View Full Version : Chiller's kW/ton... How to calculate?

ksthen

16-05-2012, 02:33 AM

Hi all,

I have here evaporator's flow = 818.3 cubic meter/hr

chilled water in = 52.9 F

chilled water out = 46.5 F

Chiller running kilowatts (from switchboard) = 950 kW or running amps = 96

How do I calculate my chiller's performance (kW/ton). Preferably in unit SI:confused:

sterl

16-05-2012, 04:13 PM

I don't believe your numbers. Or I need to buy a whole lot of chillers just like yours.....

COP in consistent units is (KW extracted/KW Input) and high numbers represent good performance.

KW/Ton is the inverse: It is the input divided by the output, so low numbers are good. But the units don't quite work out if that Ton is TR as in the effect of melting one ton of ice.

A cubic meter of water weighs 1000 Kg. So your mass flow is 818000 kg/hr. The sensible heat of water at these temperatures is 4.18 kJ/kG-Deg C. 52.9 Deg F. is equiv of 11.6 Deg. C. 46.5 Deg F is equiv of 8 deg C.

A Kilowatt is a Kilojoule per Second. So the Arithmetic is:

Q= Mdot Cp Delta-T

=818000*4.18*(11.6-8)

=12167298 Kilojoules per hour....

so

To bring it to KW: Divide by 3600 so its Kilojoules per Second

=3377 KW or about 960 TR

That gives you a COP of about 36 which is far too high. Sure you are not looking at the power input to the Chilled Water Pumps? Are those temperatures being produced by a cooling tower without the compressor operating?

Magoo

17-05-2012, 03:53 AM

Your running amps at 96, and 950 Kw drawn power, what is the voltage supply to drive motor.

Using sterl's calcs the COP would be 3.6 not 36, probably a typo error,anyway way to high.

ksthen

17-05-2012, 10:46 AM

Voltage supply = 6.6kV

centrifs

17-05-2012, 09:30 PM

I calc COP at 3.16 and kw/ton at 1.10. (if I'm correct) that machine is not running too well. The YK chillers are rated at about 5 COP>. Elec kw/hr 1081 at .97 PF, Refrig kw/hr 3428

ksthen

18-05-2012, 10:15 AM

I don't believe your numbers. Or I need to buy a whole lot of chillers just like yours.....

COP in consistent units is (KW extracted/KW Input) and high numbers represent good performance.

KW/Ton is the inverse: It is the input divided by the output, so low numbers are good. But the units don't quite work out if that Ton is TR as in the effect of melting one ton of ice.

A cubic meter of water weighs 1000 Kg. So your mass flow is 818000 kg/hr. The sensible heat of water at these temperatures is 4.18 kJ/kG-Deg C. 52.9 Deg F. is equiv of 11.6 Deg. C. 46.5 Deg F is equiv of 8 deg C.

A Kilowatt is a Kilojoule per Second. So the Arithmetic is:

Q= Mdot Cp Delta-T

=818000*4.18*(11.6-8)

=12167298 Kilojoules per hour....

so

To bring it to KW: Divide by 3600 so its Kilojoules per Second

=3377 KW or about 960 TR

That gives you a COP of about 36 which is far too high. Sure you are not looking at the power input to the Chilled Water Pumps? Are those temperatures being produced by a cooling tower without the compressor operating?

Sorry, how actually to convert kW to TR?

chemi-cool

18-05-2012, 10:38 AM

As simple as 123, sterl gave you the right way, if you look at the bottom, you will see that 3377KW = 960 TonR, now you divide 3377 by 960 = 3.51.

So, 3.51 Kw=1 TonR

sterl

21-05-2012, 07:50 PM

MY BAD

Somebody caught the error: That COP should be 3.6 way up there. I mis-read the KW from the original post as 96....And that input for 3377 KW of output would be an amazing machine.

So in input over output terms: its just less than 1-KW/TR.

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