VFD FORMULA 1

[maxgolder]

Dear All,

I use a volt meter to get the actual input amperage on my VFD.
And all time, after calculation I get the saving energy value like the saving by theory or more higher.

How is it possible?
Someone have a great experience and can explain me how and why there are a difference between my bill and the consumption obtained by calculation?

thank you
maxximo

Also, I dont have a kw hour meter inside.


Below is the calculation that I find in internet:

KW= (volts(avg) x amps(avg) x power factor x 1.732) divided by 1,000

If you don't have a power factor use 0.9. If possible use a volt meter to read the actual voltage between each phase. Use the average of the three voltage readings (V1+V2+V3 / 3). Take the amperage reading from each phase and average them (A1+A2+A3 / 3). To find KWh, just multiply KW by the amount of hours your system will be serving the load.

Example:

An operating three phase motor has voltages measured with a voltmeter on each phase of 453, 458, and 461 volts, amperage measured on each phase with an ammeter are 14.1, 13.9, and 13.8 amps, power factor was measured as 0.82.

(453 + 458 + 461) / 3 = 457V

(14.1 + 13.9 + 13.8) / 3 = 13.9A

(457V x 13.9A x .82pf x 1.732) / 1000 = 9.02 Kwatts

To calculate the kWh for one days use:

9.02 Kw x 24hrs = 216.48 kWh

 

[jraef]

Unless you paid a lot of money for a very sophisticated meter, most portable meters are incapable of accurately measuring anything related to VFDs because of the complex wave forms created. This accounts for 99% of the confusion I see in people attempting to verify VFD energy performance from the outside. That's why better VFDs will include internal power monitoring, they already have the ability to measure and properly interpret the data because it's part and parcel to what they need to do anyway. If your drive does not have this capability it's either too cheap or too old (or both). You will likely need a new meter or a new VFD if it's that important to you.

"Dear future generations: Please accept our apologies. We were rolling drunk on petroleum."
— Kilgore Trout (via Kurt Vonnegut)
For the best use of Eng-Tips, please click here -> faq731-376

 

[maxgolder]

thank you very much!
but we dont use inside the vfd a kilowatt hour meter.

Other thing, the meter used to grab the amperage and the voltage is perfectly calibrated and certified (brand flux, power logger 3 phases).

But my question is how and why we get a big saving with the calculation . it is not possible.
maybe my formula used is wrong?
help?
We have 17 VFD and at different frequency we save the big part of energy.

at 25 hz 80 % of saving
at 40hz 50% of saving

thank you

maxximo

 

[itsmoked]

Your meter can be calibrated all it wants to be for sinusoid waveforms. The problem is that there are NO sinusoids anywhere near a VFD. Your meters are lying to you.

You need to get a submetering watthour meter and install it in the power leads.


Alternatively you can turn off everything else that shares the power meter you have and use it and ONLY the VFD to see what the draw it.

You cannot measure the current and the voltage and 'do some math'.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[maxgolder]

Thank you very much also,

How I can convince my engineer because I am not a engineer but with my job and my position, I understand how work a VFD and I have aquired some knowledge.

But when I talk with my engineer, he learned this formula and this calculation method to know the consumption or the saving.
But all time I explain him it is not possible and he dont believe me.
I show him all calculation done with different vfd at different frequency and all time we get the high level of savings.

I wish to find some technical arguments to show him.
It is not possible to do this calculation and tell this is the savings.

thank you for all support.

maxximo

 

[KllrWolf]

Get him to put an oscilloscope on the VFD. You would then see how non-sinusoidal the waveforms are. Then explain to him how the measurement tools are designed to work with a sinusoidal waveform.

 

[waross]

What are you driving with the motors? The type of load has a lot to do with the performance of a VFD at reduced speeds.

On you KWHr meter you will find a number labeled kH.
Some typical kH numbers are 7.2, 3.6 but there are lots of other values in use. This is the Watt Hours per revolution of the meter disk. (Or cycle of the moving dots on an electronic meter.)
Record the time in seconds for ten consecutive revolutions. Discard any extremely high or low times. If you discard more than 2 or three times, try again, possibly at a different time of the day. Take an average of your good time readings. Divide that into 3600. That will give you the average number of revolutions per hour. Multiply that by the kH. Now you have the average number of watt hours the meter is seeing at the time of your test.
Multiply that by your metering multiplier (1 for a direct connected meter.) Now divide by 1000
Now you have the number of KWHrs per Hr or the KW demand.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[maxgolder]

Hi All friends,

Thank you for your great support,

So if I understand, It is better to install a kw hour meter analog and no digital in the panel for each vfd.
With this installation, we get proprely the information about the consumption and accurately.

thank you friend.

maxximo

 

[jraef]

"Accuracy" is relative. What you REALLY want to know is what the utility is going to be billing you for, regardless of whether it is accurate or not. If you think it is inaccurate in their favor, then it may be worth buying a really good meter and tracking it, or at least hiring someone to make the case for you (I've been hired for that, proved that the utility was indeed over charging). If you think it is inaccurate in your favor, then keep it to yourself!

"Dear future generations: Please accept our apologies. We were rolling drunk on petroleum."
— Kilgore Trout (via Kurt Vonnegut)
For the best use of Eng-Tips, please click here -> faq731-376

 

[maxgolder]

Hi,
Thank you for all.

Dear itsmoked and all, when you told me:" The problem is that there are NO sinusoids anywhere near a VFD. Your meters are lying to you".

anywhere means also in the input?, because my engineer tell me now it is thrue but only in the output.

He dont believe in the input we dont get a sinusoidal wave in the vfd.

Also Killrwolf, told me :"Get him to put an oscilloscope on the VFD'
You mean put an oscilloscope in the input?

You think it is enough to show him no-sinusoidal wave there?

thank you for all.

maxximo

 

[waross]

The input current is non-sinusoidal. The magnitude of the voltage distortion is influenced by the impedance of the source.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[maxgolder]

thank you.

I forgot to tell you also, the meter used is a true rms meter (flux)

So, there are some conditions should be respected to mesure the amp in the input?
Also, the voltage used in our calculation come from the namplate motor ?


KW= (volts(avg) x amps(avg) x power factor x 1.732) divided by 1,000

thank you

maxximo

 

[LionelHutz]

If you don't have a power factor use 0.9.

Are you measuring power factor? Your engineer should know that you can't get an accurate answer when you assume one of the numbers you use in the calculation.

I've never heard of the meter manufacturer Flux and a quick search found no info. Do you have an English link?

Also, don't assume that slowing down the VFD and having it use fewer kW's means you're automatically saving energy. You need to look at kW used per unit work done.

As an example - water pump. The kW may go down, but if the gallons pumped per kW goes down more, then the kw/gallon moved is actually worse and you're using more energy to move each gallon of water. In the end, you take longer to pump each gallon of water and use more energy to pump each gallon of water even though the kW at the input to the VFD is less. Oops, that VFD is costing you money!

Anyone that only looks at the kW without considering the complete process is likely not calculating the energy savings correctly.