| Date post: | 07-Mar-2016 |
| Category: |
Documents |
| Upload: | tranh-van-thuong |
| View: | 16 times |
| Download: | 0 times |
of 70
2Case Studies
Series Compensated Transmission Network . . . . . 2-3
Chopper-Fed DC Motor Drive . . . . . . . . . . . . 2-21
Synchronous Machine and Regulators . . . . . . . . 2-32
Variable-Frequency Induction Motor Drive . . . . . . 2-40
HVDC System . . . . . . . . . . . . . . . . . . . 2-52
2 Case Studies
2-2
The case studies in this chapter were built to provide examples of uses for thePower System Blockset. They are:
Series Compensated Transmission Network Chopper-Fed DC Motor Drive Synchronous Machine and Regulators Variable-Frequency Induction Motor Drive HVDC System
Cases 1 and 5 are studies of AC and DC transmission on power systems. Cases2 and 4 illustrate typical applications of the Power System Blockset to motordrives. Case 3 demonstrates the performance of a nonlinear voltage regulatoron a synchronous alternator.
Series Compensated Transmission Network
2-3
Series Compensated Transmission NetworkThe example described in this section illustrates phenomena related tosubsynchronous resonance in a series-compensated AC transmission network.
Description of the Transmission NetworkThe single diagram shown in Figure 2-1 represents a three-phase, 60 Hz, 735kV power system transmitting power from a power plant consisting of six 350MVA generators to an equivalent network through a 600 km transmission line.The transmission line is split in two 300 km lines connected between buses B1,B2, and B3.
Figure 2-1: Series and Shunt Compensated Network
In order to increase the transmission capacity, each line is series compensatedby capacitors representing 40% of the line reactance. Both lines are also shuntcompensated by a 330 Mvar shunt reactance. The shunt and seriescompensation equipment is located at the B2 substation where a 300 MVA-735/230 kV transformer feeds a 230 kV-250 MW load through a 25 kV tertiarywinding.
equivalent
330 MvarB1
B2735 kV
line 1; 300 km
330 Mvar
Generators
6*350 MVA
30 000 MVA
CB1 CB2
B3
D/Yg
6*350 MVA 13.8 kV/735 kV
Yg/Yg/D 735 /230 /25 kV300 MVA
50 MW
100 MW
40%
13.8 kV
40%
MOV1
MOV2line 2; 300 km
Van ThanhHighlighthin tng
2 Case Studies
2-4
Each series compensation bank is protected by metal oxide varistors (MOV1and MOV2). The two circuit breakers of line 1 are shown as CB1 and CB2.
We want to study the transient behavior of this circuit when faults are appliedon line 1 and at bus B2.
This network is available in the psb3phseriescomp.mdl demonstration file.Load the psb3phseriescomp system and save it in your working directory ascase1 in order to allow further modifications to the original system.
Compare the circuit modeled in the Power System Blockset (Figure 2-2) withthe schematic diagram of Figure 2-1. The generators are simulated with aSimplified Synchronous Machine block. A Three-Phase Transformer(Two-Windings) block and a Three-Phase Transformer (Three-Windings) blockare used to model the two transformers. Saturation is implemented on thetransformer connected at bus B2.
B1 and B3 blocks are 3-phase V-I Measurement blocks taken from theMeasurements library of powerlib_extras. B2 is a similar block that has beenmodified to accommodate two 3-phase inputs and one 3-phase output.Theseblocks have been reformatted and given a black background color to give themthe appearance of bus bars. They output the three line-to-ground voltagesmultiplexed on output 4 and the three line currents multiplexed on output 5.Open the dialog boxes of B1 and B2. See how the blocks are programmed tooutput voltages in p.u. and current in pu/100 MVA.
The fault is applied on line 1, on the line side of the capacitor bank. Note thatthe fault and the two line circuit breakers are simulated with blocks from thethree-phase library. Open the dialog boxes of the 3-Phase Fault block and of the3-Phase breakers CB1 and CB2. See how the initial breaker status andswitching times are specified. A line-to-ground fault is applied on phase A att=1 cycle. The two circuit breakers that are initially closed are then open at t=5cycles, simulating a fault detection and opening time of 4 cycles. The fault iseliminated at t=6 cycles, one cycle after the line opening.
Series Compensated Transmission Network
2-5
Figure 2-2: Series Compensated Network (psb3phseriescomp.mdl)
2 Case Studies
2-6
Series Compensation1 SubsystemNow, open the Series Compensation1 subsystem of the psb3phseriescompsystem. The three-phase module consists of three identical subsystems, one foreach phase. A note indicates how the capacitance value and the MOVprotection level have been calculated. Open the Series Compensation1/ PhaseA subsystem. You can see the details of the connections of the series capacitorand the Surge Arrester block (MOV block). The transmission line is 40% seriescompensated by a 62.8 F capacitor. The capacitor is protected by the MOVblock. If you open the dialog box of the MOV block, you will notice that itconsists of 60 columns and that its protection level (specified at a referencecurrent of 500 A/column or 30 kA total) is set at 298.7 kV. This voltagecorresponds to 2.5 times the nominal capacitor voltage obtained at a nominalcurrent of 2 kA rms.
A gap is also connected in parallel with the MOV block. The gap is fired whenthe energy absorbed by the surge arrester exceeds a critical value of 30 MJ. Inorder to limit the rate of rise of capacitor current when the gap is fired, adamping RL circuit is connected in series. Open the Energy & Gap firingsubsystem. It shows how the energy dissipated in the MOV is calculated byintegrating the power (product of the MOV voltage and current). When theenergy exceeds the 30 MJ threshold, a closing order is sent to the breaker blocksimulating the gap.
Series Compensated Transmission Network
2-7
Figure 2-3: Series Compensation Module
SeriesCompensation1 Subsystem
SeriesCompensation1/PhaseA Subsystem
SeriesCompensation1/PhaseA Subsystem/Energy & Gap firing
2 Case Studies
2-8
Three-Phase Saturable Transformer ModelOpen the 300 MVA 735/230 kV Transformer dialog box and notice that thecurrent-flux saturation characteristic has been set at:
[0 0 ; 0.0012 1.2; 1 1.45] in pu
This data is the current and flux values at points 1, 2, and 3 of the piece-wiselinear approximation to the flux linkage curve shown in Figure 2-4.
Figure 2-4: Saturable Transformer Model
The flux-current characteristic is approximated by two segments (see Figure2-4). The saturation knee point is 1.2 p.u. The first segment corresponds to themagnetizing characteristic in the linear region (for fluxes below 1.2 pu). At 1p.u. voltage, the inductive magnetizing current is 0.0010/1.0= 0.001 pu,corresponding to 0.1% reactive power losses.
R2L2L1R1
Ls
nonlinear inductance (reactive power losses)
Bus B2
Ideal transformer
LS Lac L1=slope
1.2pu
LmRm
slope
point 2
0.0012pu
point 3
point 1
flux
linka
ge (p
u)
current (pu/300 MVA)
linear resistance (active power losses)
1.45pu
1pui
Lm=1000pu
Series Compensated Transmission Network
2-9
The iron core losses (active power losses) are specified by the magnetizationresistance Rm=1000 pu, corresponding to 0.1% losses at nominal voltage.
The slope of the saturation characteristic in the saturated region is 0.25 p.u.Therefore, taking into account the primary leakage reactance (L1=0.15 pu), theair core reactance of the transformer seen from the primary winding is 0.4pu/300 MVA).
Setting the Initial Load Flow and ObtainingSteady-StateBefore performing transient tests on your case1 system you must initializeyour system for the desired load flow. Use the load flow utility of the powerguito obtain an active power flow of 1500 MW out of the machine with a terminalvoltage of 1 p.u. (13.8 kV).
Open the powergui. In the Tools menu, select Load Flow and MachineInitialization. A new window appears. In the upper right window you have athe name of the only machine appearing in your system: 6*350*MVA 13.8 kV.Its Bus Type should be PV Generator and the desired Terminal Voltage shouldbe already set to the default value (nominal voltage of 13800 V). In the ActivePower field, enter 1500e6 (1500 MW) as the desired output power. Click on theExecute Load Flow button. Once the load flow is solved, the phasors of AB andBC machine voltages as well as currents flowing in phases A and B are updatedin the left window. The required mechanical power to drive the machine isdisplayed in watts and in p.u. and the required excitation voltage E isdisplayed in pu:
Pmec :1.5159e9 W [0.72184 pu]E/Vf : 1.0075 pu
Notice that constant blocks containing these two values are already connectedto the Pm and E inputs of the machine block. If you open the machine dialogbox, you will see that the machine initial conditions (initial speed deviationdw=0; internal angle theta, current magnitudes and phase angles) have beenautomatically transferred in the last line.
Once the load flow is performed, you can obtain the corresponding voltage andcurrent measurements at the different buses. In the Tools menu, select SteadyState Voltages and Currents. You can observe, for example, the phasors forphase A voltages at buses B1, B2 and B3 and the current entering line 1 at busB1.
2 Case Studies
2-10
B1/Va : 6.088e5 V ; 18.22 degreesB2/Va : 6.223e5 V ; 9.26 degreesB3/Va : 6.064e5 V ; 2.04 degreesB1/Ia : 1560 A ; 30.50 degrees
The active power flow for phase A entering line 1 is therefore
corresponding to a total of 464*3=1392 MW for the three phases.
