Question
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?
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?
Answer: Option C
Answer: (c).1745.8 kJ/m² hr
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Answer: (c).1745.8 kJ/m² hr
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