Calculating Heat Loss
In buildings, our interest in heat loss is in sizing our HVAC equipment, sizing windows for solar gain, and of course estimating what a building's yearly energy cost will be. Since the exact conditions are complex and changing, we ignore all the subtleties of heat loss and generate approximations using approximate U values and historical weather conditions: for example the typical worst case cold day in any year.
However, keep in mind that the actual energy used for heating a building is very dependent on occupant behavior: what temperature the house is kept at, and whether, for example, they leave bedroom windows open all night in the winter. Energy consultants and energy code officials sometimes seem to treat the heat loss values as gospel, when in fact they are just very good guesses. It would not be surprising to find, for example, that after an energy retrofit, the amount of energy used didn't go down as much as predicted, because the occupants are now more comfortable.
There are numerous software packages that calculate heat loss (including some free ones), some of which can do some very sophisticated modeling. Even with the most sophisticated modeling, actual energy use is dependent enough on user behavior and specific weather conditions, so the actual energy use can be quite different from what is modeled. Since modeling is quite accurate for specific conditions, we can generate best case, worst case, and typical case values to get an idea of how the building will perform.
This page is for those who don't have software, or want to do ballpark modeling.
There are two primary methods of heat loss in building, conduction thru the building envelope (ie the exterior surface, floor, walls, roof, windows, etc) and via air infiltration (or rather warm air escaping the building being replaced by cold outside air). Other methods, like radiant loss from the roof in cold nights, and convection losses due to wind are considered minor factors, even though in some situations they can be significant.
Heat loss thru the envelope
The general heat loss formula is: Q=U*A*ΔT, or in plain words, the heat loss of an area of size A is determined by the U value of the materials and the difference in temperature between inside and out (that is the difference in temperature of the two surfaces, not the two air temperatures, which might not be quite the same. Below is an adjustment for air temperatures.)
To get the heat loss of an entire building, you divide the building into areas that have the same U value, and then add them all up to get the total heat loss. So typically you will end up with four different areas: walls, windows & doors, roof and floor. If one of those areas had parts that have a different U value (for example a wall bump out that is constructed differently), you will end up breaking that into its own category also.
Heat loss thru an
assembly:
Because walls, roofs etc are typically made of
a layers of different materials, and those layers are not identical throughout
the entire assembly (typically, most of an assembly is a cavity filled with
insulation, but other parts are more solid (and less insulating) . In the
case of walls, the assembly contains areas of very different materials (eg
windows and doors).
The R value of a material is either found in a table for the entire material (eg an R11 fiberglass batt which is 3.5" thick), or by using the R value per inch of material (eg R3.5/inch) and multiplying by the actual thickness (R3.5/inch*3.5inches=R12.2).
The R value of multiple layers of material is equal to the sum of the R values of all the materials. The R value of an assembly with different sections is the inverse of the sum of the U values of each section. This is easily seen by example:
The
example wall section at right consists of two different cross sections:
(A) where there are no 2x4 studs, is sheetrock-insulation-plywood-siding, and (B)
the section where there are studs, is sheetrock-2x4-insulation-2x4-plywood-siding.
In this example, the 2x4s are 24" apart, that means every 24" section of wall
consists of 22.5" of assembly A and 1.5" of assembly B.
The R value for section A is: .6 (sheetrock@ R1 per inch) + 33.3 (cellulose@ R3.7 per inch) + .5 (plywood@ R1 per inch) + .5 (siding+air barrier: estimate )=34.9. The R value for section B is .6 ( sheetrock) +3.5(2x4)+7.4 (cellulose) + 3.5(2x4) + .5 (plywood) + .5 (siding) = 16.
To get the R value of the complete wall, we add up the U values of each section
multiplied by what percentage of the overall assembly they represent, and then
take the inverse. For our sample wall, section A is 94% (ie 22.5" out of
24"), and so section B is 6%. The basic formula is: Uwall
= Ua*Pa + Ub+Pb + Uc*Pc
+
.... where Ux is the U value of a section and Px is
the section's percentage of the whole.
For our wall, Uwall = (1/34.9)*.94+(1/16)*.06 = .0307, which is a R
value of about 32.5.
In a actual wall there are significant differences from this simple wall
section, for example there are often double or triple studs in the corners,
there are top and bottom plates, headers of various sizes on windows, fire
blocking, electrical outlets, plumbing, vents etc. An accurate value would
require breaking the wall up into each different component section, while a good
ballpark would be to double (or maybe 1.5x) the effective size of the lesser
insulated area. In any case the effect isn't that large:
Doubling the less insulated area is (1/34.9)*.88+(1/16)*.12=R30.6
while increasing it by 1.5 times is (1/34.9)*.91+(1/16)*.9=R31.5
R value anomalies:
The R value of a material is calculated at steady state (ie a constant
temperature different between inside and out), with no wind, no sun, no dark
cold winter sky, and in fact none of the various real life weather conditions
that a building is regularly exposed to. Luckily these cases represent
only a minority of time.
