Control and Dynamics in Power Systems and Microgrids (Solutions, Instructor Solution Manual) [1 ed.] 1138034991, 9781138034990

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Solution Manual for Control and Dynamics in Power Systems and Microgrids Lingling Fan January 30, 2017

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Chapter 2 Dynamic Simulation

1. A voltage source is serving an RLC series connected circuit. Let R = 0.01Ω, L = 0.01 H, C = 0.001 F. The compensation degree of the system is Xc /XL , approximately 70.36%. Find its current response for a step response of the voltage source, and a sinusoidal 60 Hz input (amplitude 1 V)of the voltage source. Use Laplace transformation to find the current in Laplace domain and current in time domain. Solution: We will first find the transfer function from the source voltage to the current by imple1 menting the impedance model. The impedance model of the circuit is R + Ls + Cs . Therefore, the transfer function is I(s) 1 (1) = 1 . Vs (s) R + Ls + Cs Next, we consider two cases. Case 1, the source voltage is a step response and its Laplace transform is 1s . Case 2, the source voltage is sinusoidal. Case 1: Vs (s) = 1s for a step response. Therefore, we may find the current’s Laplace transform as 1 Cs 0.001 100 I(s) = = −5 2 = 2 (2) s LCs2 + RCs + 1 10 s + 10−5 s + 1 s + s + 105 A few algebraic manipulations will lead to I(s) =

100 . (s + 0.5)2 + 3162

(3)

Since the inverse Laplace transform for (s−a)b2 +b2 is eat sin(bt), we then find i(t) in the following expression. i(t) = 0.3162e−0.5t sin(316t) A (4) Case 2: vs (t) = sin(377t). Its Laplace transform is Vs (s) = Laplace transform as I(s) =

s2

377 . s2 +3772

We may find the current’s

377 0.001s 37700s = 2 2 −5 2 −5 + 377 10 s + 10 s + 1 (s + 3772 )(s2 + s + 105 )

(5)

The denominator has four roots: ±j377 and −0.5 ± j316. In order to find i(t), we need to carry out inverse Laplace transform. The technique used is Partial Fraction Expansion to split up a complicated fraction into forms that are in the Laplace transform table. For the above I(s), its fraction expansion expression is I(s) =

k3 s + k4 k1 s + k2 + s2 + 3772 s2 + s + 105

(6)

where ki s are real numbers. To find k1 and k2 , let (5) and (6) both be multiplied by (s2 +3772 ). Then we evaluated the right 2 3 s+k4 side terms of the two equations at ±j377. Note that s2k+s+10 5 (s + 377) will be 0 if it is evaluated

2

at s = ±j377. Therefore the following equations will be found. 37700s k1 s + k2 = 2 5 s + s + 10 −j377 −j377 37700s k1 s + k2 = 2 5 s + s + 10 j377 j377

(7a) (7b)

=⇒ k1 = −0.8948, k2 = 3.0187 Similarly, we can find k3 = 0.8948 and k4 = −2.1239. The inverse Laplace transform is k2 k4 −0.5t i(t) = k1 cos(377t) + sin(377t) + k3 e−0.5t sin(316t) + e cos(316t) 377 316 = −0.8948 cos(377t) + 0.008 sin(377t) + 0.8948e−0.5t sin(316t) − 0.0067e−0.5t cos(316t)

(8)

The above expression can be further arranged as follows. i(t) = 0.8948 sin(377t − 1.5618) + 0.8948e−0.5t sin(316t − 0.0075)

(9)

Remarks: We may examine the above expression and know that the steady-state response of the current is 0.8948 sin(377t − 1.5618). This time-domain function can also be obtained by phasor-based calculation. 1 The impedance at 60 Hz or 377 rad/s is Z = R + j377L − j 377C = 1.1175ej1.5618 . Therefore the current phasor magnitude should be the voltage source magnitude divided by |Z| = 1.1175 and the phase shift should be −1.5618 radian. The steady-state current should be 1 i(t) = sin(377t − 1.5618) = 0.8948 sin(377t − 1.5618). 1.1175

2. Use MATLAB linear system analysis tools to define a linear system for the above RLC circuit. Treat the voltage source as the input while the current as the output. Give a set of Bode plots of the system by varying R. Notate the plot properly. Use MATLAB function step to examine the dynamic response of the current with a step response of the voltage source. Use Matlab function lsim to examine the dynamic response of the current with a sinusoidal input. Solution: The MATLAB codes are as follows. R = [0.01, 0.05, 0.1] ; L = 0.01; C = 0.001; s = tf(’s’); for i=1: length(R) plant(i) = 1/(R(i) + L*s + 1/(C*s)); bode(plant(i)); hold on; end figure step(plant(1)); grid on; t=0:0.001:1; u = sin(377*t); y = lsim(plant(1), u, t, 0); figure plot(t,y); xlabel(’Time (s)’); grid on;

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Magnitude (dB)

Bode Diagram R = 0.01 R = 0.05 R =0.1

30 20 10 0

Phase (deg)

90 45 0 −45 −90 2.4 10

2.5

2.6

10 Frequency (rad/s)

10

(a) Bode plots for three different Rs. Step Response 0.4 0.3

Amplitude

0.2 0.1 0 −0.1 −0.2 −0.3 −0.4

0

2

4

6 Time (seconds)

8

10

12

(b) Problem 2: step output for a step input.

