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Description of the Transmission System Setting the Initial Load Flow and Obtaining Steady State |
The example described in this section illustrates modeling of series compensation and related phenomena such as subsynchronous resonance in a transmission system.
The single-line diagram shown here represents a three-phase, 60 Hz, 735 kV power system transmitting power from a power plant consisting of six 350 MVA generators to an equivalent system through a 600 km transmission line. The transmission line is split into two 300 km lines connected between buses B1, B2, and B3.
Series and Shunt Compensated Transmission System
To increase the transmission capacity, each line is series compensated by capacitors representing 40% of the line reactance. Both lines are also shunt compensated by a 330 Mvar shunt reactance. The shunt and series compensation equipment is located at the B2 substation where a 300 MVA-735/230 kV transformer feeds a 230 kV-250 MW load.
Each series compensation bank is protected by metal-oxide varistors (MOV1 and MOV2). The two circuit breakers of line 1 are shown as CB1 and CB2.
This power system is available in the power_3phseriescomppower_3phseriescomp model. Load this model and save it in your working directory as case1 to allow further modifications to the original system.
Compare the SimPowerSystems™ circuit model (Series-Compensated System (power_3phseriescomp)) with the schematic diagram above (Series and Shunt Compensated Transmission System). The generators are simulated with a Simplified Synchronous Machine block. A Three-Phase Transformer (Two Windings) block and a Three-Phase Transformer (Three Windings) block are used to model the two transformers. Saturation is implemented on the transformer connected at bus B2.
The B1, B2, and B3 blocks are Three-Phase V-I Measurement blocks taken from the Measurements library. These blocks are reformatted and given a black background color to give them the appearance of bus bars. They output the three line-to-ground voltages and the three line currents. Open the dialog boxes of B1 and B2. Note how the blocks are programmed to output voltages in pu and currents in pu/100 MVA. Notice also that the voltage and current signals are sent to internal Goto blocks by specifying signal labels. The signals are picked up by the From blocks in the Data Acquisition subsystem.
The fault is applied on line 1, on the line side of the capacitor bank. Open the dialog boxes of the Three-Phase Fault block and of the Three-Phase Breaker blocks CB1 and CB2. See how the initial breaker status and switching times are specified. A line-to-ground fault is applied on phase A at t = 1 cycle. The two circuit breakers that are initially closed are then open at t = 5 cycles, simulating a fault detection and opening time of 4 cycles. The fault is eliminated at t = 6 cycles, one cycle after the line opening.
Series-Compensated System (power_3phseriescomp)
Now, open the Series Compensation1 subsystem of the power_3phseriescomp model. The three-phase module consists of three identical subsystems, one for each phase. A note indicates how the capacitance value and the MOV protection level are calculated. Open the Series Compensation1/Phase A subsystem. You can see the details of the connections of the series capacitor and the Surge Arrester block (renamed MOV). The transmission line is 40% series compensated by a 62.8 µF capacitor. The capacitor is protected by the MOV block. If you open the dialog box of the MOV block, notice that it consists of 60 columns and that its protection level (specified at a reference current of 500 A/column or 30 kA total) is set at 298.7 kV. This voltage corresponds to 2.5 times the nominal capacitor voltage obtained at a nominal current of 2 kA RMS.
A gap is also connected in parallel with the MOV block. The gap is fired when the energy absorbed by the surge arrester exceeds a critical value of 30 MJ. To limit the rate of rise of capacitor current when the gap is fired, a damping RL circuit is connected in series. Open the Energy & Gap firing subsystem. It shows how you calculate the energy dissipated in the MOV by integrating the power (product of the MOV voltage and current).
When the energy exceeds the 30 MJ threshold, a closing order is sent to the Breaker block simulating the gap.
Series Compensation Module
Series Compensation1/PhaseA Subsystem
Series Compensation1/PhaseA Subsystem/Energy and Gap Firing
Open the 300 MVA 735/230 kV Transformer dialog box and notice that the current-flux saturation characteristic is set at
[0 0 ; 0.0012 1.2; 1 1.45] in pu
These data are the current and flux values at points 1, 2, and 3 of the piecewise linear approximation to the flux linkage curve shown here.
Saturable Transformer Model
The flux-current characteristic is approximated by the two segments shown in the graph here. The saturation knee point is 1.2 pu. The first segment corresponds to the magnetizing characteristic in the linear region (for fluxes below 1.2 pu). At 1 pu voltage, the inductive magnetizing current is 0.0010/1.0 = 0.001 pu, corresponding to 0.1% reactive power losses.
The iron core losses (active power losses) are specified by the magnetization resistance Rm = 1000 pu, corresponding to 0.1% losses at nominal voltage.
The slope of the saturation characteristic in the saturated region is 0.25 pu. Therefore, taking into account the primary leakage reactance (L1 = 0.15 pu), the air core reactance of the transformer seen from the primary winding is 0.4 pu/300 MVA.
Before performing transient tests, you must initialize your model for the desired load flow. Use the load flow utility of the Powergui to obtain an active power flow of 1500 MW out of the machine with a terminal voltage of 1 pu (13.8 kV).
