T-866-MODE Assignment 2€¦ · T-866-MODE Assignment 2 The Synchronous Machine (Autumn2015) Nicholas Mark Randall 13.12.2015

T-866-MODEAssignment 2

The Synchronous Machine(Autumn 2015)

Nicholas Mark Randall

13.12.2015

Contents

0.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1 Determining per unit values of the involved parameters 41.1 Q.1.1 and Q.1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Q.1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Writing the dynamic model of the synchronous machine 72.1 Q.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3 Steady State Analysis of the Synchronous Machine 83.1 Q.3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83.2 Q.3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83.3 Q.3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

4 Electrical Transient Analysis the Machine 124.1 Q.4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

5 Calculate the Short Circuit Current By the Classical Method 165.1 Q.5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165.2 Q.5.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

6 Conclusion 19

A MatLab functions 20A.1 Q1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20A.2 Q2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21A.3 Q3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22A.4 Q4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23A.5 Q5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24A.6 Q5.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

1

List of Figures

1.1 Mutual Inductance: Φm1 = The mutual Inductance of winding 1 acting by itself,Φm2 = The mutual Inductance of winding 2 acting by itself,Φl1 = Leakage fluxthat is contributed to coil 1,Φl2 = Leakage flux that is contributed to coil 2,Φm= The combined mutual inductance of Φm1 and Φm2 . . . . . . . . . . . . . . . . 6

3.1 Phasor Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

4.1 Plot of the field winding current in the rotor . . . . . . . . . . . . . . . . . . . . . 134.2 Plot of the field winding current in the rotor broken into three part . . . . . . . . 144.3 Armature winding currents in the stator . . . . . . . . . . . . . . . . . . . . . . . 15

5.1 The calculate short circuit current in phase a by using the classical method . . . 175.2 A plot of the difference between the current classical method calculated and the

MATLAB model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175.3 A plot of the difference between the current classical method calculated and the

MATLAB model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2

LIST OF FIGURES

0.1 IntroductionThis assignment will cover creating a MATLAB model of a synchronous generator. Once themodel has been constructed the following tests will be run through it. First covering the conceptsof self, leakage, and mutual inductance. Then applying the steady state conditions to the model.Once the model was run with the steady state and results were compared with a documentprovided by the professor. The model was then verified and the dynamical test on the modelwere run. After that the model was run with a short circuit condition. All the results can beseen in the following chapters.

Page 3 of 26

Chapter 1

Determining per unit values of theinvolved parameters

1.1 Q.1.1 and Q.1.2In section 1.1 the assignment was to write an expression for matrix LD_mat that makes a con-nection between the flux linkage component and current acting on the d-axis. This was donethrough the use of a matrix and equations. The following are equations for the LD_mat matrix:

Ψd = −Ldid + Lafdifd + Lakdikd (1.1)

Ψfd = Lffdifd + Lfkdikd − Lfdaid (1.2)

Ψkd = Lkdf ifd + Lkkdikd − Lkdaid (1.3)

The equations for the LQ_mat are:

Ψq = −Ldid + Lafdifd + Lakdikd (1.4)

Ψd = Lffdifd + Lfkdikd − Lfdaid (1.5)

Then with the equations put into matrices:

The LD_mat. ΨdΨfdΨkd

=

−Ld Lafd Lakd−Lfda Lffd Lfkd−Lkda Lkdf Lkkd

idifdikd

The LQ_mat.[

ΨqΨkq

]=

[−Ld Lakq−Lkqa Lkkq

] [iqikq

]

1.2 Q.1.3Explaining the concepts of self, leakage, and mutual inductance. Self inductance happens whena coil as current in it and the current produces a magnetic field which in turn produces anothercurrent. In a sense this is a feedback loop. Resulting in electromotive force in the coil of wire.Also can be described as flux that links the currents. Meaning linking the original current along

4

CHAPTER 1. DETERMINING PER UNIT VALUES OF THE INVOLVED PARAMETERS

with the current generated by the magnetic field or self generated current hence the name selfinductance. This can be represented by the following equations:

L11 =N1(Φm1 + Φl1)

i1(1.6)

L22 =N2(Φm2 + Φl2)

i2(1.7)

Leakage is the magnetic flux that results in a miss alignment of magnetic flux linkage anddoes not contribute to the mutual inductance. Leakage creates self inductance in the core, whichcontributes to currents being created in the primary coil. This means leakage can be consideredto be acting as an inductor connected in series with the circuit. There is always going to be leak-age when mutual inductance is use do to the magnetic flux never completely linking with the coils.

