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The inner and outer surfaces of a furnace wall, 25 cm thick, are at 300 degree Celsius and 30 degree Celsius. Here thermal conductivity is given by the relation K = (1.45 + 0.5 * 10-5 t2) KJ/m hr deg Where, t is the temperature in degree centigrade. Calculate the heat loss per square meter of the wall surface area?

The inner and outer surfaces of a furnace wall, 25 cm thick, are at 300 degree Celsius and 30 degree Celsius. Here thermal conductivity is given by the relation K = (1.45 + 0.5 * 10-5 t2) KJ/m hr deg Where, t is the temperature in degree centigrade. Calculate the heat loss per square meter of the wall surface area? Correct Answer 1745.8 kJ/m2 hr

Q = -k A d t/d x, Q d x = – k A d t = – (1.45 + 0.5 * 10-5 t2) A d t. Integrating over the wall thickness δ, we get Q = 436.45/0.25 = 1745.8 kJ/m2 hr.

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