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Contingency Ranking by Time Domain Simulations ECE 422/522 Russell Patterson Micah Till Terry Jones

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Project Goals Identify top five critical contingencies in Zones 2-A and 2-B

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Project Goals Rank the top five contingencies by the stability criterion. Rank the top five contingencies by their Critical Clearing Time and compare the ranking with that based on values of Determine the effect of varying load models

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Task 1: Rank N-1 Contingencies Contingency(η)(η)Faulted LineZone LM to 83 circuit 12-A LM to 173 circuit 12-A LM to 168 circuit 12-A LM to 114 circuit 12-A LM to 170 circuit 12-A

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Task 1: Rank N-1 Contingencies Bus 83 is the major bottleneck for power flow between Zones 1-A and 2-A As seen, all of the buses in Table 2 are electrically close to bus 83 Many of these branches are negative reactance, which means they represent series capacitors In reality, some of the series capacitors would be bypassed during faults almost 4GW

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Briefly - Series Capacitors Used to increase the MW transfer capability of the path (X L – X C ) 60% series compensation means X C = 60% of X L Too much compensation can lead to problems like overvoltage and subsynchronous resonance (SSR)

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Briefly - Series Capacitors subsynchronous resonance (SSR) can result if too much series compensation is applied More X C means higher f R SSR usually in the range of 10 to 50Hz rad/s Hz IEEE definition: “Subsynchronous resonance (SSR) is an electric power condition where the electric network exchanges energy with a turbine/generator at one or more of the natural frequencies of the combined system below the synchronous frequency of the system.” Reference: EPRI Power Systems Dynamics Tutorial

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Briefly - Series Capacitors Series capacitors are bypassed when current through produces voltage across of % Initially bypassed by arrester (non-linear) Hard bypassed prior to exceeding MOV energy capability V I

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Briefly - Series Capacitors Series capacitors directly in faulted line will be hard bypassed Series capacitors farther away will likely just be partially bypassed as their voltage peaks Those not hard bypassed will be quickly re-inserted after fault is cleared Hard bypassed will take as long as 5-cycles to reinsert (if at all)

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Briefly - Series Capacitors Series compensation at Dafang, China Ref: “Series Capacitor By-pass Switch – ABB”

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Briefly - Series Capacitors How can series capacitor bypassing be handled in transient stability analysis? TSAT – create user defined model which can be accessed in dynamic editor as SERUDM (TSAT has a test case of a thyristor-controlled series compensator (TCSC) model) PSS/E has model called SCGAP2 PSLF has model called SCGAP

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Task 1: Rank N-1 Contingencies To test the assumption that properly modeled series capacitor bypassing would change the results, a 3-phase fault was place close-in to bus 83 and the rotor angle of generator 112 was recorded. This was repeated with the capacitive branches shorted out (bypassed). Note that while the results will eliminate the effects of series capacitors, they still will not accurately reflect system response

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Task 1: Rank N-1 Contingencies Initial run in black, run with series capacitors bypassed in red Note that initial case loses stability shortly after fault inception

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Task 2: Critical Clearing Time TSAT was able to identify the CCT by gradually increasing the clearing time until the stability criterion became negative. The closer the faulted line is to generator 112, the smaller the CCT ContingencyCCT (cycles)Faulted Line LM to 83 circuit 1 LM to 173 circuit 1 LM to 168 circuit 1 LM to 114 circuit 1 LM to 170 circuit 1

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Task 2: Critical Clearing Time Cannot transmit power through zero voltage bus

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Task 3: Load Model Comparison The load models were varied as shown: Note that conversion to a constant impedance load was required below a certain threshold for TSAT to calculate solutions for all contingencies Results for LM2, LM3, and LM4 were compared to the CCT values from LM1 Load ModelDescriptionConversion LM1 P as 100% constant I Q as 100% constant Z V = 0.7 p.u. LM2P and Q as 100 % constant ZN/A LM3P and Q as 100 % constant IV = 0.35 p.u. LM4P and Q as 100 % constant PV = 0.85 p.u.

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Task 3: Load Model Comparison Constant impedance (LM2) performs better than the mixed model (LM1) Constant current (LM3) and power (LM4) cannot solve unless certain buses are converted back to constant impedance (LM2) Even after this, LM3 and LM4 underperform LM1 Contingency CCT (cycles) LM1LM2LM3LM4 LM LM LM LM LM

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Task 3: Load Model Comparison Thank You!

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