Transient Performance for a Line FaultIn order to speed up the simulation, you need to discretize the network. Noticethat your case1 contains the Discrete System block. The sample time isspecified in the block dialog as a variable Ts. This sample time Ts is also usedin the integrator block of the MOV energy calculator controlling the gap.
In the MATLAB window define the variable
Ts=50e-6;
Check that the simulation parameters are set as follows:
Stop time : 0.2Solver options Type: Fixed-step; discrete( no continuous state)Fixed step size: Ts
Line-to-Ground Fault Applied on Line 1Check that the fault breaker is programmed for a line-to-ground fault on phaseA. Start the simulation and observe the waveforms on the three scopes. Thesewaveforms are reproduced on Figure 2-5.
Pa Va Ia a( )cos 608.8kV
2-----------------------
1.56kA2
-------------------- 30.50 18.22( )cos 464MW= = =
Series Compensated Transmission Network
2-11
Figure 2-5: Simulation Results for a 4 Cycle Line-to-Ground Fault at the Endof Line 1
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.22
0
2
4
Vabc
(pu)
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2100
50
0
50
100
Iabc (
pu/10
0 MVA
)
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.21
0
1
2x 104
Ia Fa
ult (A
)
Time (s)
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.24
2
0
2
4x 105
U Csa
(V)
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.21
0.5
0
0.5
1x 104
I MOV
a (A)
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.20
5
10
15x 106
Ener
gy M
OVa
(J)
Time (s)
2 Case Studies
2-12
The simulation starts in steady state. At the t=1 cycle, a line-to-ground fault isapplied and the fault current reaches 10 kA (trace 3). During the fault, theMOV conducts at every half cycle (trace 5) and the energy dissipated in theMOV (trace 6) builds up to 13 MJ. At t=5 cycles the line protection relays openbreakers CB1 and CB2 (see three line currents on trace 2) and the energy staysconstant at 13 MJ. As the maximum energy does not exceed the 30 MJthreshold level, the gap is not fired. At the breaker opening, the fault currentdrops to a small value and the line and series capacitance starts to dischargethrough the fault and the shunt reactance. The fault current extinguishes atthe first zero crossing after the opening order given to the fault breaker (t=6cycles). Then the series capacitor stops discharging and its voltage oscillatesaround 220 kV (trace 4).
Three-Phase-to-Ground Fault Applied on Line 1Open the 3-Phase Fault block dialog box. Check the Phase B Fault and PhaseC Fault, so that you now have a three-phase-to-ground fault.
Restart the simulation.The waveforms are reproduced on Figure 2-6.
Series Compensated Transmission Network
2-13
Figure 2-6: Simulation Results for a 4-Cycle Three-Phase-to-Ground Fault atthe End of Line 1
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.24
2
0
2
4x 105
U Csa
(V)
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.21
0.5
0
0.5
1x 104
I MOV
a (A)
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.20
1
2
3
4x 107
Energ
y MOV
a (J)
Time (s)
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.22
1
0
1
2
Vabc
(pu)
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2200
100
0
100
Iabc (
pu/10
0 MVA
)
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.22
1
0
1
2x 104
Ia Fa
ult (A
)
Time (s)
Gap firing
2 Case Studies
2-14
Notice that during the fault the energy dissipated in the MOV (trace 6) buildsup faster that in the case of a line-to-ground fault. The energy reaches the 30MJ threshold level after three cycles, one cycle before opening of the linebreakers. As a result, the gap is fired and the capacitor voltage (trace 4) quicklydischarges to zero through the damping circuit.
Frequency AnalysisOne particular characteristic of series compensated systems is the existence ofsubsynchronous modes (poles and zeros of the system impedance below thefundamental frequency). Dangerous resonances can occur if the mechanicaltorsion modes of turbine/generator shafts are in the vicinity of the zeros of thesystem impedance. Also, high subsynchronous voltages due to impedancespoles at subsynchronous frequencies will drive transformers into saturation.The transformer saturation due to subsynchronous voltages is illustrated atthe end of this case study. The torque amplification on a thermal machine isillustrated in another demonstration (see psbthermal.mdl file).
Now measure the positive-sequence impedance versus frequency seen from busB2.
We have already explained in Session 2 of the Tutorial chapter how theImpedance Measurement block allows you to compute the impedance of alinear system from its state-space model. However, your case1 system containsseveral nonlinear blocks (machine and saturation of transformers). If youconnect the Impedance Measurement block to your system, all nonlinear blockswill be ignored. This is correct for the transformer, but you would get theimpedance of the system with the machine disconnected. Before measuring theimpedance, you must therefore replace the machine block with an equivalentlinear block having the same impedance.
Delete the Simplified Synchronous Machine block from your case1 circuit andreplace it with the Inductive source with neutral block from the Three-Phaselibrary of powerlib_extras. Open the block dialog box and set the parametersas follows in order to get the same impedance value (L=0.22 pu/ (6*350 MVA)Quality factor=15):
Voltage magnitude (V) :13.8e3/sqrt(3)*sqrt(2)Phase angle : 0Frequency (Hz) : 60Source resistance (Ohms) : 13.8^2/(6*350)*0.22/15Source inductance (H) : 13.8^2/(6*350)*0.22/(2*pi*60)
Series Compensated Transmission Network
2-15
Save your modified circuit as case1Zf.
Open the Measurements library of powerlib and copy the ImpedanceMeasurement block in your model. This block contains a current source and avoltage measurement that will be used to perform the impedancemeasurement. Connect the two inputs of this block between phase A and phaseB of B2 bus. Measuring the impedance between two phases will give two timesthe positive-sequence impedance. Therefore you must apply a factor of 1/2 onthe impedance in order to obtain the correct impedance value. Open theImpedance Measurement dialog box and set the Multiplication factor to 0.5.
Now open the powergui. In the Tools menu select Impedance vs FrequencyMeasurement. A new window opens, showing your Impedance Measurementblock name. Fill in the frequency range by typing 0:500. Select the linear scalesto display Z magnitude vs frequency plot. Check the Save data to workspacebutton and enter Zcase1 as the variable name that will contain the impedancevs frequency. Click on the Display button.
When the calculation is finished a graphic window appears with the magnitudeand phase as a function of frequency. If you check in your workspace, youshould have a variable named Zcase1. It is a two-column matrix containingfrequency in column 1 and complex impedance in column 2.
The impedance as function of frequency (magnitude and phase) is shown inFigure 2-7.
2 Case Studies
2-16
Figure 2-7: Impedance vs. Frequency Seen From Bus B2
You can observe three main modes: 9 Hz, 175 Hz, and 370 Hz. The 9 Hz modeis mainly due to a parallel resonance of the series capacitor with the shuntinductors. The 175 Hz and 370 Hz modes are due to the 600 km distributedparameter line. These three modes are likely to be excited at fault clearing.
If you zoom on the impedance in the 60 Hz region, you can find the system'sshort circuit level at bus B2. You should find a value of 58 at 60 Hz,corresponding to a three-phase short circuit power of (735kV) 2/58 = 9314 MVA.
Transient Performance for a Fault at Bus B2The configuration of the substation circuit breakers normally allows to clear abus fault without losing lines or transformers. You will now modify your case1
0 50 100 150 200 250 300 350 400 450 5000
500
1000
1500
2000
Mag
. (ohm
s)
0 50 100 150 200 250 300 350 400 450 500100
50
0
50
100
Phas
e (de
grees
)
Frequency (Hz)
9Hz
58 Ohms at 60Hz
Subsynchronous resonance
Series Compensated Transmission Network
2-17
circuit in order to perform a three-cycle, three-phase-to-ground fault at bus B2as shown on Figure 2-9:
1 Disconnect the 3-Phase Fault block and reconnect it as shown on Figure 2-9so that the fault is now applied on bus B2.
2 Open the 3-Phase Fault dialog box and make the following modifications:
Phase A, Phase B, Phase C, Ground Faults : all checkedTransition times :[2/60 5/60]Transition status [1, 0, 1...]: (0/1)
You have now programmed a three-phase-to-ground fault applied at t=1cycle.
3 Open the dialog boxes of circuit breakers CB1 and CB2 and make thefollowing modifications:
Switching of Phase A: not checkedSwitching of Phase B: not checkedSwitching of Phase C: not checked
The circuit breakers will not be switched anymore. They will stay at theirinitial state (closed).
4 Insert a Selector block (from the Simulink/Signals & Systems library) in theVabc output of bus B2 connected to the Scope. Set its Elements parameterto 1. This will allow you to see clearly the phase A voltage on the scope.
5 You will now add blocks to read the flux and the magnetization current ofthe saturable transformer connected at bus B2.
Copy the Multimeter block from the Measurement library into your case1circuit. Open the Transformer dialog box. In the Measurements pop-upmenu, select Flux and magnetization Current. Open the Multimeter block.Verify that you have six signals available. Select flux and magnetizationcurrent on phase A, and click on OK.
6 You have now two signals available at the output of the Multimeter block.Use a Demux block to send these two signals on a two trace scope (SeeFigure 2-9).