Open the Powergui block and select Machine Initialization. A new window appears. In the upper right window you have the name of the only machine present in your system. Its Bus type should be PV Generator and the desired Terminal Voltage should already be set to the nominal voltage of 13800 V. In the Active Power field, enter 1500e6 as the desired output power. Click the Compute and Apply button. Once the load flow is solved, the phasors of AB and BC machine voltages as well as currents flowing in phases A and B are updated in the left window. The required mechanical power to drive the machine is displayed in watts and in pu, and the required excitation voltage E is displayed in pu.
Pmec | 1.5159e9 W [0.72184 pu] |
E/Vf | 1.0075 pu |
Notice that Constant blocks containing these two values are already connected to the Pm and E inputs of the machine block. If you open the Machine block dialog box, you see that the machine initial conditions (initial speed deviation dw = 0; internal angle theta, current magnitudes, and phase angles) are automatically transferred in the last line.
Once the load flow is performed, you can obtain the corresponding voltage and current measurements at the different buses. In the Powergui block, select Steady State Voltages and Currents. You can observe, for example, the phasors for phase A voltages at buses B1, B2, and B3 and the current entering line 1 at bus B1.
B1/Va | 6.088e5 V ; 18.22 degrees |
B2/Va | 6.223e5 V ; 9.26 degrees |
B3/Va | 6.064e5 V ; 2.04 degrees |
B1/Ia | 1560 A ; 30.50 degrees |
The active power flow for phase A entering line 1 is therefore
$${P}_{a}={V}_{a}\cdot {I}_{a}\cdot \mathrm{cos}\left({\phi}_{a}\right)=\frac{608.8\text{}kV}{\sqrt{2}}\cdot \frac{1.56\text{}kA}{\sqrt{2}}\cdot \mathrm{cos}\left(\mathrm{30.50.}-18.22\right)=464\text{}MW$$
corresponding to a total of 464 * 3 = 1392 MW for the three phases.
To speed up the simulation, you need to discretize the power system. The sample time is specified in the Powergui block as a variable Ts. The sample time Ts=50e-6 has already been defined in the Model Initialization function in the Callbacks of the Model Properties. The sample time Ts is also used in the Discrete Integrator block of the MOV energy calculator controlling the gap.
Ensure that the simulation parameters are set as follows.
Stop time | 0.2 |
Solver options type | Fixed-step; discrete (no continuous state) |
Fixed step size | Ts |
Ensure that the fault breaker is programmed for a line-to-ground fault on phase A. Start the simulation and observe the waveforms on the three scopes. These waveforms are shown here.
Simulation Results for a Four-Cycle Line-to-Ground Fault at the End of Line 1
The simulation starts in steady state. At the t = 1 cycle, a line-to-ground fault is applied and the fault current reaches 10 kA (a: trace 3). During the fault, the MOV conducts at every half cycle (b: trace 2) and the energy dissipated in the MOV (b: trace 3) builds up to 13 MJ. At t = 5 cycles the line protection relays open breakers CB1 and CB2 (see the three line currents on trace 2) and the energy stays constant at 13 MJ. As the maximum energy does not exceed the 30 MJ threshold level, the gap is not fired. At the breaker opening, the fault current drops to a small value and the line and series capacitance starts to discharge through the fault and the shunt reactance. The fault current extinguishes at the first zero crossing after the opening order given to the fault breaker (t = 6 cycles). Then the series capacitor stops discharging and its voltage oscillates around 220 kV (b: trace 1).
Double-click the Three-Phase Fault block to open the Block Parameters dialog box. Select the Phase B Fault and Phase C Fault check boxes, so that you now have a three-phase-to-ground fault.
Restart the simulation. The resulting waveforms are shown.
Simulation Results for a Four-Cycle Three-Phase-to-Ground Fault at the End of Line 1
Note that during the fault the energy dissipated in the MOV (b: trace 3) builds up faster than in the case of a line-to-ground fault. The energy reaches the 30 MJ threshold level after three cycles, one cycle before the opening of the line breakers. As a result, the gap is fired and the capacitor voltage (b: trace 1) quickly discharges to zero through the damping circuit.
One particular characteristic of series-compensated systems is the existence of subsynchronous modes (poles and zeros of the system impedance below the fundamental frequency). Dangerous resonances can occur if the mechanical torsion modes of turbine/generator shafts are in the vicinity of the zeros of the system impedance. Also, high subsynchronous voltages due to impedance poles at subsynchronous frequencies drive transformers into saturation. The transformer saturation due to subsynchronous voltages is illustrated at the end of this case study. The torque amplification on a thermal machine is illustrated in another example (see the power_thermalpower_thermal model).
Now measure the positive-sequence impedance versus frequency seen from bus B2.
The section Analyze a Simple Circuit explains how the Impedance Measurement block allows you to compute the impedance of a linear system from its state-space model. However, your case1 model contains several nonlinear blocks (machine and saturation of transformers). If you connect the Impedance Measurement block to your system, all nonlinear blocks are ignored. This is correct for the transformer, but you get the impedance of the system with the machine disconnected. Before measuring the impedance, you must therefore replace the machine block with an equivalent linear block having the same impedance.
