Question

Consider an ordinary house with \(R-13\) walls (walls that have an \(R\)-value of $\left.13 \mathrm{~h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F} / \mathrm{B} t \mathrm{u}\right)\(. Compare this to the \)R$-value of the common double-door windows that are double pane with \(\frac{1}{4}\) in of airspace and have aluminum frames. If the windows occupy only 20 percent of the wall area, determine if more heat is lost through the windows or through the remaining 80 percent of the wall area. Disregard infiltration losses.

Short answer

Heat loss comparison: $$ \frac{0.2A \Delta T}{2} \,\,\, \text{versus} \,\,\, \frac{0.8A \Delta T}{13} $$ Ratio: $$ \frac{\frac{0.2A \Delta T}{2}}{\frac{0.8A \Delta T}{13}} = \frac{13}{4} $$ In this case, more heat is lost through the windows than through the remaining 80 percent of the wall area.

Step-by-step solution

Define the heat loss function and known parameters

The equation for the heat loss from a surface is given by: $$ Q=\frac{A(T_{in} - T_{out})}{R} $$ where: - \(Q\) is the heat loss (in units of \(\mathrm{B}t\mathrm{u}\) per hour); - \(A\) is the surface area (in square feet); - \(T_{in}\) is the inside temperature (in degrees Fahrenheit); - \(T_{out}\) is the outside temperature (in degrees Fahrenheit); - \(R\) is the \(R\)-value of the surface (in units of \(\mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F} / \mathrm{B}t\mathrm{u}\)). We know the walls have an \(R\)-value of \(13 \,\mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F} / \mathrm{B}t \mathrm{u}\), and that windows cover \(20 \%\) of the total wall area. We need to determine the \(R\)-value of windows and compare the heat loss through windows and the rest of the wall.

Determine the R-value of windows

To determine the R-value of windows, you can refer to a table of \(R\)-values for various materials and situations. For our specific case, let's assume you've looked this up and found the \(R\)-value of a typical residential double-pane window with a quarter-inch airspace and aluminum frames to be approximately \(2 \,\mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F} / \mathrm{B}t \mathrm{u}\).

Calculate heat loss through windows

To determine the heat loss through windows, we'll first set the windows' area as \(0.2A\) (since the windows cover \(20\%\) of the wall area). Let \(T_{in} - T_{out}\) be a constant difference in temperature, denoted as \(\Delta T\). Now we can plug this information into the heat loss equation: $$ Q_{windows} = \frac{0.2A \Delta T}{2} $$

Calculate heat loss through remaining wall area

Since windows cover \(20\%\) of the wall area, the remaining wall area is \(80\%\), or \(0.8A\). Utilizing the heat loss equation with the given \(R\)-value of \(13 \,\mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F} / \mathrm{B}t \mathrm{u}\) for the walls, we have: $$ Q_{wall} = \frac{0.8A \Delta T}{13} $$

Compare heat loss through windows and remaining wall area

Now, we can compare the heat loss values for the windows and the remaining wall area. Our goal is to determine whether more heat is lost through the windows or the remaining 80 percent of the wall area. To do this, we will compare \(Q_{windows}\) and \(Q_{wall}\): $$ \frac{0.2A \Delta T}{2} \,\,\, \text{versus} \,\,\, \frac{0.8A \Delta T}{13} $$ Next, we can divide the equations to see if one is greater than the other: $$ \frac{\frac{0.2A \Delta T}{2}}{\frac{0.8A \Delta T}{13}} = \frac{13}{4} $$ Since the ratio is greater than \(1\), it means that more heat is lost through the windows than through the remaining \(80\%\) of the wall area.