AME 60634 Intermediate Heat Transfer
Homework solutions
Updated May 4, 2015
Example 1
What is the steady state temperature of a resistor through which a
current is flowing?
Assume: (a) Only convection heat transfer to air. (b) Resistor at
uniform temperature. (c) No heat conduction through electrical wires.
(d) Joule heating in resistor. (e) Convection governed by Newton’s law
of cooling.
i
Figure 1: Example 1.
Variables: Qg = rate of heat generation, i = electrical current, R =
resistance, Qconv = rate of heat lost by convection, h = convection heat
transfer between resistor and surrounding air, A = heat transfer area of
resistor, TR = temperature of resistor, T∞ = temperature of surrounding
air. (d) Joule heating in wire.
The heat generation and heat loss are
Qg = i2 R,
Qconv = h A (TR − T∞ ),
respectively. At steady state, Qg = Qconv , so that
i2 R = h A (TR − T∞ ),
i2 R
.
TR = T∞ +
hA
1
Example 2
A saturated liquid enters a heated pipe, Fig. 2, and leaves as a liquidvapor mixture. How much energy has been transferred to the fluid from
the pipe?
Variables: Q = heat transfer rate to liquid, ṁ = mass flow rate, h2
and h1 are the specific enthalpies at the outlet and inlet, respectively.
Figure 2: Example 2.
The heat transfer rate is
Q = ṁ(h2 − h1 ),
Of course, being a phase change process, the temperature difference is
not proportional to the enthalpy change. From thermodynamics, h1 =
hf and h2 = (1 − x)hf + xhg where subscripts f and g refer to the
saturated liquid and vapor states, respectively, and x is the quality factor
(ratio by weight of gas to liquid).
Example 3
A thick plate, shown in Fig. 13, has a lower surface kept at constant
temperature T0 , and has convection to a fluid at temperature T∞ at its
upper surface. Find the temperature distribution in the plate.
2
c
Tinf
T0
Figure 3: Example 3.
Assume: (a) Infinite in two dimensions. (b) Fourier conduction with
constant thermal conductivity in plate. (c) Convective heat transfer
coefficient h known.
Variables: T (x) = temperature distribution, x = coordinate measured upward from the bottom of the plate, L = thickness of plate.
Steady-state temperature distribution T (x) is governed by
k
d2 T
= 0,
dx2
Integrating twice
dT
= a,
dx
T = ax + b.
(1)
At bottom and top
T0 = b,
T (L) = aL + b,
so that
T (L) − T0
,
L
b = T0 .
a=
(2)
(3)
Equating the conductive and convective heat flux at the top surface
T (L) − T0
= h(T (L) − T∞ ),
L
¾
½
kT0
k
− hT∞ ,
T (L) − − h = −
L
L
kT0 + hLT∞
T (L) =
.
k + hL
−k
3
(4)
The temperature distribution is Eq. (1), with a, b and T (L) given by
Eqs. (2), (3) and (4).
Example 4
The temperatures of the inner and outer surfaces of a wall composed
of two concentric cylindrical layers of different thermal conductivities are
known. What is the steady-state heat flux? See Fig. 4.
Figure 4: Example 4
Assume: (a) Fourier conduction with constant thermal conductivity
in each material. (b) No thermal resistance at interface between cylinders.
Variables: q = conductive heat rate, T1 and T2 = the temperatures
of the inner and outer surfaces, respectively; r1 , ri and r2 = the radii of
the inner surface, the interface, and the outer surfaces, respectively; L =
length of the cylinders; k1 and k2 = thermal conductivities of the inner
and outer cylinders, respectively; R1 and R2 are the thermal resistances
of the inner and outer cylinders, respectively.
4
Then
T1 − T2
,
R1 + R2
ln(ri /r1 )
R1 =
,
2πLk1
ln(r2 /ri )
R2 =
.
2πLk2
q=
Thus
q=
T1 − T2
.
ln(ri /r1 ) ln(r2 /ri )
+
2πLk1
2πLk2
Example 5
What is the steady-state temperature distribution in a fin with a
radiation boundary condition at the tip?
Assume: (a) Base temperature is known. (b) Tip is blackbody.
Variables: T (x) = temperature distribution; x = coordinate along
fin measured from the base.
The steady-state fin equation is
d2 T
− m2 (T − T∞ ) = 0,
2
dx
where m2 = hA/(kA) with the solution
T = C1 emx + C2 e−mx .
Boundary conditions are
T (0) = Tb ,
dT ¯
4
].
−k ¯x=L = σ[T (L)4 − T∞
dx
5
so that
¡
−k C1 me
mL
Tb = C1 + C2 ,
¢
£
¤
4
= σ (C1 emL + C2 e−mL )4 − T∞
,
−mL
− C2 e
which can be numerically solved for C1 and C2 if values of m, L and T∞
are given.
Example 6
Show that there is a critical insulation radius which minimizes heat
transfer from an insulation-covered pipe.
[cf. Example 3.5, 7th Ed.]
Variables: ri = inner radius of insulation, r = outer radius of insulation, k = thermal conductivity of insulation material, h = external
convective heat transfer coefficient.
The thermal resistance per unit length of the pipe is
′
Rtot
=
ln(r/r1 )
1
+
,
2πk
2πrh
and the heat rate per unit length is
q′ =
T∞ − Ti
′
Rtot
The maximum heat rate is obtained by minimizing q ′ or maximizing
′
Rtot
. With the latter
′
dRtot
,
drcr
1
1
=
−
,
2 h
2πrcr k 2πrcr
0=
from which rcr = k/h. This is the critical radius of the insulation above
and below which the heat rate is higher.
6
Example 7
If the temperature distribution inside a wall (0 ≤ x ≤ L) is T (x) =
a + bx + cx2 , find the rate of heat generation in the wall.
Assume: (a) One-dimensional Fourier conduction. (b) Steady state.
(c) Uniform thermal conductivity.
Variables: T (x) = temperature distribution; x = coordinate through
the wall; g = heat generation per unit volume; k = thermal conductivity
of wall.
The temperature distribution is given by
d2 T
g
+ =0
2
dx
k
from which
d2 T
g = −k 2 ,
dx
= −2ck.
Example 8
Find the steady state temperature distribution in a square plate with
uniform heat generation and uniform temperature at the boundary.
Variables: T (x, y) = temperature distribution; (x, y) = coordinates
of a point on the plate with x and y coordinates parallel to the sides
and the origin at one corner; g = heat generation per unit volume; k =
thermal conductivity of plate.
7
The problem
∂2T
g
∂2T
+
=
−
,
∂x2
∂y 2
k
BCs: T (0, y) = T (L, y) = T (x, 0) = T (x, L) = T0 ,
can be written as
∂2θ
∂2θ
+
= −1,
∂X 2 ∂Y 2
BCs: θ(0, Y ) = θ(1, Y ) = θ(X, 0) = θ(X, 1) = 0,
where
X=
x
,
L
Y =
y
,
L
θ=
g
(T − T0 ).
k
(a) First method
The equation can be written as
∂2Θ ∂2Θ
+
= 0,
∂X 2 ∂Y 2
where
Θ=θ+
g 2
X .
2k
with boundary conditions
Θ(0, Y ) = 0, Θ(X, 0) =
g 2
g
g 2
X , Θ(L, y) =
, Θ(X, 1) =
X ,
2k
2k
2k
This can be split up into four problems: the same equation with each
one of the following boundary conditions.