Mutual inductance is when the magnetic field from a voltage in coil one induces a voltage incoil two. This will only happen when the magnetic field is constantly being in change or whenthe current is being changed. This phenomenon can be represented by the following equation:

L12 = N1Φm2

i2(1.8)

L12 = N2Φm1

i1(1.9)

L11 = L22 = Self InductanceL12 = L21 = Mutual InductanceN1 = N2 = number of turn in the coilΦm1 = The mutual Inductance of winding 1 acting by itself.Φm2 = The mutual Inductance of winding 2 acting by itself.Φl1 = Leakage flux that is contributed to coil 1Φl2 = Leakage flux that is contributed to coil 2i1 = Current in coil 1i2 = Current in coil 2

After writing code for representing the LD_mat and LQ_mat see the m code below in section 1.2:

1 LD_mat = [−(L_al + L_ad), L_ad, L_ad;2 −L_ad, (L_fdl+L_ad), L_fkd;3 −L_ad, L_fkd, (L_kdl+L_ad)];4

5 LQ_mat = [−(L_al + L_aq), L_aq;6 −L_aq, (L_kql+L_aq)];

Page 5 of 26

CHAPTER 1. DETERMINING PER UNIT VALUES OF THE INVOLVED PARAMETERS

Figure 1.1: Mutual Inductance: Φm1 = The mutual Inductance of winding 1 acting by itself, Φm2

= The mutual Inductance of winding 2 acting by itself,Φl1 = Leakage flux that is contributed tocoil 1,Φl2 = Leakage flux that is contributed to coil 2,Φm = The combined mutual inductanceof Φm1 and Φm2

Page 6 of 26

Chapter 2

Writing the dynamic model of thesynchronous machine

The next part of this assignment was to construct a model for simulating the synchronousmachine. This model will be constrained by voltages, the known values, where the other valuesare unknowns. The model was made up of several parameters. The parameters were loaded intoa function called Load_parameters. First the current value (id,ifd,ikd,iq,ikq) will need to besolved for using the matrices, LD_mat and LQ_mat in section 1.1. Then the state equations canbe rewritten as:

dΨd

dt= ω0(ωpuΨq +Raid + ed)

dΨq

dt= ω0(−ωpuΨd +Raiq + eq)

dΨfd

dt= ω0(−Rfdifd + efd)

dΨkd

dt= ω0(−Rkdikd)

dΨkq

dt= ω0(−Rkqikq)

In order to do this in MatLab the built in function ode23 was used. This function needs aformatted input where the right side of the differential equations evaluated in the state equationssee the equations above. This was done though using the following function dPsi_dt see section2.1.

2.1 Q.2.1In this question the goal was to write function that can be used in the built in MatLab functionode23. This function was call dPsi_dt and returns a vector x′ where the input was vector xand time t. The code for this function is in Appendix ??.

7

Chapter 3

Steady State Analysis of theSynchronous Machine

In this chapter the steady state analysis was performed for providing the initial condition forthe simulation. In order to do this a function had to be written that would calculate the initialstates of x0 for a given solution.This function was given a set of values and the values were checkwith the given solution to see if they matched. The value did not match at first but then thecode was looked over and some errors found. Once the errors were fix the solution matched.This function initialize_state_vector() calculated the following parameters and equation:

