PDA

View Full Version : Refrigeration nameplate tonnage capacity vs actual KW drawn







Ron
20-07-2016, 10:11 PM
I'm perplexed trying to understand the ton rating capacity vs the actual KW drawn by the chiller at our facility. The chiller unit is a Trane Model: CVHF 570. I believe this means the nominal capacity is 570 tons. This converts to ~ 2004 KW. Yet it typically draws around 230 amps when loaded up substantially. Since it is a 575 volt unit, the actual KW drawn are approximately 230 KW, which is nowhere near the nominal capacity of the chiller. I would love if someone could solve this discrepancy for me.

Brian_UK
20-07-2016, 10:44 PM
I think that you are trying to get two different items to produce the same result.

A ton of refrigeration refrs to the heat of fusion performed by the refrigerant during its cycle. This will be dependant on the equipment efficiency and the type of refrigerant.

The kW rating of the unit input is relating to the electrical motive input required to process the refrigerant cycle of the machine.

Scramjetman
26-08-2016, 02:49 PM
Machine tonnage is an imperial measure of the heat it can move. To translate, the formula is: tons of refrig x 3.517 = kW refrigeration. As Brian mentioned, kW refrigeration is a measure of heat. Electrical kW is measure of input electrical energy. So a machine can have 2 x kW figures - one that measures electrical input power and a second figure that is a measure of the heating or cooling done by the machine.

The ratio of the two figures is the COP (Coefficient of Performance) For your machine this is calculated as follows: 2004kW (refrig kW) / 230kW(input kW) = 8.7. For each kW of input power, we can get 8.7kW of cooling kW.

Greek_engineer
27-08-2016, 09:24 AM
I cannot understand that: 230 amps * 575 volt =132.2 kw input

Scramjetman
27-08-2016, 11:04 AM
I cannot understand that: 230 amps * 575 volt =132.2 kw input

For 3 phase power:
kWe = (Amps x 1000)/ (volts x power factor x √3)

If power factor is 1 then 230A reveals approx. 230kW. Power factor for chillers is generally less - anywhere between 0.85 and 0.95 which will translate to slightly higher electrical kW, say up to 270kW.

The formula Watts = Volts x Amps is for single phase resistance only circuits.

Magoo
28-08-2016, 01:28 AM
Kw e = Volts x Amps x 1.73 x Pf x motor efficiency / 1000.
the variables are Power factor and motor efficiency, good power factor 0.98 >0.99 and motor efficiency 0.8 > 0.9, power factor automatic correction systems can save huge amounts of energy costs.

water chiller capacities are generally based on specific conditions and flow rates,
Kw R = Mass flow rate x specific heat value x temp difference as delta T ' K / 1000
Kw R = L/ sec x 4.186 Kj/kg x TD 'C / 1000.
With refrig capacity based on tons, can be be confusing as flow rates could be IMP GPM or US GPM.

magoo

Scramjetman
29-08-2016, 12:48 PM
Damn Magoo! You passed me in thermodynamics equations. Now I've got to study Quantum theory or something to catch up! Bugger!

Magoo
30-08-2016, 04:10 AM
Hi Ron and Scramjetman.
all good fun. Generally North American water chiller manufacturers publish water flow rates in SI and Imperial, and delta P across each HX., which makes commissioning fairly straight forward, includes a nominal fouling factor.
Another thing to check apart from Pf and motor efficiency, is supply voltage balance between phases, total max imbalance is 10%, if higher get local power supply authority to re-balance site transformer, may require a tapping change, and with Pf balancing will save client huge amounts on operating costs.
Its all about efficiency these days and saving a dollar.
magoo

Brian_UK
30-08-2016, 04:12 PM
Shame the OP hasn't bothered to come back.

mrr
02-09-2016, 07:47 AM
My 2$...

1- Power consumption: If you measure the current drawn for each phase and you measure the phase-to-phase voltage ("line voltage"). Provided the system is balance and supposing an ideal motor efficiency:

P = 3/sqrt(3) * V * I * cos(phi) = sqrt(3) * V * I * power_factor, (W)

where P is the active power (W).

In your case, you made 230*575*sqrt(3)/1000 =229 kW, which would be the apparent power. You would have to multiply it by the power factor to get the active power.

2- Cooling capacity:

You "use" these 229 kW to "move" energy from a "cold" place to a "hot" place.

EER (Energy Efficiency Ratio) = Cooling capacity / power consumption

In other words, let's say EER = 2, that means for each 1 electric Joule your system consumes it moves 2 Joules from the "cold" place to the "hot" place.

For a detailed explanation of this you would need to read more about thermodynamics and vapor compression systems.

Hope it helps.