# Steady state heat flow

1. Aug 18, 2010

### bobred

1. The problem statement, all variables and given/known data
There is a wall 100 mm thick, find the U-value. Then a plasterboard lining is added of thickness x and a gap between it and the wall of 20 mm

wall thickness $$b=0.1$$m, thermal conductivity $$\kappa_1=0.5 W m^{-1} K^{-1}$$
plasterboard thickness $$x$$m, thermal conductivity $$\kappa_2=0.1 W m^{-1} K^{-1}$$
Air gap $$g=0.02$$m,
Heat transfer coeff inside $$h_{in}=10 W m^{-2} K^{-1}$$
Heat transfer coeff outside $$h_{out}=100 W m^{-2} K^{-1}$$
Combined heat transfer coeff air gap $$h_{c}=10 W m^{-2} K^{-1}$$
Temp inside $$\Theta_1 = 300 K$$
Temp outside $$h_{c}= 270 K$$
Cross sectional area $$A$$

2. Relevant equations
For just the wall
rate of heat transfer $$q=\frac{UA(\Theta_1-\Theta_2)}{b}$$ where
$$U=(\frac{1}{h_{in}}+\frac{b}{\kappa_1}+\frac{1}{h_{out}})^{-1}$$

For wall and plasterboard
rate of heat transfer $$q=\frac{UA(\Theta_1-\Theta_2)}{b}$$ where
$$U=(\frac{1}{h_{in}}+\frac{b}{\kappa_1}+\frac{1}{h_{c}}+\frac{x}{\kappa_2}+\frac{1}{h_{out}})^{-1}$$

3. The attempt at a solution
I can work out the U-value for the wall, but I am asked then to find the U-value for the wall after lining, I can't see how to get this without having x.

Any ideas, Thanks

Last edited: Aug 18, 2010
2. Aug 20, 2010

### bobred

Mistake above
$$h_{c}= 270 K$$ should read $$\Theta_{2}= 270 K$$ and both $$q=\frac{UA(\Theta_1-\Theta_2)}{b}$$ should be $$q=UA(\Theta_1-\Theta_2)$$

Any ideas?