Power Measurement 3 Phase AC

Well, energy savings will depend on how your using the motor in your plant and what you are currently using for speed control. The inverters themselves are up to (around) 98% efficient (application dependent of course). What type of motor control equipment is the variable frequency drive (the inverter) replacing?

If you measure current at input of inverter is 130A and applying voltage on it is 420V,

tghe power will be 95.6KW, and the current you measure at output termial(motor input) is 370A and voltage is 425V so the power will be 275.2KW. So that you can easily get the coefficient will be 0.347. you will save 65% power.
(from the number, your motor seems a bit little selected)
The model of the most of inverter is all converted into DC at first by rectifier and

then converted again to AC by different frenquency, says, from 5HZ to 400Hz. and

voltage can also be from 20v to full load. The ouput situation depends on your load

condition, as said by Jack#1.(some of them is directly from AC to AC, but they have a small adjust or control range)
The devices used in the equipment are almost MOSFET or IGBT. fro they are using easily and reliable.
Simply speaking, there are two main load for your using, one is constant power load like fan and pumpers and another is constant torque load, like hoist etc. YOu can chose defferent inverter corresponding to the load.

Of cause most of mordern inverter can be chose for both of the loads, because they can set up by costumer theirself to comport with the loads. inside there is a CPU for control and instruction.
They can work quadrant, when the load is braked, it can be sused for a generator, as

well.
We can discuss it and relevant products more detail at

info@ndt.cn

Hope this simple description can be help.

I guess you are using some constant power products like fan or pumpers or compressors. Because you get so much saving energy. If your load is constant torque class, you can only get 20% saving at very most.
So if you have still lots of such load in using. I propose you'd better use inverter as early and quickly as possible.
If not, you should take account of save energy and price ratio. sometimes, if you use inadequately inverter you would increase energy dissipation.

If you really can have your motor runing and delivering 100KW of mechanical power, and only use 65% of that in electrical power. please let me know. We'll be rich in a split second !

You might be able to reduce the power supply when on moments you don't need the full 100% from your motor. OK, then you save overall power. But believe me : milions of engeneers searched for getting more power from less power, but until now, nobody succeded this.

Remember : whatever we do, we only can convert power. e.g. convert from electrical to mechanical (a motor) or convert mechanical to electrical (a generator) or from heat to mechanical (a turbine) etc ... The result of a convertion will ALWAYS be less than what we started with.

What could be your mistake is that your measuring instrments are build to measure in a certain frequency range. Your frequency converter might deliver an AC that is out of the range of your measuring tool, so the reading on that tool could be completely wrong...

PS. I'm sorry that I didn't login. But this program doesn't ALLOW me to login. (just try to type in my login name : paulvandenbossche@monti.be )

It seems I hvnt got an error at power description on above paragraph.
The energy is conservation, it can neither increase nor disappear. But here we discuss is power not work.
From the top describe, the power is 275.2kw before inverter added in. After added the inverter, the power is 96kw. why couldnt we say it save 65%?
May the measure nmuber not be tolally right, because the inverter generally save 30--50% for pumpers class machine. as the power is propotion direct to speed square or quartic for such machine. the saving is very clear and evident.
Thus, the engineer's proposal is right.
The last point what I want to say here is power not work.
We can say certain facility input 1kw power and can output 5kw or ever more, says 1000kw power! its possible , no error. because we describe here is poer not work.

But we cant say certain facility imput 1kj work and output 2kj work. its wrong! it disobeys energy conservation law.
This is my opinion. I wish its right.
I wish also more and more plants and mills that using these class facilities can use

inverter to save energy, in other words, thats meaning save money.
If there is something wrong in the paragraph, welcome to criticize and intercommunicate at
info@ndt.cn

Whether you are referring to work or power your inverter will NOT add to the total. The efficency of the inverter will always be less than 1.0 because the inverter itself requires power to operate.

The inverter will provide a total power savings if compared to "across the line" pumps, since the inverter is capable of running at speeds less than 100% (base speed of motor) , when the operation allows for it.

Also note that most inverters today are PWM (Pulse width modulators) operating in the 10-20 Kilohertz range. You cannot measure output current with a clamp on meter unless it is a 'fundamental current probe" capable of distinguishing the true "work amps" from the harmonics.

Today's inverters average 95% efficiency when output frequency equals input frequency. Efficiency reduces as speed is decreased, however there still is a net benefit to the user since it is drawing less input power by reducing total work.

Hi!!

