A/C to 7.2 Nuh=hcH/k=q'(H/k)/(Ts-Tb) q'=-kdT/dx T=Ts+CT[1-(2x/H)2] dT/dx=-8CTx/H2
hcH/k=(8CTx/H)/(Ts-Tb) hc=8CTkx/(Ts-Tb) H A/C TO 7.3 mcpdT=hc(Ts-T)pdx m = mass flowrate
cp=heat capacity of water incase of 7.3 H=diameter p= perimeter Tb=bulk temperature =(ti+tf)/2
.5*3600*4180*(dT/(Ts-T))=[8CTkx/H(80-40)]pdx on integrating 7524000ln[(Ts-ti)/(Ts-tf)]=[3.14CTkd/5d](x2/2) L to 0
L=5130.76/(CTk).5
Please solve question 7.3 and it is depends on 7.2 so I upload the two questions...
Please be clear with steps. I am trying to understand the process and concept. Thank you in advance. 8.12 water enters a tube at 27 0 with a llow rate of 450 kg/h. The heat transfer from the tube wall to the fluid is given as q(W/m)-ax, where the coefficient a is 20 W/m2 and x (m) is the axial distance from the tube entrance (a) Beginning with a properly defined differential con trol volume in the tube, derive an...
just now i sent this questions. this is the answer given. however the answer i afraid that he used formula that is not for constant surface temperature and noncircular formula. this is the formula foe the noncircular tube. because the question ask about triangle. my problem is, i cannot answer question 1(b) that ask the heat transfer coefficient, h. please help me. thank you. this pic is a note on constant surface temperature. page 482 ref: HEAT AND MASS TRANSFER:...
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