temperature, (10) ideal
gas, (11) continuum, slip flow regime conditions apply, and (12) fully developed flow.
(ii) Analysis.
(a) Mass flow rate m. Equation (9.39) gives the mass flow rate through the channel:
m
where
º
pi
1 W H 3 po2 ª pi2
1)»
« 2 1 12 Kno (
24 P LRT «¬ po
po
»¼
(9.39)
PROBLEMT 9.12 (continued)
H channel height 6.7 P m 6.7 u 10 -6 m
Kno outlet Knudsen number = Kn( po)
L channel length = 0.008 m
pi inlet pressure 510 kPa 510,000 kg/s 2 - m
po
outlet pressure 102 kPa 102,000 kg/s 2 - m
W
channel width
R gas constant
6
P 18.65 u 10
400 u 10 6 m
287 J/kg - K
287 m 2 / s 2 - K
kg
s-m
Equation (9.34) gives the outlet Knudsen number
Kno
where To
O
H
Kn( po )
P
H
S
1
RTo
2
po
(9.34)
Tmo is the mean outlet temperature.
(b) Heat transfer coefficient at inlet, h(0) and outlet, h(L). The Nusselt number is used to
determine the heat transfer coefficient. Nusselt number for channel flow is defined as
Nu
2hH
k
(a)
where
k
thermal conductivity
0.02638
W
m o C
Applying (a) at the inlet, x = 0 and solving for h(0)
h(0)
k
Nu (0)
2H
(b)
h( L )
k
Nu ( L)
2H
(c)
Similarly, at the outlet, (a) gives
The Nusselt number for air at various Peclet numbers and Knudsen numbers is given in Fig.
9.11. The Peclet number is defined as
Pe
RePr
(d)
U um 2H
P
(e)
where the Reynolds number is defined as
Re
Here U is density and u m is the mean velocity obtained from continuity
PROBLEMT 9.12 (continued)
U um
m
HW
(f)
Re
2m
PW
(g)
(f) into (e)
The Knudsen number at the inlet is given by
Kni
Kn( pi )
O
P
S
H
H
2
RTi
1
pi
(h)
(c) Mean outlet temperature, Tmo . The local mean temperature Tm (x) for channel flow at
uniform surface temperature is given by equation (6.13):
Tm ( x)
Ts (Tmi Ts ) exp[
Ph
x]
mcp
(6.13)
where
cp
specific heat 1006.4 J/kg o C
P = channel perimeter = 2( H W ) 2(6.7 400) u 10 6 8.134 u 10 6 m
Tmi
Ts
um
mean inlet temperature Ti
30 o C
50 o C
mean velocity, m/s
U density, kg/m 3
h is the average heat transfer along the channel between inlet and section x, defined in (6.12)
x
h
1
h( x)dx
x
³
(6.12)
0
The mean outlet temperature is obtained by setting x = L in (6.13)
Tmo
Tm ( L) Ts (Tmi Ts ) exp[
Ph
L]
mcp
(i)
(d) Surface heat flux at the outlet, q csc (L). Application of Newton’s law at the outlet gives
surface heat flux
q csc h( L)(Ts Tmo )
(j)
(iii) Computations.
(a) Mass flow rate m. Equation (9.39) for m is based on the assumption that the flow is
isothermal. Since the outlet temperature To is not yet determined, as a first approximation we
assume To { Ti
303K in (9.34)
PROBLEMT 9.12 (continued)
18.65 u 10 6 (kg/s - m)
Kno
-6
S
2
2
6.7 u 10 (m)(102,000)(kg/s - m)
(287)(m 2 / s 2 - K )(303)(K)
0.01009
Substituting into (9.39)
m
>
@
2
400 u 10 6 (m)(6.7 u 10 6 ) 3 (m) 3 (102,000) 2 kg/s 2 - m
1
24 18.657 u 10 6 (kg/s - m)(0.008)(m)287(m 2 / s 2 - K )(303)(K)
ª§ 510000 · 2
º
510000
Ǭ
1)»
¸ 1 12 (0.01009)(
102000
«¬© 102000 ¹
»¼
m 0.098368 u 10 6 kg/s
(b) Heat transfer coefficient at the outlet, h(0) and h(L). To use Fig. 9.11 for the determination
of the Nusselt number, the Prandtl and Reynolds number, Peclet number, and Knudsen number
are needed. Air properties give
Pr
0.712
(g) gives the Reynolds number
Re
(2)0.098368 u 10 6 (kg/s)
18.65 u 10 6 (kg/s - m)400 u 10 6 (m)
26.372
Thus the Peclet number is
Pe 26.372 u 0.712 18.777
At this Peclet number the curve corresponding to Pe f gives the approximate Nusselt number
for this case. The Knudsen number at the inlet is computed using (h)
Kni
18.65 u 10 6 (kg/s - m)
-6
S
2
6.7 u 10 (m)(510,000)(kg/s - m)
2
(287)(m 2 / s 2 - K )(303)(K)
At this value of Knudsen number Fig. 9.11 gives
Nu (0) # 7.5
Substitute into (b)
h(0)
0.02638( W/m o C)
2(6.7 u 10 -6 )(m)
At the outlet where Kno
W
(7.5) 14,765
2 o
m C
0.01009 , Fig. 9.11 gives
Nu ( L) # 7.25
(b) gives
h( L)
0.02638( W/m o C)
2(6.7 u 10 -6 )(m)
(7.25) 14,273
W
2 o
m C
0.00218
PROBLEMT 9.12 (continued)
(c) Mean outlet temperature, Tmo . The average heat transfer coefficient, h , is needed to
determine Tmo . Using (6.12) to determine h requires the numerical integration of the local heat
transfer coefficient. However, since the change in h between inlet and outlet is very small, the
arithmetical mean can be used to approximate h . Thus
h
Tmo
h(0) h( L)
2
14,273 14,765
W
14,519
2
2
m o C
50( o C) (30 50)( o C) exp[
8.134 u 10 6 (m)14,519( W/m 2 o C)
1006.4(J/kg o C)(0.098368) u 10 -6 )(kg/s)
0.008(m)]
Tmo # 50.00215 o C
(d) Surface heat flux at the outlet, q csc (L). Use (j) to compute q csc (L).
q csc ( L) 14,273( W/m 2 o C)>50.00215 50@ 30.7
W
m2
(iv) Checking.
Dimensional check: computations showed that equations (9.34), (9.39), (6.13), and (j) are
dimensionally correct.
(5) Comments. (i) The flow field solution (velocity, mass flow rate and pressure) is based on
the assumption that the flow is isothermal. Computation showed that the outlet temperature is
Tmo # 50 o C . To improve the solution, an iterative procedure can be followed by repeating the
computation assuming an arithmetical average of mean temperature in the channel equal to
[30( o C) 50( o C)]/2 40 o C. (ii) The heat transfer coefficient at the outlet is high compared to
values for air encountered in typical macrochannels applications. (iii) The Nusselt number for
no-slip theory and negligible axial conduction is obtained from Fig. 9.11 at Kn = 0 and Pe f .
This gives
Nu o
7.5407
This is close to the inlet Nusselt number when rarefaction is included, indicating a small
rarefaction effect. (iv) Unlike the Nusselt number for fully developed flow in macrochannels, the
Nusselt number is not constant along microchannels.
PROBLEM 9.13
Consider isothermal Poiseuille flow of gas in a microtube of radius ro . Taking into
consideration velocity slip, show that the axial velocity is given by
vz
ro2 dp ª
r2 º
«1 4 Kn 2 »
4 P dz ¬«
ro ¼»
(1) Observations. (i) Cylindrical coordinates should be used to solve this problem. (ii)
The axial component of the Navier-Stokes equations must be solved to determine the
axial velocity v z . (iii) The procedure
r
r
ro
and simplifying assumptions used in
the solution of the corresponding
Couette flow between parallel plates,
z
detailed in Section 9.6.2, can be
applied to this case.
(2) Problem Definition. Solve the axial component of the Navier-Stokes equations of
motion.
(3) Solution Plan. Start axial component of the governing Navier-Stokes equations of
motion, introduce simplifying assumptions, write down the slip velocity boundary
conditions and solve the governing equation.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) two-dimensional (axial and
radial), (4) slip flow regime (0.001 < Kn < 0.1), (5) ideal gas, (6) constant viscosity (7)
negligible radial variation of density and pressure, (8) negligible gravity, (9) the velocity
accommodation coefficients is equal to unity, V u 1.0, (10) isothermal flow, (11) the
dominant viscous force is P
1 w § wv z ·
¨r
¸ , and
r wr © wr ¹
(12) negligible inertia forces;
v wv
wv ·
§ wv
U¨ vr z T z v z z ¸
wr
r wT
wz ¹
©
(ii) Analysis. Following the analysis of Section 9.6.2, we begin with the axial
component of the Navier-Stokes equations
§
U ¨ vr
©
wv z vT wv z
wv
wv ·
vz z z ¸
wr
wz
wt ¹
r wT
ª 1 w § wv z · 1 w 2 v z w 2 v z º
wp
P«
Ug z ¸ 2
¨r
»
2
wz
wz 2 ¼»
¬« r wr © wr ¹ r wT
(2.11z)
Introducing the above assumptions, this equation simplifies to
1 w § wv z ·
¨r
¸
r wr © wr ¹
1 wp
P wz
(a)
PROBLEM 9.13 (continued)
The boundary conditions are
wv z (0, z )
wr
u (ro , z )
O
0
(b)
wv z
wr
(c)
Since pressure is assumed independent of r, this equation can be integrated directly to
give the axial velocity v z . Thus
vz
1 wp r 2
C1 ln r C 2
P wz 4
(d)
Application of boundary conditions (b) and (c) give
0 , C2
C1
ro2 dp ª 1 O º
P dz «¬ 4 2ro »¼
(e)
Susbtituting (e) into (d)
vz
ro2 wp ª
O r2 º
»
«1 4
4 P wz ¬«
2r0 ro2 »¼
(f)
Introducing the definition of Knudsen number for tube flow
Kn
O
2ro
(g)
Substituting (g) into (f)
vz
ro2 wp ª
r2 º
«1 4 Kn 2 »
4 P wz ¬«
ro »¼
(iii) Checking. Dimensional check: (h) is dimensionally correct.
Governing equation check: (h) satisfies (a).
Boundary conditions check: (h) satisfies conditions (b) and (c).
(5) Comments. It is important to note the assumptions leading to solution (h).
(h)
PROBLEM 9.14
Consider fully developed isothermal Poiseuille flow through a microtube. Follow the analysis of
Section 9.6.3 and use the continuity equation in cylindrical coordinates to derive the following:
(a) The radial velocity component v r
vr
ro3 1 w ­° dp
®p
4 P p wz °̄ dz
ª1 r 1 r3
º
r
2 Kn( p )»
«
3
ro
¬« 2 ro 4 ro
¼»
½°
¾
°¿
where Kn(p) is the local Knudsen number.
(b) The local pressure p(z)
p( z )
po
2
ª
ª
pi º
pi2
pi º z
8Kno «8Kno )»
» «(1 2 ) 16 Kno (1 po ¼
p o »¼ L
po
«¬
¬
(9.78)
where pi is inlet pressure, p o outlet pressure, and Kno is the outlet Knudsen number.
(1) Observations. (i) This a pressure driven Poiseuille flow through a microtube. (ii) The
procedure for determining the radial velocity component and axial pressure distribution is
identical to that for slip Poiseuille flow between parallel plates. (iii) The solution to the axial
velocity is given by equation (9.74). (iv) Continuity equation gives the radial velocity
component. (v) Axial pressure is determined by setting the radial velocity component equal to
zero at the surface. (vi) Cylindrical coordinates should be used to solve this problem.
(2) Problem Definition. Determine the radial velocity component of slip flow through a tube.
(3) Solution Plant. Use the continuity equation in radial coordinates and the solution to the axial
velocity component to determine the radial component. Set the radial component equal to zero at
the surface to determine the axial pressure distribution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) two-dimensional, (4) slip flow
regime (0.001 < Kn < 0.1), (5) ideal gas, (6) constant viscosity, conductivity, and specific heats,
(7) negligible radial variation of density and pressure, (8) negligible dissipation (9) negligible
gravity, and (10) The velocity accommodation coefficient is equal to unity, V u 1.0.
(ii) Analysis.
(a) To determine the radial velocity component, we follow the derivation of Section 9.6.3 for the
analogous problem of Poiseuille flow between parallel plates. The axial velocity v z for tube flow
is derived in Section 9.6.5
r 2 dp ª
r2 º
vz o
Kn
1
4
(9.74)
«
»
4 P dz «¬
ro2 »¼
The Knudsen number, Kn, for the flow through tubes is defined as
Kn
O
2 ro
P 1 S
2 ro p
2
RT
(a)
The radial component is determined using the continuity equation for compressible flow in
cylindrical coordinates
PROBLEM 9.14 (continued)
wU 1 w
1 w
w
U rv r U vT Uv z
wt r wr
wz
r wT
0
(2.4)
This simplifies to
1 w
w
U r vr U vz
r wr
wz
0
(b)
Using the ideal gas law, (9.31), to express U in terms of pressure and rearranging, the above is
written as
w
1 w
pr vr
pv z
0
r wr
wz
Substituting (9.74) into the above
ro2 w ª dp
r2 º
« p (1 4 Kn 2 )»
4P wz ¬« dz
ro ¼»
1 w
pr vr
r wr
(c)
Integration of (c) gives the radial velocity component v r . Boundary conditions on v r are
(d)
v r (0, z ) 0
v r (ro , z )
(e)
0
Multiplying (c) by rdr , integrating with respect to r and using boundary condition (d)
³
vr
0
w
pr vr
wr
r
³
0
ro2 w ª dp
r2 º
« p (1 4 Kn 2 )» rdr
4P wz «¬ dz
ro »¼
Evaluating the integrals and noting that the integrand on the right hand side is a function of z
only, yields
ro2 d ª dp r 2
r4 º
2
pr vr
« p ( 2 Kn r 2 )»
4P d z «¬ dz 2
4ro »¼
Solving for v r
vr
ro2 1 d ª dp r 2
r4 º
2
(
)»
p
2
Kn
r
«
4P r p d z «¬ dz 2
4ro2 »¼
(f)
(b) To determine axial pressure distribution, (f) is applied to boundary condition (e)
0
1 d ª dp 1
º
p ( 2 Kn )»
«
p d z ¬ dz 4
¼
Using (a) to eliminate Kn
º
P S
d ª dp 1
RT )» 0
«p ( dz ¬ dz 4 ro p 2
¼
Integration with respect to z twice gives
1 2 P
p 8
ro
S
2
RT p
Cz D
(g)
PROBLEM 9.14 (continued)
where C and D are constants of integration. Noting that pressure is positive, the solution to this
quadratic equation is
p
4P
ro
S
2
RT 16
P2 S
ro2 2
RT 8 (Cz D)
(h)
The boundary conditions on pressure are
p (0)
pi and p( L)
po
(i)
Using (i),the constants C and D are determined
C
1
P S
( po2 pi2 ) RT ( po pi )
8L
ro L 2
D
1 2
P
pi ) 8
ro
S
2
RT pi
(j)
(k)
Substituting (j) and (k) into (h)
p(z)
P
4
po
ro po
S
2
RT 16P 2 S RT ª pi2 8P
«1 ro2 2 po2 «¬ po2 ro po
p º z p 2 8P
p
S
2 RT i
RT (1 i )» i2 2
2
po »¼ L po ro po
po
(l)
S
Using (a), this result is expressed in terms of the Knudsen number at the outlet
p( z )
po
2
ª
ª
p º
p2
p ºz
8Kno «8Kno i » «(1 i2 ) 16 Kno (1 i )»
po ¼
p o ¼» L
po
«¬
¬
(m)
(iii) Checking. Dimensional check: Each term in (h) has units of velocity. Each term in (l) is
dimensionless.
