To get the most out of your room air conditioner, you'll need to consider the size of the unit that best suits your needs.
The cooling capacity of room air conditioners is measured in British thermal units per hour, or Btu/h. To figure out the maximum cooling capacity for a room, you will first need to calculate the size of the room where the air conditioner will be placed.
To do this, simply multiply the width of the room by the length of the room. Use the following information and formulars to help you with calculating your air conditioning requirements:
Heat Load
The amount of heat generated is known as the heat gain or heat load. Heat is measured in either British Thermal Units (BTU) or Kilowatts (KW). 1KW is equivalent to 3412BTUs. The heat load depends on a number of factors, by taking into account those that apply in your circumstances and adding them together a reasonably accurate measure of the total heat can be calculated. Factors include:
The floor area of the room
The size and position of windows, and whether they have blinds or shades
The number of room occupants (if any)
The heat generated by equipment
The heat generated by lighting
Area of Room
The amount of cooling required depends on the area of the room. To calculate the area in square metres:
Room Area BTU = Length (m) x Width (m) x 337
Window Size and Position
If, your room has no windows, you can ignore this part of the calculation. If, however there are windows you need to take the size and orientation into account.
South Window BTU = South Facing window Length (m) x Width (m) x 870
North Window BTU = North Facing windows Length (m) x Width (m) x 165
If there are no blinds on the windows multiply the result(s) by 1.5.
Add together all the BTUs for the windows.
Windows BTU = South Window(s) BTU + North Window(s) BTU
Occupants
You will have to take that into account people who normally working in the space. The heat output is around 400 BTU per person.
Total Occupant BTU = Number of occupants x 400
Equipment
Clearly most heat in a room is generated by the equipment. This is trickier to calculate that you might think. The wattage on equipment is the maximum power consumption rating, the actual power consumed may be less. However it is probably safer to overestimate the wattage than underestimate it.
Equipment BTU = Total wattage for all equipment x 3.5
Lighting
Take the total wattage of the lighting and multiply by 4.25.
Lighting BTU = Total wattage for all lighting x 4.25
Total Cooling Required
Add all the BTUs together.
Total Heat Load = Room Area BTU + Windows BTU + Total Occupant BTU + Equipment BTU + Lighting BTU
This is the amount of cooling required so you need one or more air conditioning units to handle that amount of heat.
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General thumb rule
100 sqr ft area with roof height of 9 - 10 ft will need 1 Ton AC unit.
2006-11-11 14:17:14
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answer #1
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answered by Anonymous
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While bb's answer is very comprehensive, there are simpler approximations:
For large offices, shops etc allow 100-120W per square meter
For smaller offices, allow 140-180W per square meter
This is for temperate climates, with average occupancy and equipment in the rooms. If there are a load of computers, or people, up the average rating. But rememebr these are approximations only - consult a professional to get the equipment properly sized
2006-11-14 02:53:17
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answer #2
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answered by donhoban 2
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Most air conditioner's booklets and guides have this data in them.
And the governing factor is the area of the room, the area exposed directly to sunlight, and the weather conditions of the place where you live.
2006-11-11 13:26:59
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answer #3
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answered by Jason 1
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What? an in the window unit or central?
You can do a heat loss calculation (which is gain in the summer) from many heat loss sites. 10,000 BTUH is one ton of cooling. Go to manufacturers sites for sizing, voltage, etc
2006-11-11 12:30:18
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answer #4
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answered by ? 6
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Typically, residential areas need something on the order of 12,000 BTUs for every 400 to 500 sq.ft.
2006-11-11 14:05:59
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answer #5
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answered by arbiter007 6
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the unit should have a maximum area coverage on the unit.
2006-11-11 12:27:10
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answer #6
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answered by Anonymous
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