Capitulo 6

Prévia do material em texto

PROBLEM 6.1 
K NOWN: Form of the velocity and temperature profiles for flow over a surface. 
F IND: Expressions for the friction and convection coefficients. 
SCHEMATIC: 
 
 
 
A NALYSIS: The shear stress at the wall is 
 2s
y=0y=0
 u A 2By 3Cy A .
 y
∂τ µ µ∂
⎤ ⎡ ⎤= = + −⎥ ⎢ ⎥⎣ ⎦⎦ µ= 
 
H ence, the friction coefficient has the form, 
 sf 2 2
2AC
 u / 2 u
τ µ
ρ ρ∞ ∞
= = 
 
 f 2
2AC
u
.ν
∞
= < 
 
T he convection coefficient is 
 
( ) 2ff y=0 y=0
s
k E 2Fy 3Gyk T/ y
h
T T D T
∂ ∂
∞ ∞
⎡ ⎤− + −− ⎢ ⎥⎣ ⎦= =− − 
 
 fk Eh
D T∞
−= − . < 
 
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. 
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PROBLEM 6.2 
KNOWN: Surface temperatures of a steel wall and temperature of water flowing over the 
all. w 
FIND: (a) Convection coefficient, (b) Temperature gradient in wall and in water at wall 
urface. s 
SCHEMATIC: 
 
 
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer in x, (3) 
onstant properties. C 
PROPERTIES: Table A-1, Steel Type AISI 1010 (70°C = 343K), ks = 61.7 W/m⋅K; Table 
-6, Water (32.5°C = 305K), kA
 f
 = 0.62 W/m⋅K. 
A NALYSIS: (a) Applying an energy balance to the control surface at x = 0, it follows that 
 x,cond x,convq q′′ ′′− = 0
 
and using the appropriate rate equations, 
 ( )s,2 s,1s s,T Tk h TL 1 T .∞− = − 
Hence, 
 s,2 s,1 2s
s,1
T Tk 61.7 W/m K 60 Ch 705 W/m K.
L T T 0.35m 15 C∞
− ⋅= = =−
D
D ⋅ < 
 
( b) The gradient in the wall at the surface is 
( ) s,2 s,1s T T 60 CdT/dx 171.4 C/m.L 0.35m
−= − = − = −
D D 
 
I n the water at x = 0, the definition of h gives 
( ) ( )s,1f,x=0 f
hdT/dx T T
k ∞= − − 
 
( ) ( )2f,x=0 705 W/m KdT/dx 15 C 17,056 C/m.0.62 W/m K⋅= − = −⋅ D D < 
COMMENTS: Note the relative magnitudes of the gradients. Why is there such a large 
difference? 
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PROBLEM 6.3 
K NOWN: Boundary layer temperature distribution. 
F IND: Surface heat flux. 
SCHEMATIC: 
 
 
 
P
 
ROPERTIES: Table A-4, Air (Ts = 300K): k = 0.0263 W/m⋅K. 
A NALYSIS: Applying Fourier’s law at y = 0, the heat flux is 
 
( )
( )
( )
y=0
s s
y=0
s s
s
 T uq k k T T Pr exp Pr
 y
uq k T T Pr
q 0.0263 W/m K 100K 0.7 5000 1/m.
∂
∂ ν
ν
∞ ∞∞
∞∞
⎡ ⎤ ⎡ ⎤′′ = − = − − −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
′′ = − −
′′ = − ⋅ ×
u y
ν
.
 
 
 < 2sq 9205 W/m′′ = − 
C OMMENTS: (1) Negative flux implies convection heat transfer to the surface. 
(2) Note use of k at Ts to evaluate sq′′ from Fourier’s law. 
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PROBLEM 6.4 
K
 
NOWN: Variation of hx with x for laminar flow over a flat plate. 
FIND: Ratio of average coefficient, xh , to local coefficient, hx, at x. 
SCHEMATIC: 
 
 
 
ANALYSIS: The average value of hx between 0 and x is 
 
 
x x
0 0
-1/2
x x
1/2 -1/2
x
x x
1 Ch h dx x
x x
Ch 2x 2Cx
x
h 2h .
= ∫ = ∫
= =
=
dx
 
 
Hence, x
x
h 2.
h
= < 
 
COMMENTS: Both the local and average coefficients decrease with increasing distance x 
from the leading edge, as shown in the sketch below. 
 
 
 
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PROBLEM 6.5 
KNOWN: Variation of local convection coefficient with x for free convection from a 
ertical heated plate. v 
F IND: Ratio of average to local convection coefficient. 
SCHEMATIC: 
 
A NALYSIS: The average coefficient from 0 to x is 
 
x x
0 0
-1/4
x x
3/4 -1/4
x x
1 Ch h dx x dx
x x
4 C 4 4h x C x h
3 x 3 3
= =
= = =
∫ ∫
.
 
 
Hence, x
x
h 4 .
h 3
= < 
 
The variations with distance of the local and average convection coefficients are shown in the 
sketch. 
 
 
 
COMMENTS: Note that h / h 4 / 3x x = is independent of x. Hence the average coefficient 
for an entire plate of length L is L
4h h
3
= L , where hL is the local coefficient at x = L. Note 
also that the average exceeds the local. Why? 
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PROBLEM 6.6 
KNOWN: Expression for the local heat transfer coefficient of a circular, hot gas jet at T∞ 
directed normal to a circular plate at Ts of radius ro. 
F IND: Heat transfer rate to the plate by convection. 
SCHEMATIC: 
 
 
 
ASSUMPTIONS: (1) Steady-state conditions, (2) Flow is axisymmetric about the plate, (3) 
or h(r), a and b are constants and n ≠ -2. F 
ANALYSIS: The convective heat transfer rate to the plate follows from Newton’s law of 
cooling 
 
 ( ) ( )
A A
conv conv sq dq h r dA T∞= = ⋅ ⋅ −∫ ∫ T . 
 
The local heat transfer coefficient is known to have the form, 
 
 ( ) nh r a br= +
 
and the differential area on the plate surface is 
 
 dA 2 r dr.= π 
 
Hence, the heat rate is 
 
 
( ) ( )
( )
ro
0
o
n
conv s
r
2 n+2
conv s
0
q a br 2 r dr T T
a bq 2 T T r r
2 n 2
π
π
∞
∞
= + ⋅ ⋅ −
⎡ ⎤= − +⎢ ⎥+⎣ ⎦
∫
 
 
 (2 n+2conv o o sa bq 2 r r T T2 n 2π ∞
⎡ ⎤= + −⎢ ⎥+⎣ ⎦ ). < 
 
COMMENTS: Note the importance of the requirement, n ≠ -2. Typically, the radius of the 
jet is much smaller than that of the plate. 
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PROBLEM 6.7 
KNOWN: Distribution of local convection coefficient for obstructed parallel flow over a flat 
late. p 
F IND:Average heat transfer coefficient and ratio of average to local at the trailing edge. 
SCHEMATIC: 
 
 
 
A NALYSIS: The average convection coefficient is 
 ( )( )
L L
0 0
2
L x
2 3
L
1 1h h dx 0.7 13.6x 3.4x dx
L L
1h 0.7L 6.8L 1.13L 0.7 6.8L 1.13L
L
= = + −
= + − = + −
∫ ∫
2
 
 
 ( ) ( ) 2Lh 0.7 6.8 3 1.13 9 10.9 W/m K.= + − = ⋅ < 
 
T he local coefficient at x = 3m is 
 ( ) ( ) 2Lh 0.7 13.6 3 3.4 9 10.9 W/m K.= + − = ⋅
 
H ence, 
 L Lh / h 1.0.= < 
COMMENTS: The result L Lh / h 1.0= is unique to x = 3m and is a consequence of the 
existence of a maximum for The maximum occurs at x = 2m, where 
 
h xxb g.
( ) ( )2 2x xdh / dx 0 and d h / dx 0.= <
E x c e r p t s f r o m t h i s w o r k m a y b e r e p r o d u c e d b y i n s t r u c t o r s f o r d i s t r i b u t i o n o n a n o t - f o r - p r o f i t b a s i s f o r t e s t i n g o r i n s t r u c t i o n a l p u r p o s e s o n l y t o s t u d e n t s e n r o l l e d i n 
c o u r s e s f o r w h i c h t h e t e x t b o o k h a s b e e n a d o p t e d . A n y o t h e r r e p r o d u c t i o n o r t r a n s l a t i o n o f t h i s w o r k b e y o n d t h a t p e r m i t t e d b y S e c t i o n s 1 0 7 o r 1 0 8 o f t h e 1 9 7 6 
U n i t e d S t a t e s C o p y r i g h t A c t w i t h o u t t h e p e r m i s s i o n o f t h e c o p y r i g h t o w n e r i s u n l a w f u l .
PROBLEM 6.8 
 
KNOWN: Temperature distribution in boundary layer for air flow over a flat plate. 
 
F
 
IND: Variation of local convection coefficient along the plate and value of average coefficient. 
SCHEMATIC: 
 
A
 
NALYSIS: From Eq. 6.5, 
 ( )
( )
( )
y 0
s s
k T y k 70 600x
h
T T T T
∂ ∂ =
∞ ∞
×= − = +− − 
 
where Ts = T(x,0) = 90°C. Evaluating k at the arithmetic mean of the freestream and surface 
temperatures, T = (20 + 90)°C/2 = 55°C = 328 K, Table A.4 yields k = 0.0284 W/m⋅K. Hence, with 
Ts - T = 70°C = 70 K, ∞ 
 
( ) ( )20.0284 W m K 42,000x K mh 170 K⋅= = 7x W m K⋅ < 
 
and the convection coefficient increases linearly with x. 
 
