A furnace wall is composed of three layers, 10 cm of firebrick (k = 1.560 W/m·K), followed by 23 cm of kaolin insulating brick (k = 0.073 W/m·K), and finally 5 cm of masonry brick (k = 1.0 W/m·K). The temperature of the inner wall surface is 1370 K and the outer surface is at 360 K. What are the

Question

A furnace wall is composed of three layers, 10 cm of firebrick (k = 1.560 W/m·K), followed by 23 cm f kaolin insulating brick (k = 0.073 W/m·K), and finally 5 cm of masonry brick (k = 1.0 W/m·K). The temperature of the inner wall surface is 1370 K and the outer surface is at 360 K. What are the temperatures at the contacting surfaces?

Accepted Answer

The individual material thermal resistances per m² of area are

[latex]R_{1}, ext { firebrick }=\frac{L_{1}}{k_{1} A_{1}}=\frac{0.10 \mathrm{~m}}{(1.560 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\left(1 \mathrm{~m}^{2} ight)}=0.0641 \mathrm{~K} / \mathrm{W}[/latex]

[latex]R_{2}, ext { kaolin }=\frac{L_{2}}{k_{2} A_{2}}=\frac{0.23}{(0.073)(1)}=3.15 \mathrm{~K} / \mathrm{W}[/latex]

[latex]R_{3}, ext { masonry }=\frac{L_{3}}{k_{3} A_{3}}=\frac{0.05}{(1.0)(1)}=0.05 \mathrm{~K} / \mathrm{W}[/latex]

The total resistance of the composite wall is equal to 0.0641 + 3.15 + 0.05 = 3.26 K/W. The total temperature drop is equal to [latex]\left(T_{1}-T_{4} ight)[/latex] = 1370 - 360 = 1010 K.

Using equation (15-16), the energy transfer rate is

[latex]q=\frac{\Delta T}{\sum R_{\mathrm{T}}}[/latex] (15-16)

[latex]q=\frac{T_{1}-T_{4}}{\sum R}=\frac{1010 \mathrm{~K}}{3.26 \mathrm{~K} / \mathrm{W}}=309.8 \mathrm{~W}[/latex]

As this is a steady-state situation, the energy transfer rate is the same for each part of the transfer path (i.e., through each wall section). The temperature at the firebrick–kaolin interface, [latex]T_{2}[/latex], is given by

[latex]T_{1}-T_{2}=q\left(R_{1} ight)[/latex]

[latex]=(309.8 \mathrm{~W})(0.0641 \mathrm{~K} / \mathrm{W})=19.9 \mathrm{~K}[/latex]

giving

[latex]T_{2}=1350.1[/latex]

Similarly,

[latex]T_{3}-T_{4}=q\left(R_{3} ight)[/latex]

[latex]=(309.8 \mathrm{~W})(0.05 \mathrm{~K} / \mathrm{W})=15.5 \mathrm{~K}[/latex]

giving

[latex]T_{3}=375.5 \mathrm{~K}[/latex]

Question 17.1: A furnace wall is composed of three layers, 10 cm of firebri...

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