Neutral load calculation ?

[jaehlee99]

When calculating neutral load current, I was wondering how to calculate following samples:

In a 3-wire, 240/120V, single phase circuit with some unbalanceed loads, what is the neutral current? This sample is only part of the panel board.


Phase A Phase B

Lighting Load 500W

Receptacle Load 360W

Clothes Dryer 2200W 2200W

I did like that : In = 500/120 + 360/120 = 7.2A

In a 4-wire, 208/120V, three phase circuit with some unbalanceed loads, what is the neutral current? This sample is only part of the panel board.

Phase A Phase B Phase C

Condening Unit 550W 650W

LOBBY 1500W

E-Elev 1100W 1100W

DHW PUMP 600W 600W 600W

I did like that : In = 550/120 + 650/120 + 1500/120 = 22.5A

Many thanks in advance

 

[david luchini]

When calculating neutral load current, I was wondering how to calculate following samples:

In a 3-wire, 240/120V, single phase circuit with some unbalanceed loads, what is the neutral current? This sample is only part of the panel board.


Phase A Phase B

Lighting Load 500W

Receptacle Load 360W

Clothes Dryer 2200W 2200W

I did like that : In = 500/120 + 360/120 = 7.2A

A better approximation would be In=500/120 - 360/120 = 1.2A, or In=Ia - Ib. I say approximation, because the loads would need to have the same power factor to provide an accurate result.

In a 4-wire, 208/120V, three phase circuit with some unbalanceed loads, what is the neutral current? This sample is only part of the panel board.

Phase A Phase B Phase C

Condening Unit 550W 650W

LOBBY 1500W

E-Elev 1100W 1100W

DHW PUMP 600W 600W 600W

I did like that : In = 550/120 + 650/120 + 1500/120 = 22.5A

Many thanks in advance

An approximation for 3 phase would be In=SQRT (Ia^2 + Ib ^2 + Ic^2 - Ia*Ib - Ia*Ic - Ib*Ic). Again, this would require loads with the same power factor to provide an accurate result.

Why didn't you include E-Elev in your calculation?

 

[jaehlee99]

Many thanks David

An approximation for 3 phase would be In=SQRT (Ia^2 + Ib ^2 + Ic^2 - Ia*Ib - Ia*Ic - Ib*Ic). Again, this would require loads with the same power factor to provide an accurate result. Why didn't you include E-Elev in your calculation?

I did again, based on your comment:

Phase A Phase B Phase C

Condening Unit 550W 650W

LOBBY 1500W

E-Elev 1100W 1100W

DHW PUMP 600W 600W 600W

Again, I did like that : In = SQRT [(550/120)^2 + (650/120)^2 + (1500/120)^2 + (1100/120)^2 + (1100/120)^2 ] = ?? Is that right?

 

[david luchini]

I did again, based on your comment:

Phase A Phase B Phase C

Condening Unit 550W 650W

LOBBY 1500W

E-Elev 1100W 1100W

DHW PUMP 600W 600W 600W

Again, I did like that : In = SQRT [(550/120)^2 + (650/120)^2 + (1500/120)^2 + (1100/120)^2 + (1100/120)^2 ] = ?? Is that right?

Its a little hard to follow your loads, but you seem to have 550+1100+600W on A, 650+1100+600W on B and 1500+600W on C.

So, 2250W on A, 2350W on B, and 2100W on C; or Ia=2250/120=18.75A, Ib=19.58A, Ic=17.5A.

So In=SQRT(18.75^2 + 19.58^2 + 17.5^2 - (18.75*19.58) - (18.75*17.5) - (19.58*17.5)) = 1.81A.

 

[jaehlee99]

David, I appreciated your time for me.

Even I put the every load with space in the sentence, space would not be applied what I did. Sorry about that.

A better approximation would be In=500/120 - 360/120 = 1.2A, or In=Ia - Ib. I say approximation, because the loads would need to have the same power factor to provide an accurate result.

I can not quite understand that you applied In=500/120 - 360/120 instead of In=500/120 + 360/120 under 3 wire, single phase system.

 

[david luchini]

I can not quite understand that you applied In=500/120 - 360/120 instead of In=500/120 + 360/120 under 3 wire, single phase system.

I would generally write that In=Ia + Ib in this case, but don't forget that current has both Magnitude and Direction. These simplified neutral current equations generally ignore the Direction, and Ia and Ib are basically looked at as the current flowing "from" the source "to" the load.

But current will not be flowing "from" the source "to" the load on both A and B at the same time. When current is flowing from the source to the load on A, it will be flowing from the load to the source on B, and vice versa. Because of this, if you had a 10A load connected A-N and a 10A load connected B-N, you would expect to see Zero amps flowing on the neutral wire, not 20A.

Hope this is understandable.

 

[jaehlee99]

Everything is clear.

Thanks again..