E5.1 Composite Layer

Example 5.1: The wall of an oven consists of three layers of brick. The inside of the wall is built of 8" of fire brick surrounded by 4" of insulating brick and an outside layer of 6" of building rock. The oven operates at 1600 °F and it is anticipated that the outer side of the wall can be maintained at 125 °F by the circulation of air. Assume an air gap of 0.25 " was left between the insulating brick and the fire brick. How much heat will be lost per square foot of surface and what are the temperatures at the interfaces? Use the following data.

Fire brick, k₁ = 0.68 Btu/(hr·ft·°F); Air, k₂ = 0.0265 Btu/(hr·ft·°F);

Insulating brick, k₃ = 0.15 Btu/(hr·ft·°F); Building rock, k₄ = 0.4 Btu/(hr·ft·°F);

Solution: A = 1 ft²

t₁ = 1600 °F, t₅ = 125 °F

L₁ = 0.667 ft, L₂ = 0.0208 ft, L₃ = 0.333 ft, L₄ = 0.5 ft

The heat transfer resistance through different rectangular sections of the wall is given by Eq. 5.2, R = L/(k·A). The area normal to the heat transfer flow is 1 ft², so thermal resistance can be found to be

R₁ = 0.98 (hr·°F)/Btu; R₂ = 0.788 (hr·°F)/Btu;

R₃ = 2.22 (hr·°F)/Btu; R₄ = 1.25 (hr·°F)/Btu

And the overall resistance, R, is

Now heat flow through this composite wall can be found by dividing the overall driving force by the overall resistance.

This value of Q is the same for whole body and is equal to a ratio of temperature difference to resistance for any segment. We can find temperature difference between any two points provided we know the resistance.

Temperature difference across different walls are found to be:

Δt₁ = 276 °F, Δt₂ = 221 °F, Δt₃ = 626 °F, Δt₄ = 352 °F

And interface temperatures are calculated as

t₂ = 1324.4 °F, t₃ = 1102.4 °F, t₄ = 476.9 °F