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Hot humid east TX; Large subdivision pool is warm ,92F. A plan is being make to cool it with evaportive spray . One data point is : dry bulb= 82.0 , wet bulb = 79.7 F. This is likely typical average of 24 hours. The cost of pump and spray may be $ 3000 ( to purchase). How much cooling could be expected ? My thought is at this high humidity it will not be cost effective but I can't put numbers to it.

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  • $\begingroup$ Second data ;Dry 27.5 C , wet 26,5 C , about 90% humidity. $\endgroup$ – blacksmith37 Jul 6 '17 at 15:33
  • $\begingroup$ This just looks like an information dump. Also, you can edit additional information into your question instead of putting it into a comment. $\endgroup$ – hazzey Jul 6 '17 at 23:55
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I'm going to take the formal question statement as "Is evaporative cooling for a pool effective?"

There are a handful of factors that need to be considered. First and foremost though, is wet bulb temperature. The wet bulb temperature is theoretically the coldest temperature that you will ever be able to achieve utilizing a wet cooling system (a system that evaporates water to reject heat). Taking this into consideration, the absolute coldest that you will be able to make the pool given your stated data points and method is the wet bulb temperature of 80 Fahrenheit. Realistically, you will probably only be able to come within 2-5 degrees F of this (82-85 Fahrenheit) due to mass transfer limitations and lack of crossflow packing & drift elimination in a small residential unit. For sake of the argument, assume that 80 F is the best you can do using wet cooling.

If 80 F is suitable and you would like to pursue this, the next step is to determine total capacity required. Capacity when talking about heat rejection is measured in MegaWatts, kiloWatts, Watts, or BTU/hr. To help understand this concept, think about the total energy coming into the pool that's heating it up; likely the predominant source is sunlight being absorbed into the pool (there may be some heat transfer from the dry bulb temp of the air and the water, but odds are it's negligible, or possibly even cooling the pool off a small amount via some natural evaporation). So if we want to keep the pool at a constant, cooler temperature, we must have enough capacity to remove all of the heat that the sun is putting into the pool. Because we have a very small amount of knowledge about your pool, I'm going to make some considerable assumptions utilizing previously calculated values. So we're going to assume a crude measurement of a peak solar energy absorption that induces a 0.75 F / hr (0.41 C / hr) increase in pool temperature. Note that this is NOT accurate if your pool is considerably different geometry, or has a different level of reflection on the pool floor. Additionally I will use a per-unit-of-water calculation so that you can extrapolate this to your pool size as this was not provided. An example was provided for a pool size of 10,000 US gallons

Heat capacity of water: 4.184 kJ/kg*k; Q = heat flow rate; 1 US gallon water = 3.785 kg

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So, what this is saying: for a 10,000 US gallon pool, at the same conditions described in the link, we should expect to need at least 18 kW of heat removal ability to keep the pool cool.

Remember that the coldest we can get is going to be the wet bulb temp; but to get to that wet bulb temp, we have to be able to reject the same amount of heat that's coming into the pool from the sun that we previously calculated (18 kW). To do this, you need to size your evaporative cooler. Unfortunately, evaporative coolers are sized and sold on a "CFM" basis, as their BTU performance is variable depending on inlet conditions (namely temperature and humidity). You can find some manufacturers that will rate their units based on very tight design conditions; but note that a 15 kW (50,000 BTU/hr) unit at 0.3 IWG conditions may not be able to achieve 15 kW regularly. Accurate sizing of the evaporative cooler will require measurement of heating degree days, historic wet bulb temperatures, and then analysis of this against said design criteria. Alternatively, you could just way over shoot your kW target (if you need 18, maybe go with 30 kW worth, or two 50,000 BTU units). Note: if you try to size an air conditioner using BTU's or tons and translate that to a comparable evaporative cooler by using the CFM required to cool the same square footage/volume, you will undersize the evaporative cooler - The cooler is only designed to hit its design wet bulb temperature approach, NOT to match the same BTU output of an air conditioner.

I do not know which model(s) you're looking at, but just looking at some home depot styles and some on the link above, it seems like most coolers are around $100-$200 USD/1000 CFM. If we extrapolate very crudely based on the Australian design conditions used by breezair, you can achieve ~1.5 kW / 1000 CFM. To hit 18 kW for a 10,000 US gallon pool would require 12,000 CFM, or roughly $ 1200 to $ 2400 in capital expense. This doesn't include the power to run the fan (~1kW * (12 hrs / day) * $.15/kWh =~ $2/day to operate at peak load), or water costs ($0.10/gal * ~100 Gal / 12 hrs operation =~ $10/day peak operation)

One other concern with this is cooling water source. It would generally not be recommended to utilize pool water for the evaporative effect, as all the minerals in the water will concentrate up over time as you evaporate water out and recirculate it back into the pool - this will eventually cause scale buildup in the pool water, making it undesirable to swim in. If one discharges the water somewhere else as opposed to back into the pool, then you'd be wasting the treatment chemicals (bleach etc), which would be an additional cost.

Alternatives to utilizing a swamp cooler / evaporative cooler is to simply remove the heat source in the first place. If all the heat is coming from the sun, then building a way to provide shade to the water may provide a substantial benefit.

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