Chemical Engineering - Heat Transfer - Discussion
Discussion Forum : Heat Transfer - Section 9 (Q.No. 15)
15.
Three material A, B and C of equal thick-nes and of thermal conductivity of 20, 40 & 60 kcal/hr. m. °C respectively are joined together. The temperature outside of A and C are 30°C and 100°C respectively. The interface between B and C will be at a temperature of __________ °C.
Discussion:
8 comments Page 1 of 1.
Ahmad said:
1 year ago
Qt = dt/Rt.
Qt = (100-30)/(1/20+1/40+1/60).
Qt = 763.6.
Qt = QA = QB = QC = 763.6.
763.6 = T-100/(1/40+1/60).
T = 70.
Qt = (100-30)/(1/20+1/40+1/60).
Qt = 763.6.
Qt = QA = QB = QC = 763.6.
763.6 = T-100/(1/40+1/60).
T = 70.
(3)
Smriti Baroi said:
1 year ago
Q/A = (100-70)/(l/20+l/40+1/60) = 70 * 120/11l
Q/A = (100-T)/60/l.
T = 87.27.
Q/A = (100-T)/60/l.
T = 87.27.
(1)
Ravi Kumar maurya said:
2 years ago
K1.dT = K2 dT = K3dT.
3t'-3t'' = 303.
-2t'+5t'' = 900.
t'' = 300.33kelvin(27.54degree centigrade).
B/w B&C layers dT = 100-27.54.
= 72.46°..
3t'-3t'' = 303.
-2t'+5t'' = 900.
t'' = 300.33kelvin(27.54degree centigrade).
B/w B&C layers dT = 100-27.54.
= 72.46°..
Paaji said:
4 years ago
I too got the same answer. Agree @Philip.
(1)
Yogesh said:
5 years ago
Yes, exactly right @Philip.
Aashi said:
8 years ago
Yes, I agree @Philip.
(1)
Philip said:
8 years ago
Can someone please recheck the answer on this problem. thank you. As for my knowledge.
q1 = q2 = q3 = qtot.
Rtot = summation of individual R.
qtot = (100-30)/Rtot.
I got 87.27deg C. please explain.
q1 = q2 = q3 = qtot.
Rtot = summation of individual R.
qtot = (100-30)/Rtot.
I got 87.27deg C. please explain.
(8)
Dhruv said:
9 years ago
Please Explain it.
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