Homebrew A/C testing

 I have moved on to making some measurements on my first version of a DIY A/C system.  My main interest at this point is to get a rough indicator of how much heat energy the unit is capable of absorbing.  I currently don't have two temperature sensors to measure the air's temperature drop (nor do I have an anemometer to measure the air flow through the fan), so I tried doing it by measuring the amount of water consumed by the cooling tower.

Heat energy can be derived from water consumption using the fact that it takes 4184  2,451, 824 joules to convert 1 gram of water to vapor (at the same temperature).  To get the volume change of water, I measured the dimensions of my water reservoir so I can calculate the volume as a function of height change of water.  The test setup looks like this:

The metal ruler that's at an angle is used to measure the change in water level (in the photo, it's about parallel to the house's shadow line).  It's at about a 50 degree angle, which helps to improve the effective resolution of the measurement.  To convert the distance on the ruler to actual height change, I multiply the measurement (in decimal numbers, not fractions) by sin(50), about .776.  If I laid the ruler along the long axis of my reservoir I could get even better resolution, with a multiplier of .545.  Multiplying the volume change (in mililiters) by 4184 results in the number of joules absorbed by that amount of evaporated water.  Dividing that by the elapsed time gives me the watts of heat energy involved

So far I have two data points, starting earlier in the day when the ambient temperature was relatively low (76F) and the relative humidity was relatively high (61%).  In about 34 minutes the system moved approximately 2.2 megaJoules (!), which works out to an average heat power close to 1KW.  A second measurement taken 2 hours 10 minutes later showed an average heat power around 1.9KW.  Since the second number was calculated for the start of the experiment, as the outside air warmed up the effective power went a bit higher than 1.9KW (2.15KW, to be precise).

At the same time, the fan and pump consumed about 90 watts, so it can be seen that the system is capable of absorbing at least 10 times the amount of heat than the system needs to operate.  Not bad!

I need to point out that, as it is now, the system is not suitable for cooling our house.  The exit air is pretty humid so any improvement in comfort due to lower temperature is more than offset by the increased humidity.  I have gotten a 12x12 heat exchanger, a second fan and higher-volume water pump so I can try the indirect-cooled approach, where the evaporatively-cooled water in the reservoir will be pumped through the heat exchanger.  I still need to get some bulkhead fittings to make a clean system for pumping water out of the reservoir.  Desertsun02 has an indirect-cooled setup he shows on a youtube video, but he just snaked the added tubes up and over the side of the reservoir, which permits uncooled outside air to get into the tower.  While I don't really care what the exit air temperature is, that air leakage also reduces the evaporation rate, which is a concern.  Using two bulkhead fittings will get around this problem, as long as the hoses never have to go above the surface of the water.

It also will be necessary to make an enclosure for the heat exchanger so I can pull air through it with my second 12 volt fan.  So I likely will miss the opportunity to try the whole thing during our current heat wave (100F predicted today, 104F or higher tomorrow).

Update:  I let the system run until almost 5PM (a total runtime of about 60,000 seconds).  The overall heat transfer rate came to 2.125KW, for a total of around 12KWH. (51 mega joules).  

I need to think about this some.  The actual heat transfer may be substantially more because the cooling tower cooled a whole lot of air, in addition to evaporating a bunch of water (about 12 liters).  I could calculate what this is, assuming that the fan really moved 1700CFM and (based on ambient vs exit air temp) the temperature drop.  But I believe I can't add the two results -- the heat absorbed by the evaporating water was pulled out of the air (and the residual water in the cooling pads).  So using just the latent heat in the evaporated water probably is correct.  But my calculation produces a result that is about an order of magnitude higher than the water-evaporation calculation.  I definitely need to think about this.....