LJP1
Electrical
- Nov 1, 2018
- 18
Need help with help with NGR rating for Option#1 re: EPRI Technical Brief_Effective Grounding and Inverter-Based Generation. How to properly calculate current through NGR when connected to neutral of Wye winding of Wye-g/delta(not a zig-zag) grounding transformer. What is valid impedance model and calculation to yield correct current rating.
Background (See DG layout PDF)
A new 8 MVA distributed generation (DG#2) wind plant will be connected to an existing substation DG feeder. The new DG will share a parallel connection on this feeder with a previously existing 4.5 MVA DG wind plant (DG#1). Due to the significant total amount of parallel generation resulting on this feeder and the substation due to the addition of DG#2, Utility commissioned an engineering study to model and analyze the impact of various DG grounding configurations and their zero-sequence current sources to better understand overvoltage risk and relay protection coordination issues during a ground fault. The results of this study indicate both a risk for overvoltage greater than 1.39 PU and a risk of relay miscoordination during ground fault scenarios where the utility source and system ground reference may be lost.
Furthermore, the study results show that DG#2 is reliant on DG#1’s Yg-Yg step-up transformer and other distribution ground sources to fully mitigate these risks. Specifically, the DG#2 fails to meet an acceptable Coefficient of Grounding (CoG) of <0.80 while solely relying on the capacity of its own, single, 500 KVA grounding transformer (GT). As such, DG#2 will require a larger capacity GT solution to be designed and installed before it will be allowed to operate independently of other distribution system ground sources in-service.
Utility is requesting DG#2 install the equivalence of 2 MVA of GT capacity and install the equivalence of 1-ohm neutral grounding resistor(s) (NGR), in combination with the GT impedance, to improve its own dedicated CoG for the DG#2 wind plant and to mitigate overvoltage risk during ground fault scenarios.
GT-NGR design considerations and options:
1. One (1) 2 MVA GT with 1 Ω NGR unit:
a. Grounding transformer impedance is 6% at 2 MVA.
b. Converting to ohms at 13.8 kV; 0.06 pu * 13.8 kV * 13.8 kV / 2 MVA = 5.7 Ω.
c. The equivalent impedance = 5.7 Ω + 1 Ω = 6.7
d. The estimated the overall ampere rating will be 13800 V / √3 / (5.7 + 1) Ω ~ 1200 A.
e. Rated for 10 secs per IEEE C62.92.3 and C57.32
2. Four (4) - 0.5 MVA GT with 4 Ω NGR unit, connected in parallel
(equivalent circuit to Option 1 and shown in attached PDF DG layout):
a. Grounding transformer impedance is 6% at 0.5 MVA.
b. Converting to ohms at 13.8 kV; 0.06 pu * 13.8 kV * 13.8 kV / 0.5 MVA = 22.9 Ω.
c. Impedance on each path = 22.9 Ω + 4 Ω = 26.9 Ω
d. The equivalent impedance = 1 / ((1/26.9)*4) = 6.7 Ω
e. The estimated overall ampere rating will be 13800 V / √3 / (23.9 + 4) Ω ~ 300A
f. Rated for 10 secs per IEEE C62.92.3 and C57.32
Alternative calculation approach yields different results that seems too high of current rating through resistor.
• Let positive-sequence Thevenin impedance be Z1.
• Let negative-sequence Thevenin impedance be Z2.
• In the zero-sequence circuit, we will have the full impedance of the ground bank in series with 3 * Z_NGR. Zero-sequence Thevenin = Z0_GroundBank + 3*Z_NGR Note that in the zero-sequence circuit, 1/3 of the NGR value will be present and not the full NGR
• Let VF be the prefault voltage. This is a L-N quantity, and equal to 8 kV (13.8 kV LL)
The fault current in each sequence can be calculated as follows:
Ia1 = Ia2 = Ia0 = VF / (Z1 + Z2 + Z0_GroundBank + 3 * Z_NGR)
This leads to (assuming Z1 = Z2 = 0):
Ia = 3I0 = 3VF / (Z0_Groundbank + 3 * Z_NGR)
Ia = 3I0 = VF / (Z0_Groundbank / 3 + Z_NGR)
• Current with 2.0 MVA, 1 Ω NGR = 8000 V / (22.9 Ω / 4 / 3 + 1 Ω) = 2750 A.
