Click here for a transcript of Annual or Seasonal Heat Loss Problem #2 video.

### Lesson 7a, Screen 33: Calculating annual or Seasonal Heat Loss

#### Example 8 (formerly example 3-12)

*In Fargo, ND, the heating season lasts about 220 days and the average outside temperature is around 27°F. How much heat is lost through an 8 ft by 6 ft window (R=1) during the heating season?*

This problem is very similar to the last one, 3-11. Now this 3-12 involves again heat loss through a season. We need to calculate the HDD’s. The heating days are 220 days. Now the temperature difference is, inside, 65°F as usual and outside, the temperature, the average temperature happens to be 27°F for all 220 days. Therefore, Heating Degree Days (HDD) can be calculated like this: 220 days times (65°F minus 27°F). And this is equal to 220 days times 38°F which is equal to 8,360°F days.

$$\begin{array}{c}220days\\ Inside=65\xb0F,outside=27\xb0F\\ HDD=220\times \left(65\xb0F-29\xb0F\right)\\ =220days\times 38\xb0F=8360\xb0Fdays\end{array}$$
Now, do we know the area? Area is given, actually. The dimensions of the window are 8ft by 6ft. Therefore the area is 8ft times 6ft, which is 48ft^{2}

$$\begin{array}{c}Area\text{}=\text{}8ft\text{}\times \text{}6ft\text{}\\ \text{=48}f{t}^{2}\end{array}$$
We know the temperature difference, we know the HDD, and we can calculate the heat loss provided we have the R value. R-value is also given. 1 ft^{2} °F h over BTU.

$$R\text{}=\text{}\frac{1\text{}f{t}^{2}\text{}\xb0F\text{}h}{BTU}\text{}$$
Now it is easy to calculate heat loss. Heat loss for the season is equal to area (and I am trying to repeat this), area times HDD times 24 over R. So in this case it is, area is 48 ft^{2} and we have 8360 degree days times 24 hours divided by a day over R value of 1 ft^{2} °F h/BTU.

$$\begin{array}{c}Heat\text{}loss\text{}in\text{}a\text{}season\text{}=\frac{Area\text{}\times \text{}HDD\text{}\times \text{}24\text{}hrs/day}{R}\\ =\frac{\text{}48\text{}f{t}^{2}\text{}\times \text{}8360\text{}\xb0F\text{}days\times 24\text{}hrs/day}{1\text{}f{t}^{2}\xb0F\text{}h/BTU}\end{array}$$
So ft^{2}/ft^{2}, °F/°F will be canceled, hours and hours are canceled, days and days are canceled here.

$$\frac{48\text{}f{t}^{2}\times \text{}8360\text{}\xb0F\text{}days\text{}\times \text{}24h/days}{1f{t}^{2}\text{}\xb0F\text{}h/BTU}\text{}=\text{}9,630,720\text{}BTUs\text{}$$
So, when you do this math, it comes out to be 9,630,720 BTUs. This is in one full season.