Θ(0, Y ) = 0, Θ(X, 0) = 0, Θ(L, y) = 0, Θ(X, 1) = 0,
g 2
X , Θ(L, y) = 0, Θ(X, 1) = 0,
Θ(0, Y ) = 0, Θ(X, 0) =
2k
g
Θ(0, Y ) = 0, Θ(X, 0) = 0, Θ(L, y) =
, Θ(X, 1) = 0,
2k
g 2
Θ(0, Y ) = 0, Θ(X, 0) = 0, Θ(L, y) = 0, Θ(X, 1) =
X .
2k
8
Each of the four problems can be solved by separation of variables, and
the results added.
(b) Second method 1
The solution can be written as
θ(X, Y ) = θh (X, Y ) + θp (X, Y ),
where
∂ 2 θh ∂ 2 θh
+
= 0,
∂X 2
∂Y 2
∂ 2 θp ∂ 2 θp
+
= −1,
∂X 2 ∂Y 2
θh = homogeneous solution
θp = particular solution.
The particular solution is not unique. For example one can take
1
θp = (1 − x2 ).
2
From separation of variables, the homogeneous solution is
θh =
∞
X
An cosh(λy) cos(λx).
n=0
The complete solution is
∞
X
1
2
θ = (1 − x ) +
An cosh(λy) cos(λx),
2
n=0
where
λ=
π
(n + 1).
2
From the Fourier expansion of the boundary condition at y = 1
An = −
16h2 sin(nπ/2)
n3 π 3 cosh(nπ/2)
1
From P.A. Ramachandran, Advanced Transport Phenomena: Analysis, Modeling, and
Computations, Cambridge University Press, Cambridge, U.K., 2014.
9
(b) Third method 2
∞
X
gL2
T (x, y) =
{Cn sin λn x cosh λn y} +
2k
n=1
½
¾
x ³ x ´2
−
L
L
Example 9
Computer problem: Use a finite difference method to calculate the
temperature field in the previous problem (choose specific numerical values of the constants).
Example 10
Using a lumped approximation, determine the time variation of the
temperature of a body that is suddenly exposed to convection heat transfer.
Assume: Newton’s law of cooling.
Variables: T (t) = temperature of body, t = time; T∞ = ambient temperature; m = mass of body; c = specific heat of body; h = convective
heat transfer coefficient; h = convection surface area.
The governing equation is
dT
+ hA(T − T∞ ) = 0,
dt
with T (0) = T0 . The solution is
µ
¶
hA
T = (T0 − T∞ ) exp − t + T∞ .
mc
mc
2
From D.W. Hahn and M.N. ´’Ozişik, Heat Conduction, John Wiley, New York, 2012.
10
Example 11
From the governing equations for an incompressible fluid (continuity, Navier-Stokes, energy), derive the ODEs for the hydrodynamic and
thermal boundary layers for laminar flow over a flat plate.
Assume: steady state and boundary layers.
Governing equations:
∂u ∂v
+
=0
∂x ∂y
∂u
∂ 2u
∂u
+v
=ν 2
u
∂x
∂y
∂y
∂T
∂T
∂2T
u
+v
=α 2
∂x
∂y
∂y
continuity
momentum,
energy.
To obtain the ODEs, define a stream function ψ(x, y) as
∂ψ
,
∂y
∂ψ
v=− ,
∂x
u=
which will satisfy the continuity equation. Then define
r
ψ
u∞
f (η) =
,
u∞ νx
where
η=y
r
11
u∞
.
νx
Using chain rule we get
∂ψ
,
∂y
∂ψ ∂η
=
,
∂η ∂y
r
r
νx df u∞
= u∞
,
u∞ dη νx
df
= u∞ ,
dη
u=
and
∂ψ
,
∂y
¶
µ r
r
νx ∂f
ν
u∞
+
f ,
= − u∞
u∞ ∂x
2
u∞ x
r
µ
¶
1 νu∞
df
=
η −f .
2
x
dη
v=−
Differentiating these
u∞ d2 f
∂u
=−
η
,
∂x
2x dη 2
r
∂u
u∞ d2 f
= u∞
,
∂y
νx dη 2
u2∞ d3 f
∂2u
=
.
∂y 2
νx dη 3
Substituting back into the momentum equation and we get
d3 f
d2 f
df
df
2 3 + f 2 = 0, with f =
= 0 at η = 0,
= 1 at η → ∞.
dη
dη
dη
dη
Use
T∗ =
T − Ts
T∞ − Ts
and T ∗ = T ∗ (η) in the energy equation to get
d2 T ∗ Pr dT ∗
+
f
= 0, with T ∗ (0) = 0, T ∗ (∞) = 1.
dη 2
2
dη
12
Example 12
(From Incropera et al.) 20 ◦ C air at 40 m/s flows parallel to a 0.2 m
by 0.2 m thin, flat plate which is at 100 ◦ C. The air flows both above
and below the plate. The drag force parallel to the plate is measured to
be 0.075 N. What is the rate of heat transfer from the plate?
Assume: Modified Reynolds analogy.
Variables: ν = the dynamic viscosity; Ts = the temperature of the
plate; T∞ = the temperature of the air flow; U∞ = the velocity of the
air flow; Fd = the drag force; Cf = the friction coefficient; Pr = Prandtl
number; h = convective heat transfer coefficient.
The friction coefficient can be determined by the following relation
that as
2
ρU∞
A,
2
2Fd
.
Cf =
2 A
ρU∞
Fd = Cf
Then the convective heat transfer coefficient h can be determined through
Cf
= St Pr2/3 ,
2
µ
¶
h
=
Pr2/3 ,
ρU∞ cp
Cf ρU∞ cp
,
h=
2Pr2/3
2Fd ρU∞ cp
=
,
2 A
ρU∞
2Pr2/3
Fd cp
.
=
U∞ APr2/3
Then the heat rate is given by
Q = 2hA(Ts − T∞ ).
13
Example 13
For fully-developed laminar flow determine the relation between the
maximum and the bulk temperature in (a) a circular pipe, and (b) flow
between flat plates.
Example 14
Consider the wing of an aircraft as a flat plate of length 2.5 m in the
flow direction. The plane is moving at 100 m/s in air at 0.7 bar, −10◦ C.
The top surface of the wing absorbs solar radiation at a rate of 800
W/m2 . Estimate the steady-state temperature of the wing, assuming it
to be uniform.
[Clarify this.]
Variables: ν = the dynamic viscosity; Ts = the temperature of the
plate; T∞ = the temperature of the air flow; U∞ = the velocity of the
air flow; h = convective heat transfer coefficient.
The Reynolds number is
u∞ L
Re =
= 1.404 × 107 ;
ν
so that the flow is turbulent. Using
NuL = 0.037Re4/5 Pr1/3 ;
and
NuL k
;
L
the convective heat coefficient hL can be determined.
Then considering the energy balance :
h=
Qabsorb = Qconv = 2hl A(Ts − T∞ );
the temperature of the plate Ts can be determined.
14
(5)
(6)
Example 15
A beverage can 150 mm long and 60 mm in diameter is initially
at 27 ◦ C and is to be cooled by placement in a refrigerator at 4 ◦ C.
In the interest of maximizing the cooling rate, should the can be laid
horizontally or vertically?
