双语试卷1答案.pdf

Solutions Convective Heat Transfer Final Exam A 卷卷 2003 12 24 PROMLEM 1 ANALYSIS a At the trailing edge 6 1 10Re xu Therefore the flow is turbulent From Table 7 9 3154 1 Pr871Re037 0 uN 1464 By definition L k uNh 14 6 W m2 K Newton s law of cooling TTAhq s yields Ts 368 K b At the trailing edge the local heat transfer coefficient is given by 3154 1 PrRe0296 0 Nu The equation yields h 18 7 W m2 K Therefore the local heat flux TThq s 1270 W m2 At the fluid solid interface the heat flux can also be calculated by 0 y y T kq Hence the constant b is obtained from the equation kbcq Ans b 68 K PROMLEM 2 ANALYSIS a Since 127 4 Re D m D the flow is laminar From Eq 8 55 66 3 DD uNNu fully developed By definition D k uNh D 36 6 W m2 K The mean temperature variation is then computed by Eq 8 43 h Cm Px TT xTT pmis ms exp where DP and x 10 m Substitution yields Tm1 321 K b From energy balance 12mmps TTCmDLqq Therefore Tm2 352 K Equation 8 53 gives 4 36 and h 43 6 W m D Nu 2 K at x 20 m Finally the surface temperature at the exit is calculated by the relation 22msx TThq Ans Ts2 467 K PROMLEM 3 PROMLEM 4 Solution a K TTTT TTTT T cihocohi cihocohi CF 7 135 42 0 57 108 165ln 108165 ln KWCmC pCCC 1043541745 2 Since hhCC tCtCq We know that 143250 3585 10435 h C C and min 4876CKW h max CCC Therefore 47 010435 4876 maxmin CC r C WTTCq Cihi 1048340 35250 4876 minmax WtCq CC 5217505010435 Hence 5 0 max qq It follows from figure that min 75 0CUANTU Or 2 2 19 190 487675 0 mA PROMLEM 4 Solution a Assume fully developed laminar flow in the tube Since the temperature of tube surface is constant we know that 66 3 D Nu Therefore KmW D kNu h D2 32 7 1 0 2 066 3 For the constant surface temperature h cm Px TT xTT pims ms exp Substitute 32 7 1000 21 0 exp 50200 100200 m mskg 0114 0 Therefore 726 1021 014 3 0114 044 Re 4 D m D We can conclude that the assumption laminar flow is reasonable So KmWh 2 32 7 and skgm 0114 0 b At the outlet h cm Px TT xTT pims ms exp We get 32 7 10000114 0 41 0 exp 50200 200 o T T 133 o c 67133200 o T15050200 i T Therefore 103 150 67ln 15067 ln io io lm TT TT T d The total heat transfer rate from the tube