(5) Comments. Unlike fully developed Poiseuille flow in macrochannels, the radial velocity
component does not vanish (streamlines are not parallel) and axial pressure distribution is not
linear.
PROBLEM 9.15
Taking into consideration velocity slip, show that the mass flow rate for laminar, fully developed
isothermal Poiseuille flow in a microtube is give by
º
pi
S ro4 po2 ª pi2
1)»
m
« 2 1 16 Kno (
16 P LRT ¬« po
po
»¼
(9.798a)
(1) Observations. (i) Cylindrical coordinates should be used to solve this problem. (i) Axial
velocity component is needed to determine mass flow rate. (iii) Equation (9.74) gives the axial
velocity for this case. (iv) Since axial velocity vary with radial distance, mass flow rate requires
integration of the axial velocity over the flow cross section area. (v) The procedure and
simplifying assumptions used in the solution of the corresponding Couette flow between parallel
plates, detailed in Section 9.6.2, can be applied to this problem.
(2) Problem Definition. Integration of the axial velocity over flow cross section.
(3) Solution Plan. Formulate the flow rate integral, use the axial velocity for the Poiseuille flow
through tubes and carry out the integration of over tube radius.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) two-dimensional (axial and radial),
(4) slip flow regime (0.001 < Kn < 0.1), (5) ideal gas, (6) constant viscosity, (7) negligible
radial variation of density, (8) negligible gravity, (9) the velocity accommodation coefficients is
1 w § wv z ·
equal to unity, V u 1.0, (9) isothermal flow, (10) the dominant viscous force is P
¨r
¸,
r wr © wr ¹
v wv
wv ·
§ wv
and (11) negligible inertia forces; U¨ vr z T z v z z ¸
wr
r wT
wz ¹
©
(ii) Analysis. The mass flow rate is given by
ro
m
2S
³
U v z rdr
(a)
0
Based on the above assumptions, the axial velocity component is given in Section 9.6.2 as
vz
ro2 dp ª
r2 º
«1 4 Kn 2 »
4 P dz «¬
ro »¼
(9.74)
Substituting (9.74) into (a) and recalling that density U is assumed constant along the radial
distance r, gives
ro
ª
ro2 dp
r2 º
m SU
(b)
«1 4 Kn 2 » rdr
2 P dz
ro »¼
«¬
³
0
Evaluating the integral in (b)
m S U ro4 dp
>1 8Kn@
8 P dz
(c)
PROBLEM 9.15 (continued)
The density of an ideal gas is given by (9.31)
p
RT
U
(9.31)
The Knudsen number for tube flow is
P 1 S
Kn
2ro p
2
RT
(d)
Substituting (9.31) and (d) into (c)
m º
4P S
S ro4 dp ª
RT »
«p ro 2
8 PRT dz ¬
¼
(e)
This result gives the mass flow rate in terms of pressure. The solution to the pressure distribution
p(z ) is
p( z )
po
2
ª
ª
pi º
pi2
pi º z
8Kno «8Kno )»
» «(1 2 ) 16 Kno (1 po ¼
p o ¼» L
po
«¬
¬
(9.78)
where Kno is the Knudsen number at the discharge. Evaluating (d) at the discharges where
p po
Kno
P 1
S
2ro po
2
RT
(f)
Rewriting (9.78) as
A B Cz
(g)
8 Kno po
(h)
>8Kno po pi @2
(i)
p( z )
where
A
B
C
º
p o2 ª pi2
pi
1)»
« 2 1 16 Kno (
L ¬« po
po
»¼
(j)
dp
dz
(k)
Differentiating (g)
C
( B Cz ) 1 / 2
2
Substituting (g) and (k) into (e)
m ª
º
4P S
S ro4 C
( B Cz ) 1 / 2 « A B Cz RT »
8 PRT 2
ro 2
¬
¼
(l)
This simplifies to
m Substituting (j) into (m)
ro4
C
16 PRT
S
(m)
PROBLEM 9.15 (continued)
m
º
pi
S ro4 po2 ª pi2
1)»
« 2 1 16 Kno (
16 P LRT ¬« po
po
»¼
(9.79a)
(iii) Checking. Dimensional check: Equation (9.79a) has the correct mass flow units of kg/s.
Limiting check: If pi po , no flow takes place and thus the mass flow rate should be zero.
Setting pi p o in (9.78a) gives m = 0.
(5) Comments. The Knudsen number in (9.79a) represents the effect of rarefaction. Neglecting
rarefaction ( Kno 0) , (9.79a) reduces to
S ro4 p o2
m
16 P LRT
ª pi2 º
« 2 1»
»¼
¬« po
(n)
If both rarefaction and compressibility are neglected, the flow rate is given by (9.79b)
m
S ro4 p o2
8 P LRT
ª pi
º
1»
«
«¬ po
»¼
(9.70b)
PROBLEM 9.16
Pressure distribution for fully developed Poiseuille flow through tubes is given by
p( z )
po
2
ª
ª
p º
p2
p ºz
8Kno «8Kno i » «(1 i2 ) 16 Kno (1 i )»
po ¼
p o ¼» L
po
«¬
¬
(9.78)
Derive this equation using the condition that, for steady state, the mass flow rate is invariant
with axial distance z. That is
ro
º
dm d ª
«2S U v z r dr » 0
dz dz ¬«
0
¼»
³
(1) Observations. (i) To use the proposed approach, the solution to the axial velocity distribution
must be known. (ii) The velocity distribution for Poiseuille flow through tubes is given by
equation (9.74) of Section 9.6.5. (iii) Cylindrical coordinates should be used to solve this
problem.
(2) Problem Definition. Determine the mass flow rate.
(3) Solution Plan. Use the solution to the axial velocity for Poiseuille flow, equation (9.74), to
determine the mass flow rate.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) constant viscosity (4) no radial
variation of density and pressure, (5) no gravity, (6) V u 1.0, (7) the velocity field is
independent of temperature, (8) ideal gas, and (9) continuum, slip flow regime conditions apply.
(ii) Analysis. To derive pressure solution (9.78), it is proposed to use the following conservation
of mass equation
ro
º
dm d ª
2
S
U v z r dr » 0
(a)
«
dz dz «¬
»
0
¼
³
The solution to the axial velocity v z is given by (9.743)
vz
d ª
«
dz «
¬
ro
ro2 dp ª
r2 º
«1 4 Kn 2 »
4 P dz «¬
ro »¼
(9.74)
Substituting (9.743) into (a)
dm
dz
³
0
U
º
ro2 dp ª
r2 º
«1 4 Kn( p ) 2 » rd r »
4P d z ¬«
ro »¼
»¼
0
(b)
Noting that density U and pressure p vary along z and are assumed constant along r, and that the
viscosity P is constant, (b) is written as
PROBLEM 9.16 (continued)
dm
dz
d ª dp
«U
dz « dz
¬
³
ro ª
º
r2 º
«1 4 Kn( p) 2 » rd r »
ro »¼
¬«
¼»
0
0
(c)
Evaluating the integral in (c)
dm
dz
·º
d ª dp §¨ ro2 ro2
2 Kn( p ) ro2 ¸»
«U
¸»
d z « d z ¨© 2
4
¹¼
¬
0
This simplifies to
d ª dp § 1
·º
¨ 2 Kn( p ) ¸»
«U
dz ¬ dz ©4
¹¼
(d)
0
To proceed, the density and Knudsen number in (c) must be expressed in terms of pressure. Ideal
gas law (9.31) gives
p
RT
U
(9.31)
The Knudsen number for tube flow is
Kn
P 1 S
2ro p
2
d ª p dp §¨ 1 P
«
d z ¬« RT dz ¨© 4 ro
S
RT
(e)
(9.31) and (e) into (d)
2
RT
1 ·¸º
»
p ¸¹¼»
0
Since the flow is assumed isothermal, the above simplifies to
d ª dp §¨ 1 P
«p
d z «¬ dz ¨© 4 ro
S
2
RT
1 ·¸º
»
p ¸¹»¼
0
(f)
Integrating (f) once
ª dp § 1 P
« p ¨¨ «¬ dz © 4 ro
S
1 pdp P
4 d z ro
S
2
RT
1 ·¸º
» C1
p ¸¹»¼
Rewriting the above as
2
RT
dp
dz
C1
Integrating again
1 2 P
p 8
ro
S
p (0)
pi ,
2
RT p C 1 z C 2
(g)
The boundary conditions on p are
p ( L)
po
(h)
Here L is tube length. Boundary conditions (h) give C1 and C 2
C1
P S
1
( p o2 pi2 ) RT ( p o pi )
8L
ro L 2
(i)
PROBLEM 9.16 (continued)
C2
1 2 P
pi 8
ro
S
2
RT pi
(j)
The solution to quadratic equation (g) is
p
4P
ro
S
2
RT 16S
P2
ro2
RT 8(C 1 z C 2 )
(k)
Substituting (i) and (j) into (k), normalizing the pressure by p o , and introducing the Knudsen
number (e), we obtain
p( z )
po
2
ª
ª
p º
p2
p ºz
8Kno «8Kno i » «(1 i2 ) 16 Kno (1 i )»
po ¼
p o »¼ L
po
«¬
¬
(9.78)
(iii) Checking. Dimensional check: Each term in (k) has units of pressure.
Limiting check: If pi p o , axial velocity will vanish and pressure should be uniform throughout
the tube. Setting pi p o in (9.78) gives
p( z )
po
8 Kn o >8Kn o 1 @ 1
(4) Comments. This approach for determining p(z) is simpler than that proposed in Section 9.6.4
where it is necessary to first determine the radial velocity component v r .
PROBLEMT 9.17
5 P m and length L
Air is heated in a microtube of radius ro
pressure are Ti
o
20 C and
Outlet pressure is p o
2 mm. Inlet temperature and
100 kPa. Uniform surface flux,
2
q csc 1500 W/m , is used to heat the air. Taking into consideration velocity slip and
temperature jump and assuming fully developed flow and temperature, compute:
(a) Mass flow rate, m.
qcsc
r
(b) Mean outlet temperature, Tmo .
(c) Heat transfer coefficient at outlet, h(L).
(d) Surface temperature at the outlet, Ts (L).
r
ro
z
qcsc
(1) Observations. (i) This is a pressure driven
Poiseuille flow through a microtube. (ii) Tube surface is heated with uniform flux. (iii) The
solution to mass flow rate, temperature distribution and Nusselt number for fully developed
Poiseuille flow through a tube with uniform surface flux is presented in Section 9.6.5.
(2) Problem Definition. Determine flow and heat transfer characteristics of fully developed
Poiseuille flow through a tube with uniform surface heat flux.
(3) Solution Plan. Apply the analysis and results of Section 9.6.5.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) constant conductivity, specific heat
and viscosity, (4) no radial variation of density and pressure, (5) V u V T 1.0, (6) no gravity,
(7) the velocity field is independent of temperature, (8) ideal gas, (9) continuum, slip flow
regime conditions apply, and (10) fully developed flow.
(ii) Analysis.
(a) Mass flow rate m. Equation (9.79a) gives the mass flow rate through the tube:
m
ro4 po2
16 P LRTo
S
ª pi2
º
pi
1)»
« 2 1 16 Kno (
po
«¬ po
»¼
where
Kno outlet Knudsen number = Kn( p o )
L channel length = 0.002 m
pi inlet pressure 600 kPa 600,000 kg/s 2 - m
100,000 kg/s 2 - m
5 u 10 -6 m
po
outlet pressure 100 kPa
ro
tube radius = 5 P m
R
gas constant
287 J/kg - K
287 m 2 / s 2 - K
(9.79a)
PROBLEMT 9.17 (continued)
o
To | Ti
20 C
P 18.17 u 10 6 kg/s - m
The outlet Knudsen number is given by
Kno
Kn( po )
O
P
S
2 ro
2ro
2
RTo
1
po
(a)
(b) Mean outlet temperature, Tmo . Conservation of energy between inlet and outlet, gives
2S ro Lq csc
Tmi
mc p
Tm o
(b)
where
cp
specific heat 998.3 J/kg o C
q csc
surface heat flux 1500 W/m 2
Tmi
mean inlet temperature Ti
20 o C
(c) Heat transfer coefficient at the outlet, h(L). The Nusselt number is used to determine the
heat transfer coefficient. Nusselt number for channel flow is defined as
Nu
2ro h
k
(c)
where
k
thermal concuctivity
0.02564
W
m o C
Applying (c) at the outlet, z = L, and solving for h(L)
h( L)
k
Nu ( L)
2ro
(d)
The Nusselt number is given by (9.98)
Nu
2
14
7 º 4J 1
4
1
ª
( Kn 3 ) 16 Kn 2 Kn » Kn
«
2
3
24 ¼ J 1 Pr
(1 8 Kn)
16
(1 8 Kn) ¬
(9.98)
where
Kn local Knudsen number
Pr Prandtl number 0.713
J = specific heat ratio = 1.4
Evaluation (9.98) at z = L where Kn
Nu
Kno
2
4J 1
14
7º
4
1
ª
( Kno 3 ) 16 Kno2 Kno » Kno
«
2
3
24 ¼ J 1 Pr
(1 8 Kn)
16
(1 8 Kno ) ¬
(e)
PROBLEMT 9.17 (continued)
(d) Surface temperature at the outlet, Ts (L). Surface temperature distribution is given by
(9.97):
4q csc ro ª
4J q csc ro
3º
(9.97)
Ts ( z )
Kn » Kn g ( z )
«
k (1 8 Kn) ¬
16 ¼ J 1 kPr
where g(z) is given by (9.96):
g ( z)
Tmi 2q csc
q cscro
z
U c p ro v z m
k (1 8 Kn) 2
7º
ª
2 14
«¬16 Kn 3 Kn 24 »¼
(9.96)
where v zm is the mean velocity. Continuity equation gives v zm in terms of mass flow rate m
v zm
m
S ro2 U
(f)
To determine surface temperature at the outlet, Ts (L), the Knudsen number in (9.96) and (9.97) is
evaluated at outlet pressure and g(z) is evaluated at z = L. Equation (9.97) becomes
Ts ( L )
4q csc ro
k (1 8 Kno )
Using (f), and setting z = L and Kn
g ( L) Tmi 4J q csc ro
3º
ª
« Kno 16 » J 1 kPr Kno g ( L)
¬
¼
(g)
Kno in (9.96) gives g(L)
2S ro q csc
q cscro
L
mc p
k (1 8 Kno ) 2
14
7º
ª
2
«16 Kn o 3 Kno 24 »
¬
¼
(h)
(iii) Computations.