2 Case Studies
2-18
7 In the Simulation/Parameters menu, change the stop time to 0.5. Thislonger simulation time will allow you to observe the expected low frequencymodes (9 Hz). Start the simulation.
Waveforms of interest are plotted on Figure 2-8.
Figure 2-8: Simulation Results for a 3-Cycle 3-Phase-to-Ground Fault at BusB2
The 9 Hz subsynchronous mode excited at fault clearing is clearly seen on thephase A voltage at bus B2 (trace 1) and capacitor voltage (trace 3). The 9 Hzvoltage component appearing at bus B2 drives the transformer into saturation
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.55
0
5V
a (p
u)
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.52
0
2x 104
Ia F
ault
(A)
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.55
0
5x 105
UC
sa
(V)
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.52000
0
2000
Imag
A (A
)
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.55000
0
5000
Flu
xA (V
.s)
Time (s)
Series Compensated Transmission Network
2-19
as shown on the transformer magnetizing current (trace 4). The flux in phaseA of the transformer is plotted on trace 5. At fault application the voltage attransformer terminals drops to zero and the flux stays constant during thefault. At fault clearing, when the voltage recovers, the transformer is driveninto saturation as a result of the flux offset created by the 60 Hz and 9Hzvoltage components. The pulses of the transformer magnetizing current appearwhen the flux exceeds its saturation level. This current contains a 60 Hzreactive component modulated at 9 Hz.
2 Case Studies
2-20
Figure 2-9: S Series Compensated Network Used for a Fault at Bus B2
Chopper-Fed DC Motor Drive
2-21
Chopper-Fed DC Motor DriveThe example described in this section illustrates application of the PowerSystem Blockset to the operation of a DC motor drive in which the armaturevoltage is controlled by a GTO thyristor chopper.
The objective of this example is to demonstrate the use of electrical blocks, incombination with Simulink blocks, in the simulation of an electromechanicalsystem with a control system. The electrical part of the DC motor driveincluding the DC source, the DC motor, and the chopper is built using blocksfrom the Elements, Machines, and Power Electronics libraries. The DCMachine block of powerlib models both electrical and mechanical dynamics.The load torque-speed characteristic and the control system are built usingSimulink blocks.
Description of the Drive SystemA simplified diagram of the drive system is shown in Figure 2-10. The DCmotor is fed by the DC source through a chopper that consists of the GTOthyristor, Th1, and the free-wheeling diode D1. The DC motor drives amechanical load that is characterized by the inertia J, friction coefficient B, andload torque TL (which can be a function of the motor speed).
Figure 2-10: Chopper-Fed DC Motor Drive
In this diagram, the DC motor is represented by its equivalent circuitconsisting of inductor La and resistor Ra in series with the counterelectromotive force (emf) E.
Th1
D1
+-
Ra
E
DC Motor
+
-
+
-
VaVdc
La
Ia
Mechanical load
J, B, TL
Tm
2 Case Studies
2-22
The back EMF is proportional to the motor speed
where KE is the motor voltage constant and is the motor speed.
In a separately excited DC machine, the motor voltage constant KE isproportional to the field current if
where Laf is the field-armature mutual inductance.
The torque developed by the DC motor is proportional to the armature currentIa
where KT is the motor torque constant.
The DC motor torque constant is equal to the voltage constant
Thyristor Th1 is triggered by a pulse width modulated (PWM) signal to controlthe average motor voltage. Theoretical waveforms illustrating the chopperoperation are shown in Figure 2-11.
E K E=
K E Laf I f=
Tm KTIa=
KT K E=
Chopper-Fed DC Motor Drive
2-23
Figure 2-11: Waveforms Illustrating the Chopper Operation
The average armature voltage is a direct function of the chopper duty cycle :
Note that this relation is valid only when the armature current is continuous.In steady-state, the armature average current is equal to:
The peak-to-peak current ripple is
where is the duty cycle and r is the ratio between the chopper period and theDC motor electrical time constant:
Th1
T
T
Va
t
Ia
t
Va(avg)
Ia(avg)
Vdc
Va avg( ) Vdc=
Ia avg( )Va avg( ) E
Ra----------------------------------=
i VdcRa
-----------
1 e r e r e 1 ( )r+( )1 e r
-----------------------------------------------------------------------=
2 Case Studies
2-24
In this case study, we consider a variable-speed DC motor drive using a cascadecontrol configuration. A block diagram of this drive is shown in Figure 2-12.
Figure 2-12: Variable-Speed DC Motor Drive
The motor torque is controlled by the armature current Ia, which is regulatedby a current control loop. The motor speed is controlled by an external loop,which provides the current reference Ia* for the current control loop.
Modeling the DC DriveOpen the psbdcdrive.mdl file of the powerlib library by typing psbdcdrive inMATLAB command window. A circuit diagram titled psbdcdrive will appear.Before running the example, save this circuit as case2.mdl in your workingdirectory so that you can make further modifications without altering theoriginal file.
The drive system diagram is built using electrical blocks contained in thepowerlib library combined with Simulink blocks. Voltage Measurement andCurrent Measurement blocks are used as the interface between the two blocktypes. The system diagram is shown in Figure 2-13.
r TLa Ra( )
-----------------------=
ChopperVdc
CurrentController
SpeedController
*Speed Reference
Ia*
Ia
SpeedSensor
CurrentSensor
+
-
+
-
+
-+
-
DCMotorVa
Chopper-Fed DC Motor Drive
2-25
Figure 2-13: DC Motor Drive Using Power System Blockset (psbdcdrive.mdl)
The DC motor represented by the DC Machine block is modeled in two separateparts: electrical and mechanical. To view the Simulink model of the DC motor,click on the DC Machine block and use the LookUnderMask command in theEdit menu.
2 Case Studies
2-26
The armature circuit is represented by an RL circuit in series with a controlledvoltage source, the value of which is KE.
The field circuit is represented by an RL circuit.
The mechanical part is represented by Simulink blocks, which implement thefollowing equation:
The DC machine parameters are set to the desired values by using the dialogmask of the DC Machine block.
The load torque-speed characteristic can be implemented by a Simulink Fcnblock.
The motor used in this case study is a separately excited 5 HP/240 V DC motorhaving the following parameters: Ra = 0.5 , La = 10 mH, KE =1.23 V/(rad/s),KT = 1.23 N.m/A.
3m
2
A
1
F
+i
if
Laf20e6s+1
Sign
Rf
Lf
Ra
La
Mux
Mux
s
1
Integrator
+i
Ia_mot1/J
signal+
E_FCEM
Coulomb (Tf) &Viscous ( Bm*w) Friction Torques
3TL
2
A+
1
F+
Te
Te
w
w
IfIa
Tm Jddt-------- B Tsgn L+ +=
Chopper-Fed DC Motor Drive
2-27
A 10mH inductor (Ls) is connected in series with the DC motor to smooth outthe armature current. The constant excitation is implemented by connecting aDC Voltage Source block to the field winding.
The required trigger signal for the GTO thyristor is generated by a hysteresiscurrent controller, which forces the motor current to follow the reference within+h/2 and -h/2 limits (h is the hysteresis band).
The current controller is a masked block that contains:
The speed control loop uses a proportional-integral controller, which isimplemented by Simulink blocks:
Simulation of the DC DriveRun the simulation by selecting Start from the Simulation menu in Simulink.Set the simulation parameters in the Simulation Parameters menu asfollows:
Simulation time: Start Time:0, Stop time: 1.2Solver Type: Variable-step ode23tb (stiff/TR-BDF2)Max Step Size: autoInitial Step Size: autoRelative Tolerance: 1e-3Absolute Tolerance: 1e-3
The motor voltage, current waveforms and motor speed are displayed on threeaxes of the scope connected to the variables Vd, Ia and .
1g
Relay
2Ia
1Iref
1Iref
s
1
Kp
Ki2
wref
1wm
2 Case Studies
2-28
Once the simulation is completed, you can return to the MATLAB commandwindow to examine the results with more details by using the plot function.
Drive StartingIn this test, we simulate the starting transient of the DC drive. The inertia ofthe mechanical load is small in order to bring out the details of the choppercommutation details. The speed reference is stepped from 0 to 120 rad/s att=0.0 s and we observe the DC motor speed and current.
The transient responses for the starting of the DC motor drive are shown inFigure 2-14.
Note that the final system state vector xFinal can be saved by selectingWorkspace I/O/Save to workspace/Final state in the SimulationParameters window. It can be used as initial state in subsequent simulationso that the simulation can start under steady-state conditions.
Chopper-Fed DC Motor Drive
2-29
Figure 2-14: Starting of the DC Motor Drive
Steady-State Voltage and Current WaveformsWhen the steady-state is attained, you can stop the simulation and plot thecurrent and voltage waveforms using the variables Va and Ia sent back inMATLAB workspace by the scope.
The DC motor current and voltage waveforms obtained at the end of thestarting test are shown in Figure 2-15.
0 0.2 0.4 0.6 0.8 1 1.20
50
100
150
Time , s
Mot
or s
peed
, ra
d/s
DC motor drive Starting transient
0 0.2 0.4 0.6 0.8 1 1.20
5
10
15
20
25
30
35
Time , s
Arm
atur
e cu
rrent
, A
2 Case Studies
2-30
Figure 2-15: Steady-State Motor Current and Voltage Waveforms
Speed Regulation Dynamic PerformanceWe can study the drive dynamic performance, (speed regulation performanceversus reference and load torque changes), by applying two successivechanging operating conditions to the DC drive: a step change in speed referenceand a step change in load torque.
