Calculating Heat Loss of Windows

Example 1

A house in State College, PA has 380 ft ² of windows (R = 1.1), 2750 ft ² of walls and 1920 ft ² of roof (R = 30). The composite R-Value of the walls is 19. Calculate the heating requirement for the house for the heating season (HDD=6000). What is the percentage of heat that is lost through the windows?

Solution:

Heat loss in a heating season is given by

$$HeatLoss=\frac{Area\times HDD\times 24}{R-value}$$

Heat Loss through windows =

$$\frac{\text{380 }\cancel{{\text{ft}}^{\text{2}}}\times \text{ 6,000 }\bcancel{°\text{F}}\cancel{\text{days}}\times \text{ 24 }\cancel{\text{h}}/\cancel{\text{day}}}{\text{1}\text{.1 }\frac{\cancel{{\text{ft}}^{\text{2}}}\bcancel{°\text{F}}\cancel{\text{h}}}{\text{Btus}}}\text{ = 49,745,455 Btus}$$

Heat loss through walls =

$$\frac{\text{2,750 }\cancel{{\text{ft}}^{\text{2}}}\times \text{ 6,000 }\bcancel{°\text{F}}\cancel{\text{days}}\times \text{ 24 }\cancel{\text{h}}/\cancel{\text{day}}}{\text{19}\frac{\cancel{{\text{ft}}^{\text{2}}}\bcancel{°\text{F}}\cancel{\text{h}}}{\text{Btus}}}\text{ = 20,842,105 Btus}$$

Heat loss through roof =

$$\frac{\text{1,920 }\cancel{{\text{ft}}^{\text{2}}}\times \text{ 6,000 }\bcancel{°\text{F}}\cancel{\text{days}}\times \text{ 24 }\cancel{\text{h}}/\cancel{\text{day}}}{\text{30}\frac{\cancel{{\text{ft}}^{\text{2}}}\bcancel{°\text{F}}\cancel{\text{h}}}{\text{Btus}}}\text{ = 9,216,000 Btus}$$

Total heat loss = 79,803,560 BTUS

Percentage of heat loss through the windows =

$$HeatLoss=\frac{49.74MMBtus}{79.8MMBtus}\times 100=62.3\%$$

Example 2

Windows in the house described in Example 1 are upgraded at a cost of \$1,550. The upgraded windows have an R-value of 4.0.

• What is the percent savings in the energy and the heating bill if the energy cost is 11.15/MMBTUs.
• What is the pay back period for this modification?

Solution:

a) New heat loss for the same window size with the new R-value is

$$\frac{\text{380 }\cancel{{\text{ft}}^{\text{2}}}\times \text{ 6,000 }\bcancel{°\text{F}}\cancel{\text{days}}\times \text{ 24 }\cancel{\text{h}}/\cancel{\text{day}}}{\text{4}\text{.0}\frac{\cancel{{\text{ft}}^{\text{2}}}\bcancel{°\text{F}}\cancel{\text{h}}}{\text{Btus}}}\text{ = 13,680,000 Btus}$$

Annual energy savings = 49.745 MMBTUs -13.680 MMBTUs = 36.06 MMBTUs

The percent savings is $$\frac{36.06MMBTU}{79.84MMBTU}\times 100=45.1\%$$

The old heating bill would be $$79.803MMBtu*\frac{\$11.15}{MMBtu}=\$889.80$$

The new heating bill would be $$43.743MMBtu*\frac{\$11.15}{MMBtu}=\$487.73$$

The monetary savings = \$402.06 per year.

The Pay Back Period =

$$\frac{AdditionalInvestment}{Savingsperyear}=\frac{\$1550.00}{\$402.06}=3.85years$$