PSCAD Power System Lab Manual

PSCAD Power System Lab Manual

Laboratory Experiment1: Electromagnetic Transients

Objective:

Source, Inductor and capacitance

Laboratory Tasks:

1. Calculate the natural frequency of oscillation for the given circuit

2. Set the time step, plot step and duration of run and analyze the effect of time step, plot step on the results

3. Study the Damping effect of switch resistance on the oscillations Circuit: R = 0 1.0 [H] 0 .0 5 [u F ] Vc 1 0 e 4 [o h m ] 0.001 [ohm] V o lt a g e S o u rc e M o d e l 2 Figure 1: Simple RLC circuit

Source: IEEE PES

Natural frequency of oscillation for the given circuit is 712Hz

F 0

1

2 pi⋅ ⋅ LC

:=

Build the above the circuit, right click on the blank space of the design editor and select project settings, enter the time step=plot step= 1.5ms (1/F0) and duration of run=0.1s. Run the

simulation and comment on the results.

Change the project settings as below and comment on the results 1. Time step=Plot step=0.15ms (1/(10*F0)), duration of run=0.1s

2. Time step=Plot step=0.015ms (1/(100*F0)), duration of run=0.1s

With the selection of appropriate project settings run the simulation for different value of switch resistance and comment on the sensitivity and damping nature of resistance on oscillations.

Waveforms: Main : Graphs Time _{0.0000 0.0050 0.0100 0.0150 0.0200 0.0250 0.0300 0.0350 0.0400 0.0450} 0 25 50 75 100 125 150 175 200 V o lt a g e a /c c a p a c it o r

Voltage A/c Capacitor

Figure 2: Voltage across capacitor with 15us time step=plot step

Main : Graphs Time _{0.0000 0.0050 0.0100 0.0150 0.0200 0.0250 0.0300 0.0350 0.0400 0.0450} 0 25 50 75 100 125 150 175 200 V o lt a g e a /c c a p a ci to r

Voltage A/c Capacitor

Figure 3: Voltage across capacitor with 150us time step=plot step

Main : Graphs Time _{0.0000 0.0050 0.0100 0.0150 0.0200 0.0250 0.0300 0.0350 0.0400 0.0450} 0 25 50 75 100 125 150 175 200 V o lt a g e a /c c a p a ci to r

Voltage A/c Capacitor

Laboratory Experiment2: PF Improvement of a lagging load

Objective:

Laboratory Tasks:

1. Learn breaker operation

2. Calculate the reactive power requirement of the load theoretically and match with the simulated value

3. Calculate theoretically capacitance requirement to make the power factor unity 4. Learn how to measure instantaneous and average power

Circuit: BRK1 2 2 .6 1 [o h m ] 1 9 .7 2 e -3 [H ] 3 5 .9 [u F ] BRK2 In d u c tiv e lo a d w ith P F 0 .9 7 C a p a c iti v e L o a d IL VL R = 0

Figure 1: Voltage source connected to RL load with capacitor across the load

* Va Ia * * Vb Ib Vc Ic Pa ins t Pb ins t Pc inst B + D + F + Pt ins t G 1 + sT Pa avg BRK1 Breaker1 Breaker2 BRK2

Figure 2: Calculation of Instantaneous active power, Breaker control panel and switch Main ... Breaker2 0 open cl ose Main ... Breaker1 0 open cl ose

Capacitor Design: Qd :=37.6 e⋅ 6 _Var Volts Xc (Vlg) 2 Qd =88.652 := Cap 1 2 π⋅ ⋅50⋅Xc ( ) 3.591 10 5 − × = :=

Keep breaker 1 closed and breaker 2 in open position, run the simulation with default runtime settings and observe the current lagging the voltage on the plot and also note down the reactive power demand displayed on the breaker1

Calculate the capacitance required to compensate for the reactive demand, connect the calculated capacitance across the lagging load through breaker2

Keep the breaker 1 and breaker 2 in closed position and run the simulation with default runtime settings and observe the current in phase with the voltage on the plot and also observe the reactive power demand displayed on the breaker 1

Calculate the instantaneous active power of each phase by using the relation Pa(t)=Va(t)*Ia(t) and

total instantaneous active power by using Pt(t)= Pa(t)+ Pb(t)+ Pc(t). Comment on the per phase

and total three phase active power waveforms. Obtain the average power of each phase by connecting the smoothing filter.