(a) Mass flow rate m. Equation (9.79a) for m is based on the assumption that the flow is
isothermal. Since the outlet temperature To is not yet determined, as a first approximation we
assume To { Ti 293 K in (9.79a). The outlet Knudsen number is computed using (a)
Kno
18.17 u 10 6 (kg/s - m)
S
(287)(m 2 / s 2 - K )(293)(K)
-6
2
2
10 u 10 (m)(100,000)(kg/s - m)
0.0066054
Substituting into (9.79a)
m
>
@
2
ª§ 600000 · 2
º
600000
Ǭ
1)»
¸ 1 16 (0.0066054(
6
2
2
16 18.17 u 10 (kg/s - m) (0.002)(m)287(m / s - K )(293)(K) «© 100000 ¹
100000
»¼
¬
S
600 u 10 6 (m)(5 u 10 6 ) 3 (m) 3 (100,000) 2 kg/s 2 - m
m 1.426 u 10 8 kg/s
(b) Mean outlet temperature, Tmo . Equation (b) gives
Tmo
2S (1500 ) ( W/m 2 )5 u 10 6 (m)(0.002)(m)
20( o C)
o
-8
998.3(J/kg C)(1.426 u 10 )(kg/s)
Tmo
26.6204 o C
PROBLEMT 9.17 (continued)
(c) Heat transfer coefficient at the outlet, h(L). Substituting into (e), gives the Nusselt number
at the outlet
Nu ( L)
2
­
4
3
1
14
7 º ½ 2(1.4) (0.0044054)
ª
16(0.0066054) 2 (0.0066054) » ¾ ®0.0066054 «
1 8(0.0066054) ¯
16 1 8(0.0066054) ¬
3
24 ¼ ¿ 1.4 1 (0.713)
Nu ( L)
4.278
Equation (d) gives h(L)
h( L )
0.02564( W/m o C)
4.278 10,969 W/m 2 o C
-6
2(5 u 10 )(m)
(d) Surface temperature at the outlet, Ts (L). Use (h) to compute g(L)
g ( L)
20( o C) 2(1500) ( W/m 2 )S (5 u 10 6 )(m)(0.002)(m)
998.3(J/kg o C)(1.426 u 10 -8 )(kg/s)
1500 ( W/m 2 )(5 u 10 -6 )(m)
14
7º
ª
16(0.0066054) 2 (0.0066054) »
«
3
24 ¼
0.02564( W/m C)(1 8 u 0.0066054) ¬
o
g ( L)
2
26.53507 o C
Substitute into (g)
Ts ( L)
3 º 4(1.4) (1500) ( W/m 2 )(5 u10-6 )(m)
ª
(
0
.
0066054
)
(0.0066054) 23.04(o C)
«
16 »¼ 1.4 1 0.02564( W/mo C)(0.713)
0.02564( W/mo C)(1 8 u 0.0066054) ¬
Ts ( L)
4(1500) ( W/m 2 )(5 u10-6 )(m)
26.7571 o C
(iv) Checking.
Dimensional check: computations showed that equations (9.79a), (d), (g),
dimensionally correct.
Surface temperature check: Application of Newton’s law at the outlet gives
q csc
h( L)>Ts ( L) Tmo @
Solving for Ts (L)
q csc
Tmo
h( L )
Using this equation to compute Ts (L) , we obtain
Ts ( L)
and (h) are
PROBLEMT 9.17 (continued)
Ts ( L)
(1500) ( W/m 2 )
2
o
10969( W/m C)
26.6204 o C
26.7572 o C
This is close to the value determined above.
(5) Comments. (i) The flow field solution (velocity, mass flow rate and pressure) is based on
the assumption that the flow is isothermal. Computation showed that the outlet temperature is
Tmo 26.62 o C . Since the outlet is 6.62 o C above the inlet temperature, it follows that the
assumption of isothermal flow is reasonable. (ii) The heat transfer coefficient at the outlet is very
high compared to values for air encountered in typical macrochannels applications. (iii) In
computing Tmo and Ts , results are presented showing four decimal points. This is done to avoid
errors where temperature differences are small. (iv) The Nusselt number for slip theory for fully
developed macrotube flow is obtained by setting Kno 0 in (e). This gives
Nu o
4.364
Thus macrochannel theory overestimates the Nusselt number if applied to microchannels.
PROBLEM 9.18
Determine the axial variation of the Nusselt number and heat transfer coefficient of the
microtube in Problem 9.17.
(1) Observations. (i) The problem is a pressure
driven Poiseuille flow through microtube with
uniform surface heat flux. (ii) The Nusselt number
depends on the Knudsen number, Kn. Since Kn
varies along the tube due to pressure variation, it
follows that pressure distribution along the tube
must be determined. (iii) Assuming fully developed
velocity and temperature, the analysis of Section
9.6.5 gives axial pressure and Nusselt number
variation along tube. (iv) The definition of Nusselt
number gives the heat transfer coefficient.
qcsc
r
r
ro
z
qcsc
(2) Problem Definition. Determine the flow and temperature fields for fully developed
Poiseuille tube flow with uniform surface flux.
(3) Solution Plan. Apply the results of Section 9.6.5 for pressure and Nusselt number
distribution.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) two-dimensional (no angular
variation), (4) slip flow regime (0.001 < Kn < 0.1), (5) ideal gas, (6) constant viscosity and
conductivity, (7) negligible radial variation of density and pressure, (8) the accommodation
coefficients are assumed equal to unity, V u V T 1.0, (9) negligible dissipation, (10) uniform
surface flux, (11) negligible axial conduction, and (12) no gravity.
(ii) Analysis. The Nusselt number for tube flow heat is defined as
2 ro h
k
Nu
Solving the above for the heat transfer coefficient h
h
Nu
k
2 ro
(a)
The Nusselt number, Nu , is given by (9.99)
Nu
2
4
1
14
7 º 4J 1
ª
( Kn 3 ) 16 Kn 2 Kn » Kn
«
2
(1 8 Kn)
16
3
24 ¼ J 1 Pr
(1 8 Kn) ¬
(9.98)
The local Knudsen number, Kn, depends on the local pressure p(x) according to
Kn
Kn( p )
O
P
S
2 ro
2 ro
2
RT
1
p
(b)
PROBLEM 9.18 (continued)
Evaluating (b) at the outlet
Kno
Kn( po )
O
2 ro
P
2ro
S
1
RT
2
po
(c)
Equation (9.78) gives p(z )
2
ª
ª
p º
p2
p ºz
8Kno «8Kno i » «(1 i2 ) 16 Kno (1 i )»
po ¼
p o ¼» L
po
«¬
¬
p( z )
po
(9.78)
Thus, (9.78) is used to determine p(z), (b) to determine Kn( p), (c) gives Kn o , (9.98) to determine
the variation of the Nusselt number along the tube, and (a) the heat transfer coefficient.
(iii) Computations. Air properties are determined at 20 o C. To compute p(x), Kn(x),
and Nu , the following data is used
L channel length = 0.002 m
pi inlet pressure 600 kPa
100,000 kg/s 2 - m
5 u 10 -6 m
po
outlet pressure 100 kPa
ro
tube radius = 5 P m
R
gas constant
T | To | Ti
600,000 kg/s 2 - m
287 J/kg - K
287 m 2 / s 2 - K
20 o C
P 18.17 u 10 6 kg/s - m
Pr 0.713
R 287 J / kg K 287 m 2 / s 2 K
T # Ti # To
J 1.4
20 o C
P 18.17 u 10 6 kg /s m
Substituting into(c)
Kno
18.17 u 10 6 (kg/s - m)
S
(287)(m 2 / s 2 - K )(293)(K)
-6
2
10 u 10 (m)(100,000)(kg/s - m) 2
0.0066054
Using (9.78) and noting that p i / p o 6 gives axial pressure variation
p ( x)
po
>
8 u 0.0066054 (8 u 0.0066054 6)2 1 (6)2 16 u 0.0066054(1 6)
p( x)
po
0.0528432 36.63691 35.528432
z
L
@ Lz
(d)
PROBLEM 9.18 (continued)
The local Knudsen number Kn(p) is computed by taking the ratio of (b) and (c)
Kn( p )
Kn o
p / po
0.0066054
p / po
(e)
Using (d), (e) and Nusselt number, and (9.98), results are tabulated below.
z/L
p/ po
Kn
Nu
0
0.2
6.0
5.3814
0.001109
0.0012275
4.3501
11,153
0.4
0.6
4.6827
3.8612
11,150
11,144
0.8
1.0
2.8132
1.0
0.0014106
0.0017107
0.002348
0.0066054
4.3486
4.3462
4.3429
4.3344
4.2780
h( W/m 2 o C)
11,134
11,113
10,969
(iii) Checking. Dimensional check: Computations showed that units of equations (a), (b), and
(9.98) are consistent.
Limiting check: No-slip macrochannel Nusselt number is obtained by setting Kn
This gives Nu = 4.364. This agrees with the value given in Table 6.2.
0 in (9.98).
(5) Comments. (i) No-slip Nusselt number for fully developed Poiseuille flow between parallel
plates with uniform surface heat flux is Nu = 4.364 (Table 6.2). Thus, no-slip theory
overestimates the Nusselt number if applied to microtubes.
(ii) The Nusselt number for fully developed flow is constant along channels. This example shows
that in microtubes the Nusselt number vary slightly along the tube.
(iii) It should be noted that the equations used to compute p( x), and Nu are based on the
assumptions of isothermal conditions in the determination of the flow field. Determining the
outlet temperature will give an indication of the validity of this assumption.
PROBLEM 9.19
A micro heat exchanger uses micro tubes of
radius ro 3 P m and length L 6 mm. Inlet air
temperature and pressure are Ti 20 o C and
pressure pi 600 kPa. Outlet pressure is
po 100 kPa. Each tube is maintained at
uniform surface temperature Ts 60 o C. Taking
into consideration velocity slip and temperature
jump and assuming fully developed flow and
temperature, determine the following:
r
r
ro
z
Ts
Ts
(a) Heat transfer coefficient at the inlet, h(0), and outlet, h(L).
(b) Mean outlet temperature Tmo .
(1) Observations. (i) This is a pressure driven Poiseuille flow through a tube at uniform surface
temperature. (ii) Since the flow field is assumed independent of temperature, it follows that the
velocity, mass flow rate and pressure distribution for tubes at uniform surface flux, presented in
Section 9.6.6, are applicable to tubes at uniform surface temperature.. (iii) The heat transfer
coefficient can be determined if the Nusselt number is known. (iv) The variation of the Nusselt
number with Knudsen number for air is shown in Fig. 9.16. (v) The determination of Knudsen
number at the inlet and outlet and Fig. 9.16 establish the Nusselt number at these locations. (vi)
The use of Fig. 9.16 requires the determination of the Peclet number. (vii) Mean temperature
variation along the tube is given by equation (6.13). Application of this equation requires the
determination of the average heat transfer coefficient.
(2) Problem Definition. Determine the Nusselt number at the inlet and outlet and the average
hat transfer coefficient.
(3) Solution Plan. Compute the Knudsen number at the inlet and outlet, compute the Peclet
number, and use Fig. 9.16 to determine the Nusselt number. Use (6.13) to compute the outlet
temperature.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) two-dimensional (no angular
variation, (4) slip flow regime (0.001 < Kn < 0.1), (5) ideal gas, (6) constant viscosity,
conductivity and specific heats, (7) negligible radial variation of density and pressure, (8) the
accommodation coefficients are assumed to be equal to unity, V u V T 1.0, (9) negligible
dissipation, (10) uniform surface temperature, and (11) negligible gravity.
(ii) Analysis.
(a) Heat transfer coefficient at inlet, h(0) and outlet, h(L). The Nusselt number is used to
determine the heat transfer coefficient. Nusselt number for tube flow is defined as
Nu
where
2ro h
k
(a)
PROBLEM 9.19 (continued)
h heat transfer coefficient, W/m 2 o C
k thermal conductivity 0.02564W/m o C
Nu Nusselt number
ro = tube radius = 3 u 10 6 m
Applying (a) at the inlet, z = 0 and solving for h(0)
h ( 0)
k
Nu (0)
2 ro
(b)
h( L)
k
Nu ( L)
2ro
(c)
Similarly, at the outlet, (a) gives
L tube length = 0.006 m
The Nusselt number for air at various Peclet numbers and Knudsen numbers is given in Fig.
9.16. The Peclet number is defined as
Pe
RePr
(d)
where
Pr = 0.713
The Reynolds number is defined as
Re
U v z m 2 ro
P
(e)
where
v zm
P
mean velocity, m/s
viscosity = 18.17 W/m o C
U density, kg/m 3
Continuity gives
m
U u zm
S ro2
(f)
(f) into (e)
Re
2 m
S P ro
(g)
The mass flow rate m is given by equation (9.79a)
m
where
S ro4 p o2
16 P LRT
ª pi2
º
pi
1)»
« 2 1 16 Kno (
po
«¬ p o
»¼
(9.798a)
PROBLEM 9.19 (continued)
600,000 kg/s 2 - m
pi
inlet pressure
po
outlet pressure 100 kPa 100,000 kg/s 2 - m
287 J/kg - K
R
T | Ti
To
600 kPa
287 m 2 / s 2 - K
20 o C
The Knudsen number at the inlet and outlet is given by
P
O
Kni
Kn( pi )
Kno
Kn( p 0 )
S
2 ro
2 ro
2
O
P
S
2 ro
2 ro
2
RTi
1
pi
(h)
RTo
1
po
(i)
(b) Mean outlet temperature, Tmo . The local mean temperature Tm (x) for channel flow at
uniform surface temperature is given by equation (6.13):
Tm ( z ) Ts (Tmi Ts ) exp[
Ph
z]
mc p
(6.13)
where
cp
specific heat 1006.4 J/kg o C
P = channel perimeter = 2S ro , m
Tmi
Ts
mean inlet temperature Ti
20 o C
60 o C
h is the average heat transfer along the tube between inlet and outlet, defined in (6.12)
h
1
L
L
³
(6.12)
h( z ) d z
0
An accurate method for computing h requires the determination of the variation of local heat
transfer coefficient, h(z), and evaluating the integral in (6.12) numerically. An approximate
approach is to use the arithmetical average. This is justified if h(z) is linear or its change is small.
(iii Computations.
The inlet and outlet Knudson numbers are computed first
Kni
Kno
18.17 u 10 6 (kg/s - m)
S
(287)(m 2 / s 2 - K )(293)(K)
-6
2
2(3 u 10 )(m)(600,000)(kg/s - m) 2
18.17 u 10 6 (kg/s - m)
S
(287)(m 2 / s 2 - K )(293)(K)
-6
2
2(3 u 10 )(m)(100,000)(kg/s - m) 2
0.001835
0.01101
PROBLEM 9.19 (continued)
Substitute into (9.79a)
m
>
@
2
(m)(3 u 10 6 ) 3 (m) 3 (10,000) 2 kg/s 2 - m
S
16 18.17 u 10 6 (kg/s - m)(0.006)(m)287(m 2 / s 2 - K )(293)(K)
m
º
ª§ 600000 · 2
600000
1)»
Ǭ
¸ 1 12 (0.01101)(
100000
»¼
«¬© 100000 ¹
0.622154 u 10 9 kg/s
(g) gives the Reynolds number
(4)0.622154 u 10 6 (kg/s)
Re
S18.17 u 10 6 (kg/s - m)6 u 10 6 (m)
7.7266
Thus the Peclet number is
7.7266 u 0.713 5.181
Pe
At the above values of Kni and Pe, Fig. 9.16 gives
Nu (0) | 3.75
Substitute into (b)
0.02546( W/m o C)
W
(3.75) 15,910 2 o
6
m C
2(3 u 10 )(m)
h(0)
At the outlet Fig. 9.16 gives
Nu ( L) | 3.68
(b) gives
h( L)
0.02546( W/m o C)
W
(3.68) 15,620 2 o
m C
2(3 u 10 - 6 )(m)
Since the variation of h between inlet and outlet is small, using the arithmetical average heat
transfer coefficient in (6.13) is justified
h|
1
>h(0) h( L)@ 1 (15,910 15,620) 15,765 2Wo
2
2
m C
substituting into (6.13) and setting z = L
Tmo
o
o
60( C) (20 60)( C) exp[
Tmo | 60 o C
(iv) Checking.
