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Heat Transfer - Engineering

Q1:

A composite flat wall of a furnace is made of two materials 'A' and 'B'. The thermal conductivity of 'A' is twice of that of material 'B', while the thickness of layer of 'A' is half that of B. If the temperature at the two sides of the wall are 400 and 1200°K, then the temperature drop (in °K) across the layer of material 'A' is

A 125

B 133

C 150

D 160

ANS:D - 160

To find the temperature drop across the layer of material 'A', we can use the equation for heat conduction through a composite wall: 2AQ​=L1​k1​ΔT1​​+L2​k2​ΔT2​​ Where:

  • Q is the rate of heat transfer through the wall,
  • A is the cross-sectional area of the wall,
  • 1k1​ and 2k2​ are the thermal conductivities of materials 'A' and 'B' respectively,
  • Δ1ΔT1​ and Δ2ΔT2​ are the temperature differences across materials 'A' and 'B' respectively,
  • 1L1​ and 2L2​ are the thicknesses of materials 'A' and 'B' respectively.
Given that:
  • 1=22k1​=2k2​ (the thermal conductivity of 'A' is twice that of 'B'),
  • 1=122L1​=21​L2​ (the thickness of layer 'A' is half that of 'B'),
  • Δ1=Δ�ΔT1​=ΔT (the temperature drop across material 'A' that we want to find),
  • Δ2=1200−400=800ΔT2​=1200−400=800 °K (the temperature difference across material 'B').
Plugging these values into the equation, we get: 2⋅8002AQ​=21​L2​2k2​ΔT​+L2​k2​⋅800​ =42Δ+8002AQ​=4k2​ΔT+800k2​ Now, as Q/A is the same for both materials: 42Δ+8002=1Δ14k2​ΔT+800k2​=k1​ΔT1​ Since 1=22k1​=2k2​: 42Δ+8002=22Δ14k2​ΔT+800k2​=2k2​ΔT1​ 4Δ+800=2Δ14ΔT+800=2ΔT1​ 800=2Δ1−4Δ800=2ΔT1​−4ΔT 800=2(Δ1−2Δ)800=2(ΔT1​−2ΔT) 400=Δ1−2Δ400=ΔT1​−2ΔT 400=Δ1−Δ1/2400=ΔT1​−ΔT1​/2 400=Δ1/2400=ΔT1​/2 Δ1=800ΔT1​=800 Therefore, the temperature drop across the layer of material 'A' is Δ1=800ΔT1​=800 °K.