Replace the block named wref(rad/s) and the block named Load_torque(N.m) inthe diagram by two Simulink step blocks with different starting times. Thespeed reference steps from 120 rad/s to 160 rad/s at t = 0.4 s and the load torquesteps from 5 N.m to 25 N.m at t = 1.2 s. The final state vector obtained with theprevious simulation can be used as initial condition so that the simulation willstart from steady-state. Load the psbdcdrive_init.mat file, which will createthe xInitial variable. Select Workspace I/O/Load from workspace/Initialstate in the Simulation Parameters window and restart the simulation.
1.19 1.191 1.192 1.193 1.194 1.195 1.196 1.197 1.198 1.199 1.20
2
4
6
8
10
Time , s
Arm
atur
e cu
rrent
, A
Steadystate DC motor current and voltage
1.19 1.191 1.192 1.193 1.194 1.195 1.196 1.197 1.198 1.199 1.2100
0
100
200
300
Time , s
Arm
atur
e vo
ltage
, V
Chopper-Fed DC Motor Drive
2-31
The obtained response of the DC motor drive to successive changes in speedreference and load torque is shown in Figure 2-16.
Figure 2-16: Dynamic Transient of the DC Motor Drive
References[1] Leonhard,W. Control of Electrical Drives. Springer-Verlag, Berlin 1996.
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2110
120
130
140
150
160
170
180
Time , s
M
otor
spee
d , r
ad/s
DC Motor Drive Speed regulator transient response
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
5
10
15
20
25
30
35
Time , s
Ar
mat
ure
curre
nt ,
A
2 Case Studies
2-32
Synchronous Machine and RegulatorsThis case study investigates the application of a multi-input multi-outputnonlinear controller to a system consisting of a hydraulic turbine and asynchronous generator connected to an infinite bus. The complete system ismodeled using the Power System Blockset and Simulink blocks.
The objective of this case study is to demonstrate the use of the SynchronousMachine block connected to a complex control system implemented withSimulink blocks. The controller is based on a feedback linearization scheme. Itsmain goal is to control the rotor angle as well as the terminal voltage, toimprove the stability properties, and to obtain good dynamic response.Simulation results will show that the nonlinear controller is able to replace thestandard linear controllers and give better performance.
IntroductionTraditionally, stabilization of power systems was ensured by linear regulatorssuch as the automatic voltage regulator (AVR), the speed governor, or thepower system stabilizer (PSS). These compensators assume a linearized modelof the power system around an operating point.
The demand for improved performance has created the need to operate powersystems closer to the limits and therefore well outside the linear domain.Nonlinearities begin to have a significant effect then, especially afterimportant disturbances that lead to a large variation of the operating point.
This case study will allow you to step through the design of a nonlinearcontroller that takes into account all the nonlinearities of the model. Theobjectives of the controller are to regulate both the terminal voltage and theinternal power angle. The control inputs are the field excitation voltage and thegate opening of the turbine.
Mathematical ModelThe model considered is a single machine infinite bus (SMIB) system, as shownin Figure 2-17. The machine is a synchronous generator driven by a hydraulicturbine.
Synchronous Machine and Regulators
2-33
Figure 2-17: Diagram of Case Study
The dynamic equations of the machine that are used to derive the linearfeedback controller are for the three-phase Synchronous Machine block, andthe Hydraulic Turbine and Governor block (see Chapter 4, Block Reference).Since the synchronous machine is connected to an infinite bus, the dq terminalvoltages vd and vq are constrained by the load equations. In thePark-transformed coordinates (rotor reference frame), we can write:
This equation can be combined with the complete model of the SMIB system inthe nonlinear state-space form:
F(x) and G(x) are given by:
Line
Le Re
Turbine
Synchr.Machine
InfiniteBus
FaultBreaker
vdvq
Reidiq
Leid
iq
Leiqi d
V a( )cos a( )sin
+ +=
x F x( ) G x( )u+=
2 Case Studies
2-34
The explicit expressions of the coefficients A and g can be found in the Chapter4, Block Reference and are omitted here for simplicity. The other terms of thestate-space equation are x, the vector of state variables, and u, the vector ofcontrol inputs. They are defined as follows:
The currents id, iq and voltages vd, vq are the projection of the actual linecurrents and terminal voltages on the direct and quadrature axes (d-q frame).ifd and vfd represent the field current and voltage. ikq and ikd represent thedamper windings currents, and the angular speed of the machine. is theelectrical angle measured from a synchronously rotating frame. G and q arerespectively the opening of the gate and the flow rate of the turbine. Finally,uG is the voltage applied to the gate servo-motor.
F x( )
A11x1 A12x3 A13x2x7 A14x4 A15x5x7+ + + + A16 x6 a( )cos+A21x1x7 A22x2 A23x3x7 A24x4x7 A25x5+ + + + A26 x6 a( )sin+A31x1 A32x3 A33x2x7 A34x4 A35x5x7+ + + + A36 x6 a( )cos+A41x1 A42x3 A43x2x7 A44x4 A45x5x7 A46 x6 a( )cos+ + + + +A51x1x7 A52x2 A53x3x7 A54x4x7 A55x5 A56 x6 a( )sin+ + + + +x7 1( )r
A71x1x2 A72x2x3 A73x2x4 A74x1x5 A75x7 A76x8
3
x7x92
------------+ + + + +
A81 A82x8
2
x92
------
A91x9
=
G x( ) g11 0 g31 g41 0 0 0 0 00 0 0 0 0 0 0 0 g92
T
=
x id iq ifd ikd ikq q GT
=
u vfd uGT
=
Synchronous Machine and Regulators
2-35
Feedback Linearization DesignThe input-output feedback linearization technique consists in the exactcancellation of the nonlinearities of the system in order to obtain a linearrelationship between inputs and outputs in closed-loop. The nonlinear controllaw is deduced by successively differentiating each of the outputs until at leastone input appears. Consider the first output as the terminal voltage Vt:
The terminal voltage is a complicated function of the state variables in whichthe control input Vfd appears explicitly with a multiplying factor of a very smallorder of magnitude. We therefore ignore this direct dependence between Vt andVfd and compute the time derivate of output y1:
The second output y2 is the angle that has to be differentiated three timesbefore the inputs appear. This yields:
y1 Vt Vd2 Vq
2+= =
tddy1
1 x( ) 11 x( )u1 12 x( )u2, where+ +=
1 x( ) xVt F x( )=
12Vt---------- 2Vd x
Vd 2Vq xVq
+
F x( )=
11 x( ) xVt G1 x( )=
12Vt---------- 2Vd x
Vd 2Vq xVq
+
G1 x( )=
12 x( ) xVt G2 x( )=0=
2 Case Studies
2-36
Combining the equations of the outputs, the following input-output nonlinearsystem is obtained
and the nonlinear control law is easily deduced.
This will yield, in closed-loop, the exactly linearized input-output system
Once the system has been linearized, any linear control design can be appliedto regulate the outputs. Here, the pole placement method was chosen and thefollowing linear control law is proposed:
The last term in the equation of v2 is introduced in order to stabilize theinternal dynamics. These dynamics come about because the original nonlinearsystem is a ninth order while the linearized system is a fourth order. This iscalled partial linearization and we must ensure that the remaining dynamics
t3
3
d
d y2r x
F7 F x( ) r xF7G1 x( )u1 r x
F7G2 x( )u2+ +=
2 x( ) 21 x( )u1 22 x( )u2+ +=
y11( )
y23( )
1 x( )2 x( )
11 x( ) 021 x( ) 22 x( )
u1u2
+=
u1u2
11 x( ) 021 x( ) 22 x( )
11 x( )2 x( )
v1v2
+=
y11( )
y23( )
v1v2
=
v1 k11 Vt Vtref( )=v2 k21 ref( ) k22 ref( ) k23 ref( ) k24 x9 x9ref( )+ + +=
Synchronous Machine and Regulators
2-37
are asymptotically stable. A complete treatment of this question can be foundin reference [1].
Simulation ResultsThe performance of the nonlinear controller is tested on the nonlinearturbine-generator system. The controller and turbine are simulated usingSimulink blocks while the generator is represented by the SynchronousMachine block from the powerlib library. A three-phase short-circuit issimulated on the load busbar and the fault is cleared after 100 ms. Theperformance of the nonlinear controller is analyzed.
The case study is included in the psbregulator.mdl file. The system is illustratedin Figure 2-18 below. Before running the simulation, make sure that thesimulation parameters are set as follows:
Solver: ode23tb; Maximum order: 5 Stop time: 1.0 Max step size: auto; Initial step size: auto Relative tolerance: 1e-3; Absolute tolerance: auto Workspace I/O: Load initial states: psbregulinit
Figure 2-18: Simulink Diagram of Case Study (psbregulator.mdl)
Due to nonlinearities present in this system, computation of initial conditionswas not carried out. Instead, a long simulation (10 s) was executed and the finalstates saved in file psbreguldata.mat. These final states are used as initial
2 Case Studies
2-38
states in this case study. The simulation consequently starts in steady-state.At t=0.1s, the fault is suddenly applied and removed after 100 ms (6 cycles).The post-fault transient is then observed.
