kW and kVA Calculations

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Jody Boehs

Jody Boehs

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Location
Fairview, Oklahoma, USA
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Controls and Electrical Manager
Hello,

I have a 600 kW Caterpillar GenSet with a 600 kVA UPS system that carry the load until the Generator is spooled up and the load is transferred with an ATS.

The kVA and kW ratings is a bit confusing to me. Below are some calculations for you review. We are needing to add 250 amps of current to our system and I am trying to decide if our generator and UPS is capable of handling the load.

We are currently running at approximately 286 kW which is 344 kVA. This is 60% of the 600 kVA UPS capacity.

Here is what I've come up with.

kW = √3 x PF x I x V / 1000
kW = 1.73 x 0.8 x 250 x 480 / 1000
kW = 166.08

Now to covert that to kVA…

kVA = kW / PF
kVA = 166.08 / 0.8
kVA = 207.6

Adding 207.6 kVA to this UPS that is already running approximately 344 kVA, that equals 551 kVA. Will that be overloading the UPS and/or the GenSet?

Any thoughts would be appreciated!
 
KVA doesn't necessarily "add" conventionally. You can add it conventionally for a conservative total, but to understand how it really accumulates, you'd need to determine the kVA vector of each load, and add them up as vectors. If they all have a the same power factor and then same sign of kVAR, then they do add up conventionally.

The kW is always the same sign as long as everything is a load, as opposed to a mix of loads and sources. However, the kVAR component can cancel out among loads, if there is a mix of inductive and capacitative loads. Some loads may have a leading power factor (capacitative), while other loads may have a lagging power factor (inductive). Some loads may have unity power factor, where all of the kVA is kW.

What kVA is, is the product of the root-mean-square values of both the voltage and current waveforms. That is, you take a special kind of average of both the voltage and current waveforms separately, and multiply the "special kind of average" of each waveform together. Real kW power, is what you get when you take instantaneous values of voltage and current, multiply them together, and then average over the entire cycle. Voltage and current are not necessarily synchronized, otherwise kW would equal kVA. Since both voltage and current individually affect the sizing of power conditioning equipment, it is important to know the kVA value for this purpose.
 
Another thing to consider is whether there are peak currents that can exceed your calculated values when starting relatively large motors, etc. If so, can the UPS handle them and how much of a dip in voltage will occur due to the impedance of the generator. Also, can the loads tolerate such a voltage dip.
 
So how do I find the kVA values for each piece of equipment to be added? Can that be figured off of the nameplate information? This 250 amps of load consists of 7 different pieces of equipment.

2 Holding Tanks that are heated. 26 Amps each = 52 total
2 Mixing Tanks that are also heated. 14 Amps each = 28 total
1 Enrober Machine that has a heated holding tank inside it. 56 Amps
1 Cooling Tunnel. 28 Amps
1 Glycol Chiller. 90 Amps

This actually totals 254 Amps.
 
Are all of those 3-phase loads with balanced line currents? Just checking, because that could make a difference if it was otherwise.
 
First things first: If you know the amps and the system voltage, then you know the kVA being added. The added kW will be lower then this.

kVA = √3 x I x V / 1000

You can think of kVA as simply another way of describing the amp load or capacity of a system. That 600 kVA UPS can supply 722A, doesn't matter what the PF is; you are limited to 722A.

In your formulas you are assuming a power factor of 0.8; this may or may not be correct. Resistive heaters will have a PF closer to 1; motors will be anywhere but probably 0.6 to 0.8, depending on lots of factors.

So the first thing you need to check is the statement 'We are currently running at approximately 286 kW which is 344 kVA '. Find out the actual amp loading on the system. That 286 kW might be anywhere from 318 to 480 kVA.

As Carultch says, simply adding your amp loads up doesn't quite work, since an accurate result requires vector math and consideration of harmonics; but simply adding up the amp loads will get you in the ballpark.

-Jon
 
Jody,

Consider all the ratings of the generator and UPS.

If this is an older UPS, the 600kVA rating will be associated with a kW rating less than 600.

Like the generator, you can't exceed the kVA or kW rating.

The generator is rated at 600kW and 750kVA. An older UPS may have ratings as separate, like 600kVA and 480kW.

The 480kW, if it is an older UPS, may be your bottleneck.
 
ron said:
Jody,

Consider all the ratings of the generator and UPS.

If this is an older UPS, the 600kVA rating will be associated with a kW rating less than 600.

Like the generator, you can't exceed the kVA or kW rating.

The generator is rated at 600kW and 750kVA. An older UPS may have ratings as separate, like 600kVA and 480kW.

The 480kW, if it is an older UPS, may be your bottleneck.
Click to expand...
Well put. Thank you.
 
winnie said:
First things first: If you know the amps and the system voltage, then you know the kVA being added. The added kW will be lower then this.

kVA = √3 x I x V / 1000

You can think of kVA as simply another way of describing the amp load or capacity of a system. That 600 kVA UPS can supply 722A, doesn't matter what the PF is; you are limited to 722A.

In your formulas you are assuming a power factor of 0.8; this may or may not be correct. Resistive heaters will have a PF closer to 1; motors will be anywhere but probably 0.6 to 0.8, depending on lots of factors.

So the first thing you need to check is the statement 'We are currently running at approximately 286 kW which is 344 kVA '. Find out the actual amp loading on the system. That 286 kW might be anywhere from 318 to 480 kVA.

As Carultch says, simply adding your amp loads up doesn't quite work, since an accurate result requires vector math and consideration of harmonics; but simply adding up the amp loads will get you in the ballpark.

-Jon
Click to expand...
So you are telling me that kVA is the number I should be focused on? I have always thought that kW is the true power and that's what I needed to figure off of. Forgive my ignorance here!
 
ron said:
Jody,

Consider all the ratings of the generator and UPS.

If this is an older UPS, the 600kVA rating will be associated with a kW rating less than 600.

Like the generator, you can't exceed the kVA or kW rating.

The generator is rated at 600kW and 750kVA. An older UPS may have ratings as separate, like 600kVA and 480kW.

The 480kW, if it is an older UPS, may be your bottleneck.
Click to expand...
This UPS was installed in 2015 so i'm not sure if that is considered "older" or not. But, you are saying exactly what my UPS people are telling me... "It is a 600 kVA UPS which is 480 kW. The 600 kW generator is sized to handle the 480 kW and then uses the additional 120 kW to recharge the UPS if it is already under a load. The UPS is definitely my bottleneck it sounds like!
 
Jody Boehs said:
So you are telling me that kVA is the number I should be focused on? I have always thought that kW is the true power and that's what I needed to figure off of. Forgive my ignorance here!
Click to expand...

You have multiple things that you need to be focused on.

kW is the actual power consumed by the load. It is limited by things such as the power rating of the generator engine. No matter what you do, you cannot exceed the kW of your power source or something has to give.

kVA is simply the supply voltage multiplied by the supply current. It is a good way of expressing the current handling limitations of your system.

You have to be below both your kW limits _and_ your kVA limits.

You cannot simply assume that kVA is kW/0.8. That is why I suggested that you actually measure the load current or load kVA, rather than stating 'We are currently running at approximately 286 kW which is 344 kVA'

-Jon
 
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