 
The average coefficient over the range 0 ≤ x ≤ 5 m is 
 
 
52L 5 2
0 0
0
1 17 17 xh hdx xdx 42.5W m K
L 5 5 2
= = = =∫ ∫ ⋅ < 
 
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PROBLEM 6.9 
 
KNOWN: Variation of local convection coefficient with distance x from a heated plate with a 
uniform temperature Ts. 
FIND: (a) An expression for the average coefficient 12h for the section of length (x2 - x1) in terms of 
C, x1 and x2, and (b) An expression for 12h in terms of x1 and x2, and the average coefficients 1h and 
2h , corresponding to lengths x1 and x2, respectively. 
SCHEMATIC: 
 
hx = Cx-1/2
dq’ 
ASSUMPTIONS: (1) Laminar flow over a plate with uniform surface temperature, Ts, and (2) Spatial 
variation of local coefficient is of the form 1/ 2xh Cx
−= , where C is a constant. 
 
ANALYSIS: (a) The heat transfer rate per unit width from a longitudinal section, x2 - x1, can be 
expressed as 
 ( )(12 12 2 1 sq h x x T T )∞′ = − − (1) 
where 12h is the average coefficient for the section of length (x2 - x1). The heat rate can also be 
written in terms of the local coefficient, Eq. (6.11), as 
 (2) ( ) ( )2
1 1
x x
12 x s s xx
q h dx T T T T h∞ ∞′ = − = −∫ 2x dx∫
Combining Eq. (1) and (2), 
 ( ) 21
x
12 xx2 1
1h
x x
= − ∫ h dx (3) 
and substituting for the form of the local coefficient, 1/ 2xh Cx
−= , find that 
 ( )
2
2
1
1
x 1/ 2 1/ 21/ 2x 1/ 2 2 1
12 x2 1 2 1 2 1x
x x1 C xh Cx dx 2C
x x x x 1/ 2 x x
− ⎡ ⎤ −= = =⎢ ⎥− − ⎢ ⎥⎣ ⎦∫ − (4)< 
(b) The heat rate, given as Eq. (1), can also be expressed as 
 ( ) ( )12 2 2 s 1 1 sq h x T T h x T T∞′ = − − − ∞ (5) 
 
which is the difference between the heat rate for the plate over the section (0 - x2) and over the section 
(0 - x1). Combining Eqs. (1) and (5), find, 
 2 2 1 112
2 1
h x h xh
x x
−= − (6)< 
 
COMMENTS: (1) Note that, from Eq. 6.6, 
 
x x 1/ 2 1/ 2
x x0 0
1 1h h dx Cx dx 2Cx
x x
− −= = =∫ ∫ (7) 
 
or xh = 2hx. Substituting Eq. (7) into Eq. (6), see that the result is the same as Eq. (4). 
 
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PROBLEM 6.10 
 
KNOWN: Expression for face-averaged Nusselt numbers on a cylinder of rectangular cross 
section. Dimensions of the cylinder. 
 
FIND: Average heat transfer coefficient over the entire cylinder. Plausible explanation for 
variations in the face-averaged heat transfer coefficients. 
 
SCHEMATIC: c = 40 mm
d = 30 mm
Air
V = 10 m/s
T∞ = 300 K
c = 40 mm
d = 30 mm
Air
V = 10 m/s
T∞ = 300 K
 
 
 
 
 
 
 
 
 
 
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties. 
 
PROPERTIES: Table A.4, air (300 K): k = 0.0263 W/m⋅K, ν = 1.589 × 10-5 m2/s, Pr = 0.707. 
 
ANALYSIS: 
For the square cylinder, c/d = 40 mm/30 mm = 1.33 
 
-3
d -5 2
Vd 10 m/s × 30 × 10 mRe = = = 18,880
1.589 × 10 m /sν 
 
Therefore, for the front face C = 0.674, m = ½. For the sides, C = 0.107, m = 2/3 while for the 
back C = 0.153, m = 2/3. 
 
Front face: 
 1/2 1/3d,fNu = 0.674 × 18,880 × 0.707 = 82.44
 2df -3
kNu 0.0263 W/m K × 82.44h = = = 72.27 W/m K
d 30 × 10 m
⋅ ⋅ 
 
Side faces: 
 2/3 1/3d,sNu = 0.107 × 18,880 × 0.707 = 67.36
 d,s 2s -3
kNu 0.0263 W/m K × 67.36h = = = 59.05 W/m K
d 30 × 10 m
⋅ ⋅ 
 
Back face: 
 2/3 1/3d,bNu = 0.153 × 18,880 × 0.707 = 96.43
d,b 2
b -3
kNu 0.0263 W/m K × 96.43h = = = 84.54 W/m K
d 30 × 10 m
⋅ ⋅ 
 
Continued… 
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PROBLEM 6.11 
KNOWN: Radial distribution of local convection coefficient for flow normal to a circular 
isk. d 
F IND: Expression for average Nusselt number. 
SCHEMATIC: 
 
 
 
A SSUMPTIONS: Constant properties. 
A NALYSIS: The average convection coefficient is 
 ( )
( )
As
ro
0
o
s
s
n
o o2
o
r2 n+2
o
3 n
o o 0
1h hdA
A
1 kh Nu 1 a r/r
D r
kNu r arh
2r n 2 r
ππ
=
⎡ ⎤= +⎢ ⎥⎣ ⎦
⎡ ⎤⎢ ⎥= +⎢ ⎥+⎣ ⎦
∫
∫ 2 rdr 
 
w
 
here Nuo is the Nusselt number at the stagnation point (r = 0). Hence, 
 
( )
( )
( )
o
D
D
rn 22
o
o
o
0
o
r/rhD a rNu 2Nu
k 2 n+2 r
Nu Nu 1 2a/ n 2
+⎡ ⎤⎛ ⎞⎢ ⎥= = + ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
⎡ ⎤= + +⎣ ⎦
 
 
 ( ) 1/2 0.36D D1 2a/ n 2 0.814Re Pr .Nu ⎡ ⎤= + +⎣ ⎦ < 
COMMENTS: The increase in h(r) with r may be explained in terms of the sharp turn which 
the boundary layer flow must make around the edge of the disk. The boundary layer 
accelerates and its thickness decreases as it makes the turn,causing the local convection 
coefficient to increase. 
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PROBLEM 6.12 
 
KNOWN: Convection correlation and temperature of an impinging air jet. Dimensions and initial 
temperature of a heated copper disk. Properties of the air and copper. 
 
F
 
IND: Effect of jet velocity on temperature decay of disk following jet impingement. 
SCHEMATIC: 
 
ASSUMPTIONS: (1) Validity of lumped capacitance analysis, (2) Negligible heat transfer from sides 
nd bottom of disk, (3) Constant properties. a
 
ANALYSIS: Performing an energy balance on the disk, it follows that 
(st s conv radE VcdT dt A q qρ ′′ ′′= = − +� ) . Hence, with V = AsL, 
 
 
( ) ( )r suh T T h T TdT
dt cLρ
∞− + −= − r
sur
 
 
where, and, from the solution to Problem 6.11, ( )( )2 2r surh T T T Tεσ= + +
 
 1/ 2 0.36D D
k k 2ah Nu 1 0.814 Re Pr
D D n 2
⎛ ⎞= = +⎜ ⎟+⎝ ⎠ 
 
W
 
ith a = 0.30 and n = 2, it follows that 
 ( ) 1/ 2 0.36Dh k D 0.936Re Pr= 
where ReD = VD/ν. Using the Lumped Capacitance Model of IHT, the following temperature histories 
were determined. 
 
 Continued ….. 
E x c e r p t s f r o m t h i s w o r k m a y b e r e p r o d u c e d b y i n s t r u c t o r s f o r d i s t r i b u t i o n o n a n o t - f o r - p r o f i t b a s i s f o r t e s t i n g o r i n s t r u c t i o n a l p u r p o s e s o n l y t o s t u d e n t s e n r o l l e d i n 
c o u r s e s f o r w h i c h t h e t e x t b o o k h a s b e e n a d o p t e d . A n y o t h e r r e p r o d u c t i o n o r t r a n s l a t i o n o f t h i s w o r k b e y o n d t h a t p e r m i t t e d b y S e c t i o n s 1 0 7 o r 1 0 8 o f t h e 1 9 7 6 
U n i t e d S t a t e s C o p y r i g h t A c t w i t h o u t t h e p e r m i s s i o n o f t h e c o p y r i g h t o w n e r i s u n l a w f u l .
 
PROBLEM 6.12 (Cont.) 
 
0 500 1000 1500 2000 2500 3000
Time, t(s)
300
400
500
600
700
800
900
1000
Te
m
pe
ra
tu
re
, T
(K
)
V = 4 m/s
V = 20 m/s
V = 50 m/s 
 
The temperature decay becomes more pronounced with increasing V, and a final temperature of 400 K is 
reached at t = 2760, 1455 and 976s for V = 4, 20 and 50 m/s, respectively. 
 
