0
\$\begingroup\$

I get from a Landis+Ggyr E650 electricity meter data via a serial IR head (62056-21 / D0) about every 20s. This meter provides a lot of data, but not the current actual active power (kW).

Some of the meter data is voltage, current and phase angles shown as followed.

CurrentL1 = 3.43 A
CurrentL2 = 3.35 A
CurrentL3 = 3.36 A
CurrentN  = 0.04 A

VoltageL1_N = 241.1 V
VoltageL2_N = 241.4 V
VoltageL3_N = 241.3 V

Angle U (L1) to U (L1) = 0*Deg
Angle U (L2) to U (L1) = 120*Deg
Angle U (L3) to U (L1) = 240*Deg
Angle I (L1) to U (L1) = 200*Deg
Angle I (L2) to U (L1) = 321*Deg
Angle I (L3) to U (L1) = 80*Deg

I know in general the formula P = U * I * cos phi * sqrt(3), but I am not sure how I can use these single phase values shown above to do that calculation.

How can I calulcate with these values above the 3-phase actual active power in kW?


PS: I also know that it is possible to do a power calculation by two kWh readings and time measurement, but it looks the meter is not updating the kWh counter very regularily and it takes a too long time difference (~2min) to get stable atual values.

\$\endgroup\$
3
  • \$\begingroup\$ Page 63 shows it will display Active and reactive power: manualslib.com/manual/1632363/… \$\endgroup\$
    – Solar Mike
    Apr 3, 2022 at 15:55
  • \$\begingroup\$ Unfortunatly does it not get send via the communication. I shared the received data there in a forum if you are interested in it. \$\endgroup\$ Apr 3, 2022 at 16:23
  • \$\begingroup\$ For the one we used, we would get the data online with a 20min time lag due to the Power company data "loop" but we got all the values... \$\endgroup\$
    – Solar Mike
    Apr 3, 2022 at 16:25

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Browse other questions tagged or ask your own question.