Delete the Simplified Synchronous Machine block from your case1 model and replace it with the Three-Phase Source block from the Electrical Sources library. Open the block dialog box and set the parameters as follows to get the same impedance value (L = 0.22 pu/ (6 * 350 MVA) Quality factor = 15).
Phase-to-phase rms voltage | 13.8e3 |
Phase angle of phase A | 0 |
Frequency (Hz) | 60 |
Internal connection Yg | Specify impedance using short-circuit level |
3-phase short-circuit level | 6*350e6 |
Base voltage | 13.8e3 |
X/R ratio | 15 |
Save your modified model as case1Zf.
Open the Measurements library of powerlib and copy the Impedance Measurement block into your model. This block is used to perform the impedance measurement. Connect the two inputs of this block between phase A and phase B of the B2 bus. Measuring the impedance between two phases gives two times the positive-sequence impedance. Therefore you must apply a factor of 1/2 to the impedance to obtain the correct impedance value. Open the dialog box and set the multiplication factor to 0.5.
In the Powergui block, select Impedance vs Frequency Measurement. A new window opens, showing your Impedance Measurement block name. Fill in the frequency range by entering 0:500. Select the linear scales to display Z magnitude vs. frequency plot. Click the Save data to workspace button and enter Zcase1 as the variable name to contain the impedance vs. frequency. Click the Display button.
When the calculation is finished, the magnitude and phase as a function of frequency are displayed in the two graphs on the window. If you look in your workspace, you should have a variable named Zcase1. It is a two-column matrix containing frequency in column 1 and complex impedance in column 2.
The impedance as a function of frequency (magnitude and phase) is shown here.
Impedance vs. Frequency Seen from Bus B2
You can observe three main modes: 9 Hz, 175 Hz, and 370 Hz. The 9 Hz mode is mainly due to a parallel resonance of the series capacitor with the shunt inductors. The 175 Hz and 370 Hz modes are due to the 600 km distributed parameter line. These three modes are likely to be excited at fault clearing.
If you zoom in on the impedance in the 60 Hz region, you can find the system's short-circuit level at bus B2. You should find a value of 58 Ω at 60 Hz, corresponding to a three-phase short-circuit power of (735 kV)^{2} / 58 = 9314 MVA.
The configuration of the substation circuit breakers normally allows clearing a fault at the bus without losing the lines or the transformers. You now modify your case1 model to perform a three-cycle, three-phase-to-ground fault at bus B2:
Disconnect the Three-Phase Fault block and reconnect it so that the fault is now applied on bus B2.
Open the Three-Phase Fault block and make the following modifications in its dialog box:
Phase A, Phase B, Phase C, Ground Faults | All selected |
Transition times | [2/60 5/60] |
Transition status [1, 0, 1...] | (0/1) |
You have now programmed a three-phase-to-ground fault applied at the t = 2 cycles.
Open the dialog boxes of circuit breakers CB1 and CB2 and make the following modifications:
Switching of Phase A | Not selected |
Switching of Phase B | Not selected |
Switching of Phase C | Not selected |
The circuit breakers are not switched anymore. They stay at their initial state (closed).
In the Data Acquisition subsystem, insert a Selector block (from the Simulink^{®} Signals & Systems library) in the Vabc_B2 output of bus B2 connected to the scope. Set the Elements parameter to 1. This allows you to see the phase A voltage clearly on the scope.
You now add blocks to read the flux and the magnetization current of the saturable transformer connected at bus B2.
Copy the Multimeter block from the Measurements library into your case1 model. Open the Transformer dialog box. In the Measurements list, select Flux and magnetization current. Open the Multimeter block. Verify that you have six signals available. Select flux and magnetization current on phase A, and click OK.
You now have two signals available at the output of the Multimeter block. Use a Demux block to send these two signals on a two-trace scope.
In the Simulation > Model Configuration Parameters dialog box, change the stop time to 0.5. This longer simulation time allows you to observe the expected low-frequency modes (9 Hz). Start the simulation.
The resulting waveforms are plotted here.
Simulation Results for a Three-Cycle Three-Phase-to-Ground Fault at BusB2
The 9 Hz subsynchronous mode excited at fault clearing is clearly seen on the phase A voltage at bus B2 (trace 1) and capacitor voltage (trace 3). The 9 Hz voltage component appearing at bus B2 drives the transformer into saturation, as shown on the transformer magnetizing current (trace 5). The flux in phase A of the transformer is plotted on trace 4. At fault application the voltage at transformer terminals drops to zero and the flux stays constant during the fault.
At fault clearing, when the voltage recovers, the transformer is driven into saturation as a result of the flux offset created by the 60 Hz and 9 Hz voltage components. The pulses of the transformer magnetizing current appear when the flux exceeds its saturation level. This current contains a 60 Hz reactive component modulated at 9 Hz.