Determine the overall heat-transfer coefficient for the composite wall of Illustrative Problem 11.4 if h on the hot side is 0.9 Btu/(h· $ft .^{2}$ ·°F) and h on the cold side is 1.5 Btu/(h· $ft .^{2}$ ·°F). The temperatures given are to be the respective air temperatures (see Figure 11.7).

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For a plane wall, the areas are all the same, and if we use 1 ft.2 of the wall surface as the reference area,

[latex]\begin{array}{r} ext { brick resistance }=\frac{\Delta x}{k A}=\frac{6 / 12}{0.4 imes 1}=1.25 \ ext { concrete resistance }=\frac{\Delta x}{k A}=\frac{\frac{1}{2} / 12}{0.8 imes 1}=0.052 \ ext { plaster resistance }=\frac{\Delta x}{k A}=\frac{\frac{1}{2} / 12}{0.3 imes 1}=0.139 \ ext { "hot film" resistance }=\frac{1}{h A}=\frac{1}{0.9 imes 1}=1.11 \ ext { "cold film" resistance }=\frac{1}{h A}=\frac{1}{1.5 imes 1}=0.67 \ \ \hline \ ext { total resistance }= 3.22\end{array}[/latex]

The overall conductance (or overall heat-transfer coefficient) U = 1/(overall resistance) = 1/3.22 = 0.31 Btu/h·[latex]ft .^{2}[/latex]. In Illustrative Problem 11.24, the solution is straightforward, because the heat-transfer area is constant for all series resistances.

Question 11.24: Determine the overall heat-transfer coefficient for the comp...

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