A building or room gains heat from many sources. Inside occupants, computers, copiers, machinery, and lighting all produce heat. Warm air from outside enters through open doors and windows, or as ‘leakage’ though the structure. However the biggest source of heat is solar radiation from the sun, beating down on the roof and walls, and pouring through the windows, heating internal surfaces.

The sum of all these heat sources is know as the heat gain (or heat load) of the building, and is expressed either in

For an air conditioner to cool a room or building its output must be greater than the heat gain. It is important before purchasing an air conditioner that a heat load calculation is performed to ensure it is big enough for the intended application.

The sum of all these heat sources is know as the heat gain (or heat load) of the building, and is expressed either in

**BTU**(British Thermal Units) or**kW**(Kilowatts).For an air conditioner to cool a room or building its output must be greater than the heat gain. It is important before purchasing an air conditioner that a heat load calculation is performed to ensure it is big enough for the intended application.

**Quick calculation for offices**For offices with average insulation and lighting, 2/3 occupants and 3/4 personal computers and a photocopier, the following calculations will suffice:

Heat load (BTU) = Length (ft.) x Width (ft.) x Height (ft.) x 4

Heat load (BTU) = Length (m) x Width (m) x Height (m) x 141

For every additional occupant add 500 BTU.

If there are any additional significant sources of heat, for instance floor to ceiling south facing windows, or equipment that produces lots of heat, the above method will underestimate the heat load. In which case the following method should be used instead.

**A more accurate heat load calculation for any type of room or building**The heat gain of a room or building depends on:

- The size of the area being cooled.
- Sze and position of windows, and whether they have shading./li>
- number of occupants.
- generated by equipment and machinery.
- generated by lighting .

By calculating the heat gain from each individual item and adding them together, an accurate heat load figure can be determined.

**Step One**Calculate the area in square feet of the space to be cooled, and multiply by 31.25

Area BTU = length (ft.) x width (ft.) x 31.25

**Step Two**Calculate the heat gain through the windows. If the windows don’t have shading multiply the result by 1.4.

North window BTU = Area of North facing windows (m. sq.) x 164

If no shading, North window BTU = North window BTU x 1.4

South window BTU = Area of South facing windows (m. sq.) x 868

If no shading, South window BTU = South window BTU x 1.4

Add the results together.

Total window BTU = North window + South window

**Step Three**Calculate the heat generated by occupants, allow 600 BTU per person.

Occupant BTU = number of people x 600

**Step Four**Calculate the heat generated by each item of machinery - copiers, computers, ovens etc. Find the power in watts for each item, add them together and multiply by 3.4

Equipment BTU = total equipment watts x 3.4

**Step Five**Calculate the heat generated by lighting. Find the total wattage for all lighting and multiply by 4.25

Lighting BTU = total lighting watts x 4.25

**Step Six**Add the above together to find the total heat load.

Total heat load BTU = Area BTU + Total Window BTU + Occupant BTU + Equipment BTU + Lighting BTU

**Step Seven**Divide the heat load by the cooling capacity of the air conditioning unit in BTU, to determine how many air conditioners are needed.

Number of a/c units required = Total heat load BTU / Cooling capacity BTU

You can download a demo version of HVAC calculation software here.

Download ComfortAir HVAC Software

## 6 comments:

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GREAT, Quick calculation for heat loads.

Thanks alot for the info

i'd just want to ask a few questions about the calculations. the multipliers in Flooe Area Calculations 31.25, Window Size and Position 164 and 868...Occupants, Equipments, and Lightings...do you call this a "safety factor"? is this the standard multipliers or safety factors in the Heat Loads Computations? because i've seen other calculations like this in the internet, but it is somewhat different. could you pls help me?

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