COMMENTS: The maximum Biot number, Bi = ( )r Ch h L k+ u , is associated with V = 50 m/s 
(maximum h of 169 W/m2⋅K) and t = 0 (maximum hr of 64 W/m2⋅K), in which case the maximum Biot 
number is Bi = (233 W/m2⋅K)(0.025 m)/(386 W/m⋅K) = 0.015 < 0.1. Hence, the lumped capacitance 
approximation is valid. 
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PROBLEM 6.13 
KNOWN: Local convection coefficient on rotating disk. Radius and surface temperature of disk. 
emperature of stagnant air. T 
F IND: Local heat flux and total heat rate. Nature of boundary layer. 
SCHEMATIC: 
 
 
 
A SSUMPTIONS: (1) Negligible heat transfer from back surface and edge of disk. 
ANALYSIS: If the local convection coefficient is independent of radius, the local heat flux at every 
point on the disk is 
 
 < ( ) ( )2 2sq h T T 20 W / m K 50 20 C 600 W / m∞′′ = − = ⋅ − ° =
 
Since h is independent of location, 2h h 20 W / m K= = ⋅ and the total power requirement is 
 
 ( ) ( )2elec s s o sP q hA T T h r T Tπ∞ ∞= = − = − 
 
 < ( ) ( ) ( )22elecP 20 W / m K 0.1m 50 20 C 18.9 Wπ= ⋅ − ° =
 
If the convection coefficient is independent of radius, the boundary layer must be of uniform thickness 
δ. Within the boundary layer, air flow is principally in the circumferential direction. The 
circumferential velocity component uθ corresponds to the rotational velocity of the disk at the surface 
(y = 0) and increases with increasing r (uθ = Ωr). The velocity decreases with increasing distance y 
from the surface, approaching zero at the outer edge of the boundary layer (y → δ). 
 
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PROBLEM 6.14 
 
KNOWN: Air flow over a flat plate of known length, location of transition from laminar to 
turbulent flow, value of the critical Reynolds number. 
 
FIND: (a) Free stream velocity with properties evaluated at T = 350 K, (b) Expression for the 
average convection coefficient, lamh (x) , as a function of the distance x from the leading edge in 
the laminar region, (c) Expression for the average convection coefficient turbh (x) , as a function 
of the distance x from the leading edge in the turbulent region, (d) Compute and plot the local and 
average convection coefficients over the entire plate length. 
 
SCHEMATIC: 
T∞, u∞
hlam=Clamx-0.5
hturb=Cturbx-0.2
TurbulentLaminarx
xc
T∞, u∞
hlam=Clamx-0.5
hturb=Cturbx-0.2
TurbulentLaminarx
xc 
 
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties. 
 
PROPERTIES: Table A.4, air (T = 350 K): k = 0.030 W/m⋅K, ν = 20.92 × 10-6 m2/s, Pr = 0.700. 
 
ANALYSIS: 
(a) Using air properties evaluated at 350 K with xc = 0.5 m, 
5c
x,c
u x
Re 5 10ν
∞= = × 
5 5 6
c
2u 5 10 x 5 10 20.92 10 m / s 0.5 m 20.9 m s−∞ = × = × × × =ν < 
(b) From Eq. 6.13, the average coefficient in the laminar region, 0 ≤ x ≤ xc, is 
 
( ) ( ) lam lamx x -0.5 0.5C Clam lam lam lam0 o1 1 1 -0.5h x = h x dx = x dx = x 2C 2x x x = x h (x)=∫ ∫ (1) < 
 
(c) The average coefficient in the turbulent region, xc ≤ x ≤ L, is 
 ( ) ( ) ( )
c
c
c
c
x x0.5 0.8x x
turb lam turb lam turb0 x
0 x
1 x
h x h x dx h x dx C C
x 0.
= + = + x
5 0.8
⎡ ⎤⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦
∫ ∫ 
Continued… 
 
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 PROBLEM 6.14 (Cont.) 
 
 
( ) ( )0.5 0.8 0.8turb lam c turb c1h x 2C x 1.25C x xx= + −⎡⎢⎣ ⎤⎥⎦ (2) < 
 
(d) The local and average coefficients, Eqs. (1) and (2) are plotted below as a function of x for the 
range 0 ≤ x ≤ L. 
 
0 0.5 1
Distance from leading edge, x (m)
0
50
100
150
C
on
ve
ct
io
n 
co
ef
fic
ie
nt
 (W
/m
^2
.K
)
Local - laminar, x <= xc
Local - turbulent, x => xc
Average - laminar, x <= xc
Average - turbulent, x => xc 
 
 
 
 
 
 
 
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PROBLEM 6.15 
K NOWN: Air speed and temperature in a wind tunnel. 
FIND: (a) Minimum plate length to achieve a Reynolds number of 108, (b) Distance from 
eading edge at which transition would occur. l 
SCHEMATIC: 
 
 
 
A
 
SSUMPTIONS: (1) Isothermal conditions, Ts = T∞. 
PROPERTIES: Table A-4, Air (25°C = 298K): ν = 15.71 × 10-6m2/s. 
 
A NALYSIS: (a) The Reynolds number is 
 x
 u x u xRe .ρ µ ν
∞ ∞= = 
 
o achieve a Reynolds number of 1 × 108, the minimum plate length is then T
 
 
( )8 6x
min
1 10 15.71 10 m / sReL
u 50 m/s
ν −
∞
× ×
= =
2
 
 
 < minL 31.4 m.= 
(b) For a transition Reynolds number of 5 × 105
 
 
( )5 -6x,c
c
5 10 15.71 10 m / sRe
x
u 50 m/s
ν
∞
× ×
= =
2
 
 
 < cx 0.157 m.= 
C OMMENTS: Note that 
 x,cc
L
Rex
L Re
= 
 
This expression may be used to quickly establish the location of transition from knowledge of 
 Re .x,c L and Re
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PROBLEM 6.16 
KNOWN: Transition Reynolds number. Velocity and temperature of atmospheric air, 
ngine oil, and mercury flow over a flat plate. e 
F IND: 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
 
P ROPERTIES: For the fluids at T = 300 K and 350 K: 
 ν(m2/s) 
 Fluid Table T = 300 K T = 350 K 
 Air (1 atm) A-4 15.89 × 10-6 20.92 × 10-6
 Engine Oil A-5 550 × 10-6 41.7 × 10-6 
 Mercury A-5 0.1125 × 10-6 0.0976 × 10-6 
 
A NALYSIS: The point of transition is 
 
5
c x,c
5 10x Re
u 1 m/s
ν .ν
∞
×= = 
 
S ubstituting appropriate viscosities, find 
 xc(m) 
 Fluid T = 300 K T = 350 K < 
 Air 7.95 10.5 
 Oil 275 20.9 
 Mercury 0.056 0.049 
 
 
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. 
 
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PROBLEM 6.17 
 
KNOWN: Pressure dependence of the dynamic viscosity, thermal conductivity and specific heat. 
 
FIND: (a) Variation of the kinematic viscosity and thermal diffusivity with pressure for an 
incompressible liquid and an ideal gas, (b) Value of the thermal diffusivity of air at 350 K for 
pressures of 1, 5 and 10 atm, (c) Location where transition occurs for air flow over a flat plate 
with T∞ = 350 K, p = 1, 5 and 10 atm, and u∞ = 2 m/s. 
 
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Transition at Rex,c = 5 
× 105, (4) Ideal gas behavior. 
 
PROPERTIES: Table A.4, air (350 K): µ = 208.2 × 10-7 N⋅s/m2, k = 0.030 W/m⋅K, cp = 1009 
J/kg⋅K, ρ = 0.995 kg/m3. 
 
ANALYSIS: 
 
(a) For an ideal gas 
 
p = ρRT or ρ = p/RT (1) 
 
while for an incompressible liquid, ρ = constant (2) 
 
The kinematic viscosity is ν = µ/ρ (3) 
 
Therefore, for an ideal gas 
 ν = µRT/p or (4) < -1ν p∝
and for an incompressible liquid 
 
 ν = µ/ρ or ν is independent of pressure. < 
 
The thermal diffusivity is 
 
 k / cα = ρ
 
Therefore, for an ideal gas, 
 (6) < -1α = kRT/pc or α p∝
For an incompressible liquid α = k/ cρ or α is independent of pressure < 
 
(b) For T = 350 K, p = 1 atm, the thermal diffusivity of air is 
 
 -6 23
0.030 W/m Kα = = 29.9 × 10 m /s
0.995 kg/m × 1009 J/kg K
⋅
⋅ < 
 
Using Equation 6, at p = 5 atm, 
Continued… 
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 PROBLEM 6.17 ( Cont.) 
 
 
-6 2α = 29.9 × 10 m /s/5 = 5.98 × 10-6 m2/s < 
 
At p = 10 atm, 
 = 2.99 × 10-6 2α = 29.9 × 10 m /s/10 -6 m2/s < 
 
(c) For transition over a flat plate, 
 5cx,c
x u
Re = 5 × 10
ν
∞ = 
Therefore 
 5cx = 5 × 10 (ν/u )∞
 
For T∞ = 350 K, p = 1 atm, 
 
 -7 2 3 -6 2ν = µ/ρ = 208.2 × 10 N s/m 0.995 kg/m = 20.92 × 10 m /s⋅ 
 
Using Equation 4, at p = 5 atm 
 
 -6 2 -6 2ν = 20.92 × 10 m /s 5 = 4.18 × 10 m /s 
 
At p = 10 atm, 
 
 -6 2 -6 2ν = 20.92 × 10 m /s 10 = 2.09 × 10 m /s 
 
Therefore, at p = 1 atm 
 < 5 -6 2cx = 5 × 10 × 20.92 × 10 m /s/(2m/s) = 5.23 m
 
At p = 5 atm, 
 < 5 -6 2cx = 5 × 10 × 4.18 × 10 m /s/(2m/s) = 1.05 m
 
At p = 10 atm 
 < 5 -6 2cx = 5 × 10 × 2.09 × 10 m /s/(2m/s) = 0.523 m
 
COMMENT: Note the strong dependence of the transition length upon the pressure for the gas 
(the transition length is independent of pressure for the incompressible liquid). 
 
