## Explain 3 phase generator power

Paul Steichen Member
Join Date: Feb 2014
Posts: 15
Explain 3 phase generator power

I have a 3-phase generator that I want to understand better. I know that I can draw 120V single phase off of any of the 3 legs (yeah, I know, balanced loads). But how many Amps? The generator is rated at 40KW. Would I be able to draw 40K ÷ 120 = 333A? (W ÷ V = A)

Or if supplying 220V equipment from two out of phase legs (208V differential, once again balanced across the 3 legs), would I say I can supply 40K ÷ 208 = 192A?

Or how would I even begin to describe the Amperage when supplying 3 phases? Or the voltage for that matter? Would I say it supplies 120V 3-phase power?

Really would like to understand this thing better. I use it all the time on the farm, but sound like an idiot when I try to talk about it!

AllanJ Member
Join Date: Mar 2010
Location: USA
Posts: 4,422
The generator is described as a 40 KW 120/208V three phase wye generator. (abbreviated if you wish to 3, a zero with a slash through it, and the letter Y). It supplies 120 volts single phase, 208 volts single phase, and 208 volts 3 phase.

Assuming that the 40KW rating is continuous, yes you can draw 192 amps divided among 208 volt (some 240 volt equipment will also work) single phase equipment connected to any two legs each and with the load balanced among all 3 legs.

Yes you can draw 333 amps divided among 120 volt single phase equipment connected to any one leg plus the neutral each and with the loads balanced.

The power is 3 phase only when the equipment draws from all 3 legs and the 3 power draws interact within the equipment.

Less of a total load if not perfectly balanced.

Less of a total load if a lot of mechanized (motorized) equipment is being used.

It takes special math to describe what combinations of 120 volt loads and 208 volt loads will max out the generator at its 40 KW.

Last edited by AllanJ; 04-18-16 at 03:57 PM.
PJmax Group Moderator
Join Date: Oct 2012
Location: Jersey
Posts: 63,138
You should provide a make and model number of your generator.

Tolyn Ironhand Group Moderator
Join Date: Nov 2007
Location: United States
Posts: 14,093
To calculate amps from 3 phase power:

KWx1000 / E (volts) x PF(power factor) x 1.73

We know the KW is 40, lets assume the volts is 208, and the power factor is 1 (not likely unless the entire load is resistive)

40 x 1000 / 208 x 1 x 1.73
40,000 / 359.84 = 111.16 amps

Last edited by Tolyn Ironhand; 04-19-16 at 03:57 AM.
Paul Steichen Member
Join Date: Feb 2014
Posts: 15
Tolyn: Your math must work. The generator has a 110A main breaker (can draw 135A for up to 2 hours without tripping though, 145A trips immediately).

Paul Steichen Member
Join Date: Feb 2014
Posts: 15
Thanks AllanJ, that helps a lot. It's a bit confusing how the amperage available changes from 333 to 192 to 111 depending on what voltage and phase one is referring to. At least I know I was on the right track. I'll have to do some more reading!

Paul Steichen Member
Join Date: Feb 2014
Posts: 15
So, this thing has a selector switch that let's me monitor the amperage on each of the 3 hot legs and the neutral. I'm assuming it's proper to keep each of the hot legs below 111A, and as balanced as possible, and then assuming I'm drawing all that as 120V single phase, that the neutral is capable of sinking the full 333A? I've never drawn that much yet, just wondering if I'm understanding correctly...

Tolyn Ironhand Group Moderator
Join Date: Nov 2007
Location: United States
Posts: 14,093
The current between each phase will cancel the current imposed on the neutral. If you have three 10 amp 120 volt loads, one on each phase, the current on the neutral would be zero.

The formula to find neutral current is: I = current on each phase (a,b, or c)
N = neutral current

Paul Steichen Member
Join Date: Feb 2014
Posts: 15
Tolyn: Ah, thanks. That would explain why the neutral is the same guage as the hots. It will only carry unbalanced load, which will never exceed the current carried by one hot.

Upon closer inspection, the rotary knob for monitoring the amperage has 4 positions, but the 4th is OFF, not neutral.

CasualJoe Member
Join Date: Jan 2010
Location: United States
Posts: 9,827
Would I be able to draw 40K ÷ 120 = 333A? (W ÷ V = A)
That formula doesn't work with 3-phase power. Tolyn's post #4 is the correct one.

Last edited by CasualJoe; 04-19-16 at 04:19 PM.
Paul Steichen Member
Join Date: Feb 2014
Posts: 15
Joe: I may be a bit confused. Wouldn't each of the three phases in a wye configuration be capable of producing 1/3 of the rated output? If each of the 3 hot legs were wired to independent loads drawing 111A each as 120V single phase, would that not result in a 333A total draw? And then 120V x 333A = ~40KW?

AllanJ Member
Join Date: Mar 2010
Location: USA
Posts: 4,422
Yes, the 40 KW 120/208 volt 3 phase wye supply wired to independent single phase (hot to neutral) 120 volt loads will be delivering peak power of about 111 amps on each leg.

The legs are 120 degrees out of phase. Draw two (straight) arrows of equal length, head of one to the tail of the other and making a 120 degree angle between them. Practice drawing several sets of such arrows until you get a pair where the distance from tail of the first to head of the second is 111 mm. (I could say 111 inches except that would be too big for an ordinary sheet of paper.)

How long in mm would each of the arrows be? Answer, about 64 mm. So using 208 volt (phase to phase two hots at a time) single phase loads you max out at 64 amps between (pick a phase) the A phase and the B phase and 64 amps between the A phase and the C phase to have 111 amps flowing on the A phase. For a balanced maxed out total load each phase is doing the same thing for a grand total of 192 amps at 208 volts.

Using trigonometry you can compute the answer much more quickly than trial and error drawing of arrows.

Or use your original shortcut (Law of Conservation of Energy?) that 40 KW divided by 208 volts gives a grand total of about 192 amps.

This is an example of vector arithmetic. Many rough calculations for 3 phase power can be done using vector arithmetic. The amount (volts, amps, or whatever) is represented by the length of the arrow and the phase is represented by the tilt of the arrow. For a single phase result or answer, you don't care what the tilt of the "result" arrow is.

Last edited by AllanJ; 04-19-16 at 07:04 PM.
CasualJoe Member
Join Date: Jan 2010
Location: United States
Posts: 9,827
If each of the 3 hot legs were wired to independent loads drawing 111A each as 120V single phase, would that not result in a 333A total draw?
I don't see that example any different than a 3-phase load of 111 amps. The draw would be 111 amps. For 3 sparate single phase loads of 111 amps each I still would call the total current drawn 111 amps X 3 loads.

My previous post probably should have been just this:

(W ÷ V = A)

That formula doesn't work with 3-phase power. Tolyn's post #4 is the correct one.

Tolyn Ironhand Group Moderator
Join Date: Nov 2007
Location: United States
Posts: 14,093
No matter what, you can only draw a max load on any one leg no more then 111 amps ( limited to 110 amps due to the over current device installed.) As Joe said, if you have 111 amps on each leg, the load is 111 amps, not 330. It is not additive.

The math still works:

111 amps x 120 volt (voltage between one leg and neutral) = 13,320 watts

If you had 111 amps on all three legs:

13,320 watts x 3 = 39,960 watts or 39.960 KW

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