2 1.5 1 0.5 0 −0.5 −1 −1.5 −2

0

0.2

0.4

0.6

0.8

Time (s)

(c) Problem 2: lsim output for a sinusoidal input.

Figure 1: Problem 2.

4

1

The generated figures are shown in Figs. 1a, 1b, and 1c. 3. For the above RLC circuit, build a two-order state-space model. The state variables are the current and the voltage across the capacitor. Use MATLAB function ode to simulate the dynamic response of the current for a step input and a sinusoidal input. Solution: The state-space model has been built in (2.19). The MATLAB codes to conduct the two simulation case studies are as follows. R = 0.1 ; L = 0.01; C = 0.001; dxdt_step = @(t, x, R, L, C)[(1-R*x(1)-x(2))/L; x(1)/C] ; dxdt_sin = @(t, x, R, L, C)[(sin(377*t)-R*x(1)-x(2))/L; x(1)/C] ; [T1,y1]= ode23(@(t,x)dxdt_step(t, x, R, L, C), [0 1],[0; 0]); [T2,y2]= ode23(@(t,x)dxdt_sin(t, x, R, L, C), [0 1],[0; 0]); figure plot(T1,y1); grid on; xlabel(’Time (s)’); figure plot(T2,y2); grid on; xlabel(’Time (s)’);

2

5 capacitor voltage

capacitor voltage 1.5

1 0

current

0.5

0

−0.5

0

0.2

0.4

0.6

0.8

−5

1

0

0.2

0.4

0.6

Time (s)

Time (s)

(a)

(b)

0.8

1

Figure 2: Problem 3: (a) Dynamic response for a step input; (b) Dynamic response for a sinusoidal input.

2

Chapter 3 Frequency Control

1. Use parameters in Example 1 Fig. 3.29. For a single generator load serving system, derive the linear system model and build the model in MATLAB/Simulink. Find the droop to make ∆ω = −0.2 for ∆PL = 0.1. • Find the bandwidth of the system with only primary frequency control. • Provide the dynamic simulation of the system frequency due to a step response of load increase 0.1. • Specify ∆Pc to bring ∆ω back to zero.

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• Design the secondary frequency control to bring the system frequency back to nominal when load varies. Choose a gain of the integration block to make the frequency return to nominal in less than 100 seconds. Find the value of ∆Pc at steady-state and see if this value matches your previous calculation. Provide the dynamic responses of ∆ω and ∆Pc . • Increasing the gain of the integration block will lead to instability. Find the marginal value of the gain. Designate an analysis procedure to analytically find the marginal gain. Hint: you can use root loci method. Solution: a) The droop parameter R is not given. The electromechanical dynamics is given 1 10 as 2Hs+D = 10s+1 . This indicates that H = 0.5 and D1 = 0.1. At steady-state, we know that 1 −1 ∆ω = D1 +1/R ∆PL . Apply the condition: ∆PL = 0.1, ∆ω = −0.2. We may find D1 + 1/R = 0.5. Therefore, R = 1/(0.5 − 0.1) = 2.5. The bandwidth of the system can be found by examining the closed-loop transfer function. The closed-loop transfer function is 1 − s+0.1 ∆ω = 1 1 1 ∆PL 1 + s+0.1 R s+1 s+1 = (s + 0.1)(s + 1) + 1/R s+1 = 2 s + 1.1s + 0.5

G=

(10)

Since the steady-state gain of G is 2. The bandwidth can be √ found by examining the Bode plot of G/2 and the corresponding frequency when the gain is 1/ 2. The Bode plot of G/2 is shown in Fig. 3. Bode Diagram

Magnitude (dB)

10

System: untitled1 Frequency (rad/s): 0.812 Magnitude (dB): −3

0 −10 −20

Phase (deg)

−30 0

−45

−90 −2 10

−1

0

10 10 Frequency (rad/s)

1

10

Figure 3: Bode plot of G/2. From the Bode plot, we can find the bandwidth is 0.812 rad/s. b) and c) are examined by simulation. ∆Pc should match ∆PL to bring ∆ω back to zero. The Simulink model is shown in Fig. 4. The dynamic simulation results are shown in Fig. 5. The simulation results show that given a 0.1 increase in the load, there is 0.2 pu decrease in the 6

Figure 4: Simulink model for Problem 1.

0.1

∆ PL

0.05 ∆ Pc

0 0

5

10

15

20

0

5

10 Time (s)

15

20

0.1

∆ω

0 −0.1 −0.2 −0.3

Figure 5: Dynamic simulation results. At t = 1 s, ∆PL changes from 0 to 0.1. At t = 10 s, ∆Pc changes from 0 to 0.1. frequency. Further, the frequency can be brought back to nominal by increasing the reference power to match the load. d) and e) examine secondary frequency control design and identify the limit of the gain of the integral controller. The Simulink model is shown in Fig. 6 and the simulation results are shown in Fig. 7. It can be seen that when ki is 0.2, the secondary frequency control can bring the frequency back to nominal. ∆Pc will eventually become 0.1 to match the load change. This observation confirms our computation in a). When ki is 0.6, the system loses stability. The magnitude of the oscillation keeps increasing. We may use root locus method to show the limit of ki . The open-loop transfer

