Heat Transfer - Engineering
Q1: A wall has two layers of materials A and B; each made of a different material. Both the layers have the same thickness. The thermal conductivity of materialA is twice that of B. Under the equilibrium, the temperature difference across the wall is 36°C. The temperature difference across the layer A is __________ °C.A 6
B 12
C 18
D 24
ANS:B - 12 To solve this problem, we can use the concept of thermal resistance. The thermal resistance (R) of a material is calculated as the thickness (d) divided by the product of the material's thermal conductivity (k) and its area (A): R=k⋅Ad For each layer of the wall, the thermal resistance can be calculated as: RA=kA⋅Ad =RB=kB⋅Ad Given that the thermal conductivity of material A (kA) is twice that of material B (kB), and both layers have the same thickness and area, we can simplify the equation to: =12RA=21RB Now, let's denote the temperature difference across layer A as ΔTA and across layer B as ΔTB. According to the equation for thermal resistance in series: Δtotal=Δ+ΔΔTtotal=ΔTA+ΔTB Given that the total temperature difference across the wall (ΔtotalΔTtotal) is 36°C and =12RA=21RB, we can set up the equation: 36°=Δ+2Δ36°C=ΔTA+2ΔTA 36°=3Δ36°C=3ΔTA Solving for ΔTA: Δ=36°3=12°ΔTA=336°C=12°C So, the temperature difference across layer A is Δ=12°ΔTA=12°C. |
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