Question

Air at 20°C blows over a hot plate 50 by 75cm maintained at 250°C.The convection transfer coefficient,h, is 25W/Km² .Assuming the plate is made of carbon steel (1%) 2cm thick and that 300W is lost from the plate surface by radiation, calculate the inside plate temperature?

Answer 1

Thermal conductivity of carbon steel (1%) = 43 W/m-K

Heat lost by convection Qconvection = hAT

A = area available for heat transfer = 0.5 m x 0.75 m = 0.375 m²

.: Qconvection = 25 W/m²-K x 0.375 m² ?x (250 - 20) K = 2156.25 W

Total heat lost = Qconvection + Qradiation = 2156.25 + 300 = 2456.25 W

For steady state operation,

heat gained by conduction = heat lost by convection and radiation

.: Qconduction = Qconvection + Qradiation = 2456.25 W

Heat transfer by conduction in a plate in given by,

Qconduction = kAT / x

x = thickness of the plate = 2 cm = 0.02 m

A = area normal to heat flow = 0.5 x 0.75 = 0.375 m²

.: 2456.25 W = 43 x 0.375 x (Tinner - 250) / (0.02)

.: Tinner = 253.0565 C...........................................Answer