## Example of Residential Heat Load Estimate/ Calculations

written by: Haresh Khemani โข edited by: Lamar Stonecypher โข updated: 10/19/2009

Now that we have seen the various heat loads inside the room and also surveyed the room, let us see one example heat load calculations for the residential building using the heat load calculations form.

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### Example Heat Load Calculations

Now that we have seen the various heat loads inside the room and also surveyed the room, let us see one example heat load calculations for the residential building using the heat load calculations form shown below. To start with, fill the details given at the top of the form. These are given below:

Customer: Mr. Allan Smith

Installation by: Mr. Garry and Mr. Ronny

Estimate number: 0022

Estimate by: Ms. Sheena Roy

Equipment Selected: Manufacturer, Model and Size (to be filled at the end of heat load estimate): 2.5 TR Split type.

Direction House Faces: North

Gross Floor area (of the house): 1500 sq ft.

Gross inside volume (of the room for which heat load calculations are being done): 300 sq ft

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### Sample Heat Load Calculations for Residential Building

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### Design Conditions:

Dry Bulb Temp (DBT) F

Outside 100

Inside 78

Difference 22

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### Direct Solar Heat Gain by the Windows

There are three windows in the room each of the size 6 x 4 = 24 sq ft. There is one window each in east, south and west direction. The glass of the all windows is single, there is no shading and no outside awnings.

Fill the details of areas in the heat load calculations form as shown in the attached form. Round off the proper associated factor with window in each direction for from no shading option. If the windows has shades or outside awning, one has to round of the factors from those columns. For this particular example the rounded factors have been shown in the form. For window in east direction it is 100, while for window in south and west directions it is 75 and 150 respectively.

Now multiply the area of each of the window by factor associated with it as shown in the form. For window in east direction it is 24 x 100 = 2400, for window in south it is 24 x 75 = 1800, for window in west it is 24 x 150 = 3600. The highest of all these, 3600 has to be selected and filled in the last column. Thus the total solar heat gained by the window is 3600 BTU/HR.

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### Solar Heat Gained by the Windows due to Designed Conditions (Internal and External Temperature Difference):

The total area of three windows is 24 + 24 + 24 = 72 sq ft and they are all of single glass. Fill this in the area column for single glass window as shown in the form. Since the difference between external and internal dry bulb temperature is 22F, the factor associated with it would be 27, so it has to be rounded of. The product of 72 and 27 is 1944. Thus the solar heat gained by the windows due to design temperature is 1944 BTU/HR.

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### Heat Gained by the Walls

Let the size of the room is 20 ft x 15 ft = 300 sq ft, which is the total floor area of the room. Let us suppose the height of each wall is 12 ft and none of them are insulated. Two walls of this room if length 20 ft and 15 ft are exposed directly to the sun, while remaining two are partitions.

The total area of walls exposed directly to the sun is 20 x 12 + 15 x 12 = 420 sq ft. Since the designed temperature difference is 22F and there is no insulation, the factor associated with it is 7. The product of 420 and 7 is 2940, which is the total BTU/HR gained by the walls exposed directly to the sun.

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One ton of AC = 12000 BTU/HR. Thus total tonnage required in the room is 29281.5/12000 = 2.44 tons, which can be taken as 2.5 TR. The total recommended tonnage for the room is 2.5. For this tonnage split air conditioner is the best option. One can install wall mounted split air conditioners of 1.5 ton and 1.0 ton at two different inside the room.
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### Heat Gained by the Partitions

There are two partitions in the room of size 20 x 12 = 240 sq ft and 15 x 12 = 180 sq ft. The first one is with air conditioned room and the other with non-air conditioned room. For heat load calculations we have to consider only the second one. The factor associated with designed temperature difference of 22F is 4. Hence the total heat gained by partition is 180 x 4 = 540 BTU/HR.

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### Heat Gained by the Roof

The size of roof is same as the size of the floor, which is 20 x 15 = 300 sq ft. The roof is exposed directly to the sun, it is flat with no vented air and it is non-insulated. For 22F of design temperature difference, the factor associated with it is 34. Thus the total heat gained by the roof is 300 x 34 = 10200 BTU/HR.

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### Heat Gained by the Ceiling

Since the roof is directly exposed to the sun, there is no ceiling for the room, hence there is no heat gained by the ceiling.

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### Heat Gained by the Floor

The size of the floor is 20 x 15 = 300 sq ft. Let us consider that the room is located over other non-conditioned room, so it gains some heat from it. For the designed temperature difference of 22F, the associated factor is 4. Thus the heat gained by the floor is 300 x 4 = 1200 BTU/HR.

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### Heat Gained by the Room Air from the Outside Air

The total amount of outside air or the infiltrated air inside the room is proportional to the floor area of the room, thus the total floor area, 300 sq ft, of the room has to be considered. The factor associated with it for designed temperature difference of 22F is 3. Thus the heat gained by the room air from the outside air is 300 x 3 = 900 BTU/HR.

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### Heat Gained by the Room Air from the People or Occupants

Let us suppose the average number of people inside the room would be six. Thus the heat gained by the room air from people is 6 x 200 = 1200 BTU/HR.

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### Subtotal of all the Heat Gained by the Room Air

The subtotal heat gained by the room air is total of all the heat gains as mentioned. The total heat gained is 3600 + 1944 + 2940 + 540 + 10200 + 1200 + 900 + 1200 = 22524 BTU/HR.

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### Latent Heat Allowance

The latent heat allowance includes heat absorbed from the moisture and other small sources. The latent heat allowance is 30% of subtotal, which is 0.3 x 22525 = 6757.5 BTU/HR.

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### Total Heat Load inside the Room

Thus the total heat load inside the room is 22524 + 6757.5 = 29281.5 BTU/HR.

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### Recommended Total Tonnage of AC and Type of AC

One ton of AC = 12000 BTU/HR. Thus total tonnage required in the room is 29281.5/12000 = 2.44 tons, which can be taken as 2.5 TR. The total recommended tonnage for the room is 2.5. For this tonnage split air conditioner is the best option. One can install wall mounted split air conditioners of 1.5 ton and 1.0 ton at two different locations inside the room.

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### Image and Heat Load Calculations Form Courtesy

Principles of Air Conditioning by V. Paul Lang Published in India by D. B. Taraporevala Sons & Co. Private Limited.

#### How to Do Residential Heat Load Calculations

This is the series of articles that explains how to perform heat load calculations for a residential building or other smaller buildings in easy manner by using the ready-made form.