Assume: (a) Heat transfer from the can is by natural convection. (b)
neglect heat transfer from the sides. (c) Heat transfer from a vertical
cylinder is assumed to be as from a vertical flat plate.
Variables: g = acceleration due to gravity; β = coefficient of volumetric expansion; Ts = exterior temperature of cylinder; T∞ = refrigerator
temperature; ν = coefficient of dynamic viscosity; α = thermal diffusivity; D = diameter of cylinder; L = length of cylinder; P r = ν/α =
Prandtl number; k = thermal conductivity of air.
Horizontal cylinder
gβ(Ts − T∞ )D3
,
να
(
)
1/6
0.387RaD
N uD = 0.60 +
,
8/27
[1 + (0.559/P r)9/16 ]
RaD =
N uD k
,
D
Qh = hπDL(Ts − T∞ )
h=
Vertical cylinder
15
gβ(Ts − T∞ )L3
,
να
(
)2
1/6
0.387RaL
N uL = 0.825 +
,
4/9
[1 + (0.492/P r)9/16 ]
RaL =
N uL k
,
L
Qv = hπDL(Ts − T∞ )
h=
The optimum orientation of the can depends on whether Qh or Qh
is greater.
Example 16
The steady-state temperatures at different distances along a fin (6061
aluminum, 0.5 in diameter, 12 in long, ambient air temperature 20.8 ◦ C)
are measured to be the following.
x [m]
0.2794
0.2540
0.2286
0.2032
0.1778
0.1524
0.1270
0.1016
0.0762
0.0508
0.0254
T [◦ C]
42.3
43.8
47.2
52.7
59.1
65.1
72.2
86.0
104.5
128.3
154.7
Find the convective heat transfer coefficient.
Assume: Convective heat transfer along side and at tip.
16
1
0.9
0.8
0.7
θ
0.6
0.5
0.4
0.3
0.2
0.1
0
0.1
0.2
0.3
0.4
0.5
ξ
0.6
0.7
0.8
0.9
1
Figure 5: Points: measured temperature distribution; continuous line: theoretical temperature distribution for h = 450 W/m2 K.
Variables: x = distance from base, x0 = x-coordinate of origin, L =
distance between origin and tip of fin, θ = normalized temperature,
T∞ = ambient temperature, ξ = normalized distance.
At a distance of 0.0254 m from the base the temperature is 154.7 ◦ C.
We will take this point to be x = x0 and the temperature there to be
T = T0 . We define
x − x0
,
L
T − T0
θ=
.
T0 − T∞
ξ=
The experimental data are plotted in Fig. 5. The theoretical temperature
distribution
θ
cosh mL(1 − ξ) + (h/mk) sinh mL(1 − ξ)
=
θ0
cosh mL + (h/mk) sinh mL
is also plotted in the figure. We can see that h ≈ 450 W/m2 K fits the
data well.
17
Example 17
1.5 kg/s of water enters a 10 cm diameter, 10 m long pipe at 40 ◦ C.
What is the exit temperature of the water if the inner wall of the pipe
is uniformly at 20 ◦ C?
Assume: (a) Heat rate depends on local temperature difference between the bulk temperature of the water and the temperature of the
wall.
Variables: ṁ = mass flow rate; c = specific heat; T (x) = bulk temperature of water as a function of x; x = distance along the pipe; Tin =
bulk temperature of water at inlet x = 0; Tw = inner wall temperature;
D = diameter of pipe; h = convective heat transfer coefficient for heat
transfer from water to pipe wall.
The governing equation is
ṁc dT = hπD(Tw − T ) dx,
dT
hπD
hπD
+
T =
Tw .
dx
ṁ
ṁ
The solution is
T (x) = Tw + Ae−hπDx/ṁ
The boundary condition is T (x) = Tin at x = 0. This gives
Tin = Tw + A,
A = Tin − Tw ,
so that
T (x) = Tw + (Tin − Tw )e−hπDx/ṁ .
At x = L, the exit temperature Tout is
Tout = Tw + (Tin − Tw )e−hπDL/ṁ .
18
Example 18
A 1 mm diameter, 20 mm long nichrome wire of resistivity 10−6 Ω·m
carries a current of 0.5 A. What is the surface temperature of the wire
if it is placed (a) in a transverse flow of 20 ◦ C air at 1 m/s, or (b)
horizontally in quiescent air at 20 ◦ C?
Variables: i = electrical current; R = electrical resistance; Ts =
surface temperature of wire; Q = heat generated in wire; σ = resistivity;
L = length of wire; A = cross sectional area of wire.
σL
,
A
Q = i2 R,
= hAs (Ts − T∞ ).
R=
Thus
Q
+ T∞ ,
hAs
i2 σL
=
+ T∞ ,
hAs A
Ts =
Example 19
Measurements show the following Reynolds and Nusselt numbers for
convection from a heated object.
19
3.2
3
y = 0.47*x - 0.5
2.8
2.6
ln Nu
2.4
2.2
2
1.8
1.6
1.4
4
4.5
5
5.5
6
ln Re
6.5
7
7.5
8
Figure 6: Linear regression between x = ln Re and y = ln Nu.
Re
40.0
126.5
400
1265
4000
Nu
3.38
5.79
9.89
16.92
28.93
Show that the data fit a power law of the form Nu = CRem , and find
C and m.
Since ln N u = ln C + m ln Re, we expand the table to get
Re
40.0
126.5
400
1265
4000
ln Re
3.6889
4.8402
5.9915
7.1428
8.2940
Nu
3.38
5.79
9.89
16.92
28.93
20
ln N u
1.2179
1.7561
2.2915
2.8285
3.3649
A2 , T2 = 500 K, ε2 = 0.5
A3 , T3 = 800 K, ε3 = 0.8
D = 0.3 m
L = 0.3 m
A1 , T1 = 400 K, ε1 = 0.4
Figure 7
A linear regression of ln Re and ln Nu obtained using MATLAB (shown
in Fig. 6) gives: ln C = −0.5, and m = 0.47, so that the linear best fit is
ln Nu = −0.5 + 0.47 ln Re,
Nu = 0.6065 Re0.47 .
Example 20
Computer problem: Write a computer program to determine the temperature profile in a flat plate laminar boundary layer for Pr = 0.7.
Example 21
Consider a cylindrical furnace that is 0.3 m long and 0.3 m in diameter, as shown in Fig. 7. The two ends, A1 and A2 , have diffuse, gray
surfaces that are maintained at 400 and 500 K with emissivities of 0.4
and 0.5, respectively. The lateral surface, A3 , is also diffuse and gray
with an emissivity of 0.8 and a temperature of 800 K. Determine the net
radiative heat transfer from each of the surfaces, A1 , A2 and A3 .
21
Example 22
Consider two large parallel plates at different temperatures with
emissivities of 0.8 and 0.4, respectively. A radiation shield of emissivity
0.05 is placed between them. Calculate the percentage reduction in the
heat transfer rate due to the shield.
Example 23
The six internal walls of a 3 m × 3 m × 3 m cubical enclosure are
opaque, diffuse and gray. The two opposing walls are at temperatures
and emissivities of 300K, 0.8 and 400K, 0.9, respectively. The other four
walls are well insulated so there is no heat flow through them. Determine
the heat rate by radiation from the hot wall to the cold.
