Energy Modeling

Historically the only purpose for modeling was to size heating and cooling systems, but now its used to tradeoff insulation amount, window efficiency and air tightness with HVAC/solar array sizes. Modeling also allows your to compare to a standard such as LEED, PassiveHouse, or standard construction via a HERS rating, if you happen to be interested in such comparisons, as well as determine how much PV you'll need if you want to be a zero-energy house. Modeling will also tell you how much passive solar gain you'll get, as well as whether you're getting too much in the summer. You can also get an estimate of what you're yearly energy consumption will be.

Modeling energy use is theoretically simple, but practically complex, particularly over longer time periods. Occupant behavior such as thermostat setting, opening windows, and equipment use have a very large effect on building energy use. Likewise, energy use literally changes with the weather. On top of all this, add in radiant heat/loss gain (see discussion in the R-values section), and that fact that real-world assembly R-values vary somewhat from theoretical ones (see next section).

There are numerous software packages that calculate heat loss (including
some free ones), some of which can do some very sophisticated modeling in an
attempt to deal with the practical complications. Different
programs have different capabilities, so the best one depends on what you
want out of it. Even if you do buy a package, keep in mind that annual
energy use is so dependent on occupant behavior and weather, no model can
possibly predict the end result accurately.^{1} The more
sophisticated packages will take more effects into account and hence get a
better estimate, but you can still get useful estimates with just a simple
model done in a spreadsheet.

If the concern is only ballpark yearly energy use, or only worst case energy use (for example to size a backup heating system), or typical case energy use, a simple spreadsheet will give a good result. The rest of the document describes how to do this, and explains the limitations of this approach.

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 factors, such as radiant loss/gain really
only affect the temperature difference from inside to out. Those factors can be
quite significant for short periods of time, and may even significantly affect
the yearly amount, but are ignored here.^{2}

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 assemblies of different materials, calculating
heat loss thru that assembly requires combining the R-values of the various
materials to calculate an effective R-value for the assembly.

First, divide the assembly into sections that are uniform from inside to outside, for example in a 2x4 wall, there is the part where insulation fills the cavity and the part where there is a 2x4 and no insulation.

Second, calculate the R-value of each section by adding of the R-values of each of its layers. For example, a typical 2x4 wall would be: R.5(wood siding)+R.5( 1/2" wood sheathing)+R11(insulation)+R.5(sheetrock)=R12.5. 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 R3.1/inch) and multiplying by the actual thickness (R3.1/inch*3.5inches=R11).

Third, calculate the U value of the assembly as the sum of the weighted U values of each section. To do this, you will first need to calculate what percentage of the total area each of the different sections occupy.

U_{assembly} = U_{1}*%area_{1}+U_{2}*%area_{2}+...

The R value of the assembly is then just the inverse of its U-value.

Here is an example:

The example wall section at right consists of two different cross sections: (A) where there are no 2x4 studs: it's sheetrock-insulation-plywood-siding, and (B) the section where there are studs: it's 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 )=R34.9.

The R value for section B is: .6 (sheetrock@ R1 per inch ) +3.5(2x4)+7.4(cellulose) + 3.5(2x4) + .5 (plywood) + .5 (siding) = R16.

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: U_{wall}
= U_{a}*P_{a} + U_{b}+P_{b} + U_{c}*P_{c}
+

.... where U_{x} is the U value of a section and P_{x} is
the section's percentage of the whole.

For our wall, U_{wall} = (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 use standard framing factors (which is the percentage of the wall that is solid framing). Standard framing factors are much higher than the sample wall section shown above--the range from 12% to 20%--that is two to three times more than the sample above. While this does reduce the whole wall R-value, the effect isn't terrible:

Doubling the framing to 12% results in: (1/34.9)*.88+(1/16)*.12=R30.6

Increasing the framing to 20% results in : (1/34.9)*.8+(1/16)*.2=R28.2

**Real
world issues**: There are two complexities here: (1) that conditions
the wall is under are different from the assumptions behind this equation
(mostly due to the radiant temperature being different than the air
temperature, see R-values for the full
discussion) and (2) that the insulation material itself maybe degraded are
installed improperly. Both can be big factors, and less-than-perfect
installation is probably more than norm than a rarity. Batt
insulation is notoriously hard to install so it fills the cavity evenly.
Blown in loose fill is easier, but still not trivial, and if not installed
dense enough, it will settle and leave voids at the top of walls. In this case,
the final R-value depends on the fill being installed to the specified
density. With loose fill, the R-value increases with increasing
density up to a point, but goes down after that. It is generally quite
difficult to pass that point, at is requires compressing the insulation.

