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Unformatted text preview: PROBLEM 6.4 KNOWN: Distribution of local convection coefficient for obstructed parallel flow over a flat
plate. FIND: Average heat transfer coefficient and ratio of average to local at the trailing edge.
SCHEMATIC: 7;,V hx:0.7+13.6x—5.4x3
@ 3‘7 EOE 3% :6; E
‘<——vL=3m——> ANALYSIS: The average convection coefficient is ill 3—1 Lhd—1 L07136 342d
mejo Xx—EI0(.+ .x~.x)x i L(0.7L+6.8L2 —1.13L3)= 0.7+6.8L—1.13L2 EL: EL=0.7+6.8(3)—1.13(9)=10.9 W/mZ‘K. <
The local coefficient at x = 3m is hL = 0.7 +13.6(3)—3.4(9)= 10.9 W/m2 K.
Hence, EL th =1.0. r. < COMMENTS: The result hL / hL : 1.0 is unique to x : 3m and is a consequence of the
existence of a maximum for hX The maximum occurs at x = 2m, where (ahX /dx) = 0 and (c1211)( 1de < o.) PROBLEM 6.29 KNOWN: Experimental measurements of the heat transfer coefficient for a square bar in
cross flow. FIND: (a) h for the condition when L = lm and v = lSm/S, (b) h for the condition when L
: 1m and V : 30m/s, (c) Effect of defining a side as the characteristic length. SCHEMATIC: V D r——TL=0.5m ResetH5
—i> V1=20m/s 111:50 W/mzK —_D V3=15mls ZZ=40W/mz~K ASSUMPTIONS: (1) Functional form E = CRemPrn applies with C, m, n being
constants, (2) Constant properties. ANALYSIS: (a) For the experiments and the condition L = tin and V = lSrn/s, it follows
that Pr as well as C, m, and n are constants. Hence EL 05 (VL)m, Using the experimental results, ﬁnd m. Substituting values 51L, _ V1L1 m 50x0.5_ 20x05 m
3sz sz2 40x05 15x05 giving m = 0.782. It follows then for L = 1m and V : lSrn/s, in 0.782
Ezhlh— V'L :50 W x0—515X1'0 234.3W/m2K. <
L VlLl m2.K 1.0 20x05
(b) For the condition L = 1m and V = 30m/s, find
m 0.782
hehlﬂ VL :50 W x93 30x”) 259.0W/m2»K. <
L VlLl m2.K 1.0 20x05 (c) If the characteristic length were chosen as a side rather than the diagonal, the value of C
would change. However, the coefficients m and it would not change. 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. PROBLEM 6.35 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 x 4
mm chip located 120m from the leading edge. FIND: Surface temperature of the chip surface, T5. SCHEMATIC:
Appropria‘f'e correlaﬁon: _ . 1/3
Wuzzgoc [‘4mm Nux=004Rex085Pr
W V=10m/s i I Ch_
'P
I Board
I—>x L=120mm ASSUMPTIONS: (l) Steadystate 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
for the chip surface is equivalent to the local value at x = L. PROPERTIES: TableA4, Air (assume TS : 450C, Tf= {45 + 25)/2 = 35°C = 308K, latm): V = ‘ 16.69 x 10'6m2/s, k = 26.9 x 10'3 W/mK, Pr = 0.703. ANALYSIS: From an energy balance on the chip (see above), qCOIN 2Eg :30W. (1)
Newton’s law of cooling for the upper chip surface can be written as
Ts : Too + ciconv I h Achip (2) where Achip 2 [2. Assume that the average heat transfer coefficient over the chip surface is equivalent to the local coefficient evaluated at x = L. That is, hemp x hX (L) where the local
coefﬁcient can be evaluated from the special correlation for this situation, 0.85
Nux : hxx :004 E Pr“3
k v and substituting numerical values with x = L, find 0.85
hX —0.045[£] FYI/3 L v 0.85
0120111 16.69x10' m Is
The surface temperature of the chip is from Eq. (2), T5 = 250c+ 30x10'3 W/107 W/m2 .Kx(0.004m)2 = 425°C. < COMMENTS: (1) Note that the estimated value for Tf used to evaluate the air properties was
reasonable. (2) Alternatively, we could have evaluated hemp by performing the integration of the local value, h(x). ...
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This note was uploaded on 08/08/2008 for the course ME 364 taught by Professor Rothamer during the Fall '08 term at University of Wisconsin.
 Fall '08
 Rothamer
 Heat Transfer

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