Air layers: in calculating the heat flow thru an assembly (wall, ceiling etc), the assumption is that the temperature of the inside surface is equal to the inside temperature, and the outside surface is equal to the outside temperature, but this is not necessarily the case, especially in poorly insulated walls. A more accurate calculation accounts for the fact that a thin barrier of air clings to each surface, and that barrier itself has an R value of between .1 and .7, depending on whether the air is moving (eg wind), and whether the surface is vertical or horizontal. Since, in real life, temperature is always change, and no year is exactly like another, we are only ever doing ballpark calculations, and so these factors aren't that relevant.
Radiant effects: the heat loss equations are primarily about conduction only, or more specifically about about the heat transfer to the outside in the case where the mean radiant temperature is the same as air temperature. As this is often not the case, the concept of sol-air temperature5 was created to express what the actual effective temperature is, at least for the given surface (a surface exposed to direct sunlight, especially a highly absorbent one like a dark roof, will have a much higher sol-air temperature than one in the shade, and a surface exposed to the cold night sky, colder than one facing warmer earth. Both will have effective temperatures much different than air temperatures)
Radiant heat transfer effects heat loss in two ways: (1) by making the outside surface colder or warmer than air temperature (2) by changing the amount of radiant heat loss/gain.
The effect of sol-air temperature can be quite significant, but on a yearly basis, in most climates it will apparently not skew the results too much.
Convective effects: A windy day will remove heat faster than a still day, but the effect is probably not relevant compared to the additional loss due to increased air infiltration.
If there is significant temperature stratification in a room (forced air, vaulted ceiling being the classic case, but also with wood stoves), there will be increased heat loss thru the ceiling due to the higher temperature that exists there.
Thermal mass: a high mass wall (for example, adobe, rammed earth, or even concrete) that is exposed to daily sunshine will lose much less heat than the R value would suggest because it not only experiences radiant effects during the day, but stores that heat, and releases it during the night, dramatically reducing the heat loss from interior air. Of course on cloudy days, this effect disappears, and on days when the day is hot and the night is warm, will result in excessively warm interior temperatures.
If you live in a suitable climate (sunny days in the winter, cool nights in the summer), and have a limited budget, thermal mass walls make a very sustainable and functional wall. However, an insulated wall will always perform better, dramatically so in most climates. You can still put mass on the inside if you need it (see passive solar).
Heat loss thru a
slab, basement wall or crawl space:
Calculating the heat loss thru a slab involves two additional difficulties;
first that soil has a high specific heat and second that the R value of
different soils varies considerably. Further complicating the issue is
that the ground under the middle of a slab stays at a nearly constant
temperature all year, while the ground near the edge changes with the weather.
To get around this problem, engineers have calculated approximations for the heat loss thru an entire slab, which is given as a factor F, which is a heat loss amount per linear foot of the perimeter of a slab (whereas R and U are per square foot). So for a slab or basement, rather than calculating the surface area and multiplying by the U value, you calculate the length of the perimeter and multiply by F. These values are given in tables which depend on specific construction (walls versus slab), and amount of insulation (an example, from Seattle's energy code website, for slabs is here.)
Because heat loss thru the perimeter is assumed to dominate the heat loss thru a slab, builders have traditionally only installed insulation under the perimeter of a slab, with a 2-4' width around the perimeter being typical. Green builders have begun to question whether this is sufficient, especially when the slab is heated, or is a solar thermal mass slab, and it is now more common to put insulation under the entire slab, typically 2". Because insulation also raises the temperature of the slab, extra slab insulation will also increase the room's mean radiant temperature. In the future, it is likely that not only will full slab insulation become standard, but much thicker amounts seem likely also.
(The following is speculation) By insulating to a high level (potentially including significant vertical insulation to prevent perimeter losses), it is possible to get a better estimate of heat loss without using F values. Since virtually no one currently does this, you won't find F values for the situation in tables. Calculate downward losses, using the difference in temperature between inside and average ground temperature. Calculate perimeter losses by assuming an average depth to which the losses occur, and use the temperature difference between inside and outdoors.
Finally, here is a quick summary of why calculating heat loss thru a slab is difficult...
