Calculating Annual Heat Loss Examples

Example 1

Please watch the following 3:29 presentation on Example Problem #1. For the 150-day heating season in Roanoke, VA, the average temperature was 47° F. How much heat is lost through a 176 ft² wall (R=16) during the entire season?

Click here for a transcript of Annual or Seasonal Heat Loss Problem #1 video.

Lesson 7a, Screen 32: Calculating Annual or Seasonal Heat Loss

Example 7 (formerly Example 3-11)

For the 150-day heating season in Roanoke, VA, the average temperature was 47°F.

How much heat is lost through a 176 ft2 wall (R=16) during the entire season?

Ok. In this problem, 3-11, we are trying to calculate the heat loss for a 150-day season in Roanoke, VA. It is a 150-day heating season. The outside temperature is given as 47°F. Inside, is obviously 65°F. Ok?

Outside: 47°F
Inside: 65°F

Degree Days can be calculated like this. Heating Degree Days are equal to the number of days, which is 150 in a season here, times the temperature difference. 65 minus 47°F.

H D D = 150 d a y s \times (65 - 47 ° F)

Now we can calculate that. That is equal to 150 days times 18. So that turns out to be 2700 °F days.

H D D = 150 d a y s \times 18 = 2700 ° F d a y s

Now, what else do we need? We need the area. Area is given. Area is equal to 176 ft^{2, and} we are also given the R-value. R-value is 16.

\begin{array}{l} A r e a = 176 f t^{2} \\ R = 16 \end{array}

Therefore, we can calculate the heat loss, heat loss through a season. Heat loss through the season is equal to Area times HDD times 24 over R.

\begin{array}{l} Heat loss thru season = \frac{A r e a \times H D D \times 24 hours in a day}{R} \\ = 176 f t^{2} \times \frac{2700 ° F d a y s \times 24 h o u r s}{16 f t^{2} ° F h / D a y s} \\ = 712, 800 B t u s \end{array}

Now, what can we cancel? ft²/ ft², °F/°F, hour/hour and days/days are canceled. So when we do this calculation here, we get the answer of 712,800 Btus in this 150 day heating season.

Example 2

Please watch the following 3:42 presentation on Example Problem #2. 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?

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{matrix} 220 d a y s \\ I n s i d e = 65 ° F, o u t s i d e = 27 ° F \\ H D D = 220 \times (65 ° F - 29 ° F) \\ = 220 d a y s \times 38 ° F = 8360 ° F d a y s \end{matrix}

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

\begin{matrix} A r e a = 8 f t \times 6 f t \\ =48 f t^{2} \end{matrix}

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.

R = \frac{1 f t^{2} ° F h}{B T U}

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.

\begin{matrix} H e a t l o s s i n a s e a s o n = \frac{A r e a \times H D D \times 24 h r s / d a y}{R} \\ = \frac{48 f t^{2} \times 8360 ° F d a y s \times 24 h r s / d a y}{1 f t^{2} ° F h / B T U} \end{matrix}

So ft²/ft², °F/°F will be canceled, hours and hours are canceled, days and days are canceled here.

\frac{48 f t^{2} \times 8360 ° F d a y s \times 24 h / d a y s}{1 f t^{2} ° F h / B T U} = 9, 630, 720 B T U s

So, when you do this math, it comes out to be 9,630,720 BTUs. This is in one full season.

Example 1

Lesson 7a, Screen 32: Calculating Annual or Seasonal Heat Loss

Example 7 (formerly Example 3-11)

Example 2

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

Example 8 (formerly example 3-12)

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