Question

A furnace wall is composed of three layers: 15 cm of firebrick [k = 1.5 W/(m*K)], followed by 22.5 cm of kaolin insulation brick [ k = 0.075 W/(m*K)], and finally followed by 4 cm of masonry brick [ k = 1.0 W/(m*K)]. The temperature of the inner wall surface (firebrick) is 1400 K and the outer surface (masonry brick) is at 350 K. Assume each layer has an area normal to the direction of heat flow equal to 1 m^2.

a.) What is the temperature at the firebrick-kaolin interface?

b.) What is the temperature at the kaolin-masonry interface?

c.) Across what material is the temperature drop the highest?

Answer 1

Heat transfer rate through each segment of layers must be same (to maintan the enrgy balance)

Now we know that for three layers of conductivity k1, k2, k3, effective conductivity for series combination si

1/keff = 1/k1 +1/k2 + 1/k3 sismillarly for two layers 1/keff = 1/k1 + 1/k2

We know that Q = kA(DeltaT/distnace) where Q = heat transfer rate, A = area of crosssection =1

T1 = temp at inlet of firebrick

T2 = temp at outlet of firbric or at inlet of kaolin = 1400K

T3 = temp at outlet of kaolin or at inlet of masonary brick

T4 = temp at outlet of masonary brick =350

a) temp at firebrick/kaolin interface = T2 = 1400K

b) applying balace for heat transfer equations or K(Delta T/distance) same and same as effective heat transfer between point 2 & 4 (firebrick-kaolin surface to outer of masonary brick)

1.5(T2-T1)/15 = 0.075 (T3-T2)/22.5 = 1*(T4-T3)/4 = ((1/k2)+(1/k3)^-1*(350-1400)/(4+22.5) = -26.5

thus 1*(350-T3)/4 = -26.6 or T3= 456.4 K

c) temperature drop will be the highest across lowest conductivity/length material and that material is kaolin insulation because it has the lowest conductivity & highest thickness so its conductivity/thickness ratio will be lowest