The R value of soil is determined by how dense it is (sand versus clay), but more importantly by how much water is in it. With a low moisture content, it is similar to concrete (about R1 per foot). Because soil is often not homogenous, and its moisture content changes, you are stuck using either worst case, or a guesstimate average.
Because soil can store a significant quantity of heat (assuming it is either dry, or its water content isn't moving, both of which can be bad assumptions in some climates), the area under a slab or basement will tend to warm up, forming a bubble of heated soil under the house that is warmest near the slab, and cools further away. The size of the heat bubble affects the effective width of the soil that adds to the insulating value of the slab. It is common practice to assume the heat loss directly down is negligible, and instead calculate loss only thru the edge.
Crawl spaces:
While the ideal crawl space is ventilated so that its at outside temperature, in
reality this never happens. As a result the heat loss to a crawl space is
less than to the outside. Energy codes will typically have an adjustment
factor that you are required to use, or you can estimate by just increasing the
effective R value by some amount.
(Aside: since humidity problems come along with this reduced heat loss, some people recommend that crawl spaces be part of the conditioned space (ie heated), although it seems like in that case, you'd be better just not building them at all, or leaving the underside completely open, as in a home on poles.)
In addition to heat loss thru the envelope via conduction, all buildings leak air. This leakage comes thru places like fan vents, seals around windows and doors, and leaks thru small cracks in the walls, ceiling and floor. The amount of leakage is dependent on the difference in pressure between outside and inside, which is due to three factors:
- Wind: wind raises the pressure on the windward side of the house, and lowers the pressure on the leeward side, resulting in air being both pushed and pulled thru the building.
- Stack effect: when the outdoor air is cooler than the indoor air, the indoor air wants to rise thru the ceiling via convective pressure. The greater the temperature difference, and the taller the building, the greater the pressure. Stack effect is generally smaller than the effect of wind.
- Imbalanced mechanical ventilation: dryer vents, range hoods, and especially downdraft cook tops force air outside without bringing any back in, resulting in lower pressure inside than outside, which in turn, sucks air in from outside.
Because the conditions that cause infiltration vary significantly, infiltration is measured a specific pressure difference via a device called a Blower Door. With this device placed in an exterior doorway, the house is depressurized to be 50 Pascals less than outside (which is much higher than normal weather would create: its the equivalent of a steady 20mph wind blowing at all sides of the building). Once this pressure is achieved, the devices measures the airflow it needed to produce this pressure, which of course is the same as the airflow leaking into the house thru all of its various cracks. Using a formula, this leakage amount can be converted to an average amount during a year, or more importantly an average amount during the heating season. (See this in Home Energy for background on this conversion). Essentially, depending on the climate and site of the building, you take the ACH50/CFM50 numbers and divide by somewhere between 10 and 20.4
There are three ways infiltration is specified: as CFM, as ACH, and as equivalent hole size (often called leakage area), with the latter being interesting as a visualization, but not generally used in other calculations. The actual results of the blower door test are written as CFM50 or ACH50, for example 300CFM50, or 2.3ACH50, to indicate they are the values are at 50 Pascals pressure. Since CFM50 depends on the size of the building, the ACH50 is more useful in comparing one building with another, and thus for stating a buildings relative tightness. Based on the current typical green building standards, an ACH50 of 2 or less is extremely tight (with less than 1 being incredibly tight), while an ACH50 between 2 and 3 or so is tight, and greater than 5 or so is not so tight. By comparison, it is not unusual in really leaky buildings that the blower door can't pump enough air to even get an ACH50 value, and when it can ACH50 values of over 10 are not uncommon.3
Once you've converted the CFM50 rate to a "natural" flow rate that represents typically heating season, you can now calculate the heat loss due to infiltration. To do so, you use the heat capacity of air (which is .018 BTU/ft3-°F), times volume, V (in CFM) of air movement, times the difference in temperature: Q=.018*V*ΔT. This is because the heat loss is all due to how much heat is contained (and carried off) by the air leakage; or alternatively its how much heat is necessary to heat the outside air that has replaced the warm air that leaked out.
For more information on infiltration and ventilation see the ventilation section.
The following is an example house, to show a complete heat loss calculation. This house is 25'x40' (1000SF) on the interior with an 8' ceiling, is built with the double stud wall shown in the example above, double glaze low-E windows (U .3), R5 doors, and has a unheated crawlspace. To simplify things, rather than using F values for floor loss, it is assumed an un-vented crawlspace has an average temperature of around 55°F1. Assume the floor and the ceiling are built with 12" TGIs (or equivalent), and that the insulation is blown in cellulose (R3.7/inch - or equivalent).The following are the east, south, west and north elevations for this hypothetical house (which is simplified to make the calculations easy: its intended to be realistic enough, but it isn't a real house). Assume that the house measured 2ACH50, which corresponds to a .2ACH natural ventilation rate (which in turn is about 27CFM). Because this is quite small, assume another 25CFM mechanical ventilation.