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Figure 6: Simulink model for a system with secondary frequency control. 0.2 ki=0.2

∆ Pc

0.15 0.1

0.3 0.2

∆ PL

0.1 0

0

5

10

15

−0.1

20

0.2

5

10

15

20

0

5

10 Time (s)

15

20

0.1 ∆ω

∆ω

0

0.2

0.1 0 −0.1 −0.2

∆P

L

0.05 0

ki=0.6

∆ Pc

0 −0.1

0

5

10 Time (s)

15

−0.2

20

Figure 7: Dynamic simulation results for a system with secondary frequency control. function is first obtained by decoupling the secondary frequency control loop and it is: 1

1

1 1 ∆ω s+1 s+0.1 = G = 1 1 1 s ∆Pc s 1 + s+1 s+0.1 R 1 = s(s2 + 1.1s + 0.5) 0

(11)

The root loci for G0 are shown in Fig. 8. It can be seen that the gain cannot exceed 0.559. 2. Use the parameters in Example 3. In Matlab/Simulink, build the linearized model for a two-area interconnection system. Each area consists of a generator with a load. A load change of 100 MW occurs in Area 1. • Demonstrate the ACE steady-state values, steady-state frequency in Hz, change in tie-line flow in MW for each area through time-domain simulation for the system with primary frequency control only. Observe if the values match the calculation. • Design secondary frequency control for each area and demonstrate that the frequency and tie-line power flow will go back to nominal. Please present simulation results. Solution: This problem is for students to repeat the example presented in Section 3.5.3. 8

Root Locus 2

−1

Imaginary Axis (seconds )

1.5 1 0.5

0.559

0 −0.5 −1 −1.5 −2 −2.5

−2

−1.5

−1

−0.5

0

0.5

1

−1

Real Axis (seconds )

Figure 8: Root loci for G0 .

3

Chapter 4 Synchronous Generator Models

1. A round-rotor generator (Xs = 1.0, r = 0.1) is synchronized to a bus whose voltage is 1∠00 . At synchronization iF = 1000A (actual). The generator is then adjusted until SG = 0.8 + j0.6. (SG is the power supplied to the generator bus.) a. Find iF and the generator efficiency (assuming no generator loss except I 2 r). b. With the same iF , what is the maximum active power the generator can deliver. Solution: Ea is proportional to the rotor excitation current iF . At the synchronization condition, the generator is sending out 0 power and the stator current is also 0. Therefore, Ea = Va , the internal voltage is equal to the terminal voltage. So we find that when Ea is 1, iF is 1000 A. We can then look at the second operation condition and use phasor diagram or circuit analysis to find Ea . Then from Ea , we find iF . For the second operating condition, the generator is sending out complex power SG = 0.8 + j0.6 while the terminal bus voltage is 1∠00 . Thus the stator current I a can be found as  ∗ SG Ia = = 0.8 − j0.6 = 1∠−36.870 . Va Based on the Thevenin equivalent circuit, we can find E a as E a = V a + (r + jXs )I a . Thus E a = 1.0 + (0.1 + j1.0)(0.8 − j0.6) = 1.836∠23.770 . a) When Ea = 1, iF is 1000 A. Therefore, when Ea = 1.836, iF will be 1836 A. The generator efficiency can be found by computing the real power loss on the stator. The 0.8 stator current magnitude is 1. Then the loss is I 2 r = 0.1. The efficiency is 0.8+0.1 = 88.9%.

9

b) If iF does not change, then Ea will be kept at 1.86. The angle δ may change to have a different output power. The expression of the complex power to the grid is as follows.  ∗ Ea − V a ∗ SG = V a I a = V a r + jXs (12) Va Ea −j(δ−θz ) Va2 jθz e − e = |Z| |Z| p r where |Z| = r2 + Xs2 = 1.005 and θz = cos−1 |Z| = 84.290 . Take the real part of SG , we have Pg =

Va Ea V2 cos(δ − θz ) − a cos θz . |Z| |Z|

When δ = θz , PG will be maximized and PGmax =

Va Ea V2 1.0 × 1.836 1.02 − a cos θz = − cos 84.290 = 1.7279. |Z| |Z| 1.005 1.005

2. Consider a salient-pole generator delivering power through a short transmission line to an infinite bus. V ∞ = 1∠00 , Ea = 1.4. The active power delivered to the infinite bus is 0.6. We are given the generator reactances Xd = 1.6 and Xq = 1.0 and the line reactance XL = 0.4. Neglect resistances and find E a and I a . Solution: A most straightforward approach for this problem is to have the active power expression in terms of δ and Ea ready. We know that ! 2 Ea V∞ 1 1 V∞ P = sin δ + − sin(2δ). ed eq ed 2 X X X ed = Xd + XL , X eq = Xq + XL . where X Therefore, we can find δ from that equation.   2 1.4 × 1 1 1 V∞ 0.6 = sin δ + − sin(2δ) 2.0 1.4 2.0 2 0.6 = 0.7 sin δ + 0.1071 sin(2δ)

(13)

We may use MATLAB’s function f zero to find roots for a nonlinear equation 0.7 sin(x) + 0.1071 sin(2x) − 0.6. An initial guess should be given as the second argument of this function. Here we may select 0 as the initial guess. fzero(@(x) 0.7*sin(x)+0.1071*sin(2*x)-0.6, 0)

