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Assignment 4 :PPCM

by: Shammya Saha

Assignment 4 :PPCM EEE 598

Shammya Saha
GPA 3.83

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This is the solution for assignment 4, EEE 598 : Power Plant Control & Monitoring
Power System Control and Monitoring
Dr. Undrill
Class Notes
Speed Governor
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This 13 page Class Notes was uploaded by Shammya Saha on Sunday April 3, 2016. The Class Notes belongs to EEE 598 at Arizona State University taught by Dr. Undrill in Fall 2015. Since its upload, it has received 17 views. For similar materials see Power System Control and Monitoring in Electrical Engineering at Arizona State University.

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Date Created: 04/03/16
Shammya Saha Assignment 4 120 950 8778 The system is this assignment contains auxiliary that is supplied by generator 1. Here generator 1 is feeding the auxiliary motor through a transformer, the auxiliary motor may be responsible for running the lubricating oil supply or cooling fan. As the dynamics of generator related to motor tripping is not modeled, the system output might look stable while the motor is tripped. This does not necessary mean that the plant is stable at that situation as tripping of motor might cause the generator to overheat which will cause the generator to be tripped. Considering both the generators are producing 175MW a three phase to ground fault on one of the transmission lines is simulated in the PSLF. The motor parameters used in the simulation are provided in Table 1. Table 1: Motor Parameters Variable Description Value Synchronous reactance 3.5 Transient reactance 0.19 Stator resistance, p.u. 0.02 Transient rotor time constant 1.83 Inertia constant, sec. 0.3 Damping factor, p.u 2.0 Saturation factor at 1 p.u. flux 0.01 Saturation factor at 1.2 p.u. flux 0.1 Voltage threshold for tripping, p.u 0.85 Voltage trip pickup time sec. 2.5 Frequency threshold for tripping ,Hz 0.1 Frequency trip pickup time sec. 99 Voltage at which reconnection is permitted p.u. 1 Time delay for reconnection sec. 99 Acceleration factor for initialization 0.5 Sub-transient reactance, p.u 0.165 Stator leakage reactance, p.u 0.12 Sub-transient rotor time constant, sec. 0.02 Time step subdivision factor. 10 Speed threshold for subdividing time step, p.u. 0.8 The 13.8KV / 4.16 KV transformer which is serving the motor is considered to have an impedance of 5.59% [1]. Power factor for the motor is considered to be 0.8 [4]. Considering the motor is driving the lubricating oil pump motor or cooling fan load, the inertia constant is considered to be 0.3. Considering the fault scenario given in the question statement a PSLF simulation is run to analyze the effect of the fault on the plant when both the generators are producing 175 MW. Figure 1: Generator & Motor States during Fault 2 From Fig. 1, it is evident that in the event of the fault when the breaker clears the fault at 12.5 cycle due to backup protection, the auxiliary of the plant is going to stop (from motor speed plot). Moreover, both the generator have oscillatory terminal voltage and power plot which indicates the system instability. The following figure shows the effect of motor inertia constant on motor speed drop when the fault Figure 2: Motor Speed Plot for Different H value occurs and cleared finally at 12 cycles. It is clear from Fig. 2 is that any reasonable value of the inertia constant (0.3s to 0.6s) does not refrain the motor from stopping. So it can be clearly stated, if bad switching event occurs, the plant is not going to survive this event. Possible solutions to keep the Plant Auxiliary Running: The objective of the following analysis provided below is to keep the auxiliary running so that the plant does not go to full shutdown mode. Analysis has been done changing possible parameters and PSLF simulation is run for individual case. 3 Solution -1: Considering the bus 3 breaker opens at 12 cycles by backup protection, simulation has been run by tripping generator 2 just after 12 cycles i.e. 12.6 cycles (a time delay of 0.1 seconds from the backup breaker protection) considering the motor inertia to be 0.5, slightly increased from the previous assumed value of 0.3 so now the motor deaccelerates a little slow. Motor has it’s under voltage protection setting set to be at 0.85 pu and voltage trip pickup time is set to be 2.5 seconds. From Fig. 3 it can be mentioned that the plant auxiliary is still running which ensures the plant survives this event. There will be loss of power which is being transmissted through tranmission linebuttheplantwillnot gointoashutdownmode.Generator1powerhasanoscillatorybehaviour but it reaches to a stable value. Figure 3: Motor Generator Output for Solution 1 (H= 0.5, ????????