It: Magnitude of the terminal current in per unit 3.1 id: d-axis component of the terminal current3.6φ: Power angle of the load 3.2 iq: q-axis component of the terminal current3.7δ: The machine rotor angle 3.3 ifd: The field winding current 3.10ed: d-axis component of the terminal voltage 3.4 efd: The field winding voltage 3.11eq: q-axis component of the terminal voltage 3.5 x: The state vectorΨd: Stator circuit flux linkages 3.8 x0: The initial state vectorΨfd: Rotor circuit flux linkages 3.12 Ψq: Stator circuit flux linkages 3.9Ψkq: Rotor circuit flux linkages ?? Ψkd: Rotor circuit flux 3.13

Table 3.1: Parameter

3.1 Q.3.1Question 3.1 was to write the function that would take in the input of terminal per unit voltageEt and per unit active and reactive power flowing from the machines terminal (Pt and Qt) andoutput in terms of efd, ed, and eq. These output values correspond to the global variable andoutput the vector x0. The vector x0 contains the intimal values of the state vector x. Used inthe function dPsi_dt discussed in section 2.1 above.

3.2 Q.3.2Used the function in section 3.1]to calculate the initial conditions of the synchronous machinewhere using the following values: Et=1.0pu, Pt=0.5pu, and Qt=0.3pu. The results of thisfunction included the parameters in section 3.1. This results were found by using the followingequations in the function initialize_state_vector(). These equations are:

8

CHAPTER 3. STEADY STATE ANALYSIS OF THE SYNCHRONOUS MACHINE

It =

√P 2t +Q2

t

Et(3.1)

Φ = acos(Pt

It ∗ Et) (3.2)

δi = atan((Lq ∗ It ∗ cos(φ) −Ra ∗ It ∗ sin(φ)

Et +Ra ∗ It ∗ cos(φ) + Lq ∗ It ∗ sin(φ)) (3.3)

ed = Et ∗ sin(δ) (3.4)

eq = Et ∗ cos(δ) (3.5)

id = It ∗ sin(δ + φ) (3.6)

iq = It ∗ cos(δ + φ) (3.7)

Ψd = eq +Ra ∗ iq (3.8)

Ψq = −ed −Ra ∗ id (3.9)

ifd =eq +Ra ∗ iq + (Lad + Lal) ∗ id

Lad(3.10)

efd = Rfd ∗ ifd (3.11)

Ψfd = (Lfdl + Lad) ∗ ifd− Lad ∗ id (3.12)

Ψkd = Lad ∗ (ifd − id) (3.13)

Ψkq = −Laq ∗ iq (3.14)

All these equations were used in the function initialize_state_vector() and calculated thefollowing results:

--------------------------------------Initial values--------------------------------------It = 6.009e-01 puphi = 5.880e-01 rad, 33.690 degdelta = 5.156e-01 rad, 29.542 dege_d = 5.917e-01 pue_q = 1.044e+00 puid = 5.365e-01 puiq = 2.706e-01 puPsi_d = 1.046e+00 puPsi_q = -5.954e-01 pui_fd = 1.060e+00 pue_fd = 1.696e-03 puPsi_fd = 1.545e+00 puPsi_kd = 1.100e+00 puPsi_kq = -5.684e-01 pux0 =1.0459-0.59541.54481.0995-0.5684

Note: x0 is the initial state and was calculated using the following equations: 3.14, 3.13,3.8,3.9, and 3.12.

Page 9 of 26

CHAPTER 3. STEADY STATE ANALYSIS OF THE SYNCHRONOUS MACHINE

3.3 Q.3.3The figure 3.3 shows how the terminal voltage Et and current It are related to the internal voltageEq. The phasor diagram shows the rotor angle, and the angle between the terminal voltage andthe current for the data given in section 3.2. Interpreting the phasor diagram in figure 3.3. Itcan be seen that terminal It lags the terminal voltage by 30.7°. It can also be seen that the rotorangle δi or load angle is 29.5°, which means the magnetic field of the internal voltage Eq (therotor’s magnetic field) leading the terminal voltage Et (the stator’s magnetic field) by an angleof 29.5°.