Vlg 100e 3

(

)

Waveforms: Main : Graphs Time _{0.170 0.180 0.190 0.200 0.210 0.220 0.230 0.240} -200 -150 -100 -50 0 50 100 150 200 V ol ta ge a nd C ur re nt w /o c ap ac ito r Current Voltage

Figure 3: Voltage and Current waveforms without capacitor

Main : Graphs Time _{0.170 0.180 0.190 0.200 0.210 0.220 0.230 0.240} -200 -150 -100 -50 0 50 100 150 200 V ol ta ge a nd C ur re nt w ith c ap ac ito r Current Voltage

Laboratory Experiment3: Transformer Inrush

Objective:

Motivation

Transformers are essential parts in power system as they provide voltage step up/step down. This exercise demonstrates the Transformer model available in Transformer section of PSCAD Mater Library and the effect of core saturation and inrush current in the transformer when energized

PSCAD model

Two types of transformer models are available in the PSCAD master library as classical model and the UMEC (Unified Magnetic Equivalent Circuit) model. In this exercise, we are only interested in the classical model. Depending upon the number of winding and whether it is a single phase or a three phase, many configurations of transformer models are available from these two types.

Three phase two winding transformer model based on classical approach is used in this exercise.

a). Single line view b).Three phase view

Figure 1 : Three phase two winding transformer model in PSCAD

Figure 1 shows the corresponding transformer model available in PSCAD Master Library under

“Transformers”. Options are provided so that the user may choose between either a magnetizing branch (linear core), or a current injection routine to model magnetizing characteristics. If desired, the

magnetizing branch can be eliminated altogether, leaving the transformer in 'ideal' mode, where all that remains is a series leakage reactance and losses.

This component is the equivalent of three, 1-Phase, 2-Winding Transformers connected in a 3-phase bank, where the user can select the winding interconnections to be Y or Δ on either side. Inter-phase coupling is not represented in the classical transformer models. An equivalent circuit is shown below, using 1-phase transformers:

Figure 2 : Equivalent 3 phase circuit based on single phase transformers

#1 #2 A B C A B C 230.0 [kV] #2 #1 230.0 [kV] 100.0 [MVA]

If inter-phase coupling is essential for your study, then you should choose the equivalent UMEC transformer model.

• The Classical Approach

The theory of mutual coupling can be easily demonstrated using the coupling of two coils as an example. This process can be extended to N mutually coupled windings as shown in References [1], [2] and [3]. For our purpose, consider the two mutually coupled windings as shown below:

Figure 3 : Two Mutually Coupled Windings

Where,

L11 - Self inductance of winding 1 L22 - Self inductance of winding 2

L12 - Mutual inductance between windings 1 & 2

The voltage across the first winding is V1 and the voltage across the second winding is V2. The following equation describes the voltage-current relationship for the two, coupled coils:

(1) In order to solve for the winding currents, the inductance matrix needs to be inverted:

(2)

Where,

For 'tightly' coupled coils, wound on the same leg of a transformer core, the turns-ratio is defined as the ratio of the number of turns in the two coils. In an 'ideal' transformer, this is also the ratio of the primary and secondary voltages. With voltages E1 and E2 on two sides of an ideal transformer, we have:

E1/E2 = a (3)

And

Making use of this turns-ratio 'a' Equation 1 may be rewritten as:

(5)

Figure 4 : Equivalent Circuit of Two Mutually Coupled Windings

Where,

Now the inductance matrix parameters of Equation 1 can be determined from standard transformer tests, assuming sinusoidal currents. The self inductance of any winding 'x' is determined by applying a rated RMS voltage Vx to that winding and measuring the RMS current Ix flowing in the winding (with all other windings open-circuited). This is known as the open-circuit test and the current Ix is the

magnetizing current. The self-inductance Lxx is given as:

(6) Where,

ω - The radian frequency at which the test was performed

Similarly, the mutual-inductance between any two coils 'x' and 'y' can be determined by energizing coil 'y' with all other coils open-circuited. The mutual inductance Lxy is then:

(7)

Transformer data is often not available in this format. Most often, an equivalent circuit, as shown in Figure 4, is assumed for the transformer and the parameters L1, L2 and aL12 are determined from open and short-circuit tests.

For example if we neglect the resistance in the winding, a short circuit on the secondary side (i.e. V2 = 0) causes a current to flow (assuming aL12 >> L2). By measuring this current we may calculate

the total leakage reactance L1 + L2. Similarly, with winding 2 open-circuited the current flowing is,

from which we readily obtain a value for L1 + aL12.

