Two & Three Winding Transformers and Circulating Fault Currents By: Abdur Rehman 2019 Hands on Relay School PSE.com // GeneralPAC.com 1
Purpose at the 2019 HRS • Intention is to share our knowledge and passion for Power Systems Engineering & Protection • NOT intended to replace any lecture… • Intention is to supplement other lectures PSE.com // GeneralPAC.com
Assumed Knowledge • Basic understanding of symmetrical components - Phase & Ground faults • Working knowledge of Transformer Phase Shifts • Fundamental understanding of power systems PSE.com // GeneralPAC.com
Outcomes & Takeaways • Better understanding of transformers under faulted conditions • Practical application of sequence network diagrams • Visual display of circulating sequence currents for 2 & 3 winding transformers • Introduction to help better understand the handouts PSE.com // GeneralPAC.com
Outcomes & Takeaways • Sit back… Relax… and Enjoy the show. • The step-by-step details are in the accompanying PDF for your review… Presentation = Introduction • Make sure to send me an email for the accompanying PDF document, presentation, and videos… abdur.rehman@pse.com PSE.com // GeneralPAC.com
Video Tutorial available on Youtube • Search for “General PAC” or visit generalpac.com • Topic “Fault Analysis in Power System” PSE.com // GeneralPAC.com
D AB Y-G 2 Winding Transformer 30 MVA 115 kV 13.8 kV MVA SC = 600 MVA Example Y AB Line A Line a Line B Line b Line C Line c PSE.com // GeneralPAC.com
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6 Step Process to calculate fault current Step 1: Convert the system quantities into per-unit quantities Step 2: Identify the type of fault Step 3: Draw the unfaulted sequence networks Step 4: Interconnect the faulted sequence networks Step 5: Hand Calculate Sequence Currents & Voltages Step 6: Convert to Phase currents & voltages PSE.com // GeneralPAC.com
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Sequence Components Source Generator Transformer System 115kV 13.8kV (1) (1) Z G Z T (1) (1) Z G Z T j0.10 j0.05 j0.05 j0.10 Positive + = 1pu 1pu Sequence 115kV 13.8kV (2) (2) Z G Z T (2) (2) Z G Z T Negative j0.05 j0.10 j0.05 j0.10 + = Sequence 115kV Open Open 13.8kV (0) (0) Z G (0) (0) Z T Z G Z T Zero j0.05 j0.10 j0.05 j0.10 = + Sequence PSE.com // GeneralPAC.com
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3PH Fault 1LG Fault 115kV 13.8kV (1) Z G (1) Z T 115kV 13.8kV (1) Z G (1) Z T 1 PU 1 pu 115kV 13.8kV (2) Z G (2) Z T Open 115kV 13.8kV (0) Z G (0) Z T PSE.com // GeneralPAC.com
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3 Phase Fault Big Picture 1 of 5 PSE.com // GeneralPAC.com
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3 Phase Fault Big Picture Comparison 1004|-60 8367|-90 8367|150 1004|180 8367| 30 1004| 60 PSE.com // GeneralPAC.com
1LG Fault Big Picture PSE.com // GeneralPAC.com
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Step #4 PSE.com // GeneralPAC.com
Step #5 PSE.com // GeneralPAC.com
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1LG Fault Big Picture Comparison PSE.com // GeneralPAC.com
LL Fault Big Picture 1 of 1 PSE.com // GeneralPAC.com
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2LG Fault Big Picture 1 of 1 PSE.com // GeneralPAC.com
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3 Winding Transformer Example Y-D-Y G 30 MVA 115 kV 13.8 kV MVA SC = 600 MVA PSE.com // GeneralPAC.com
Sequence Components Transformer System Generator (1)) 115kV 13.8kV (1) (1) (1) (1) Z H Z M Z H Z M Z G (1) Z G (1) (1) Z L Z L Positive + = 1pu 1pu Sequence T T (2) (2) Z H Z M (2) 115kV 13.8kV (2) (2) Z H Z M (2) Z G Z G Negative (2) Z L (2) + Z L = Sequence T T 115kV Open 13.8kV 13.8kV (0) (0) (0) Z G Z T Z M Open (0) (0) Z H Z M (0) Z G (0) = Z L (0) Z L Zero + Sequence T T PSE.com // GeneralPAC.com
3 Phase Fault Big Picture 1 of 1 PSE.com // GeneralPAC.com
1LG Fault Big Picture 1 of 1 PSE.com // GeneralPAC.com
LL Fault Big Picture 1 of 1 PSE.com // GeneralPAC.com
Questions? Abdur.rehman@pse.com Generalpac.com PSE.com // GeneralPAC.com
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