2S(3 u 10 6 )(m)15,765( W/m 2 o C)
998.3(J/kg o C)0.622154 u 10 - 9 (kg/s)
0.006(m)]
PROBLEM 9.19 (continued)
Dimensional check: computations showed that equations (b), (c), (9.79a), and (6.13) are
dimensionally correct.
(5) Comments. (i) The flow field solution (velocity, mass flow rate and pressure) is based on
the assumption that the flow is isothermal. Computation showed that the outlet temperature is
Tmo # 60 o C . To improve the solution, an iterative procedure can be followed by repeating the
computation assuming an arithmetical average of mean temperature in the channel equal to
[20( o C) 60( o C)]/2 40 o C. (ii) The heat transfer coefficients are very high compared to values
for air encountered in typical macrotubes applications.
PROBLEM 9.20
Air enters a microtube at temperature Ti 20 o C and pressure pi 600 kPa. Outlet pressure is
po 100 kPa. Tube radius is ro 1 P m and its length is L 1.2 mm. The surface is maintained
at uniform temperature Ts 40 o C. Taking into
consideration velocity slip and temperature jump
and assuming fully developed conditions,
determine the variation along the tube of the
following:
r
r
ro
z
(a) Nusselt number, Nu (z ).
(b) Heat transfer coefficient, h(z ).
(c) Mean temperature, Tm (z ).
Ts
Ts
(1) Observations. (i) The problem is a pressure driven Poiseuille flow through microtube at
uniform surface temperature. (ii) The Nusselt number depends on the Knudsen number, Kn.
Since Kn varies along the tube due to pressure variation, it follows that pressure distribution must
be determined. (iii) Assuming fully developed velocity and temperature, the analysis of Section
9.6.6 gives axial pressure and Nusselt number variation along the tube. (iv) The definition of
Nusselt number gives the heat transfer coefficient. (v) The variation of the Nusselt number with
Knudsen number and Peclet number for air is shown in Fig. 9.16. (vii) Mean temperature
variation along the tube is given by equation (6.13). Application of this equation requires the
determination of the average heat transfer coefficient.
(2) Problem Definition. Determine the flow and temperature fields for fully developed
Poiseuille tube flow with uniform surface temperature.
(3) Solution Plan. Apply the results of Section 9.6.6 for pressure and Nusselt number
distribution. Use Fig. 9.16 to determine Nusselt number variation along the tube.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) two-dimensional (no angular
variation), (4) slip flow regime (0.001 < Kn < 0.1), (5) ideal gas, (6) constant viscosity,
conductivity, and specific heat, (7) negligible radial variation of density and pressure, (8) the
accommodation coefficients are assumed equal to unity, V u V T 1.0, (9) negligible
dissipation, (10) uniform surface temperature, and (11) no gravity.
(ii) Analysis.
(a) Nusselt number Nu (z ), and (b) Heat transfer coefficient, h(z ).
tube flow is defined as
2 ro h
Nu
k
The Nusselt number for
Solving the above for the heat transfer coefficient h
h
Nu
k
2 ro
(a)
PROBLEM 9.20 (continued)
The Nusselt number for air at various Peclet numbers and Knudsen numbers is given in Fig.
9.16. The local Knudsen number, Kn, depends on the local pressure p(z) according to
Kn
Kn( p )
O
P
S
2 ro
2 ro
2
RT
1
p
(b)
Equation (9.78) gives p( z )
p( z )
po
2
ª
ª
p º
p2
p ºz
8Kno «8Kno i » «(1 i2 ) 16 Kno (1 i )»
po ¼
p o »¼ L
po
«¬
¬
(9.78)
The Peclet number is defined as
Pe
RePr
(c)
where
Pr = 0.713
The Reynolds number is defined as
Re
U v z m 2 ro
P
(d)
where
v zm
P
mean velocity, m/s
viscosity = 18.17 u 10 -6 kg/m s
U density, kg/m 3
Continuity gives
m
U u zm
S ro2
(e)
(e) into (d)
Re
2 m
S P ro
(f)
The mass flow rate m is given by equation (9.79a)
m
S ro4 p o2
16 P LRT
ª pi2
º
pi
1)»
« 2 1 16 Kno (
po
«¬ p o
»¼
where
inlet pressure
po
outlet pressure 100 kPa 100,000 kg/s 2 - m
R
287 J/kg - K
T | Ti
To
20 o C
600 kPa
600,000 kg/s 2 - m
pi
287 m 2 / s 2 - K
(9.79a)
PROBLEM 9.20 (continued)
(b) Mean outlet temperature, Tm . The local mean temperature Tm ( z ) for channel flow at
uniform surface temperature is given by equation (6.13)
Tm ( z ) Ts (Tmi Ts ) exp[
Ph
L]
mc p
(6.13)
where
c p specific heat = 998.3 J/kg o C
P = channel perimeter = 2S ro , m
Tmi
Ts
mean inlet temperature Ti
20 o C
40 o C
h is the average heat transfer along the tube between inlet and location z, defined in (6.12)
z
h
1
L
³
h( z ) d z
(6.12)
0
An accurate method for computing h requires the determination of the variation of local heat
transfer coefficient, h(z), and evaluating the integral in (6.12) numerically. An approximate
approach is to use the arithmetical average. This is justified if h(z) is linear or its change is small.
Thus, (9.78) is used to determine p(z), (b) to determine Kn( p), (c) gives the Peclet number, Fig.
9.16 gives the variation of the Nusselt number along the tube, (a) the heat transfer coefficient,
and (6.13) the mean temperature.
(iii) Computations. Air properties are determined at 20 o C. To compute p(z), Kn( z ),
and Nu , the following data is used
L channel length = 0.0012 m
pi inlet pressure 600 kPa 600,000 kg/s 2 - m
ro
outlet pressure100 kPa 100,000 kg/s 2 - m
tube radius = 1P m 1u 10 -6 m
R
gas constant
po
T | To | Ti
287 J/kg - K
287 m 2 / s 2 - K
20 o C
P 18.17 u 10 6 kg/s - m
Pr 0.713
R 287 J / kg K
T # Ti # To
J 1.4
287 m 2 / s 2 K
20 o C
P 18.17 u 10 6 kg /s m
Evaluating (b) at the outlet
PROBLEM 9.20 (continued)
18.17 u 10 6 (kg/s - m)
S
(287)(m 2 / s 2 - K )(293)(K)
-6
2
2
2 u 10 (m)(100,000)(kg/s - m)
Kno
Noting that pi / po
m
m
S
0.033027
6 , (9.79a) gives m
>
(1 u 10 6 ) 4 (m)(100,000) 2 kg/s 2 - m
6
2
16 18.17 u 10 (kg/s - m)(0.0012)(m)287(m / s
2
@
2
>(6)
- K )(293)(K)
2
1 16 (0.033027)(6 1)
@
0.403 u 10 10 kg/s
Substituting into (f) gives the Reynolds number
Re
(4)0.403 u 10 10 (kg/s)
S18.17 u 10 6 (kg/s - m)2 u 10 6 (m)
1.412
Thus the Peclet number is
Pe 1.412 u 0.713 1.007
Using (9.78) and noting that p i / p o 6 gives axial pressure variation
p( x)
po
>
8 u 0.033027 (8 u 0.033027 6) 2 1 (6) 2 16 u 0.033027(1 6)
p( x)
po
0.264216 39.240402 37.64216
z
L
@ Lz
(g)
The local Knudsen number Kn(p) is given by
Kn( p )
Kno
p / po
0.033027
p / po
(h)
With Pe | 1 , equations (g), (h), Fig. 9.16, and (a) are used to compute and tabulate pressure,
Knudsen number, Nusselt number, and heat transfer coefficient, respectively, as functions of
z/L.
z/L
p/ po
Kn
Nu h( W/m 2 o C)
6.0
0.0055505 4.01
0
0.2 5.3814 0.0061536 4.0
0.4 4.6827 0.0070972 3.99
0.6 3.8612 0.0086528 3.97
0.8 2.8132 0.0119812 3.95
1.0 1.0
0.033027 3.75
T ( z )o C
51,410
40
51,280
51,150
40
40
51,900
40
50,640
40
48,080
40
PROBLEM 9.20 (continued)
Since the variation of h between inlet and outlet is small, using the arithmetical average heat
transfer coefficient in (6.13) is justified. For determining the outlet temperature, we set
h|
1
>h(0) h( L)@ 1 (51,410 48,080)
2
2
49,745
W
m o C
2
To determine the highest mean temperature, (6.13) is applied at z = L
Tmo
40( o C) (20 40)( o C) exp[
2S (1u 10 6 )(m)49,745( W/m 2 o C)
0.0012(m)]
998.3(J/kg o C)0.403 u 10 -10 (kg/s)
Tmo | 40 o C
Application (6.13) at other values of z gives Tm ( z ) | 40 o C . At z = 0, (6.13) gives Tmo
20 o C .
(iii) Checking. Dimensional check: Computations showed that units of equations (a), (f), (g), (h),
and (9.79a) are dimensionally consistent.
(5) Comments. (i) No-slip Nusselt number for fully developed Poiseuille flow through tubes at
uniform surface temperature is Nu = 3.656 (Table 6.2). Thus, no-slip theory overestimates the
Nusselt number if applied to microtubes.
(ii) The Nusselt number for fully developed flow in macrotubes is constant. This example shows
that for microtubes the Nusselt number varies along the tube.
(iii) It should be noted that the equations used to compute p(x), and Nu are based on the
assumptions of isothermal conditions in the determination of the flow field. Determining the
outlet temperature will give an indication of the validity of this assumption.
(v) The fluid equilibrates with surface temperature very close to the inlet and remains essentially
at constant temperature, equal to throughout the tube. At a distance z 1P m , fluid temperature
increases to 39.98 o C.
Problem 1.1
x Heat is removed from the surface by convection. Therefore, Newton's law of cooling is
applicable.
x
Ambient temperature and heat transfer coefficient are uniform
x
Surface temperature varies along the rectangle.
Problem 1.2
x Heat is removed from the surface by convection. Therefore, Newton's law of cooling
may be helpful.
x Ambient temperature and surface temperature are uniform.
x Surface area and heat transfer coefficient vary along the triangle.
Problem 1.3
x Heat flux leaving the surface is specified (fixed).
x Heat loss from the surface is by convection and radiation.
x Convection is described by Newton's law of cooling.
x Changing the heat transfer coefficient affects temperature distribution.
x Surface temperature decreases as the heat transfer coefficient is increased.
x Surface temperature gradient is described by Fourier’s law
x Ambient temperature is constant.
Problem 1.3
x Metabolic heat leaves body at the skin by convection and radiation.
x Convection is described by Newton's law of cooling.
x Fanning increases the heat transfer coefficient and affects temperature distribution,
including surface temperature.
x Surface temperature decreases as the heat transfer coefficient is increased.
x Surface temperature is described by Newton’s law of cooling.
x Ambient temperature is constant.
Problem 1.4
x Metabolic heat leaves body at the skin by convection and radiation.
x Convection is described by Newton's law of cooling.
x Fanning increases the heat transfer coefficient and affects temperature distribution,
including surface temperature.
x Surface temperature decreases as the heat transfer coefficient is increased.
x Surface temperature is described by Newton’s law of cooling.
x ) Ambient temperature is constant.
Problem 1.5
x Melting rate of ice depends on the rate of heat added at the surface.
x Heat is added to the ice from the water by convection.
x Newton's law of cooling is applicable.
x Stirring increases surface temperature gradient and the heat transfer coefficient. An
increase in gradient or h increases the rate of heat transfer.
x Surface temperature remains constant equal to the melting temperature of ice.
x Water temperature is constant.
Problem 1.6
x This problem is described by cylindrical coordinates.
x For parallel streamlines v r vT 0 .
x Axial velocity is independent of axial and angular distance.
Problem 1.7
x This problem is described by cylindrical coordinates.
x Streamlines are concentric circles. Thus the velocity component in the radial direction
vanishes ( vr 0 ).
x For one-dimensional flow there is no motion in the z-direction ( v z
0 ).
x The T -velocity component, vT , depends on distance r and time t.
Problem 1.8
x This problem is described by Cartesian coordinates.
x For parallel streamlines the y-velocity component v
0.
x For one-dimensional flow there is no motion in the z-direction (w = 0).
x The x-velocity component depends on distance y and time t.
Problem 1.9
x This problem is described by Cartesian coordinates.
x For parallel streamlines the y-velocity component v
0.
x For one-dimensional flow there is no motion in the z-direction (w = 0).
x
The x-velocity component depends on distance y only.
Problem 1.10
x Heat flux leaving the surface is specified (fixed).
x Heat loss from the surface is by convection and radiation
x Convection is described by Newton's law of cooling.
x Changing the heat transfer coefficient affects temperature distribution.
x Surface temperature decreases as the heat transfer coefficient is increased.
x Surface temperature gradient is described by Fourier’s law
x Ambient temperature is constant.
Problem 1.11
x Heat is removed from the surface by convection. Therefore, Newton's law of cooling is
applicable.
x Ambient temperature and heat transfer coefficient are uniform.
x Surface temperature varies along the area.
x The area varies with distance x.
Problem 2.2
x The fluid is incompressible.
x Radial and tangential velocity components are zero.
x Streamlines are parallel.
x Cylindrical geometry.
Problem 2.3
x The fluid is incompressible.
x axial velocity is invariant with axial distance.
x Plates are parallel.
x Cartesian geometry.
Problem 2.4
x The fluid is incompressible.
x Radial and tangential velocity components are zero.
x
Streamlines are parallel.
x Cylindrical geometry.
Problem 2.5
x Shearing stresses are tangential surface forces.
x W xy and W yx are shearing stresses in a Cartesian coordinate system.
x Tangential forces on an element result in angular rotation of the element.
x If the net external torque on an element is zero its angular acceleration will vanish.
Problem 2.6
x Properties are constant.
x Cartesian coordinates.
x Parallel streamlines: no velocity component in the y-direction.
x
Axial flow: no velocity component in the z-direction.
x The Navier-Stokes equations give the three momentum equations.
Problem 2.7
x Properties are constant.
x Cylindrical coordinates.
x Parallel streamlines: no velocity component in the r-direction.
x Axial flow: no velocity component in the T -direction.
x No variation in the T -direction. The Navier-Stokes equations give the three momentum
equations.
Problem 2.8
x Properties are constant.
x Cartesian coordinates.
x Two dimensional flow (no velocity component in the z-direction
x The Navier-Stokes equations give two momentum equations.
Problem 2.9
x Properties are constant.
x Cylindrical coordinates.
x Two dimensional flow (no velocity component in the T -direction.
x
The Navier-Stokes equations give two momentum equations.
Problem 2.10
x Motion in energy consideration is represented by velocity components.
x Fluid nature is represented by fluid properties.
Problem 2.11
x Properties are constant.
x Cartesian coordinates.
x Parallel streamlines: no velocity component in the y-direction.
x Axial flow: no velocity component in the z-direction.
Problem 2.12
x Properties are constant.
x Cartesian coordinates.
x Parallel streamlines: no velocity component in the y-direction.
x Axial flow: no velocity component in the z-direction.
x The fluid is an ideal gas.
Problem 2.13
x This is a two-dimensional free convection problem.
x The flow is due to gravity.
x The flow is governed by the momentum and energy equations. Thus the governing
equations are the Navier-Stokes equations of motion and the energy equation.
x
The geometry is Cartesian.
