# Example Problem with Complete Solution

 4B-2 : Heat Transfer Through the Wall of a House 3 pts An insulated frame wall of a house has an average thermal conductivity of 0.0318 Btu/h-ft-°R. The wall is 6 in thick and it has an area of 160 ft2. The inside air temperature is 70°F and the heat transfer coefficient for convection between the inside air and the wall is 1.5 Btu/h-ft2-°R. On the outside of the wall, the convection heat transfer coefficient is 6 Btu/h-ft2-°R and the air temperature is -10°F. Ignoring radiation, determine the rate of heat transfer through the wall at steady-state in Btu/h. Read : The key here is to recognize that, at steady state, the convection heat transfer rate into the wall must be equal to the rate at which heat is conducted through the wall and that must be equal to the rate at which heat is removed from the wall by convection on the outside.  We can write 3 eqns in 3 unknowns: Newton's Law of Cooling for the inside and outside surfaces and Fourier's Law of Conduction for heat transfer through the wall.  The three unknowns are the inside and outside wall surface temperatures and the heat transfer rate. Given: k = 0.0318 Btu/h-ft-°R hi 1.5 Btu/h-ft2-°R L 0.5 ft ho 6 Btu/h-ft2-°R A 160 ft2 To -10 °F Ti 70 °F Find: q = ??? Btu/h Assumptions: The system operates at steady-state. Newton's Law of Cooling applies for convection heat transfer on both the inside and outside surfaces of the wall. The thermal conductivity within the wall is constant.  This is a weak assumption, but it lets us approximate dT/dx as ΔT/Δx. Solution: The key here is to recognize that, at steady state, the convection heat transfer rate into the wall must be equal to the rate at which heat is conducted through the wall and that must be equal to the rate at which heat is removed from the wall by convection. Inside convection: Eqn 1 Outside convection: Eqn 2 Conduction through the wall: Eqn 3 Now, we have three equations in 3 unknowns: q, Twi and Two. We must algebraically solve the equations simultaneou Solve Eqn 1 for Twi : Eqn 4 Solve Eqn 2 for Two : Eqn 5 Replace Twi and Two in Eqn 3 using Eqn 4 and Eqn 5: Eqn 6 Manipulate Eqn 6 algebraically to get: Eqn 7 Solve Eqn 7 for q: Eqn 8 Plugging the given values into Eqn 8 yields: Answer: q = 773 Btu/h The positive sign of q indicates that heat flows in the positive x-direction, as defined in the diagram. We could now evaluate Twi and Two, using Eqns 4 and 5, but it is not required. Twi 66.8 oF Two -9.2 oF 