Cooling tanks with fans - evaporation equations

[Robert Smith]

There's a thread running in the Freshwater section that is getting a bit technical. It seems the summary is that fans are the best method of cooling largish tanks, especially over days or longer. For small tanks or to compensate for short spikes in temperature adding ice in sealed containers (eg. a softdrink bottle) seems an option.

The following is the most technical bit which could be useful for those who wish to plan to use fans and want to know how big and show fast.

------------- Technical Bit -----------------

You can calculate the rate of evaporation from a tank of water through the following equations:

m1 = A.(Xs-X).(25+19.V)

where:

m1 = water evaporated (kg)

A = area of the tank surface (m2)

Xs = saturated air humidity (kg/kg)

X = actual air humidity (kg/kg)

V = velocity of air (m/s)

And that can give you the temperature change in the water through:

dT = (H.m1)/(c.m2)

where:

H = evaporation heat of water (2500 kJ/kg)

m1 = water evaporated (kg) from above

c = specific heat of water (4.186 kJ/kg/degC)

m2 = total mass of the water in the tank (kg)

So assuming:

- A 100L tank 30 cm wide x 85 cm long x 40 cm high. Then the area (A) = 0.3 x 0.85 = 0.255m2 and the mass (m2) is 100kg;

- The tank water is 30 degC, giving Xs = 0.0271 kg/kg;

- The room temperature of 30 degC and humidity of about 70%RH, giving X = 0.0189 kg/kg;

- A fan pushing air over the surface at 2m/s (not especially fast, it sounds like it could be as high as 6m/s for high velocity domestic fans);

You get:

m1 = 0.255 x (0.0271 - 0.0189) x (25 + 19 x 2) = 0.132 Kg/Hr

dT = (2500 x 0.132) / (4.186 x 100) = 0.79 degC/Hr

So it seems reasonable to expect cooling of about 1 degree per hour in a smallish tank.

[HandS]

[Gannet]

i 2nd that

[tel]

in the realm of saltwater we tend to use rather different equation to work out average evaporation. i shall type this slowly so that all can follow it:

1. mark the waterline of the tank with a felt tip

2. mark the waterline again after the fan has been on for 24 hours.

3. measure the distance between the 2 marks, and then mutiply that by the length and breadth of the tank.

this is how much water has evaporated in 24 hours ( this equation is a little less accurate for those people whose demented relatives drink from the tank )

as to the average cooling over an hour we tend to turn the fan on and look at a thermometer 8)

[Robert Smith]

Tel,

That's all very pratical and all, but it assumes that you are just measuring what happens to something you've already got. If you want to plan a cooled system then you need to step back and look at tank surface area, how big a fan to buy, how fast to run the fan, typical humidity etc.

This is the Technical forum, not the Practical forum.