Flowing Water Cooling Rate and System Capacity

[TRIVEDIGB]

I have to provide refrigeration system to cool flowing water (@1.6 GPM) from 90F to 46F in short time. I am wondering how to calculate the time it will take and the capacity of the unit I will need.

Thanks.

[Magoo]

Hi
A little bit more information could help. The statement "short time is not that exact ".

[TRIVEDIGB]

Thanks Magoo.

It is 25 minutes or less when ambient is 72F and 80 Minutes or less when ambient is 90F.

[Tesla]

Hi TRIVEDIGB
If you care to search this forum you will find detailed calculation for what you have asked.

[TRIVEDIGB]

Tesla:

I was not able to find.

[nike123]

Let see if I could give you some guidance on calculation.
1,6 GPM in 25 min mean 1,6*25=40 gallons of water. That is around 180 kg of water.
1 kg of water at 90°F (32,2°C) has specific enthalpy of 134 kj/kg
1 kg of water at 46 (7,7) has specific enthalpy of 33 kj/kg
To cool down 1 kg od water from 90°F to 46°F you need to remove 133-33=101 kj of heat.
To cool down 180kg of water you need to remove 180*101= 18180 kj of heat which is around 5kWh of heat.
To remove that heat in 25 min. time, you need cooling capacity of 60/25*5 = 12kW or 3,4TR
Now you only need to ad to that heat gain in pipes and vessels where water circulate. If they are well insulated, 10% added capacity will probably do the job.
Is this for cooling milk? Some other colleagues could add their practical advices from real life.

[TRIVEDIGB]

Thanks a lot Nike123. Appreciated.

[D.D.KORANNE]

I guess it is UK gallon 1.6 / min


BTU / HR
USGPM = -------------------------
500 X ( 90-46 )

1.9 X 500X 44 = 41800 BTU/ HR = 3.48 TR ................ WATER COOLING LOAD