The angle is δ = 0.7812rad = 44.760 . Based on the phasor diagram, we may then find the dq-axis currents. ( ed Iad Ea = V∞ cos δ − X eq Iaq 0 = −V∞ sin δ + X ( Iad = cos 0.7812−1.4 = −0.3450 2 =⇒ sin 0.7812 Iaq = = 0.5030 1.4 10

(14)

(15)

Thus I a = (Iaq + jIad )ejδ = (0.5030 − j0.3450)ej0.7812 = 0.6999∠10.310 . The final answer is: E a = 1.4∠44.760 I a = 0.61∠10.310

3. For a set of balanced three-phase currents (with amplitude Im , frequency ω, initial phase a angle 0), that is,   ia = Im cos(ωt) (16) ib = Im cos(ωt − 2π 3 )   2π ic = Im cos(ωt + 3 ) a) Find its space vectors at t = 0, ωt =

2π 3 ,

and ωt =

4π 3 .

If these are the stator currents for a generator, and assume the number of windings for each phase is N , the air gap permeability is µ0 , the length of the generator shaft is l, the radius from the center of the rotor to the airgap is r and the rotor is a round rotor (or the air gap has a uniform gap distance g, g  r). b) Find the total MMF generated by the currents at a random place (α angle from the reference axis) in the airgap and a random time (F (α, t)). c) Find the magnetic flux density at angle α in the airgap B(α, t). d) Find the total flux linkage linked to stator winding bb0 (λbb0 (t)), and its corresponding EMF ebb0 (t). Solution: a) For a balanced three-phase set, the space vector is the analytic form of ia : → − i (t) = Im ejωt . → − Therefore, at t = 0, i (0) = Im . → − j 2π 3 . At ωt = 2π 3 , i (0) = Im e − j 4π 4π → At ωt = 3 , i (0) = Im e 3 . b) F (α, t) = N (ia cos(α) + ib cos(α −

2π 2π ) + ic cos(α + ) 3 3

3 = N Im cos(α − ωt) 2

(17)

c) B(α, t) =

4 µ0 3 N Im cos(α − ωt) π 2g 2

11

(18)

d) Z λbb0 = N

2π + π2 3

rlB(α, t)dα

2π − π2 3

2π 4 µ0 3 = 2rlN N Im cos(ωt − ) π 2g 2 3 | {z }

(19)

ˆ λ

ebb0

dλbb0 ˆ sin(ωt − 2π ) = ωλ =− dt 3

4. For a salient rotor, the airgap distances are gd and gq for the d-axis and the q-axis. Decompose the above current space vector into d-axis and q-axis, assuming the rotor axis position is θ. Find → − → − → − a) i , id and iq . − → − → b) MMF: Fd , Fq − → − → c) Flux density: Bd , Bq d) λd (t) and λq (t). Solution: a) → − i = Im ejωt → − id = Im cos(θ − ωt)ejθ

(20)

π → − iq = Im sin(θ − ωt)ejθ− 2

b) MMF: − → 3 Fd = N Im cos(θ − ωt)ejθ 2 π − → 3 Fq = N Im sin(θ − ωt)ejθ− 2 2

(21)

c) Flux density: − → 4 µ0 3 Bd = N Im cos(θ − ωt) ejθ π 2gd 2 | {z } Bd

π − → 4 µ0 3 Bq = N Im sin(θ − ωt) ejθ− 2 π 2gq 2 | {z }

(22)

Bq

→ − → − d) The flux linkages in stator circuit aa0 due to B d and B q are:

λaq

λad = 2rlN Bd cos(θ) π = 2rlN Bd cos(θ − ) 2 12

(23)

5. A synchronous generator in the steady-state is delivering power to an infinite bus. θ = ω0 t+ π2 +δ, 1 0 = 1 sec. δ = π4 , λq = λd = √3ω , iq = √13 , id = − √13 , r = 0, Xd = ω0 Ld = 1, ω0 kMF = 1, Td0 0 q k = 32 . a) Find the torque Te . b) Find va (t), ia (t), and iF . c) At t = 0, the generator is suddenly disconnected from the infinite bus. Assume that vF =constant, iD = iQ = 0 (ignore damping circuits on the rotor). Sketch iF (t). Hint: use the fact that flux cannot jump to find iF (0+ ). Solution: a) Given the dq-axis stator flux linkages and stator currents, we should be able to find the electromagnetic torque Te using the following equation: Te =

3 (λd iq − λq id ) . 2

Therefore, we can find 3 1 Te = 2 ω0



1 1 1 1 √ √ +√ √ 3 3 3 3

 =

1 . ω0

b) The dq-axis stator voltages, currents and flux linkages relationship is as follows. dλd dt dλq . vq = −riq + ωλd − dt

vd = −rid − ωλq −

Since the operating condition is steady-state, ω = ω0 and Therefore, we have

dλd dt

=

dλq dt

= 0. In addition, r = 0.

