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

A double-glazed window of total surface area 1.5 m21.5\text{ m}^2 consists of two glass panes, each of thickness 4.0 mm4.0\text{ mm} and thermal conductivity 0.80 W⋅m−1⋅K−10.80\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}, separated by a stagnant air gap of thickness 2.0 mm2.0\text{ mm} with thermal conductivity 0.025 W⋅m−1⋅K−10.025\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}. If a steady-state temperature difference of 18.0∘C18.0^\circ\text{C} is maintained across the window's outer boundary surfaces, what is the rate of heat transfer through the window in watts?

Answer: 300 W

Answer

The steady-state rate of heat transfer through the double-glazed window is 300 W300\text{ W}.
Heat conduction through composite layers in series is governed by the total thermal resistance. The thermal resistance per unit area of each layer is r=dkr = \frac{d}{k}. For two 4.0 mm4.0\text{ mm} glass panes (r=0.005 m2⋅K⋅W−1r = 0.005\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1} each) and one 2.0 mm2.0\text{ mm} air gap (r=0.080 m2⋅K⋅W−1r = 0.080\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}), the total unit resistance is rtotal=0.090 m2⋅K⋅W−1r_{\text{total}} = 0.090\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}. Multiplying by the area 1.5 m21.5\text{ m}^2 and temperature difference 18.0∘C18.0^\circ\text{C} gives Qt=1.5×18.00.090=300 W\frac{Q}{t} = \frac{1.5 \times 18.0}{0.090} = 300\text{ W}.

Step-by-Step Solution

1
Convert layer thicknesses to standard units (meters)
dglass=4.0 mm=0.004 md_{\text{glass}} = 4.0\text{ mm} = 0.004\text{ m}, dair=2.0 mm=0.002 md_{\text{air}} = 2.0\text{ mm} = 0.002\text{ m}
SI units are required for calculations using thermal conductivity in W⋅m−1⋅K−1\text{W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}.
2
Calculate thermal resistance per unit area for each layer
rglass=0.0040.80=0.005 m2⋅K⋅W−1r_{\text{glass}} = \frac{0.004}{0.80} = 0.005\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}, rair=0.0020.025=0.080 m2⋅K⋅W−1r_{\text{air}} = \frac{0.002}{0.025} = 0.080\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}
Thermal resistance per unit area is given by r=dkr = \frac{d}{k}.
3
Sum thermal resistances in series to obtain total unit resistance
rtotal=rglass1+rair+rglass2=0.005+0.080+0.005=0.090 m2⋅K⋅W−1r_{\text{total}} = r_{\text{glass1}} + r_{\text{air}} + r_{\text{glass2}} = 0.005 + 0.080 + 0.005 = 0.090\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}
Heat flows sequentially through all three layers in series.
4
Calculate rate of heat flow across total window area
\frac{Q}{t} = \frac{A \cdot \Delta T}{r_{\text{total}}} = \frac{1.5 \cdot 18.0}{0.090} = 300\text{ W}
Rate of thermal conduction through composite series layers is Qt=ΔTRtotal\frac{Q}{t} = \frac{\Delta T}{R_{\text{total}}} where Rtotal=rtotalAR_{\text{total}} = \frac{r_{\text{total}}}{A}.

Key Concept

Series thermal conduction through composite layers and thermal resistance
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Topics

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

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