crshears,
You may avoid a little discomfort by wearing conductive footwear when touching control cabinets. If you are not standing on a ground mat and touching a grounded structure when a high voltage ground fault occurs, you may experience fibrillation and death if you aren't wearing...
With a grd Wye - grd Wye connection to the utility, the BESS system may be furnishing zero sequence current to utility system unbalanced load. There also may be unbalanced plant load causing neutral current. If the neutral is grounded both at the service entrance and at the BESS, there could...
kW demand is the difference between consecutive 15 min kWh readings times 4.
If you have an average power use for 15 minutes of 1 kW, then the kWh usage for that period is 1 kW time 0.25 hr, or 0.25 kWh.
Some utilities provide more detail. I can go in the Duke Energy website, or use their mobile app, and download kWh readings in 30 minute intervals for my home energy meter. With this, you can calculate 30 minute demand values.
If you use another phase for the 0°, you will get a different answer, but only because the angles are rounded to zero decimals. Note that the sum of all of the phase angle differences has to equal exactly 360°, but 119 + 119 + 123 = 361.
I arbitrarily chose 0° for the R (A1) phase angle. The difference between R & Y phase angles is 119° per the meter (Phi 12), so Y phase angle is 0 - 119 = -119°. The difference between B & R phase angles is 119° (Phi 31), so B phase angle is 0 + 119 = 119°.
For Full Load
Phase Mag Angle deg Angle Rad Real Im
R 180.2 0 0 180.2 0
Y 208.1 -119 -2.08 -100.89 -182.01
B 191.4 119 2.08 -92.79 167.40
N 19.88 -13.48 -14.61
Note...
Note that B peaks considerably after the crossing of R&Y. This would occur if the angles of R&Y were 0 and -120, and the angle of B was 100 deg.
Try moving the cursor on the meter to determine the zero crossing times of the three currents.
In this case, you can't just treat it as a second remotely connected grid with a current split factor. The buried bare 250 kcmil conductors will dissipate current into the earth. If the distance between grids is large, then there will be a voltage drop and a reduction in GPR, but it won't be...
The OP's equation only works if the currents are 120 deg apart. If R is 0 deg, Y is -120, and B is 110 deg, the neutral current would be 17.15.
See attached spreadsheet.https://files.engineering.com/getfile.aspx?folder=13002f2a-9fd4-4d86-970f-19188bf6cc8c&file=Unbalanced_Currents.xlsx
So, if you have a VFD efficiency of 0.96 and a measured power into the VFD of 114.5 kW, the power out of the VFD into the motor is 114.5*0.96 = 109.9 kW. If the motor has a 0.95 efficiency, then the power into the shaft is 109.9*0.95 = 104.4 kW.
If the ground connection between the main and remote ground grids is by overhead lines or insulated cables, you could treat the remote ground as a terminal and use software such as SES FCDIST to determine how much current flows to the remote grid and the GPR at the remote terminal.
If the...
True except for the becoming obvious part. In a single phase to ground fault, there is no net current in the unfaulted phases. The I0 is offset by I1 and I2 in the unfaulted phases.
Most microprocessor based recloser controls will have an input available that you could program to trip the recloser from the differential relay. Some may require an optional module.
Non-microprocessor controlled reclosers will not have that capability.