Example 24
A small pizza in a large oven is schematically shown in Fig. 8. The
25 cm diameter pizza is at 175◦ C and the radiative heat transfer to the
pizza is 1500 W. What is the temperature of the walls assuming that
they are blackbodies at uniform temperature? Consider only radiation
with a pizza emissivity of 0.8.
22
Figure 8
Example 25
Consider a cavity in the form of a regular tetrahedron, three sides of
which have temperatures and emissivities of 700 K and 0.7, 500 K and
0.5, and 300 K and 0.3, respectively. The fourth side is well insulated
and re-radiating. Find the radiosities of the four sides.
Example 26
A space system uses a radiative heat exchanger for cooling purposes.
A liquid with mass flow rate of ṁ flows inside a thin-walled tube, while
heat is transferred from the outside of the tube to free space by radiation.
Using a differential volume element, derive the governing equation for
the mean temperature of the liquid along the length of the tube. Solving
this equation, obtain the liquid absolute temperature at the outlet of a
tube of length L given that the inlet absolute temperature is Tin . Assume
that free space is at a temperature of absolute zero and that the outside
of the tube radiates to free space as a blackbody. Assume steady state
and neglect the thickness of the tube wall.
Variables: V = mean velocity along pipe.
A differential element is shown in Fig. 9. The governing equation for
the fluid in the pipe is
ρV Ac
dT
= σǫT 4
dx
23
dx
Radiation
Figure 9: Radiative heat exchanger
from which
ρV Ac
dT
= σǫ dx.
T4
Integrating
ρV Ac
1
= σǫx + C
3T 3
3
Using the boundary condition T = Tin at x = 0, C = ρcV /3Tin
, so that
¶
µ
1
ρV Ac
1
− 3 = σǫx,
3
T 3 Tin
¶−1/3
µ
1
3σǫx
+ 3
T =
ρV Ac Tin
Example 27
Obtain Wien’s displacement and Stefan-Boltzmann laws from Planck’s
blackbody spectral intensity distribution.
Planck’s blackbody spectral intensity distribution is
Eλ,b =
λ5
C1
{exp(C2 /λT )} − 1
24
The maximum value of Eλ,b is when λ = λmax , where
dEλ,b
=0
dλ
Stefan-Boltzmann’s law is for the total radiation Eb where
Z ∞
Eλ,b dλ.
Eb =
0
Example 28
The energy flux associated with solar radiation incident on the outer
surface of the earth’s atmosphere has been accurately measured and
is known to be 1353 W/m2 . The diameters of the sun and earth are
1.39 × 109 and 1.29 × 107 m, respectively, and the distance between the
sun and the earth is 1.5 × 1011 m.
1. What is the emissive power of the sun?
2. Approximating the sun’s surface as black, what is its temperature?
3. At what wavelength is the spectral emissive power of the sun a maximum?
4. Assuming the earth’s surface to be black and the sun to be the only
source of energy for the earth, estimate the earth surface temperature.
Example 29
Derive the equations for radiation energy exchange between opaque,
diffuse, gray internal surfaces of an enclosure.
25
Example 30
Derive the expression for the view factor between two inclined plates
of equal width and a common edge.
Example 31
A piston cylinder arrangement contains gas which is expanded from
pressure 300 kPa, volume 0.01 m3 to pressure 75 kPa, volume 0.03 m3 .
What is the polytropic index of the process? What is the work done?
Draw the process on a pressure-volume diagram.
Asuume: polytropic process
Variables: p = pressure, V = volume.
For a polytropic process
pV n = C (constant).
Thus
(300)(0.01)n = (75)(0.03)n ,
µ
¶n
0.03
300
=
,
75
0.01
4 = 3n ,
so that n = ln 4/ ln 3 = 1.26.
26
400
350
300
y
250
200
150
100
50
0
0
0.005
0.01
0.015
0.02
0.025
V m3
0.03
0.035
0.04
Figure 10: Polytropic process
Work done by the system is
Z V2
W =
p dV,
V1
V2
=
Z
CV −n dV,
V1
¯V2
V 1−n ¯¯
=C
¯ ,
1 − n¯
V1
¢
C ¡ 1−n
V2 − V11−n
=
1−n
The process is shown in Fig. 10.
Example 32
An ideal Rankine water-steam cycle works between turbine inlet conditions 600 kPa, 350 ◦ C and condenser pressure 20 kPa. What is the
thermal efficiency of the cycle? Draw the cycle in a temperature-entropy
diagram.
27
Example 33
What is the excess temperature for nucleate pool boiling of water at
atmospheric pressure over a large, horizontal brass plate if the heat flux
is one-half of the critical value.
From
′′
qmax
= Chf g ρv
·
σg(ρl − ρv )
ρ2v
¸1/4
,
′′
′′
calculate qmax
and then qs′′ = qmax
/2. With this value find ∆T , where
¸1/2 µ
¶3
·
cp,l ∆Te
g(ρl − ρv
′′
.
qs = µl hf g
σ
Cs,f hf g Prnl
Example 34
Estimate the critical heat flux for pool boiling of water on a large
horizontal plate at atmospheric pressure.
Use
′′
qmax
= Chf g ρv
·
σg(ρl − ρv )
ρ2v
¸1/4
Example 35
Estimate the critical heat flux for boiling of water at atmospheric
pressure in flow at 0.1 m/s over a cylinder of diameter 0.05 cm.
28
′′
If qmax
/ρv hf g V < [(0.275/π)(ρl /ρv )1/2 + 1],
"
µ
¶1/3 #
′′
1
qmax
4
=
1+
.
ρv hf g V
π
WeD
Otherwise
′′
qmax
(ρl /ρv )3/4
(ρl /ρv )1/2
=
+
1/3
ρv hf g V
169π
19.2πWeD
What is the heat flux for dropwise condensation heat transfer between
moist air at 30 ◦ C, relative humidity 76% and a metal sheet at 10 ◦ C.
Use
hdc = 255, 510 [W/m2 K]
Example 36
Show that the log mean temperature difference for counterflow heat
exchangers is
∆Tlm =
∆T2 − ∆T1
,
ln(∆T2 /∆T1 )
where
∆T1 = Th,i − Tc,o ,
∆T2 = Th,o − Tc,i .
29
Example 37
For counterflow heat exchangers, derive the relationship of (a) effectiveness as a function of number of transfer units, and (b) vice versa.
Show that
ε=
and

1 − exp [−NTU(1 − Cr )]




 1 − Cr exp [−NTU(1 − Cr )]



NTU


1 + NTU
NTU =
µ
¶

ε−1
1


ln

 Cr − 1
εCr − 1



 ε
1−ε
for Cr < 1
for Cr = 1
for Cr < 1
for Cr = 1
Example 38
[Incropera & DeWitt] A shell-and-tube heat exchanger must be designed to heat 2.5 kg/s of water from 15 ◦ C to 85 ◦ C. The heating is to
be accomplished by passing hot engine oil, which is available at 160 ◦ C,
through the shell side of the heat exchanger. The oil is known to provide
an average convection coefficient ho = 400 W/m2 ·K on the outside of
the tubes. Ten tubes pass the water through the shell. Each tube is
thin walled, of diameter D = 25 mm, and makes eight passes through
the shell. If the oil leaves the exchanger at 100 ◦ C, what is its flow rate?