The nonlinear controller calls a MATLAB initialization function to compute thegains before the simulation. Although this process has been automated to takeinto account the parameters in the dialog boxes of the various blocks, it is notrecommended that you change any value in any block.
If you decide to change some values, a long simulation must be run and thefinal states must be in a file called psbregulinit.mat. Figure 2-19 shows theresponse of the generators terminal voltage, load angle, and the control effortof the regulator. You can observe how the stabilization of Vt is obtained in lessthan 0.25 seconds with this controller. The load angle takes longer to stabilize,because the time constant of the mechanical part of the system is much largerthan the electrical time constants. If you want to compare results with classicalregulators, replace the nonlinear controller with the same excitation systemand Hydraulic Turbine and Governor block used in the psbturbine demo. Youwill notice that the system takes longer to stabilize than in this case study.
Synchronous Machine and Regulators
2-39
Figure 2-19: Simulation Results Obtained With Case Study
References[1] Akhrif O., F.A. Okou, L.A. Dessaint, and R. Champagne, Application of aMultivariable Feedback Linearization Scheme for Rotor Angle Stability andVoltage Regulation of Power Systems. IEEE Transactions on Power Systems,Vol. 14, No. 2, May 1999, pp. 620-628.
2 Case Studies
2-40
Variable-Frequency Induction Motor DriveThis case study presents a variable-frequency AC motor drive in which a pulsewidth modulated (PWM) inverter is used as a variable-voltagevariable-frequency source to drive an induction motor in variable-speedoperation.
The drive, including the motor, the power converter, and the speed controlsystem, is modeled by using the Power System Blockset and Simulink blocks.The drive operation is studied for different operating conditions: starting,steady-state, and transients.
The objective of this example is to demonstrate the use of Machine library, andPower Electronics library blocks in combination with Simulink blocks in thesimulation of a complex electromechanical system operating at high frequency.The electrical part of the AC motor drive including the PWM inverter is builtusing the Universal Bridge block. The induction motor is represented by theAsynchronous Machine block, which models both electric and mechanicaldynamics. The control system including current and speed regulators is builtusing Simulink blocks. The interface between electrical and control systems isensured by blocks of the Measurement library of powersys.
Description of the Induction Motor DriveThe induction motor requires a variable-frequency three-phase source forvariable-speed operation. This source can be realized by using a powerconverter system consisting of a rectifier connected to an inverter through a DClink.
Figure 2-20 shows a block diagram of the power circuit of a typicalvariable-frequency induction motor drive.
Variable-Frequency Induction Motor Drive
2-41
Figure 2-20: Variable-Frequency Induction Motor Drive
The power grid AC voltage is converted into a fixed DC voltage by the rectifier.The harmonics are filtered out by an LC filter to provide a smooth DC voltage,which is then applied to the inverter input.
Figure 2-21: Three-Phase IGBT Inverter
The inverter consists essentially of six power switches that can bemetal-oxide-semiconductor-field-effect transistors (MOSFET), gate-turn-offthyristors (GTO), or insulated-gate-bipolar transistors (IGBT), depending onthe drive power capacity and the inverter switching frequency (Hz). Figure2-21 shows a simplified diagram of a three-phase IGBT inverter.
The inverter converts the DC link voltage into an adjustable three-phase ACvoltage. Different control schemes can be used to control the inverter outputvoltage and frequency. One of the most utilized schemes is pulse widthmodulation (PWM) in which three-phase variable sinusoidal voltage
InductionMotor
L
C
Filter
+
-
Vdc
Dc link
Rectifier Inverter
60 HzPowerGrid
V f
+
-
Vdc
AB
C
2 Case Studies
2-42
waveforms are obtained by modulating the on and off times of the powerswitches.
In industrial drive applications, the PWM inverter operates as a three-phasevariable-frequency, variable-voltage source with fundamental frequencyvarying from zero to three times the motor nominal frequency.
In some control schemes where a three-phase, variable-frequency currentsource is required, current control loops are added to force the motor currentsto follow an input reference (usually sinusoidal).
The inverter-fed induction motor drive can be controlled by using variousschemes depending on the application, desired performance, and controllerdesign complexity. The most utilized schemes are:
Stator V/Hz control Stator currents and open loop flux control Vector control (field-oriented control) Direct torque control (DTC)
A Field-Oriented Variable-Speed Induction MotorDriveIn this case study, we consider a variable-speed induction motor drive usingfield-oriented control. In this control scheme, a d-q coordinates reference framelocked to the rotor flux space vector is used to achieve decoupling between themotor flux and torque. They can be thus separately controlled by statordirect-axis current and quadrature-axis current respectively, like in a DCmotor. Figure 2-22 shows a block diagram of a field-oriented induction motordrive.
Variable-Frequency Induction Motor Drive
2-43
Figure 2-22: Field-Oriented Variable-Frequency Induction Motor Drive
The induction motor is fed by a current-controlled PWM inverter, whichoperates as a three-phase sinusoidal current source. The motor speed iscompared to the reference * and the error is processed by the speed controllerto produce a torque command Te*.
As shown in Figure 2-23, the rotor flux and torque can be separately controlledby the stator direct-axis current ids and quadrature-axis current iqs,respectively.
PWMInverter
InductionMotor
SpeedSensor
+
-
Vdc
ia* ib* ic*
SpeedReference
+ -
CurrentControllers
iaib
ic
abcdq
abcdq
Rotor FluxCalculation
eCalculation
ids
ids*
iqs
iqs*
e
SpeedController
Calculationiqs* |r|
Te*
|r| Calculationids*
Rotor FluxReference
TorqueReference
2 Case Studies
2-44
Figure 2-23: Field-Oriented Control Principle
The stator quadrature-axis current reference iqs* is calculated from torquereference Te* as
where Lr is the rotor inductance, Lm is the mutual inductance, and |r|est isthe estimated rotor flux linkage given by
where r = Lr/Rr is the rotor time constant.
The stator direct-axis current reference ids* is obtained from rotor fluxreference input |r|*.
The rotor flux position e required for coordinates transformation is generatedfrom the rotor speed m and slip frequency sl.
Isq
d
a
Ir
ids
iqs
e
rsl
rRotor axis
e
Rotor flux axis
iqs23---
2p----
LrLm--------
Te
r est---------------- =
r estLmids1 rs+-----------------=
idsr Lm
------------=
Variable-Frequency Induction Motor Drive
2-45
The slip frequency is calculated from the stator reference current iqs* and themotor parameters.
The iqs* and ids* current references are converted into phase current referencesia*, ib*, ic* for the current regulators. The regulators process the measured andreference currents to produce the inverter gating signals.
The role of the speed controller is to keep the motor speed equal to the speedreference input in steady-state and to provide a good dynamic duringtransients. It can be of proportional-integral (PI) type.
Modeling the Induction Motor DriveOpen the psbacdrive.mdl file of the powerlib library by typing psbacdrive inthe MATLAB command window. A circuit diagram titled psbacdrive willappear. Before running the example, save this circuit as case4.mdl in yourworking directory so that you can make further modifications without alteringthe original file.
Figure 2-24 shows the psbacdrive diagram in which blocks from the PowerSystem Blockset and Simulink are used to model the induction motor drive.
e m sl+( ) td=
slLm
r est----------------
RrLr------- iqs =
2 Case Studies
2-46
Figure 2-24: Variable-Speed Field-Oriented Induction Motor Drive(psbacdrive.mdl)
The induction motor is modeled by an Asynchronous Machine block. The motorused in this case study is a 50HP, 460V, 4 poles, 60 Hz motor having thefollowing parameters: Rs = 0.087 , Lls = 0.8 mH, Lm = 34.7 mH, Rr = 0.228, Llr = 0.8 mH.
The current-controlled PWM inverter circuit is shown in Figure 2-24. TheIGBT inverter is modeled by a Universal Bridge block in which the PowerElectronic device and Port configuration options are selected as IGBT/Diodeand ABC as output terminals respectively. The DC link input voltage isrepresented by a 780 V DC voltage source.
The current regulator, which consists of three hysteresis controllers, is builtwith Simulink blocks. The motor currents are provided by the measurementoutput of the Asynchronous Machine block.
Specify Ts= 2e6
To start in steady state with wm=120 rad/s and Tload = 0>> load psbacdrive_init
and check Initial state in Simulation/Parameters/Workspace I/O
Vector Control of a VariableFrequency Induction Motor Drive
120
wref
wrefTe*
PhirIq*
iqs* Calculation
Phir* Id*
id* Calculation
Vdc
+
v
Vab
z
1
z
1
z
1
Iq
Phir
wm
Teta
Teta Calculation
w*
wTe*
Speed_controller
Scope
0.96
Phir* ?
More Info0
Load_torque
Load_torque
A
B
C
Tm
a
b
c
m_SI
Induction Motor50 HP / 460 V
+
pulses
A
B
C
IGBT Inverter
IdPhir FluxCalc.
