Calculation of kWh with instant A and U 1

[alinoltean]

thread238-151922
Hi Guys, I'm a programmer and got an Electrical question.
I have a Frequency Inverter (3 phase 380 V)from which I can read instant values of U and A going to a motor (cos Phi = 0.75). I need to implement a kWh counter so my customer can see how much he is consuming over a period of time.
My PLC reading the data out of the Frequency Inverter can get the data at 1 second intervals. The values read vary, because the motor is accelerating and decelerating all the time.
I'm thinking something like: get some instant consumption each second which I can add up all the time, to display a counter. Can you suggest a formula for this ?

 

[jghrist]

I don't think you can get a reasonably accurate kWh usage from U and A instantaneous values every second. You don't know the rms value of either and you don't know where on the waveform you are getting the instantaneous value. You say cos Phi = 0.75, but this displacement power factor probably isn't constant. The output of the inverter also will not be a perfect sine wave either and there will be distortion.

 

[waross]

If this inverter is a VFD designed for motors RTFM. A kW function may be included in the available parameters.

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

 

[LionelHutz]

You should know that assuming the power factor is 0.75 will make the resulting number meaningless.

kWh -> watts times hours, with the k being a factor of 1000. A 1 second interval is 1/3600th of an hour. So, calcuate the watts and multiply by 1/3600 to get the Wh for each 1 second interval. Divide by 1000 to get the kWh for each second. Keep a running total to calculate the total kWh from the time you began the calculations. You'll also likely need the ability to reset this number.

 

[Skogsgurra]

I doubt very much that "I can read instant values of U and A". It is normally the RMS values that are output and not instantaneous. The use is normally to connect an analogue meter so you can read U and A comfortably.

Multiplying those values will not give you any meaningsful information (if you do not think that VA is meaningful). Do what waross says. Find the parameter that connects the power (kW) variable to one of the analogue outputs.

If your PLC is connected to the inverter via a bus system, you can read the variable directly from the inverter's control system.



Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[waross]

The power factor is a ratio that changes with the voltage and the load on the motor but not in a linear fashion.


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

 

[alinoltean]

Thanks everyone for the input.
I would like to mention that the inverter is indeed connected via a bus to the PLC, and that via this bus I read the instant values of the U and A. Also there is no kW, or kWh parameter to be read from the inverter.
I understand the calculation will not be accurate, but my client is happy also with an approximate value.
I think that LionelHutz's post provides a pretty good calculation. If anyone thinks different please advise.

Thanks guys.

 

[Skogsgurra]

That is a very unusual inverter that doesn't provide a power signal. What make and type is it?

If power is not available, you will get a much better idea if you multiply torque and speed and adjust for efficiency.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[rbulsara]

Is this a student project? Have your client buy a kwh meter and connect upstream of the inverter. He is paying for all the losses as well.

Rafiq Bulsara

http://www.srengineersct.com

 

[waross]

I agree with Rafiq, buy a kWHr meter.
But:
Volts x Amps = Volt Amps (VA)
Volts x Amps x Power Factor = Watts (W)
W/VA = PF
W² + VAR² = VA²

From the motor nameplate find full load V and A.
Use VA x PF = W to determine the Watts at full load.
Use (VA² - W²)^0.5 = VAR

Now that you have found an approximation for VAR you may use the formula:
(VA² - VAR²)^0.5 = W

This may give the most accurate estimation of the Watts consumed by the motor subject to the following caveats:
1> electricpete pointed out a while ago that the VARs are not constant with motor loading, however an approximation including a factor for VARs will be much more accurate than approximations that neglect VARs or that assume a constant PF. (thanks Pete)
2> This will give an approximation of the Watts consumed by the motor. You may want to add a factor to account for the losses in the inverter.
When approximating losses remember that some types of loss may be relatively constant over the load range while some losses vary as the square of the current.


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