How long must the tubes be to accomplish the desired heating?
30
Example 39
Water flows inside a copper tube and air flows transversely over it.
For the air: freestream velocity = 1 m/s, incoming temperature = 20 ◦ C;
for the water: inlet temperature = 40 ◦ C, average velocity = 0.1 m/s;
for the tube: inner diameter = 0.04 m, outer diameter = 0.05 m, length
= 5 m. Determine the overall heat transfer coefficient between the water
and the air. What is the outlet temperature of the water in the tube?
Example 40
Find the average velocity and bulk temperature in turbulent channel
flow. Assume that, if suitably normalized, the two distributions are both
power laws.
Variables: y = Cartesian coordinate measured from the center of
channel. U (y) = velocity distribution; T (y) = temperature distribution;
U = Uc and T = Tc at y = 0; T = Ts at surface y = ±h.
We normalize the temperature and velocity distributions as
U
,
Uc
T − Ts
,
T∗ =
Tc − Ts
y
y∗ = .
h
U∗ =
The temperature and the velocity field satisfy power laws so that
U ∗ = 1 − |y ∗ |n ,
T ∗ = 1 − |y ∗ |m .
31
We can use only one side, i.e. y ∗ ≥ 0, so that the bulk velocity can be
determined as
Z
1 h
U (y) dy,
Ubulk =
h 0
Z 1
= Uc
U ∗ (y ∗ ) dy ∗ ,
0
Z 1
= Uc
{1 − y ∗ n } dy ∗ ,
0
¶ ¯1
µ
y ∗ n+1 ¯¯
∗
,
= Uc y −
n + 1 ¯0
n
= Uc
.
n+1
Similarly, the bulk temperature is
Z h
1
Tbulk =
U (y) T (y) dy,
Ubulk h 0
Z 1
Uc
U ∗ (y ∗ ) {Tc − Ts ) T ∗ (y ∗ ) + Ts } dy ∗ ,
=
Ubulk 0
Z
n+1 1
(1 − y ∗ n ) {(Tc − Ts )(1 − y ∗ m ) + Ts } dy ∗ ,
=
n
Z0 1
n+1
(1 − y ∗ n ) {Tc − (Tc − Ts ) y ∗ m } dy ∗ ,
=
n
Z0 1
ª
©
n+1
=
Tc − (Tc − Ts ) y ∗ m − Tc y ∗ n + (Tc − Ts ) y ∗ n+m dy ∗ ,
n
½0
¾
1
1
1
n+1
Tc − (Tc − Ts )
.
− Tc
+ (Tc − Ts )
=
n
m+1
n+1
n+m+1
Example 41
A metal slab of low Biot number is convectively cooled from either
side with different heat transfer coefficients. What is the steady-state
temperature of the metal?
32
Mc
Mc
dT
= h1 A(T∞,1 − T ) + h2 A(T∞,2 − T ),
dt
dT
+ (h1 + h2 )AT = (h1 T∞,1 + h2 T∞,2 )A.
dt
Example 42
Show that, under certain circumstances, Maxwell’s equations can be
reduced to the wave equation.
[Clarify this]
Assume: Zero charge and current density.
First solution:
Maxwell’s equations are
∇ · E = 0,
∇ · B = 0,
∂B
∇×E=−
,
∂t
∂E
∇ × B = µ0 ε 0
;
∂t
From the above,
∇ × ∇ × E = −∇ ×
∂B
,
∂t
∂
∇ × B,
∂t
∂2E
= −µ0 ε0 2 ;
∂t
=−
and
∇ × ∇ × E = ∇(∇ · E) − ∇2 E,
= −∇2 E;
33
Therefore,
∂2E
;
∂t2
which satisfy the form of the wave equation. Similar operation can be
applied to show the magnetic fields obey the same rule.
Second solution:
Maxwell’s equations of electromagnetic theory are
∇2 E = µ 0 ε0
∂D
∂t
∂B
∇×E=−
∂t
∇·D=ρ
∇·B=0
(7a)
∇×H=J+
(7b)
(7c)
(7d)
Taking the curl of Eq. (7a), D = ǫE and J = gE, where ǫ and g are
constants, we have
∇×∇×H=∇×J+∇×
∂D
,
∂t
∂
(∇ × E)
∂t
Using Eq. (7b) and B = µH, with µ constant,
= g∇ × E + ǫ
∇ × ∇ × H = −gµ
∂H
∂2H
− ǫµ 2
∂t
∂t
From the vector relation
∇2 A = ∇(∇ · A) − ∇ × (∇ × A),
(8)
we get
∇(∇ · H) − ∇2 H = ∇ × ∇ × H
∂ 2H
∂H
− ǫµ 2
= −gµ
∂t
∂t
Since µ is a constant, the first term in the previous equation vanishes
because
1
∇ · H = ∇ · B = 0.
µ
34
from Eq. (7d). Thus
∇2 H − gµ
∂ 2H
∂H
− ǫµ 2 = 0.
∂t
∂t
It can be shown that E also satisfies the same equation. Taking the
curl of Eq. (7b)
∂B
,
∂t
∂E
∂ 2E
= −gµ
− ǫµ 2 .
∂t
∂t
∇ × ∇ × E = −∇ ×
Using Eq. (8), and ρ = 0 (charge-free space) gives ∇ · D = 0, and
∇2 E − gµ
∂E
∂E
− ǫµ 2 = 0.
∂t
∂t
In free space g = 0, and the equations in H and B are the wave
equations
∂H
= 0,
∂t2
∂E
∇2 E = ǫµ 2 = 0,
∂t
∇2 H = ǫµ
√
where c = 1/ ǫµ is the phase velocity of the wave.
Example 43
A cooling experiment provides the following normalized temperatures
Θ(t) that are suspected to be a result of two time constants. Find them.
35
t [s]
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
large
Θ(t)
1.0000
0.6496
0.4275
0.2854
0.1935
0.1332
0.0931
0.0661
0.0476
0.0347
0.0256
0
θ(t) =
T − T∞
,
Ti − T∞
[Clarify this]
Let
which should take the form as:
θ(t) = C1 e−λ1 t + (1 − C1 )e−λ2 t ;
Using the experimental data to fit and we can get
λ1 = 0.499049; λ1 = 0.999642;
Therefore, the time constants are,
t1 =
1
1
; t2 = ;
λ1
λ2
Example 44
For an n-generational tree with the the left branches as thermal resistances of unit value and the right thermal capacitors also of unit value,
36
plot the real and imaginary parts of the overall thermal impedance as a
function of n.
Example 45
Two masses at different temperatures that are both convectively
cooled are in contact through a thermal resistance. If one of the masses
is externally heated, find the thermal behavior of both the masses.
[Clarify this]
The governing equations are
dT
+ hA1 (T1 − T∞1 ) +
dt
dT
M2 C2
+ hA2 (T2 − T∞1 ) +
dt
M1 C1
1
(T1 − T − 2) = Q1 ,
R
1
(T2 − T − 1) = Q2 ,
R
where either Q1 or Q2 represents the external heat source. The sets of the
first order ODEs can then be solved numerically using any mathematical
tools (e.g. ode45 in matlab).
Example 46
How long must a constant-area fin have to be for the steady-state
heat rate to be within 10% of that for an infinitely long fin?
37
Example 47
Set up a mathematical model for a slightly tapered fin in the steady
state, and solve.