For this example, we assume the house is in a moderate climate, with an average heating season temperate of around 40°F, and typical coldest night is around 20°F. The heat loss at typical temperature will help calculate an approximate how what percentage of the necessary heat can be supplied with passive solar, while the heat loss at the coldest day will help size the backup heating equipment. Local codes specify this typical cold temperature, usually called the design temperature, which for Seattle is 23°F.
| Assembly | Area | U | BTU/F | ΔTtyp | BTU | ΔTcold | BTU |
|---|---|---|---|---|---|---|---|
| Windows | 141 | .3 | 42.3 | 30 | 1269 | 50 | 2115 |
| Doors | 40 | .2 | 8 | 30 | 240 | 50 | 400 |
| Walls | 859 | .03 | 26 | 30 | 780 | 50 | 1300 |
| Floor | 1000 | .02 | 21 | 15 | 315 | 15 | 315 |
| Ceiling | 1000 | .02 | 21 | 30 | 630 | 50 | 1050 |
| Conduction Total | 118.3 | 3234 | 5180 | ||||
| Volume | ACHnat | BTU/F | |||||
| Infiltration | 8000 | .2 | 28.8 | 30 | 864 | 50 | 1440 |
| Totals | 147.1 | 4098 | 6620 |
All of the Btu values are per hour. To get a daily amount, use the average daily temperature to get an hourly loss, then multiply by 24. Because this house is superinsulated, and quite small, these values for heat loss are very low compared to typical heating systems, whose maximum output is more in the 40,000 to 80,000 Btu/hr range. For a more fair comparison, we should size the heat for the most extreme cold day, say 0°F, but even then this house still uses only 10,000 Btu/hr.
An estimate of the yearly heating and cooling energy use can move be determined by using your climate's degree days data (with limitations, see discussion in the units section). To do so, take the calculated loss per degree, multiply by 24 to convert the hourly loss to daily loss, then multiply that by your HDD amount. So, for example if the example house is located in a 4000HDD climate, its seasonal loss will 147.1*24*4000, or about 14millon BTU, assuming the house is kept at 65, and that the assumptions behind the heat loss formula are the typical conditions.
There are ways to calculate your climates HDD value for indoor temperatures of other than 65, for example 68, or 70 but in my experience the resulting HDD values predicts a larger heat loss than actually happens. One issue might be that there are mild days that technically count as degree days (and the number goes up as you raise the indoor temp), but in practice don't result in heat usage because the building never gets cold enough to turn the heat on.2 Unless one wants to do significantly more complex modeling, it is best to use HDD as a ballpark estimate.
Notes
1: a vented crawlspace will presumably have a lower temperature during the heating season. There are more complex methods of calculating this downward loss, but because we are only interested in a ballpark result, this simplification is probably reasonable.
As an example of using F-factors, if instead we assumed the house was slab on grade with full R10 insulation, we would find the F factor from the table of .36 is the slab is unheated and .55 if it is heated. Since the perimeter of the house is 130ft, this gives a loss/°F of between 46.8Btu & 71.5Btu. Since these values are higher than the one in the table, it suggests a higher level of slab insulation is probably in order (and/or vertical perimeter insulation),but unfortunately Seattle's tables doesn't have F values for more highly insulated slabs.
2. my theory is that there is just less heat loss when its above 55°F, because infiltration goes way down (and its accompanying heat loss) and radiant heat loss also goes down. So the heat loss you see for the first 10-15°F temperature drop is not as much as the heat loss you see for the successive temperature drop. If so, this could explain why calculating yearly heat loss with standard 65°F HDD values gives an OK approximation even when the inside temp is closer to an average of 68 or 69°F. But this is just a wild guess...
3. this is based on a conversation with a energy retrofit specialist, who said he regularly found houses with 3ACHnat which depending on what conversion was used, corresponds to an ACH50 of between 30 and 60.
4. Like everything else in heat loss calculation, this is an educated ballpark. It has never been clear to me whether the standard adjustment is really valid for the heating season, or includes too much of the non-heating season. What is clear, is that on the coldest days, the infiltration is going to be more than the adjusted ACH50 value.
5. the definitions of sol-air temperature on the web are confusing at best, and as far as I can tell there is no readily available way to measure such a temperature. I've stated the effect in more straightforward terms. Whether I got the definition exactly right or not, the principle is still correct. When terminology like this generally makes principles clearer, this doesn't appear to be the case with sol-air temperature.