Conducting a test with winding two energized and winding one open-circuit .

The nominal turns-ratio 'a' is also determined from the open circuit tests.

PSCAD computes the inductances based on the open-circuit magnetizing current, the leakage reactance and the rated winding voltages.

• Derivation of Parameters

To demonstrate how the necessary parameters are derived for use by EMTDC, an example of a two winding, single-phase transformer is presented. The data for the transformer is as shown in Table 1:

Parameter Description Value

TMVA Transformer single-phase MVA 100

MVA

f Base frequency 60 Hz

X1 Leakage reactance 0.1 pu

NLL No load losses 0.0 pu

V1 Primary winding voltage (RMS) 100 kV

Im1 Primary side magnetizing current 1 %

V2 Secondary winding voltage (RMS) 50 kV

Im2 Secondary side magnetizing current 1 %

Table 1 : Transformer Data

If we ignore the resistances in Figure 3, we can obtain the (approximate) value for L1 + L2, from the short

circuit test, as:

(8) Where,

Zbase1 – base impedance

As no other information is available, we assume for the turns ratio 'a' the nominal ratio:

We also have for the primary and secondary base currents:

(10)

Thus, we see that by energizing the primary side with 100 kV, we obtain a magnetizing current: (11)

But we also have the following expression from the equivalent circuit: (12) Where, Therefore since, (13) Then, L1 =L2 (14)

By combining Equations 8 and 14 we obtain L1 = L2 =13.263 mH and from Equation 12 we obtain aL12 =

26.5119 H. The values for the parameters in Equation 1 are then obtained as:

L11 = L1 + a.L12 = 26. 5252 H (15)

L22 = (L2 +a.L12)/a 2

= 6.6313 H (16)

L12 = 13.2560 H (17)

System Overview

Figure 5 represents the circuit used in this exercise. Grid on the primary side of the transformer is represented using an equivalent Thevenin’s voltage source while the other side is connected to a high impedance resistance to model unloaded condition

Breaker is used to initialize the transformer at a given time value.

R L R R L Is rc B R K # 1 # 2 1 .0 e 6 Transformer Data 100 MVA, 230 kV/33 kV Impedance 10% (0.1 pu) No load losses 0.5% (0.005pu) Copper loss 0.3% (0.003) Magnetizing current (no load) 1%

230 kV BUS BRK Timed Breaker Logic Open@t0 Source Data Voltage 230 kV at 0.0 Deg. Z+ = 10 Ohms at 88 Deg. Z0 = 7 Ohms at 82 Deg. V230 V230 _Voltage Isrc Isrc E a Ea Ea

Experiment

1. Build a new case and save it as Transformer_1.pscx and build the circuit as in Figure 5. Study the Transformer parameters in the PSCAD model. Right click on the transformer model and select “Edit Parameters”. Set the transformer as ideal. Set the transformer “Saturation Enabled” as “No” under “Saturation” (Figure 6). Plot the bus voltage and source current. Discuss the results obtained.

Figure 6 : Parameter dialog box for transformer

2. Save the case as Transformer_2.psc and set the transformer as non ideal while keeping the saturation disabled. Plot the Bus voltage and source current and see the effect of saturation. Discuss the effects before and after the saturation

3. Set the transformer as an Ideal one and set the “Saturation Enabled” Yes. Plot the Bus voltage and source current and compare the results. Investigate the effect of inrush current and core saturation.

Note: Always set the transformer to 'ideal' when enabling saturation. Otherwise, both the magnetizing branch and the saturation routine will both be used. Here ideal does not mean “lossless ideal” losses are still present regardless of this parameter.

Select the “Saturation Enabled” as No.

Results:

1. Transformer Ideal = Yes, Saturation Enabled = No

Main : Graphs Time(... _{0.490 0.500 0.510 0.520 0.530 0.540 0.550 0.560} -200 -150 -100 -50 0 50 100 150 200 2 3 0 kV B u s V o lt a g e ( kV ) Voltage -30 -20 -10 0 10 20 30 T ra n sf o rm e r S e c V o lt a g e (k V ) Ea -0.0020 -0.0015 -0.0010 -0.0005 0.0000 0.0005 0.0010 0.0015 0.0020 2 3 0 kV B u s cu rr e n t( kA ) Isrc

