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Evaporative Cooler - Efficiency 1

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elpohl

Mechanical
Dec 20, 2006
8
Hello there, first thread!

My friends and I just finished building an evaporative cooler using some cellulose panels and a walmart vent. We are pumping water at ambient temperature and we need to calculate the efficiency. I know that its a ratio between the temperatures, but since we are using water vapor I believe that the relative humidity has to be included somewhere.

I'm really stumped and I'm afraid of thermodynamics, but hopefully with this project we will get a better hang of it. Any assistance will be appreciated!!!

Thanks,

Paul.
 
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Efficiency in terms of cooling effect/power in.

1. You need to measure the pump current and calc W (A).
2. Measure the inlet air dry and wet bulb temps.
3. Measure the outlet air dry and wet bulb temps.
4. Use a psychrometric chart from ASHRAE to determine enthalpy at each of the inlet and outlet.
5. Measure inflow air velocity. Calculate mass airflow.
6. Multiply the mass flow by the difference in enthalpies to get cooling effect, W.
7. Divide to get efficiency.
 
Excellent! Thank you! So I dont need to calculate the "humidity"? I'm wondering this because, in essence, the inlet and outlet temperatures of the air are not gonna be that different...

Since its a direct evaporative cooling system, open system, wont the performance be dynamic due to the changes in humidity?
 
By measuring wet and dry bulb temperatures you are, in effect, calculating relative humidity. Evaporative coolers work much better in dry climates. When I lived in New Mexico there was one in my apartment instead of an air conditioner.
 
won't be much of a change in enthalpy, except for a slight increase because of fan heat.

Work out your sensible cooling gained at the cost of humidifying the air

Entering/leaving wet bulb should be about the same, the fan heat will raise it a bit.

Evaporative cooling follows the constant wet bulb of the entering air.

Take the "V" out of HVAC and you are left with a HAC(k) job.
 
Ah dang... I studied this subject in spanish language, so it's a bit more confusing to me. I understood everything except the HVAC term... so long so good though! Its getting clearer and clearer. :)
 
its my signatire, its short for Hoover Vacuums, ignore the last sentence

Take the "V" out of HVAC and you are left with a HAC(k) job.
 
In the case of the evaporative cooler, the air will loose sensible heat to the water which will be equal to the gain moisture contents (or humidity) from the water. The untimate potential of the air passing through the cooler is to have the dry bulb temperature at the outlet equal to the web bulb temperature of the air at the inlet (which is true for infinite length counter flow cooler).

As the evaporator coolers are used in the dry arid areas to cool and add moisture to the air. As the air passes through the cooler, the dry air give sensible heat to the moisture and result in the evaporation of the water and this moisture contents is then added to the air. Thus accounting for the energy balance in the two fluids.
 
I'm not sure what it would mean to define an efficiency for an evaporative cooler. (Seems like refrigeration units have a "coefficient of performance" rather than efficiency). The problem is you are not just cooling air, you are also drastically changing the humidity. The effect you want is to FEEL cooler, and the humidity offsets that. So, for example, you might feel cooler with increased air flow and less humidity- even though temperature is slightly higher.
 
AbbyNormal: Hahaha, I feel dumb. :$

Termcool: Thanks for the explanation!

JStephen: Yes you are right, its performance, not really efficiency. And yes, the cooling effect is more a feeling than something else.

Ok, so I think I'll proceed with Cinimace steps and perform the calculations. Hopefully I can find the air flow specifications of my vent, cause otherwise its going to be quite a chore... thus, its winter here (not the best season to finish this project), so the cooling feeling will almost go unnoticed. :S
 
Ok... everything went out perfect. I even found an online psychometric calculator (ASHRAE ain't free :#). But... I'm having problems calculating the mass airflow... any ideas?

I know its Q*density, but... how can I have Q out of a store ventilator? :S
 
If you can measure the volumetric flow, then use the specific volume from the chart to work out the mass flow rate of the dry air.

Charts are set up with a pound of dry air and so much moisture for every pound of dry air.

You can then determine how much moisture 'accompanies' each pound of dry air before and after the humidification/evaporative cooling

Take the "V" out of HVAC and you are left with a HAC(k) job.
 
The efficiency of an evaporative cooler can be calculated, simply, as difference of DB/initial wetbulb depression.

For example, 90F is DB, 70F is WB before entering the cooler and 75F is DB after the cooler then Eff. = (90-75)*100/(90-70) = 75%. You should note that, like JStephen suggested, the efficiency here is figurative.

 
That would be how efficient it is in terms of cooling the air as compared to the maximum cooling that could be obtained. I was thinking in terms of amount of cooling versus electrical power used.
 
..while not thermal or electrical efficiency, you can "estimate" the cooler's Saturation Efficiency (Es) using just three temperatures:

Es = (DBD/WBD)*100% = (T1-T2)/(T2-T3)*100%

where:
DBD = Dry Bulb Depression (T1-T2)
WBD = Wet Bulb Depression (T1-T3)
T1 = Dry Air (inlet) INTO cooler, F-deg
T2 = Dry Air (outlet) FROM cooler, F-deg
T3 = Wet Bulb Temperature(officially), but the temp of the return sump water (draining from pads) can be substituted as a "close enough" approximation.

...common 2"-thick aspen-excelsior pad coolers achieve 85%-90% with new, clean pads; efficiency is highest at LOW airflow rates.
...newer Celdek/Munters, etc. celulose (sp) pads can achieve 90%-95%, but only at higher airflow rates.

Tucson, AZ.
 
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