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In a furnace the heat loss through the 150 mm thick refractory wall lining is estimated to be 50 W/m2 . If the average thermal conductivity of the refractory material is 0.05 W/mK, the temperature drop across the wall will be : (a) 140°C (b) 150°C (c) 160°C (d) 170°C |
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Answer» (b) 150°C Data given : \(\cfrac QA\) = 50 W/m2 K = 0.05 W/mK t = 150 mm = 0.15 m Q = \(\cfrac{KA\Delta \theta}t\) or, \(\cfrac QA\) = \(\cfrac{K\Delta \theta}t\) 50 = \(\cfrac{0.05\times\Delta\theta}{0.15}\) \(\Delta\theta\) = 150° |
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