2. Transformer Ideal = No, Saturation Enabled = No Main : Graphs Time(... _{0.480 0.500 0.520 0.540 0.560 0.580 0.600 0.620 0.640 0.660} -200 -150 -100 -50 0 50 100 150 200 2 3 0 kV B u s V o lt a g e ( kV ) Voltage -30 -20 -10 0 10 20 30 T ra n sf o rm e r S e c V o lt a g e (k V ) Ea -0.0060 -0.0040 -0.0020 0.0000 0.0020 0.0040 0.0060 0.0080 2 3 0 kV B u s cu rr e n t( kA ) Isrc

3. Transformer Ideal = Yes, Saturation Enabled = Yes Main : Graphs Time(... _{0.450 0.475 0.500 0.525 0.550 0.575 0.600 0.625 0.650 0.675} -200 -150 -100 -50 0 50 100 150 200 2 3 0 kV B u s V o lt a g e ( k V ) Voltage -30 -20 -10 0 10 20 30 T ra n sf o rm e r S e c V o lt a g e (k V ) Ea -0.60 -0.40 -0.20 0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 2 3 0 kV B u s cu rr e n t( kA ) Isrc

Figure 9 : Bus Voltage, Va and Isrc characteristic around T = 0.5

Main : Graphs x _{0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0} -0.20 0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 Isrc_a

Main : Graphs x _{7.400 7.420 7.440 7.460 7.480 7.500 7.520 7.540 7.560 7.580} -0.0040 -0.0030 -0.0020 -0.0010 0.0000 0.0010 0.0020 0.0030 0.0040 0.0050 Isrc_a

Figure 11 : Phase A of the source current at steady state

Both the inrush phenomena during initialization of the transformer and the saturation effect of the magnetizing are visualized in Figure 10 and Figure 11.

References

1. H. W. Dommel, Digital Computer Solution of Electromagnetic Transients in Single and

Multiphase Networks, IEEE Transactions on Power Apparatus and Systems, PAS-88, #4, pp. 388-399, April 1969.

2. H. W. Dommel, Transformer Models in the Simulation of Electromagnetic Transients, Proc. 5th Power Systems Computing Conference, Cambridge, England, September 1-5, 1975, Paper 3.1/4. 3. V. Brandwajn, H. W. Dommel, I. I. Dommel, Matrix Representation of Three Phase N-Winding

Transformers for Steady State Transient Studies, IEEE Transactions on Power Apparatus and Systems, PAS-101, #6, pp. 1369-1378, June 1982.

Laboratory Experiment4: Induction Motor Starting

Objective:

Motivation

Induction motors are the most common type of motors used in the industries. This exercise introduces the induction machine model available in the PSCAD “Master library” under “Machines” and investigates its starting characteristics.

PSCAD model

PSCAD has two fully developed Induction Machine models; namely the “Squirrel Cage Induction Machine” and the “Wound Rotor Induction Machine”. If the rotor terminals are shorted in the wound rotor machine, it is identical to the squirrel cage machine. It is recommended to use the Wound Rotor induction machine model for all the simulations which include squirrel induction machines.

PSCAD model of the wound rotor induction machine is shown in Figure 1.

S TL

I M W

Figure 1 : PSCAD model of the wound rotor induction machine

The model can be controlled by specifying either the mechanical torque (input TL) on the machine shaft or the machine speed (input W) as the input to the induction machine. The input ‘S’ is simply a switch to select between the two input modes. When S=0, input ‘TL’ is selected and if S=1 input ‘W’ is selected. Induction machine can be either run at “torque input” mode or “speed input” mode depending upon the value or input of S.

• Torque Input Mode

In the ‘torque input’ mode, the speed of the machine is calculated based on the equations of mechanical motion.

The electrical torque is calculated by the model on its terminal conditions (voltage/current). Tm is the

mechanical torque which is an input to the machine. J is the inertia due to all moving parts of the mechanical shaft. ωm is the speed of the rotor and B accounts for mechanical damping.

• Speed Input Mode

In the “Speed Input” mode, the machine will operate at the specified speed input “W”. This can be a variable or a constant. In some simulations, it may be advantageous to start the machine in the “Speed Input” mode and then switch to the “torque input” mode once steady state has been achieved. This can be done by switching input ‘S’ from 1 to 0. For a more complex representation of the mechanical system such as “multi mass torsional shaft model”, the input ‘W’ must be selected.

