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Difficulty: HardModes of Heat Transfer (Conduction, Convection, and Radiation)

A composite furnace wall consists of two tightly joined layers of equal cross-sectional area. Layer 1 has a thickness of 0.02 m0.02\text{ m} and thermal conductivity of 200 W⋅m−1⋅K−1200\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}. Layer 2 has a thickness of 0.04 m0.04\text{ m} and thermal conductivity of 100 W⋅m−1⋅K−1100\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}. The outer surface of Layer 1 is maintained at 120∘C120^\circ\text{C} and the outer surface of Layer 2 is maintained at 20∘C20^\circ\text{C}. Under steady-state conditions, what is the rate of heat transfer per unit area through the composite wall?

  1. 200 kW⋅m−2200\text{ kW}\cdot\text{m}^{-2}Answer
  2. B
    250 kW⋅m−2250\text{ kW}\cdot\text{m}^{-2}
  3. C
    500 kW⋅m−2500\text{ kW}\cdot\text{m}^{-2}
  4. D
    100 kW⋅m−2100\text{ kW}\cdot\text{m}^{-2}

Answer

The rate of heat transfer per unit area through the composite furnace wall is 200 kW⋅m−2200\text{ kW}\cdot\text{m}^{-2}.
Under steady-state conduction through composite layers in series, the total thermal resistance per unit area is the sum of the individual thermal resistances: Rtotal=d1k1+d2k2=1.0×10−4+4.0×10−4=5.0×10−4 m2⋅K⋅W−1R_{\text{total}} = \frac{d_1}{k_1} + \frac{d_2}{k_2} = 1.0 \times 10^{-4} + 4.0 \times 10^{-4} = 5.0 \times 10^{-4}\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}. Applying Fourier's law of heat conduction, the heat flux is ΔTRtotal=1005.0×10−4=200,000 W⋅m−2=200 kW⋅m−2\frac{\Delta T}{R_{\text{total}}} = \frac{100}{5.0 \times 10^{-4}} = 200,000\text{ W}\cdot\text{m}^{-2} = 200\text{ kW}\cdot\text{m}^{-2}.

Step-by-Step Solution

1
Calculate the thermal resistance per unit area (RR) for each layer using Ri=dikiR_i = \frac{d_i}{k_i}.
R1=0.02 m200 W⋅m−1⋅K−1=1.0×10−4 m2⋅K⋅W−1R_1 = \frac{0.02\text{ m}}{200\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}} = 1.0 \times 10^{-4}\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1} and R2=0.04 m100 W⋅m−1⋅K−1=4.0×10−4 m2⋅K⋅W−1R_2 = \frac{0.04\text{ m}}{100\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}} = 4.0 \times 10^{-4}\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}.
Thermal resistance quantifies a material's opposition to conductive heat flow per unit cross-sectional area.
2
Find the total thermal resistance per unit area (RtotalR_{\text{total}}) by summing the series resistances.
Rtotal=R1+R2=(1.0×10−4)+(4.0×10−4)=5.0×10−4 m2⋅K⋅W−1R_{\text{total}} = R_1 + R_2 = (1.0 \times 10^{-4}) + (4.0 \times 10^{-4}) = 5.0 \times 10^{-4}\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}.
Layers in series add their thermal resistances directly.
3
Determine the rate of heat flow per unit area Q/tA=ΔTRtotal\frac{Q/t}{A} = \frac{\Delta T}{R_{\text{total}}}.
Q/tA=120∘C−20∘C5.0×10−4 m2⋅K⋅W−1=1005.0×10−4=200,000 W⋅m−2=200 kW⋅m−2\frac{Q/t}{A} = \frac{120^\circ\text{C} - 20^\circ\text{C}}{5.0 \times 10^{-4}\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}} = \frac{100}{5.0 \times 10^{-4}} = 200,000\text{ W}\cdot\text{m}^{-2} = 200\text{ kW}\cdot\text{m}^{-2}.
At steady state, the heat flux across the composite structure is driven by the overall temperature gradient divided by total thermal resistance.

Key Concept

Thermal Resistance in Composite Layers (Conduction)
Estimated Time:2m 0s
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Topics

Modes of Heat Transfer (Conduction, Convection, and Radiation)

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