APPENDIX D:
Heat-Load Calculations for Your House

AS MENTIONED IN CHAPTER 6, heat load is a measure of how many Btus per hour are needed to heat or cool a building. Its two main components are heat loss by transmission, and by air leakage. We’ll look at these components separately, but first we need to define your house in terms of heat loss. Before starting this section, please review the information on U-factor and R-value in Chapter 6.

The House

For simplicity, let’s use a small, six-sided, stud-framed box and call it a house. The box is 16 feet wide, 24 feet long, and 8 feet tall. This gives us a 384-square-foot house with a volume of 3,072 cubic feet. We’ll put our wall studs, along with our floor and ceiling joists two feet on-center and insulate the box with R-11 fiberglass inside the framing cavities. In order to determine the heat loss of the box, we need to first figure out the surface area of all six sides.

We have two walls that are 8 × 16, two that are 8 × 24, and a roof and ceiling that are each 16 × 24. Multiplying these areas out and adding them together gives us a total surface area of 1,408 square feet through which heat is lost through the envelope of the house.

Transmission Heat Loss

The basic formula for heat loss in Btus per hour is:

U-factor × Area × Temperature Difference (delta-T, or ÄT)

Or, in short:

U × A × ∆T

We’ve insulated to R-11 and we know that U-factor is 1 ÷ R, so

1 ÷ R-11 = U-0. 0909

What is the temperature difference? Let’s make it cozy. 72°F on the inside, 20°F on the outside. Our difference, or ÄT is:

72° – 20° = 52°

Putting the equation U × A × ∆T together, gives us:

0.0909 × 1,408 × 52 = 6,655 Btus per hour
required to keep our box house warm.

There is more to our box, though. Remember the stud wall? Every stud represents a “thermal break,” or an uninsulated area that conducts heat better than through the insulated area. A kiln-dried two-by-four actually measures 1. 5" × 3.5", and wood has an R-value of around 1 per inch. So we have an R-3.5 value in all that wood which needs to be taken into account.

Our eight-foot (96-inch) -tall, 2 × 4 studs are two feet on center, and the house has a perimeter of 80 feet. That gives us 40 studs offering a thermal break through the walls.

1.5" × 96" × 40 = 5,760 square inches or 40 square feet of stud area in the walls.

Now add the roof and floor, which have 12 joists each, 16 feet (192 inches) long, or 48 square feet of framing material in the ceiling and floor. That adds up to a total area of 88 square feet in our structure with an R-value of 3.5 or a U-factor of 0.286. We need to subtract that 88 square feet from our total box area and do two separate U × A × ÄT calculations. First find the area:

1,408 – 88 = 1,320 square feet of area insulated to R-11, or U-0. 0909.

Then apply the heat loss formula:

0.0909 × 1,320 × 52 = 6,239 Btus per hour
heat lost through the insulated cavity.

Plus the framing material heat loss:

0.286 × 88 × 52 = 1,309 Btus per hour lost through framing.

Our total heat loss is now up to 7,548 Btus per hour.

How about windows and doors? Let’s add one window on each side plus a door. We’ll use ENERGY STAR-rated windows with a U-factor of 0.35. They are each 15 square feet. Our door is 20 square feet with an R-value of 7 (U-0. 143).

First we’ll subtract the 80 square feet of window and door area from our R-11 box area to get our new box U-factor of:

0.0909 × 1,240 × 52 = 5,861 Btus per hour.

Now for the windows:

We have 60 square feet of ENERGY STAR windows having a thermal value of U-0.

35.

0.35 × 60 × 52 = 1,092 Btus per hour of heat loss through the windows.

Plus the 20-square-foot door:

0.143 × 20 × 52 = 149 Btus per hour.

Our home has a total heat load of:

• R-11 walls and ceiling = 5,861 Btus per hour

• Stud walls, floor and ceiling joists = 1,309 Btus per hour

• Windows = 1,092 Btus per hour

• Door = 149 Btus per hour

Total = 8,411 Btus per hour of heat are lost through the envelope.

An alternate method of performing a heat-load calculation is to add up all the component UA values (U-factors multiplied by the area in square feet) to arrive at the total unit U-factor.

For example:

• The R-11 walls have a U-factor of 0.0909 and a total area of 1,240 square feet. The UA for this wall component is 112. 7.

• The studs and joists total 88 square feet with a U-factor of 0.286, so the UA = 25.1

• Windows occupy 60 square feet with a U-factor of 0.35 for a UA of 21.

