ECN Forum
Posted By: WARREN1 Chiller KW - 02/20/02 08:30 PM
Is there a way to convert the chiller rating in KW to an electrical equivalent?
I have seen some ways to do it and should have it written (someplace where I wouldn't forget), but can't locate it. Maybe someone out here has some good experience.
Thanks
Posted By: Redsy Re: Chiller KW - 02/21/02 12:22 PM
Aren't chillers were usually rated in either tons of refrigeration or BTU/Hr.
1 Kilowatt = 341 BTU/Hr. 1 BTU/Hr.= 0.3Watt
1 ton = 12,000 BTU/Hr.
I thought KW = KW regardless of the type of system.
Posted By: Scott35 Re: Chiller KW - 02/22/02 10:05 AM
Warren1,

If you are looking for Horsepower [HP] from the KW rating, try the simple calculation of:

  • 746 Watts = 1 Horsepower


Take the KW rating, convert it into simple Watts by moving the decimal point 3 places to the right [or multiply by 1000], then divide that number by 746. The result is the HP.

Example:

HP = [KW*1000]/746
2.984 KW = 4 HP

2.984 * 1000 = 2984
2984 / 746 = 4

If you already knew this junk, please ignore the above calcs.

Now, if your looking for KVA, from which you derive load calcs, Panel Schedule calcs and an Ampere Rating figure, that will be really tough to pull off with only a KW rating.

One must know the KVARs included with the drawn KWs, in order to find the total line KVA load.
The Reactive Power [KVARs] will vary from Machine to Machine, resulting in various KVA ratings.
In this case, you will need documentation - either from Manufacturer's Manuals, or from a Nameplate's ratings.

A good example of 2 different KVA per KW ratings I have run across recently reflect separate 400 watt High Pressure Sodium fixtures [HID lighting].

1: Most commonly used, run-of-the-mill HPS Ballast kits for the 400 watt lamp are the typical "Quadri-Volt" Ballast [per Advance Transformer Co.], which is a 4 Input Voltage type Autotransformer [CWA to be exact], and use a 55 MFD [Micro Farad] Capacitor to bring the Power Factor from Nominal - or +/- 50%, to High Power Factor - or +/- 90%.

This CWA setup draws 480 VA [0.480 KVA]. The lamp is driven with around 400 Watts, and the Lamp / Ballast combination at High PF draws an additional 80 VARs[ Volt-Amps Reactive] - making the complete Volt-Amp load on the circuit 480 VA.

2: Due to some incorrect ordering, 3 fixtures with CWI Configs were used to make up the slack.
These are to drive the same S51 lamp, but the Line VA load is now 465 VA. The PF correction Capacitor is 48 MFD.
This results in only 65 VARs while driving the lamp at aproximately 400 Watts.

As you can see, there's really no way to figure KVA from KWs, when dealing with Reactive Equipment.

If something in the fixture dissipates additional heat, this will be an increased Wattage draw, but not more VARs - so the Line KVA increases to reflect the additional Wattage drawn from the supply.

If the loads were Linear [nearly 100% Resistive, or very-very-very minimal Reactive Power], then you could apply KW's directly to HP AND to figure load Amperes.

Let me know if this was in the direction you were going with your Q [AKA does this help??].

Scott SET
Posted By: wolfdog Re: Chiller KW - 02/23/02 03:59 AM
If you are looking for capacity vs. KW, it varies depending on compressor type,suction temperature,and air cooled or water cooled.
Figures fall roughly between .6 and 1 ton per HP. I can be more specific if necessary.

[This message has been edited by wolfdog (edited 02-22-2002).]
Posted By: WARREN1 Re: Chiller KW - 02/25/02 03:17 PM
Thanks to all for your responses.
Actually, here is the problem. In doing design, the electrical group has to know at least some approximate loads for a building and the process before we can begin to size the services into the building/plant. Usually the HVAC can give estimated tonnage for the process and building cooling. They won't (until they consult with the local Trane rep.) give anything more. So for planning purposes, we the electrical group must make some conversion from tons to KW, KVA, HP or Amps. Someone before me has calculated that an 800 ton unit is 492KW. Now I must verify that information.
If a 0.6 factor is used, the answer is 480KW.

Wolfdog, can you be more specific? Thanks.
© ECN Electrical Forums