Left out of this calculation is the effect of the air layers on the inside and outside of the assembly (for a more detailed explanation see "Air Layers"). Typical values are R.7 for the inside layer, and R.2 for the outside. These are averages, with built in assumptions about the conditions the wall experiences. Note that ceiling/roof values are somewhat lower. Because they are relatively small compared to super-insulated assemblies, and because real-world assemblies tend to have more structural material, more pipes, more wires, and less-than-perfect installations of insulation, the air-layer R-value is ignored throughout the sensiblehouse documents.

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
(albeit, somewhat slower than daily temperature changes).

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. Dry sand might be R2 per foot, depending on how dense it is, while dry clay is likely similar to concrete (about R1 per foot), while wet clay might have almost no R-value at all. 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. Another alternative is a fully conditioned basement.)

In addition to heat loss thru the envelope via conduction, all buildings leak air and this mechanism is described in detail in the infiltration section. At issue is how to use these measured CFM50/ACH50 numbers to calculate the hourly heat loss for some typical or maximum actual conditions the building will experience. Ideally we'd like a table or formula that would allow us to know the value of ACH at various temperatures so that we could get a more accurate total heat loss for any outdoor temperature, but because wind speed is typically such an important component of ACH, its really impractical to do this hence we use the estimates.

The issue here is that we know that the infiltration rate is higher when
its cold and windy, so unless the typical weather conditions are that it's
windier when the temperatures are moderate than cold, ACH_{nat} will
result in under estimating the heat loss due to infiltration. If
you're calculating the worst case heat loss, for example for equipment
sizing, rather than using ACH_{nat} in the heat loss formula, you
might want to use ACH50/10 or even ACH50/5, which are both just different
fudge factors than those used to calculate ACH_{nat} - it really
depends on your climate and whether you think ACH_{nat} is a good
estimate of infiltration at cold temperatures or not^{3}.

Once you've determined an infiltration rate, the heat loss is calculated via one of the following simple formulas:

Q=V*ACH_{n}*.018*ΔT

or

Q=CFM_{n}*1.08*ΔT

Where **Q** is the hourly infiltration heat loss,
**V** is the volume of the house, .018 is the heat capacity of
air whose units are BTU/ft^{3}-°F, **
*ΔT** is the difference in temperature and **ACH**_{n}
and **CFM**_{n} are the normalized blower door test
values for whatever conditions you want to assume; the typical assumption
being ACH_{nat}, ie the adjusted valve based on the statistical
model representing the "natural" ventilation rate. Note that
the value 1.08 is just the heat capacity of air, .018 times 60, since CFM is
a per minute rate and we're looking for a per hour rate.

Intuitively this number represents the amount of heat containing the air that leaks out, or more appropriately the amount of heat required to heat up the air that leaks in as a result of air leaking out. There is some evidence that as air leaks out thru an insulated cavity, the cavity acts a bit like an HRV, but given that you don't really want air leaking thru an insulated cavity, especially not at a slow rate where condensation can happen, its best not to assume this happens.

If you have mechanical ventilation, the calculation is essential the same, but in this case the ventilation rate is whatever the fan is rated for. If the ventilation is a HRV or ERV you need to adjust the temperature difference by the efficiency of heat recovery, so for example if ΔT is 50 degrees and the efficiency is 70%, then the effective temperature difference is only 30% of ΔT, or 15 degrees in this case.

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^{4} of around 55°F. 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 approximation of 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 | ΔT_{typ} |
BTU | ΔT_{cold} |
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 | ACH_{nat} |
BTU/F | |||||

Infiltration | 8000 | .2 | 28.8 | 30 | 864 | 50 | 1440 |

Ventilation | .1875 | 27 | 30 | 810 | 50 | 1350 | |

Totals | 174.1 | 4098 | 7970 |

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 super-insulated, 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.

