Cooling Water Requirement

How ever did you get this job?

A rule of thumb is 3 gpm/TR, or 1950 gpm in this case. This is for non-evaporative water cooling, in which case the water will warm up by ~10°F.

For evaporative water cooling, about 19.5 gpm of water would be evaporated. If bleed-off water is made equal to this, then 39 gpm of water would be needed.

The cooling water is continuous when the chiller is running, but can be stopped when the chiller is off.

For part load operation, the cooling water can be decreased proportionately.

From the peanut gallery, what is TR?

TR = tons of refrigeration (one ton being 12,000 Btu/h.)

Ahh--thanks!

....and then of course he might be bleeding water to rid the tank of build up. Or, he might have an injection water treatment system to keep all those minerals in suspension. In either case he sounds like a smart guy and he'll account for these in his calculations.

Go back and look at the OP's past questions. He's in totally over his head.

I was joking

Thanks, I feel better now.

This depend on two things (1) cooling water inlet & outlet temp and (2) power consumption to produce 650 TR refrigeration.

Assuming COP approximately 4 which will give power consumption, you may require about 250 cu.m/hour cooling water if increase in temperature is 10 degC. If increase is less you require more cooling water in same inverse proportion. Basically you have to remove heat from condenser which is equivalent to heat added at evaporator and compressor both (see PH chart).

I want to correct myself in certain assumptions. It is not known from OP about the type of chiller or refrigerant. I assume it as R134a centrifugal chiller for which COP is above 5. Cooling water temperature rise in condenser can be taken maximum 5 deg C for two pass exchanger and not 10 deg C as earlier. So cooling water requirement taking 10% allowance can be calculated as follows.

Heat added at evaporator = 650 x 3000 kCcal/h

Heat added at compressor= (650/5) x 3000 kCal/h

Total Heat added = 2,340,000 kCal/h

Cooling water required = 2,340,000/5 kg/h (for 5 degC temp. rise)

= 515 cu.m/h