Example 48
A thin wire is bent in the form of a circular loop and exposed to a
flow of air. What is the temperature distribution along the wire if there
is nonuniform internal heat generation in the wire?
Example 49
Show that the time-dependent temperature distribution in a fin that
is exposed to convection and radiation is stable using a (a) linear and
(b) nonlinear analysis.
Example 50
What is the long-time response of the temperature distribution in a
convective fin if the external temperature is oscillatory?
Assume: Constant cross-section, constant heat transfer coefficient.
Variables: T = average temperature over cross-section, T0 = mean
external temperature, ∆T = amplitude of external temperature oscillation.
38
The governing equation is
∂2T
∂T
+
+ m2 T = T0 + ∆T sin ωt.
2
∂t
∂x
Let the particular integral be
T = a(x) + b(x) sin(ωt + φ(x)).
Substituting
Example 51
Develop and analyze a mathematical model for on-off control of a
lumped body with internal heating undergoing convection to an air
stream with oscillatory temperature.
Section (3.6.2) of Notes.
Example 52
Fig. 11 shows a water tank with convective losses whose temperature
is being controlled by a heated inlet at constant mass flow rate. The
controller for the heating, however, is at a distance from the tank. Set up
a mathematical model for the control system using proportional control,
and solve.
Assume: One-dimensional flow, complete mixing in tank, no heat
loss in pipe.
39
T (t)
Q(t)
Figure 11: Tank and inlet pipe
Variables: ṁ = mass flow rate in pipe; T (t) = average temperature
in the tank; Tin (t) = inlet and outlet temperatures of tank, Tp = temperature in pipe leaving the heater; c = specific heat of water; Q(t) = heat
input from heater; t = time, L = distance between heater and entrance
of tank, V = average velocity in pipe; τ = time delay.
Heat balance in the the tank is
Mc
dT
= ṁ (Tin − T ) .
dt
Because of mixing of the water in the tank, the water comes out of it at
temperature T . Heat balance in the pipe is
ṁ(Tp − T0 ) = Q
Delay equation is
Tin (t) = Tp (T − τ )
where τ = L/V .
Example 53
Consider a large number of particles starting from the origin and
undergoing one-dimensional random walks. Write a computer program
to determine their positions and graph their probability distribution after
some time.
40
Example 54
The Maxwell-Boltzmann distribution for a gas is3
r³
mv 2
m ´3
4πv 2 e− 2kT ,
f (v) =
2πkT
where m is the mass of the atom, v is its velocity, and T is the absolute
temperature. Show that
1. The distribution satisfies
′
2
kT vf (v) + f (v)(mv − 2kT ) = 0,
f (1) =
r
2 − m ³ m ´3/2
e 2kT
π
kT
2. The most probable speed is
vp =
r
2kT
m
3. The mean speed is
hvi =
r
2
8kT
= √ vp
πm
π
4. The rms speed is
p
hv 2 i =
3
r
3kT
=
m
r
3
vp
2
See http://en.wikipedia.org/wiki/Maxwell-Boltzmann_distribution
41
Example 55
For a dispersive wave, show that the group velocity cg and the phase
velocity c are related by4
cg = c + k dc/dk
Differentiating
c=
ω
k
w.r.t. k, where c = c(k), we have
d ³ω ´
dc
=
,
dk
dk k
µ ¶
1 dω
d 1
=
,
+ω
k dk
dk k
1 dω
ω
=
− 2,
k dk
k
cg
c
=
− ,
k
k
so that
cg = c + k
dc
.
dk
Example 56
Show that Cattaneo’s law of heat conduction combined with conservation of energy gives a telegraph equation for the temperature.
4
See http://www3.nd.edu/ msen/Teaching/Waves/WavesBook.pdf
42
Cattaneo’s equation is (one-dimensional)
τ
∂T
∂q
+ q = −k
.
∂t
∂x
Conservation of energy is
ρc
∂T
∂q
+
= 0.
∂t
∂x
Differentiate the first w.r.t t and the second w.r.t x
τ
∂ 2 q ∂q
∂ 2T
+
=
−k
,
∂t2
∂t
∂t∂x
∂2T
∂2q
ρc
+ 2 = 0.
∂x∂t ∂x
Dividing the first by k, the second by ρc and subtracting, the term
∂ 2 T /∂t∂x is eliminated to give
τ ∂ 2 q 1 ∂q
1 ∂2q
+
=
.
k ∂t2
k ∂t
ρc ∂x2
On the other hand, on differentiating Cattaneo’s equation w.r.t. x
gives
τ
∂q
∂2T
∂2q
+
= −k 2 .
∂x∂t ∂x
∂x
From the energy equation
∂q
∂T
= −ρc ,
∂x
∂t
2
∂ q
∂ 2T
τ
= −τ ρc 2 ,
∂x∂t
∂t
so that
τ ρc
∂2T
∂ 2T
∂T
=
k
+
ρc
.
∂t2
∂t
∂x2
The equations for q and T are both telegraph equations.
43
Example 57
From one-dimensional motion of a gas molecule in a box, derive the
ideal gas law5 .
Variables: τ = travel time between rebounds at a wall, (vx , vy , vz ) =
average velocity components of a molecule in x, y, z directions, v = magnitude of average velocity of a molecule, A = area normal to x-direction,
V = volume,
For a box of length L
2L
,
vx
vx
1
.
f= =
τ
2L
τ=
The momentum change of a molecule rebounding from a wall is
∆J = mvx − (−mvx ) = 2mvx .
The average force on the wall due to this single molecule is
F = ∆J f = (2mvx )
³v ´
x
2L
=
mvx2
L
Thus the pressure due to a single molecule rebounding from a wall is
p1 =
F
mvx2
=
A
V
where V = AL. If there are N molecules bouncing from the wall, then
the pressure due to these is
p=
N mvx2
V
Since molecules have velocity components in all three directions
v 2 = vx2 + vy2 + vz2
5
http://quantumfreak.com/derivation-of-pvnrt-the-equation-of-ideal-gas/
44
Assuming that vx2 = vy2 = vz2 , vx2 = v 2 /3, and
p=
N mv 2
3V
The kinetic energy of each molecule is proportional to the absolute
temperature so that
EKE =
mv 2
3
3kT
= kT =⇒ v 2 =
2
2
m
where k is a constant. Thus
Nm 2
v ,
µ3V ¶ µ
¶
Nm
3kT
=
,
3V
m
RT
,
=
V
p=
where R = N k.
Example 58
The following questions relate to the paper:
On the steady-state velocity of the inclided toroidal thermosyphon, M.
Sen. E. Ramos, C. Treviño, ASME J. Heat Transfer, Vol. 107, pp. 974977, 1985.
http://www3.nd.edu/~msen/Teaching/IntHT/Sen1985a.pdf
1. Derive Eq. (6) from the previous equations.
2. Derive Eq. (7) from Eq. (6).
3. Replot Fig. 4 using the appropriate parameter values.
45
The independent variable is the angular coordinate θ, and the two dependent variables in the two equations are the temperature distribution
φ(θ) and the fluid velocity w which is a constant.
