Water cooling and mixing problem

[Hircus]

Hi all,

 

This is my first post so please bear with me. I run closed recirculating fish tanks for my research, but have always struggled with maths and 'moving' equations around. Basically the scenario is this:\

 

I have a closed 2000 ltr system, but have no way of cooling the water. I have access to plenty of freshwater from a nearby creek which runs at outside ambient temperature. How do I work out how much colder creek water to add to the inside tank system to meet a specific temperature. I would like to generate myself a quick reference chart so I do not have to do constant calculations everyday.

 

For example:

 

2000 ltr fish tank is at 20C. Creek outside is at 8C. How much creek water do I need to add to make the fish tank run at 15C? Bear in mind - I have to replace the 20C water with 8C - I cannot add more or the tank will overflow. How is the equation Q=mcT rewritten for this problem - or is there another equation? I will repeat the addition of cold water every few days as the temperature creeps back up again (as the pump adds heat), so will have to constantly make calculations depending on tank temp., and creek temp.

 

Any help would be greatfully received.

 

Thanks

[CaptainPanic]

I'm assuming that this is not some homework assignment (if it is, you're not learning anything from simply getting the answer).

 

Total volume = 2000

 

(Volume warm water) + (volume cold water) = (Total volume)

(Volume warm water) = (Total volume) - (volume cold water)

 

T_final = { (Volume cold water)*T_cold + (Volume warmwater)*T_warm } / { Total volume }

 

And combining two formulas:

 

T_final = { (Volume cold water)*T_cold + ( (Total volume) - (volume cold water) )*T_warm } / { Total volume }

 

And that last formula now has to be rearranged:

 

(Volume cold water) = (Total volume) * { (T_final - T_warm) / (T_cold - T_warm) }

 

If the creek is 8 degrees, and the warm water is 20, and you want to arrive at 15, then you need:

 

(Volume cold water) = 2000 * { (15 - 20) / (8 - 20) } = 2000 * {-5 / -12} = 2000 * {5 / 12} = 833.3 liter cold water

 

 

 

A better (long term) solution might be a thermostat and a pump. If the temperature gets above a certain value, the pump will start until the temperature drops below a certain value. (Make sure that whatever shop you get it understands that it's for cooling, not for heating).

[Bignose]

Aquarium chillers are commercially available. They are sold to people in very warm climates for marine tanks that have to maintain pretty tight temperature control for the inhabitants.

 

See http://www.fosterandsmithaquatics.com/fish-supplies/aquarium-water-temperature-chiller/ps/c/3578/4900 for one example.