Flux CalculationDiscrete System
m
is_abc
wm
Te
DemuxId*
Iq*
Teta
Iabc*
DQABC
Iabc*
IabcPulses
Current Regulator
Teta
Iabc
Id
Iq
ABCDQ
Vab (V)
Variable-Frequency Induction Motor Drive
2-47
The conversions between abc and dq reference frames are executed by theABC-DQ and DQ-ABC blocks of Figure 2-24:
The rotor flux is calculated by the Flux_Calculation block of Figure 2-24:
The rotor flux position (e) is calculated by the Teta Calculation block of Figure2-24:
1Pulses
Mu
em
em2Iabc
1Iabc*
2Iq
1Id
f(u)iq
f(u)id
Mu2/3
2/3cos(u)
sin(u)
2Iabc
1Teta
1Iabc
ic
f(u)ib
f(u)ia
MuMu
1
1
1
cos(u)
sin(u)
3Teta
2Iq*1
Id*
ABC-DQ
DQ-ABC
1Phir 0.1557s+1
34.7e3 1Id
1Teta
Mu
Muxs
1
2
134.7e3*u[1]/(u[2]*0.1557+1e3)
3wr
2Phir
1Iq
2 Case Studies
2-48
The stator quadrature-axis current reference (iqs*) is calculated by theiqs*_Calculation block of Figure 2-24:
The stator direct-axis current reference (ids*) is calculated by theid*_Calculation block of Figure 2-24:
The speed controller is of proportional-integral (PI) type and is implementedusing Simulink blocks:
Simulating the Induction Motor DriveRun the simulation by selecting Start from the Simulation menu in Simulink.The simulation parameters in the Simulation Parameters menu can be set asfollows:
Simulation time: Start Time:0, Stop time: 1.5Solver option: Type: Variable-step ode23tb (stiff/TR-BDF2)Max Step Size: autoInitial Step Size: autoRelative Tolerance: 1e-3Absolute Tolerance: 1e-3
The motor voltage and current waveforms as well as the motor speed andtorque are displayed on three axes of the scope connected to the variables Vab,Is, Te, and .
1Iq*
Mux u[1]*0.341/(u[2]+1e3)2
Phir
1Te*
1Id*
1/34.7e
KF
1Phir*
1Te*
s
1
Kp
Ki2w
1w*
Variable-Frequency Induction Motor Drive
2-49
Once the simulation is complete, return to the MATLAB command window toexamine the results with more details by using the plot instruction.
Drive StartingYou can start the drive from a standstill by specifying 0 initial conditions forall state variables in the powergui interface and specifying[1,0,0,0,0,0,0,0] as the initial conditions for the Asynchronous Machineblock. In this example, the speed reference is stepped from 0 to 120 rad/s att = 0 and we observe the motor speed, torque, and current.
The transient responses for the starting of the induction motor drive are shownin Figure 2-25.
Note that you can save the final system state vector by previously selectingWorkspace I/O/Save to work space/Final state in the SimulationParameters window.
Figure 2-25: Starting of the Induction Motor Drive
0 0.5 1 1.50
50
100
150Induction motor drive starting
Mot
or sp
eed ,
rad/s
0 0.5 1 1.5
0
100
200
300
400
Mot
or to
rque
, N.m
0 0.5 1 1.5500
0
500
Stato
r cur
rent
(Ia) ,
A
Time , s
2 Case Studies
2-50
Steady-State Voltage and Current WaveformsWhen the steady state is attained, you can stop the simulation and plot thevoltage and current waveforms using the variables Vab and Ia sent back in theMATLAB workspace by the scope.
Figure 2-26 shows the motor current and voltage waveforms obtained when theload torque is 50 Nm.
Figure 2-26: Steady-State Motor Current and Voltage Waveforms
Speed Regulation Dynamic PerformanceWe can study the drive dynamic performance (speed regulation performanceversus reference and load torque changes) by applying two changing operatingconditions to the drive: a step change in speed reference and a step change inload torque.
Replace the Constant wref and Load_torque blocks in the psbacdrive.mdldiagram by two Simulink step functions with different starting times.
0.2 0.25 0.3 0.35 0.4 0.45 0.5
500
0
500
Induction motor drive Current and voltage waveforms
Stato
r volt
age (
Vab)
, V
0.2 0.25 0.3 0.35 0.4 0.45 0.550
0
50
Time , s
Stato
r cur
rent
(Ia) ,
A
Variable-Frequency Induction Motor Drive
2-51
The speed reference steps from 120 rad/s to 160 rad/s at t = 0.2 s and the loadtorque steps from 0 N.m to 200 N.m at t = 1.4 s. The final state vector obtainedwith the previous simulation can be used as the initial condition so that thesimulation will start from steady-state. Load the psbacdrive_init.mat file,which will create the xInitial variable. Select Workspace I/O/Load fromworkspace/Initial state in the Simulation Parameters window and restartthe simulation.
The obtained response of the induction motor drive to successive changes inspeed reference and load torque is shown in Figure 2-27.
Figure 2-27: Dynamic Performance of the Induction Motor Drive
References[1] Leonhard W., Control of Electrical Drives. Springer-Verlag, Berlin 1996.
[2] Murphy J. M. D. and Turnbull F. G., Power Electronic Control of AC Motors.Pergamon Press, Oxford 1985.
[3] Bose B. K., Power Electronics and AC Drives. Prentice-Hall. EnglewoodCliffs 1986.
0 0.5 1 1.5 2 2.5100
120
140
160
180
Mot
or sp
eed ,
rad/s
Speed controller transient response
0 0.5 1 1.5 2 2.5
0
100
200
300
400
Mot
or to
rque
, N.m
0 0.5 1 1.5 2 2.5100
50
0
50
100
Stato
r cur
rent
(Ia) ,
A
Time , s
2 Case Studies
2-52
HVDC SystemThe final example described in this section illustrates modeling of a highvoltage direct current (HVDC) transmission link [1]. Perturbations are appliedin order to examine the system performance [2]. The objectives of this exampleare to demonstrate the use of the Universal Bridge block,and the Three-PhaseTransformer (Three-Windings) block in combination with Simulink blocks inthe simulation of a complete pole of a 12-pulse HVDC transmission system.The electrical part representing the AC network is built using three-phaseblocks. The Discrete 12-Pulse HVDC control system is a generic controlavailable in the Controls library of powerlib_extras.
Description of the HVDC Transmission SystemThis network is available as a demo entitled Complete 12-Pulse HVDCTransmission System. Open this demo and save the psbhvdc12pulse.mdl filein your working directory as case5 in order to allow further modifications tothe original system.This system is shown on Figure 2-28.
A 1000 MW (500 kV, 2kA) DC interconnection is used to transmit power froma 500 kV, 5000 MVA, 60Hz network to a 345 kV, 10 000MVA, 50Hz network.The AC networks are represented by damped L-R equivalents with an angle of80 degrees at fundamental frequency (60 Hz or 50 Hz) and at the thirdharmonic.
The rectifier and the inverter are 12-pulse converters using two UniversalBridge blocks connected in series. Open the two converter subsystems to seehow they are built. The converters are interconnected through a 300 kmdistributed parameter line and 0.5 H smoothing reactors.The convertertransformers (Wye grounded /Wye/Delta) are modeled with Three-PhaseTransformer (Three-Windings). The transformer tap changers are notsimulated. The tap position is rather at a fixed position determined by amultiplication factor applied on the primary nominal voltage of the convertertransformers (0.90 on rectifier side; 0.96 on inverter side).
From the AC point of view, an HVDC converter acts as a source of harmoniccurrents. From the DC point of view, it is a source of harmonic voltages. Theorder n of these characteristic harmonics are related to the pulse number p ofthe converter configuration: n = kp 1 for the AC current, and n = kp for thedirect voltage, k being any integer. In our example, p = 12, so that injectedharmonics on the AC side are: 11, 13, 23, 25, and on the DC side are: 12, 24.
HVDC System
2-53
Figure 2-28: HVDC System
2 Case Studies
2-54
AC filters are used to prevent the odd harmonic currents from spreading out onthe network. The filters are grouped in two subsystems. These filters alsoappear as large capacitors at fundamental frequency, thus providing reactivepower compensation for the rectifier consumption due to the firing angle . For=30 degrees, the converter reactive power demand is approximately 60% ofthe power transmitted at full load. Look under the AC filters subsystem maskto see the high Q (100) tuned filters at the 11th and 13th harmonic and the lowQ (3), or damped filter, used to eliminate the higher order harmonics, e.g., 23rdand up. Extra reactive power is also provided by capacitor banks.
Two circuit breakers are used to apply faults on the rectifier AC and DC sides.
The rectifier and inverter control systems use the Discrete 12-pulse HVDCControl block of the Control library of powerlib_extras.
The power system and the control system are both discretized with the samesample time Ts.
Define parameter Ts=50e-6 in your workspace prior to start simulations.
Frequency Response of the AC and DC SystemsYou will now measure the frequency response of the AC systems (rectifier andinverter sides) and of the DC line.
We have already explained in Session 2 in the Tutorial chapter how theImpedance Measurement block allows you to compute the impedance of alinear system from its state-space model. As the thyristor valves of theconverters are nonlinear blocks, they will be ignored in the impedancecalculation and you will get the impedances with the valves open.