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PROBLEM 6.18 
KNOWN: Characteristic length, surface temperature and average heat flux for an object 
laced in an airstream of prescribed temperature and velocity. p 
FIND: Average convection coefficient if characteristic length of object is increased by a 
actor of five and air velocity is decreased by a factor of five. f 
SCHEMATIC: 
 
A SSUMPTIONS: (1) Steady-state conditions, (2) Constant properties. 
ANALYSIS: For a particular geometry, 
 ( )L LNu f Re , Pr .= 
T he Reynolds numbers for each case are 
Case 1: ( ) 21 1L,1
1 1 1
100m/s 1mV L 100 m / sRe ν ν ν= = = 
 
Case 2: ( ) 22 2L,2
2 2 2
20m/s 5mV L 100 m / sRe .ν ν ν= = = 
 
Hence, with ν1 = ν2, 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
=
= = 
F or Case 1, using the rate equation, the convection coefficient is 
 
( )
( )
( ) ( ) ( )
1 1 1 s 1
211 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
∞
∞ ∞
= −
′′= = = =− − − ⋅
 
 
H ence, it follows that for Case 2 
 22h 0.2 200 W/m K 40 W/m K.= × ⋅ = ⋅2 < 
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. 
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PROBLEM 6.19 
K NOWN: Heat transfer rate from a turbine blade for prescribed operating conditions. 
F IND: 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 
eometrically similar. g 
A NALYSIS: For a prescribed geometry, 
 ( )LhLNu f Re , Pr .k= = 
T he Reynolds numbers for the blades are 
 ( ) ( )L,1 1 1 L,2 2 2Re V L / 15 / Re V L / 15 / .ν ν ν= = = = ν 
 
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
.
∞
=
=
= = −
 
 
H ence, the heat rate for the second blade is 
 
( ) ( )( )( )
( ) ( )
s,21 2
2 2 2 s,2
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
∞∞ ∞
∞
∞
−= − = −
− −= =− −
1
.
 
 
 < 2q 2066 W= 
COMMENTS: The slight variation of ν 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. 
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PROBLEM 6.20 
KNOWN: Experimental measurements of the heat transfer coefficient for a square bar in 
ross flow. c 
FIND: (a) h for the condition when L = 1m and V = 15m/s, (b) h for the condition when L 
 1m and V = 30m/s, (c) Effect of defining a side as the characteristic length. = 
SCHEMATIC: 
 
 
 
ASSUMPTIONS: (1) Functional form m nNu CRe Pr= applies with C, m, n being 
onstants, (2) Constant properties. c 
ANALYSIS: (a) For the experiments and the condition L = 1m and V = 15m/s, it follows 
hat Pr as well as C, m, and n are constants. Hence t 
 ( )mhL VL .∝ 
 
U sing the experimental results, find m. Substituting values 
 
m m
1 1 1 1
2 2 2 2
h L V L 50 0.5 20 0.5 
h L V L 40 0.5 15 0.5
⎡ ⎤ × ×⎡ ⎤= =⎢ ⎥ ⎢ ⎥× ×⎣ ⎦⎣ ⎦
 
 
g iving m = 0.782. It follows then for L = 1m and V = 15m/s, 
 
m 0.782
21
1 21 1
L V L W 0.5 15 1.0h h 50 34.3W/m K.
L V L 1.0 20 0.5m K
⎡ ⎤⋅ ×⎡ ⎤= = × =⎢ ⎥ ⎢ ⎥⋅ ×⎣ ⎦⋅⎣ ⎦
⋅ < 
 
( b) For the condition L = 1m and V = 30m/s, find 
 
m 0.782
21
1 21 1
L V L W 0.5 30 1.0h h 50 59.0W/m K.
L V L 1.0 20 0.5m K
⎡ ⎤⋅ ×⎡ ⎤= = × =⎢ ⎥ ⎢ ⎥⋅ ×⎣ ⎦⋅⎣ ⎦
⋅ < 
 
(c) If the characteristic length were chosen as a side rather than the diagonal, the value of C 
ould change. However, the coefficients m and n would not change. w 
COMMENTS: The foregoing Nusselt number relation is used frequently in heat transfer 
analysis, providing appropriate scaling for the effects of length, velocity, and fluid properties 
on the heat transfer coefficient. 
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PROBLEM 6.21 
K NOWN: Local Nusselt number correlation for flow over a roughened surface. 
F IND: Ratio of average heat transfer coefficient to local coefficient. 
SCHEMATIC: 
 
 
 
A NALYSIS: The local convection coefficient is obtained from the prescribed correlation, 
 
0.9 1/3
x x x
0.9 0.9
1/3 -0.1
x 1
k kh Nu 0.04 Re Pr
x x
V xh 0.04 k Pr C x .
xν
= =
⎡ ⎤= ≡⎢ ⎥⎣ ⎦
 
 
T o determine the average heat transfer coefficient for the length zero to x, 
 
x x
0 0
-0.1
x x 1
0.9
-0.11
x 1
1 1h h dx C x dx
x x
C xh 1.11 C x
x 0.9
≡ ∫ = ∫
= = .
 
 
Hence, the ratio of the average to local coefficient is 
 
-0.1
x 1
-0.1x 1
h 1.11 C x 1.11.
h C x
= = < 
COMMENTS: Note that x xNu / Nu is also equal to 1.11. Note, however, that 
x
x 0 x
1Nu Nu dx.
x
≠ ∫ 
 
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PROBLEM 6.22 
KNOWN: Freestream velocity and average convection heat transfer associated with fluid 
low over a surface of prescribed characteristic length. f 
FIND: Values of L L HNu ,Re , Pr, j for (a) air, (b) engine oil, (c) mercury, (d) water. 
SCHEMATIC: 
 
 
 
PROPERTIES: For the fluids at 300K: 
 
Fluid Table ν(m2/s) k(W/m⋅K) α(m2/s) Pr 
 
 Air A.4 15.89 × 10-6 0.0263 22.5 × 10-7 0.71 
 Engine Oil A.5 550 × 10-6 0.145 0.859 × 10-7 6400 
 Mercury A.5 0.113 × 10-6 8.54 45.30 × 10-7 0.025 
Water A.6 0.858 × 10-6 0.613 1.47 × 10-7 5.83 
 
ANALYSIS: The appropriate relations required are 
 
L
L
2/3
L H
L
NuhL VLNu Re Pr j StPr St
k R
ν
ν α= = = = = e Pr 
 
 Fluid LNu LRe Pr Hj < 
 
 
 Air 3802 6.29 × 104 0.71 0.068 
 Engine Oil 690 1.82 × 103 6403 0.0204 
 Mercury 11.7 8.85 × 106 0.025 4.52 × 10-6
 Water 163 1.17 × 106 5.84 7.74 × 10-5
 
COMMENTS: Note the wide range of Pr associated with the fluids. 
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PROBLEM 6.23 
K
 
NOWN: Variation of hx with x for flow over a flat plate. 
FIND: Ratio of average Nusselt number for the entireplate to the local Nusselt number at x 
 L. = 
SCHEMATIC: 
 
 
 
A NALYSIS: The expressions for the local and average Nusselt numbers are 
 
( )
L
-1/2 1/2
L
L
L
CL Lh L CLNu
k k
h LNu
k
= = =
=
k 
 
where 
 
L L
0 0
-1/2 1/2 -1/2
L x
1 C 2Ch h dx x dx L 2 CL
L L L
= ∫ = ∫ = = . 
 
Hence, 
 ( )L
-1/2 1/22 CL L 2 CLNu
k k
= = 
 
and 
 L
L
Nu
2.
Nu
= < 
 
COMMENTS: Note the manner in which LNu is defined in terms of Lh . Also note that 
 
L
L 0 x
1Nu Nu dx.
L
≠ ∫ 
 
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PROBLEM 6.24 
KNOWN: Laminar boundary layer flow of air at 20°C and 1 atm having t 1.13 .δ δ= 
FIND: Ratio t/δ δ when fluid is ethylene glycol for same conditions. 
SCHEMATIC: 
 
 
 
A SSUMPTIONS: (1) Laminar flow. 
PROPERTIES: Table A-4, Air (293K, 1 atm): Pr = 0.709; Table A-5, Ethylene glycol 
293K): Pr = 211. ( 
ANALYSIS: The Prandtl number strongly influences relative growth of the velocity, ,δ and 
thermal, t ,δ boundary layers. For laminar flow, the approximate relationship is given by 
 n
t
Pr δδ≈ 
 
w here n is a positive coefficient. Substituting the values for air 
 ( )n 10.709
1.13
= 
 
f ind that n = 0.355. Hence, for ethylene glycol it follows that 
 0.355 0.355
t
Pr 211 6.69.δδ = = = < 
 
COMMENTS: (1) For laminar flow, generally we find n = 0.33. In which case, t/ 5.85.δ δ = 
(2) Recognize the physical importance of ν > α, which gives large values of the Prandtl 
number, and causes t .δ δ> 
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PROBLEM 6.25 
K NOWN: Air, water, engine oil or mercury at 300K in laminar, parallel flow over a flat plate. 
F IND: Sketch of velocity and thermal boundary layer thickness. 
A SSUMPTIONS: (1) Laminar flow. 
P ROPERTIES: For the fluids at 300K: 
Fluid Table Pr 
 
