Power Measurement (Electrical Engineering)

What is source voltage and phase and what voltage and phase are you putting out at 3 phase.

The input should be 5-10% more than the output depending on the efficiency and how close each is running to their designed maximum.

An invertered designed to use 250KW to supply 240 KW load will have both running at their max efficiencies(more or less).

In addition there will be some phase factor adjustment needed in the way of a capacitive reactance for get as close to a unity power factor as possible as thsi will minimize circulating currents. A good AC power EE wiil advise you better than I could once you supply the in/out parameters

Hello Guest,

According to mfg, for a voltage system of 380V 3phase, the 200kW motor should has FLC about 360A, this is the current when motor running at its full speed of rpm, so the output frequency of drive must be the same of system, say 50Hz.

Are you sure about this? Have you measured the output frequency too, and voltage?

And rpm? I just want to know how do you confirm the FLC condition?

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The AC drives are configured of build-in converter and inverter. Converter convert AC to DC, and inverter convert DC back to variable voltage and frequency of AC. If your drive capacity is 130A, so it should be 75kW in 380V system.

If you adjust to nominal voltage&Frequency, the drive or in-line protection devices should trip, because your 75kW of drive will never accomodate your 200kW of motor.

Otherwise you set your drive to a certain value of frequency and voltage, you may run your motor but never get the full speed, or full load current/FLC.

I think there's a mistake in selecting your drive.

Or maybe there're things that I miss to catch in your description.

Regards,

Abu Khansa

How are you measuring the FLC of the motor? Are you using your clamp-on ammeter or are you reading the value from the VFD? Most clamp-on ammeters are NOT capable of reading the output of a VFD, the values that they read are meaningless. The VFD measured the current using Hall effect transducers and has a special algorithm that it uses to interpolate the actual motor working current based on the output flux vectors since it is in control of everything anyway. Without that, an external meter has a hard time differentiating between harmonic currents and working current.

I have a sneaking suspicion that your load current is actually the sum of the currents on all three phases so it you break that down its 120A per phase which is close to what you are reading.

For JRaef I think you will find that most AC clamp on meters, sometimes called tong testers, use a current transformer not hall effect sensors. The hall effect ones are used on DC current meters not AC ones. The theory is that you have 1 turn on the primary side and N turns on the meter side the measured current on the meter side will be 1 ÷ N of the actual current. So if you have 1,000 turns on the meter side Amps in the conductor will register as mA on the meter. Keep in mind though if the secondary winding ever goes open circuit the voltage will be N times that on the input side and that is why current transformer are usually shipped with the outputs shorted out.

1 Your motor may be connected star where as inverter may be connected delta.

2 the clamp tester are generally designed for sine wave. where as your inveter may be producing diffrent waveform. you should check both the current by CRO and then Calculate the RMS on the basis of type of waveform.

using standard clamp meters on inverter output side is useless as inverters have high freq output. you may measure at the input side with the motor running at rated speed that is at 50 hz output freq. even then the motor may be running at a different speed than the speed at which it will run when connected to a utility supply of 50 hz.the reason for this is that the drive will partly try to compensate the slip as a standard control method as it senses the speed from the waveforms of the output lines.

best would be to have a power analyser put on the input of the inverter and measure power

We need a few more details to figure this one out so I'll try to point you in the right direction. The kva of the input side of the inverter should roughly match the kva of the load side. Are you using a single phase input inverter to drive a three phase load? If so the input kva is simply the Line to line (L-L) voltage times the line current (both rms values). If you have a three phase input the input kva will be the L-L voltage times the line current divided by the square root of 3. The output kva will be the same in either case just the L-L voltage times the line current divided by the square root of three. The inverter should be fairly efficient so these numbers should match to within... oh say 75 to 80 percent... Keep in mind what everyone else has said about the use of clamp on meters too, everything should be kept at the same frequency during the tests 50-60 Hz.

dear friend, im an electrician here in saudi arabia, ARAMCO project, i have some explanation in my mind, it may not a solid answer but it will give you an idea.

inverter must be more than 100% than its load, the inverter cannot give the capacity what the load demand, same as a child carrying a load more that his weight, it is better to buy a 500 amperes inverter to supply the 360 amperes motor, consider the inrush current of a motorits full load current, perhaps, it is better for you to read a table regarding of motor.your inverter will burn and damaged im sure of that.

RAC

Power measurment of Inverters [if frequency is HF] is a tidious job can only be done with professional Eqpt because 50/60Hz Measuring Eqpt cannot do correctly. It involves Voltage/ current & Phase angle measurements. Even on line frequency it may be misleading.

To read output current from a VFD you need to use a True RMS meter to get an accurate reading such as a Fluke 337. Also the reading you get will not equal the rating of the motor or the drive because they are 3 phase readings and you are only measuring one leg. However you can not merely multiply by three, instead it is actually the square root of 3 to get the total current.

Your multiplication by 3^½ depends on how you have the motor wired. If it connected in star format then the current is multiplied by three. You only introduce the 3^½ when it is connected in delta format and the reason is that the phase to phase voltage is 3^½ times the phase to neutral voltage.

When you talk about three phase you need to take into account the direction of each of the phases. As you can see in the diagram above you can't just add phase A and B or use the phase to neutral voltage. If you do the trigonometry you will find that the phase to phase voltage is 3^½ times the phase to neutral voltage. By the way 3^½ means Square Root of Three not 3.5.

In reply to #11

Hello Masu

What you say is true (for calculating power) but only if phase-neutral voltage is used. Then power = 3 x phase current x voltage (x PF and eff which aren't relevant to this). But it might cause confusion as the quoted voltage of a 3-phase system is phase-phase. Because phase-phase is √3 x phase-neutral voltage as you say, it gives the same answer as √3 x phase-phase voltage x phase current.

If you have a motor running and don't know whether it's connected star or delta, you don't need to find out in order to calculate the power.