ION 7650 energy calculation

[joel_mora_soati]

Hello experts,

I have a customer asking about how ION Meter do energy calculations. He made a manual calculation to compare energy calculated manually to meter calculation as image attached.

Just the "error" and "Intervalo de energía activa CALCULADA" columns are calculated, the other ones are using ION meter variables outputs took from PME.

The formula used by customer for manual energy was:

((SUM of three phase currents* AVG Three Phase Voltage * ^3 * Power Factor)/4000) It is also a image attached "formula"

The customer now is asking me why is a big difference between calculations and measured value. I wonder if you can help me to justificate technically that ION meter is calculating in the proper way.

Thanks!

[Charles_Murison · Accepted solution]

Hello @joel_mora_soati,

Moved the question to the metering forum

The direct question, how does the ION7650 calculate energy? This is done by integrating the power every second. Have a look at the ION Reference Integrator module for more information.

Regarding your equation, there are a few unknowns like what volts mode does the ION7650 meter have, what version of firmware, what template, what setting for the kVA summation. I will assume you have a 4w wye system.

 
In a Wye system, kW total = kW a + kW b + kW c, where kW a ~= Ia * VLN a * PF, since we do not have VLN a, I will estimate VLN A ~= VLL AB / SQRT(3), repeat for each phase. 

It looks to me like formula should be /RAIZ(3) and not *RAIZ(3)

If you have a Delta system the math would be different as there would be a 2 watt meters. to calculate power.

Regards,

Charles

[joel_mora_soati]

Thanks @Charles_Murison 

It is very clear for me.

Refering to the manual calculation, to calculate the active power which of the meter measurements are correct to use:

- Current X Mean (sum the 3 phases) or Current Avg Mean? =>>>>> (Let say this is I in the formula below)
- Voltage X Mean (Average of 3 phases) or Voltage LL Avg Mean? =>>>>(Let say this is V in the formula below)

As you said, the proper way to calculate will be:

kW = I*(V/SQRT(3))*PF

I really apreciate your help, thanks.

[Charles_Murison]

Hello @joel_mora_soati ,

The example provided is very simplified, this assumes steady-state, sinusoidal signals. The calculation in general and voltage specifically would depend on what volts mode the system is. Note that the meter likely also logged the power demand. 

In general

P = S x PF

where S = V x I, and PF = P / S

P = active power

S = apparent power

Regards,

Charles