 Air A.4 0.71 
 Water A.6 5.83 
 Engine Oil A.5 6400 
 Mercury A.5 0.025 
ANALYSIS: For laminar, boundary layer flow over a flat plate. 
 n
t
~ Prδδ 
w here n > 0. Hence, the boundary layers appear as shown below. 
Air: 
 
Water: 
 
Engine Oil: 
 
Mercury: 
 
COMMENTS: Although Pr strongly influences relative boundary layer development in laminar 
flow, its influence is weak for turbulent flow. 
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PROBLEM 6.26 
KNOWN: Expression for the local heat transfer coefficient of air at prescribed velocity and 
temperature flowing over electronic elements on a circuit board and heat dissipation rate for a 4 × 4 
m chip located 120mm from the leading edge. m 
IND: Surface temperature of the chip surface, Ts. F 
SCHEMATIC: 
 
 
ASSUMPTIONS: (1) Steady-state conditions, (2) Power dissipated within chip is lost by convection 
across the upper surface only, (3) Chip surface is isothermal, (4) The average heat transfer coefficient 
or the chip surface is equivalent to the local value at x = L, (5) Negligible radiation. f 
PROPERTIES: Table A-4, Air (assume Ts = 45°C, Tf = (45 + 25)/2 = 35°C = 308K, 1atm): ν = 
6.69 × 10-6m2/s, k = 26.9 × 10-3 W/m⋅K, Pr = 0.703. 1
 
ANALYSIS: From an energy balance on the chip (see above), 
 (1) conv gq E 30= =� W.
Newton’s law of cooling for the upper chip surface can be written as 
 s convT T q / h A∞= + chip
.
 (2) 
where Assume that the average heat transfer coefficient 2chipA = A ( )h over the chip surface is 
equivalent to the local coefficient evaluated at x = L. That is, ( )chip xh h≈ L where the local 
coefficient can be evaluated from the special correlation for this situation, 
 
0.85
1/ 3x
x
h x VxNu 0.04 Pr
k ν
⎡ ⎤= = ⎢ ⎥⎣ ⎦ 
and substituting numerical values with x = L, find 
 
0.85
1/ 3
x
k VLh 0.04 Pr
L ν
⎡ ⎤= ⎢ ⎥⎣ ⎦ 
 
 ( )
0.85
1/ 3 2
x -6 2
0.0269 W/m K 10 m/s 0.120 mh 0.04 0.703 107 W/m K.
0.120 m 16.69 10 m / s
⎡ ⎤⋅ ×⎡ ⎤= =⎢ ⎥⎢ ⎥⎣ ⎦ ×⎣ ⎦
 ⋅ 
The surface temperature of the chip is from Eq. (2), 
 < ( )2-3 2sT 25 C 30 10 W/ 107 W/m K 0.004m 42.5 C.⎡ ⎤= + × ⋅ × =⎢ ⎥⎣ ⎦D D
COMMENTS: (1) Note that the estimated value for Tf used to evaluate the air properties was 
reasonable. (2) Alternatively, we could have evaluated chiph by performing the integration of the 
local value, h(x). 
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PROBLEM 6.27 
 
KNOWN: Expression for the local heat transfer coefficient of air at prescribed velocity and 
temperature flowing over electronic elements on a circuit board and heat dissipation rate for a 4 × 
4 mm chip located 120 mm from the leading edge. Atmospheric pressure in Mexico City. 
 
FIND: (a) Surface temperature of chip, (b) Air velocity required for chip temperature to be the 
same at sea level. 
 
SCHEMATIC: 
 
 
 
 
 
 
 
p = 76.5 kPap = 76.5 kPa
 
ASSUMPTIONS: (1) Steady-state conditions, (2) Power dissipated in chip is lost bey convection 
across the upper surface only, (3) Chip surface is isothermal, (4) The average heat transfer 
coefficient for the chip surface is equivalent to the local value at x = L, (5) Negligible radiation, 
(6) Ideal gas behavior. 
 
PROPERTIES: Table A.4, air (p = 1 atm, assume Ts = 45 °C, Tf = (45 °C + 25 °C)/2 = 35 °C): k 
= 0.0269 W/m⋅K, ν = 16.69 × 10-6 m2/s, Pr = 0.706. 
 
ANALYSIS: 
(a) From an energy balance on the chip (see above), 
 (1) conv gq E 30= =� W. 
Newton’s law of cooling for the upper chip surface can be written as 
 s convT T q / h A∞= + chip
.
 (2) 
where From Assumption 4, 2chipA = A ( )chip xh h≈ L where the local coefficient can be 
evaluated from the correlation provided in Problem 6.35. 
 
0.85
1/ 3x
x
h x VxNu 0.04 Pr
k ν
⎡ ⎤= = ⎢ ⎥⎣ ⎦ (3) 
The kinematic viscosity is 
 µν = 
ρ
 (4) 
while for an ideal gas, 
 pρ = 
RT
 (5) 
 
Combining Equations 4 and 5 yields 
 (6) -1ν p∝
 
Continued… 
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PROBLEM 6.27 (Cont.) 
 
The Prandtl number is 
 ν µρc µcPr = = = 
α ρk k
 (7) 
 
which is independent of pressure. 
 
Therefore, at sea level (p = 1 atm) 
 -6 2k = 0.0269 W/m K, ν = 16.69 × 10 m /s, Pr = 0.706⋅
 
0.85
1/3
x
0.85
1/3 2
x -6 2
k VLh = 0.04 Pr
L ν
0.0269 W/m K 10 m/s × 0.120 mh = 0.04 (0.706) = 107 W/m K
0.120 m 16.69 × 10 m /s
⎡ ⎤⎢ ⎥⎣ ⎦
⋅ ⎡ ⎤⎡ ⎤ ⋅⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦
 
 
-3
s 2 2
30 × 10 WT = 25°C + = 42.5°C
107 W/m K × (0.004 m)⋅ 
 
In Mexico City (p = 76.5 kPa) 
 -6 2 -6 2101.3lPaν = 16.69 × 10 m /s × = 22.10 × 10 m /s
76.5kPa
⎡ ⎤⎢ ⎥⎣ ⎦ 
k = 0.0269 W/m K, Pr = 0.706⋅ 
 
0.85
1/3 2
x -6 2
0.0269 W/m K 10 m/s × 0.120 mh = 0.04 (0.706) = 84.5 W/m K
0.120 m 22.10 × 10 m /s
⋅ ⎡ ⎤⎡ ⎤ ⋅⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦ 
 
-3
s 2 2
30 × 10 WT = 25°C + = 47.2°C
84.5 W/m K × (0.004 m)⋅ < 
 
(b) For the same chip temperature, it is required that hx = 107 W/m2·K. Therefore 
 
 
0.85
2 1/3
x -6 2
0.0269 W/m×K V × 0.120 mh = 107 W/m K = 0.04 (0.706) 
0.120 m 22.10 × 10 m /s
⎡ ⎤⎡ ⎤⋅ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦ 
 
From which we may find V = 13.2 m/s < 
 
COMMENTS: (1) In Part (a), the chip surface temperature increased from 42.4 °C to 47.2 °C. 
This is considered to be significant and the electronics packaging engineer needs to consider the 
effect of large changes in atmospheric pressure on the efficacy of the electronics cooling scheme. 
(2) Careful consideration needs to be given to the effect changes in the atmospheric pressure on 
the kinematic viscosity and, in turn, on changes in transition lengths which might dramatically 
affect local convective heat transfer coefficients. 
 
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PROBLEM 6.28 
 
KNOWN: Location and dimensions of computer chip on a circuit board. Form of the convection 
correlation. Maximum allowable chip temperature and surface emissivity. Temperature of cooling air 
nd surroundings. a 
FIND: Effect of air velocity on maximum power dissipation, first without and then with consideration of 
adiation effects. r 
SCHEMATIC: 
 
ASSUMPTIONS: (1) Steady-state, (2) Negligible temperature variations in chip, (3) Heat transfer 
exclusively from the top surface of the chip, (4) The local heat transfer coefficient at x = L provides a 
ood approximation to the average heat transfer coefficient for the chip surface. g 
PROPERTIES: Table A.4, air ( ( )cT T T∞= + 2
conv
 = 328 K): ν = 18.71 × 10-6 m2/s, k = 0.0284 W/m⋅K, 
r = 0.703. P
 
ANALYSIS: Performing an energy balance for a control surface about the chip, we obtain P = q + 
q
c
rad, where qconv = (s chA T T∞− ) , qrad = ( )r s c surh A T T− , and . With ( )( )2 2r c sur c surh T T T Tεσ= + +
Lh h≈ , the convection coefficient may be determined from the correlation provided in Problem 6.26 
(NuL = 0.04 Pr0.85LRe
1/3). Hence, 
 ( ) ( ) ( )( )( )2 0.85 1/ 3 2 2c L c c sur c sur c surP 0.04 k L Re Pr T T T T T T T Tεσ∞= − + + +⎡ ⎤⎢ ⎥⎣ ⎦A − 
where ReL = VL/ν. Computing the right side of this expression for ε = 0 and ε = 0.85, we obtain the 
following results. 
0 5 10 15 20 25
Velocity, V(m/s)
0
0.05
0.1
0.15
0.2
0.25
0.3
C
hi
p 
po
w
er
, P
c(
W
)
epsilon = 0.85
epsilon = 0 
Since hL increases as V0.85, the chip power must increase with V in the same manner. Radiation exchange 
increases Pc by a fixed, but small (6 mW) amount. While hL varies from 14.5 to 223 W/m2⋅K over the 
rescribed velocity range, hr = 6.5 W/m2⋅K is a constant, independent of V. p 
COMMENTS: Alternatively, h could have been evaluated by integrating hx over the range 118 ≤ x ≤ 
122 mm to obtain the appropriate average. However, the value would be extremely close to hx=L. 
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PROBLEM 6.29 
KNOWN: Form of Nusselt number for flow of air or a dielectric liquid over components of a circuit 
ard. c 
FIND: Ratios of time constants associated with intermittent heating and cooling. Fluid that provides 
faster thermal response. 
 