The energy equation is
(
−Dφ for θ0 ≤ θ ≤ θ0 + π
dφ
=
πw
.
dθ
D
for θ0 − π ≤ θ ≤ θ0
The solution is
µ
¶

D

a exp −
θ
πw
φ(θ) =

 D θ+b
πw
for θ0 ≤ θ ≤ θ0 + π
.
for θ0 − π ≤ θ ≤ θ0
At the common point θ = θ0 , the temperature must be the same, so that
¶
µ
D
D
θ0 =
θ0 + b,
a exp −
πw
πw
and at the other common point θ = θ0 + π
¶
µ
D
D
(θ0 + π) =
(θ0 + π) + b.
a exp −
πw
πw
Subtracting one from the other
µ
¶
µ
¶
D
D
D
a exp −
(θ0 + π) − a exp −
θ0 =
πw
πw
w
so that
a=
D/w
µ
¶
µ
¶
D
D
exp −
(θ0 + π) − exp −
θ0
πw
πw
46
Substituting the temperature in the momentum equation
Z π
π
φ cos θ dθ,
w=
4D −π
µZ θ0 +π
¶
Z θ0
π
=
φ cos θ dθ ,
φ cos θ dθ +
4D
θ0 −π
θ0
¾
½ Z θ0 +π
Z θ0
D
π
−Dθ/πw
θ cos θ dθ ,
a
e
cos θ dθ +
=
4D
θ0
θ0 −π πw
"
(µ
)#θ0 +π
¶2
·
¸θ0
aπ
e−Dθ/πw
D
1
=
cos θ + θ sin θ
+
cos θ + sin θ
.
4D e−2Dθ/πw + 1
πw
4w
θ0 −π
θ0
..
.
where we have used
Z
ecx
(c cos x + sin x),
ecx cos x dx = 2
c +1
Z
x cos x dx = cos x + x sin x
In the end
¶2
µ
D
D
π sin θ0
cos θ0 +
1
D
w
πw
2
(
)
w − cos θ0 −
= 0.
coth
µ
¶
2
2
2w
D
4 1+
πw
If D/w ≪ 1, then6
(
1+
µ
D
πw
¶2 )−1
coth
6
=1−
µ
D
πw
¶2
+ h.o.t ,
D
2w
D
=
+
+ h.o.t.,
2w
D
6w
h.o.t. = “higher-order terms”
47
(9)
where, for the moment, terms of order up to (D/w)2 have been kept7 .
Thus, Eq. (9) becomes
(
)½
µ
¶2 ) (
µ
¶2
¾
1
D
D
2w
D
D
1
2
1−
cos θ0 +
+
= 0,
π sin θ0
w − cos θ0 −
2
4
πw
w
πw
D
6w
½
¾
2D
1
1
2
w − cos θ0 −
2 cos θ0 +
sin θ0 = 0, up to order D/w,
2
4
πw
µ ¶
D
3
w − w cos θ0 −
sin θ0 = 0.
2π
Example 59
The following questions are from:
The toroidal thermosyphon with known heat flux, M. Sen. E. Ramos,
C. Treviño, Int. J. Heat Mass Transf., Vol. 28, No. 1, pp. 219-233, 1985.
http://www3.nd.edu/~msen/Teaching/IntHT/Sen1985b.pdf
1. Derive Eqs. (8) and (9) from (6) and (7).
2. Derive Eqs. (17) and (18).
3. Derive Eqs. (26)–(28) from (8) and (9).
4. Plot a 3D version of Fig. 8(a–c).
Example 60
Consider the change in temperature of a lumped system with convective heat transfer where the ambient temperature, T∞ (t), varies with
48
time in the form shown in Fig. 12. Find the amplitude of oscillation of
the system temperature, T (t), for a small period δt.
Assume: (a) The period is equally divided into T∞ = Tmin and T∞ =
Tmax . (b) The solution is periodic.
We divide the period into two parts: one when the ambient temperature is high (T∞ = Tmax ), and another when it is low (T∞ = Tmin ). The
solution of the equation
(
Tmin low
dT
,
+ αT = α
dt
Tmax high
where α = hA/M c, is
(
a e−αt + Tmin
T (t) =
b e−αt + Tmax
low
.
high
If we let the temperatures at the beginning of each part of the period to
be T1 for low, and T2 for high, then at t = 0
T1 = a + Tmin ,
T2 = b + Tmax .
Correspondingly, at the end of each part
T2 = a e−α
δt/2
+ Tmin ,
T1 = b e−α
δt/2
+ Tmax .
Eliminating a and b
T2 = (T1 − Tmin ) e−α
δt/2
+ Tmin ,
−α δt/2
T1 = (T2 !la − Tmax ) e
+ Tmax .
Subtracting
T1 − T2 = {(T2 − T1 ) − (Tmax − Tmin )} e−α
©
ª
(T1 − T2 )(1 + e−α δt/2 ) = (Tmax − Tmin ) 1 − e−α δt/2 ,
T1 − T2 = (Tmax − Tmin )
7
1 − e−α δt/2
1 + e−α δt/2
Note that coth x = x−1 + x/3 − x3 /45 + . . . has been used.
49
δt/2
+ (Tmax − Tmin ),
For δt ≪ 2M c/hA,
1 − e−α δt/2
1
1
= ( α δt)(2 − α δt)−1 + . . . ,
−α
δt/2
1+e
2
2
1
1
= ( α δt) 2 (1 + α δt) + . . . ,
2
4
= α δt + . . . ,
where only the lowest order terms have been kept8 . Thus
T1 − T2 = (Tmax − Tmin ) α δt + . . . .
Example 61
Consider a thin, rectangular flat plate of thickness δ and constant
thermal conductivity k, shown in Fig. 13. Both sides of the plate are
subjected to convective heat transfer with a constant heat transfer coefficient h. Show that the temperature field in the steady state, T (x, y),
is governed by
∂2T
∂2T
+
− m2 (T − T∞ ) = 0,
∂x2
∂y 2
where m2 = 2h/kδ, and T∞ is the temperature of the fluid surrounding
the plate.
8
Note that
ex = 1 + x +
x2
+ ...
2
and
(1 + x)−1 = 1 − x + x2 + . . .
50
δt
T (t)
Tmax
T1
T1
T2
T2
Tmin
t
Figure 12: Ambient temperature variation.
x
y
Figure 13: Thin plate with convection on each side
qy+dy
y
δ
dy
dx
qy
x
Figure 14: Element of two-dimensional fin.
51
An element of the fin is shown in Fig. 14. Consider a small area
dx×dy on the surface of the plate. We will write the energy conservation
equation for the prism of height δ and cross-section dx × dy.
(a) The mass of the prism is ρ dx dy δ, and its rate of increase in internal
energy is
ρ dx dy δ c
∂T
.
∂t
(b) The heat rate through a surface of area (dx δ) is qy = −k (dx δ) (∂T /∂y),
and that through an equal surface a distance dy away is qy+dy . Thus,
from a Taylor series expansion
qy+dy = qy +
∂qy
dy,
∂y
∂qy
dy,
∂y
∂2T
= k dx δ
dy.
∂y 2
qy − qy+dy = −
The net energy coming in by conduction through the four faces is
∂2T
∂2T
dx
+
k
dx
δ
dy
∂x2
∂y 2
k dy δ
(c) The convective heat transfer coming in through the upper and lower
faces is
2 h dx dy (T∞ − T )
Energy balance gives
ρ dx dy δ c
∂T
∂2T
∂ 2T
= k dy δ
dx
+
k
dx
δ
dy − 2 h dx dy (T − T∞ ).
