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ch09


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Location where boundary layer becomes turbulent.
SCHEMATIC:
ASSUMPTIONS: (1) Isothermal, vertical surface in an extensive, quiescent medium, (2) Boundary
layer assumptions valid.
PROPERTIES: Table A-4, Air ( )( )T T T / 2 350K, 1 atm :f s= + =¥ n = 20.92 ´ 10-6 m2/s, k =
0.030 W/m×K, Pr = 0.700.
ANALYSIS: (a) From the similarity solution results, Fig. 9.4 (see above right), the boundary layer
thickness corresponds to a value of h » 5. From Eqs. 9.13 and 9.12,
( ) 1 / 4y x Gr / 4xh -= (1)
( ) ( ) ( )2m 13 2 3 6 2 9 3Gr g T T x / 9.8 130 25 K x / 20.92 10 m / s 6.718 10 xx s 2 350Ksb n
-= - = ´ - ´ = ´¥ (2)
( ) ( )
1 / 439 2y 5 0.25m 6.718 10 0.25 / 4 1.746 10 m 17.5 mm.
- -æ ö» ´ = ´ =ç ÷è ø
 (3) <
(b) From the similarity solution shown above, the maximum velocity occurs at h » 1 with
( )f 0.275.h¢ = From Eq.9.15, find
( ) ( )
6 2 1 / 22 2 20.92 10 m / s 31/2 9u Gr f 6.718 10 0.25 0.275 0.47 m/s.xx 0.25m
n
h
-´ ´ æ ö¢= = ´ ´ =ç ÷è ø
 <
The maximum velocity occurs at a value of h = 1; using Eq. (3), it follows that this corresponds to a
position in the boundary layer given as
( )y 1/5 17.5 mm 3.5 mm.max = = <
(c) From Eq. 9.19, the local heat transfer coefficient at x = 0.25 m is
( ) ( ) ( )
1 / 41/4 39Nu h x/k Gr / 4 g Pr 6.718 10 0.25 / 4 0.586 41.9x x x
æ ö= = = ´ =ç ÷è ø
2h Nu k/x 41.9 0.030 W/m K/0.25 m 5.0 W/m K.x x= = ´ × = × <
The value for g(Pr) is determined from Eq. 9.20 with Pr = 0.700.
(d) According to Eq. 9.23, the boundary layer becomes turbulent at xc given as
( )
1 / 39 9 9Ra Gr Pr 10 x 10 /6.718 10 0.700 0.60 m.x,c x,c c
é ù= » » ´ =ê úë û
<
COMMENTS: Note that b = 1/Tf is a suitable approximation for air.
PROBLEM 9.7
KNOWN: Thin, vertical plates of length 0.15m at 54°C being cooled in a water bath at 20°C.
FIND: Minimum spacing between plates such that no interference will occur between free-
convection boundary layers.
SCHEMATIC:
ASSUMPTIONS: (a) Water in bath is quiescent, (b) Plates are at uniform temperature.
PROPERTIES: Table A-6, Water (Tf = (Ts + T¥)/2 = (54 + 20)°C/2 = 310K): r = 1/vf = 993.05
kg/m
3
, m = 695 ´10
-6
 N×s/m
2
, n = m/r = 6.998 ´ 10
-7
 m
2
/s, Pr = 4.62, b = 361.9 ´ 10
-6
 K
-1
.
ANALYSIS: The minimum separation distance will be twice the thickness of the boundary layer at
the trailing edge where x = 0.15m. Assuming laminar, free convection boundary layer conditions, the
similarity parameter, h, given by Eq. 9.13, is
( )y 1 / 4Gr / 4xxh =
where y is measured normal to the plate (see
Fig. 9.3). According to Fig. 9.4, the boundary
layer thickness occurs at a value h » 5.
It follows then that,
( ) 1 / 4y x Gr / 4bl xh -=
where
( ) 3g T T xsGrx 2
b
n
- ¥=
( ) ( ) ( )232 6 1 7 2 8Gr 9.8 m/s 361.9 10 K 54 20 K 0.15m / 6.998 10 m / s 8.310 10 .x - - -= ´ ´ - ´ ´ = ´ <
Hence, ( ) 1 / 48 3y 5 0.15m 8.310 10 / 4 6.247 10 m 6.3 mmbl - -= ´ ´ = ´ =
and the minimum separation is
d 2 y 2 6.3 mm 12.6 mm.bl= = ´ = <
COMMENTS: According to Eq. 9.23, the critical Grashof number for the onset of turbulent
conditions in the boundary layer is Grx,c Pr » 10
9. For the conditions above, Grx Pr = 8.31 ´ 10
8 ´
4.62 = 3.8 ´ 109. We conclude that the boundary layer is indeed turbulent at x = 0.15m and our
calculation is only an estimate which is likely to be low. Therefore, the plate separation should be
greater than 12.6 mm.
PROBLEM 9.8
KNOWN: Square aluminum plate at 15°C suspended in quiescent air at 40°C.
FIND: Average heat transfer coefficient by two methods \u2013 using results of boundary layer similarity
and results from an empirical correlation.
SCHEMATIC:
ASSUMPTIONS: (1) Uniform plate surface temperature, (2) Quiescent room air, (3) Surface
radiation exchange with surroundings negligible, (4) Perfect gas behavior for air, b = 1/Tf.
PROPERTIES: Table A-4, Air (Tf = (Ts + T¥)/2 = (40 +15)°C/2 = 300K, 1 atm): n = 15.89 ´ 10
-6
m
2
/s, k = 0.0263 W/m×K, a = 22.5 ´ 10
-6
 m
2
/s, Pr = 0.707.
