In the previous calculations, we determined hourly and daily heat loss. How do we calculate annual or seasonal heat loss?

Since the temperature outside the house may not remain the same day after day, the heat loss will vary by the day. Thus, to obtain the heat loss for a whole year, we do the following:

- Calculate heat loss for each day.
- Add the heat loss for all days in a year that needed heating.
- Leave the R-value of the wall and the area of the wall the same – they will not change.
- Determine the difference between inside and outside temperature, since it will change for each day.

Recall that the formula for daily heat loss is:

$$\text{HeatLoss}(\frac{BTUs}{h})=\frac{\text{Area}(f{t}^{2})\times \text{TemperatureDifference}(\xb0F)}{\text{R-value}\left(\frac{f{t}^{2}\xb0Fh}{BTUs}\right)}\times \frac{24h}{day}$$

Thus, theoretically, we would need to perform this calculation for every day of the 365-day calendar year.

For example, if the average outside temperature were to be 35°F, 32°F, 28°F, and so on for each day, the heat loss for the whole year or the season can be calculated as follows:

$$\begin{array}{l}\text{HeatLoss=}\underset{{\text{day1heatloss}}}{\underbrace{\frac{480f{t}^{2}\times \left(65-\stackrel{{\text{outsidetempforday1}}}{\stackrel{{\ufe38}}{{35}}}\right)\text{\xb0F}}{13\frac{f{t}^{2}\text{\xb0F h}}{\text{BTU}}}\times 24\frac{h}{day}}}\text{}+\text{}\underset{{\text{day2heatloss}}}{\underbrace{\frac{480f{t}^{2}\times \left(65-\stackrel{{\text{outsidetempforday2}}}{\stackrel{{\ufe38}}{{32}}}\right)\text{\xb0F}}{13\frac{f{t}^{2}\text{\xb0F h}}{\text{BTU}}}\times 24\frac{h}{day}}}\text{}+\text{}\\ \underset{{\text{day3heatloss}}}{\underbrace{\frac{480f{t}^{2}\times \left(65-\stackrel{{\text{outsidetempforday3}}}{\stackrel{{\ufe38}}{{28}}}\right)\text{\xb0F}}{13\frac{f{t}^{2}\text{\xb0F h}}{\text{BTU}}}\times 24\frac{h}{day}}}\text{}+\text{}\mathrm{...}\text{andsoonforallheatingdays}\text{.}\end{array}$$

Since the area (480 ft^{2}), R-value $\frac{f{t}^{2}\text{}\xb0F\text{}BTU}{h}$,and 24 h in a day are common for ALL heating days, we can bring those out and rewrite the equation as:

$$\begin{array}{l}=\frac{480\text{}f{t}^{2\text{}}x}{13\frac{f{t}^{2}{\text{}}^{o}F\text{}h}{BTU}}\times 24\frac{h}{day}\left\{{\left(65-35\right)}^{o}F+{\left(65-32\right)}^{o}F+{\left(65-28\right)}^{o}F\mathrm{...}so\text{}on\text{}for\text{}all\text{}heating\text{}days\right\}\\ \\ =\frac{480\text{}f{t}^{2\text{}}x}{13\frac{f{t}^{2}{\text{}}^{o}F\text{}h}{BTU}}\times 24\frac{h}{day}\left\{{30}^{o}F+{33}^{o}F+{37}^{o}F+so\text{}on\text{}for\text{}all\text{}heating\text{}days\right\}\\ \\ Where\left\{{30}^{o}F+{33}^{o}F+{37}^{o}F+so\text{}on\text{}for\text{}all\text{}heating\text{}days\right\}is\text{}the\text{}sum\text{}of\text{}Heating\text{}Degree\text{}Days\end{array}$$

This equation can be even further simplified. The formula for Annual or Seasonal heat loss can be written in general terms as:

$$Heat\text{}Loss\text{}in\text{}a\text{}Season\text{}=\text{}\frac{Area(f{t}^{2})}{Rvalue\left(\frac{f{t}^{2}{\text{}}^{o}F\text{}h}{BTU}\right)}\times 24\frac{h}{day}\times (HDD\text{}for\text{}the\text{}season)$$

$$Heat\text{}Loss\text{}in\text{}a\text{}Season\text{}=\text{}\frac{Area}{Rvalue}\times 24\times HDD\text{}$$