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ch06


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Ecq k 1 2 1
L 2 \u3b7
\u3b7
=
\u2212 \u22c5\uf8ee \uf8f9
\u2032\u2032 = \u2212 \u2212 +\uf8ef \uf8fa\uf8f0 \uf8fb
( )L 00 L 0T T Pr Ecq k 1 2k T T LL 2
\u2212 \u22c5\uf8eb \uf8f6
\u2032\u2032 = \u2212 + = \u2212 \u2212\uf8ec \uf8f7\uf8ed \uf8f8 <
Continued...
PROBLEM 6.24 (Cont.)
(d) Using Eq. (1), the dimensionless temperature distribution is plotted as a function of dimensionless
distance, \u3b7 = y/L. When Pr\u22c5Ec = 0, there is no dissipation and the temperature distribution is linear, so
that heat transfer is by conduction only. As Pr \u22c5Ec increases, viscous dissipation becomes more
important. When Pr\u22c5Ec = 2, heat transfer to the upper plate is zero. When Pr\u22c5Ec > 2, the heat rate is out
of the oil film at both surfaces.
0 0.25 0.5 0.75 1
eta = y/L
0
0.5
1
1.5
2
th
et
a 
= 
(T(
y)-
T0
)/(T
L-T
0)
Pr*Ec = 0, conduction
Pr*Ec = 1
Pr*Ec = 2, adiabatic at y=L
Pr*Ec = 4
Pr*Ec = 10
PROBLEM 6.25
KNOWN: Steady, incompressible, laminar flow between infinite parallel plates at different
temperatures.
FIND: (a) Form of continuity equation, (b) Form of momentum equations and velocity profile.
Relationship of pressure gradient to maximum velocity, (c) Form of energy equation and temperature
distribution. Heat flux at top surface.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional flow (no variations in z) between infinite, parallel plates, (2)
Negligible body forces, (3) No internal energy generation, (4) Incompressible fluid with constant
properties.
ANALYSIS: (a) For two-dimensional, steady conditions, the continuity equation is
( ) ( ) u v 0.
 x y
\u2202 \u3c1 \u2202 \u3c1
\u2202 \u2202+ =
Hence, for an incompressible fluid (constant \u3c1) in parallel flow (v = 0),
 u 0.
 x
\u2202
\u2202 = <
The flow is fully developed in the sense that, irrespective of y, u is independent of x.
(b) With the above result and the prescribed conditions, the momentum equations reduce to
2
2
 p u p0 0
 x y
 y
\u2202 \u2202 \u2202µ\u2202 \u2202\u2202
= \u2212 + = \u2212 <
Since p is independent of y, \u2202p/\u2202x = dp/dx is independent of y and
2 2
2 2
 u d u dp
.
dx
 y dy
\u2202µ µ
\u2202
= =
Since the left-hand side can, at most, depend only on y and the right-hand side is independent of y,
both sides must equal the same constant C. That is,
2
2
d u C.
dy
µ =
Hence, the velocity distribution has the form
( ) 2 1 2Cu y y C y C .2µ= + +
Using the boundary conditions to evaluate the constants,
( ) ( )2 1u 0 0 C 0 and u L 0 C CL/2 .µ= \u2192 = = \u2192 = \u2212
Continued \u2026..
PROBLEM 6.25 (Cont.)
The velocity profile is ( ) ( )2Cu y y Ly .2µ= \u2212
The profile is symmetric about the midplane, in which case the maximum velocity exists at y = L/2.
Hence,
( )
2 2
max max
C L L dp
u L/2 u or u .
2 4 8 dxµ µ
\uf8ee \uf8f9
= = \u2212 = \u2212\uf8ef \uf8fa\uf8ef \uf8fa\uf8f0 \uf8fb
<
(c) For fully developed thermal conditions, (\u2202T/\u2202x) = 0 and temperature depends only on y. Hence
with v = 0, \u2202u/\u2202x = 0, and the prescribed assumptions, the energy equation becomes
22
2
 i d T dp du
 u k u .
 x dx dydy
\u2202\u3c1 µ\u2202
\uf8ee \uf8f9
= + + \uf8ef \uf8fa\uf8f0 \uf8fb
With i = e + p/\u3c1, i e 1 dp e e T e where 0.
 x x dx x T x x
\u2202 \u2202 \u2202 \u2202 \u2202 \u2202 \u2202 \u3c1
\u2202 \u2202 \u3c1 \u2202 \u2202 \u2202 \u2202 \u3c1 \u2202= + = + =
Hence, the energy equation becomes
22
2
d T du0 k .
dydy
µ \uf8ee \uf8f9= + \uf8ef \uf8fa\uf8f0 \uf8fb <
With du/dy = (C/2µ) (2y - L), it follows that
( )2 2 2 22d T C 4y 4Ly L .4kdy µ= \u2212 \u2212 +
Integrating twice,
( )
2 4 3 2 2
3 4
C y 2Ly L yT y C y C
4k 3 3 2µ
\uf8ee \uf8f9
= \u2212 \u2212 + + +\uf8ef \uf8fa\uf8ef \uf8fa\uf8f0 \uf8fb
Using the boundary conditions to evaluate the constants,
( ) ( ) ( )
2 3 1 2
2 4 2 1 3
T TC LT 0 T C T and T L T C .
24k Lµ
\u2212
= \u2192 = = \u2192 = +
Hence, ( ) ( )
2 4 3 2 2 3
2 1 2
y C y 2Ly L y L yT y T T T .
