Example Problem with Complete Solution

4B-2 : Heat Transfer Through the Wall of a House 4 pts
The walls of a well-insulated home in the U.S. are about 6 in thick and have a thermal conductivity of 0.03 Btu/hr-ft-oR. A comfortable indoor temperature is 70oF and on a cold day the outdoor temperature is -20oF.
The convection heat transfer coefficient on the inside surface of the wall is 1.7 Btu/h-ft2-oR while wind makes the convection heat transfer coefficient on the outside of the wall 5.4 Btu/h-ft2-oR.
Calculate the steady-state heat transfer rate through the wall in Btu/h. Assume radiation heat losses are negligible and the area of the wall is 100 ft2.
 
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.
Diagram:
Given: k = 0.03 Btu/h-ft-R Tin 70 oF
L 0.5 ft hin 1.7 Btu/h-ft2-R
A 100 ft2 hout 5.4 Btu/h-ft2-R
Tout -20 oF
Find: q = ??? Btu/h
Assumptions: 1 - The system operates at steady-state.
2 - Newton's Law of Cooling applies for convection heat transfer on both the inside and outside surfaces of the wall.
3 - The thermal conductivity within the wall is constant. This is a weak assumption, but it lets us approximate dT/dx as ΔT/Δx.
Equations / Data / Solve:
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 simultaneously for the three unknowns.
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
q = 516.1 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 67.0 oF Two -19.0 oF
Verify: The first two assumptions cannot be verified, but we can shed some light on the last assumption.
The wall temperature varies from -19oF on the outside to 67oF on the inside. This seems like a wide range of temperatures. I was not able to find data specific to this temperature range, but at higher temperatures, I found data that indicated the thermal conductivity of wood changed by about 20% over a range of 100C above room temperature. This leads me to believe the answer below may not be accurate to 2 significant digits.
Answers : q = 520 Btu/h