=0.85pu, ????????=2.5s) So if the motor inertia is something close to 0.3, it is evident that motor has to be repalced by a high inertia setup (minimum 0.5s) of motor considering all other paramters are same as presented in Table 1. And in the event of the tripping the circuit breaker at 12 cycles, generator 2 has to be tripped. 4 As the motor reacclerates after the fault event another PSLF simulation is run considering the inertia constant to be 0.5 but changing the motor undervoltage trip limit to be 0.7 and the trip timer to be 1 second. Figure 4: Motor Generator Output for Solution (H= 0.5, ????????=0.7pu, ????????=1s) Comapring Fig. 3 and Fig. 4 it can be concluded that in both the cases the system is stable and the auxiliary will stay running. 5 Figure 5: Comparison of Motor Parameter for Different Vt and Tv value (Blue Curve Vt=0.85, Tv=2.5; Red Curve Vt=0.7, Tv=1) Figure 5 shows a comparison for motor paramters depicted in Figure 3 and Fig. 4. The terminal current plot is higher when the voltage threshhold is set at 0.7 and trip timer is set at 1s.This is due tolowervoltagethreshholdwhichmight triggertheover currentrelayinthe motor.So,itisadvised that the plant operator runs the auxiliarymotor with under voltage protection setting set at 0.85 pu. 6 Determining the Time Delay for Tripping Generator 2: The time delay to trip generator 2 after the tripping of backup circuit breaker is also calcualted by dynamic simulation. By varying the under voltage trip setting and under voltage trip timer the maximum allowable delay is found to be 0.04 seconds. So to conclude, the generator 2 has to be tripped within 0.04 seconds of the tripping event of the circuit breaker i.e. within 14.4 cycles of the fault occurance, otherwise the motor goes into locked rotor state and starts drawing almost 500% of the rated current. This will force the over current protection relayto trip the motor. Figure 6 shows the motor and generator paramtere outputs when generator 2 is tripped at 14.4 cycles. Figure 6: Motor & Generator parameters when the generator 2 is tripped at 14.4 cycles With an intention to increase the time delay to ensure more reliable operation, the motor under voltage trip limit is set to 0.7 pu and the trip time is set to be 4 second. The time delayfor generator tripping is increased from 0.04 to 0.05 seconds. Figure 5 shows the simulated results from PSLF. 7 Figure 7: Motor & Generator Parameters when Generator 2 is tripped at greater than 14.4 cycles Figure 7 shows that both the generator have their terminal voltage stable after the fault. The oscillation in the generator 1 power output gets damped. But if the motor speed is considered, it does not stop completely, but goes into a locked rotor state. In this situation, it rquires a high reactive power and that is why, the terminal current stays at 5 pu for more than 20 seconds and does not settle down to 1 per unit. As PSLF does not consider the overcurrent protection settings in motor model, the motor does not trip in the simualtion. But in real world, the overcurrent protetion will trip the motor. Consiequently, the plant is going to lose its auxiliary and goes into shut down mode. Hence to summarize solution 1,  The motor inertia constant is to be minimum 0.5.  Generator 2 has to be tripped so that plant can survive the fault.  Generator 2 has to be tripped within 0.04 seconds of clearing the fault by the from bus breaker at 12 cycles. This gives an additional time delay of 2.4 cycles. 8 Solution-2: As the auxiliary of the plant is the concern here, an additioanl battery backup can be included in the plant so that in case of fault, the rquired power can be supplied from the batterybackup through an inverter that can produce enough real & reactive power so the auxiliary of the plant is still on. A rough battery bank calculation is provided here. As the real power requirement is 800KW bythe motor, considering the battery backup can run the auxiliary for 30 minutes, the required power output by the inverter is 400KWh. Considering an inverter effeciency to be 95%, the battery should provide 420KWh to the inverter. Considering a DC bus voltage of 840V which corresponds to 70 piece 12V lead acid battery in series, the required ampere hour rating of the batery is to be = (420 ∗ 1000)/840 =500Ah Considering a depth of discharge of 80% and the high rate [3] of the batterythe ampere hour rating has to be = 500 = 1250????