As I understand, the inverter input power is 95kW and output power is 275kW. That means, you are getting around 180kW extra from the inverter! Inverter is only a frequency changing equipment. It can not generate electric energy. It can only convert from one frequency to another. The readings given clearly shows that electric energy is generated by inverter! This is against basic principles! Energy can not be created nor destroyed, but can be converted from one form to another only. As per your readings and explanation I have read here, this inverter will solve the energy crisis our world is facing now!!! Because it is doing the impossible one - Taking 95kW and giving 275kW!!! If it is so, it is not required to install new power plants in future! Just to make inverters only!!! I feel, there is some mistake in the readings. Do not take this for granted and order more. Discuss with your electrical engineer and ask him how the inverter can generate electric energy. Discussion in presence of inverter supplier will help more.

Sorry for my error description in the classification of load in above paragraph.
the fan and pumpers are belong to variable torque load not constant power load, whereas lathe and paper making and texitle machine are constant load.

Regarding the above comments, it appears that you may have got a false reading from your clamp meter on the AC current coming out of the inverter to the motor. The inverter (by its very nature) chops up the voltage waveform and pulses it out into the motor creating a simulated AC waveform. By its very nature, a choppy pulse current waveform is going to be difficult to measure with a standard averaging clamp on current meter (well it won't work). This may be where the confusion has come from (with more power out of the inverter than is being put in).

Inverter efficiency is always less than 100%.

lets illustrate by this figure 1. I think it will be clear for everyone's

comprehension.
Part A is conneted directly to line (3Phase). The power P1 is

275kw.(1.75x425Vx370A). (at terminal of plug of motor connected by delta

method) All according to measuring value indicated above.
Part B is fig of added in an inverter and connect to line(3p). The power P2

is 96kw (1.75x420Vx130A). Now the saving will be 65%!
Right?
How much is P3? The first guest hasnt pointed out. in fact, its unnecessary. Becaue what our interested in is only power supplied from main line.
If someone is interested in its value, we could give an answer as follow:
As he runs an inverter, adn if he measures the line current of the motor at this time, he will get a still 340A current (delta method and according to the rate of 250kw the guest gives above, this will work in a resonable way

for an inverter) (or say 370A, he measures in pratice, this will be greater than his motors rate) and if he measures the voltage at the time, he will read a very low value and lower frequency if he tests it. I suggest he measures practicely. from nmubers above, we can get as follow:
96kwx0.95/370A= 246V(line voltage). I hope soon to get his reply to

info@ndt.cn
Although Im not a specialist of motor control and hve forgotten most of this knowledges, I think Im able to and try to answer anyone's problem at the field. however, Ive managed the line for some years.

Another question I want to explain is power is defferent from work. although

their relationship is too tight to distinguish them.
As I said fore. we can say an object acquires 1kw power then can even

output 1Mw power. Its really right. but we couldnt say it obtained a 1kj

and can output 1.00000...01kj work. its wrong.
Why? I hope to left the question to viewers to answer it. its not difficult

The clamp current meter we used to use daily is digital clamp meter. It has a wide respondence. from 10hz to 200khz. It can measure the inverter's output. no problem. if the output is not pure sine wave, pls multiply a coeficient. which accroding to your experience of ordinary. different wave form differnt coeficient.
Most of inverter is worked in the way of PWM and SPWM. and the more widely use for the latter here.
Frequency from 20Khz to 44KHz, some even to 100KHz.
The best bet way to measer the wave form is using an oscill0scope!
If the inverter is a higher voltage output, for example 6KV, pls add an insulation transformer and be sure to pay attatin to ensure safety!!!

I studied the case refered by you

the Variable Frequency Drives(VFD) are widely used for Variable speed applications.the power saving will be calculated on the basis of the input power measured without installing variable frequecy drives and the input power measured with installing the VFD(At input of VFD).

In your data you have not meentioned the power measurement without VFD.

So After measuring as said above you can see in saving and depending on saving and ROI you can consider for future VFD installations considering the applications .

Please reply after measuring the power without VFD.

Aruna Kumar

Mr.Abdul,

To measure input power in kW or current in Amps on the input side of the inverter, one requires True RMS power meter with PF range from 0.00 Lag to 0.00 Lead. The question of measuring power input on the output side of the inverter, using any portable power meter or tongue tester is not permitted. Only an osciloscope has to be used, which is not truly portable. Thus the power measurement on the output side of an inverter is possible only in a laboratory.

Thus it would conclude that the measurement of 370A on the output of the inverter is wrong. Please go by the power measurement on the input side of any inverter, using a True RMS meter.

J.Srinivasan