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Zorluk: ZorModes 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}Cevap
  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}

Cevap

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}.

Adım Adım Çözüm

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.

Anahtar Kavram

Thermal Resistance in Composite Layers (Conduction)
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Modes of Heat Transfer (Conduction, Convection, and Radiation)

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