Calculation of kWh with instant A and U
Calculation of kWh with instant A and U
(OP)
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 ?
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 ?






RE: Calculation of kWh with instant A and U
RE: Calculation of kWh with instant A and U
Bill
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"Why not the best?"
Jimmy Carter
RE: Calculation of kWh with instant A and U
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.
RE: Calculation of kWh with instant A and U
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
www.gke.org
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: Calculation of kWh with instant A and U
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Calculation of kWh with instant A and U
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.
RE: Calculation of kWh with instant A and U
If power is not available, you will get a much better idea if you multiply torque and speed and adjust for efficiency.
Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: Calculation of kWh with instant A and U
Rafiq Bulsara
http://www.srengineersct.com
RE: Calculation of kWh with instant A and U
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