cooling capacity calculation.

[gaber2611]

Hello everybody,

Please i need your help regarding a project i am planning to execute.

Project description:

Cooling a 5000l water tank, made of fiberglass walls, supported on four stands, exposed to sun, mounted on a building roof, drains a 1000l/day, required to cool water from 55 to 25 celsuise degrees, during the day while at night will be no-load, the volume of the tank is 5 cubic meters, cylindrical shape.

any ideas about how to calculate both of cooling capacity and/or electric power consumption?, i googled enough, but didn't help, so hopefully someone can help here.

Thanks in advance.

[gaber2611]

Any kind of help?.

[Tesla]

Hi gaber
If you search this forum you will find the formula you need to calculate the heat load of the water Q=MxSHCxDT. Then we can move on from there.

[Magoo]

With 1000 litre water usage per day, use a smaller tank. Insulate exterior as well as most of load will be solar and transmission heat gains. The smaller the tank the less heat gain, lower power running costs.

[Yuri B.]

And start cooling it at nights at cheaper electricity rates, at cooler ambient temperatue, and, to, maybe. slightly below than 25 degree C.

[Sandro Baptista]

Hi Gaber,

Please explain better how the process runs?

Have you a flow consumption of about 1 m3 per ...lets say 8 hours? And you need the water at +25ºC and returning to the tank at +55ºC?

During the night there is no water consumption and you only have to maintain the water at +25ºC?

Regards