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30/12/2021 12:55 pm
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Two identical rods are arranged as shown in the figure. D is the midpoint of rod BC. Find the heat transfer rate in W through rod AD. (given,L/KA = 10 kw-1).
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30/12/2021 1:03 pm
Given,
QBD + QAD = QDC
Ka\(\frac{200-t}{\frac{l}{2}}\) + KA\(\frac{25-t}{l}\)= \(\frac{(t -100)KA}{\frac{l}{2}}\)
= 400 - 2t + 125 - t
= 2t - 200
725 = 5t
t = 145°C
QAD = \(\frac{KA(125 - 145)}{l}\)
= \(\Big|\frac{-\frac{20}{L}}{\frac{l}{KA}}\Big|\) = \(\Big|\frac{20}{10}W\Big|\)
= 2W
This post was modified 4 years ago by admin