Efficient Cooling for Industrial Rooms: How to Calculate Energy Needs

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SUMMARY

The discussion focuses on calculating the energy needs for cooling industrial rooms to a maximum temperature of 10°C, regardless of outside conditions. It highlights that there is no simple formula for this calculation due to the complexities of heat gain through walls and the influence of equipment heat output. The CHVAC demo version is suggested as a potential tool for small projects, while a rough estimate for well-isolated rooms without ventilation is provided, indicating that the cooling load is approximately equal to the heat output from equipment. Ventilation complicates the cooling process, especially when outside temperatures exceed 20°C, leading to potential efficiency issues with cooling equipment.

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wolram
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Can anyone give me a formula for calculating the energy needed in cooling a room?
The room has to be at 10c maximum, regardless of outside temperature.
as there are several rooms all with differing volumes, can a formula for
1000 cubic meters be used.
Thanks.
 
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Unless there is something a lot simpler about your question than I realize, there is no simple formula for this. Heat gain through walls is a complicated animal. You may be able to get by using the demo version of CHVAC for such a small project.

However, when you say "industrial rooms", there is a chance that you can get a simple estimate that will get you close enough: If the rooms are well isolated from the outside and get no ventilation and the heat given off by the equipment in those rooms is substantial, then the cooling load (size of the air conditioner required) is approximately equal to the quantity of heat given off by the equipment in the room.
 
Thanks Russ

I think this is a complex problem as the rooms have to be ventilated.
the equiptment used to keep the rooms at 10c work ok until the outside
temp gets to around 20c, our cooling equiptment can be adjusted to a
lower temp, but that results in the evaps freezing up, which means more
defrost cycles and lower overall efficency.
 

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