Page 10 of 26

CHAPTER 3. STEADY STATE ANALYSIS OF THE SYNCHRONOUS MACHINE

Figure 3.1: Phasor Diagram

Page 11 of 26

Chapter 4

Electrical Transient Analysis theMachine

In this chapter, will look at a function that was constructed to calculate the initial conditions for adynamic system with another function mathematically models the mechanics of the system. Forthese simulations the machine was considered to be initially unloaded, operating at a terminalvoltage Et = 1.0pu. This was done using a part from section 3.1 for determining the initial statevector x0. Also the assumptions that were made is ed = eq = 0.

4.1 Q.4.1Ran dynamic simulation and plotted the following:1. The current in the field winding ifd see figure 4.1.2. The current in the three stator phases ia, ib, and ic see figure 4.1

In figure 4.1 is the field winding current in the rotor over time during a three-phase short circuit.This means that the assumption was ed = eq = 0. From looking at figure ?? it can be seenthat current jumps from roughly 1.2pu to about 3.4pu. This is what is expected when a short-circuit happens because there is no more resistance in the circuit. This means the voltage goesto equilibrium state and once the equilibrium is obtained the current will level out. This plotdemonstrates that the current start to level off at 0.5ms. The oscillation in the graph originatesfrom the A.C. signal at 50Hz but is transformed by electromotive force inducting more currentinto the lines when magnetic flux is collapsing. This graph can be broken down into three partSubtransient Period, Transient Period, and Steady State Period. This is shown in figure 4.1. Inthe Subtransient Period the current spike only lasts for the first few cycles caused by the currentsinduced in the damper windings Ψkd and Ψkq. After that the current enters the Transient Periodwhere the amplitude decays vary slowly due to the current that is induced in the field windings.The final stage of the current is the Steady State Period this is were the current remains constant.

This graph was generated by using an inverse of the matrices in section 1.2. These matriceswere then multiplied x, which give the results of id, iq, and ifd. The result of ifd was thenplotted versus time at 0 to 0.5 interval resulting in figure 4.1.

12

CHAPTER 4. ELECTRICAL TRANSIENT ANALYSIS THE MACHINE

Figure 4.1: Plot of the field winding current in the rotor

Page 13 of 26

CHAPTER 4. ELECTRICAL TRANSIENT ANALYSIS THE MACHINE

Figure 4.2: Plot of the field winding current in the rotor broken into three part

In the plot in figure 4.1 is of armature winding currents in the stator broken into the threedifferent phases called: ia, ib, and ic. All three of the phases have different currents depending ontheir position at the point of short circuit. This is do to the fact the all three phases are at somedifferent phase angle (usually separated by 120°). Resulting in a different magnitude in the d.c.component of current. This is why the three phases have three different start place on the graph.Then once the d.c component decays the current of each phase become closer in magnitude asseen by the graph. Order to get this graph the dp0 Transformation was preformed this meansthe equations 4.3 and 4.4 were used for solving the three phase currents. These were first derivedby using the equations for modeling the rotor circuit flux linkages and they are written as:

ψkd = Lfkdifd + Lkkdikd − Lakd[iacos(θ) + ib(θ −2π

3) + iccos(θ +

2π

3)] (4.1)

ψkq = Lkkqikq + Lakq[iasin(θ) + ibsin(θ − 2π

3) + icsin(θ +

2π

3)] (4.2)

After the transformation of the stator phase current equations 4.1 and 4.2 from the rotordomain to the stator domain become the following equations:

id = kd[iacos(θ) + ibcos(θ −2π

3) + iccos(θ +

2π

3)] (4.3)

iq = −kq[iasin(θ) + ibsin(θ − 2π

3) + icsin(θ +

2π

3)] (4.4)

Page 14 of 26

CHAPTER 4. ELECTRICAL TRANSIENT ANALYSIS THE MACHINE

Then setting the value kd and kq equal to 23 make a balanced conditions and the peak values

of id and iq will become matched with the peak value of the stator currents.To enable switch between the rotor domain and stator domain there will need to be three

equations because there are three unknowns. The third equation is found by using the zerosequence current i0. This works because the currents id and iq together make a field that isequal the original field set by the stator currents. This third equation can be written as:

i0 =1

3(ia + ib + ic) (4.5)

