### Example 1

Please watch the following 6:01 presentation about Example Problem #1. A house in Bismark, ND (HDD = 9,000) has 860 ft^{2 }of windows (R = 1), 2,920 ft^{2} of walls (R = 19), and 3,850 ft^{2} of roof (R = 22). Calculate how much heating oil is required to heat this house for the heating season. The furnace efficiency is 80%.

Click here to open a text description of the Annual Heating Cost Example 1.

### Annual Heating Cost Example 1

**A house in Bismark, ND (HDD = 9000) has 860 ft ^{2} of windows (R = 1), 2920 ft^{2} of walls (R = 19), and 3850 ft^{2} of roof (R = 22). Calculate how much heating oil is required to heat this house for the heating season. The furnace efficiency is 80%.**

Ok, this problem is very similar to what we have done in before. Which is calculating the heat loss from various surfaces and adding all of those to get the total heat loss from the house. And to calculate the oil requirement based on the furnace efficiency. The house is located in Bismark, ND and HDD there is given as 9,000 so HDD is given. This HDD remains the same for all different surfaces. And we also know the surfaces of each of these components like windows. Let’s do the calculation first for windows.

Windows area is given as 860 ft and we also know the R-value of this. R-value is given as 1 – this is ft °F h/BTUs. So we can calculate heat loss through windows. Where all we need is the area, 860 ft times HDD which is 9,000 in this case times 24 hours in a day divided by, we get the R-value here. That is 1 ft°F h/BTU. So ft/ ft cancel, °F/°F, 24 hours/24 hours, days and days so the total heat loss through windows is the number that we get here. It happens to be 185,760,000 BTUs.

$$\begin{array}{l}{\text{Windows=860ft}}^{\text{2}}\\ {\text{R-value=1ft}}^{\text{2}}\text{}\xb0\text{Fh/BTUs}\\ \text{HeatLoss=}\frac{{\text{860ft}}^{\text{2}}\text{}\times {\text{9,000ft}}^{\text{2}}\text{}\xb0\text{Fday}\times \text{24hrs/day}}{{\text{1ft}}^{\text{2}}\text{}\xb0\text{Fh/BTUs}}\\ \\ \text{=185,760,000BTUs}\end{array}$$Similarly we can calculate the heat loss - this is through windows. We can calculate heat loss through the walls. Wall area is 2,920 ft^{2} times 9,000 °F days times 24 hours per day divided by R-value of 19 for walls. And we can cancel out the units and make sure that everything makes sense and this turns out to be 33,195,789 BTUs.

Now heat loss through the roof. You can calculate it again separately and the area of the roof happens to be 3850 ft^{2}, and 9,000 degree days times 24 hours over a day divided by, the roof generally has higher R-value, 22 ft^{2} °F h/BTUs, ok? Now let’s cancel out these units and the heat loss is to be 37,800,000 BTUs. So the total heat loss is the sum of all these three and when you add these up you get 256,755,789 BTUs.

Now, the furnace efficiency is given and we are using in this case, heating oil. Heating oil is 130,000 BTUs so how much is required? How much heating oil is required? We need 256,755,789 BTUs. When we buy heating oil, we get 130,000 BTUs per every gallon and although we get 130,000, this is theoretical, the efficiency is given as 0.8 so only 80% will get really available as heat. So, that makes it, actually, the requirement little bit higher, so that is equal to now, if we do this calculation, it will be 2,468.8 gallons or roughly 2,469 gallons of oil is required to heat this place.

$$\frac{\text{256,755,789BTUs}}{\text{130,000BTUs/gal}\times \text{0}\text{.8}}\text{=2,468}\text{.8gallonsor2,469gallons}$$