# Thread: Chiller cooling output calculation in KW

1. Regular Guest
Join Date
Apr 2009
Posts
9
Post Likes

## Chiller cooling output calculation in KW

Dear all,

I have water-cooled chiller, I need to calculate its cooling output in KW. I have chilled water flow in GPM and supply & return temperature in C.

2. Professional Member
Join Date
Apr 2005
Location
Split, Croatia
Posts
37
Post Likes
You need to have very precise thermometer to calculate correctly cooling output.
Since dT is usually around 5K your imprecise temperature reading could lead to 30% (or even more) error.

Q=m*c(T2-T1)

m=mass of water
c=specific heat capacity of water

3. You can estimate cooling output in British Thermal Units per Hour (BTUH) via:

500 x GPM x deltaT = BTUH

Keep your units consistent. If you know the Gallons Per Minute, use Fahrenheit for the temperature scale.

Once you determine the BTUH, divide that number by 12,000 to get tonnage.

To get kilowatts consumed per ton, you measure what the machine is using in kilowatts to deliver the measured tonnage.

Example: process chiller flowing about 50 GPM with a 10 degree split between return and supply water...

500 x 50 x 10 = 250,000 BTUH

250,000/12,000 = 20 tons

Assume machine is drawing 12 kilowatts to achieve this output:

12 kilowatts/20 tons = .6 kilowatts per ton

Remember that there are 3.413 BTUs per watt, 3413 BTUs per kilowatt. So, on a one to one basis, you would need about 3.515 kilowatts per ton (12,000 BTU), but we can achieve a ton of refrigeration with less kilowatts per ton consumed than that, by far.

4. Professional Member
Join Date
Feb 2006
Posts
226
Post Likes
The formula for KW of refrigeration is L/s x Delta T(*C) x 4.19kj/kg (for Water)
To convert GPM to L/s GPM (div by) 60= GPS
GPS x 3.785= L/s
60 seconds in a minute
3.785 litres in a US Gallon
I have it as 3.415 BTU/ Watt

5. Regular Guest
Join Date
Apr 2009
Posts
9
Post Likes
Thanks to all of you...

6. New Guest
Join Date
Jul 2009
Posts
3
Post Likes

## Clarification...

Originally Posted by shophound
You can estimate cooling output in British Thermal Units per Hour (BTUH) via:

500 x GPM x deltaT = BTUH

Keep your units consistent. If you know the Gallons Per Minute, use Fahrenheit for the temperature scale.

Once you determine the BTUH, divide that number by 12,000 to get tonnage.

To get kilowatts consumed per ton, you measure what the machine is using in kilowatts to deliver the measured tonnage.

Example: process chiller flowing about 50 GPM with a 10 degree split between return and supply water...

500 x 50 x 10 = 250,000 BTUH

250,000/12,000 = 20 tons

Assume machine is drawing 12 kilowatts to achieve this output:

12 kilowatts/20 tons = .6 kilowatts per ton

Remember that there are 3.413 BTUs per watt, 3413 BTUs per kilowatt. So, on a one to one basis, you would need about 3.515 kilowatts per ton (12,000 BTU), but we can achieve a ton of refrigeration with less kilowatts per ton consumed than that, by far.
Dear Shophound

Good calculations. Please clarify what is the '500' used for? I mean what does it denote?
regards
RK

7. Professional Member
Join Date
Nov 2006
Location
North Carolina
Posts
136
Post Likes
500 is a constant

8. New Guest
Join Date
Jul 2009
Posts
3
Post Likes
500 is a constant
I know it is a constant. What I want to know is that how is it derived? it is just not a number which is assumed.

9. Professional Member
Join Date
Feb 2006
Posts
226
Post Likes
500 is a constant for water. The reason I believe is, you are dealing in GPM(minutes) and BTU/HR (hours)
60 Min. in an hour and the weight of 1 US gallon of water 8.31 lbs.
60 x 8.31 = 498.6
rounded to 500.
Process system with glycol depends on weight of 1 gall of dilluted glycol.

10. New Guest
Join Date
Jul 2009
Posts
3
Post Likes
Originally Posted by bertoh
500 is a constant for water. The reason I believe is, you are dealing in GPM(minutes) and BTU/HR (hours)
60 Min. in an hour and the weight of 1 US gallon of water 8.31 lbs.
60 x 8.31 = 498.6
rounded to 500.
Process system with glycol depends on weight of 1 gall of dilluted glycol.
Thanks Bertoh, I got my answer.

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•

## Related Forums

The place where Electrical professionals meet.