A 5-m-wide, 4-m-high, and 40-m-long kiln used to cure concrete pipes is made of 20 -cm-thick concrete walls and ceiling \((k=0.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The kiln is maintained at \(40^{\circ} \mathrm{C}\) by injecting hot steam into it. The two ends of the kiln, \(4 \mathrm{~m} \times 5 \mathrm{~m}\) in size, are made of a 3 -mm-thick sheet metal covered with 2 -cm-thick Styrofoam \((k=0.033 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The convection heat transfer coefficients on the inner and the outer surfaces of the kiln are \(3000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Disregarding any heat loss through the floor, determine the rate of heat loss from the kiln when the ambient air is at \(-4^{\circ} \mathrm{C}\).

For this step, we should calculate the surface area of the walls, ceiling, and the ends of the kiln. This is necessary for calculating the heat transfer through each section. Walls: \(2 * (5 \mathrm{m} * 4 \mathrm{m}) = 40 \mathrm{m}^2\) Ceiling: \(5 \mathrm{m} * 40 \mathrm{m} = 200 \mathrm{m}^2\) Kiln Ends: \(2 * (4 \mathrm{m} * 5 \mathrm{m}) = 40 \mathrm{m}^2\)

For each layer's resistance, we will use the formula: \[R = \frac{L}{kA}\], where L is the thickness, k is the thermal conductivity, and A is the surface area. For concrete walls and ceiling (we will calculate this in one step because they have the same thermal conductivity): \[R_{concrete} = \frac{0.2 \mathrm{m}}{(0.9 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}})(240 \mathrm{m}^2)} = 9.26\times 10^{-4} \frac{\mathrm{K} \cdot \mathrm{m}^2 }{\mathrm{W}}\] For sheet metal and Styrofoam ends: \[R_{metal} = \frac{3 \times 10^{-3} \mathrm{m}}{(45 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}})(40 \mathrm{m}^2)} = 1.67\times 10^{-5} \frac{\mathrm{K} \cdot \mathrm{m}^2 }{\mathrm{W}}\] \[R_{Styrofoam} = \frac{0.02 \mathrm{m}}{(0.033 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}})(40 \mathrm{m}^2)} = 1.5152\times 10^{-2} \frac{\mathrm{K} \cdot \mathrm{m}^2 }{\mathrm{W}}\]

Now, we will calculate the convection resistance on both the inner and outer surfaces of the kiln. We use the formula: \[R_{conv} = \frac{1}{hA}\], where h is the convection heat transfer coefficient and A is the surface area. For the inner surface: \[R_{conv\_in} = \frac{1}{(3000 \frac{\mathrm{W}}{\mathrm{m}^2 \cdot \mathrm{K}})(240 \mathrm{m}^2)} = 1.39\times 10^{-4} \frac{\mathrm{K} \cdot \mathrm{m}^2 }{\mathrm{W}}\] For the outer surface: \[R_{conv\_out} = \frac{1}{(25 \frac{\mathrm{W}}{\mathrm{m}^2 \cdot \mathrm{K}})(240 \mathrm{m}^2)} = 1.67\times 10^{-3} \frac{\mathrm{K} \cdot \mathrm{m}^2 }{\mathrm{W}}\]

Now we will add the resistances together: Total resistance: \[R_{total} = R_{conv\_in} + R_{concrete} + R_{conv\_out} + R_{metal} + R_{Styrofoam} = 1.39\times 10^{-4} + 9.26\times 10^{-4} + 1.67\times 10^{-3} + 1.67\times 10^{-5} + 1.5152\times 10^{-2} = 2.6167\times 10^{-2} \frac{\mathrm{K} \cdot \mathrm{m}^2 }{\mathrm{W}}\]

Now that we have the total resistance, we can calculate the rate of heat loss using the formula: \[q = \frac{\Delta T}{R_{total}}\], where q is the rate of heat loss, and \(\Delta T\) is the temperature difference between the inside and outside of the kiln. \[q = \frac{40^{\circ} \mathrm{C} - (-4^{\circ} \mathrm{C})}{2.6167\times 10^{-2} \frac{\mathrm{K} \cdot \mathrm{m}^2 }{\mathrm{W}}} = 1684.22 \mathrm{W}\] Therefore, the rate of heat loss from the kiln is approximately \(1684.22 \mathrm{W}\).