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 $1.25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\( (or \)0.22 \mathrm{Btw} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}$ ) for low-e coated, argon-filled, quadruple-pane windows to \(6.25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (or $1.1 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}$ ) for a single-pane window with aluminum frames. Determine the range for the rate of heat loss through a $1.2-\mathrm{m} \times 1.8-\mathrm{m}\( window of a house that is maintained at \)20^{\circ} \mathrm{C}\( when the outdoor air temperature is \)-8^{\circ} \mathrm{C}$.

Short answer

Answer: The range of heat loss through the given window is between 96 W and 480 W.

Step-by-step solution

Calculate Temperature Difference

To find the temperature difference between the indoors and outdoors, subtract the outdoor temperature from the indoor temperature. Temperature Difference = Indoor Temperature - Outdoor Temperature Temperature Difference = \(20^\circ C - (-8^\circ C) = 28 ^\circ C\)

Calculate Surface Area

Find the surface area of the window by multiplying its dimensions (height x width), given as \(1.2 m\) and \(1.8 m\). Surface Area = Height × Width Surface Area = \(1.2 m \times 1.8 m = 2.16 m^2\)

Calculate Minimum Heat Loss Rate

Now, calculate the minimum heat loss rate by using the lowest U-factor of \(1.25 W/m^2 \cdot K\). Minimum Heat Loss Rate = (U-factor) × (Surface Area) × (Temperature Difference) Minimum Heat Loss Rate = \((1.25 W/m^2 \cdot K) \times (2.16 m^2) \times (28 ^\circ C) = 96 W\)

Calculate Maximum Heat Loss Rate

Next, calculate the maximum heat loss rate by using the highest U-factor of \(6.25 W/m^2 \cdot K\). Maximum Heat Loss Rate = (U-factor) × (Surface Area) × (Temperature Difference) Maximum Heat Loss Rate = \((6.25 W/m^2 \cdot K) \times (2.16 m^2) \times (28 ^\circ C) = 480 W\)

Determine the Range

Finally, present the range of heat loss rates for the given window as a pair of values (minimum, maximum). Range of Heat Loss Rates = (Minimum Heat Loss Rate, Maximum Heat Loss Rate) Range of Heat Loss Rates = (96 W, 480 W) So, the range of heat loss through a \(1.2 m \times 1.8 m\) window of a house maintained at \(20^\circ C\) when the outdoor air temperature is \(-8^\circ C\) is between 96 W and 480 W.