Problem 2.15
x The flow is due to gravity.
x For parallel streamlines the velocity component v = 0 in the y-direction.
x Pressure at the free surface is uniform (atmospheric).
x Properties are constant.
x The geometry is Cartesian.
Problem 2.16
x This is a forced convection problem.
x Flow properties (density and viscosity) are constant.
x Upstream conditions are uniform (symmetrical)
x The velocity vanishes at both wedge surfaces (symmetrical).
x Surface temperature is asymmetric.
x Flow field for constant property fluids is governed by the Navier-Stokes and continuity
equations.
x If the governing equations are independent of temperature, the velocity distribution over
the wedge should be symmetrical with respect to x.
x The geometry is Cartesian.
Problem 2.18
x The geometry is Cartesian.
x Properties are constant.
x Axial flow (no motion in the z-direction).
x Parallel streamlines means that the normal velocity component is zero.
x
Specified flux at the lower plate and specified temperature at the upper plate.
Problem 2.19
x The geometry is cylindrical.
x No variation in the axial and angular directions.
x Properties are constant.
Problem 2.20
x The geometry is cylindrical.
x No variation in the angular direction.
x Properties are constant.
x Parallel streamlines means that the radial velocity component is zero.
Problem 2.21
x The geometry is cylindrical. (ii)
x No variation in the axial and angular directions.
x Properties are constant.
Problem 2.22
x This is a forced convection problem.
x The same fluid flows over both spheres.
x Sphere diameter and free stream velocity affect the Reynolds number which in turn affect
the heat transfer coefficient.
Problem 2.23
x This is a free convection problem.
x The average heat transfer coefficient h depends on the vertical length L of the plate.
x L appears in the Nusselt number as well as the Grashof number.
Problem 2.24
x This is a forced convection problem.
x The same fluid flows over both spheres.
x Sphere diameter and free stream velocity affect the Reynolds number which in turn affect
the heat transfer coefficient. (iv) Newton’s law of cooling gives the heat transfer
Problem 2.25
x Dissipation is important when the Eckert number is high compared to unity.
x If the ratio of dissipation to conduction is small compared to unity, it can be neglected.
Problem 2.26
x The plate is infinite.
x No changes take place in the axial direction (infinite plate).
x This is a transient problem.
x Constant properties.
x Cartesian coordinates.
Problem 2.27
x The plate is infinite.
x No changes take place in the axial direction (infinite plate).
x This is a transient problem.
x Constant properties.
x Cartesian coordinates.
x Gravity is neglected. Thus there is no free convection.
x The fluid is stationary.
Problem 3.1
x
x
x
x
x
Moving plate sets fluid in motion in the x-direction.
Since plates are infinite the flow field does not vary in the axial direction x.
The effect of pressure gradient is negligible.
The fluid is incompressible (constant density).
Use Cartesian coordinates.
Problem 3.2
x Moving plate sets fluid in motion in the x-direction.
x
Since plates are infinite the flow field does not vary in the axial direction x.
x The effect of pressure gradient must be included.
x The fluid is incompressible.
x Using Fourier’s law, Temperature distribution gives surface heat flux of the moving
plate.
x Use Cartesian coordinates.
Problem 3.3
x Moving plate sets fluid in motion in the x-direction.
x Since plates are infinite the flow field does not vary in the axial direction x.
x The fluid is incompressible (constant density).
x Use Cartesian coordinates.
Problem 3.4
x Moving plates set fluid in motion in the positive and negative x-direction.
x Since plates are infinite the flow field does not vary in the axial direction x.
x The fluid is incompressible (constant density).
x The fluid is stationary at the center plane y = 0.
x Symmetry dictates that no heat is conducted through the center plane.
x Use Cartesian coordinates.
Problem 3.5
x Fluid motion is driven by axial pressure drop.
x For a very long tube the flow field does not vary in the axial direction z.
x The fluid is incompressible (constant density).
x Heat is generated due to viscous dissipation. It is removed from the fluid by convection at
the surface.
x The Nusselt number is a dimensionless heat transfer coefficient.
x To determine surface heat flux and heat transfer coefficient requires the determination of
temperature distribution.
x Temperature distribution depends on the velocity distribution.
x Use cylindrical coordinates.
Problem 3.6
x Fluid motion is driven by axial pressure drop.
x For a very long tube the flow field does not vary in the axial direction z.
x The fluid is incompressible (constant density).
x Use cylindrical coordinates.
Problem 3.7
x Fluid motion is driven by axial motion of the rod. Thus motion is not due to pressure
gradient.
x For a very long tube the flow field does not vary in the axial direction z.
x The fluid is incompressible (constant density).
x Heat is generated due to viscous dissipation. It is removed from the fluid by conduction at
the surface.
x The Nusselt number is a dimensionless heat transfer coefficient.
x To determine the heat transfer coefficient require the determination of temperature
distribution.
x Temperature distribution depends on the velocity distribution.
x Use cylindrical coordinates.
Problem 3.8
x Fluid motion is driven by gravity.
x No velocity and temperature variation in the axial direction.
x The fluid is incompressible (constant density).
x Heat is generated due to viscous dissipation.
x Temperature distribution depends on the velocity distribution.
x Use Cartesian coordinates.
Problem 3.9
x Fluid motion is driven by gravity.
x No velocity and temperature variation in the axial direction.
x The fluid is incompressible (constant density).
x Heat is generated due to viscous dissipation.
x Temperature distribution depends on the velocity distribution.
x the inclined surface is at specified temperature and the free surface exchanges heat by
convection with the ambient.
x Use Cartesian coordinates.
Problem 3.11
x Fluid motion is driven by shaft rotation
x The housing is stationary.
x Axial variation in velocity and temperature are negligible for a very long shaft.
x Velocity and temperature do not vary with angular position.
x The fluid is incompressible (constant density).
x Heat generated by viscous dissipation is removed from the oil at the housing.
x No heat is conducted through the shaft.
x The maximum temperature occurs at the shaft.
x Heat flux at the housing is determined from temperature distribution and Fourier’s law of
conduction.
x Use cylindrical coordinates.
Problem 3.12
x Fluid motion is driven by sleeve rotation
x The shaft is stationary.
x Axial variation in velocity and temperature are negligible for a very long shaft.
x Velocity and temperature do not vary with angular position.
x The fluid is incompressible (constant density).
x Heat generated by viscous dissipation is removed from the oil at the housing.
x No heat is conducted through the shaft.
x The maximum temperature occurs at the shaft. (ix) Use cylindrical coordinates.
Problem 3.13
x Fluid motion is driven by shaft rotation
x Axial variation in velocity and temperature are negligible for a very long shaft.
x Velocity, pressure and temperature do not vary with angular position.
x The fluid is incompressible (constant density).
x Heat generated by viscous dissipation is conducted radially.
x The determination of surface temperature and heat flux requires the determination of
temperature distribution in the rotating fluid.
x Use cylindrical coordinates.
Problem 3.14
x Axial pressure gradient sets fluid in motion.
x The fluid is incompressible.
x The flow field is determined by solving the continuity and Navier-Stokes equations.
x Energy equation gives the temperature distribution.
x Fourier’s law and temperature distribution give surface heat flux.
x Axial variation of temperature is neglected.
Problem 4.3
x This is forced convection flow over a streamlined body.
x Viscous (velocity) boundary layer approximations can be made if the Reynolds number
Rex > 100.
x Thermal (temperature) boundary layer approximations can be made if the Peclet number
Pex = Rex Pr > 100.
x The Reynolds number decreases as the distance along the plate is decreased.
Problem 4.4
x The surface is streamlined.
x The fluid is water.
x Inertia and viscous effects can be estimated using scaling.
x If a viscous term is small compared to inertia, it can be neglected.
x Properties should be evaluated at the film temperature T f
(Ts Tf ) / 2.
Problem 4.5
x The surface is streamlined.
x The fluid is water.
x Convection and conduction effects can be estimated using scaling.
x If a conduction term is small compared to convection, it can be neglected.
x The scale for G t / L depends on whether G t ! G or G t G .
x Properties should be evaluated at the film temperature T f
(Ts Tf ) / 2.
Problem 4.6
x The fluid is air.
x Dissipation and conduction can be estimated using scaling.
x Dissipation is negligible if the Eckert number is small compared to unity.
Problem 4.7
x The surface is streamlined.
x The fluid is air.
Problem 4.9
x This is a forced convection problem over a flat plate.
x At the edge of the thermal boundary layer, the axial velocity is u | Vf .
x Blasius solution gives the distribution of the velocity components u(x,y) and v(x,y).
x Scaling gives an estimate of v(x,y).
Problem 4.11
x This is a laminar boundary layer flow problem.
x Blasius solution gives the velocity distribution for the flow over a semi-infinite flat plate.
(iii) A solution for the boundary layer thickness depends on how the thickness is defined.
Problem 4.12
x Since the flow within the boundary layer is two-dimensional the vertical velocity
component does not vanish. Thus stream lines are not parallel.
x Blasius solution is valid for laminar boundary layer flow over a semi-infinite plate.
x The transition Reynolds number from laminar to turbulent flow is 5u 105 .
x Boundary layer approximations are valid if the Reynolds number is greater than 100.
Problem 4.13
x This is an external flow problem over a flat plate.
x Blasius’s solution for the velocity distribution and wall shearing stress is assumed to be
applicable.
x Of interest is the value of the local stress at the leading edge of the plate.
Problem 4.14
x This is an external flow problem over a flat plate.
x The force needed to hold the plate in place is equal to the total shearing force by the fluid
on the plate.
x Integration of wall shear over the surface gives the total shearing force.
x Blasius’s solution for the velocity distribution and wall shearing stress is assumed to be
applicable.
Problem 4.16
x This is an external forced convection problem for flow over a flat plate.
x Of interest is the region where the upstream fluid reaches the leading edge of the plate.
x The fluid is heated by the plate.
x Heat from the plate is conducted through the fluid in all directions.
x Pohlhausen’s solution assumes that heat is not conducted upstream from the plate and
therefore fluid temperature at the leading edge is the same as upstream temperature.
Problem 4.18
x This is a forced convection problem over a flat plate.
x At the edge of the thermal boundary layer, fluid temperature is T | T f .
x Pohlhausen’s solution gives the temperature distribution in the boundary layer.
x The thermal boundary layer thickness G t increases with distance from the leading edge.
x G t depends on the Prandtl number.
Problem 4.19
x This is an external forced convection problem for flow over a flat plate.
x Pohlhausen’s solution for the temperature distribution and heat transfer coefficient is
assumed to be applicable.
x Of interest is the value of the local heat flux at the leading edge of the plate.
x Knowing the local transfer coefficient and using Newton’s law, gives the heat flux
Problem 4.20
x This is an external forced convection problem for flow over a flat plate.
x Pohlhausen’s solution for the temperature distribution is assumed to be applicable.
x Of interest is the value of the normal temperature gradient at the surface.
Problem 4.22`
x This is an external forced convection problem over two flat plates.
x
Both plates have the same surface area.
x For flow over a flat plate, the heat transfer coefficient h decreases with distance from the
leading edge.
x Since the length in the flow direction is not the same for the two plates, the average heat
transfer coefficient is not the same. It follows that the total heat transfer rate is not the
same.
x The flow over a flat plate is laminar if the Reynolds number is less than 5u105.
Problem 4.23
x This is an external forced convection problem for flow over a flat plate.
x Pohlhausen’s solution for the temperature distribution and heat transfer coefficient is
assumed to be applicable.
x Of interest is the value of the heat transfer rate from a section of the plate at a specified
location and of a given width.
x Newton’s law of cooling gives the heat transfer rate.
Problem 4.24
x This is an external forced convection problem for flow over a flat plate.
x Of interest is the variation of the local heat transfer coefficient with free stream velocity
and distance from the leading edge.
x Pohlhausen's solution applies to this problem.
Problem 4.25
x This is an external flow problem.
x At the edge of the thermal boundary layer, y G t , fluid temperature approaches free
stream temperature. That is, T
Tf and T *
(Tf Ts ) /(Tf Ts ) 1 .
x According to Pohlhausen's solution, Fig. 4.6, the thermal boundary layer thickness
depends on the Prandtl number, free stream velocity Vf, kinematic viscosity Q and
location x.
Problem 4.26
x This is an external forced convection problem for flow over a flat plate.
x The Reynolds number and Peclet number should be checked to determine if the flow is
laminar and if boundary layer approximations are valid.
x Pohlhausen's solution is applicable if 100 < Rex < 100 u105 and Pex = Rex Pr > 100.
x Thermal boundary layer thickness and heat transfer coefficient vary along the plate.
x Newton’s law of cooling gives local heat flux. (vi) The fluid is water.
Problem 4.27
x This is an external forced convection problem over a flat plate.
x Increasing the free stream velocity, increases the average heat transfer coefficient. This in
turn causes surface temperature to drop.
x Based on this observation, it is possible that the proposed plan will meet design
specification.
x Since the Reynolds number at the downstream end of the package is less than 500,000, it
follows that the flow is laminar throughout.
x Increasing the free stream velocity by a factor of 3, increases the Reynolds number by a
factor of 3 to 330,000. At this Reynolds number the flow is still laminar.
x The power supplied to the package is dissipated into heat and transferred to the
surroundings from the surface.
x Pohlhausen's solution can be applied to this problem.
x The ambient fluid is unknown.
Problem 4.28
x This is an external forced convection problem of flow over a flat plate.
x Convection heat transfer from a surface can be determined using Newton’s law of
cooling.
x The local heat transfer coefficient changes along the plate. The total heat transfer rate
can be determined using the average heat transfer coefficient.
x For laminar flow, Pohlhausen's solution gives the heat transfer coefficient.
x For two in-line fins heat transfer from the down stream fin is influenced by the upstream
fin. The further the two fins are apart the less the interference will be.
Problem 4.29
x This is an external forced convection problem for flow over a flat plate.
x Pohlhausen’s solution for the temperature distribution and heat transfer coefficient is
assumed to be applicable.
x Knowing the heat transfer coefficient, the local Nusselt number can be determined.
x the Newton’s law of cooling gives the heat transfer rate.
x Pohlhausen’s solution gives the thermal boundary layer thickness.
Problem 4.30
x This is an external forced convection problem of flow over a flat plate.
x Convection heat transfer from a surface can be determined using Newton’s law of
cooling.
x The local heat transfer coefficient changes along the plate.
x For each triangle the area changes with distance along the plate.
x The total heat transfer rate can be determined by integration along the length of each
triangle.
x Pohlhausen's solution may be applicable to this problem.
Problem 4.31
x This is an external forced convection problem of flow over a flat plate.
x
Heat transfer rate can be determined using Newton’s law of cooling.
x The local heat transfer coefficient changes along the plate.
x The area changes with distance along the plate.
x The total heat transfer rate can be determined by integration along the length of the
triangle.
x Pohlhausen's solution may be applicable to this problem.
Problem 4.32
x This is an external forced convection problem of flow over a flat plate.
x Heat transfer rate can be determined using Newton’s law of cooling.
x The local heat transfer coefficient changes along the plate.
x The area changes with distance along the plate.
x The total heat transfer rate can be determined by integration along over the area of the
semi-circle.
x Pohlhausen's solution gives the heat transfer coefficient.
Problem 4.33
x This is an external forced convection problem of flow over a flat plate.
x Heat transfer rate can be determined using Newton’s law of cooling.
x The local heat transfer coefficient changes along the plate.
x The area changes with distance along the plate.
x The total heat transfer rate can be determined by integration along the length of the
triangle.
x Pohlhausen's solution may be applicable to this problem.