Now rewriting the three equations into a matrix form and taking the equations 4.3, 4.4, and4.5 over the dq0 domain the follow matrix is derived:idiq

i0

=2

3

cos(θ) cos(θ − 2π3 ) cos(θ + 2π

3 )−sin(θ) −sin(θ − 2π

3 ) −sin(θ + 2π3 )

12

12

12

iaibic

Then solving the for the value ia, ib, and ic an inverse transformation was performed result

the following matrix: iaibic

=

cos(θ) −sin(θ) 1cos(θ − 2π

3 ) −sin(θ − 2π3 ) 1

cos(θ + 2π3 ) −sin(θ + 2π

3 ) 1

idiqi0

This inverse transformation matrix was then used in the code in the function Plot_currents seesection A.4. This function takes the values of the θ iterated through a for loop with a length(t),t is time. The result is the values of ia, ib, and ic throughout time. Then these values are plottedover some time interval see figure 4.1.

Figure 4.3: Armature winding currents in the stator

Page 15 of 26

Chapter 5

Calculate the Short Circuit CurrentBy the Classical Method

In this chapter the task was to calculated the current by a classical method obtained the currentby using Kirchhoff’s circuit laws and obtaining the following equation:

ia = −(sin(αsc)

x”de−

tTa − (

1

xd+ (

1

x′d

− 1

xd)e− t

T′d + (

1

x”d− 1

x′d

)e− t

T”d )sin(ωt+ αsc)) (5.1)

αsc is the angle between voltage phase a and the positive real axisωpu=1xd = Ld = Lad+Ll d-axes Synchronousx

′

d = 0.43pu Transient Reactancex”d=0.22pu Subtransient ReactanceT

′

d=0.9773T ”d=0.1023

5.1 Q.5.1In this section the short circuit current in phase a was plotted using the classical method alongwith the method used in section 4.1. The plot is in figure 5.1 that shows the two current plottedthey exactly equal in order show them on the same plot ia was plotted with dashed line. Theseplots show that the currents have a second harmonic component in the phase a of the statorcurrent. This second harmonic component was because the stator poles produce a stationary fieldinducing a standard frequency, which in turn produces a electromotive force in the rotor. Thisresults in rotor current fluctuating, which produces an a.c. component that causes fundamentalfrequency flux then causing a rotation producing a current in the stator winding this is the secondharmonic current [1].The graph was generated by using the equation 5.1 this equation put into a for loop. The forloop was set to the number of iterations by the length of θ. θ calculated by the following equationθ = ωt+ θ0. Where ω = 2πf , t is the interval time, θ0 is the initial θ which was set to 0, and fis the frequency which was set to 50Hz. The plot was then plotted with iasc versus t.

16

CHAPTER 5. CALCULATE THE SHORT CIRCUIT CURRENT BY THE CLASSICALMETHOD

Figure 5.1: The calculate short circuit current in phase a by using the classical method

5.2 Q.5.2

Figure 5.2: A plot of the difference between the current classical method calculated and theMATLAB model

Page 17 of 26

CHAPTER 5. CALCULATE THE SHORT CIRCUIT CURRENT BY THE CLASSICALMETHOD

Figure 5.3: A plot of the difference between the current classical method calculated and theMATLAB model

The graph in figure 5.2 was to show the difference between using the classical method and themethod in the MATLAB model. Looking at the graph it can be seen that difference between thetow current very more at first 0.5ms but then after 6.0ms current variance is minimal. This isshown in figure 5.2. The graph was produce by taking the different of iasc and ia. Then plottingthe differences versus time.

Page 18 of 26

Chapter 6

Conclusion

This Assignment included the following learning objectives:

• Describe the concepts of self, leakage, and mutual inductance.

• Apply circuit theory to analyze synchronous machines in steady state.

• Understand the theory, motivation, and how to implement the dq0-reference frame trans-formation.

• Apply the dynamical model of the synchronous machines in order to simulate the machine’sresponse to a disturbance.

• Describe the characteristics of the short circuit response in synchronous machines.

