# How much does an evaporative cooler cool down a room

So I'm trying to create a system that automatically runs evaporative coolers to keep a warehouse at a desired temperature, given the internal and external temperature and humidity. I can get the coolers to run automatically and read the data I want quite easily.

Here's what I need to know: is there a way to calculate how much a cooler cools down a room? I can find out the volume of the room and the airflow of the coolers, so an equation that requires those will be fine. Is there an equation that describes how much a cooler cools a room down?

• You mean you're trying to design the system? Obviously any thermostat solves the control problem. Commented Sep 10, 2016 at 15:05
• You need specs of the cooler for a start. The aireflow doesn't mean much without knowing the temperature of ghats air, etc. And it's not easy to come up with numbers without having figures on how well the room is insulated. But once you have specs of the cooler, you could estimate the insulation of the room by using a heater of known power and seeing how much that heats the room. Commented Sep 10, 2016 at 18:01
• I'll just point out that relative humidity makes a lot of difference in how a room feels, so if you're doing this to make it cooler for people or horses (creatures that sweat) then you might not actually be making it more comfortable, like how some people are okay with a "dry heat". Some things (electronics, sugar, etc.) may also react poorly to high humidity. Just remember that a swamp cooler trades temperature for relative humidity (though you might be able to cool one side of a heat exchanger). Commented Sep 13, 2016 at 14:21

The capacity of an evaporative cooler depends on the relative humidity. The lowest theoretical temperature that the cooler can reach is the wet bulb temperature.

The wikipedia article has some discussion on the design and shows how it works with psychometric charts.

For the design you'd look at a 'typical' hot day, compare wet bulb to dry bulb for that day and assume a certain efficiency (e.g. it reaches 80% from dry bulb to wet bulb). The energy in the supply air stream would be: $$Q = \dot{m}\Delta_Tc_p\eta$$ i.e.:

Energy = massflow x temperature drop x specific_heat x efficiency

$\Delta_T$ being the temperature difference before and after the evap cooler using your assumptions.

$Q$ would be your cooling capacity

HTH