Problem 4.34
x This is an external forced convection problem of flow over a flat plate.
x This problem involves determining the heat transfer rate from a circle tangent to the
leading edge of a plate
x Heat transfer rate can be determined using Newton’s law of cooling.
x The local heat transfer coefficient changes along the plate.
x The area changes with distance along the plate.
x The total heat transfer rate can be determined by integration along the length of the
triangle.
x Pohlhausen's solution may be applicable to this problem.
Problem 4.36
x The flow field for this boundary layer problem is simplified by assuming that the axial
velocity is uniform throughout the thermal boundary layer.
x Since velocity distribution affects temperature distribution, the solution for the local
Nusselt number can be expected to differ from Pohlhausen’s solution.
x The Nusselt number depends on the temperature gradient at the surface.
Problem 4.37
x The flow field for this boundary layer problem is simplified by assuming that the axial
velocity varies linearly in the y-direction.
x Since velocity distribution affects temperature distribution, the solution for the local
Nusselt number can be expected to differ from Pohlhausen’s solution.
x The Nusselt number depends on the temperature gradient at the surface.
Problem 4.38
x The flow and temperature fields for this boundary layer problem are simplified by
assuming that the axial velocity and temperature do not vary in the x-direction.
x The heat transfer coefficient depends on the temperature gradient at the surface.
x Temperature distribution depends on the flow field.
x The effect of wall suction must be taken into consideration.
Problem 4.39
x This is a forced convection flow over a plate with variable surface temperature.
x The local heat flux is determined by Newton’s law of cooling.
x The local heat transfer coefficient and surface temperature vary with distance along the
plate. The variation of surface temperature and heat transfer coefficient must be such that
Newton’s law gives uniform heat flux.
x The local heat transfer coefficient is obtained from the local Nusselt number.
Problem 4.40
x This is a forced convection flow over a plate with variable surface temperature.
x The Reynolds number should be computed to determine if the flow is laminar or
turbulent.
x The local heat transfer coefficient and surface temperature vary with distance along the
plate.
x The local heat transfer coefficient is obtained from the solution to the local Nusselt
number.
x The determination of the Nusselt number requires determining the temperature gradient
at the surface.
Problem 4.41
x This is a forced convection flow over a plate with variable surface temperature.
x The Reynolds number should be computed to determine if the flow is laminar or
turbulent.
x Newton’s law of cooling gives the heat transfer rate from the plate.
x The local heat transfer coefficient and surface temperature vary with distance along the
plate. Thus determining the total heat transfer rate requires integration of Newton’s law
along the plate.
x The local heat transfer coefficient is obtained from the local Nusselt number.
Problem 4.42
x This is an external forced convection problem of flow over a flat plate
x Convection heat transfer from a surface can be determined using Newton’s law of
cooling.
x The local heat transfer coefficient and surface temperature vary along the plate.
x For each triangle the area varies with distance along the plate.
x The total heat transfer rate can be determined by integration along the length of each
triangle.
Problem 4.43
x This is a forced convection boundary layer flow over a wedge.
x Wedge surface is maintained at uniform temperature.
x The flow is laminar.
x The fluid is air.
x Similarity solution for the local Nusselt number is presented in Section 4.4.3.
x The Nusselt number depends on the Reynolds number and the dimensionless temperature
gradient at the surface dT (0) / dK. (vii) Surface temperature gradient depends on wedge
angle.
Problem 4.44
x This is a forced convection boundary layer flow over a wedge.
x Wedge surface is maintained at uniform temperature.
x The flow is laminar.
x The average Nusselt number depends on the average heat transfer coefficient..
x Similarity solution for the local heat transfer coefficient is presented in Section 4.4.3.
Problem 4.45
x Newton’s law of cooling gives the heat transfer rate from a surface.
x Total heat transfer from a surface depends on the average heat transfer coefficient h .
x Both flat plate and wedge are maintained at uniform surface temperature.
x Pohlhausen’s solution gives h for a flat plate.
x Similarity solution for the local heat transfer coefficient for a wedge is presented in
Section 4.4.3.
Problem 4.46
x The flow field for this boundary layer problem is simplified by assuming that axial
velocity within the thermal boundary layer is the same as that of the external flow.
x Since velocity distribution affects temperature distribution, the solution for the local
Nusselt number differs from the exact case of Section 4.4.3.
x The local Nusselt number depends the local heat transfer coefficient which depends on
the temperature gradient at the surface.
Problem 4.47
x The flow field for this boundary layer problem is simplified by assuming that axial
velocity within the thermal boundary layer varies linearly with the normal distance.
x Since velocity distribution affects temperature distribution, the solution for the local
Nusselt number differs from the exact case of Section 4.4.3.
x The local Nusselt number depends on the local heat transfer coefficient which depends on
the temperature gradient at the surface.
Problem 5.1
x In general, to determine the Nusselt number it is necessary to determine the velocity and
temperature distribution
x Fluid velocity for Pr 1 is assumed to be uniform, u | Vf . This represents a
significant simplification.
Problem 5.2
x In general, to determine the Nusselt number it is necessary to determine the velocity and
temperature distribution.
x Fluid velocity for Pr !! 1 is assumed to be linear, u
Vf ( y / G ) .
Problem 5.3
x The velocity is assumed to be uniform, u
Vf , throughout the thermal boundary layer.
x A leading section of length xo is unheated.
x at x ! xo , surface heat flux is uniform.
x The determination of the Nusselt number requires the determination of the temperature
distribution.
x Surface temperature is unknown.
x The maximum surface temperature for a uniformly heated plate occurs at the trailing end.
Problem 5.4
x The velocity distribution is known.
x Surface temperature is uniform.
x The determination of the Nusselt number requires the determination of the temperature
distribution.
x Newton’s law of cooling gives the heat transfer rate. This requires knowing the local heat
transfer coefficient.
Problem 5.5
x The velocity distribution is known
x Total heat transfer is equal to heat flux times surface area.
x Heat flux is given. However, the distance x = L at which G t
H / 2 is unknown.
Problem 5.6
x The determination of the Nusselt number requires the determination of the velocity and
temperature distributions.
x Velocity is assumed uniform.
x Surface temperature is variable.
x Newton’s law of cooling gives surface heat flux. This requires knowing the local heat
transfer coefficient.
Problem 5.7
x The determination of the Nusselt number requires the determination of the velocity and
temperature distributions.
x Velocity is assumed linear.
x Surface temperature is variable.
x Newton’s law of cooling gives surface heat flux. This requires knowing the local heat
transfer coefficient.
Problem 5.8
x The determination of the Nusselt number requires the determination of the velocity and
temperature distributions.
x Velocity is assumed uniform.
x Surface temperature is variable.
x Newton’s law of cooling gives surface heat flux. This requires knowing the local heat
transfer coefficient.
Problem 5.9
x In general, to determine the Nusselt number it is necessary to determine the velocity and
temperature distribution.
x Surface heat flux is variable. It decreases with distance x.
x Surface temperature is unknown.
x Newton’s law of cooling gives surface temperature. This requires knowing the local heat
transfer coefficient. (v) G t / G 1.
Problem 5.10
x In general, to determine the Nusselt number it is necessary to determine the velocity and
temperature distribution.
x Fluid velocity for Pr 1 is assumed to be uniform, u | Vf . This represents a
significant simplification.
x Surface heat flux is variable. It increases with distance x.
x Surface temperature is unknown. Since flux increases with x and heat transfer coefficient
decreases with x, surface temperature is expected to increase with x. Thus maximum
surface temperature is at the trailing end x = L.
x Newton’s law of cooling gives surface temperature. This requires knowing the local heat
transfer coefficient.
Problem 5.11
x In general, to determine the Nusselt number it is necessary to determine the velocity and
temperature distribution.
x Fluid velocity for Pr 1 is assumed to be uniform, u | Vf . This represents a
significant simplification.
x Surface heat flux is variable. It increases with distance x.
x Surface temperature is unknown. Since flux increases with x and heat transfer coefficient
decreases with x, surface temperature is expected to increase with x. Thus maximum
surface temperature is at the trailing end x = L.
x Newton’s law of cooling gives surface temperature. This requires knowing the local heat
transfer coefficient.
Problem 5.12
x This problem is described by cylindrical coordinates.
x Velocity variation with y is negligible.
x Conservation of mass requires that radial velocity decrease with radial distance r.
x Surface temperature is uniform.
Problem 5.13
x This problem is described by cylindrical coordinates.
x Velocity variation with y is negligible.
x Conservation of mass requires that radial velocity decrease with radial distance r.
x Surface heat flux is uniform
x Surface temperature is unknown.
Problem 5.14
x In general, to determine the Nusselt number it is necessary to determine the velocity and
temperature distribution.
x Fluid velocity for Pr 1 is assumed to be uniform, u | Vf . This represents a
significant simplification.
x The plate is porous.
x Fluid is injected through the plate with uniform velocity.
x The plate is maintained at uniform surface temperature.
x A leading section of the plate is insulated.
Problem 5.15
x In general, to determine the Nusselt number it is necessary to determine the velocity and
temperature distribution.
x Fluid velocity for Pr 1 is assumed to be uniform, u | Vf . This represents a
significant simplification.
x The plate is porous.
x Fluid is injected through the plate with uniform velocity.
x The plate is heated with uniform surface flux
x Surface temperature is unknown, (vii) A leading section of the plate is insulated.
Problem 5.16
x There are two thermal boundary layers in this problem.
x The upper and lower plates have different boundary conditions. Thus, temperature
distribution is not symmetrical.
x The lower plate is at uniform temperature while heat is removed at uniform flux along the
upper plate.
x Fluid velocity is assumed uniform throughout the channel.
Problem 6.1
x This is an internal forced convection problem.
x Scaling gives estimates of Lh and Lt .
x Exact solutions for Lh and Lt are available for laminar flow through channels.
x Exact solutions for Lt depend on channel geometry and surface boundary conditions.
Problem 6.2
x This is an internal forced convection problem.
x Scaling gives estimates of Lh and
x Exact solutions for Lh and Lt are available for laminar flow through channels.
x Exact solutions for Lt depend on channel geometry and surface boundary conditions.
Problem 6.3
x This is an internal force convection problem.
x The channel is a long tube.
x The surface is maintained at a uniform temperature.
x Since the tube section is far away from the entrance, the velocity and temperature can be
assumed fully developed.
x Tube diameter, mean velocity and inlet, outlet and surface temperatures are known. The
length is unknown.
x The fluid is air.
Problem 6.4
x This is an internal force convection in a tube.
x The surface is heated at uniform flux.
x Surface temperature increases along the tube and is unknown.
x The flow is assumed laminar and fully developed.
x The heat transfer coefficient for fully developed flow through channels is constant.
x According to Newton’s law of cooling, surface temperature is related to mean fluid
temperature, surface heat flux and heat transfer coefficient.
Problem 6.5
x This is an internal force convection in a tube.
x The surface is heated at uniform flux.
x Surface temperature increases along the tube and is unknown.
x The flow is assumed laminar and fully developed.
x The heat transfer coefficient for fully developed flow through channels is constant.
x According to Newton’s law of cooling, surface temperature is related to mean fluid
temperature, surface heat flux and heat transfer coefficient.
Problem 6.6
x This is an internal forced convection problem in a tube.
x The surface is heated at uniform flux.
x Surface temperature changes along the tube and is unknown.
x The Reynolds number should be checked to determine if the flow is laminar or turbulent.
x If hydrodynamic and thermal entrance lengths are small compared to tube length, the
flow can be assumed fully developed throughout.
x For fully developed flow, the heat transfer coefficient is uniform.
x The length of the tube is unknown.
x The fluid is water.
Problem 6.7
x This is an internal force convection problem.
x The channel is a tube.
x The surface is maintained at a uniform temperature.
x Entrance effect is important in this problem.
x The average Nusselt number for a tube of length L depends on the average heat transfer
coefficient over the length.
Problem 6.8
x This is an internal forced convection problem.
x The fluid is heated at uniform wall flux.
x Surface temperature changes with distance along the channel. It reaches a maximum
value at the outlet.
x The Reynolds and Peclet numbers should be checked to establish if the flow is laminar or
turbulent and if this is an entrance or fully developed problem.
x The channel has a square cross-section.
x Application of Newton’s law of cooling at the outlet relates outlet temperature to surface
temperature, surface flux and heat transfer coefficient.
x Application of conservation of energy gives a relationship between heat added, inlet
temperature, outlet temperature, specific heat and mass flow rate.
Problem 6.9
x This is an internal forced convection problem in tubes.
x The flow is laminar and fully developed.
x The surface is maintained at uniform temperature.
x All conditions are identical for two experiments except the flow rate through one is half
that of the other.
x The total heat transfer rate depends on the outlet temperature.
Problem 6.10
x This is an internal forced convection problem.
x The channel has a rectangular cross section.
x Surface temperature is uniform.
x The Reynolds and Peclet numbers should be checked to establish if the flow is laminar or
turbulent and if entrance effects can be neglected.
x Channel length is unknown.
x The fluid is air.
Problem 6.11
x This is an internal force convection problem.
x The channel is a rectangular duct.
x The surface is maintained at a uniform temperature.
x The velocity and temperature are fully developed.
x The Reynolds number should be checked to determine if the flow is laminar or turbulent.
x Duct size, mean velocity and inlet, outlet and surface temperatures are known. The
length is unknown. (vii) Duct length depends on the heat transfer coefficient.
x The fluid is water.
Problem 6.12
x This is an internal forced convection problem in a channel.
x The surface is heated at uniform flux.
x Surface temperature changes along the channel. It reaches a maximum value at the outlet.
x The Reynolds number should be checked to determine if the flow is laminar or turbulent.
x Velocity and temperature profiles become fully developed far away from the inlet.
x The heat transfer coefficient is uniform for fully developed flow.
x The channel has a square cross section.
x tube length is unknown. (ix) The fluid is air.
Problem 6.13
x This is an internal forced convection problem in tubes.
x The flow is laminar and fully developed.
x The surface is maintained at uniform temperature.
x All conditions are identical for two tubes except the diameter of one is twice that of the
other.
x The total heat transfer in each tube depends on the outlet temperature.
Problem 6.14
x This is an internal forced convection problem.
x Equation (6.3) gives scaling estimate of the thermal entrance length.
x Equation (6.20b) gives scaling estimate of the local Nusselt number.
x The Graetz problem deals with laminar flow in the entrance of a tube at uniform surface
temperature.
x Graetz solutions gives the thermal entrance length (distance to reach fully developed
temperature) and local Nusselt number.
Problem 6.15
x This is an internal forced convection problem.
x Equation (6.20b) gives scaling estimate of the local Nusselt number.
x The Graetz problem deals with laminar flow in the entrance of a tube at uniform surface
temperature.
Problem 6.16
x This is an internal forced convection problem in a tube.
x
The velocity is fully developed.
x The temperature is developing.
x Surface is maintained at uniform temperature.
x The Reynolds number should be computed to establish if flow is laminar or turbulent.
x Tube length is unknown.
x The determination of tube length requires determining the heat transfer coefficient.
Problem 6.17
x This is an internal forced convection problem in a tube.
x The velocity is fully developed.
x The temperature is developing.
x Surface is maintained at uniform temperature.
x The Reynolds number should be computed to establish if flow is laminar or turbulent.
x Outlet mean temperature is unknown.
x The determination of outlet temperature
coefficient.
requires determining the heat transfer
x Since outlet temperature is unknown, air properties can not be determined. Thus a trial
and error procedure is needed to solve the problem.
