Calculating Wall Heat Loss Example Problems

Lesson 7b, Screen 14: Calculation of Wall Heat Loss

Example 15 (formerly Example 3-14)

A house in State College, PA has 580 ft² of windows (R=1), 1920 ft² of walls, and 2750 ft² of roof (R=22). The walls are made up of wood siding (R=0.81), 0.75” plywood, 3” of fiberglass insulation, 1.5” of polyurethane board, and 0.5” gypsum board. Calculate the heating requirement for the house for the heating season.

Here in this problem, we are trying to calculate the heat loss from the entire house. Heat loss through windows, heat loss through roof, heat loss through walls and add them up. We can directly calculate heat loss through windows. Area is given as 580 ft², R-value is given as 1 and HDD for State College is 6000 degree F days (°F days). So heat loss for the season is equal to 580 ft² times 6000 °F days times 24 hours in a day divided by 1 R-value (ft² (°F hour/BTU). So we can cancel these and the heat loss through windows is 83,520,000 BTUs.

Windows

$$\begin{array}{l}Area=580f{t}^2\\ R=1\\ HDD=6000°Fdays\end{array}$$

Similarly, we can calculate heat loss through the roof. And area of the roof is given as 2750 ft² times 6000 °F days times 24 hours over a day divided by 22 ft² °F hour over BTU. Now, when you do this calculation, we lose about 18 million BTUs through the roof.

Heat Loss Season

$$\frac{{\text{580ft}}^{\text{2}}\times \text{6000 }°\text{Fdays}\times \text{24hr/day}}{1f{t}^2\times \frac{°Fdays}{BTU}}$$

Heat Loss Roof

$$\begin{array}{l}=\frac{2750f{t}^2\times 6000°Fdays\times 24hr/days}{22f{t}^2\times \frac{°Fhr}{BTU}}\\ =18,000,000BTUs\end{array}$$

For the wall heat loss, we need to calculate the composite R-value because we are not given the R-value straight away, and we are given the area. So, let’s calculate the composite R-value. The first layer is wood siding, and wood siding is 0.81 in R- value and the next layer is ¾” plywood and whose R-value is 0.94. And we got 3” of fiberglass and each inch provides an R-value of 3.7 therefore 3” would be 11.10, and we have another 1 ½ of polyurethane which has 6.25 per inch R-value therefore 1.5” would provide an R-value of 9.375. At the end or inside, we have ½” drywall whose R-value is 0.45. So together, all these layers would provide an R-value of 22.7.

Wood siding = 0.81
¾ “ plywood = 0.94
3” fiberglass = 11.10
1.5” polyurethane = 9.375
.5” dry wall = 0.45
Total = 22.7

So, heat loss now can be calculated. We know the area that is 1920 ft2 given and same 6000 °F days times 24 hours per day divided by 22.67 and this comes out to be 12,119,164 BTUs.

$$\begin{array}{c}Heatloss=\frac{1920f{t}^2\times 6000°Fdaysx24hr/days}{22.67f{t}^2\frac{°Fhr}{BTU}}\\ =12,119,164BTUs\end{array}$$

So when you add all these three heat losses, the total heat loss is equal to 113,753,164 BTUs per heating season.

$$TotalHeatLoss=113,753,164BTUs$$

Wall Heat Loss #3

Problem

[ON SCREEN TEXT]: A single-story house in Anchorage, AK (HDD = 11,000) is 50 ft by 70 ft with an 8 ft high ceiling. There are six windows (R = 1) of identical size, 4 ft wide by 6 ft high. The roof is insulated to R-30. The walls consist of a layer of wood siding (R = 0.81), 2” of polyurethane board (R = 6.25 per inch), 4” of fiberglass (R = 3.70 per inch), and a layer of drywall (R = 0.45). Calculate the heat loss through the house (not counting the floor) for the season. (A good estimate for the area of the roof is 1.1 times the area of the walls, before you subtract out the area for the windows.)

