1

My brother had an idea for a DIY project and I am curious if the science works out well enough to try it.

The idea is to collect rainwater then mist it onto the air conditioners condenser coils (the hot side). The hypothesis is that the water mist will provide additional cooling because water is a better conductor of heat and that evaporating water has a cooling effect.

My question for Physics.SE is how can I work this out on paper first to determine if its remotely useful?

I think what I need is to understand how to compute heat transfer from one medium to another. That would let me compare the difference between the coil to air and to water.

I suppose the simpler solution is to actually build this contraption and measure how long the A/C runs during a given time period and noting both the indoor and outdoor temperatures.

Freiheit
  • 113
  • Here's a link to a pdf brochure a "Condenser Misting System" from an outfit in FL: http://www.hvacfl.com/Docs/Condenser%20Misting%20System%20Florida.pdf‎ – Alfred Centauri Jul 17 '13 at 17:17

1 Answers1

2

In theory it would improve efficiency. The cooler it is, the less pressure is required to re-liquify the refrigerant and the motor works less hard.

Be careful! The system may be designed to require a certain pressure to work. Also, the water use would be very high. Finally, water can corrode metal, etc.

We can approximate some things: Water will evaporate and take x energy per gram. Please don't discard liquid water! That would be a big waste. The evaporation will cool it from y degrees to z degrees. This is a more involved calculation, it requires using heat and mass transfer and is dependent on the geometry. Expected about 10-20C. The condensor pressure will drop by p (look up the vapor pressure of your refrigerant, or a typical refrigerant if you don't know). The motor has to output p*(volume flow rate of compressor gas) less power. The volume flow rate can be calculated from the heat of vaporization of the refrigerant and how much heat is going into the water. Less power means less energy, but we do not know the motor's power output-power input curve. Ideally out = in, but there are always losses.

Kevin Kostlan
  • 6,463