Unbalance 3-phase current calculation

[stevee]

Ok, probably a simple answer, or maybe not but let me ask anyhow.

On a typical 3-phase, 4-wire, 480/277v system with a 100A rated panelboard and a 100A rated main circuit breaker. The panelboard is used for lighting load operating at 480V.

The lights are single phase, 480V.

The connected load between phases A and B is 40A
The connected load between phases A and C is 30A
The connected load between phases B and C is 30A

What would the current be on the conductors supplying power to the panelboard?

In other words with the above connected loads...
A phase would draw _____ amps?
B phase would draw _____ amps?
C phase would draw _____ amps?

Is there a simple formula to calculate the individual currents on each phase?

 

[skeshesh]

Just think of it as a semi-panel schedule. 40A@480V(single phase)=19200VA and 30A@480V(single phase)=14400VA. Divide the watts up between phases like you would do in a panel schedule, i.e. if there's 19200VA between phases A & B then under each phase column would get 19200/2=9600VA. Add up the VA for each phase till you get a subtotal for each phase then divide by L-N voltage and get the load current on each phase.

 

[mithun46]

can you explain it some more clearly?????????

Please post the entire solution for the above problem if possible.

 

[kwired]

I believe the balanced portion of the load will be 30 amps / 1.73 so 17.34 per phase for the balanced load.

You then have 10 amps unbalanced between A and B

A=27.34
B=27.34
C=17.34

 

[david luchini]

I believe the balanced portion of the load will be 30 amps / 1.73 so 17.34 per phase for the balanced load.

You then have 10 amps unbalanced between A and B

A=27.34
B=27.34
C=17.34

I think you mean 30 amps * 1.73 so 51.96 per phase for the balanced load.

I come up with:

A=60.83
B=60.83
C=51.96

This presumes that the power factor is the same for all three loads so that the phase currents are 120 deg apart.

 

[stevee]

Just think of it as a semi-panel schedule. 40A@480V(single phase)=19200VA and 30A@480V(single phase)=14400VA. Divide the watts up between phases like you would do in a panel schedule, i.e. if there's 19200VA between phases A & B then under each phase column would get 19200/2=9600VA. Add up the VA for each phase till you get a subtotal for each phase then divide by L-N voltage and get the load current on each phase.

I think this method makes the most sense to me.

Using this method the current comes out to be:
40A x 480 = 19,200W / 2 = 9,600W
30A x 480 = 14,400W / 2 = 7,200
Total Load connected to Phase A = 16,800W
Current Phase A = 16,800/277 = 60.6A

Does anyone know of a reference that this calculation could be found in?

 

[david luchini]

I think this method makes the most sense to me.

Using this method the current comes out to be:
40A x 480 = 19,200W / 2 = 9,600W
30A x 480 = 14,400W / 2 = 7,200
Total Load connected to Phase A = 16,800W
Current Phase A = 16,800/277 = 60.6A

Does anyone know of a reference that this calculation could be found in?

This method provides a close approximation in this case, Ia=60.65, Ib=60.65, Ic=51.99, as compared to using the calculation Ia=Iab-Ica (etc.) which gives Ia=60.83, Ib=60.83, Ic=51.96.

However, the larger the unbalance, the larger the error. For instance, with Iab=40, Ibc=30 and Ica=10, the shortcut method would give Ia=43.32, Ib=60.65 and Ic=34.66, but the calculation method would give Ia=45.83, Ib=60.83 and Ic=36.05.

You can see with the larger imbalance the shortcut method gives a line A current which is off by 2.5A, and the line C current is off by almost 1.5A.

 

[kingpb]

The equations are derived from the standard single phase/pole power equation.

View attachment 5210

The 1.15 for a 2-pole load, some may refer to as the mythical Oregon Fudge Factor.

But it's really pretty straight forward math. Until you try the symmetrical components....

 

[kwired]

I think you mean 30 amps * 1.73 so 51.96 per phase for the balanced load.

I come up with:

A=60.83
B=60.83
C=51.96

This presumes that the power factor is the same for all three loads so that the phase currents are 120 deg apart.

You are correct. Except I don't know where 60.83 comes from shouldn't it be 61.96 for A and B?

 

[david luchini]

You are correct. Except I don't know where 60.83 comes from shouldn't it be 61.96 for A and B?

It wouldn't be 61.96. If the current from A-B = 40<0, from B-C = 30<-120 and from C-A = 30<-240, then using KCL Ia=Iab-Ica.

Or, Ia= (40<0) - (30<-240)

Ia= (40+j0) - (-15+j25.98)

Ia= 55 - j25.98

Ia= 60.83<-25.28