PROPERTIES: Prescribed. Air: k = 0.026 W/m⋅K, ν = 2 × 10-5 m2/s, Pr = 0.71. Dielectric liquid: 
 = 0.064 W/m⋅K, ν = 10-6 m2/s, Pr = 25. k
 
ANALYSIS: From Eq. 5.7, the thermal time constant is 
 
 t
s
c
hA
ρτ ∀= 
 
Since the only variable that changes with the fluid is the convection coefficient, where 
 
 L
m
m n n
L
k k k VLh Nu C Re Pr C Pr
L L L ν
⎛ ⎞= = = ⎜ ⎟⎝ ⎠ 
 
the desired ratio reduces to 
 
 
( )
( )
m n
t,air a d d a d
a a d at,dielectric d
h k Pr
h k Pr
τ ν
τ ν
⎛ ⎞ ⎛ ⎞= = ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
 
 
 
0.8 0.335t,a
6t,d
0.064 2 10 25 88.6
0.026 0.7110
τ
τ
−
−
⎛ ⎞× ⎛ ⎞⎜ ⎟= =⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
 
 
Since its time constant is nearly two orders of magnitude smaller than that of the air, the dielectric 
liquid is clearly the fluid of choice. < 
 
COMMENTS: The accelerated testing procedure suggested by this problem is commonly used to test 
the durability of electronic packages. 
 
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PROBLEM 6.30 
K NOWN: Form of the Nusselt number correlation for forced convection and fluid properties. 
F
 
IND: Expression for figure of merit FF and values for air, water and a dielectric liquid. 
PROPERTIES: Prescribed. Air: k = 0.026 W/m⋅K, ν = 1.6 × 10-5 m2/s, Pr = 0.71. Water: k = 
.600 W/m⋅K, ν = 10-6 m2/s, Pr = 5.0. Dielectric liquid: k = 0.064 W/m⋅K, ν = 10-6 m2/s, Pr = 25 0
 
ANALYSIS: With the convection coefficient may be expressed as m nL LNu ~ Re Pr ,
 
 
m m n
n
1 m m
k VL V k Prh ~ Pr ~
L Lν ν−
⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
 
 
The figure of merit is therefore 
 
 
n
F m
k PrF ν= < 
 
and for the three fluids, with m = 0.80 and n = 0.33, 
 
 ( )0.8 2.6F Air Water DielectricF W s / m K 167 64,400 11,700⋅ ⋅ < 
 
W ater is clearly the superior heat transfer fluid, while air is the least effective. 
COMMENTS: The figure of merit indicates that heat transfer is enhanced by fluids of large k, large 
Pr and small ν. 
 
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PROBLEM 6.31 
 
KNOWN: Form of the Nusselt number correlation for forced convection and fluid properties. 
Properties of xenon and He-Xe mixture. Temperature and pressure.Expression for specific heat for 
onatomic gases. m
 
FIND: Figures of merit for air, pure helium, pure xenon, and He-Xe mixture containing 0.75 mole 
fraction of helium. 
 
PROPERTIES: Table A-4, Air (300 K): k = 0.0263 W/m⋅K, ν = 15.89 × 10-6 m2/s, Pr = 0.707. Table 
A-4, Helium (300 K): k = 0.152 W/m⋅K, ν = 122 × 10-6 m2/s, Pr = 0.680. Pure xenon (given): k = 
0.006 W/m⋅K, µ = 24.14 × 10-6 N·s/m2. He-Xe mixture (given): k = 0.0713 W/m·K, µ = 25.95 × 10-6 
·s/m2. N
 
ANALYSIS: With the convection coefficient may be expressed as m nL LNu ~ Re Pr ,
m m n
n
1 m m
k VL V k Prh ~ Pr ~
L Lν ν−
⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
 
 
The figure of merit is therefore 
n
F m
k PrF ν= (1) 
 
For xenon and the He-Xe mixture, we must find the density and specific heat. Proceeding for pure 
xenon: 
3
-2 3
P 1 atm 131.29 kg/kmolρ 5.33 kg/m
T 8.205 10 m atm / kmol K 300 K
×= = =ℜ × ⋅ ⋅ ×
M
3
p
5 5 8.315 10 J/kmol Kc 158 J/kg
2 2 131.29 kg/kmol
ℜ × ⋅= = =
M
 
 
Thus ν = µ/ρ = 24.14 × 10-6 N·s/m2/5.33 kg/m3 = 4.53 × 10-6 m2/s and Pr = µcp/k = 24.14 × 10-6 N·s/m2 
× 158 J/kg/0.006 W/m·K = 0.636. 
 
For the He-Xe mixture, the molecular weight of the mixture can be found from 
mix 0.75 kmol He/kmol 4.0 kg/kmol He 0.25 kmol Xe/kmol 131.29 kg/kmol Xe
 35.82 kg/kmol
= × + ×
=
M 
from which we can calculate ρ = 1.46 kg/m3, cp = 580 J/kg·K, ν = µ/ρ = 25.95 × 10-6 N·s/m2/1.46 
kg/m3 = 1.78 × 10-5 m2/s, and Pr = µcp/k = 25.95 × 10-6 N·s/m2 × 580 J/kg/0.0713 W/m·K = 0.211. 
 
Finally, for the four fluids, with m = 0.85 and n = 0.33, we can calculate the figure of merit from 
Equation (1): 
 FF (W·s0.85/m2.7·K) < 
Air 281 
Helium 284 
Xenon 180 
He-Xe 465 
 
COMMENTS: The effectiveness of the He-Xe mixture is much higher than that of pure He, pure Xe, 
or air. By blending He and Xe, the high thermal conductivity of helium and the high density of xenon 
are both exploited in a manner that leads to a high figure of merit. 
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PROBLEM 6.32 
KNOWN: Ambient, interior and dewpoint temperatures. Vehicle speed and dimensions of 
indshield. Heat transfer correlation for external flow. w 
FIND: Minimum value of convection coefficient needed to prevent condensation on interior surface 
f windshield. o 
SCHEMATIC: 
 
 
 
A
 
SSUMPTIONS: (1) Steady-state, (2) One-dimensional heat transfer, (3) Constant properties. 
PROPERTIES: Table A-3, glass: kg = 1.4 W/m⋅K. Prescribed, air: k = 0.023 W/m⋅K, ν = 12.5 × 
0-6 m2/s, Pr = 0.70. 1
 
ANALYSIS: From the prescribed thermal circuit, conservation of energy yields 
 ,i s,i s,i ,o
i g
T T T T
1/ h t / k 1/ h
∞ ∞− −= + o
 
where oh may be obtained from the correlation 
 L
0.8 1/ 3o
L
h LNu 0.030Re Pr
k
= = 
With V = (70 mph × 1585 m/mile)/3600 s/h = 30.8 m/s, ReD = (30.8 m/s × 0.800 m)/12.5 × 10-6 m2/s 
= 1.97 × 106 and 
 ( ) ( )0.8 1/ 36 2o 0.023W / m Kh 0.030 1.97 10 0.70 83.1W / m K0.800 m ⋅= × = ⋅ 
F
 
rom the energy balance, with Ts,i = Tdp = 10°C 
 
( )
( )
1
s,i ,o
i
g o,i s,i
T T t 1h
k hT T
−∞
∞
⎛ ⎞−= +⎜ ⎟⎜ ⎟− ⎝ ⎠
 
 
 
( )
( )
1
i 2
10 15 C 0.006 m 1h
50 10 C 1.4 W / m K 83.1W / m K
−⎛ ⎞+ ° ⎜ ⎟= +⎜ ⎟− ° ⋅ ⋅⎝ ⎠
 
 
 2ih 38.3W / m K= ⋅ < 
COMMENTS: The output of the fan in the automobile’s heater/defroster system must maintain a 
velocity for flow over the inner surface that is large enough to provide the foregoing value of ih . In 
addition, the output of the heater must be sufficient to maintain the prescribed value of T∞,i at this 
velocity. 
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PROBLEM 6.33 
 
KNOWN: Characteristic length of a microscale chemical detector, free stream velocity and 
temperature, hydrogen wind tunnel pressure and free stream velocity. 
 
FIND: Model length scale and hydrogen temperature needed for similarity. 
 
SCHEMATIC: 
Sensor
Ls = 80 µm
Air
T∞ = 25°C
V = 10 m/s
p = 1 atm
Sensor
Ls = 80 µm
Air
T∞ = 25°C
V = 10 m/s
p = 1 atm
 Heated model
Lm = ? 
Hydrogen
T∞ = ?
V = 0.5 m/s
p = 0.5 atm
Heated model
Lm = ? 
Hydrogen
T∞ = ?
V = 0.5 m/s
p = 0.5 atm
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible microscale 
or nanoscale effects, (4) Ideal gas behavior. 
 
