Calculate the maximum calculated ampere load on the Neutral conductor

[julian padilla]

The loads shown below are controlled independently. The maximum calculated ampere load that conductor N will ever be required to supply in the feeder shown below is, 17, 25, 125, or 100 amps?

What would be the math/calculation to solve this problem?
Julian

 

[beanland]

Sinusoidal Loads

Sinusoidal Loads

Without knowing the harmonic content of the loads, it is not possible to compute the result. If the 100A load is entirely 3rd harmonic, the load could be 300A on the neutral. If we assume essentially sinusoidal loads, the current on the neutral will be 100A for one phase, 100A for 2 phases, and zero for 3 phases. You need just to do the vector math.

 

[julian padilla]

Without knowing the harmonic content of the loads, it is not possible to compute the result. If the 100A load is entirely 3rd harmonic, the load could be 300A on the neutral. If we assume essentially sinusoidal loads, the current on the neutral will be 100A for one phase, 100A for 2 phases, and zero for 3 phases. You need just to do the vector math.

I'm sorry, but that is confusing. Given the multiple choice answers (17, 25, 125, or 100 amps), isn't there a simple mathematical solution?
Something like the following:
Step_1: Using Ohm's Law, I(single-phase) = P/E = 12,000 / 120 V = 100 amps per load.
Step_2: Given the loads are lighting, apply the 125% rule; 100 amps x 1.25 = 125 amperes maximum load

Please advise.

 

[don_resqcapt19]

I'm sorry, but that is confusing. Given the multiple choice answers (17, 25, 125, or 100 amps), isn't there a simple mathematical solution?
Something like the following:
Step_1: Using Ohm's Law, I(single-phase) = P/E = 12,000 / 120 V = 100 amps per load.
Step_2: Given the loads are lighting, apply the 125% rule; 100 amps x 1.25 = 125 amperes maximum load

Please advise.

There is a simple mathematical solution only where the loads are linear. With the information give, the calculation is not possible. If the loads were restive the answer would be 100 amps. For that case you can use this formula:
neutral current = √ (A² + B² + C²) - (A*B + A*C + B*C)
A, B and C equal the phase currents.
Having nonlinear loads changes everything.

 

[charlie b]

If all three lighting loads were turned on at full power at the same time, the current on the neutral would be zero. The maximum neutral current would take place when two of the lighting loads were on full and the third was off. In that case, the current in the neutral would be 100 amps, as others have mentioned already.

The "125% rule" has nothing to do with how much current any given load will draw. Rather, it tells us to select a wire that can handle 125% of the current a continuous load would draw.

 

[julian padilla]

If all three lighting loads were turned on at full power at the same time, the current on the neutral would be zero. The maximum neutral current would take place when two of the lighting loads were on full and the third was off. In that case, the current in the neutral would be 100 amps, as others have mentioned already.

The "125% rule" has nothing to do with how much current any given load will draw. Rather, it tells us to select a wire that can handle 125% of the current a continuous load would draw.

Don, thank you for the formula, and Charles thank you for the logic. It all makes sense now. Thank you guys are a huge help!
Neutral Current = Square Root of (AxA + BxB + CxC - AxB - AxC - BxC).
Neutral Current = Square Root of (100x100 + 100x100 + 0x0 - 100x100 - 100x0 - 100x0).
Square Root of (10,000 + 10,000 + 0 - 10,000 - 0 - 0).
Square Root of (10,000) = 100 amperes.

 

[Johnhall30]

12kW / 120V = 100A is the line current from one of the loads, which creates a 100A current in the neutral conductor.

If two loads are connected, the current in the neutral conductor add together with 120 degree phase shift --> 100sin(x) + 100sin(x +120) = 100. The cuurrent is still 100A in the Neutral.

However, if three loads are connected, the current in the neutral conductor is added for all three phases --> 100sin(x) + 100sin(x + 120) + 100sin(x + 240) = 0. No current in the neutral

 

[Tony S]

A drawing I did for a UK site based on I_n=√((I_a²+I_b²+I_c²)-((I_a*I_b)+(I_a*I_c)+(I_b*I_c)))