The concepts of self, leakage, and mutual inductance was expanded in section 1.2. The circuittheory that was applied for synchronous machines in steady state was done by first breaking therotor and stator into to different circuits. For the analysis of rotor, started with treating the rotorcircuit as different circuit one where the flux in in line with the d-axis and the other where fluxis line with the q-axis. The stator circuit was made up of three inductors (one for each phase)connected in the y-configuration. Order to do this analysis the magnetic mutual coupling betweenthe components needed to be considered. For the theory, motivation, and how to implement thedp0-reference frame transformation. The implementation dq0-reference frame transformation wascovered in the programming of MATLAB code and analysis of dq0-reference frame transformationcovered in chapter 4. In applying the dynamical model the synchronous machines was coveredin chapter 4 showing how the system response. The description of characteristics of the shortcircuit response in synchronous machines was discussed in chapter 5.

19

Appendix A

MatLab functions

A.1 Q1.1

1 %% T−866−MODE2 % Assignment 2 − Task 13 % Einar Falur Zoega Sigurðsson4 %%5 % Script to load the machine parameters6 global LD_mat LQ_mat R_fd R_a R_kd R_kq L_ad L_fkd L_akd L_afd L_aq L_al...7 L_fdl L_kdl L_kql e_fd e_d e_q omega T_a;8 %%9 S = 1100E6; %Rated size of the machine [VA]

10 V = 26E3; %Rated line−line voltage [V]11 f = 50; %Machine frequnece [Hz]12 T_a = 0.1; %Armatur time constant [s]13 %%14 [L_ad, L_afd, L_akd, L_fkd] = deal(2.1);15 L_aq = 2.1;16 L_al = 0.1;17 L_fdl = 0.42; % = L_ffd in the formula18 L_kdl = 0.1826; % = L_kkd in the formula19 L_kql = 0.1281; % = L_kkq in the formula20 %% The per unit d−axis inductance matrix21 LD_mat = [ −(L_al + L_ad), L_ad, L_ad;22 −L_ad, (L_fdl+L_ad), L_fkd;23 −L_ad, L_fkd, (L_kdl+L_ad)];24 %% The per unit q−axis inductance matrix25 LQ_mat = [ −(L_al + L_aq), L_aq;26 −L_aq, (L_kql+L_aq) ];27 %%28 %The per unit resistance in the field winding, stator winding and in the29 %d−axis and q−axis damper windings respectively.30 R_a = 0.0070;31 R_fd = 0.0016;32 R_kd = 0.0085;33 R_kq = 0.0215;34 %%35 % omega_0 defined36 omega=2*pi*50;

20

APPENDIX A. MATLAB FUNCTIONS

A.2 Q2.1

1 %% T−866−MODE2 % Assignment 2 − Task 23 %Einar Falur Zoega Sigurðsson4 function dx = dPsi_dt(t,x)5 %%6 global LD_mat LQ_mat R_fd R_a R_kd R_kq e_fd e_d e_q omega7 x_D = [x(1);x(3);x(4)]; %x_D = [Psi_d; Psi_fd; Psi_kd;];8 x_Q = [x(2);x(5)]; %x_Q = [Psi_q; Psi_kq;];9

10 i_D = (LD_mat)^(−1)*x_D; %i_D= [id; ifd; ikd;]11 i_Q = (LQ_mat)^(−1)*x_Q; %i_Q= [iq; ikq;]12

13 i_kd = i_D(3);14 i_kq = i_Q(2);15

16 dx = zeros(5,1); %Output for ode23 determined17 %Calculating the deriative of Psi in terms of t18 dx(1) = omega * (x(2) + R_a*i_D(1) + e_d);19 dx(2) = omega * ( −x(1) + R_a*i_Q(1) + e_q);20 dx(3) = omega * (−(R_fd) * i_D(2) + e_fd);21 dx(4) = omega * (−R_kd * i_D(3));22 dx(5) = omega * (−R_kq * i_Q(2));23

24 end

Page 21 of 26

APPENDIX A. MATLAB FUNCTIONS

A.3 Q3.1

1 %% T−866−MODE2 % Assignment 2 − Task 33 % Einar Falur Zoega Sigurðsson4 %%5 % Function to initiate the state voector6

