Impedance Modeling Issues and Equivalent Impedance Examples
Explore impedance modeling challenges in generating units like battery-inverter modules and transformers, affecting system accuracy. Learn about equivalent impedance examples and step-up transformer specifications for efficient system operation and maintenance.
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Presentation Transcript
DGR (Battery) equivalent impedance expectation 7/28/2020
Simplest model Real equipment like circuit breaker is needed between line and node1 To isolate DGR to grid no pseudo-equipment is needed between DGR and node1 CLR and node1 Advantage Minimize pseudo-equipment Simplify operation model Facilitate outage coordination Line node2 Breaker node1 DGR CLR 2 PUBLIC
RARF sample RARF Planning Gen Data 3 PUBLIC
Impedance modeling issue Impedance does not reflect equivalent impedance of entire generating unit RARF Planning GEN tab one battery-inverter module Impedance does not reflect impedance of entire generating unit does not have pad-mount step up transformer data Transformer impedance derived from test report or word document which is not part of RARF data does not have GenUnit-Battery-inverter-transformer configuration Estimate based on one-line Example 1 unit with 4 batteries, each batteries connect to one inverter, each inverter to one transformer 4 PUBLIC
Impedance modeling issue contd Other generators using GEN FORM has equivalent impedance impedance data in RARF not accurately reflected in ERCOT models affecting downstream systems 5 PUBLIC
Equivalent impedance example ? ?? 11 ??? ? ?? 2.75 ??? Module1 R=0.02 p.u. X=1.02 p.u. R=0.08 p.u. X=4.08 p.u. ???????1= ?????????1 Module2 R=0.02 p.u. X=1.02 p.u. R=0.08 p.u. X=4.08 p.u. ???????2= ?????????2 Module3 R=0.02 p.u. X=1.02 p.u. R=0.08 p.u. X=4.08 p.u. ???????3= ?????????3 Module4 R=0.02 p.u. X=1.02 p.u. R=0.08 p.u. X=4.08 p.u. ???????4= ?????????4 6 PUBLIC
Equivalent impedance example contd Test Step-up Transformer Specifications Voltage Low Side: 0.5 kV High Side: 12.5 kV MVA ratings: 2.75 MVA Positive sequence impedances: 6% Zero sequence impedance: 5% X/R: 10 Available Taps: 5 ( 2 x 3%) ???= 6% 11 2.75 ???= ??? sin tan 1?/? =6% sin tan 110 4 = 0.23881 ?.?. 11 2.75 ???= ??? cos tan 1?/? =6% cos tan 110 4 = 0.023881 ?.?. 7 PUBLIC
Equivalent impedance example contd ??1= ???1+ ???????1 ??1= 0.08 + 0.023881 = 0.103881 ?.?. ??1= 4.08 + 0.23881 = 4.31881 ?.?. ??2= ???2+ ???????2 ??1= 0.08 + 0.023881 = 0.103881 ?.?. ??1= 4.08 + 0.23881 = 4.31881 ?.?. ??3= ???3+ ???????3 ??1= 0.08 + 0.023881 = 0.103881 ?.?. ??1= 4.08 + 0.23881 = 4.31881 ?.?. ??4= ???4+ ???????4 ??1= 0.08 + 0.023881 = 0.103881 ?.?. ??1= 4.08 + 0.23881 = 4.31881 ?.?. 8 PUBLIC
Equivalent impedance example contd ???= ??1 ??2 ??3 ??4 1 ???????????= 1 1 1 1 ??1+ =0.103881 ??2+ ??3+ ??4 = 0.02597 ?.?. 4 1 ???????????= 1 1 1 1 ??1+ =4.31881 ??2+ ??3+ ??4 = 1.0797 ?.?. 4 9 PUBLIC
Q & A QUESTIONS? 10 PUBLIC
