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Steady state heat flow

  1. Aug 18, 2010 #1
    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 [tex]b=0.1[/tex]m, thermal conductivity [tex]\kappa_1=0.5 W m^{-1} K^{-1}[/tex]
    plasterboard thickness [tex]x[/tex]m, thermal conductivity [tex]\kappa_2=0.1 W m^{-1} K^{-1}[/tex]
    Air gap [tex]g=0.02[/tex]m,
    Heat transfer coeff inside [tex]h_{in}=10 W m^{-2} K^{-1}[/tex]
    Heat transfer coeff outside [tex]h_{out}=100 W m^{-2} K^{-1}[/tex]
    Combined heat transfer coeff air gap [tex]h_{c}=10 W m^{-2} K^{-1}[/tex]
    Temp inside [tex]\Theta_1 = 300 K[/tex]
    Temp outside [tex]h_{c}= 270 K[/tex]
    Cross sectional area [tex]A[/tex]

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

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


    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. jcsd
  3. Aug 20, 2010 #2
    Mistake above
    [tex]h_{c}= 270 K[/tex] should read [tex]\Theta_{2}= 270 K[/tex] and both [tex]q=\frac{UA(\Theta_1-\Theta_2)}{b}[/tex] should be [tex]q=UA(\Theta_1-\Theta_2)[/tex]

    Any ideas?
     
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