7 function [x0,E_qC,E_tC,I_tC,E_iC] = initialize_state_vector(Et, Pt, Qt)8 %%9 global R_fd R_a L_aq L_al L_fdl L_ad e_fd e_d e_q i_d i_q i_fd phi ∆;

10 L_q = L_aq+L_al;11 It = sqrt(Pt^2+Qt^2)/Et;12 phi = acos(Pt/(It*Et));13 ∆ = atan((L_q*It*cos(phi)−R_a*It*sin(phi))/(Et+R_a*It*cos(phi)...14 +L_q*It*sin(phi)));15 e_d = Et * sin(∆);16 e_q = Et * cos(∆);17 i_d = It * sin(∆+phi);18 i_q = It * cos(∆+phi);19 i_fd = (e_q+R_a*i_q+(L_ad+L_al)*i_d)/L_ad;20 e_fd = R_fd*i_fd;21 %%22 x0 = zeros(5,1); % defining 5X1 collumn vector x0 ]23 x0(1,1) = e_q + R_a*i_q; % Psi_d24 x0(2,1) = −e_d − R_a *i_d; % Psi_q25 x0(3,1) = (L_fdl+L_ad)*i_fd − L_ad*i_d;% Psi_fd26 x0(4,1) = L_ad*(i_fd − i_d); % Psi_kd27 x0(5,1) = −L_aq * i_q; % Psi_kq28 %%29 disp('−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−');30 disp(' Initial values ');31 disp('−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−');32 disp(sprintf('It = %.3e pu',It));33 disp(sprintf('phi = %.3e rad, %.3f deg',phi,phi/pi*180));34 disp(sprintf('∆ = %.3e rad, %.3f deg',∆,∆/pi*180));35 disp(sprintf('e_d = %.3e pu',e_d));36 disp(sprintf('e_q = %.3e pu',e_q));37 disp(sprintf('id = %.3e pu',i_d));38 disp(sprintf('iq = %.3e pu',i_q));39 disp(sprintf('Psi_d = %.3e pu',x0(1)));40 disp(sprintf('Psi_q = %.3e pu',x0(2)));41 disp(sprintf('i_fd = %.3e pu',i_fd));42 disp(sprintf('e_fd = %.3e pu',e_fd));43 disp(sprintf('Psi_fd = %.3e pu',x0(3)));44 disp(sprintf('Psi_kd = %.3e pu',x0(4)));45 disp(sprintf('Psi_kq = %.3e pu',x0(5)));46 %%47 E_qC = 1i*(L_ad*i_fd);48 E_tC = Et*cos((pi/2)−∆)+1i*Et*sin((pi/2)−∆);49 I_tC = It*cos((pi/2)−∆−phi)+It*1i*sin((pi/2)−∆−phi);50 E_iC = E_tC+I_tC*R_a;

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APPENDIX A. MATLAB FUNCTIONS

A.4 Q4.1

1 % T−866−MODE2 % Assignment 2 − Task 43 % Einar Falur Zoega Sigurðsson4 %%5 function Plot_currents(t,x)6 global omega i_d i_q LD_mat LQ_mat i_fd i_0 i_a7 theta=omega*t;8 %%9 for i=1:length(theta)

10 i_D=(LD_mat)^(−1)*[x(i,1); x(i,3); x(i,4)];11 i_Q=(LQ_mat)^(−1)*[x(i,2);x(i,5)];12 i_d(i)=i_D(1,1);13 i_q(i)=i_Q(1,1);14 i_fd(i)=i_D(2,1);15 end16 i_0=0; %i_0 = zero when SC.17 %% Calculating the dq0 inverse to get i_a, i_b and i_c.18 for k=1:length(t)19 mat_park=[cos( theta(k) ), −sin(theta(k)), 1;20 cos(theta(k)−(2*pi/3)), −sin(theta(k)−(2*pi/3)), 1;21 cos(theta(k)+(2*pi/3)), −sin(theta(k)+(2*pi/3)), 1];22 Is=mat_park*[i_d(k);i_q(k); i_0];23 i_a(k)=Is(1);24 i_b(k)=Is(2);25 i_c(k)=Is(3);26 end27 %% Plotting the field winding current in the rotor28 figure(2)29 plot(t,i_fd,'b−')30 legend('i_f_d')31 title('Field winding current','FontSize',14)32 xlabel('Time [ms]','FontSize',12)33 ylabel('Current [pu]','FontSize',12)34 %% Armature winding currents in the stator35 figure(3)36 hold on37 plot(t,i_a,'r−')38 plot(t,i_b,'g−')39 plot(t,i_c,'b−')40 legend('i_a','i_b','i_c')41 title('Stator currents i_a, i_b and i_c','FontSize',14)42 xlabel('Time [ms]','FontSize',12)43 ylabel('Current [pu]','FontSize',12)44 end

