Answer:
a. 0.122 ft b. -70 Btu/h ft² c. 633.33 °F
Explanation:
a. Since the rate of heat loss dQ/dt = kAΔT/d where k = thermal conductivity, A = area, ΔT = temperature gradient and d = thickness of insulation.
Now [dQ/dt]/A = kΔT/d
Given that [dQ/dt]/A = rate of heat loss per unit area = -900Btu/h ft², k = 0.10 Btu/h ft ℉(for asbestos), ΔT = T₂ - T₁ = 500 °F - 1600 °F = -1100 °F. We need to find the thickness of asbestos, d. So,
d = kΔT/[dQ/dt]/A
d = 0.10 Btu/h ft ℉ × -1100 °F/-900Btu/h ft²
d = 0.122 ft
b. If the 3 in thick Kaolin is added to the outside of the asbestos, and the outside temperature of the asbestos is 250℉, the heat loss due to the Kaolin is thus
[dQ/dt]/A = k'ΔT'/d'
k' = 0.07 Btu/h ft ℉(for Kaolin), ΔT' = T₂ - T₁ = 250 °F - 500 °F = -250 °F and d' = 3 in = 3/12 ft = 0.25 ft
[dQ/dt]/A = 0.07 Btu/h ft ℉ × -250 °F/0.25 ft
[dQ/dt]/A = -70 Btu/h ft²
c. To find the temperature at the interface, the total heat flux equals the individual heat loss from the asbestos and kaolin. So
[dQ/dt]/A = k(T₂ - T₁)/d + k'(T₃ - T₂)/d' where [dQ/dt]/A = -900Btu/h ft², k = 0.10 Btu/h ft ℉(for asbestos), k' = 0.07 Btu/h ft ℉(for Kaolin), T₁ = 1600 °F, T₂ = unknown and T₃ = 250℉.
Substituting these values into the equation, we have
-900Btu/h ft² = 0.10 Btu/h ft ℉(T₂ - 1600 °F)/0.122 ft + 0.07 Btu/h ft ℉(250℉ - T₂)/0.25 ft
-900Btu/h ft² = 0.82 Btu/h ft ℉(T₂ - 1600 °F) + 0.28Btu/h ft ℉(250℉ - T₂)
-900 °F = 0.82(T₂ - 1600 °F) + 0.28(250℉ - T₂)
-900 °F = 0.82T₂ - 1312°F + 70 °F - 0.28T₂
collecting like terms, we have
-900 °F + 1312°F - 70 °F = 0.82T₂ - 0.28T₂
342 °F = 0.54T₂
Dividing both sides by 0.54, we have
T₂ = 342 °F/0.54
T₂ = 633.33 °F