1 1 = −√ 3ω0 3 1 1 =√ vq = +ω0 √ 3ω0 3

vd = −ω0 √

The space vector of the stator current can then be found as   π 1 1 1 → − jθ v = (vq + jvd )e = √ − j √ ej(ω0 t+ 2 +δ) = ejω0 t . k 3 3 Taking the real part of the space vector, we find va (t): va (t) =

1 cos(ω0 t). k

Using the space vector concept, we can also find ia (t). First, √ 2 → − j(ω0 t+ π2 +δ) i = (iq + jid )e = √ ej(ω0 t+2π) . 3 13

Then, 1 cos(ω0 t). k Finally, iF can be found from the d-axis flux linkage λd . Note that ia (t) =

λd = MF iF + Ld id . Therefore, kω0 λd = kω0 MF iF + kω0 Ld id . √1 , 2

Using the given parameter kω0 MF = 1, kω0 λd = iF =

√

kid = − √12 , we have

2.

c) Subproblem C deals with dynamics. The dynamic event is a sudden disconnection of the generator from the grid. After disconnection, the stator currents are 0. From this condition, we can see that the rotor circuit’s flux linkage λF = LF iF + 32 MF id will be contributed only by iF . We may write the system dynamics for the operation condition when the generator is disconnected from the grid as follows. diF dλF = rF iF + LF . dt dt √ = iF (0− )rF = 2rF . 0− indicates the moment just before the

vF = rF iF +

vF is constant. Therefore, vF disconnection event. The above equation becomes the following after both sides are divided by rF . √ The steady-state value of iF is iF (t) becomes:

√

2 = iF +

LF diF . rF dt

0 = 2. Note that Td0

iF (t) =

√

LF rF

2 + (iF (0+ ) −

= 1 second. Therefore, the expression of √

2)e−t .

0+ indicates the moment just when the disconnection happens. We will use the fact that flux linkages cannot change suddenly to find iF (0+ ). λd (0+ ) = λd (0− ) = √

1 . 3ω0

In addition, λd (0+ ) = MF iF (0+ ) when the stator current is 0. Therefore, 1 iF (0+ ) = √ . 2 Therefore, iF (t) =

√

√ 2+

2 −t e . 2

6. Prove that a synchronous generator can be represented by Thevenin equivalents in dq-axis. In the q-axis, the voltage source’s RMS value is Ea0 . In the d-axis, the equivalent inductance is 2 F) L0d = Ld − (kM if rF is ignored. λF is a constant. LF 14

Solution: We will apply Thevenin equivalent concept to tackle this problem. For open-circuit, the terminal voltage is the Thevenin equivalent voltage. First, assume that the stator circuits are open. Therefore, dq-axis currents are 0. The related flux linkages are: λF = LF iF λd = MF iF

(24)

λq = 0 Therefore, we have the terminal voltages: dλd MF dλF =− =0 dt LF dt ω0 MF λF vq = ω0 λd = ω0 MF iF = LF

vd = −

(25)

Therefore, the Thevenin equivalent voltage is oriented at the q-axis and the RMS value is ω0 MF Ea0 = √ λF . 2LF Next, we consider when the stator current is not zero. We will express λd by id and λF . λF − k 2 MF id + Ld id λd = MF iF + Ld id = MF LF   (kMF )2 MF = Ld − id + λF LF LF {z } |

(26)

L0d

In the d-axis circuit, vd = −ω0 λq + 0 in the d-axis is Ld .

dλd dt

= −ω0 λq + L0d didtd . Therefore, the equivalent inductance

7. The shaft of a synchronous machine is clamped or “blocked” and is thus not free to turn. A set 0 of positive-sequence voltage is applied to the motor terminals. Assume that (1)V a = Va ej0 , (2) θ(t) = π2 , (3)iF = 0, (4)damping circuits not existing. a) Find id and iq in the steady-state. b) Find an expression for the average torque Te,av . Remarks: Torque contains a sinusoidal expression. Sinusoidal component does not generate energy. Therefore, only dc component is sought. c) Suppose that r = 0. What is Te,av ? Solution: We will apply space vector’s projection on dq-axis to find vd and vq . Then, based on the stator voltage, current, flux linkage dynamic equations, we find id and iq . The space vector of the voltage is → − v = Va ejωt ,

15

where ω = 377 rad/s. The rotor is standing still and the rotor position θ is a constant at → − projections of V on the dq-axes are: π vd = Va cos( − ωt) 2 π vq = Va sin( − ωt) 2

π 2.

The

(27)

Further, the rotor excitation current iF = 0. Therefore, λd = Ld id and λq = Lq iq . The stator dq-axis voltage, current and flux linkage equations become. did dt diq −Va cos(ωt) = −riq − Lq dt Va sin(ωt) = −rid − Ld

(28)

The steady-state expressions of id and iq can be found using phasor-related calculation and convert back to time-domain expressions. Va π/2−θzd √ For example, I ad = √ 2 1 e . The time-domain expressions are 2 2 r +(ωLd )

−Va id = p sin(ωt − θzd ) 2 r + (ωLd )2 Va cos(ωt − θzq ) iq = p r2 + (ωLq )2

(29)

ωL

q −1 d where θzd = tan−1 ωL r , θzq = tan r . b) We will use this expression Te = 32 (λd iq − λq id ).

