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Most recent update 6-16-2013

Calculators for estimating heating and cooling system capacity requirements, by calculating structure heat losses (heating) and gains (cooling)

Download calculators in Excel .xls format:

Heat-loss calculator for heating systems

Heat-gain calculator for cooling systems

These Excel spreadsheets can be used with any spreadsheet program that opens Excel (.xls) files, such as OpenOffice. Just download them and have fun. The example pages are shown below as .jpg files. These calculators are not copyrighted -- use them freely as you wish.

I'm not an HVAC engineer, just a guy who needed to verify estimates from contractors and who was curious. I designed these calculators in frustration over the lack of available simple calculators that are expandable in complexity. I needed to estimate the heating and cooling loads for my church's worship space. Commercial programs work OK and are expandable, but they're not free. Free programs are OK as far as they go, but they're not expandable.

I made separate sheets for gain and for loss so that I could keep them open together and run the results independently. They can be combined on the same worksheet of course, but I wanted to keep typical runs on a single 8-1/2 x 11 page. Feel free to set them up as you desire.

Constants
1 Watt = 3.41 BTU
1 person = 100 W = 341 BTU
R
(transmission resistance) = 1 / U (transmission conductance) in BTU per hour.
Loss or gain in BTU per hour = (surface area / R) * temp difference, and also = surface area * U * temp difference.

Variables
Infiltration
Infiltration loss or gain = structure volume * air heat transfer factor * air changes per hour * temp difference. See notes on infiltration below.

R-value estimates
2 x 4 stud walls with glass-fiber insulation, exterior siding, and interior 1/2" sheet-rock, are typically rated at R = 11.
2 x 6 stud walls with the same construction as 2 x 4 walls are typically rated at R = 19.
2 x 12 joist ceilings with attic above glass-fiber insulation, and interior 1/2" sheet-rock below, are similarly rated at R = 30. Cathedral ceilings with 2 x 12 joists, glass-fiber insulation, 3/4" decking, 15 lb felt, 3-tab composition roofing, and 1/2" sheet-rock below, are also typically rated for R = 30.
Single-pane windows are typically R = 1, double-pane with 1/2" air space are typically R = 2.
Storm-windows or triple-pane windows help, typically increasing R by about +0.5 to +1.

Look at sources on the web for other construction types. General values will be fine for estimating.

Simplifying assumptions
I made simplifying assumptions, but they do not limit the usability -- you can make yours more complex by adding elements and estimating added losses and gains. The examples are for a 2,000 sq. ft. one-story house over a crawl space, 40' x 50', with 8' ceilings. It has five 3' x 4' windows on each outside wall and four 2' x 4' skylights in the ceilings. Pretty basic newer US smaller house.

Floors -- The examples assume that a crawl space, if any, is at the same temperature as the outside air, which makes calculations easy, but may not be accurate. If the floor is slab-on-grade, then it's R value will range from a low of 1 with no underlying insulation, to perhaps 8.5 with underlying R 7.5 foam insulation. If the main floor is over a basement (which is a heat-sink), then the temperature differential will obviously be the difference between the temp of the main-floor room and the temp of the basement.

The slab floor element is in the top section of factors independent of outside air temperature, because soil temps are relatively constant year-round within about 10 degrees or so. The floor over crawl space is in the bottom section of factors that depend on outside air temp. Don't use both -- put in an area = 0 for the one not used.

Attics -- The examples assume an attic. For heating, the attic is assumed to be at the outside air temp. If it's warmer, then your heating load will go down, so the assumption is worst-case. For cooling, the attic is usually much hotter than the outside air, which increases the cooling load. The example assumes an attic temp of 140 degrees for an outside air temp of 105 degrees. Shading by trees or other buildings may lower this, so the example is probably worst-case unless you live in a brutally hot area.

Outside air infiltration -- This is a major source of gain or loss and can't be neglected. The examples assume an air heating/cooling transfer factor of 0.018 -- I had forgotten the meaning of this factor, and Frank Purdy kindly emailed me that this is the BTU value to raise/lower 1 cubic foot of air 1 degree F. Likewise, various sources indicate that exchanges per hour vary from 0.2 for very tight buildings, to 0.5 or higher for less tight construction -- the examples assume 0.4.

So, in the calculator, separately calculate the house's volume, which is not computed automatically, treat 0.018 as a constant, and estimate the air changes per hour. Generally, there is said to be less infiltration in cooling season than in heating season, because people keep windows and doors more tightly closed to improve efficiency, whereas in heating conditions, windows are often cracked for ventilation.

