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 relationK = (1.45 + 0.5 * 10-5 t2) KJ/m hr deg

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1745.8 kJ/m² hr

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nWhere, t is the temperature in degree centigrade. Calculate the heat loss per square meter of the wall surface area?

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1355.3 kJ/m2 hr

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2345.8 kJ/m2 hr

TuteeHUB · Answer

1745.8 kJ/m2 hr

TuteeHUB · Answer

7895.9 kJ/m2 hr

1.	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 relationK = (1.45 + 0.5 * 10-5 t2) KJ/m hr deg
A.	nWhere, t is the temperature in degree centigrade. Calculate the heat loss per square meter of the wall surface area?
B.	1355.3 kJ/m<sup>2</sup> hr
C.	2345.8 kJ/m<sup>2</sup> hr
D.	1745.8 kJ/m<sup>2</sup> hr
E.	7895.9 kJ/m<sup>2</sup> hr
Answer» D. 1745.8 kJ/m<sup>2</sup> hr

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 relationK = (1.45 + 0.5 * 10-5 t2) KJ/m hr deg

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