Question Regarding 3-Phase Calcuation

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HI All,

For some reason I've been unable to get a definitive answer about this simple question:

So when I have a generator operating in 3-Phase and I use the formula KW x 1000/E x PF x 1.73 to get amps, the answer would be per phase or leg, correct?

For Example: 40 KW x 1000 / 480v x .8pf x 1.73 = 40,000 / 664.32 = 60 Amps @ 480 volts

This means that ALL 3 phases would be capable of 60 amps at 480, correct?

I'm pretty sure that's the case but "pretty sure" doesn't cut it when your dealing with electricity. That'd be akin to telling a pilot you're "pretty sure" it's OK if they land in about ten minutes on our super-busy runway. Ha!:happysad:

Any answers would be appreciated. Even the Snarky ones...so long as they are correct.

Many Thanks,
Smatthew
 

GoldDigger

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HI All,

For some reason I've been unable to get a definitive answer about this simple question:

So when I have a generator operating in 3-Phase and I use the formula KW x 1000/E x PF x 1.73 to get amps, the answer would be per phase or leg, correct?

For Example: 40 KW x 1000 / 480v x .8pf x 1.73 = 40,000 / 664.32 = 60 Amps @ 480 volts

This means that ALL 3 phases would be capable of 60 amps at 480, correct?

I'm pretty sure that's the case but "pretty sure" doesn't cut it when your dealing with electricity. That'd be akin to telling a pilot you're "pretty sure" it's OK if they land in about ten minutes on our super-busy runway. Ha!:happysad:

Any answers would be appreciated. Even the Snarky ones...so long as they are correct.

Many Thanks,
Smatthew
In a three phase situation where there is a neutral wye point, there are two sets of voltages and currents to be considered in calculations:

The first set is the line to line voltage and the line-to-line (phase) current.
The second set is the line to neutral voltage and the line current.

The simplest calculation to understand is based on the fact that for a balanced load with unity power factor the power is determined only by the voltage and current and does not depend on whether the load is a wye or a delta load. For a wye just multiply the line current times the line to neutral voltage (for one leg of the wye) and then multiply by 3. Since the line-to-line voltage, VLL is sqrt(3) times the line to neutral voltage, VLN, this power formula is equivalent to VLL x IL x sqrt(3), since 3 = sqrt(3) x sqrt(3)

The more difficult calculation looks at how the line current relates to the line-to-line (phase) current. Vector math shows that the sum of the two line-to-line currents, ILL, at one point of the delta, which constitutes the line current in the common line (ILN or just IL, is ILL x sqrt(3) so the power, which is 3 x VLL x ILL can also be written as sqrt(3) x VLL x IL. Same formula....

So in the calculation you gave the current will be the line current, not the phase current. This distinction is important when you are given the specified load current of a line-line load. You have to remember that this is not the same as the line current.
 

charlie b

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. . . the answer would be per phase or leg, correct?
Yes. However . . . ,

. . . it would serve your interests to refrain from speaking in terms of "per phase" or "per leg." These phrases do not accurately describe what is happening. They can also lead a person to come to the incorrect conclusion that if 60 amps is the "per phase value," then the "total amps" is 3 times that, or 180 amps. The simple truth is that it is the same current that is flowing on each phase. For example, when Phase A is at its peak positive value, current leaves the source on Phase A, splits into two separate conductors at the load, and returns to the source with some current on Phase B and the rest on Phase C. Here again, it is the same current. Perhaps the peak current on Phase A is 60, and the peak current on Phase B is 60, and the peak current on Phase C is 60, but those three peaks do not take place at the same time. So it is nonsense to speak of a total current of 180.

 

OldSparks

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Location
Vacaville CA USA
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Retired: Electrician, Submarine Electronics (21 years), Potable water system maintenance boss (21 years).
Yes. However . . . ,

. . . it would serve your interests to refrain from speaking in terms of "per phase" or "per leg." These phrases do not accurately describe what is happening. They can also lead a person to come to the incorrect conclusion that if 60 amps is the "per phase value," then the "total amps" is 3 times that, or 180 amps. The simple truth is that it is the same current that is flowing on each phase. For example, when Phase A is at its peak positive value, current leaves the source on Phase A, splits into two separate conductors at the load, and returns to the source with some current on Phase B and the rest on Phase C. Here again, it is the same current. Perhaps the peak current on Phase A is 60, and the peak current on Phase B is 60, and the peak current on Phase C is 60, but those three peaks do not take place at the same time. So it is nonsense to speak of a total current of 180.


Excellent response!
 

Shaneyj

Senior Member
Location
Katy, Texas
Occupation
Project Engineer
Yes. However . . . ,

. . . it would serve your interests to refrain from speaking in terms of "per phase" or "per leg." These phrases do not accurately describe what is happening. They can also lead a person to come to the incorrect conclusion that if 60 amps is the "per phase value," then the "total amps" is 3 times that, or 180 amps. The simple truth is that it is the same current that is flowing on each phase. For example, when Phase A is at its peak positive value, current leaves the source on Phase A, splits into two separate conductors at the load, and returns to the source with some current on Phase B and the rest on Phase C. Here again, it is the same current. Perhaps the peak current on Phase A is 60, and the peak current on Phase B is 60, and the peak current on Phase C is 60, but those three peaks do not take place at the same time. So it is nonsense to speak of a total current of 180.

Interesting this topic comes up. Just yesterday I was researching on this topic and found one of your posts, Charlie b, from 2008.
You came up with a great analogy.
I'll call it the parable of the dollar.
Do you recall this?
I found it quite clever.
In that particular thread, IIRC, you had 2 posts, that together comprised a fairly in depth response in a simple to understand format.
I'll link to it later when I get to my laptop.

Sent from my SM-G965U using Tapatalk
 
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