Problem 6.18
x This is an internal forced convection problem in a tube.
x The velocity is fully developed and the temperature is developing.
x The surface is heated with uniform flux.
x The Reynolds number should be computed to establish if the flow is laminar or turbulent.
x Compute thermal entrance length to determine if it can be neglected.
x Surface temperature varies with distance from entrance. It is maximum at the outlet. Thus
surface temperature at the outlet is known.
x Analysis of uniformly heated channels gives a relationship between local surface
temperature, heat flux and heat transfer coefficient.
x The local heat transfer coefficient varies with distance form the inlet.
x Knowing surface heat flux, the required power can be determined.
x Newton’s law of cooling applied at the outlet gives outlet temperature.
Problem 6.19
x This is an internal forced convection problem in a rectangular channel.
x The velocity is fully developed and the temperature is developing.
x The surface is maintained at uniform temperature.
x The Reynolds number should be computed to establish if the flow is laminar or turbulent.
x Compute entrance lengths to determine if they can be neglected
x Surface flux varies with distance from entrance. It is minimum at outlet.
x Newton’s law gives surface flux in terms of the local heat transfer coefficient h(x) and
the local mean temperature Tm (x) .
x The local and average heat transfer coefficient decrease with distance form the inlet.
x The local mean temperature depends on the local average heat transfer coefficient
h (x). (x) Surface temperature is unknown.
Problem 7.2
x This is an external free convection problem over a vertical plate.
x The Rayleigh number should be computed to determine if the flow is laminar or
turbulent.
x The solution for laminar flow is given in Section 7.4
x For laminar flow, Fig.7.2 gives the viscous boundary layer thickness G and Fig. 7.3 gives
the thermal boundary layer thickness G t .
x Newton’s law of cooling gives the heat transfer rate.
x Equation (7.23) gives the average heat transfer coefficient h . (vii) The fluid is water.
Problem 7.3
x This is an external free convection problem for flow over a vertical plate.
x Laminar flow solution for temperature distribution for a plate at uniform surface
temperature is given in Fig. 7.3 .
x The dimensionless temperature gradient at the surface is given in Table 7.1.
x The solution depends on the Prandtl number.
Problem 7.4
x This is a free convection problem.
x Heat is lost from the door to the surroundings by free convection and radiation.
x To determine the rate of heat loss, the door can by modeled as a vertical plate losing
heat by free convection to an ambient air.
x As a first approximation, radiation can be neglected.
x Newton’s law of cooling gives the rate of heat transfer.
x The Rayleigh number should be computed to determine if the flow is laminar or
turbulent.
x For laminar flow the solution of Section 7.4 is applicable.
Problem 7.5
x This is a free convection and radiation problem.
x The geometry is a vertical plate.
x Surface temperature is uniform.
x Newton’s law of cooling gives convection heat transfer rate while Stefan-Boltzmann law
gives radiation heat transfer rate.
x The Rayleigh number should be computed to determine if the flow is laminar or
turbulent.
x For laminar flow the solution of Section 7.4 is applicable.
x Since radiation heat transfer is considered in this problem, all temperatures should be
expressed
Problem 7.6
x This is a free convection problem.
x The power dissipated in the electronic package is transferred to the ambient fluid by
free convection.
x As the power is increased, surface temperature increases.
x The maximum power dissipated corresponds to the maximum allowable surface
temperature.
x Surface temperature is related to surface heat transfer by Newton’s law of cooling.
x The problem can be modeled as free convection over a vertical plate.
x The Rayleigh number should be computed to determine if the flow is laminar or
turbulent.
x For laminar flow the solution of Section 7.4 is applicable.
x The fluid is air.
Problem 7.7
x This is a free convection problem.
x The power dissipated in the electronic package is transferred to the ambient fluid by
free convection.
x As the power is increased, surface temperature increases.
x The maximum power dissipated corresponds to the maximum allowable surface
temperature.
x Surface temperature is related to surface heat transfer by Newton’s law of cooling.
x The problem can be modeled as free convection over a vertical plate.
x The Rayleigh number should be computed to determine if the flow is laminar or
turbulent.
x For laminar flow the solution of Section 7.4 is applicable.
x The fluid is water.
Problem 7.8
x This is a free convection problem.
x The surface is maintained at uniform temperature.
x Newton’s law of cooling determines the heat transfer rate.
x Heat transfer rate depends on the heat transfer coefficient.
x The heat transfer coefficient decreases with distance from the leading edge of the plate.
x The width of each triangle changes with distance from the leading edge.
x For laminar flow the solution of Section 7.4 is applicable.
Problem 7.9
x This is a free convection problem over a vertical plate.
x The surface is maintained at uniform temperature.
x Local heat flux is determined by Newton’s law of cooling.
x Heat flux depends on the local heat transfer coefficient
x Free convection heat transfer coefficient for a vertical plate decreases with distance from
the leading edge. Thus, the flux also decreases.
x The Rayleigh number should be computed to determine if the flow is laminar or
turbulent. For
x Laminar flow the solution of Section 7.4 is applicable.
x The fluid is air.
Problem 7.10
x This is a free convection problem over a vertical plate.
x The power dissipated in the chips is transferred to the air by free convection.
x This problem can be modeled as free convection over a vertical plate with constant
surface heat flux.
x Surface temperature increases as the distance from the leading edge is increased. Thus,
the maximum surface temperature occurs at the top end of the plate (trailing end).
x The Rayleigh number should be computed to determine if the flow is laminar or
turbulent.
x For laminar flow the analysis of Section 7.5 gives surface temperature distribution.
x The fluid is air.
x Properties depend on the average surface temperature Ts . Since Ts is unknown, the
problem must be solved by trail and error.
Problem 7.11
x This is a free convection problem over a vertical plate.
x The power dissipated in the chips is transferred to the air by free convection
x This problem can be modeled as free convection over a vertical plate with constant
surface heat flux.
x Surface temperature increases as the distance from the leading edge is increased. Thus,
the maximum surface temperature occurs at the top end of the plate (trailing end).
x The Rayleigh number should be computed to determine if the flow is laminar or
turbulent.
x For laminar flow the analysis of Section 7.5 gives surface temperature distribution.
x The fluid is air.
x Properties depend on the average surface temperature Ts . Since Ts is unknown, the
problem must be solved by trail and error.
Problem 7.12
x This is a free convection problem over a vertical plate at uniform surface temperature.
x In general, to determine the Nusselt number it is necessary to determine the velocity and
temperature distribution.
x The integral method can be used to determine the velocity and temperature distribution.
x Application of the integral method reduces to determining the velocity and temperature
boundary layer thickness.
Problem 7.13
x This is a free convection problem over a vertical plate at uniform surface heat flux.
x In general, to determine the Nusselt number it is necessary to determine the velocity and
temperature distribution.
x The integral method can be used to determine the velocity and temperature distribution.
x Application of the integral method reduces to determining the velocity and temperature
boundary layer thickness.
Problem 8.1
x This is an external forced convection problem.
x The geometry can be modeled as a flat plate.
x Surface temperature is uniform.
x Newton’s law of cooling gives heat transfer rate from the surface to the air.
x The average heat transfer coefficient must be determined.
x The Reynolds number should be evaluated to establish if the flow is laminar, turbulent or
mixed.
x Analytic or correlation equations give the heat transfer coefficient.
Problem 8.2
x This is an external forced convection problem.
x The geometry can be modeled as a flat plate.
x Surface temperature is uniform.
x To determine the heat flux at a given location, the local heat transfer coefficient must be
determined.
x The average heat transfer coefficient is needed to determine the total heat transfer rate.
x Newton’s law of cooling gives surface flux and total heat transfer rate.
x The Reynolds number should be checked to establish if the flow is laminar, turbulent or
mixed.
x Analytic or correlation equations give the heat transfer coefficient.
Problem 8.3
x This is an external forced convection problem of flow over a flat plate.
x Surface temperature is assumed uniform.
x The heat transfer coefficient in turbulent flow is greater than that in laminar flow.
Thus higher heat transfer rates can be sustained in turbulent flow than laminar flow.
x The Reynolds number should be checked to establish if the flow is laminar, turbulent or
mixed.
x Heat loss from the surface is approximately equal to the power dissipated in the package.
x Newton’s law of cooling gives a relationship between heat transfer rate, surface area, heat
transfer coefficient, surface temperature and ambient temperature.
x The fluid is air.
Problem 8.4
x This is an external forced convection problem.
x The geometry is a flat plate.
x Surface temperature is uniform.
x Newton’s law of cooling gives the heat transfer rate.
x The Reynolds number should be checked to establish if the flow is laminar, turbulent or
mixed.
x Analytic or correlation equations give the heat transfer coefficient.
x If the flow is laminar throughout, heat transfer from the first half should be greater than
that from the second half.
x Second half heat transfer can be obtained by subtracting first half heat rate from the heat
transfer from the entire plate.
x The fluid is water.
Problem 8.5
x The chip is cooled by forced convection.
x This problem can be modeled as a flat plate with an unheated leading section.
x Newton's law of cooling can be applied to determine the rate of heat transfer between the
chip and the air.
x Check the Reynolds number to establish if the flow is laminar or turbulent.
Problem 8.6
x Heat transfer from the collector to the air is by forced convection.
x This problem can be modeled as a flat plate with an unheated leading section.
x Newton's law of cooling can be applied to determine the rate of heat transfer between the
collector and air.
x The heat transfer coefficient varies along the collector.
x The Reynolds number should be computed to establish if the flow is laminar or turbulent.
Problem 8.7
x This is an external forced convection problem.
x The flow is over a flat plate.
x Surface temperature is uniform.
x Plate orientation is important.
x Variation of the heat transfer coefficient along the plate affects the total heat transfer.
x The heat transfer coefficient for laminar flow decreases as the distance from the leading
edge is increased. However, at the transition point it increases and then decreases again.
x Higher rate of heat transfer may be obtained if the wide side of a plate faces the flow. On
the other hand, higher rate may be obtained if the long side of the plate is in line with the
flow direction when transition takes place
x The fluid is water.
Problem 8.8
x This is an external forced convection problem.
x The flow is over a flat plate.
x The problem can be modeled as flow over a flat plate with uniform surface heat flux.
x Surface temperature varies with distance along plate. The highest surface temperature is
at the trailing end.
x Tripping the boundary layer at the leading edge changes the flow from laminar to
turbulent. This increases the heat transfer coefficient and lowers surface temperature.
x Newton’s law of cooling gives surface temperature.
Problem 8.9
x This is an external forced convection problem.
x The flow is normal to a tube.
x Surface temperature is uniform.
x Tube length is unknown.
x Newton’s law of cooling can be used to determine surface area. Tube length is related to
surface area.
x The fluid is water.
Problem 8.10
x Heat is removed by the water from the steam causing it to condense.
x The rate at which steam condenses inside the tube depends on the rate at which heat is
removed from the outside surface.
x Heat is removed from the outside surface by forced convection.
x This is an external forced convection problem of flow normal to a tube. (v) Newton’s law
of cooling gives the rate of heat loss from the surface.
Problem 8.11
x Electric power is dissipated into heat and is removed by the water.
x
This velocity measuring concept is based on the fact that forced convection heat transfer
is affected by fluid velocity.
x Velocity affects the heat transfer coefficient which in term affects surface temperature.
x Newton’s law of cooling relates surface heat loss to the heat transfer coefficient, surface
area and surface temperature.
x This problem can be modeled as external flow normal to a cylinder.
x The fluid is water.
Problem 8.12
x This is an external forced convection problem.
x The flow is normal to a rod.
x Surface heat transfer rate per unit length is known. However, surface temperature is
unknown.
x In general, surface temperature varies along the circumference. However, the rod can be
assumed to have a uniform surface temperature.
x This problem can be modeled as forced convection normal to a rod with uniform surface
flux or temperature.
x Newton’s law of cooling gives surface temperature.
x The fluid is air.
Problem 8.13
x Electric power is dissipated into heat and is removed by the fluid.
x This velocity measuring instrument is based on the fact that forced convection heat
transfer is affected by fluid velocity.
x velocity affects the heat transfer coefficient which in term affects surface temperature and
heat flux.
x Newton’s law of cooling relates surface heat loss to the heat transfer coefficient, surface
area and surface temperature.
x This problem can be modeled as external flow normal to a cylinder.
x The fluid is air.
Problem 8.14
x The sphere cools off as it drops. Heat loss from the sphere is by forced convection.
x The height of the building can be determined if the time it takes the sphere to land is
known.
x Time to land is the same as cooling time.
x Transient conduction determines cooling time.
x If the Biot number is less than 0.1, lumped capacity method can be used to determine
transient temperature.
x Cooling rate depends on the heat transfer coefficient.
Problem 8.15
x The electric energy dissipated inside the sphere is removed from the surface as heat by
forced convection.
x This problem can be modeled as external flow over a sphere.
x Newton’s law of cooling relates heat loss from the surface to heat transfer coefficient,
surface area and surface temperature. (iv) The fluid is air.
Problem 8.16
x The sphere cools off as it drops. Heat loss from the sphere is by forced convection.
x This is an external flow problem with a free stream velocity that changes with time.
x This is a transient conduction problem. The cooling time is equal to the time it takes the
sphere to drop to street level.
x If the Biot number is less than 0.1, lumped capacity method can be used to determine
transient temperature.
x Cooling rate depends on the heat transfer coefficient.
Problem 8.17
x This is an internal forced convection problem.
x The channel is a tube.
x The outside surface is maintained at a uniform temperature.
x Neglecting tube thickness resistance means that the inside and outside surface
temperatures are identical.
x Fluid temperature is developing.
x Inlet and outlet temperatures are known.
x The Reynolds number should be determined to establish if the flow is laminar or
turbulent.
x The required tube length depends on the heat transfer coefficient.
x The fluid is water.
Problem 8.18
x This is an internal force convection problem.
x The channel is a tube.
x The surface is maintained at a uniform temperature.
x The velocity is fully developed.
x The temperature is developing.
x The outlet temperature is unknown.
x
The Reynolds number should be checked to establish if the flow is laminar or turbulent.
x The fluid is air.
Problem 8.19
x This is an internal forced convection problem
x Tube surface is maintained at uniform temperature.
x The velocity is fully developed.
x The length of tube is unknown.
x The temperature is developing. However, depending on tube length relative to the
thermal entrance length, temperature may be considered fully developed throughout.
x The Reynolds number should be checked to determine if the flow is laminar or turbulent.
x The fluid is water.
Problem 8.20
x This is an internal forced convection problem.
x Tube surface is maintained at a uniform temperature.
x The velocity and temperature are developing. Thus, entrance effects may be important.
x The outlet temperature is unknown.
x The fluid is air.
Problem 8.21
x This is an internal forced convection problem.
x Tube surface is maintained at uniform temperature.
x The section of interest is far away from the inlet. This means that flow and temperature
can be assumed fully developed and the heat transfer coefficient uniform.
x It is desired to determine the surface flux at this section. Newton’s law of cooling gives a
relationship between local flux, surface temperature and heat transfer coefficient.
x The Reynolds number should be checked to determine if the flow is laminar or turbulent.
x The fluid is water.
Problem 8.22
x This is an internal forced convection problem in a tube.
x Both velocity and temperature are fully developed.
x Tube surface is maintained at uniform temperature.
x The Reynolds number should be computed to establish if flow is laminar or turbulent.
x Mean velocity, mean inlet and outlet temperatures and tube diameter are known.
x The fluid is air.