3 3 (λd iq − λq id ) = (Ld − Lq )id iq 2 2 h Va2 π π i −3 (Ld − Lq ) p cos(θzd − θzq + ) + cos(2ωt − θzd − θzq − ) = 4 2 2 (r2 + (ωLd )2 )(r2 + (ωLq )2 ) (30)

Te =

The average torque is the dc component of Te . 3 Va2 Tav = (Ld − Lq ) p sin(θzd − θzq ) 4 (r2 + (ωLd )2 )(r2 + (ωLq )2 )   ωLq r 3 Va2 ωLd r = (Ld − Lq ) − 4 |Zd ||Zq | |Zd ||Zq | |Zd ||Zq | 3 Va2 r = ω(Ld − Lq )2 4 |Zd |2 |Zq |2

(31)

d) When r = 0, Tav = 0. This problem shows that when energy conservation law. When the rotor is blocked, no mechanical energy is produced. Without considering the loss, the electrical energy input to the machine should also be 0. 0 = 2 sec. 8. An SMIB system and assume that Xd = 1.15, Xq = 0.6, Xd0 = 0.15, XL = 0.2, r = 0, Td0 0 0 The generator is in steady-state with Ef d = 1 and Ea = 1∠15 . The infinity bus’s voltage angle is

16

0. At t = 0, Ef d is changed to a new constant value: Ef d = 2. Assume that the rotation is still uniform. Find Ea0 (t) for t ≥ 0. Solution: The dynamic model to be adopted is 0 Td0

dEa0 + Ea = Ef d . dt

We need to find Ea ’s expression in terms of Ea0 for the SMIB system first to arrive at a first-order ODE. Based on the phasor diagram, we have ed 1.35 Ea − V∞ cos δ X = = . 0 e0 Ea − V∞ cos δ 0.35 X d δ is given as 150 . We need to further find V∞ . Since at steady-state, Ef d = 1 and Ef d = Ea , we have Ea = 1 at steady-state. Further at steady-state Ea0 = 1 as well. This shows that the d-axis current must be zero since Ea = Ea0 − (Xd − Xd0 )Iad . Therefore, V ∞ ’s projection on the q-axis must be 1: V∞ cos(150 ) = 1. The dynamics of Ea0 becomes: 0 Td0

dEa0 1.35 0 + (E − 1) + 1 = Ef d = 2. dt 0.35 a

=⇒

dEa0 + 1.9286Ea0 = 2.4286. dt We can find the steady-state Ea0 (∞) = 1.2598. The dynamic equation becomes Ea0 (t) = Ea0 (∞) + (Ea0 (0+ ) − Ea0 (∞))e−1.9286t . Note that Ea0 cannot jump. Therefore, we can use Ea0 (0− ) = 1. Finally, Ea0 (t) = 1.2598 − 0.2598e−1.9286t .

9. In MATLAB or Simulink, build a dynamic simulation model of a synchronous generator model with electromagnetic dynamics ONLY. This generator is connected to an infinitive bus. The initial condition is: V¯∞ = 1∠0o , I¯a = 1∠00 , Electromechanical dynamics can be ignored, i.e., you can assume the rotor speed is constant at ω0 = 377 rad/s and you don’t need to put the swing dynamics into the model. You can opt to have just one-order dynamics (only rotor flux dynamics) or you can opt to have a third-order dynamic model by considering the dynamics of both rotor flux and qd stator flux linkages. You can even build a fifth-order dynamic model by considering the dynamics of rotor excitation circuit flux, D winding flux, Q winding flux, stator qd flux. The parameters of the machine are given as follows: kMF can be found based on the relationship between L0d , Ld , LF and MF . a) Find the steady-state internal voltage E¯a , stator currents id , iq , various flux linkages λd , λq , λF , as well as the rotor position θ against the stationary reference frame, and the position of q-axis 17

Xd Xq Xd0 Xline rF

1 0.8 0.2 0.5 0.5/ω0

Ld Lq L0d

1/ω0 0.8/ω0 0.2/ω0

Lf

1/ω0

relative to the phasor reference frame (rotating at ω0 and at t=0, the phasor reference frame is aligned with the stationary reference frame). b) Show that the model has flat run for 1 second. Give a plot to show Ea0 (t) for 1 second or three subplots horizontally to show λd , λq and λF for 1 second. c) Show the system dynamic response Ea0 (t) or flux linkages for a step change (10% increase) in Ef d or vF . Show the dynamic response of the terminal voltage RMS magnitude Va (t). Plot the flat run (9.2) and the following dynamic responses in one plot. Remarks: This exercise gives you an opportunity to learn dynamic model building. It includes two essential steps: initialization or flat run and dynamics due to differential equations. Initialization helps to calibrate the initial state variable values x(0). Solution: The MATLAB code that carries out initial state calculation and dynamic simulation is shown as follows. %build 3rd EMT simulation model clear; clc %% parameters u.omega = 377; % compute Mf based on Ld’: Ld’ = Ld - (kMf)^2/LF w0 =377; Ld1 =0.2/w0; Ld =1/w0; LF = 1/w0; k=sqrt(3/2); parameter.MF = sqrt((Ld-Ld1)*LF)/k; % MF = 0.7163/w0; parameter.LF = 1.0/377; parameter.Ld = 1.0/377 +0.5/377; % line inductance has been added parameter.Lq = 0.8/377 +0.5/377; % line inductance has been aded. %parameter.MF = 0.58/377; parameter.rF = 0.5/377; parameter.r = 0.0; %% carry out initial calculation Vinf = 1; Ia = 1; Xq = 0.8; Xd = 1.0; Xline = 0.5; a_prime = Vinf + (parameter.r+j*(Xq + Xline))*Ia; % find a’ delta = angle(a_prime); % find delta iq = sqrt(2)*cos(delta); id = sqrt(2)*sin(-delta); Ea = Vinf*cos(delta)+parameter.r*iq/sqrt(2) -id/sqrt(2)*(Xd+Xline); iF = sqrt(2)*Ea/377/parameter.MF; %Ea = 1/sqrt(2)*w0*MF*iF u.vF = parameter.rF*iF; u.vq = sqrt(2)*cos(delta); u.vd = sqrt(2)*sin(-delta);

