The walls of a wellinsulated
home in the U.S. are about 6 in
thick and have a thermal conductivity of 0.03 Btu/fft^{o}R. A comfortable indoor temperature is 70^{o}F and on a cold day the outdoor temperature is 20^{o}F. 

The convection heat transfer
coefficient on the inside
surface of the wall is 1.7
Btu/hft^{2}^{o}R while wind makes the convection heat transfer coefficient
on the outside of the
wall 5.4 Btu/hft^{2}^{o}R. 

Calculate the steadystate heat transfer rate through the wall in Btu/h. Assume radiation heat losses are negligible and the area of the wall is 100 ft^{2}. 





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/hft°R 


T_{in} 
70 
^{o}F 



L 
0.5 
ft 


h_{in} 
1.7 
Btu/hft^{2}°R 


A 
100 
ft^{2} 


h_{out} 
5.4 
Btu/hft^{2}°R 







T_{out} 
20 
^{o}F 


Find: 
q = 
??? 
Btu/h 


















Assumptions: 
1
 The system operates at steadystate. 



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, T_{wi} and T_{wo}. 


We must algebraically
solve the equations simultaneously for the three unknowns. 




Solve Eqn 1 for T_{wi} : 



Eqn 4 


Solve Eqn 2 for T_{wo} : 




Eqn 5 













Replace T_{wi} and T_{wo} 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 xdirection, as defined in
the diagram. 













We could now evaluate T_{wi} and T_{wo}, using Eqns 4 and 5,
but it is not required. 





T_{wi} 
67.0 
^{o}F 


T_{wo} 
19.0 
^{o}F 


Verify: 
The first two
assumptions cannot be verified, but we can shed some light on the last
assumption. 


The wall temperature
varies from 19^{o}F on the outside to 67^{o}F 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 100°C above room
temperature. This leads me to believe the answer
below may not be accurate to 2 significant digits. 












Answers : 
q = 
520 
Btu/h 

