PROPERTIES: Table A.4, air (T = 25 °C): Prs = 0.707, νs = 15.71 × 10-6 m2/s, hydrogen (250 K) 
Pr = 0.707, ν = 81.4 × 10-6 m2/s. 
 
ANALYSIS: For similarity we require Rem = Res and Prm = Prs. For the sensor, 
 
 
-6
s s
s -5 2
s
V L 10 m/s × 80 × 10 mRe = = = 50.93
ν 1.571 × 10 m /s
 
 Prs = 0.707 
 
For the model, Prm = Prs = 0.707. 
From Table A.4, we note Prs = 0.707, ν = 81.4 ×10-6 m2/s at T∞ = 250 K and p = 1 atm. < 
 
The value of the Prandtl number is independent of pressure for an ideal gas. The kinematic 
viscosity is pressure-dependent. Hence, 
 µ µ ρ(at 1.0 atm)ν(at 0.5 atm) = = × 
ρ(at 0.5 atm) ρ(at 1.0 atm) ρ(at 0.5 atm)
 
For an ideal gas, 
 1.0 atmν(at 0.5 atm) = ν(at 1.0 atm)× = 2ν(at 1.0 atm)
0.5 atm
 
Therefore, 
 νm = 81.4 × 10-6 m2/s × 2 = 163 × 10-6 m2/s 
 
For similarity, 
 
Continued… 
 
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PROBLEM 6.33 (Cont.) 
 
Rem = Res = 50.93 = m m m-6 2
m
V L 0.5 m/s × L = 
ν 163 × 10 m /s
 
 
or Lm = 16.6 × 10-3 m = 16.6 mm < 
 
COMMENTS: (1) From Section 2.2.1, we know that the mean free path of air at room 
conditions is approximately 80 nm. Since Ls is three orders of magnitude greater than the mean 
free path, the air may be treated as a continuum. (2) Hydrogen can leak from enclosures easily. 
By keeping the wind tunnel pressure below atmospheric, we avoid possible leakage of flammable 
hydrogen into the lab. Also, if leaks occur, air must enter the wind tunnel. It is much easier to seal 
against air leaks than hydrogen leaks. (3) Prm = 0.707 at 100 K also. However, the operation of 
the hydrogen wind tunnel at such a low temperature would be much more difficult than at 250 K. 
 
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PROBLEM 6.34 
K NOWN: Drag force and air flow conditions associated with a flat plate. 
F IND: Rate of heat transfer from the plate. 
SCHEMATIC:A SSUMPTIONS: (1) Chilton-Colburn analogy is applicable. 
PROPERTIES: Table A-4, Air (70°C,1 atm): ρ = 1.018 kg/m3, cp = 1009 J/kg⋅K, Pr = 0.70, 
ν = 20.22 × 10-6m2/s. 
 
A NALYSIS: The rate of heat transfer from the plate is 
 ( ) ( )2 sq 2h L T T∞= − 
 
where h may be obtained from the Chilton-Colburn analogy, 
 ( ) ( )
( )
2/3 2 / 3f
H
p
2
4sf
2 23
C hj St Pr Pr
2 u c
0.075 N/2 / 0.2mC 1 1 5.76 10 .
2 2 2 u / 2 1.018 kg/m 40 m/s / 2
ρ
τ
ρ
∞
−
∞
= = =
= = = ×
 
 
Hence, 
 ( ) ( ) ( )
-2/3f
p
2 / 3-4 3
2
Ch u c Pr
2
h 5.76 10 1.018kg/m 40m/s 1009J/kg K 0.70
h 30 W/m K.
ρ ∞
−
=
= × ⋅
= ⋅
 
 
The heat rate is 
 ( ) ( ) ( )22q 2 30 W/m K 0.2m 120 20 C= ⋅ − D
 < q 240 W.=
COMMENTS: Although the flow is laminar over the entire surface (ReL = u∞L/ν = 40 m/s 
× 0.2m/20.22 × 10-6m2/s = 4.0 × 105), the pressure gradient is zero and the Chilton-Colburn 
analogy is applicable to average, as well as local, surface conditions. Note that the only 
contribution to the drag force is made by the surface shear stress. 
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PROBLEM 6.35 
KNOWN: Air flow conditions and drag force associated with a heater of prescribed surface 
emperature and area. t 
F IND: Required heater power. 
SCHEMATIC: 
 
 
 
ASSUMPTIONS: (1) Steady-state conditions, (2) Reynolds analogy is applicable, (3) 
ottom surface is adiabatic. B 
PROPERTIES: Table A-4, Air (Tf = 350K, 1atm): ρ = 0.995 kg/m3, cp = 1009 J/kg⋅K, Pr = 
.700. 0 
A NALYSIS: The average shear stress and friction coefficient are 
 
( )
2D
s 2
2
3s
f 2 23
F 0.25 N 1 N/m
A 0.25 m
1 N/mC 8
 u / 2 0.995 kg/m 15m/s / 2
τ
τ
ρ .93 10 .
−
∞
= = =
= = = ×
 
 
F rom the Reynolds analogy, 
 2 /3f
p
h CSt Pr .
 u c 2ρ
−
∞
= = 
 
Solving for h and substituting numerical values, find 
 ( ) ( ) ( ) 2 / 33 -3
2
h 0.995 kg/m 15m/s 1009 J/kg K 8.93 10 / 2 0.7
h 85 W/m K.
−= ⋅ ×
= ⋅
 
 
H ence, the heat rate is 
 ( ) ( ) ( )2 2sq h A T T 85W/m K 0.25m 140 15 C∞= − = ⋅ − D 
 
 < q 2.66 kW.=
 
COMMENTS: Due to bottom heat losses, which have been assumed negligible, the actual 
power requirement would exceed 2.66 kW. 
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PROBLEM 6.36 
KNOWN: Heat transfer correlation associated with parallel flow over a rough flat plate. 
elocity and temperature of air flow over the plate. V 
F IND: Surface shear stress l m from the leading edge. 
SCHEMATIC: 
 
 
 
A SSUMPTIONS: (1) Modified Reynolds analogy is applicable, (2) Constant properties. 
PROPERTIES: Table A-4, Air (300K, 1atm): ν = 15.89 × 10-6m2/s, Pr = 0.71, ρ = 1.16 
kg/m3. 
 
A NALYSIS: Applying the Chilton-Colburn analogy 
 
0.9 1/ 3
2 / 3 2 / 3 2 / 3f x x
x
x x
-0.1f
x
C Nu 0.04 Re PrSt Pr Pr Pr
2 Re Pr Re Pr
C 0.04 Re
2
= = =
=
 
 
w here 
 6x -6 2
u x 50 m/s 1mRe 3.15 10 .
15.89 10 m / sν
∞ ×= = = ×
×
 
 
H ence, the friction coefficient is 
 ( ) ( )0.16 2f sC 0.08 3.15 10 0.0179 / u / 2τ ρ− ∞= × = = 
 
a nd the surface shear stress is 
 ( ) ( )22 3s fC u / 2 0.0179 1.16kg/m 50 m/s / 2τ ρ ∞= = ×
 
 < 2s 25.96 kg/m s 25.96 N/m .τ = ⋅ = 2 
COMMENTS: Note that turbulent flow will exist at the designated location. 
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PROBLEM 6.37 
 
KNOWN: Dimensions and temperature of a thin, rough plate. Velocity of air flow parallel to 
plate (at an angle of 45° to a side). Heat transfer rate from plate to air. 
 
FIND: Drag force on plate. 
 
SCHEMATIC: 
L = 0.2 m 
Air
T∞ = 20°C
u∞ = 30 m/s
Ts = 80°C 
L = 0.2 m 
Air
T∞ = 20°C
u∞ = 30 m/s
Ts = 80°C 
 
 
 
 
 
 
 
 
 
 
ASSUMPTIONS: (1) The modified Reynolds analogy holds, (2) Constant properties. 
 
PROPERTIES: Table A-4, Air (50°C = 323 K): cp = 1008 J/kg·K, Pr = 0.704. 
 
ANALYSIS: The modified Reynolds analogy, Equation 6.70, combined with the definition of the 
Stanton number, Equation 6.67, yields 
 (1) -1/3fC /2= (Nu/Re)Pr
 
The drag force is related to the friction coefficient according to 
 (2) 2D s s f sF = τ A = C ρu A / 2∞⋅
 
Combining Equations (1) and (2) 
 -1/3 2D s
NuF = Pr ρu A
Re ∞
 
 
Substituting the definitions of Nu and Re, we find 
 -1/3 2 -1/3 2/3cD s s
c p p
hL ν h ν hF = Pr ρu A = Pr u A = Pr u A
k u L c α c∞ ∞∞
s∞ 
Where Lc is a characteristic length used to define Nu and Re. With hAs = q/ ∆T we have 
 
2/32/3
D 
p
qu Pr 2000 W × 30 m/s × (0.704)F = = 0.785 N 
c ∆T 1008 J/kg K × 60 K
∞ =⋅ < 
 
COMMENTS: (1) Heat transfer or friction coefficient correlations for this simple configuration 
apparently do not exist. (2) Experiments to measure the drag force would be relatively simple to 
implement and measured drag forces could be used to determine the heat transfer coefficients 
using the Reynolds analogy. (3) The solution demonstrates advantages associated with working 
the problem symbolically and only introducing numbers at the end. First, the length scale in Nu 
and Re did not have to be defined because it cancelled out. Second, the properties k, ν, and ρ also 
cancelled out. 
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PROBLEM 6.38 
KNOWN: Nominal operating conditions of aircraft and characteristic length and average friction 
oefficient of wing. c 
F IND: Average heat flux needed to maintain prescribed surface temperature of wing. 
SCHEMATIC: 
 
 
 
ASSUMPTIONS: (1) Applicability of modified Reynolds analogy, (2) Constant properties. 
 
