assignchap6-4-29-35 - PROBLEM 6.4 KNOWN Distribution of...

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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 Reset-H5 -—i> V1=20m/s 111:50 W/mz-K —_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/m2-K. < L Vl-Ll 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 Vl-Ll 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) 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 for the chip surface is equivalent to the local value at x = L. PROPERTIES: TableA-4, 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/m-K, 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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assignchap6-4-29-35 - PROBLEM 6.4 KNOWN Distribution of...

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