ANALYSIS: Calculate the Rayleigh number to determine the boundary layer flow conditions,
3Ra g T L / L b n a= D
( ) ( ) ( ) ( ) ( )2 6 2 6 2 73Ra 9.8 m/s 1/300K 40 15 C 0.2m / 15.89 10 m / s 22.5 10 m / s 1.827 10L - -= - ° ´ ´ = ´
where b = 1/Tf and DT = T¥ - Ts. Since RaL < 10
9
, the flow is laminar and the similarity solution of
Section 9.4 is applicable. From Eqs. 9.21 and 9.20,
( ) ( )h L 4 1/4LNu Gr / 4 g PrLL k 3= =
( )
1/20.75 Pr
g Pr
1 / 41/20.609 1.221 Pr 1.238 Pr
=
+ +é ùê úë û
and substituting numerical values with GrL = RaL/Pr, find
( ) ( ) ( )
1/41/2 1/2g Pr 0.75 0.707 / 0.609 1.22 0.707 1.238 0.707 0.501= + + ´ =é ùê úë û
1/470.0263 W/m K 4 1.827 10 /0.707 2h 0.501 4.42 W/m K.L 0.20m 3 4
× ´
= ´ ´ = ×
æ öæ ö ç ÷ç ÷ ç ÷è ø è ø
<
The appropriate empirical correlation for estimating hL is given by Eq. 9.27,
( )
1/40.670 Rah LL LNu 0.68L 4 / 9k 9/161 0.492/Pr
= = +
+é ùê úë û
( ) ( ) ( ) 4 / 91/4 9/167h 0.0263 W/m K/0.20m 0.68 0.670 1.827 10 / 1 0.492/0.707L = × + ´ +é ùé ùê úê úë ûê úë û
2h 4.42 W/m K.L = g <
COMMENTS: The agreement of hL calculated by these two methods is excellent. Using the
Churchill-Chu correlation, Eq. 9.26, find h 4.87 W/m K.L = × This relation is not the most accurate
for the laminar regime, but is suitable for both laminar and turbulent regions.
PROBLEM 9.9
KNOWN: Dimensions of vertical rectangular fins. Temperature of fins and quiescent air.
FIND: (a) Optimum fin spacing, (b) Rate of heat transfer from an array of fins at the optimal spacing.
SCHEMATIC:
ASSUMPTIONS: (1) Fins are isothermal, (2) Radiation effects are negligible, (3) Air is quiescent.
PROPERTIES: Table A-4, Air (Tf = 325K, 1 atm): n = 18.41 ´ 10
-6
 m
2
/s, k = 0.0282 W/m×K, Pr =
0.703.
ANALYSIS: (a) If fins are too close, boundary layers on adjoining surfaces will coalesce and heat
transfer will decrease. If fins are too far apart, the surface area becomes too small and heat transfer
decreases. Sop » dx=H. From Fig. 9.4, the edge of boundary layer corresponds to
( ) ( )1 / 4/ H Gr / 4 5.Hh d= »
Hence,
( ) ( ) ( )
( )
33 2g T T H 9.8 m/s 1/325K 50K 0.15ms 7Gr 1.5 10H 2H2 6 218.41 10 m / s
b
n
- ¥= = = ´
-´
( ) ( ) ( )1 / 47H 5 0.15m / 1.5 10 / 4 0.017m 17mm S 34mm.opd = ´ = = » <
(b) The number of fins N can be found as
( )N W/ S t 355/35.5 10op= + = =
and the rate is ( ) ( )q 2 N h H L T T .s= × - ¥
For laminar flow conditions
( )
4 / 99/161/4Nu 0.68 0.67 Ra / 1 0.492/PrH L
é ù= + +ê úë û
( ) ( ) 4 / 91 / 4 9/167Nu 0.68 0.67 1.5 10 0.703 / 1 0.492/0.703 30H é ù= + ´ ´ + =ê úë û
( ) 2h k Nu / H 0.0282 W/m K 30 /0.15 m 5.6 W/m KH= = × = ×
( ) ( ) ( )2q 2 10 5.6 W/m K 0.15m 0.02m 350 300 K 16.8 W.= × ´ - = <
COMMENTS: Part (a) result is a conservative estimate of the optimum spacing. The increase in
area resulting from a further reduction in S would more than compensate for the effect of fluid
entrapment due to boundary layer merger. From a more rigorous treatment (see Section 9.7.1), Sop »
10 mm is obtained for the prescribed conditions.
PROBLEM 9.10
KNOWN: Interior air and wall temperatures; wall height.
FIND: (a) Average heat transfer coefficient when T¥ = 20°C and Ts = 10°C, (b) Average heat
transfer coefficient when T¥ = 27°C and Ts = 37°C.
SCHEMATIC:
ASSUMPTIONS: (a) Wall is at a uniform temperature, (b) Room air is quiescent.
PROPERTIES: Table A-4, Air (Tf = 298K, 1 atm): b = 1/Tf = 3.472 ´ 10
-3
 K
-1
, n = 14.82 ´ 10
-6
m
2
/s, k = 0.0253 W/m×K, a = 20.9 ´ 10
-6
 m
2
/s, Pr = 0.710; (Tf = 305K, 1 atm): b = 1/Tf = 3.279 ´
10
-3
 K
-1
, n = 16.39 ´ 10
-6
 m
2
/s, k = 0.0267 W/m×K, a = 23.2 ´ 10
-6
 m
2
/s, Pr = 0.706.
ANALYSIS: The appropriate correlation