SARMA PISUPATI: OK, for this problem, 3.15. We do, more or less, the same thing as the last house, except this house is in Anchorage, Alaska, where HDD happens to be 11,000 degree days. OK. Now we have to calculate here, heat loss through windows, through walls, and through roof, to calculate the total heat loss from the house, the same way that we have done before.

Heat loss through windows. We need the area of the windows first. So area of windows equal to-- Each window is 4 feet by 6 feet. So this is 24 square feet. So we have six of them, so the total area is 144 square feet. So heat loss equal to 144 feet square times HDDs 11,000 degree days times 24 hours in a day, and divided by r value, r value happens to be 1. So 1 feet square, degrees Fahrenheit, hour over BTU.

$$\begin{array}{l}HDD=11,000°Fdays\\ WindowsArea=4f{t}^2\times 6f{t}^2\\ 24f{t}^2\times 6=144f{t}^2\\ Heatloss=\frac{144f{t}^2\times 11,000days\times 24hrs/day}{1f{t}^2°F\frac{hrs}{BTU}}\end{array}$$

So you could cancel out some of these. We'll get BTUs heat loss. This is in a year. OK. So we would get this one to be approximately 38 million. 38,016,000 BTUs. This is per season or per year. OK. Now the second thing that we have to do is walls.

Here we have to calculate the area and also the r value. r value because it is a composite wall consisting of four layers. It has wood siding, which has an r value of 0.81 and 2 inches of polyurethane. And 2 times 6.25, so that is 12.5 r value. And we have a third layer, 4 inches of fiberglass. 4 inches times 3.7, that's equal to 14.8.

And then we also have 1/2 inch drywall. And we know that the drywall r value is straight away 0.45, so now the total r value of this wall happens to be 28.56. OK. Now we have to calculate the area of the walls. Area of the walls we can calculate.

Let me draw a little picture here and show you. If we have-- Let's say this is the area, and we have-- and this side is about 70, this side is 50. And we will have under there, 50 this side, and 70 this side. So total times this height is 8 feet. So the total area would be 2 times 120. 120 is 70 plus 50 times 2, because twice these values, times 8 feet is the height. So we get about 1,920 feet square.

Of course, this also has some windows, about six windows as we originally looked at, but at this point, we will take this as 1920 and subtracting windows from this. Windows is 144 feet square, so 1920 minus 144, would give us 1776 feet square.

$$1920f{t}^2-144f{t}^2=1776f{t}^2$$

OK, so now we have to calculate the area of-- Now, if we calculate the area of 1776 feet square times 11,000 times, this is degrees Fahrenheit days, times 24 hours per day, divided by-- we have 28.56. So the heat loss, in here, comes out to be 16,416,806 BTUs per year.

$$\begin{array}{c}\text{Heat loss}=1776f{t}^2\times 11,000°F day\times 26 hrs/day\\ =16,416,806 BTUs/yr\end{array}$$

OK, now we have to calculate roof. Roof happens to be 1.1 times the area of the walls. Area of the walls was 1920 feet square. Again, we have to take this without the windows, actual area of the walls, times 1.1, that happens to be 2112 feet square.

$$\begin{array}{l}Roof=1.1\times 1920f{t}^2\\ =2112f{t}^2\end{array}$$

So heat loss 2112 feet square times 11,000 times 24, divided by-- r value of this is 30-- degrees Fahrenheit, feet square, hour, over BTU. So this comes out to be  18,585,600 BTUs per year.

$$\begin{array}{l}\text{Heat loss}=\frac{2112f{t}^2\times 11,000\times 24}{30°Ff{t}^2\frac{h}{BTU}}\\ =18,585,600BTUs/yr\end{array}$$

So the total heat loss now, would be the addition of all these three together. So that is, first we have, how much? If we look at, we have 38,016,000, so 38,016,000 plus 16,416,806 plus 18,585,600, so that is equal to 73, 019,000 BTUs. So that is the total heat loss from the house.

$$\begin{array}{c}\text{Total heat loss}=38,016,000+16,416,806+18,585,600\\ =73,019,000BTUs\end{array}$$