The heat loss model described here is for steady state heat loss under ideal conditions. In the real world, these conditions are at least as uncommon as they are common (detailed discussion in the section on r-values). How much the actual loss will vary from the modeled loss is unclear. There are really only two significant factors: the sunny surfaces of the building will likely have a lower heat loss due to radiant gain, and roof heat loss will be higher when the night sky is clear and dry. Under a cloudy sky at night, the model should be pretty accurate, and on a cloudy day it will also be relatively close (although there will still be some radiant effect), but at all other times real loss will vary from the model.

While the heat transfer on a given day isn't likely to be much different from what is calculated, a small error in the daily amount will result in a significant error in the yearly amount. In addition any given years weather will vary from the average, so the annual heat loss calculation should be viewed as a ballpark estimate. Finally nightime setback (or for that matter any setback) in the thermostat setting may change your heat loss. Still, the yearly heat loss calculation can be used to compare one building to another fairly accurately, and will still provide a decent estimate of annual energy use (although you will have to factor in internal gain and solar gain, see the passive solar section).

To estimate the yearly heating and cooling energy, you calculate the
building's heat loss per degree (ie just **Q=A*U**), multiply
by 24 to convert hourly loss to daily, then multiply by the number of degree
day for your location (for a discussion on degree days and its limitations,
see the units section). So, for example if the
example house from above is located in a 4000HDD climate, its seasonal loss will
174.1*24*4000, or about 16.7millon BTU. The cooling energy is calculated
the same only using the CDD number instead.

The other catch is that if that buildings are usually kept at 68 to 70F,
not 65F, so the actual heat loss is likely more than 16.7million BTU.
HHD numbers for indoor temperatures other than 65F are available, but my
experiments indicates they give too large a result--at least for
Seattle--and I'm assuming its because almost every day in Seattle has an
average temperature below 70F, yet clearly many of them are close enough to
70 (and with enough sunshine) that no external energy is used--yet you can't
weight the solar gain against the heat loss because its likely to be greater
than the heat loss, and the windows are likely to be open to exhaust all the
extra gain.^{5}

Note also that real world buildings are much more complex than the simple model presented in the above example: the R value of a wall varies by how much lumber is actually in it; some wall sections often end up getting built differently than others; pipes, ducts and other voids reduce R-value to less than the nominal value; and most buildings have quite a few more than four wall surfaces. The more accurately all these things are accounted for, the more accurate the result will be.

1: In particular, when upgrading to super-insulation, there is some tendency to keep a house warmer than it would have previously been, so the net savings is sometimes smaller than expected.

2: No source I could find dealt with these factors at all, nor could I find any data to indicate how big they are. The typical response is to just put more insulation in the attic for instance, because the summer roof is often hotter than air temp, and the winter roof is often colder.

3: My take is that it doesn't in many climates, simply because winters often seem windier, but even if that not the case, stack effect is clearly greater when its cold, so taking an average will underestimate heat loss when its cold, and over estimate when its not. But because the formula multiplies by temperature difference, the cold underestimate will be greater than the warm overestimate, leaving a bias. At least that's the thought.

4: 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.

5: I've spent a bunch of time trying to make my model of the Seattle house match the energy use I actually see--unfortunately we have a gas stove, a gas dryer and gas hot water, so there is extra work in separating those out from heating energy. Using HDD (65F), I get a loss of 45mBTU/yr, which is the actual, but I also calculate that I have 10mBTU of internal gain (electrical load) and anywhere from 7-12mBTU of solar gain. When I tried HDD (70) I got a loss of more like 60mBTU, which is too high...but then the house is typically at 68F

Either there is a sizeable error in my model, or HDD just doesn't give a good result. While my model does likely have errors, I'm convinced HDD doesn't give that good a result. See this article http://www.energylens.com/articles/degree-days for a detailed discussion. In particular HDD (65F) underestimated because the building is typically warmer, and any other HDD values are too large because the model assumptions are wrong when the temperature is over 60F outside--the heat is usually off, and an excess night heat loss just results in indoor temperatures going below 68F, then climbing back to it or above during the day.