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?
1355.3 kJ/m² hr
2345.8 kJ/m² hr
1745.8 kJ/m² hr
7895.9 kJ/m² hr
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Q. 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 *...
Darshan : September 05, 2023
The inner and outer surfaces of furnace wall, 25 cm thick, are at 300°C and 30°C
respectively. Relation for thermal conductivity given below
k = (1.45+(0.5 x 10-5 t2) KJ/m-hr-deg
t is temperature in centigrade
Calculate heat loss per square meter of the wall surface area
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