1. Junior Member
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## Three Phase panel single phase loads

I'm trying to work out the load of 14 10kw 208v loads on a three phase service. When I do the math I come up with a different number than when I actually draw out the panel board. Please see the attached calculation work sheet.

Tom

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Mar 2004
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When you have Ø-Ø connected loads, the current of resistive loads is in phase with the Ø-Ø voltage. This messes up calculations involving adding phase currents together. If you use kVA in each position (with 1/2 of total for each circuit on each phase), the totals will be correct. Then use the Ø-N voltage to calculate phase current for each phase.
 A B C 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 Total kVA 50 45 45 Volts 120 120 120 Amps 416.67 375.00 375.00

3. You will also note that when you average the three phase currents (i.e., add 416.75 + 375 + 375, then divide by 3), you get the same 388 that you originally had shown. My view is that you should not attempt to deal with currents at all, until the very end, even if the answer sometimes comes out right anyway. Add everything up in units of VA, then use the total VA to calculate current.

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## Voltage - Line to Line and Line to Neutral

Here's another way to think about it...

If you have (2) 120V single phase loads at 5000 VA each, the current for each circuit is:
5000/120 = 41.67 Amps
If you have (1) 208V single phase load at 10000 VA, the current for this circuit is:
10000/208 = 48.1 Amps

This is what your calculations dilema is showing in this case. The difference between the phase-to-phase voltage and phase-to-neutral voltage.

And to Charles' point, you should use the VA method for adding all the loads together and then calculate total amperage from the total VA.

5. In the OP the calculation used to compute amps, i.e. 14000/208/1.732 = 388A is correct ONLY for a completely balanced three phase system/panel.

VA = VLL*I*cos(ang)

To demonstrate the validity of the formula, you can seen by this calculation in the OP, that a breaker sized accordingly would not be sufficient for Phase A, as the connected load could exceed its capability, assuming this was the only load, and no spare capacity is planned.

What consistently gets forgotten or is not understood, is that when switching from line-line voltage to line-neutral voltage and taking multiphase loads and splitting them between phases, is that there is more at stake then simply the square root of three for voltage, i.e.:

Vab = sqrt3* Van

This is because the line current will either lead by 30 or lag by 30 degrees the phase voltage depending on which phase your looking at. For a load connected across ab phases the individual phase "a" power is determined by:

Pa = Va*Ia*cos(ang - 30deg)
Qa = Va*Ia*sin(ang -30deg)

If the power factor is 0, then the formula reduces to:

Pa = Va*Ia*cos(-30deg); where the cos(-30) = 0.866

The load on Phase B is:
Pb = Vb*Ib*cos(30deg); where the cos(30) = 0.866

Two things can be realized by this formula. The first is that the total load can be divided in half and placed on each phase (as posted earlier). The second is that when solving for the load current it is determined by algebra and solving for "I" which results in the same current as determined by:

BTW: some people may have run across a "fudge factor" or "Oregon Factor" as being 1.155 that is used when calculating the load current from the phase load. It's not really a fudge factor, it's simply the inverted value of the cos(30deg) = 0.866. When you solve for "I" from the above formula, the 0.866 ends up in the denominator, and 1/0.866 = 1.155.

IMO the only way to determine breaker/panel sizing is to take the largest phase VA, multiply by 3, then solve for current using the balanced formula, this will guarantee the service/breaker/panel can handle the total load without overloading an individual phase, plus you will have some spare capacity.

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Originally Posted by kingpb
IMO the only way to determine breaker/panel sizing is to take the largest phase VA, multiply by 3, then solve for current using the balanced formula, this will guarantee the service/breaker/panel can handle the total load without overloading an individual phase, plus you will have some spare capacity.
Isn't this the same as sizing for the highest phase amps as calculated using the VA method?

In the example, the largest phase VA is 5000. 5000·3/208/sqrt(3) = 41.64 A. This would be the same as Phase A amps if we used 208/sqrt(3) = 120.0888 instead of 120 V.

7. Originally Posted by jghrist
Isn't this the same as sizing for the highest phase amps as calculated using the VA method?

In the example, the largest phase VA is 5000. 5000·3/208/sqrt(3) = 41.64 A. This would be the same as Phase A amps if we used 208/sqrt(3) = 120.0888 instead of 120 V.
Not quite following your thinking with the calc. Keep in mind that the 5000VA for the load divided by 120V is not the correct load current. the load current is 10,000VA/208V = 48A. (cos(30) difference)

In your load schedule, which I agree with, you got 50KVA, 45KVA and 45KVA for Phase A, B, and C respectively. If you add those together the value is 140KVA. Taking 140KVA/208/sqrt3 = 388A, by going to the next breaker up you are at 400A. That however is not going to be large enough since you have Phase A at over 400A.

By using the 50KVA*3 = 150KVA, then 150KVA/208/sqrt 3 = 416A (no surprise). So, yes you can use the largest phase amps, it would be the same thing. I like to stay in KVA as much as possible, especially if there is a transformer involved because there are standard transformer sizes that would be the driving factor in size selection, and they are all rated in KVA.

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I don't know if I am on the right track. But I would do this similar to the range calculations in 220.55 and the example in annex D(5)(a).

A- B 5
B- C 5
C- A 5 take twice the max connected between any two phases = 2 x 5 = 10 x 10,000 =

100,000/208 = 481 amps between two phases. 481/ 2 = 240.5 amps per phase 240.5 x 3 = 721.5 for the whole system.

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Originally Posted by CONDUIT
I don't know if I am on the right track. But I would do this similar to the range calculations in 220.55 and the example in annex D(5)(a).

A- B 5
B- C 5
C- A 5 take twice the max connected between any two phases = 2 x 5 = 10 x 10,000 =

100,000/208 = 481 amps between two phases. 481/ 2 = 240.5 amps per phase 240.5 x 3 = 721.5 for the whole system.
Nah! It's never a problem with balanced loading of phases (5kW X 3).

Iline = (3X5,000)/(208 x sqrt(3) ) = 41.64 amps.

10. Originally Posted by CONDUIT
I don't know if I am on the right track. But I would do this similar to the range calculations in 220.55 and the example in annex D(5)(a).

A- B 5
B- C 5
C- A 5 take twice the max connected between any two phases = 2 x 5 = 10 x 10,000 =

100,000/208 = 481 amps between two phases. 481/ 2 = 240.5 amps per phase 240.5 x 3 = 721.5 for the whole system.
I'm afraid you are off base with this approach. As you can see in post 2, the currents will be 417A, 375A and 375A. 721.5A is quite a bit high from these values, and 240.5 is quite a bit low (I'm not sure which you are saying is line current.)

220.55 is used for applying a demand factor to the load calculation for multiple ranges on the same feeder. For instance, if you had 14 ranges rated at 12kW ea (168kW total) then by 220.55, you would use twice the max connected between any two phases = 2*5=10. 10 ranges has a max demand of 25kW per T220.55. 25kW divide by two phases = 12.5kW/phase, times 3 phases = 37.5kW demand (vs. 168kW connected) on the feeder. There is no similar applicable demand factor being applied for these loads.

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