Choosing proper wire gauge

Question

I have a machine (a machine with seven three-phase motors) with this information on its nameplate:

3 Phases

400 V, 50 Hz

100 A, 40 kW

I am wondering:

1. Is 100 A for all three phases or only for one phase?
2. What does it mean by 400V? Line to line voltage? Or the phase one?
3. Is 40 kW the maximum power consumption? Or I should consider the reactive power and provide a larger power source such as 50 kVA? Also is it for all phases?
4. What size of wire should I choose?

Edited:
5. How was the power, 40 kW, calculated? 400V x 100A = 40kW? So what about P = sqrt(3) * VL * IL?

Environment: My machine has 7 three-phase motors. Local electricity line: 220V, 50Hz, each phase.

Answer 1

Is 100 A for all 3 phases or only for one phases?

That will be each phase.

What does it mean by 400V? Line to line voltage? or the phase one?

It is line to line voltage.

Is 40 kW the maximum power consuming? Or I should consider the reactive power and provide a larger source power such as 50 kW?

An easy way to calculate is to use the phase to neutral voltage and calculate for one phase.

• One phase will have \$ P = \frac {P_{TOT}}{3} = \frac {40}{3} = 13.3 \ \text{kW} \$.
• The phase to neutral voltage will be \$ {V_{L-L}}{\sqrt 3} = {400}{\sqrt 3} = 234\ \text V\$.
• If the load is purely resistive then \$ I = \frac {P}{V} = \frac {13300}{234} = 57 \ \text A \$.

The 100 A rating, therefore, must be to cover start-up current and reactive power.

Also is it for all phases?

Yes. The power rating is for the whole machine.

What size of wire shall I choose?

Check your local regulations which should have tables to guide you. The distance from the supply will be a factor and whether any power-factor correction capacitors are placed at the feed or load end of the cable.

Answer 2

The "100A" is the breaker rating for each. (phase)
The "400V" is line-line.
The "40kW" is max. steady load rating.
The size of the wire is AWG 6 max. ( 101A rated @ some temp )

The choice of wire gauge depends on local industrial codes and feed length. The voltage drop depends on the resistance of the wire and the load in amps.

Power consumption in Watts and VARs depends on pf, efficiency, mechanical load and acceleration which briefly exceeds by 300% steady-state max on start up or more.

Answer 3

100A is the current for each of the three phases. V X A X sqrt(3) = 69.3 kVA. V X A X sqrt(3) X power factor = input kW. 40 kW is the rated mechanical output power assuming that the marking on the motor conforms to international standards. Input power X efficiency = output power.

If the 40 kW rating is marked on the machine, it is the input power for the entire machine including the sum of the power inputs to all of the motors plus whatever else uses electric power in the machine. The power factor is then 40/69.3 = 0.577, a rather low power factor, but not unreasonable.

Answer 4

kW is definitely a power rating of some sort, because otherwise it would say kVA.

Hard to say WHAT power, though. Maybe it is output power? The output power of a three phase motor is:

P = sqrt(3) * I * V * PF * eff

Where P is the output power of the motor, I is the nameplate current, V is the nameplate voltage, PF is the power factor, and eff is the efficiency. Let's re-arrange it so solve for PF * eff.

PF * eff = P / (sqrt(3) * I * V) PF * eff = 40,000 / (1.73 * 100 * 400) = 1/1.73 = 0.577

Is that plausible? It is a bit on the low side. With 7 motors, the motor size must be around 7.5 HP. I would expect efficiency for a motor that size to be 90% and power factor maybe 0.8, so that the product would be 0.72, not 0.577. So, it is a mystery.

In order to make the math work, the power factor and efficiency would be something like 75 or 76 percent each.

The other explanation is to say that the input power really is 40,000W, and the kVA really is sqrt(3) * 400 * 100, in which case the power factor would be a miserable 0.577. This is possible. It is just kind of a low power factor.