PROBLEM 6.1
KNOWN: Temperature distribution at x2 in laminar thermal boundary layer.
FIND: (a) Whether plate is being heated or cooled, (b) Temperature distributions at two other x
locations. Locations of largest and smallest heat fluxes, (c) Temperature distribution at x 2 for lower
and higher free stream velocities. Which velocity condition causes the largest heat flux.
SCHEMATIC:
T∞
T∞
Free stream
Thermal
boundary
layer
δt
Ts
x1
x2
x3
x
ASSUMPTIONS: (1) Steady-state conditions, (2) Laminar, incompressible flow.
ANALYSIS: (a) Since the sketch indicates that the free stream temperature is greater than the
surface temperature, the plate is being heated by the fluid. This is consistent with the fact that the
surface heat flux in the positive y-direction is given by Eq. 6.3:
qs′′ = − k f
∂T
∂y
y =0
From the sketch, the temperature gradient is positive, therefore the heat flux is negative. The heat
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transfer is in the negative y-direction, the plate is being heated by the fluid.
(b) At every location in the boundary layer, the temperature must vary from T s at the surface to T ∞ in
the free stream. This change must occur within the thermal boundary layer thickness, as shown in the
sketch below.
T∞
T∞
T∞
T∞
Ts
x1
x2
x3
x
Continued...
PROBLEM 6.1 (Cont.)
The magnitude of the heat flux is proportional to the temperature gradient at the surface, ∂T / ∂y y =0 ,
which is shown schematically as a dashed line. The temperature gradient is steeper (larger) at x 1
where the thermal boundary layer is thinner and less steep (smaller) at x 3 where the thermal boundary
layer is thicker. Therefore, the magnitude of the local heat flux is largest at x 1 and smallest at x 3 . <
(c) As the free stream velocity increases the boundary layer becomes thinner. Sketches for a low and
high free stream velocity are shown below.
T∞
Free stream
T∞
Thermal
boundary
layer
Ts
x2
Low freestream velocity case.
T∞
T∞
Free stream
Thermal
boundary
layer
Ts
x2
x
High freestream velocity case.
The temperature gradient, shown as the dashed line, is steeper for the higher free stream velocity case.
Therefore the higher free stream velocity case has the higher convective heat flux.
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COMMENTS: It is important to understand how the temperature gradient at the surface varies as the
thickness of the boundary layer changes.
PROBLEM 6.2
KNOWN: Form of the velocity and temperature profiles for flow over a surface.
FIND: Expressions for the friction and convection coefficients.
SCHEMATIC:
ANALYSIS: The shear stress at the wall is
τs = µ
∂ u
= µ  A + 2By − 3Cy 2 
= Aµ .

 y=0
∂ y  y=0
Hence, the friction coefficient has the form,
τs
2Aµ
=
2 / 2 ρ u2
ρ u∞
∞
=
Cf
Cf =
2Aν
2
u∞
.
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The convection coefficient is
2

−k f ( ∂ T/∂ y ) y=0 −k f  E + 2Fy − 3Gy  y=0
h =
=
Ts − T∞
D − T∞
h=
−k f E
.
D − T∞
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COMMENTS: It is a simple matter to obtain the important surface parameters from
knowledge of the corresponding boundary layer profiles. However, it is rarely a simple matter
to determine the form of the profile.
PROBLEM 6.3
KNOWN: Boundary layer temperature distribution.
FIND: Surface heat flux.
SCHEMATIC:
PROPERTIES: Table A-4, Air (T s = 300K): k = 0.0263 W/m⋅K.
ANALYSIS: Applying Fourier’s law at y = 0, the heat flux is
∂ T
∂ y
u y
 u 

