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.
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 48ft2
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 ft2 °F h over BTU.
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 ft2 and we have 8360 degree days times 24 hours divided by a day over R value of 1 ft2 °F h/BTU.
So ft2/ft2, °F/°F will be canceled, hours and hours are canceled, days and days are canceled here.
So, when you do this math, it comes out to be 9,630,720 BTUs. This is in one full season.