∂t
∂x2
∂y 2
Dividing by the volume of the element dx dy δ gives
¶
µ 2
2h
∂2T
∂ T
∂T
−
=k
+
(T − T∞ )
ρc
2
2
∂t
∂x
∂y
δ
52
In the steady state
¶
µ 2
2h
∂2T
∂ T
−
+
(T − T∞ ) = 0,
k
2
2
∂x
∂y
δ
∂ 2T
∂2T
2h
+
−
(T − T∞ ) = 0,
2
2
∂x
∂y
kδ
∂ 2T
∂2T
2h
+
− m2 (T − T∞ ) = 0, where m2 =
.
2
2
∂x
∂y
kδ
Example 62
Fig. 15 shows a closed loop with two reservoirs and piping in which
fluid is continuously being pumped around. Q̇ is the rate at which heat
is put in one reservoir, and is taken out of the other. The pipes between
the reservoirs are not insulated, and heat is lost there to the environment
by convection. Find the steady-state temperatures, T1 and T2 , of the two
reservoirs.
Assume: (a) The reservoirs are insulated.
Variables: T1′ = temperature at outlet of reservoir 1, T2′ = temperature at outlet of reservoir 2, TU (x) = temperature in upper pipe, TL (x) =
temperature in lower pipe, T∞ = ambient temperature, θ = T − T∞ , x =
coordinate along the pipe in direction of flow, L = pipe length from one
reservoir to the other, U = overall heat transfer coefficient from fluid to
ambient, P = perimeter of pipe, ṁ =mass flow along pipes, c = specific
heat of liquid.
For each of the pipes
ṁc
dθ
+ U P θ = 0,
dx
with the solutions
θU (x) = ae−U P x/ṁc ,
θL (x) = be−U P x/ṁc ,
53
for the upper and lower pipes, respectively.
At the beginning (x = 0) and the end (x = L) of the upper pipe:
θ2′ = a,
θ1 = ae−U P L/ṁc .
Similarly for lower pipe:
θ1′ = b,
θ2 = be−U P L/ṁc .
In reservoirs 1 and 2:
ṁc(θ1′ − θ1 ) = Q̇, =⇒ θ1′ = θ1 + Q̇/ṁc
ṁc(θ2′ − θ2 ) = −Q̇ =⇒ θ2′ = θ2 − Q̇/ṁc.
respectively.
Thus
Ã
!
Q̇
θ2 = θ1 +
e−U P L/ṁc ,
ṁc
!
Ã
Q̇
e−U P L/ṁc ,
θ1 = θ2 −
ṁc
(Ã
!
)
Q̇
Q̇
=
θ1 +
e−U P L/ṁc −
e−U P L/ṁc ,
ṁc
ṁc
so that for reservoir 1
¢
¡
¢
¡
Q̇
θ1 1 − e−2U P L/ṁc = − e−U P L/ṁc 1 − e−U P L/ṁc ,
ṁc
Q̇ 1 − e−U P L/ṁc −U P L/ṁc
e
.
θ1 = −
ṁc 1 − e−2U P L/ṁc
Similarly, in reservoir 2
θ2 =
Q̇ 1 − e−U P L/ṁc −U P L/ṁc
e
,
ṁc 1 − e−2U P L/ṁc
54
θ2 =
Q̇ 1 − e−U P L/ṁc −U P L/ṁc
e
.
ṁc 1 − e−2U P L/ṁc
Thus
Q̇
ṁc
Q̇
T2 = T∞ +
ṁc
T1 = T∞ −
1 − e−U P L/ṁc −U P L/ṁc
e
,
1 − e−2U P L/ṁc
1 − e−U P L/ṁc −U P L/ṁc
e
,
1 − e−2U P L/ṁc
and
1
(T1 + T2 ) = T∞ .
2
In the steady state the mean temperature of the two reservoirs is the
same as that of the ambient. The temperature in reservoir 1 is lower
than ambient and that in the reservoir 2 is higher.
Example 63
A natural circulation loop is heated uniformly from one side and
cooled from the other, as shown in Fig. 16. Determine the steady-state
velocity of the fluid inside. Neglect the lengths AB and CD, i.e. the
horizontal parts of the loop.
Assume: Wall shear stress proportional to mean flow rate.
Variables: s = longitudinal coordinate that starts from D and goes
around the loop in a counterclockwise direction, L = total loop length,
q0 = heat rate per unit length on right leg, −q0 = heat rate per unit
length on left leg, T (s) = steady state temperature distribution, u =
steady state velocity, ρ0 = density, A = cross-sectional area, cp = specific
heat of the fluid, TD , TB = temperatures at D and B, respectively, P =
inner perimeter of tube, α = constant in relation between shear stress
and average velocity, β = coefficient of thermal expansion, g̃(s) = gravity
in direction opposite to increasing s.
55
TU
Q̇
T1
T2′
P
Q̇
T1′
T2
TL
Figure 15: Pumped loop; P is a pump.
B
A
C
D
Figure 16: Natural circulation loop with vertical pipes
56
The energy equation is
u
q(s)
dT
=
,
ds
ρ0 Acp
from which
1
T (s) =
ρ0 Acp u
Z
s
q(s′ ) ds′ + TD .
0
so that
T (s) =
 q0 s

+ TD


 ρ0 Acp u



−
right leg 0 ≤ s ≤ L/2
q0
(s − L2 ) + TB
ρ0 Acp u
left leg L/2 ≤ s ≤ L
Since TB = TA , we get
TB = T (s) with s = L/2 in right leg solution
q0
(L/2) + TD
=
ρ0 Acp u
Also TD = TC so that
TD = T (s) with s = L in left leg solution
q0
(L/2) + TB .
=−
ρ0 Acp u
These two conditions give the same result
TB − TD =
q0
(L/2)
ρ0 Acp u
The momentum equation is
Z
β L
Pα
u=
T (s)g̃(s) ds.
ρ0 A
L 0
The gravity function is
(
g
g̃(s) =
−g
right leg 0 ≤ s ≤ L/2
left leg L/2 ≤ s ≤ L
57
so that
"Z
¾
¾ #
Z L ½
q0 (s − L/2)
q0 s
+ TD ds −
+ TB ds ,
−
ρ0 Acp u
ρ0 Acp u
0
L/2
"
¯L/2
¯
¯L/2 #
¯L/2
′2 ¯L/2
¯
¯
q0
βg
s2 ¯¯
s
q
0
¯ − TB s′ ¯
¯ +
=
+
T
s
,
D
¯
¯
L ρ0 Acp u 2 ¯0
ρ0 Acp u 2 ¯0
0
0
βg
Pα
u=
ρ0 A
L
L/2
½
(where s′ = s − L/2)
¯L/2 µ
·
µ 2¶
¶ ¯L/2 ¸
¯
¯
q
(L/2)
q0
L
βg
0
¯ − T½
,
T½
+½
s′ ¯¯
=
D s¯
½D +
L ρ0 Acp u
4
ρ0 Acp u
0
0
·
µ 2¶
µ 2¶¸
βg
q0
q0
L
L
=
−
,
L ρ0 Acp u 4
ρ0 Acp u
4
= 0.
u = 0 since the temperatures on each leg at every height is the same,
so that there is no net buoyancy force.
58