Calculating three phase power usage.

[Chris1234]

I’ve been asked to calculate power usage of a machine to works out how much it costs to run.

Current draw is:

L1: 9.71
L2: 9.61
L3: 11.14

N: 7.78

Obviously when designing a circuit you would base it in the highest phase ampage.
But is power usage worked out the same way? Or are three phases added together because of the unbalanced load?

KVA= 11.14 x 400 x 1.732 x / 1000 x unit price

Or is it:

KVA= 3x 10.15 x 400 x 1.732 / 1000 x unit price

Never had to do this before!
Thanks.

 

[Pete999]

I’ve been asked to calculate power usage of a machine to works out how much it costs to run.

Current draw is:

L1: 9.71
L2: 9.61
L3: 11.14

N: 7.78

Obviously when designing a circuit you would base it in the highest phase ampage.
But is power usage worked out the same way? Or are three phases added together because of the unbalanced load?

KVA= 11.14 x 400 x 1.732 x / 1000 x unit price

Or is it:

KVA= 3x 10.15 x 400 x 1.732 / 1000 x unit price

Never had to do this before!
Thanks.

I’ve been asked to calculate power usage of a machine to works out how much it costs to run.

Current draw is:

L1: 9.71
L2: 9.61
L3: 11.14

N: 7.78

Obviously when designing a circuit you would base it in the highest phase ampage.
But is power usage worked out the same way? Or are three phases added together because of the unbalanced load?

KVA= 11.14 x 400 x 1.732 x / 1000 x unit price

Or is it:

KVA= 3x 10.15 x 400 x 1.732 / 1000 x unit price

Never had to do this before!
Thanks.

https://www.dataforth.com/catalog/pdf/an110.pdf

I’ve been asked to calculate power usage of a machine to works out how much it costs to run.

Current draw is:

L1: 9.71
L2: 9.61
L3: 11.14

N: 7.78

Obviously when designing a circuit you would base it in the highest phase ampage.
But is power usage worked out the same way? Or are three phases added together because of the unbalanced load?

KVA= 11.14 x 400 x 1.732 x / 1000 x unit price

Or is it:

KVA= 3x 10.15 x 400 x 1.732 / 1000 x unit price

Never had to do this before!
Thanks.

https://www.dataforth.com/catalog/pdf/an110.pdf

 

[Lucien Nunes]

The power consumption is the total of the three phases - it doesn't matter whether it's balanced or not. So:
P=(9.71+9.61+11.14) x 400/1.73=7.0kVA
This is the apparent power in kVA; you would need to know (or guess) the power factor in order to obtain the real power in kW, from which to calculate the running cost.

Worst case, at unity pf, the machine uses 7.0kW and therefore 7.0 x unit price per hour. If the load is mainly heating this will be close. If it is mainly induction motor load, the pf might be more like 0.8, in which case the meter will register only 0.8 x 7.0=5.6kW. Not knowing the machine, it looks to me from the fact that two lines have similar current, that the main 3-phase load is around 9.7A, to which 1.5A of single-phase control system is added on one phase. In which case, go with 0.8 or 1.0 according to whether that main load is motors or heaters, or hedge your bets and call it 0.9 giving a real power of 6.3kW.

 

[static zap]

Do you know how you are billed ,is there any mention of Peak Demand ?
Adding new equipment could influence this in an industrial setting.

 

[davesparks]

What is the nature of the load? That neutral current seems a bit odd for a ‘normal’ load.

 

[Lucien Nunes]

Good point, I think I read that as 1.78A by mistake. Looks like there's significant harmonic distortion as it's high relative to the difference between the lines.