% vF = rF*iF at steady-state

18

% compute the initial flux linkages for F, d, q. x(:,1) = [parameter.LF*iF + 1.5*parameter.MF*id; parameter.Ld*id + parameter.MF*iF; parameter.Lq*iq]; print(’Ea, id, iq, lambda_d, lambda_q, lambda_F’); [Ea, id, iq, x(2,1), x(3,1), x(1,1)] %% Numerical integration using Trapesoidal method % step size: 0.001 second. % total simulation time: 20 seconds. % at t=10, vF becomes 1.1 of its original value. i=1; for t=0:0.001:20 i=i+1; if(t 0 and the other for K5 < 0). 20

−3

x 10

6.5

−3

λF

9

x 10

6 Te

8.5

5.5

0

5

10

15

20

8

λd

2 7.5

0 −2

0

5

10

15

20

0

5

10 Time (s)

15

20

15

20

3.3

0

5

10

15

20

2 λq

Pe

3.2

0 −2

3.1

0

5

10 Time (s)

15

3

20

iF

(a)

(b)

4

1.54

3.5

1.52

3

0

5

10

15

1.5

20

−1.1 Va

id

1.48

−1.2

1.46

iq

−1.3

0

5

10

15

20

2

1.44

0

1.42

−2

0

5

10 Time (s)

15

1.4

20

0

(c)

5

10 Time (s)

(d)

Figure 9: Problem 9. Solution: a). The code to compute K1 to K6 for a given δ is shown as follows. Xd=1.15; Xq=0.6; Xdprime=0.15; XL=0.2; r=0; Tdoprime=2; Efd=1; Eaprime=1;%exp(j*15*pi/180) rad; Vinf=1; %exp(j*0) delta=-pi/3:pi/18:pi/3; %rad deltadeg=delta*180/pi; %degree %equations required Xdz=Xd+XL; Xqz=Xq+XL; Xdprimez=Xdprime+XL; Iad=(Vinf*cos(delta)-Eaprime)./Xdprimez; Iaq=(Vinf*sin(delta))./Xqz; Vaq=XL/Xdprimez*Eaprime+Xdprime/Xdprimez*Vinf*cos(delta); Vad=-Xq/Xqz*Vinf*sin(delta); Va=(Vad.^2+Vaq.^2).^(1/2); %Ki K1=T

21

K1=Eaprime*Vinf/Xdprimez.*cos(delta)+Vinf^2*(1/Xqz-1/Xdprimez)*cos(2*delta); K2=Vinf/Xdprimez*sin(delta); K3=Xdprimez/Xdz; K4=(1/K3-1)*Vinf*sin(delta); K5=-Vinf*(Vaq.*Xdprime./(Xdprimez*Va).*sin(delta)+Vad.*Xq./(Xqz*Va).*cos(delta)); K6=Vaq*XL./(Va*Xdprimez); figure subplot(511) plot(deltadeg,K1); ylabel(’K1’,’FontName’,’Times New Roman’,’FontSize’,13);%axis([xmin xmax ymin ymax]) grid on subplot(512) plot(deltadeg,K2); ylabel(’K2’,’FontName’,’Times New Roman’,’FontSize’,13); grid on subplot(513) plot(deltadeg,K4); ylabel(’K4’,’FontName’,’Times New Roman’,’FontSize’,13); grid on subplot(514) plot(deltadeg,K5); ylabel(’K5’,’FontName’,’Times New Roman’,’FontSize’,13); grid on subplot(515) plot(deltadeg,K6); ylabel(’K6’,’FontName’,’Times New Roman’,’FontSize’,13); xlabel(’\delta(degree)’,’FontName’,’Times New Roman’,’FontSize’,13) grid on

K1

3 2

K2

1 -60 5

K4

-20

0

20

40

60

-40

-20

0

20

40

60

-40

-20

0

20

40

60

-40

-20

0

20

40

60

-40

-20

0

20

40

60

0 -5 -60 5 0

-5 -60 0.05

K5

-40

0

K6

-0.05 -60 0.6 0.5 0.4 -60

δ(degree) Figure 10: Varying delta’s influence on Ki b) From Fig. 11, we can see that K5 becomes negative when δ is over 50◦ . Note a large δ also indicates heavy power transfer. The influence of heavy power transfer on K5 and further system stability is explained using two scenarios. Scenario 1: δ = 30◦ (where K5 > 0). Scenario 2: δ = 80◦ (where K5 < 0). The system block diagram is shown in Fig. 12. 22