ℎ 0.8∗0.5 So, for a 30 minute backup time 70 1250Ah Lead Acid batteries can provide backup through an inverter. Econominc Analysis: A sample battery pricing is used [2] to do the economic analysis for the battery bank setup. Each battery price is 3968$ which results an total investment of 4.96 million dollar for the battery. An inverter that has a maximum input DC voltage range of 1000V can provide the required battery backup (maximum allowable time delay of 0.24 seconds) during the fault which does not require tripping generator 2. Using the above technique the ampere hour rating of battery required corresponding to back up time for a fixed DC bus voltage of 840V is provided in the following table. Backup Time Required Ah Rating of Battery 10 minute 418 15 minute 627 30 minute 1250 Solution 3: According to the question statement, both the generators are producing 175 MW when the fault occurs in the system. A PSLF simulation is run by changing the generation to 160MW to 175 MW and applying the same fault. The motor ineria constant for this simulation is considered 0.5s where the under voltage limit is 0.85 pu and the trip pickup time is 2.5 seconds. It has been found that system shows a stable behaviour at this generation limit and no generator has to be tripped to achive this stability[Fig 8]. 9 Figure 8: Motor & Generator Parameters when both the generators are producing 160MW From Fig. 8 it can be concluded that if the power plant generators are producing 160MW each, the plant can stay stable though the fault is cleared by backup protection. The advantage of this solution is there is no requirement of tripping generator 2. On the other hand, as the power plant is not operating at its maximum capacity the plant has less revenue. Moreover, it is tough to determine the probability of this fault occuring and clearing issue simulataneously. To summarize solution 3, the plant can stay safe in this event if the plant limits her total generation to 320MW (each generator contributing equally to the production). In this solution, no generator has to be tripped. But this solution causes a constant loss of revenue. Solution 4: The main reason of the fault is the from bus breaker made a delay of opening the lines. So if the plant owners owns the transmission (verticallyintegrated system), changing the backup protection time setting of the circuit breaker from 12 cycles to 9.6 cycles can solve the problem. A PSLF simulation has been run in order to address the scenario, while keeping the motor inertia constant at 0.3, under voltage trip limit to be 0.85 and time setting to be 2.5 seconds. 10 It is found that the system is stable even both the generators are producing 175MW. Figure 9: Motor & Generator Parameters for fault clearing at 9.6 cycles Keeping the switching time at 9.6 cycles, the minimum value for the motor trip time is found to be 0.68 seconds. That means, the system can stay stable in this fault event if the motor trip time is set to be 0.68 seconds considering the backup protection works at 9.6 cycles. Solution 5: Solution 5 provides a PSLF simulation where the motor inertia constant is kept to be 0.3 where the transient reactance of motor is changed from 0.19 to 0.16. bot hthe generators are producing 175 MW. It has been found that generator 2 has to be tripped to keep the system stable but now generator 2 can be tripped within a maximum limit of 28.8 cycles from the fault occurance i.e. generator 2 has to be tripped within 16.8 cycles after the backup protection works. Figure 10 shows the simulated results of motor & generator parameters where Fig. 11 shows the motor speed value for different generator 2 tripping time. 11 Figure 10: Motor & Generator Parameters when Transient Reactance is 0.16 and Generator 2 is tripped Figure 11: Motor Speed Plot for Generator Trip Time when Transient Reactance is 0.16 12 Reference : [1] [2] battery-bank [3] [4] “Load Modelling for Power Flow and Transient Stability Computer Studies, Volume 2” by EPRI 13


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