Question

ED - EXPERIMENTAL DESIGN RATE LIMITING STEP IN HEAT TRANSFER SYSTEM WITH SERIES RESISTANCES A water stream at a mass flow rate of 0.3 kg/s and temperature of 70°C flows through an aluminum horizontal pipe with an inner diameter of 0.025 m, wall thickness of 3 mm and length of 2 m. The ambient air is at 25 °C and the outlet temperature of water is 65 °C. Assume steady-state conditions are reached and temperature of air is constant. Given data: 47 Re= umD Nup= 1.86 ( Kepe RePr) 1/3 ; Re < 2300 Nup=0.023 Re4/5 Pr0.3 ; Re >10 000 hair = 12 W/m2K kal = 237 W/mK kwater = 0.66 W/mK Cp.water= 4190 J/kg K Hwater= 0.4125.10-3 kg/ms a) Determine the rate limiting step for this system by comparing the thermal resistances. (10 pts) b) Calculate the change in the total resistance when one of the resistances is halved while the others are held constant? Repeat your calculation for each resistance. (10 pts)

Answer 1

From the data of the nusselt number and Reynolds number we can find out the heat transfer coefficient.

Mass flow Rate in a 0.3kg/8 die cha= 10.025m = di call thicknessa 3mm 0.003 m 80 doz dit axt= do= = 0.031m volumetric flow Rate= ela 0.3 kg/6 3.067X104 m3/ 978 kg / 3 3:067x10 m/ 7 (0.025m) 2 80 V velocity (W) - 7² The U= 0.62 m/s SV Di q 998 Kg/m3 x 0.62 m/s X 0.025m 0.41258103 - 37039.7 kg/m-4 so Re> 10ooo so 415 0.3 Pr NUDE 0.023 Re Pr= Cp 4, K 4190 I x 0.412 5X103 Kg/om- Как 0.66 w MK Pr= 2.62 u15 0.3 80 b (0.02 sm) 0.66 W 0.023 (310397) (262) h=3662032 W mk mek

so there are three theemal Resulance Timore QmwW mo To= 25°C Ri= hiti Ro In Polpi Roz ho to Ai= TDIL 2TKL AozTDOL 80 Ri= 1 3662132 W_XT X 0.025mx2m moc Ri= 0.00174 °C W Rea In (0.63/0.025) 0.0000 722 °C - Rc W 2KX 237 W x 2 m то К ho to 1 - Ro= 0.428 °C 12W.XTX0:03) m) X 2m W m2K 80 MM mo Ri= 0.0074 R = 0.0000722 RO=0.428 80 bere the Ro>> Ri Ro » Rc 80 the external cir resistance will control the heat Transfer Rate. • R = Ri t Ret Ro= 0.4298 n Ro So Rate Imiting step is from external wall to dio

Because, let see, 9- AT qate (70-25) "c 0.4298 oc/w 104.7 W Rtotal and let see temperature drop across the outer wall and Air Twal - Taid Tw-25 9 104.7= Ro 0.428°/w Tw= 69.81 80 here the BT= 44.81c) which is highest Rate homitung step is cuter wall to his heat Toarustéo Dolf R Rih Ptotal 0.00174 +0.0000722+0.420 2 if R = 0.4289 so it does not affect overall Ro since he is very low, it does not afect Rol2 if Ro- cah Rt= 0.00174 + 0.0000"722 +0,428 Rot = 0.216 °C/W 'This will decrease the overall Resitance but still Rois Neally equal to Rt so Bate trening step is outer surface Air

Rate controlling step will have highest temperature drop.