Aquaponic Gardening

A Community and Forum For Aquaponic Gardeners


Thank sylvia for your advise.

how can I cool down the water tuperature coming out of the tilapia tank, around 85F to to 77F the aptimum temperature for growing vegetable. I'm talking industrial size operation.

thank you

Views: 109

Replies are closed for this discussion.

Replies to This Discussion

What is your average air temperature. You can use evaporative cooling to cool and aerate the water. You can do this by running a pump and using a sprinkler head.

items that work a solar heaters during the day can (if the nights are cooler and clear) also work as radiant chillers, especially if they are not protected by glass. Like a solar pool heater panel will act as a radiant chiller if water is run through them on a cool clear night. This might not be much use in your situation though but probably of more use in a dry climate than in a humid one like mine.
Thank Kobus.
I'm planing of building my system in a tropical place so the air temperature at night is around 75F and around 96F during the day.
Once I figure out the best way to do it, I'll let you know.
Best regards br/>

Kobus Jooste said:
Take a look at these two calculations (I'm sure the temperature has to be in Celcius though). It gives you an idea of how much heat energy went into your water to begin with, which will also clue you in on the amount of effort from your side that will be required to get that heat energy out of the water again:

1) How much energy will be required to rid your system of the temperature:

kW required to heat/cool a flow of water = (DT x Q) / 14.4

where DT = rise in temperature required, and Q = flow in litres per minute

2) How much energy is stored in that water for you to try to get out again:

kW required to heat body of water = ((V/1440) x DT) / 14.4

where DT = rise in temperature required, and V = Volume in litres

If I look at the second one, my heat gain will look like this: System volume is around 3800 liters, and to go from 85 F to 77F is roughly the equivalent of a 5 degree Celcius change. Thus ((3800/1440) x 5) /14.4 = 0.92 kW of energy stored in the water body. That is a lot of energy. Getting rid of it requires the following (assuming that the flow rate in my system is 3800 liters per hour just for argument sake - translated to flow per minute that is 63): (5 x 63) / 14.4 = a scary 21.8 kW. This sounds impossible, but if you look at the thermal energy storage capacity of water, you will see why. Each cubic meter of water is capable of storing 93 kWh (334 million joules or 317 000 BTU's) of energy.

Perhaps the best thing to consider is to try and insulate your system (tanks and beds) to try and reduce the amount of energy that the water is able to absorb. This will be a once-off expense in stead of constantly trying to cool a large body of water in a hot climate. If you look at nature, desert animals in particular, the best way of cooling heat exchange where you do not use evaporation is counter current heat exchange (tuna use the same method to have near warm-blooded animal body temp in cold water). You need to pass your hot water flow in close proximity to a very cool substance over a prolonged length of flow. This will gradually pull energy out of the water, but as your exchange budget balances (the water heats the coolant up) this process will slow down. There are also a range of solar-power cooling devices on the market if you want a more compact exchange surface.


© 2022   Created by Sylvia Bernstein.   Powered by

Badges  |  Report an Issue  |  Terms of Service