• Equivalent circuit

In the Wound Rotor Induction Machine, the rotor terminals are accessible to the user, and can be connected to an external resistance or an electrical circuit. In addition to the stator and rotor windings, there is a provision in the model to include up to three additional windings to model the effects of rotor bars (if any).The d-axis equivalent circuit for the wound rotor induction machine with one squirrel cage in effect is shown in Figure 2. This is derived in a manner similar to that for the synchronous machine. Similar equivalent circuits are applicable to the q axis as well as to the squirrel cage machine.

Figure 28: d axis equivalent circuit

NOTE: All reactance and resistance values are those referred to the stator. Where,

R1= Stator resistance R2= wound rotor resistance R3= first cage resistance Xa= Stator leakage reactance Xkd1= wound rotor leakage reactance Xkd2= first cage leakage reactance Xmd= Magnetizing reactance

System Overview

Part 1. Studying start up characteristics of the induction machine (Case Ind_Motor_Starting_1.pscx) Figure 3 represents a system where induction machine is connected to a power system which comprises of an equivalent Thevinin’s voltage source and connected through transformers.

0.0 WIN S TL I M W Rrotor Ix 0 .1 [H ] BRK #1 #2 BRK Timed Breaker Logic Open@t0 INDUCTION MOTOR 13.8 kV, Wound rotor type

External rotor resistance 0 V A Three Phase Breaker Tload # 1 # 2 + R ro to r + R ro to r + R ro to r

Figure 3: PSCAD system for Torque input mode of the induction machine

The breaker (BRK) is initially closed and open at t = 0.5 s. The motor is started from zero speed and the applied mechanical torque is varied as a function of speed as shown in Figure 4.

D + F + Tload k b * W X2

0.0 WIN S TL I M W Rrotor + R ro to r + R ro to r + R ro to r A B Ctrl Ctrl = 1 Tload 1.8 DIST Ix #1 #2 0 .1 [H ] BRK #1 #2 BRK # 1 # 2 Tim ed Breaker Logic Open@t0 INDUCTION MOTOR 13.8 kV, Wound rotor type TIN Passive load 0 V A 0.24 [MVAR] 0.6 [MW] V A External rotor resistance

Figure 5 : PSCAD system (Induction machine starting characteristics (Case : Ind_Motor_Starting_2.pscx)

Part 2. Consider the system shown in figure 5 where load torque is changed from Tload to 1.8 at t = 3s. If electrical torque is larger than mechanical torque, machine will accelerate.

Experiment

Part 1. Build a new case and save it as Ind_Motor_Starting_1.pscx. Build the case as shown in Figure 3. Model the load torque as shown in Figure 4, Run the case and plot the current, electrical torque, reactive power drawn by the motor and the speed of the motor.

Part 2. Save the case as Ind_Motor_Starting_2.pscx. Modify the system as shown in Figure 5. Run the case and observe the variation of system parameters as similar to part 1.

Results

At the instant breaker is closed, large starting current is drawn by the induction machine as shown in figure 7. The speed builds up slightly and settles at a speed slightly lower than the synchronous speed. The motor starting up process take a large amount of reactive power which causes a voltage dip in the system. Figure 6 shows the electrical and load torque characteristics of the induction motor. The motor take few seconds to settle down.

Part 1 Main : Graphs x _{0.00 0.50 1.00 1.50 2.00 2.50 3.00} -5.0 0.0 5.0 10.0 15.0 20.0 ( p u )

Electric torque Tload

0.00 0.50 1.00 1.50 2.00 2.50 3.00 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 S p e e d ( p u ) Speed -1.00 -0.50 0.00 0.50 1.00 M o to r C u rr e n t (k A ) Current -0.50 0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 R e a c ti ve P o w e r (p u ) Q -15.0 -10.0 -5.0 0.0 5.0 10.0 15.0 ( kV ) TERMINAL VOLTAGE

Part 2: Initial transients are similar to the part 1. After t =3 s, the system settle after some transients. 0.0 1.0 2.0 3.0 4.0 5.0 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 S p e e d ( p u ) Speed -1.00 -0.50 0.00 0.50 1.00 C u rr e n t (k A ) Current -0.50 0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 R e a c ti v e P o w e r (p .u .) Q -15.0 -10.0 -5.0 0.0 5.0 10.0 15.0 V o lt a g e ( kV ) TERMINAL VOLTAGE

Main : Graphs x _{0.0 1.0 2.0 3.0 4.0 5.0} -5.0 -2.5 0.0 2.5 5.0 7.5 10.0 12.5 15.0 17.5 20.0 ( p u )

Electric torque Torque Input