I’ve received a few questions about the process by which a swamp cooler takes hot air and water and creates cooler air rather than warmer water. The key process involved is evaporation, and a way to quantify this is through the enthalpy of vaporization, which describes how much energy it takes to turn a specific amount of a liquid into a vapor at a constant temperature. For water, this is roughly 970 BTU / lb. Since a BTU is defined as the amount of energy needed to raise 1 lb of water by 1 degree Fahrenheit, this means that evaporation takes 970 times more energy than raising the temperature of a given amount of water by 1 degree F.
In a swamp cooler, hot and relatively dry air brings thermal energy into the system. For reasons associated with the discipline of heat and mass transfer (and encouraged by the design of the swamp cooler), the primary energy transfer involved as the hot air meets the thin film of water on the cooling pad is evaporation. During this process, 970 BTU’s of energy are transferred out of the hot air to evaporate every pound of water used by the swamp cooler. This results in cooler, moister air as the energy used to evaporate the water lowers the air temperature and the evaporated water increases the humidity of the air.
It may be easier to see this as an example. Suppose it’s a hot and dry day at sea level with an outside temperature of 110 degrees F at 10% relative humidity. Using a psychrometric chart identifies the following:
• Specific enthalpy of dry air: 33 BTU / lb dry air
• Specific volume of dry air: 14.55 cu ft / lb dry air
• Wet bulb temperature: 68 degrees F
• Dewpoint: 41 degrees F
• Specific humidity: 0.0055 lb water / lb dry air
For the sake of illustration, we’ll consider a steady state analysis and ignore transient effects either outside or inside.
A swamp cooler works along lines of constant enthalpy. Since the wet bulb temperature equates to a relative humidity of 100% at the same enthalpy as the starting point, that identifies the lowest a swamp cooler can possibly cool the air to. For most cases 100% relative humidity would not be desirable, so taking a more comfortable (particularly in the desert) indoor relative humidity target of 55% gives a temperature of 80 degrees. The specific humidity at that condition is 0.012 lb water / lb dry air, so 0.0065 pounds of water would need to be added to every pound of air to cool to that condition. To put that in perspective a 2,500 square foot house holds about 20,000 cubic feet of air, or a bit under 1400 pounds of air. That would require about 9 pounds of water, or a touch over a US gallon.
Compare that with an air conditioner. Whereas on a psychrometric chart the swamp cooler moves along lines of constant enthalpy (increasing humidity while decreasing temperature at constant energy), an air conditioner moves along lines of constant moisture content while changing energy to decrease temperature. For the same case of cooling to 80 degrees, an air conditioned space only increases the relative humidity to 25% and decreases the specific enthalpy to approximately 25 BTU / lb dry air. That requires 8 BTU / lb dry air energy removal, which for the example 2,500 square foot house would need 11,200 BTU of heat removal. Air conditioner cycle efficiency in the US is denoted by the SEER / EER system (cooling BTU / Overall power input (W)) and in the desert environment the EER is the more applicable value. The minimum EER currently allowed for new construction in this area is 12.2, so 11,200 / 12.2 = 918 Watts are required.
In the real world this example analysis is only a starting point. Outside temperatures and humidities are constantly changing. Houses naturally heat up in the daytime and cool at night. There are blowers and ducting to consider. A swamp cooler needs to exhaust the air displaced by that which it pulls in, so unless the system is completely optimized a fair bit of the cool air gets blown outside before it has exhausted it’s capability for cooling. Air conditioner coils can become less effective due to blockages and debris buildup. Activities in the house increase the amount of cooling needed, ….. Despite this, it is helpful to have an understanding of the physical processes involved in even an ideal case.