The rate of heat loss through a unit surface area of a window per unit temperature difference between the indoors and the outdoors is called the U-factor. The value of the U-factor ranges from about 1.25 W / m ^2 · °C (or 0.22 Btu / h · ft ^2 · °F) for low-e coated, argon-filled, quadruple-pane

Question

The rate of heat loss through a unit surface area of a window per unit temperature difference between the indoors and the outdoors is called the U-factor. The value of the U-factor ranges from about [latex] 1.25 W / m ^{2} \cdot{ }^{\circ} C [/latex] (or [latex] 0.22 Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F [/latex]) for low-e coated, argon-filled, quadruple-pane windows to [latex] 6.25 W / m ^{2} \cdot{ }^{\circ} C [/latex] (or [latex] 1.1 Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F [/latex]) for a singlepane window with aluminum frames. Determine the range for the rate of heat loss through a [latex] 1.2- m imes 1.8- m[/latex] window of a house that is maintained at 20°C when the outdoor air temperature is 8°C.

Accepted Answer

The range of U-factors for windows are given. The range for the rate of heat loss through the window of a house is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat losses associated with the infiltration of air through the cracks/openings are not considered.

Analysis The rate of heat transfer through the window can be determined from

[latex] \dot{Q}_{ ext {window }}=U_{ ext {overall }} A_{ ext {window }}\left(T_{ ext {in }}-T_{ ext {out }} ight) [/latex]

where [latex] T_{ i }[/latex] and [latex] T_{ o }[/latex] are the indoor and outdoor air temperatures, respectively, [latex] U_{ ext {overall }} [/latex] is the U-factor (the overall heat transfer coefficient) of the window, and [latex] A_{ ext {window }}[/latex] is the window area. Substituting,

Maximum heat loss:

[latex] \dot{Q}_{ ext {window, } \max }=\left(6.25 W / m ^{2} \cdot{ }^{\circ} C ight)\left(1.2 imes 1.8 m ^{2} ight)[20-(-8)]{ }^{\circ} C = 3 78 W[/latex]

Minimum heat loss:

[latex] \dot{Q}_{ ext {window, } \min }=\left(1.25 W / m ^{2} \cdot{ }^{\circ} C ight)\left(1.2 imes 1.8 m ^{2} ight)[20-(-8)]^{\circ} C = 7 6 W [/latex]

Discussion Note that the rate of heat loss through windows of identical size may differ by a factor of 5, depending on how the windows are constructed.

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