Question

The figure shows the cross section of a wall made of three layers. The thicknesses of the layers are L1, L2 0.600 L1, and L3 0.400 L1. The thermal conductivities are k1, k2 0.820 k1, and k3 0.660 k1. The temperatures at the left and right sides of the wall are TH = 30 °C and Tc =-10 °C, respectively. Thermal conduction is steady. (a) What is the temperature difference AT2 across layer 2 (between the left and right sides of the layer)? If k2 were, instead, equal to 1.100 ki, (b) would the rate at which energy is conducted through the wall be greater than, less than, or the same as previously and (c) what would be the value of AT2? ko lk T1 1-9 (a) AT2 c) AT2

Answer 1

Thermal resistance for whole layer, R = R₁+R₂+R₃

R = L₁/k₁ + L₂/k₂ + L₃/k₃

R = L₁/k₁ + (0.600)L₁/(0.820)k₁ + (0.400)L₁/(0.660)k₁

R = [(0.820*0.660)L₁ + (0.660)(0.600)L₁ + (0.820)(0.400)L₁] / (0.820*0.660)k₁

R = 1.2652 L₁ / 0.5412 k₁

R = 2.338 (L₁ / k₁)

ΔT/R = ΔT₂/R₂

ΔT₂ = ΔT R₂/R

ΔT₂ = (Th - Tc)(0.600/0.820)*(L₁/k₁) / 2.338 (L₁ / k₁)

ΔT₂ = (30-(-10)*(0.600/0.82) / 2.338

ΔT₂ = 12.52°C

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b) Since new k₂ > original k₂, the energy conducted is greater than previous

c) If k₂=1.1 k₁

R = L₁/k₁ + L₂/k₂ + L₃/k₃

R = L₁/k₁ + (0.600)L₁/(1.1)k₁ + (0.400)L₁/(0.660)k₁

R = [(1.1*0.660)L₁ + (0.660)(0.600)L₁ + (1.1)(0.400)L₁] / (1.1*0.660)k₁

R = (1.562/0.726) (L₁ / k₁)

R = 2.152 (L₁ / k₁)

ΔT/R = ΔT₂/R₂

ΔT₂ = ΔT R₂/R

ΔT₂ = (Th - Tc)(0.600/1.1)*(L₁/k₁) / 2.152 (L₁ / k₁)

ΔT₂ = (30-(-10)*(0.600/1.1) / 2.152

ΔT₂ = 11.14°C