Question 17.1: A furnace wall is composed of three layers, 10 cm of firebri...
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?
The individual material thermal resistances per m² of area are
R_{1}, \text { 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}\right)}=0.0641 \mathrm{~K} / \mathrm{W}R_{2}, \text { kaolin }=\frac{L_{2}}{k_{2} A_{2}}=\frac{0.23}{(0.073)(1)}=3.15 \mathrm{~K} / \mathrm{W}
R_{3}, \text { masonry }=\frac{L_{3}}{k_{3} A_{3}}=\frac{0.05}{(1.0)(1)}=0.05 \mathrm{~K} / \mathrm{W}
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 \left(T_{1}-T_{4}\right) = 1370 – 360 = 1010 K.
Using equation (15-16), the energy transfer rate is
q=\frac{\Delta T}{\sum R_{\mathrm{T}}} (15-16)
q=\frac{T_{1}-T_{4}}{\sum R}=\frac{1010 \mathrm{~K}}{3.26 \mathrm{~K} / \mathrm{W}}=309.8 \mathrm{~W}
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, T_{2}, is given by
T_{1}-T_{2}=q\left(R_{1}\right)=(309.8 \mathrm{~W})(0.0641 \mathrm{~K} / \mathrm{W})=19.9 \mathrm{~K}
giving
T_{2}=1350.1Similarly,
T_{3}-T_{4}=q\left(R_{3}\right)=(309.8 \mathrm{~W})(0.05 \mathrm{~K} / \mathrm{W})=15.5 \mathrm{~K}
giving
T_{3}=375.5 \mathrm{~K}