ROPERTIES: Prescribed, Air: ν = 16.3 × 10-6 m2/s, k = 0.022 W/m⋅K, Pr = 0.72. P
 
ANALYSIS: The average heat flux that must be maintained over the surface of the air foil is 
( sq h T T∞′′ = − ) , where the average convection coefficient may be obtained from the modified 
Reynolds analogy. 
 
 L L2 /3 2 /3f 1/3L L
Nu NuC St Pr Pr
2 Re Pr Re Pr
= = = 
 
Hence, with ( ) 6 2 7LRe VL / 100 m / s 2m /16.3 10 m / s 1.23 10 ,ν −= = × = ×
 
 ( )( )L 1/ 370.0025Nu 1.23 10 0.72 13,7802= × = 
 
 ( )L 2k 0.022 W / m Kh Nu 13,780 152 W / m KL 2m
⋅= = = ⋅ 
 
 ( )2q 152 W / m K 5 23 C 4260 W / m′′ ⎡ ⎤= ⋅ − − ° =⎣ ⎦ 2 < 
 
COMMENTS:If the flow is turbulent over the entire airfoil, the modified Reynolds analogy 
provides a good measure of the relationship between surface friction and heat transfer. The relation 
becomes more approximate with increasing laminar boundary layer development on the surface and 
increasing values of the magnitude of the pressure gradient. 
 
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PROBLEM 6.39 
 
KNOWN: Average frictional shear stress of sτ = 0.0625 N/m2 on upper surface of circuit board with 
ensely packed integrated circuits (ICs) d
 
FIND: Allowable power dissipation from the upper surface of the board if the average surface 
emperature of the ICs must not exceed a rise of 25°C above ambient air temperature. t
 
SCHEMATIC: 
 
 
 
ASSUMPTIONS: (1) Steady-state conditions, (2) The modified Reynolds analogy is applicable, (3) 
Negligible heat transfer from bottom side of the circuit board, and (4) Thermophysical properties 
equired for the analysis evaluated at 300 K, r
 
PROPERTIES: Table A-4, Air (Tf = 300 K, 1 atm): ρ = 1.161 kg/m3, cp = 1007 J/kg⋅K, Pr = 0.707. 
ANALYSIS: The power dissipation from the circuit board can be calculated from the convection rate 
equation assuming an excess temperature (Ts - T∞) = 25°C. 
 (s sq h A T T∞= − ) (1) 
 
The average convection coefficient can be estimated from the Reynolds analogy and the measured 
average frictional shear stress s.τ 
 2 /3 sf f 2 p
C hSt Pr C St
2 VV / 2
τ
ρρ= = c= (2,3,4) 
With V = u∞ and substituting numerical values, find h. 
 2 /3s 2 p
h Pr
V cV
τ
ρρ = 
 
 s p 2 / 3
c
h Pr
V
τ −= 
 
 ( )2 2 / 3 20.0625 N / m 1007 J / kg Kh 0.707 39.7 W / m K
2 m / s
−× ⋅= = ⋅ 
 
Substituting this result into Eq. (1), the allowable power dissipation is 
 < ( )2 2q 39.7 W / m K 0.120 0.120 m 25 K 14.3 W= ⋅ × × × =
COMMENTS: For this analysis using the modified or Chilton-Colburn analogy, we found Cf = 
0.0269 and St = 0.0170. Using the Reynolds analogy, the results are slightly different with 
2h 31.5 W / m K= ⋅ and q = 11.3 W. 
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PROBLEM 6.40 
K NOWN: Evaporation rate of water from a lake. 
FIND: The convection mass transfer coefficient, mh . 
SCHEMATIC: 
 
 
 
ASSUMPTIONS: (1) Equilibrium at water vapor-liquid surface, (2) Isothermal conditions, 
3) Perfect gas behavior of water vapor, (4) Air at standard atmospheric pressure. ( 
PROPERTIES: Table A-6, Saturated water vapor (300K): pA,sat = 0.03531 bar, ρA,sat = 
1/vg = 0.02556 kg/m
3. 
 
ANALYSIS: The convection mass transfer (evaporation) rate equation can be written in the 
orm f 
 ( )Am A,s A,
nh ρ ρ ∞
′′= − 
 
w here 
 A,s A,sat ,ρ ρ= 
 
t he saturation density at the temperature of the water and 
 A, A,satρ φρ∞ = 
 
which follows from the definition of the relative humidity, φ = pA/pA,sat and perfect gas 
behavior. Hence, 
 ( )Am A,sat
nh
1ρ φ
′′= − 
and substituting numerical values, find 
 ( )
2
3
m 3
0.1 kg/m h 1/3600 s/hh 1 
0.02556 kg/m 1 0.3
−⋅ ×= =
−
.55 10 m/s.× < 
COMMENTS: (1) From knowledge of pA,sat, the perfect gas law could be used to obtain the 
saturation density. 
 ( )
A,sat A 3
A,sat -2 3
p 0.03531 bar 18 kg/kmol 0.02548 kg/m .
T 8.314 10 m bar/kmol K 300K
×= = =ℜ × ⋅ ⋅
ρ M 
T his value is within 0.3% of that obtained from Table A-6. 
(2) Note that psychrometric charts could also be used to obtain ρA,sat and ρA,∞. 
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PROBLEM 6.41 
KNOWN: Evaporation rate from pan of water of prescribed diameter. Water temperature. Air 
emperature and relative humidity. t 
FIND: (a) Convection mass transfer coefficient, (b) Evaporation rate for increased relative humidity, 
c) Evaporation rate for increased temperature. ( 
SCHEMATIC: 
 
 
ASSUMPTIONS: (1) Water vapor is saturated at liquid interface and may be approximated as a 
erfect gas. p 
PROPERTIES: Table A-6, Saturated water vapor (Ts = 296K): -1A,sat gvρ = = (49.4 m3/kg)-1 = 
0.0202 kg/m3; (Ts = 320 K): ( ) 1-1 3 3A,sat gv 13.98 m / kg 0.0715 kg/m .ρ −= = = 
 
ANALYSIS: (a) Since evaporation is a convection mass transfer process, the rate equation has the 
form (evap m A,s A,m h A ρ ρ ∞= −� ) and the mass transfer coefficient is 
 ( )( ) ( )( )
5evap
m 22 3
A,s A,
m 1.5 10 kg/sh 0.0179 m/s
 D / 4 /4 0.23 m 0.0202 kg/mπ ρ ρ π
−
∞
×= = =
−
�
 < 
w
 
ith Ts = T∞ = 23°C and φ∞ = 0. 
(b) If the relative humidity of the ambient air is increased to 50%, the ratio of the evaporation rates is 
 
( )
( )
( ) ( )
( )
( )
( )
m A,s s A,sevap A,s
evap m A,s s A,s s
h A T Tm 0.5 T
1 .
m 0 h A T
ρ φ ρφ ρφφ ρ
∞ ∞∞ ∞∞∞
⎡ ⎤−= ⎣ ⎦= ==
�
� Tρ− 
Hence, ( ) 35 5evap 30.0202 kg/mm 0.5 1.5 10 kg/s 1 0.5 0.75 10 kg/s.0.0202 kg/mφ
− −∞
⎡ ⎤= = × − = ×⎢ ⎥⎢ ⎥⎣ ⎦
� 
(c) If the temperature of the ambient air is increased from 23°C to 47°C, with φ∞ = 0 for both cases, 
the ratio of the evaporation rates is 
 
( )
( )
( )
( )
( )
( )
evap s m A,s A,s
evap s m A,s A,s
m T T 47 C h A 47 C 47 C
.
m T T 23 C h A 23 C 23 C
ρ ρ
ρ ρ
∞
∞
= =
= =
= =
D D
D D
�
�
D
D 
Hence, ( ) 35 5evap s 30.0715 kg/mm T T 47 C 1.5 10 kg/s 5.31 10 kg/s0.0202 kg/m− −∞= = = × = ×D� . < 
 
COMMENTS: Note the highly nonlinear dependence of the evaporation rate on the water 
temperature. For a 24°C rise in increases by 350%. s evapT ,m�
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PROBLEM 6.42 
KNOWN: Water temperature and air temperature and relative humidity. Surface recession 
ate. r 
F IND: Mass evaporation rate per unit area. Convection mass transfer coefficient. 
SCHEMATIC: 
 
 
ASSUMPTIONS: (1) Water vapor may be approximated as a perfect gas, (2) No water 
nflow; outflow is only due to evaporation. i 
PROPERTIES: Table A-6, Saturated water: Vapor (305K), 
Liquid (305K), 
-1 3
g gv 0.0336 kg/m ;ρ = =
-1 3
f fv 995 kg/m .ρ = = 
A NALYSIS: Applying conservation of species to a control volume about the water, 
 ( ) ( )
A,out A,st
evap f f f
M M
d dm A V AH A
dt dt dt
ρ ρ ρ
− =
′′− = = =
� �
� dH . 
 
S ubstituting numerical values, find 
 ( ) ( )3 4evap f dHm 995kg/m 10 m/h 1/3600 s/hdtρ −′′ = − = − −