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APPENDIX A. MATLAB FUNCTIONS

A.5 Q5.1

1 %%2 function Plot_currents2(t)3 %%4 global L_ad L_al theta_0 ∆ theta T_a omega i_asc i_a5 theta_0=06 theta = omega*t + theta_0;7 alfa_sc = (pi/2) − ∆ + theta_0;8 X_d = L_ad + L_al;9 Xp_d = 0.43;

10 Xpp_d = 0.22;11 %%12 Tp_d = 0.9773;13 Tpp_d = 0.1023;14 %%15 i_asc = zeros(1,length(theta));16 for k=1:length(theta)% The short circuit currents can be calculated17 % by a classical method, where18 %the current can be calculated19 i_asc(k) = −((sin(alfa_sc)/Xpp_d)*exp(−t(k)/T_a)...20 −(1/X_d+(1/Xp_d−1/X_d)*exp(−t(k)/Tp_d)+(1/Xpp_d−1/Xp_d)...21 *exp(−t(k)/Tpp_d))*sin(theta(k)+alfa_sc));22 end23 figure(5)24 hold on25 plot(t,i_asc,t,i_a,'r−')26 title('The Plot of Phase "a" With Short Circuit Classical Method and MATLAB ...

Method','FontSize',14)27 xlabel('Time [ms]','FontSize',12)28 ylabel('Current [pu]','FontSize',12)29 legend({'i_asc','i_a'},'Location','best')

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APPENDIX A. MATLAB FUNCTIONS

A.6 Q5.2

1 %% main 5.22 global i_asc i_a3 dif = i_asc − i_a;4 %% figure plot5 figure(6)6 plot(t,dif)7 title('The difference between the current classical method calculated and the ...

MatLab model','FontSize',14)8 xlabel('Time [ms]','FontSize',12)9 ylabel('Current [pu]','FontSize',12)

1 % A main script to execute assignment 22 %%3 clearvars ; close all; clc;4 % Define global variables that are to be used in the simulation5 global LD_mat LQ_mat R_fd R_a R_kd R_kq e_fd e_d e_q omega6 Load_parameters; % script file that loads the machine parameters and assigns7 % the relevant machine parameters to the global parameters8 Et = 1.0; Pt =1E−15; Qt =0;9 [x0,E_qC,E_tC,I_tC,E_iC] = initialize_state_vector(Et,Pt,Qt); % A function that ...

calculates the steady10 % state pre−fault conditions11 e_d =0; e_q = 0; %Apply a three phase short circuit at the terminal and..12 [t,x] = ode23(@dPsi_dt,0:0.001:7,x0);%Perform the simulation by using ode23()13 Plot_currents(t,x) %plot the currents from the simulation results.14 Plot(E_qC,E_tC,I_tC,E_iC);15 Plot_currents2(t)16 %% %x017 global i_asc i_a18 dif = i_asc − i_a;19 % figure plot20 figure(6)21 plot(t,dif)22 title('The difference between the current classical method calculated and the ...

MatLab model','FontSize',14)23 xlabel('Time [ms]','FontSize',12)24 ylabel('Current [pu]','FontSize',12)

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Bibliography

[1] Mayuresh V. Bakshi Uday A. Bakshi. Electrical Machines III. Technical Publications Pune,1 edition, 2008.

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