Insolation -- The sun shining into the house through windows and skylights, and/or heating up the exterior, has a major effect in reducing heating loads and increasing cooling loads. The examples do not account for this except for attic air heating effect on cooling load. If you have cathedral ceilings, for example, the gain from insolation might exceed the high differential air temps of an attic space -- or they may not. These gains are just hard to model, and accounting for them is why commercial programs tend to be expensive. For ballpark use, just treat cathedral ceilings as though there's an attic and use the example's 140 degree attic temp to set a differential.

Multi-story -- Heat rises, so upper floors tend to be warmer than lower floors, just as air near the ceiling is warmer than air near the floor. If the floor between stories is well-insulated, then assume a very low transfer differential and ignore it. If there is no insulation, then take a stab at a value for variation in the upper story by lowering the differential to outside air temp in cooling season by 5 degrees and raising the differential to outside by 5 degrees in heating season, and just ignore the differential between floors.

Lighting and electrical power -- Lighting and power radiating devices such as ovens, stoves and water heaters can be ignored in heating season because they reduce the heat load, and ignoring them results in worst-case figures. In cooling season, these gains need to be estimated because they can be high, as in cooking a big meal using the oven and several burners, or if the water heater is in the cooled space. For lighting, total all the wattages of lamps and fixtures and assume half are on (because maximum cooling is in the daytime) to get a fair worst-case estimate. Assume that water heaters have a 25% duty cycle and divide their total element wattage by 4 as an estimate. Don't forget about TVs and computers as additional sources of heat.

People -- people can be ignored as a factor in heat gain for family homes where occupancy is under 10. If you're doing a meeting space, definitely include expected occupancy as noted on the calculator.

Data input
The examples are for a single-floor house of 2,000 sq. ft., with 8' ceilings, and crawl space and attic. This is intentionally the simplest case. All rooms are heated or cooled to the same extent, and all have the same construction, meaning that only the area of outside walls is considered. Each outside wall has five windows of the same type, 3' x 4' in size, with double-pane glass. Four skylights are assumed, each 2' x 4' in size, also double-pane. Note that the wall areas have their window areas removed, since window losses are calculated separately

The calculators can be expanded to be as complex as you want them to be -- just add rows for rooms, dormers, etc. Ignore losses/gains through interior walls unless the adjoining space is unheated or uncooled -- in these cases, calculate the area of the joining wall and estimate the temp difference.

Temperatures -- Enter your desired inside temperature in the "Inside temp = " row, where the example value is 68, and enter the assumed worst-case outside temperature typical of your environment in the "Outside temp = " row, where the example value is 17.

Floor and ceiling -- They usually will have the same area, but use the outside measurements of the structure to figure these areas. If you have cathedral ceilings, they will be larger than the floor area by 10 to 30% depending on pitch, so measure carefully.

Walls -- Subtract window areas from any wall area calculation and treat them separately.

Windows and skylights -- Use the rough-in dimensions to calculate area. The examples lump them all together, but windows or skylights in unheated/uncooled areas change the temp differentials. Skylights are usually a small part of ceiling area and I don't bother subtracting them out.

Good practice
In general it seems that having heating or cooling unit output capacity near to the actual worst-case need is a good idea, because the units run more efficiently the longer they're on. Having a gas furnace with 5 times more BTU/hour than you need worst-case means that the unit will "short-cycle" and be less efficient -- nothing has time to heat up and the losses are higher in the unit. The same is true for heat pumps, especially in cooling mode.

Heat pumps for heating should be de-rated from nameplate capacity, I think, by about 40%, then that value added to auxiliary heat output to get an idea of total capacity. Many installers recommend that enough auxiliary heat capacity be provided to keep you warm even if the heat pump itself fails completely. I think that's a good idea, particularly if you live in a cold to very-cold climate.

In cooling, heat pump capacity should probably be increased by 40% over actual intermediate-case calculated vales, especially since there's no "auxiliary cooling" available. Humidity raises all kinds of problems about "sensible heat" and more capacity may be essential to comfort. But in many areas, peak high temperatures may only last for a few days a year -- does it make financial sense to buy enough capacity to cover the absolute-worst-case and then pay every month for inefficiencies due to short-cycling? More isn't always better. For example, in my area temps are in the 90's perhaps up to 10-15 days a year, and in the 100's almost never. I can live with a capacity that keeps us at 72 when it's 90 out, and I'm willing to live with a 5-10 degree rise in those hot days for the sake of lower overall cost, both in unit cost and usage cost over a year.

Metric units
These calculators can be used for gains and losses based on Watts per square meter per degree centigrade if the correct U or R values based on metric units are used too. Just plug the numbers in, and change the labels of the headings to help you remember that the results are in Watts per hour, not BTU per hour.

Example pages
Heat-loss sample calc's Heat-gain sample calc's

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