How to convert kVA from one voltage to another

[hohohohenry]

Hi there! Please go easy on me, I am a maintenance service technician, not an electrician.

I have a customer who has some robotic machinery that runs at 480 volts, but the service is 208 volts so each machine has a step up transformer attached to it. I would like to find the maximum kVA and Amperage for each machine from the perspective of the feeder panel (which is 208 volts)

For example, machine 1 has a nameplate which gives maximum values of kVA at 480v of 25kVA. First, am I right in calculating Amperage of about 17 Amps at 480 volts?

Next, how do I calculate the current flowing through the panel for this machine? Does the kVA rating of 25 hold true regardless of voltage? I know current will be more than 17 amps since the voltage is less. That darn 1.73 keeps hanging me up.

Help please!

 

[Jraef]

The reason they give you kVA is because it is going to remain constant regardless of incoming voltage. So assuming it is 3 phase, kVA = V x A x 1.732, so the current at 480V is going to be 25,000 / (480 x 1.732) = 30A

The 208V amperage will be higher by the voltage ratio; so 480 / 208 = 2.31, therefore the amps will be 30A x 2.31 = 69A. Double check it by the original kVA formula; 208 x 69 x 1.732 = 24,858 VA, within sloppy tolerance of 25kVA.

 

[chaterpilar]

if you know the kva of the load (whatever be the voltage) you can be assured that the supply side of the transformer will see the same kva.

Mind you transformer only transforms the voltage and not the kva, and hence your 208 volts panel will have a load of 69 amps @ 208 volts for each machine.

Cheers.

 

[Jraef]

The reason they give you kVA is because it is going to remain constant regardless of incoming voltage. So assuming it is 3 phase, kVA = V x A x 1.732, so the current at 480V is going to be 25,000 / (480 x 1.732) = 30A

The 208V amperage will be higher by the voltage ratio; so 480 / 208 = 2.31, therefore the amps will be 30A x 2.31 = 69A. Double check it by the original kVA formula; 208 x 69 x 1.732 = 24,858 VA, within sloppy tolerance of 25kVA.

As was pointed out to me, even though my math works out correctly, I made a slight blunder in depicting the formula and need to clarify.

"3 phase, kVA = V x A x 1.732"
Should read either;
3 phase, kVA = (V x A x 1.732) / 1000 or
3 phase, VA = V x A x 1.732

Also just to clarify, V = line-to-line, A = per phase

 

[hohohohenry]

Ok thanks guys or gals I made myself a little Excel spreadsheet to keep track of all the machines (there are 10 of them, with different kVA ratings) and I put in your formula to figure out Amperage at 208v for all of the machines

Am I correct in understanding that to translate that to kW, I would need to know the power factor of the machines (which I do not)

Finally, I heard about 'in rush' current when a motor turns on. These machines have motor drives. I was wondering if there is inrush current constantly while the machine is running, or each and every time one of the motors turns on--which could be every few seconds?

Basically, the customer needs to purchase 2 or 3 new machines, and we are trying to figure out if they have enough electrical service, and if there is enough 'headroom' to account for this inrush..??

Thanks!

 

[tom_gonzalez@verizon.net]

Inrush Current

Inrush Current

As the name implies, inrush current occurs when the motor starts. The current spike reaches about 3-5 times the FLA rating. It is only spiked for fractions of a second unless the motor is starting under a heavy load. The inrush is caused by the rotor not turning and supplying any CEMF which limits the current flow in a running motor..

 

[Jraef]

Ok thanks guys or gals I made myself a little Excel spreadsheet to keep track of all the machines (there are 10 of them, with different kVA ratings) and I put in your formula to figure out Amperage at 208v for all of the machines

Am I correct in understanding that to translate that to kW, I would need to know the power factor of the machines (which I do not)

Correct, but for the purposes of determining circuit size, I would just use a .8 on systems such as that with drives involved. Technically, drives can preset a .95 pf to the line, but that is if you measure displacement pf only. Drives also create what is called "distortion" pf which, although much lower than displacement values, does warrant a little consideration when discussing circuit capacity. The .8 pf will be plenty of "fudge".

Finally, I heard about 'in rush' current when a motor turns on. These machines have motor drives. I was wondering if there is inrush current constantly while the machine is running, or each and every time one of the motors turns on--which could be every few seconds?

If EVERY motor load on the machine is fed through drives of one sort or another, you do not need to worry about starting current. One of the benefits of drives, be they AC VFDs, DC servos or steppers, is that the motor power flow is completely controlled by the drive, so the AC source is nothing more than the "raw material" it needs. The drive's current draw will end up being pretty much linear to motor loading, in other words, no "inrush" (which technically means something else, but I know what you meant to say).

Basically, the customer needs to purchase 2 or 3 new machines, and we are trying to figure out if they have enough electrical service, and if there is enough 'headroom' to account for this inrush..??
Thanks!

Unless there is something else you have not mentioned, you should not need to be overly concerned with starting current. Just slap the .8pf into your equations and you should be good to go with plenty of headroom.