Question

1. Heat transfer analysis of a special-purpose panel is to be performed. The temperatures at the inner and outer surfaces of the panel measure 30°C and 22°C, respectively. The dimension of the panel is 2.9 m wide, 1.3 m high and 0.40 cm thick. Its thermal conductivity is 0.13 W/m·°C. Heat is lost from the outer surface by convection and radiation. The convective heat transfer coefficient is 23 W/m2 ·°C and an ambient temperature measures 13°C. Determine (a) the fraction of heat loss from the panel by radiation and (b) the emissivity of the panel.

Answer 1

The heat transferred due to the conduction due to the temperature difference between the inner surface and the outer surface is travelled from the inner surface to the outer surface. The heat reached from conduction to the outer surface is taken away due to the convection and radiation.

The conductive heat transfer is given by :

The convective heat transfer is given by:

Now the heat balance will be:

Fraction of heat loss by radiation= 199.81/980.2 = 0.20385 = 20.385%

The radiative heat transfer is given by :

Note: As we know that the range of emissivity lies between 0 and 1 but from these values the emissivity is coming out to be greater than 1.