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A wall has inner and outer surface temperatures of 16 and 6C6^{\circ} \mathrm{C}, respectively. The interior and exterior air temperatures are 20 and 5C5^{\circ} \mathrm{C}, respectively. The inner and outer convection heat transfer coefficients are 5 and 20W/m2K20 \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}, respectively. Calculate the heat flux from the interior air to the wall, from the wall to the exterior air, and from the wall to the interior air. Is the wall under steady-state conditions?

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Known:\textbf{Known:}

  • inner surface of wall temperature Ts1=16T_{s1} = 16^{\circ}C
  • outer surface of wall temperature Ts2=6T_{s2} = 6^{\circ}C
  • interior air temperature T1=20T_{\infty1} = 20^{\circ}C
  • exterior air temperature T2=5T_{\infty2} = 5^{\circ}C
  • inner convection heat transfer coefficient h1=5h_1 = 5 W/m2^2K
  • outer convection heat transfer coefficient h2=20h_2 = 20 W/m2^2K

Use Newton's law of cooling:

q=h(TsT), or q=h(TTs)heat flux from interior air to wall: q=h1(T1Ts1)=5(2016)=20W/m2heat flux from wall to exterior air: q=h1(Ts2T2)=20(65)=20W/m2heat flux from wall to interior air: q=h1(Ts1T1)=5(1620)=20W/m2\begin{align*} q'' = h (T_s - T_\infty) \text{, or } q'' &= h (T_\infty - T_s )\\\\ \text{heat flux from interior air to wall: } q'' &= h_1 (T_{\infty1} - T_{s1} ) = 5 (20 - 16 ) = 20 \text{W/m}^2\\ \text{heat flux from wall to exterior air: } q'' &= h_1 (T_{s2} - T_{\infty2}) = 20 (6 - 5 ) = 20 \text{W/m}^2\\ \text{heat flux from wall to interior air: } q'' &= h_1 (T_{s1} - T_{\infty1}) = 5 (16 - 20 ) = - 20 \text{W/m}^2\\ \end{align*}

Wall is in steady state condition, because heat flux from interior air to wall is equal to heat flux from wall to the exterior air so there is no change in temperature of the wall.

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