• If we go with 4 x 0.5 MVA banks, with 4 Ω NGR:
o Current through each NGR = 8000 V / (22.9 Ω / 3 + 4 Ω) = 688 A
Background (See DG layout PDF)
A new 8 MVA distributed generation (DG#2) wind plant will be connected to an existing substation DG feeder. The new DG will share a parallel connection on this feeder with a previously existing 4.5 MVA DG wind plant (DG#1). Due to the significant total amount of parallel generation resulting on this feeder and the substation due to the addition of DG#2, Utility commissioned an engineering study to model and analyze the impact of various DG grounding configurations and their zero-sequence current sources to better understand overvoltage risk and relay protection coordination issues during a ground fault. The results of this study indicate both a risk for overvoltage greater than 1.39 PU and a risk of relay miscoordination during ground fault scenarios where the utility source and system ground reference may be lost.
Furthermore, the study results show that DG#2 is reliant on DG#1’s Yg-Yg step-up transformer and other distribution ground sources to fully mitigate these risks. Specifically, the DG#2 fails to meet an acceptable Coefficient of Grounding (CoG) of <0.80 while solely relying on the capacity of its own, single, 500 KVA grounding transformer (GT). As such, DG#2 will require a larger capacity GT solution to be designed and installed before it will be allowed to operate independently of other distribution system ground sources in-service.
Utility is requesting DG#2 install the equivalence of 2 MVA of GT capacity and install the equivalence of 1-ohm neutral grounding resistor(s) (NGR), in combination with the GT impedance, to improve its own dedicated CoG for the DG#2 wind plant and to mitigate overvoltage risk during ground fault scenarios.
GT-NGR design considerations and options:
1. One (1) 2 MVA GT with 1 Ω NGR unit:
a. Grounding transformer impedance is 6% at 2 MVA.
b. Converting to ohms at 13.8 kV; 0.06 pu * 13.8 kV * 13.8 kV / 2 MVA = 5.7 Ω.
c. The equivalent impedance = 5.7 Ω + 1 Ω = 6.7
d. The estimated the overall ampere rating will be 13800 V / √3 / (5.7 + 1) Ω ~ 1200 A.
e. Rated for 10 secs per IEEE C62.92.3 and C57.32
2. Four (4) - 0.5 MVA GT with 4 Ω NGR unit, connected in parallel
(equivalent circuit to Option 1 and shown in attached PDF DG layout):
a. Grounding transformer impedance is 6% at 0.5 MVA.
b. Converting to ohms at 13.8 kV; 0.06 pu * 13.8 kV * 13.8 kV / 0.5 MVA = 22.9 Ω.
c. Impedance on each path = 22.9 Ω + 4 Ω = 26.9 Ω
d. The equivalent impedance = 1 / ((1/26.9)*4) = 6.7 Ω
e. The estimated overall ampere rating will be 13800 V / √3 / (23.9 + 4) Ω ~ 300A
f. Rated for 10 secs per IEEE C62.92.3 and C57.32
Alternative calculation approach yields different results that seems too high of current rating through resistor.
• Let positive-sequence Thevenin impedance be Z1.
• Let negative-sequence Thevenin impedance be Z2.
• In the zero-sequence circuit, we will have the full impedance of the ground bank in series with 3 * Z_NGR. Zero-sequence Thevenin = Z0_GroundBank + 3*Z_NGR Note that in the zero-sequence circuit, 1/3 of the NGR value will be present and not the full NGR
• Let VF be the prefault voltage. This is a L-N quantity, and equal to 8 kV (13.8 kV LL)
The fault current in each sequence can be calculated as follows:
Ia1 = Ia2 = Ia0 = VF / (Z1 + Z2 + Z0_GroundBank + 3 * Z_NGR)
This leads to (assuming Z1 = Z2 = 0):
Ia = 3I0 = 3VF / (Z0_Groundbank + 3 * Z_NGR)
Ia = 3I0 = VF / (Z0_Groundbank / 3 + Z_NGR)
• Current with 2.0 MVA, 1 Ω NGR = 8000 V / (22.9 Ω / 4 / 3 + 1 Ω) = 2750 A.
• If we go with 4 x 0.5 MVA banks, with 4 Ω NGR:
o Current through each NGR = 8000 V / (22.9 Ω / 3 + 4 Ω) = 688 A