Open the Measurements library and copy three Impedance Measurementblocks in your model and rename them Zrec, Zinv, and ZDC. Connect the twoinputs of Zrec and Zinv between phase A and phase B of the AC system on therectifier and inverter sides. Measuring the impedance between two phases willgive two times the positive-sequence impedance. Therefore you must apply afactor of 1/2 on the impedance in order to obtain the correct impedancevalue.Open the two Impedance Measurement blocks and set the Multiplicationfactor to 0.5. Finally, connect input 1 of the ZDC block between the DC lineterminal and the rectifier smoothing reactor and input 2 to ground. Save yourmodified circuit as case5Zf.
HVDC System
2-55
Now open the powergui. In the Tools menu select Impedance vs FrequencyMeasurement. A new window opens showing the three Zmeter block names.Fill in the Frequency range by typing 10:2:1500. Select the lin scale to displaythe Z magnitude and lin scale for the frequency axis. Check the Save data toworkspace button and enter Zcase5 as the variable name that will contain theimpedance vs frequency. Click on the Display button.
When the calculation is finished, three graphic windows appear with themagnitude and phase as function of frequency measured by the threeImpedance Measurement blocks. If you check in your workspace, you shouldhave a variable named Zcase5. It is a four-column matrix containing frequencyin column 1 and the three complex impedances in columns 2, 3, and 4 with thesame the same order as in the window displaying the block names.
The magnitudes of the three impedances as function of frequency are shown inFigure 2-29.
Figure 2-29: Positive-Sequence Impedances of the Two AC Networks and ofthe DC Line
0 500 1000 15000
100
200
300
Impe
danc
e (O
hm) Zinv 50Hz
0 500 1000 15000
200
400
600
800
Impe
danc
e (O
hm) Zrec 60 Hz
0 500 1000 15000
1000
2000
3000
4000
Frequency (Hz)
Impe
danc
e (O
hm) ZDC
660Hz 780Hz
220 Hz resonance
188 Hz resonance
56.75 Ohms
| 60 Hz
| 50 Hz
12.65 Ohms 550Hz 650Hz
240 Hz
2 Case Studies
2-56
Notice the two minimum impedances on the Z magnitudes of the AC systems.These series resonances are created by the 11th and 13th harmonic filters.They occur at 660 Hz and 780 Hz on the 60 Hz system. Notice also that theaddition of 600 Mvar capacitive filters on the inductive systems createsresonances (around 188 Hz on the rectifier side and 220 Hz on the inverterside). Zoom in on the impedance magnitude in the 60 Hz region. You shouldfind a magnitude of 56.75 for the 60 Hz system, corresponding to an effectiveshort circuit level of 5002/56.75= 4405 MVA on the rectifier side (5000 MVA -600 Mvar of filters).
For the DC line, note the series resonance at 240 Hz, which corresponds to themain mode likely to be excited on the DC side, under large disturbances.
Description of the Control SystemThe control systems of the rectifier and of the inverter use the same 12-pulseHVDC Control block from the Controls library of powerlib_extras. The blockcan operate either in rectifier or inverter mode. Use the Look Under Mask tosee how this block is built.
HVDC System
2-57
Figure 2-30: 12-Pulse HVDC Control B lock and Its Dialog Box for InverterMode
Inputs and OutputsInput 1 (Vabc) is a vectorized signal of the three line-to-ground voltagesmeasured at the primary of the converter transformer. These three voltagesare used to synchronize the pulse generation on the line voltages. Inputs 2 and3 are the DC line voltage (VdL) and current (Id). Note that the measured DCcurrents (IdR and IdI in A) and DC voltages (VdLR and VdLI in V) are scaledto p.u (1pu current = 2 kA; 1pu voltage = 500 kV) before they are used in thecontrollers.
2 Case Studies
2-58
Inputs 4 and 5 (Vd_ref and Id_ref) are the Vd and Id reference values in p.u.The VdL and Id inputs are filtered before being processed by the regulators. Afirst order filter is used on Id input and a second order filter is used on VdLinput. The filter parameters are shown in the dialog box of Figure 2-30.
Input 6 (Block) accepts a logical signal (0-1) used to block the converter whenBlock=1. Input 7 is also a logical signal that can be used for protectionpurposes. If this signal is high (1), the firing angle will be forced at the valuedefined in the block dialog box.
The first two block outputs (PulseY and PulseD) contain the vectorized signalsof the six pulses to be sent to each of the six-pulse converter connected to thewye and delta windings of the converter transformer. The third output (alpha)is the firing delay angle in degrees ordered by the regulator.The fourth output(Id_ref_lim) is the actual reference current value (value of Id_ref limited bythe VDCOL function as explained below).
Synchronization SystemThe Discrete 12-Pulse HVDC Control block uses the primary voltages (input 1)to synchronize and generate the pulses according to Vd_ref and Id_ref setpoints (inputs 4 and 5). The synchronizing voltages are measured at theprimary side of the converter transformer because the waveforms are lessdistorted. The firing command pulse generator is synchronized to thefundamental frequency of the AC source. At the zero-crossings of thecommutating voltages (AB, BC, CA), a ramp is reset. A firing pulse, isgenerated whenever the ramp value becomes equal to the desired delay angleprovided by the regulator. In order to improve the commutating voltages usedby the pulse generator, the primary voltages (Vabc) are filtered by a low Qsecond-order band-pass filter centered at the fundamental system frequency.The base system frequency and the filter bandwidth are defined in the blockdialog box.
Steady-State V-I CharacteristicThe Discrete 12-Pulse HVDC Control block implements the steady-statecharacteristic shown on Figure 2-31.
HVDC System
2-59
Figure 2-31: Rectifier and Inverter Steady-State Characteristics and VDCOLFunction
In normal operation, the rectifier controls the current at the Id_ref referencevalue whereas the inverter controls the voltage at the Vd_ref reference value.The Id_margin and Vd_margin parameters are defined in the inverter dialogbox. They are set respectively at 0.1 p.u. and 0.05 p.u. The system normallyoperates at point 1 as shown on the figure. However, during a severecontingency producing a voltage drop on the AC network 1 feeding the rectifier,the operating point will move to point 2. The rectifier will therefore be forcedto minimum mode and the inverter will be in current control mode.
Note In industrial controllers, the angle at the inverter is normally limitedin order to keep a minimum angle, where: = extinction angle = 180 - -= commutation or overlap angleThe control required to avoid commutation failures is not implemented inthis version of the HVDC control.
Id_ref=1pu
Vd_ref=1pu
Rectifiermin. (normal AC rect. voltage)
DC current (Id)
DC voltage (Vd)
Vd margin (0.05pu)
Id margin(0.1 pu)
low AC rect. voltage
operating
operating point 2(low AC voltage
point 1
Inverter
at rectifier)
2 Case Studies
2-60
VDCOL FunctionAnother important control function is implemented to change the referencecurrent according to the value of the DC voltage. This control named VoltageDependent Current Order Limiter (VDCOL) automatically reduces thereference current (Id_ref) set point when VdL decreases (as for example, duringa DC line fault or a severe AC fault). Reducing the Id reference currents alsoreduces the reactive power demand on the AC network, helping to recover fromfault. The VDCOL parameters of the Discrete 12-Pulse HVDC Control blockdialog box are explained in Figure 2-32.
Figure 2-32: VDCOL Characteristic; Id_ref =f(VdL)
The Id_ref value starts to decrease when the Vd line voltage falls below athreshold value VdThresh (0.6 pu). The actual reference current used by thecontrollers is available at the fourth controller output named Id_ref_lim.IdMinAbs is the absolute minimum Id_ref value set at 0.08 p.u. When the DCline voltage falls below the VdThresh value, the VDCOL reducesinstantaneously Id_ref. However, when the DC voltage recovers, VDCOLlimits the Id_ref rise time with a time constant defined by parameter Tup (80ms in our example).
VdThresh DC line Voltage (VdL)
Id_ref actual (Id_ref_lim)
Id_ref=1.0 pu
VdMin
1.0 pu
IdMinAbs
Id_ref=0.8 pu
Id_ref=0.027 pu
(0.6 pu)(0.18 pu)
(0.3 pu*1.0)
(0.08 pu)
(0.3 pu*0.8)
IdMin*Id_ref
Mininum Id_ref value
HVDC System
2-61
Current and Voltage RegulatorsThe rectifier and the inverter controls both have a voltage and a currentregulator operating in parallel calculating firing angles v and i. The effective angle is the minimum value of v and i. This angle is available at the thirdblock output named alpha (deg). Both regulators are of the proportional andintegral type (PI). They should have high enough gains for low frequencies (
2 Case Studies
2-62
Figure 2-33: Start Up of the DC System and Step Applied on the ReferenceCurrent
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.5
0
0.5
1
1.5
VdL
(pu)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.5
0
0.5
1
1.5
Id &
Idre
f (pu)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 180
100
120
140
160
alpha
I (deg
)
Time (s)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.5
0
0.5
1
1.5
VdL (
pu)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.5
0
0.5
1
1.5Id
& Idr
ef (pu
)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
50
100
alpha
R (de
g)
Time (s)
Rectifier
Inverter
= 18 degrees
= 142 degrees
HVDC System
2-63
The reference current follows a ramp from zero to 1 p.u. (2kA) in 0.4 s. Observethat the DC current starts to build up at t=20 ms, time at which the controllerand the pulse generators are deblocked.The DC current and voltages start fromzero and reach steady-state in approximately 0.5 s.The rectifier controls thecurrent and the inverter controls the voltage. Trace 1 of both rectifier andinverter scopes shows the DC line voltage (1pu=500 kV). Trace 2 shows thereference current and the measured Id current (1pu=2 kA). Once steady-stateis attained, the a firing angles are 18 degrees and 142 degrees respectively onthe rectifier and inverter side.
