A flat heater is sandwiched between two solids (A and B) of equal area (0.1 m^2)

Question

A flat heater is sandwiched between two solids (A and B) of equal area (0.1 m^2) with different thermal conductivities and thicknesses. The heater operates at a uniform temperature and provides a constant power of 290 W. The external surface temperature of each solid is 300 K. Assume that there is perfect thermal contact at each interface.

Data: Solid A: Thermal conductivity: 35 W/mk; Thickness: 60 mm

Solid B: Thermal conductivity: 9 W/mk; Thickness: 30 mm

a. Calculate the heat flux through each solid in J/m^2 s

b. What is the operating temperature of the heater?

Explanation / Answer

Let the Heater operate at Temperature T

heat rate for Solid A=k*A*(temperature difference)/thickness=35*0.1*(T-300)/(0.060)

heat rate for Solid B=k*A*(temperature difference)/thickness=9*0.1*(T-300)/(0.03)

9*0.1*(T-300)/(0.03)+35*0.1*(T-300)/(0.060)=290

solving for T

T=303.28 K

flux = heat rate /area

a.flux through solid A=1913.3 J/m^2.s

flux through solid B=984 J/m^2

(b) operating temperature = 303.28 K

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