0.1 0 -0.1

K5

-0.2 -0.3 -0.4 -0.5 -0.6 -0.7 0

20

40

60

80

100

120

140

160

180

δ(degree)

Figure 11: Varying delta’s influence on K5

ΔVa

K5

+

K6 ΔPm=0

+ ΔVref

+

+

KA

ΔEfd

-

K3 1+K3T’dos

+ K2

ΔE’a

-

1 Ms2+Ds+T

Δδ

K4

Figure 12: Block diagram of linearized system without PSS. Fig. 12 will be aggregated into a single loop system. Then the open-loop transfer function will be derived. We will use the open-loop transfer function to plot root loci and check if the closed-loop system is stable or not. First, the loop will be decoupled at the point before the block of K2 . For the open-loop system, we will compute the transfer function from ∆δ to ∆Ea0 . Then we close the loop and compute the closed-loop transfer function by considering the feedforward block from ∆Ea0 to ∆δ. The feedforward transfer function is notated as −K2 TEM where TEM = 1/(M s2 + Ds + T ). ∆Ea0 =

K3 0 Tdo K3 s

[−K4 ∆δ − KA (K5 ∆δ + K6 ∆Ea0 )] +1 ∆Ea0 −K3 (K4 + KA K5 ) = 0 ∆δ Tdo K3 s + K3 KA K6 + 1

(32)

The open-loop transfer function becomes: L(s) =

−K3 (K4 + KA K5 ) 0 Tdo K3 s + K3 KA K6 +

1

M s2

K2 + Ds + T

(33)

The closed-loop system is a negative feedback system. We will plot the root loci of L(s) and check when the gain is 1 if the close-loop poles are located at LHP. 23

KA = 25; Ge=K3*(K4+KA*K5)/(1+K3*K6*Ke+K3*Tdoprime*s); Gp=1/(M*s^2+D*s+T); G=-Ge.*K2.*Gp; rlocus(G); hold on; p = rlocus(G,1); plot(real(p), imag(p),’+’)

For the two scenarios, we compute Ki s again and use the above code to conduct root locus plotting and also mark the closed-loop system poles using “+”. Root Locus

Root Locus

30

40 δ = 40

20

Imaginary Axis (seconds−1)

Imaginary Axis (seconds−1)

0

10 0 −10 −20 −30 −20

0

δ = 80

30 20 10 0 −10 −20 −30

−15

−10

−5

0

5

−40 −20

10

Real Axis (seconds−1)

−15

−10

−5

0

5

10

Real Axis (seconds−1)

Figure 13: Root loci for two scenarios. a) δ = 400 . b) δ = 800 . The root loci shown in Fig. 13 show that when δ = 400 , the system is stable while it is unstable when δ = 800 . 2. In MATLAB/Simulink, build a linearized model shown in Fig. 5.19. • Compare the system dynamic performance due to a step response from ∆Vref (use 10% change) with and without PSS. • Use MATLAB rlocus function to validate Fig. 5.18. Please write the transfer function of the loop gain that rlocus will be applied. • Use MATLAB function “linmod” to obtain the state-space model and the system matrix A. Eigenvalues of A are the poles of the closed-loop system. You can also find the closed-loop transfer function by manual derivation. Identify the closed-loop eigenvalues for the system with and without PSS. Compare your eigenvalues and linear system simulation results and state if they corroborate with each other. Please refer Chapter 5 section 5.5 for model building.

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5

Chapter 7 Large-Signal Stability

1. Use the SMIB system parameter presented in the time-domain simulation example in Section 7.4. For that SMIB system, if the fault clearing time is 0.2 seconds, the initial rotor angle is 300 , Pe = 2 sin δ, please compute the rotor angle at the moment when the fault is cleared δ1 , the maximum rotor angle that can be achieved δ2 , and the minimum rotor angle that can be achieved δmin . Validate your analysis by examining the simulation results. Solution: Given that the fault clearing time is 0.2 seconds, we can find δ1 using integration. Assume that D1 = 0, we can find δ1 = ω4π0 t21 + δ0 = 1.2776 rad/s. Next we apply the equal-area method to find δ2 . The accelerating area A1 is: Z

δ1

(Pm − 0)dδ.

A1 = δ0

The decelerating area A2 is: Z

δ2

(2 sin δ − Pm )dδ.

A2 = δ1

Making A1 = A2 and use the condition that Pm = Pe (δ0 ) = 1, we arrive at the following equation: f (δ2 ) = δ2 + 2 cos δ2 − 1.1016 = 0. δ2 can be found using MATLAB function fzero or using Newton-Raphson method. If we opt to Newton-Raphson method, the iteration procedure is δ2k+1 = δ2k −

!−1 df f (δ2k ). dδ2 δk 2

cos(1.3)−1.1016 If we choose an initial value of δ20 = 1.3, then δ21 = 1.3 − 1.3+21−2 = 2.0911. After a sin(1.3) few steps, we find δ2 converges to 2.0846. Therefore, δ2 = 2.0846 rad. Next, we use the equal-area method again to find δmin . Instead of identify accelerating area and decelerating area, we use an aggregated integral:

Z

δmin

(Pm − 2 sin δ)dδ = 0. δ2

This leads to the following equation: δmin + 2 cos δmin − 1.1016 = 0. Apply Newton-Raphson again, we find δmin = −0.5759 = −32.970 .

25