Then, at t=0.6 s a step is applied on the reference current to observe thedynamic response of the regulators.
The main equations governing the steady-state operation of the DC system aregiven here so that you can check the theoretical values against the simulationresults.
The following expression relates the mean direct voltage Vd of a 12-pulsebridge to the direct current Id and firing angle
where Vdo is the ideal no-load direct voltage for a 6-pulse bridge
Vc is the line to line rms commutating voltage that is dependent on the ACsystem voltage and the transformer ratio.
Rc is the equivalent commutating resistance
Xc is the commutating reactance or transformer reactance referred to the valveside.
The following rectifier parameters were used in the simulation.
The Vc voltage must take into account the effective value of the voltage on the500kV bus and the transformer ratio. If you look at the waveforms displayedon the V_I_Rect scope, you will find 0.96 p.u. If you open the rectifiertransformer dialog box, you will find a multiplication factor of 0.91 applied onthe primary nominal voltage. The voltage applied to the inverter is thereforeboosted by a factor 1/0.91.
Vd 2 Vdo ( )cos Rc Id( )=
Vdo 3 2 pi( ) Vc=
Rc 3 pi( ) Xc=
2 Case Studies
2-64
Vc =0.96* 200 kV/0.90= 213.3 kV;Id = 2 kA = 18;Xc = 0.24 p.u. based on 1200 MVA and 222.2 kV = 9.874
Therefore:
This theoretical voltage corresponds well with the expected rectifier voltagecalculated from the inverter voltage and the voltage drop in the DC line:
The commutation or overlap angle can be also calculated. Its theoreticalvalue depends on , the DC current Id and the commutation reactance Xc.
Now verify the commutation angle by plotting the currents in two valves,showing for example current extinction in valve 1 and current build up in valve3 of one six-pulse bridge of the rectifier.
Open the rectifier subsystem. Then, open the upper bridge dialog box and theselect All voltages and currents for the Measurement parameter. Now, copythe Multimeter block from the Measurements library into your case5 circuit.Double-click on the Multimeter block. A window showing all the bridgevoltages and currents appears. Select the following signals
uSw1: Rectifier/Universal Bridge iSw1 : Rectifier/Universal Bridge iSw3 : Rectifier/Universal Bridge
Vdo 3 2 pi( ) 213.2 287.9kV= =Rc 3 pi( ) 9.874= 9.429=Vd 2 288 1kV, 18( )cos 9.429 2( )= 510kV=
Vd VdLinverter RDCline Rinduc cetan+( ) Id+=Vd 500kV 4.5 1+( ) 2+ 511kV= =
( )cos Xc Id 2 Vc
------------------------------acos =
18( )cos 9.874 2 2 213.3
----------------------------------acos 18 16.9==
HVDC System
2-65
The number of signals (3) is displayed in the multimeter icon. Using a Demuxblock, send the three multimeter output signals to a two-trace scope. (Trace 1:uSw1 Trace 2: iSw1 and iSw3). Restart the simulation. The waveformsillustrating two cycles are shown on Figure 2-34. The measured commutationangle is 14 steps of 50s or 15.1 of a 60 Hz period. Knowing that the resolutionwith a 50 s time step is 1.1, this angle compares reasonably well with thetheoretical value.
Figure 2-34: Valve Voltage and Currents (Commutation from Valve 1 to Valve3)
Response to a Step of Reference CurrentAt t=0.6 s, a 0.2 p.u. step is applied on the reference current (decrease from 1p.u. to 0.8 pu). At t=0.75s, another step is applied to set the reference back to 1
0.5 0.505 0.51 0.515 0.52 0.525 0.53
4
2
0
2x 105
uSw
1 (V
)
0.5 0.505 0.51 0.515 0.52 0.525 0.531000
0
1000
2000
3000
i S
w1 &
Sw3
(A)
0.5 0.505 0.51 0.515 0.52 0.525 0.5310
15
20
25
alp
haR
(deg)
Time (s)
=15.1
2 Case Studies
2-66
p.u. Observe the response of the current regulator. It stabilizes inapproximately 0.1 s.
Figure 2-35: Response to a 0.2 p.u. Step of the Reference Current
DC line FaultDisconnect the Step Up & Down block in order to eliminate the stepdisturbance applied on the reference current. In the DC Fault Timer andForced Delay blocks of the psbhvdc12pulse circuit, change the multiplicationfactor of 100 in the Transition Times to 1, so that a fault is now applied att=0.6s. Open the I_DCfault scope to observe the fault current. Restart thesimulation.
0.5 0.6 0.7 0.8 0.9 1
0.8
1
1.2
VdL
(pu)
0.5 0.6 0.7 0.8 0.9 1
0.8
1
1.2
VdL
(pu)
0.5 0.6 0.7 0.8 0.9 10.7
0.8
0.9
1
1.1
Id &
Idre
f (pu)
0.5 0.6 0.7 0.8 0.9 10.7
0.8
0.9
1
1.1
Id &
Idre
f (pu)
0.5 0.6 0.7 0.8 0.9 110
15
20
25
30
time (s)
alp
haR
(pu)
0.5 0.6 0.7 0.8 0.9 1140
141
142
143
144
145
time (s)
alp
ha (d
eg)
Rectifier Inverter
HVDC System
2-67
Figure 2-36: DC Line Fault on the Rectifier Side
0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.32
0
2
0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.30.5
00.5
1
1.5
0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.380
100120140160
0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.30
2000400060008000
time (s)
ifalut
DC
(A)alp
haI (d
eg)
VdL (
pu)
Id &
Idref
(pu)
0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.30.5
0
0.5
1
1.5
VdL (p
u)
0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.31
0
1
2
3
Id & I
dref (p
u)
0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.30
50
100
150
200
alphaR
(deg)
Time (s)
RECTIFIER
INVERTER
2 Case Studies
2-68
At fault application (t=0.6 s), the DC current quickly increases to 2.3 p.u. andthe DC voltage falls to zero at the rectifier. This DC voltages drop is seen by theVoltage Dependant Current Order Limiter (VDCOL), which reduces thereference current to 0.3 p.u. at the rectifier. A DC current still continues tocirculate in the fault. Then, a t=0.65 s, the rectifier firing angle is forced to165 degrees when the signal applied to the ForcedAlpha input goes high. Thissignal would normally be provided by the protection system not simulatedhere. The rectifier now operates in inverter mode. The DC line voltage becomesnegative and the energy stored in the line is returned to the AC network,causing rapid extinction of the fault current at its next zero-crossing. Then, is released at t=0.7 s and the normal DC voltage and current recover inapproximately 0.5 s
AC Line-to-Ground Fault at the RectifierYou will now modify the timer blocks in order to apply a line-to-ground fault.In the DC Fault Timer and Forced Delay blocks of psbhvdc12pulse, change themultiplication factor of 1 in the Transition Times to 100, so that the DC faultis now eliminated. In the AC Fault Timer block, change the multiplicationfactor in the transition times to 1, so that a 6 cycle line-to-ground fault is nowapplied at the rectifier. Restart the simulation.
HVDC System
2-69
Figure 2-37: Rectifier and Inverter Signals for an AC Line Fault on the RectifierSide
0.5 0.6 0.7 0.8 0.9 1 1.1 1.2
1
0
1
2
VdL (
u)
0.5 0.6 0.7 0.8 0.9 1 1.1 1.2
0
1
2
3Id
& Id
ref (pu
)
0.5 0.6 0.7 0.8 0.9 1 1.1 1.2
0
50
100
150
(g)
Time (s)
0.5 0.6 0.7 0.8 0.9 1 1.1 1.2
1
0
1
2
INVERTER signals
VdL (
u)
0.5 0.6 0.7 0.8 0.9 1 1.1 1.2
0
1
2
3
Id &
Id ref
(pu)
0 5 0 6 0 7 0 8 0 9 1 1 1 1 2
100
120
140
160
180
2 Case Studies
2-70
Figure 2-38: Voltages and Currents on the 60Hz Side for an AC Line Fault onthe Rectifier Side
Notice the 120 Hz oscillations in the DC voltage and currents during the fault.When the fault is cleared at t=0.7 s, the VDCOL operates and reduces thereference current to 0.3 p.u. The system recovers in approximately 0.4 s afterfault clearing.
References[1] Arrilaga J., High Voltage Direct Current Transmission, IEEE PowerEngineering Series 6, Peter Peregrinus Ltd. 1983.
[2] Electromagnetic Transients Program (EMTP), Workbook IV (TACS),EL-4651, Volume 4, Electric Power Research Institute, 1989.
0.5 0.6 0.7 0.8 0.9 1 1.1 1.21.5
1
0.5
0
0.5
1
1.5
Vabc
(pu)
0.5 0.6 0.7 0.8 0.9 1 1.1 1.2
50
0
50
100
Iabc
(A)
Time (s)