Problem 8.23
x This is an internal forced convection problem.
x The surface of each tube is maintained at uniform temperature which is the same for both.
x The velocity and temperature are fully developed. Thus, the heat transfer coefficient is
uniform.
x Air flows through each tube at different rates.
x The Reynolds number should be computed to establish if the flow is laminar or turbulent.
x Surface heat flux depends on the heat transfer coefficient.
Problem 8.24
x This is an internal forced convection problem.
x The geometry consists of two concentric tubes.
x Air flows in the inner tube while water flows in the annular space between the two tubes.
x The Reynolds number should be computed for both fluids to establish if the flow is
laminar or turbulent.
x Convection resistance depends on the heat transfer coefficient.
Problem 8.25
x This is an internal forced convection problem.
x The geometry consists of a tube concentrically placed inside a square duct,.
x Water flows in the tube and the duct.
x The Reynolds number should be computed for the two fluids to establish if the flow is
laminar or turbulent.
x Far away from the inlet the velocity and temperature may be assumed fully developed.
Problem 8.26
x Heat is lost from the door to the surroundings by free convection and radiation.
x
To determine the rate of heat loss, the door can by modeled as a vertical plate losing
heat by free convection to an ambient air and by radiation to a large surroundings.
x Newton’s law of cooling gives the rate of heat transfer by convection and StefanBoltzmann relation gives the heat loss by radiation.
Problem 8.27
x This is a free convection and radiation problem.
x The geometry is a vertical plate.
x Surface temperature is uniform.
x Newton’s law of cooling gives convection heat transfer rate while Stefan-Boltzmann law
gives radiation heat transfer rate.
x The Rayleigh number should be computed to determine if the flow is laminar or
turbulent.
x Since radiation heat transfer is considered in this problem, all temperatures should be
expressed in degrees kelvin.
x The fluid is air.
Problem 8.28
x This is a free convection problem.
x The power dissipated in the electronic package is transferred to the ambient fluid by
free convection.
x As the power is increased, surface temperature increases.
x The maximum power dissipated corresponds to the maximum allowable surface
temperature.
x Surface temperature is related to surface heat transfer by Newton’s law of cooling.
x The problem can be modeled as free convection over a vertical plate.
x The Rayleigh number should be computed to determine if the flow is laminar or
turbulent.
x The fluid is air.
Problem 8.29
x This is a free convection problem.
x The power dissipated in the electronic package is transferred to the ambient fluid by
free convection.
x As the power is increased, surface temperature increases.
x The maximum power dissipated corresponds to the maximum allowable surface
temperature.
x Surface temperature is related to surface heat transfer by Newton’s law of cooling.
x The problem can be modeled as free convection over a vertical plate.
x The Rayleigh number should be computed to determine if the flow is laminar or
turbulent.
x The fluid is water.
Problem 8.30
x This is a free convection problem.
x The surface is maintained at uniform temperature.
x The heat transfer coefficient decreases with distance from the leading edge of the plate.
x The heat transfer rate from the lower half 1 is greater than that from the upper half 2.
x Total heat transfer from each half can be determined using the average heat transfer
coefficient.
x Heat transfer from the upper half is equal to the heat transfer from the entire plate minus
heat transfer from the lower half.
Problem 8.31
x This is a free convection problem.
x The surface is maintained at uniform temperature.
x The heat transfer coefficient decreases with distance from the leading edge of the plate.
x The width of each triangle changes with distance from the leading edge.
Problem 8.32
x This is a free convection problem over a vertical plate.
x The surface is maintained at uniform temperature.
x Local heat flux is determined by Newton’s law of cooling.
x Heat flux depends on the local heat transfer coefficient.
x Free convection heat transfer coefficient for a vertical plate decreases with distance from
the leading edge. Thus, the flux also decreases.
x The Rayleigh number should be computed to select an appropriate Nusselt number
correlation equation.
x The fluid is air.
Problem 8.33
x This is a free convection problem over a vertical plate.
x The power dissipated in the chips is transferred to the air by free convection.
x This problem can be modeled as free convection over a vertical plate with constant
surface heat flux.
x Surface temperature increases as the distance from the leading edge is increased. Thus,
the maximum surface temperature occurs at the top end of the plate (trailing end).
x Newton’s law of cooling relates surface temperature to heat flux and heat transfer
coefficient.
x The fluid is air.
Problem 8.34
x Power supply to the disk is lost from the surface to the surroundings by free convection
and radiation.
x To determine the rate of heat loss, the disk can by modeled as a horizontal plate losing
heat by free convection to an ambient air and by radiation to a large surroundings.
x Newton’s law of cooling gives the rate of heat transfer by convection and StefanBoltzmann relation gives the heat loss by radiation.
x Free convection correlations give the heat transfer coefficient.
x Conservation of energy at the surface gives the emissivity, if it is the only unknown.
Problem 8.35
x This is a free convection problem.
x The geometry is a flat plate.
x Heat transfer from two plates is to be compared. One plate is vertical and the other is
inclined. Both plates fit in the same vertical space. Thus, the inclined plate is longer than
the vertical plate.
x Both plates are maintained at uniform surface temperature.
x Heat transfer depends on surface area and average heat transfer coefficient.
Problem 8.36
x This is a free convection problem.
x The kiln has four vertical sides and a horizontal top.
x All surfaces are at the same uniform temperature.
x Newton’s law of cooling gives the heat transfer rate.
x The sides can be modeled as vertical plates and the top as a horizontal plate.
x The fluid is air.
Problem 8.37
x Heat transfer from the surface is by free convection and radiation.
x The burner can be modeled as a horizontal disk with its heated side facing down.
x Newton’s law of cooling gives heat transfer by convection and Stefan-Boltzmann
relations gives heat transfer by radiation.
x Both convection and radiation depend on surface temperature.
x If the burner is well insulated at the bottom heated surface and its rim, then the electric
power supply is equal to surface heat transfer.
Problem 8.38
x Heat transfer from the surface is by free convection and radiation.
x The sample can be modeled as a horizontal disk with its heated side facing down or up.
x Newton’s law of cooling gives heat transfer by convection and Stefan-Boltzmann relation
gives heat transfer by radiation.
x Radiation depends on surface emissivity.
x If the disk is well insulated at the heated surface and its rim, then the electric power
supply is equal to surface heat transfer.
x Since the electric power is the same for both orientations, it follows that surface heat
transfer rate is also the same.
x Each orientation has its own Nusselt number correlation equation.
Problem 8.39
x This is a free convection problem.
x Heat is transferred from the cylindrical surface and top surface of tank to the ambient air.
x Under certain conditions a vertical cylindrical surface can be modeled as a vertical plate.
x Newton’s law of cooling gives the heat transfer rate from tank.
x The fluid is air.
Problem 8.40
x This is a free convection problem.
x The geometry is a horizontal round duct.
x Heat is transferred from duct surface to the ambient air.
x According to Newton’s law of cooling, the rate of heat transfer depends on the heat
transfer coefficient, surface area and surface and ambient temperatures.
Problem 8.41
x This is a free convection problem.
x The geometry is a horizontal pipe.
x Heat is transferred from pipe surface to the ambient air.
x Adding insulation material reduces heat loss from pipe.
x According to Newton’s law of cooling, the rate of heat transfer depends on the heat
transfer coefficient, surface area and surface and ambient temperatures
x Heat transfer coefficient and surface area change when insulation is added.
x The fluid is air.
Problem 8.42
x This is a free convection problem.
x The geometry is a horizontal wire (cylinder).
x Under steady state conditions the power dissipated in the wire is transferred to the
surrounding air.
x According to Newton’s law of cooling, surface temperature is determined by the heat
transfer rate, heat transfer coefficient, surface area and ambient temperature.
x The fluid is air.
Problem 8.43
x This is a free convection problem.
x The geometry is a horizontal tube.
x Under steady state conditions the power dissipated in the neon tube is transferred to the
surrounding air.
x According to Newton’s law of cooling, surface temperature is determined by the heat
transfer rate, heat transfer coefficient, surface area and ambient temperature.
x The fluid is air.
Problem 8.44
x This is a free convection problem.
x The geometry is a round horizontal round duct.
x Heat is transferred from the ambient air to the duct.
x According to Newton’s law of cooling, the rate of heat transfer to the surface depends on
the heat transfer coefficient, surface area and surface and ambient temperatures.
x The fluid is air.
Problem 8.45
x This is a free convection and radiation problem
x The geometry is a sphere.
x Under steady state conditions the power dissipated in the bulb is transferred to the
surroundings by free convection and radiation and through the base by conduction.
x According to Newton’s law of cooling and Stefan-Boltzmann radiation law, heat loss
from the surface depends on surface temperature.
x The ambient fluid is air.
Problem 8.46
x At steady state, power supply to the sphere must be equal to the heat loss from the surface
x Heat loss from the surface is by free convection.
x The surface is maintained at uniform temperature.
Problem 8.47
x Heat is transferred from the ambient air to the water in the fish tank.
x Adding an air enclosure reduces the rate of heat transfer.
x To estimate the reduction in cooling load, heat transfer from the ambient air to the water
with and without the enclosure must be determined.
x ) Neglecting the thermal resistance of glass, the resistance to heat transfer form the air to
the water is primarily due to the air side free convection heat transfer coefficient.
x Installing an air cavity introduces an added thermal resistance.
x The problem can be modeled as a vertical plate and as a vertical rectangular enclosure.
x The outside surface temperature of the enclosure is unknown.
x Newton’s law of cooling gives the heat transfer rate.
x The Rayleigh number should be determined for both vertical plate and rectangular
enclosure so that appropriate correlation equations for the Nusselt number are selected.
However, since the outside surface temperature of the enclosure is unknown, the
Rayleigh number can not be determined. The problem must be solved using an iterative
procedure.
Problem 8.48
x Heat is transferred from the inside to the outside.
x Adding i an air enclosure reduces the rate of heat transfer.
x To estimate the savings in energy, heat transfer through the single and double pane
windows must be determined.
x The double pane window introduces an added glass conduction resistance and a cavity
convection resistance.
x the problem can be modeled as a vertical rectangular enclosure.
x Newton’s law of cooling gives the heat transfer rate
x The aspect ratio and Rayleigh number should be determined for the rectangular enclosure
so that an appropriate correlation equation for the Nusselt number can be selected.
Problem 8.49
x Heat is transferred through the door from the inside to the outside.
x Newton’s law of cooling gives the heat transfer rate.
x The aspect ratio and Rayleigh number should be determined for the rectangular enclosure
so that an appropriate correlation equation for the Nusselt number can be selected.
x The baffle divides the vertical cavity
Problem 8.50
x Heat is transferred through the skylight from the inside to the outside.
x Newton’s law of cooling gives the heat transfer rate.
x The aspect ratio and Rayleigh number should be determined for the rectangular enclosure
so that an appropriate correlation equation for the Nusselt number can be selected.
Problem 8.51
x Power requirement is equal to the heat transfer rate through the enclosure.
x The problem can be modeled as a rectangular cavity at specified hot and cold surface
temperatures.
x The inclination angle varies from 0 o to 175 o .
x Newton’s law of cooling gives the heat transfer rate.
x The aspect ratio and critical inclination angle should be computed to determine the
applicable correlation equation for the Nusselt number.
Problem 8.52
x The absorber plate is at a higher temperature than the ambient air. Thus heat is lost
through the rectangular cavity to the atmosphere
x The problem can be modeled as an inclined rectangular cavity at specified hot and cold
surface temperatures.
x Newton’s law of cooling gives the heat transfer rate.
x The aspect ratio and critical inclination angle should be computed to determine the
applicable correlation equation for the Nusselt number.
Problem 8.53
x Heat is transferred through the annular space from the outer cylinder to the inner.
x Newton’s law of cooling gives the heat transfer rate.
x
The Rayleigh number should be determined for the enclosure formed by the concentric
cylinders so that an appropriate correlation equation can be selected.
x The cylinders are horizontally oriented.
Problem 9.1
x Definitions of Knudsen number, Reynolds number, and Mach number are needed.
x Fluid velocity appears in the definition of Reynolds number and Mach number.
Problem 9.2
x The definition of friction factor shows that it depends on pressure drop, diameter, length
and mean velocity.
x Mean velocity is determined from flow rate measurements and channel flow area.
Problem 9.3
x The determination of the Nusselt number requires the determination of the temperature
distribution.
x Temperature field depends on the velocity field.
x The velocity field for Couette flow with a moving upper plate is give in Section 9.6.2.
x The solution to the energy equation gives the temperature distribution.
Problem 9.4
x Temperature distribution depends on the velocity field.
x The velocity field for Couette flow with a moving upper plate is give in Section 9.6.2.
x The solution to the energy equation gives the temperature distribution.
x Two temperature boundary conditions must be specified.
x Temperature distribution and Fourier’s law give surface heat flux.
Problem 9.5
x To determine mass flow rate it is necessary to determine the velocity distribution.
x Velocity slip takes place at both boundaries of the flow channel.
x Because plates move in opposite directions, the fluid moves in both directions. This
makes it possible for the net flow rate to be zero.
Problem 9.6
x Model channel flow as Couette flow between parallel plates.
x Apply Fourier’s law at the housing surface to determine heat leaving the channel.
x Apply the Navier-Stokes equations and formulate the velocity slip boundary conditions.
Follow the analysis of Section 9.6.2 and Example 9.1.
x Use the energy equation to determine the temperature distribution
Problem 9.7
x To determine the temperature of the lower plate, fluid temperature distribution must be
known.
x Temperature distribution depends on the velocity field.
x The velocity field for Couette flow with a moving upper plate is given in Section 9.6.2.
x The solution to the energy equation gives the temperature distribution.
x Two temperature boundary conditions must be specified.
Problem 9.8
x To use the proposed approach, the solution to the axial velocity distribution must be
know.
x The velocity distribution for Poiseuille flow between parallel plates is given by equation
(9.30) of Section 9.6.3.
Problem 9.9
x This is a pressure driven microchannel Poiseuille flow between parallel plates.
x The solution to mass flow rate through microchannels is given in Section 9.6.3.
x Channel height affects the Knudsen number.
Problem 9.10
x This is a pressure driven microchannel Poiseuille flow.
x Since channel height is much smaller than channel width, the rectangular channel can be
modeled as Poiseuille flow between parallel plates.
x Channel surface is heated with uniform flux.
x The solution to mass flow rate, temperature distribution, and Nusselt number for fully
developed Poiseuille channel flow with uniform surface flux is presented in Section
9.6.3.
Problem 9.11
x The problem can be modeled as pressure driven Poiseuille flow between two parallel
plates with uniform surface flux.
x Assuming fully developed velocity and temperature, the analysis of Section 9.6.3 gives
the mass flow rate and Nusselt number.
x The Nusselt number depends on the Knudsen number, Kn. Since Kn varies along the
channel due to pressure variation, it follows that pressure distribution along the channel
must be determined.
Problem 9.12
x This is a pressure driven microchannel Poiseuille flow.
x Since channel height is much smaller than channel width, the rectangular channel can be
modeled as Poiseuille flow between parallel plates.
x Channel surface is maintained at uniform temperature.
x The solution to velocity, pressure, and mass flow rate is presented in Section 9.63.
x The solution to the temperature distribution and Nusselt number for fully developed
Poiseuille channel flow with uniform surface temperature is presented in Section 9.6.4.
x Surface heat flux is determined using Newton’s law.
Problem 9.13
x Cylindrical coordinates should be used to solve this problem.
x The axial component of the Navier-Stokes equations must be solved to determine the
axial velocity v z .
x The procedure and simplifying assumptions used in the solution of the corresponding
Couette flow between parallel plates, detailed in Section 9.6.2, can be applied to this
case.
Problem 9.14
x This a pressure driven Poiseuill