=
−k ( T∞ − Ts )  Pr ∞  exp  −Pr ∞ 
ν  y=0
y=0
 ν 

u
−k ( T∞ − Ts ) Pr ∞
q′′s =
−k
q′′s =
ν
−0.0263 W/m ⋅ K (100K ) 0.7 × 5000 1/m.
q′′s =
q′′s = −9205 W/m 2 .
COMMENTS: (1) Negative flux implies convection heat transfer to the surface.
(2) Note use of k at T s to evaluate q′′s from Fourier’s law.
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PROBLEM 6.5
KNOWN: Variation of h x with x for laminar flow over a flat plate.
FIND: Ratio of average coefficient, h x , to local coefficient, h x , at x.
SCHEMATIC:
ANALYSIS: The average value of h x between 0 and x is
1 x
C x -1/2
∫ h x dx =
∫ x
hx =
dx
x 0
x 0
C 1/2
=
hx =
2x
2Cx -1/2
x
h x = 2h x .
Hence,
hx
= 2.
hx
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COMMENTS: Both the local and average coefficients decrease with increasing distance x
from the leading edge, as shown in the sketch below.
PROBLEM 6.18
KNOWN: Transition Reynolds number. Velocity and temperature of atmospheric air, engine
oil, and mercury flow over a flat plate.
FIND: Distance from leading edge at which transition occurs for each fluid.
SCHEMATIC:
T = 27ºC or
77ºC
ASSUMPTIONS: Transition Reynolds number is Re x,c = 5 × 105.
PROPERTIES: For the fluids at T = 300 K and 350 K:
2
ν(m /s)
Fluid
Air (1 atm)
Table
A-4
T = 300 K
-6
15.89 × 10
Engine Oil
A-5
550 × 10
Mercury
A-5
0.1125 × 10
-6
T = 350 K
20.92 × 10-6
41.7 × 10-6
-6
0.0976 × 10-6
ANALYSIS: The point of transition is
ν
=
x c Re
=
x,c
u∞
5 ×105
ν.
1 m/s
Substituting appropriate viscosities, find
x c (m)
Fluid
Air
Oil
Mercury
T = 300 K
7.95
275
0.056
T = 350 K
10.5
20.9
0.049
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COMMENTS: (1) Note the great disparity in transition length for the different fluids. Due to
the effect which viscous forces have on attenuating the instabilities which bring about
transition, the distance required to achieve transition increases with increasing ν. (2) Note the
temperature-dependence of the transition length, in particular for engine oil. (3) As shown in
Example 6.4, the variation of the transition location can have a significant effect on the
average heat transfer coefficient associated with convection to or from the plate.
PROBLEM 6.30
KNOWN: Flow over a flat plate. Velocity and temperature of two fluids. Variation of boundary
layer thickness with x for laminar flow.
FIND: (a) Location where transition to turbulence occurs for each fluid, (b) Plot of velocity boundary
layer thickness for 0 ≤ x ≤ x c for each fluid, (c) Plot of thermal boundary layer thickness over the same
range. Which fluid has the largest local temperature gradient at the surface, Nusselt number, and heat
transfer coefficient.
SCHEMATIC:
u∞ = 2 m/s
T∞ = 300K (air)
T∞ = 380K (oil)
xc
Ts > T∞
x
ASSUMPTIONS: (1) Steady-state conditions, (2) Incompressible flow, (3) Transition occurs at a
critical Reynolds number of 5 × 105.
PROPERTIES: Table A.4, Air (T = 300 K): ν = 15.89 × 10-6 m2/s, k = 0.0263 W/m⋅K, Pr = 0.707.
Table A.5, Engine Oil (T = 380 K): ν = 16.9 × 10-6 m2/s, k = 0.136 W/m⋅K, Pr = 233.
ANALYSIS: (a) Transition occurs at Re x,c = u ∞ x/ν = 5 × 105. Therefore, for air
xc =
5 × 105
ν
u∞
5 × 105
=
15.89 × 10−6 m 2 / s
4.0 m
=
2 m/s
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The value for engine oil is 4.2 m.
(b) The velocity boundary layer thickness is given by
δ
=
δ
x
=
5
. Thus, for air
Rex
5x
15.89 × 10−6 m 2 / s × x
= 5 xν=
/ u∞ 5
= 0.0141x1/ 2
2 m/s
Rex
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where x and δ are expressed in meters. The corresponding result for engine oil is δ = 0.0145x1/ 2 .
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These results are plotted below.
Continued…
PROBLEM 6.30 (Cont.)
0.03
Engine oil
Air
delta (m)
0.02
0.01
0
0
0.5
1
1.5
2
2.5
x (m)
3
3.5
4
4.5
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(c) From Eq. 6.55 with n = 1/3, δ t = δ⋅Pr-1/3. Thus, for air
δt =
δ Pr −1/3 =
0.0141x1/ 2 × (0.707) −1/3 =
0.0158 x1/ 2
Similarly for engine oil δ t = 0.00586 x1/ 2 .
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The results for the two fluids are shown below.
0.03
Air
deltat (m)
0.02
0.01
Engine oil
0
0
0.5
1
1.5
2
2.5
x (m)
3
3.5
4
4.5
<
Continued…
PROBLEM 6.30 (Cont.)
The two fluids are subjected to the same temperature difference between the surface and the free
stream. Since the thermal boundary layer thickness is the distance over which the temperature varies
from the surface temperature to the free stream temperature, the fluid with the smaller value of δ t must
have a larger temperature gradient, −∂T/∂y y = 0 .
Therefore, engine oil has the larger temperature gradient at the surface.
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The local Nusselt number is given by Nu = hx/k, where h is defined in Equation 6.5. Therefore,
Nu =
−∂T / ∂y y =0
Ts − T∞
x
At a given x location, since T s – T ∞ is the same for both fluids, the fluid with the larger temperature
gradient has the larger local Nusselt number.
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Engine oil has the larger local Nusselt number.
The heat transfer coefficient is given by Equation 6.5:
h=
− k f ∂T / ∂y
y =0
Ts − T∞
Since engine oil has a larger temperature gradient and a larger thermal conductivity, it is associated
with a larger heat transfer coefficient.
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COMMENTS: (1) Since the kinematic viscosity of the two fluids is nearly the same, their local
Reynolds numbers, transition locations, and velocity boundary layer thicknesses are comparable. (2)
The much higher Prandtl number of the engine oil results in a much thinner thermal boundary layer
and consequently a larger temperature gradient at the surface and higher heat transfer coefficient.