Basics

The use of (power in watt) = (volt) * (ampere) works great to determine the electrical power requirements. However, it neglects the fact that every system has some amount of loss when energy is transformed from one state to another.

If you're really monitoring a 169 ampere current draw for a machine rated 60 kW at 400 V, then you have one of three conditions:
A) The motor efficiency is extremely low (roughly 88 percent). Note this value SHOULD be in the 93-95 ...

All electrical machines have some amount of dis-symmetry to them, just because they are constructed of individual parts. Sometimes this is in the steel portions (segmental laminations, or unequal airgaps, etc) and sometimes in the winding (unequal resistance paths, different connections and/or lead lengths, lead routing, etc).

As soon as there is a difference in the magnetic path (or the magnetic strength), some amount of voltage will be induced on the motor shaft. The purpose of a ...

I have a question regarding protecting a variable frequency drive (VFD). I have read that the important part to protect of the variable frequency drive is the active front end (prior to the IGBT's). I also am under the impression that the VFDs themselves protect the motors from faults. So my questions are as follows:

1. Do you need an overload to protect the motor? And if so do you put in on the PWM output side?
2. How should I protect the variable frequency drive and Motor? ...

1. Explain why the efficiency of the induction motor increases when the mechanical load increases.
2. Explain why the input power factor of the induction motor increases when the mechanical load increases.

Explain:
1. The efficiency of the induction increases when the mechanical load increases because as the motors load increases, its slip increases, and the rotor speed falls. Since the rotor speed is slower, there is more relative motion between the rotor and the stator ...

Dealing with the change in power factor. A 10HP (7.5kW), 400v, IE2, B3, Induction motor at full load will have a power factor of 0.86. At 75% power the power factor is 0.81 and at 50% it is 0.71. Assuming full load then your 10HP is 7.5kW at power factor 0.86 has a reactive power of 4.45kVAr. If the load drops to 75% and the capacitors are still in circuit, the machine only requires 4.07kVAr so it is now 0.38kVAr leading or +0.99. at 50%, 3.82kVAr leading or +0.98.

Dealing with the ...

There was a query in all this about how to measure knee point on a transformer. Knee point relates to the ability of the secondary winding of a current transformer to produce enough voltage to drive the required current around the connected burden and what excitation current is "lost" inside the transformer to do that.

So for a current transformer, leave the primary winding open circuit. Apply a small increasing voltage to the secondary and measure the current that flows into the s ...