• The door’s UA is 2.9

• Total UA for the house is 161.7

Multiplying total UA by the temperature difference gives us a heat load of 8,408 Btus per hour (this is off by three Btus from the previous calculation due to rounding errors). What else can this number do for us? First, it will give us average U-factors and R-values. You could not have done the above calculations with R-values as they cannot be added together for different component areas as UA values can. R-values can only be added together if they are part of the same building component such as putting R-10 rigid insulation on top of R-30 fiberglass in the ceiling, offering a total of R-40. To calculate the average U-factor of an area, the formula is:

U x A ÷ Total Area

or, commonly:

UA/A

For example, our home has a total surface area of 1,408 square feet and a UA of 161.7. This gives the building an average U-factor of:

161.7 ÷ 1,408 = 0. 115

and so an average R-value of:

1 ÷ 0.115 = 8. 7

UA formulas are most useful when adding up different parts of a similar component with different insulation values. For example, if a ceiling is insulated to R-30 in one area and R-38 in another, the UA of each part of the ceiling can be added together to arrive at an average U-factor. If half our example building was R-20 and the other half R-40, you might think that the average R-value is R-30. It is, in fact R-26. 7.

Heat Loss Through Air Leakage

Nothing is airtight. Our example home is fairly well put together though, and our energy auditor measured the air leakage rate with a blower door at .30 air changes per hour. This means that every hour, 30 percent of the heated air in the house exchanges with cold outdoor air due to natural leaks in the house. All this fresh, cold air needs to be heated, and we can predict how much heat will be required by determining the volume of air to be conditioned.

Our box is:

16 × 24 × 8 = 3,072 cubic feet

With an ACH of 0.30, how many cubic feet of air are exchanged every hour?

3,072 cubic feet × 0.30 = 922 cubic feet of heated air lost every hour

Air has a heat capacity of 0.018 Btus per cubic foot per °F. The heat capacity of a material tells us how much energy (in Btus) is required to raise the temperature of one cubic foot of that material 1°F. Air has a heat capacity of 0.018 meaning that it takes 0.018 Btus of heat energy to raise the temperature of one cubic foot of air by 1°F. Water has a heat capacity of 62.4, so it takes 62. 4 Btus to raise the temperature of one cubic foot of water (7.48 gallons, or 62. 4 pounds) 1°F.

To every cubic foot of cold, incoming air, we need to supply 0.018 Btus for every degree of temperature difference. In our case, that works out to be:

0.018 × 922 CF × 52° = 863 Btus per hour

That brings our total heat losses up to 9,274 Btus per hour. This is the total heat load of the house, and if 20° is the coldest temperature we’ll ever see outdoors in this climate, that’s how big our heating system needs to be.

Sizing the Heating Plant

Now that we know the heat load of the house (9,274 Btus per hour) and the region’s design temperature (20°F), we can choose our heating system. This is such a tiny place that a small gas space heater will do the job nicely. If we install a space heater rated at 10,000 Btu output, with an efficiency of 80%, it will consume fuel at the rate of:

10,000 ÷ 0.80 = 12,500 Btus per hour

On this 20° day the house will require:

9,274 Btus per hour × 24 hours = 222,576 Btus.

To keep the place warm, the heater will run for:

222,576 Btus ÷ 10,000 Btus = 22.25 hours.

If it’s an LP gas heater, it will have consumed:

222,576 Btus ÷ 91,690 Btus per gallon ÷ 0.80 = 3 gallons of fuel.

Finally, if our cabin is in a climate of 5,000 heating degree days, how much heat energy would the building require over the heating season? Here’s where we can use our average UA from above.

Total seasonal heat load is:

UA × HDD = Btus per hour
or
161.7 × 5,000 × 24 (hours per day) = 19.4 million Btus.

Our cabin needs 264 gallons of propane gas over the heating season.

Head-load calculations get more involved as we consider more building details. A wall section may have more studs for window framing, siding, sheathing, insulation, and plasterboard, all affecting the overall thermal value of the wall and the heat load. To determine air leakage, it’s best to have a blower-door test performed, but you can use these defaults as a general guide:

• 0.50 ACH is an average place to start.

• 0.40 ACH if you live in a new home.

• 0.20 ACH if you know it’s an extremely tight home (i.e., built with stress-skin panels or double-stud walls).

• If the house was built before 1960, it might be closer to 0.75 ACH.

• An old, drafty farmhouse might measure 1 or 2 ACH.