Question

# A wall has inner and outer surface temperatures of 16 and $6^{\circ} \mathrm{C}$, respectively. The interior and exterior air temperatures are 20 and $5^{\circ} \mathrm{C}$, respectively. The inner and outer convection heat transfer coefficients are 5 and $20 \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}$, 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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$\textbf{Known:}$

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

Use Newton's law of cooling:

\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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