L 4k 3 3 2 6µ
\uf8ee \uf8f9\uf8ee \uf8f9
= + \u2212 \u2212 \u2212 + \u2212\uf8ef \uf8fa\uf8ef \uf8fa\uf8f0 \uf8fb \uf8ef \uf8fa\uf8f0 \uf8fb
<
From Fourier\u2019s law,
( ) ( )
2 33 3 3
2 1
y=L
 T k C 4 Lq L k T T L 2L L
 y L 4 3 6
\u2202
\u2202 µ
\uf8ee \uf8f9
\u2032\u2032 = \u2212 = \u2212 + \u2212 + \u2212\uf8ef \uf8fa\uf8ef \uf8fa\uf8f0 \uf8fb
( ) ( )
2 3
2 1
k C Lq L T T .
L 24µ
\u2032\u2032 = \u2212 + <
COMMENTS: The third and second terms on the right-hand sides of the temperature distribution
and heat flux, respectively, represents the effects of viscous dissipation. If C is large (due to large µ
or umax), viscous dissipation is significant. If C is small, conduction effects dominate.
PROBLEM 6.26
KNOWN: Pressure independence of m, k and cp.
FIND: Pressure dependence of n and a for air at 350K and p = 1, 10 atm.
ASSUMPTIONS: Perfect gas behavior for air.
PROPERTIES: Table A-4, Air (350K, 1 atm): n = 20.92 ´ 10-6m2/s, a = 29.9 ´ 10-6m2/s.
ANALYSIS: The kinematic viscosity and thermal diffusivity are, respectively,
p/ k/ c .n m r a r= =
Hence, n and a are inversely proportional to r.
For an incompressible liquid, r is constant.
Hence n and a are independent of pressure. <
For a perfect gas, r = p/RT.
Hence, r is directly proportional to p, in which case n and a vary inversely with
pressure. It follows that n and a are inversely proportional to pressure. <
To calculate n or a for a perfect gas at p ¹ 1 atm,
( ) ( )
( ) ( )
1
p 1 atm
p
1
p 1 atm
p
n n
a a
= ×
= ×
Hence, for air at 350K,
p(atm) n(m2/s) a(m2/s)
 1 20.92 ´ 10-6 29.9 ´ 10-6
10 2.09 ´ 10-6 2.99 ´ 10-6
COMMENTS: For the incompressible liquid and the perfect gas, Pr = n/a is independent of
pressure.
PROBLEM 6.27
KNOWN: Characteristic length, surface temperature and average heat flux for an object
placed in an airstream of prescribed temperature and velocity.
FIND: Average convection coefficient if characteristic length of object is increased by a
factor of five and air velocity is decreased by a factor of five.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties.
ANALYSIS: For a particular geometry,
( )L LNu f Re , Pr .=
The Reynolds numbers for each case are
Case 1: ( ) 21 1L,1
1 1 1
100m/s 1mV L 100 m / sRe
\u3bd \u3bd \u3bd
= = =
Case 2: ( ) 22 2L,2
2 2 2
20m/s 5mV L 100 m / sRe .
\u3bd \u3bd \u3bd
= = =
Hence, with \u3bd1 = \u3bd2, ReL,1 = ReL,2. Since Pr1 = Pr2, it follows that
L,2 L,1Nu Nu .=
Hence,
2 2 2 1 1 1
1
2 1 1
2
h L / k h L / k
Lh h 0.2 h .
L
=
= =
For Case 1, using the rate equation, the convection coefficient is
( )
( )
( ) ( ) ( )
1 1 1 s 1
21 1 21
1
s s1 1
q h A T T
q / A q 20,000 W/mh 200 W/m K.
T T T T 400 300 K
\u221e
\u221e \u221e
= \u2212
\u2032\u2032
= = = = \u22c5
\u2212 \u2212 \u2212
Hence, it follows that for Case 2
2 2
2h 0.2 200 W/m K 40 W/m K.= × \u22c5 = \u22c5 <
COMMENTS: If ReL,2 were not equal to ReL,1, it would be necessary to know the specific
form of f(ReL, Pr) before h2 could be determined.
PROBLEM 6.28
KNOWN: Heat transfer rate from a turbine blade for prescribed operating conditions.
FIND: Heat transfer rate from a larger blade operating under different conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Surface area A is
directly proportional to characteristic length L, (4) Negligible radiation, (5) Blade shapes are
geometrically similar.
ANALYSIS: For a prescribed geometry,
( )LhLNu f Re ,Pr .k= =
The Reynolds numbers for the blades are
( ) ( )L,1 1 1 L,2 2 2Re V L / 15 / Re V L / 15 / .\u3bd \u3bd \u3bd \u3bd= = = =
Hence, with constant properties, L,1 L,2Re Re .= Also, Pr Pr .1 2= Therefore,
( ) ( )
( )
2 1
2 2 1 1
1 1 1
2 1
2 2 1 s,1
Nu Nu
h L / k h L / k
L L qh h .
L L A T T
\u221e
=
=
= =
\u2212
Hence, the heat rate for the second blade is
( ) ( )( )
( )
( ) ( )
s,21 2
2 2 2 s,2 1
2 1 s,1
s,2
2 1
s,1
T TL Aq h A T T q
L A T T
T T 400 35
q q 1500 W
T T 300 35
\u221e
\u221e
\u221e
\u221e
\u221e
\u2212
= \u2212 =
\u2212
\u2212
\u2212
= =
\u2212 \u2212
2q 2066 W.= <
COMMENTS: The slight variation of \u3bd from Case 1 to Case 2 would cause ReL,2 to differ
from ReL,1. However, for the prescribed conditions, this non-constant property effect is
small.
PROBLEM 6.29
KNOWN: Experimental measurements of