279 39 Solutions Instructor Manual Chapter 11 Power System Stability

8/19/2019 279 39 Solutions Instructor Manual Chapter 11 Power System Stability

1/19

CHAPTER – 11: Power System Stability

11.1 Assume that three machines whose ratings and inertia constants are respectively

given by 332211 ,,,,, H Gand H G H G are operating in synchronism, that is

( )

( )

( )s

ui

s

s

ui

s

s

ui

s

G

GPP

dt

d

f G

H G

G

GPP

dt

d

f G

H G

G

GPP

dt

d

f G

H G

3332

32

33

2222

22

22

1112

12

11

−=

−=

−=

δ

π

δ

π

δ

π

Since the machines are swinging coherently, δ δ δ δ === 321 . Thus ,

( ) ( )

( ) ( )

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ++=

−=

++−++=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ++

sss

eq

puueq puieq

eq

s

uuuiii

sss

G

H G

G

H G

G

H G H where

PPdt

d

f

H

G

GPPPPPP

dt

d

f G

H G

G

H G

G

H G

332211

2

2

11113212

2332211 1

δ

π

δ

π

⎟⎟ ⎠ ⎞⎜⎜

⎝ ⎛ ++=

sss

eqG H G

G H G

G H G H 332211

» 200*4/1000+500*3/1000+750*5/1000 = 6.05 MVJ/MVA

M = f

GH

×180 = 1000*6.05/(180*50) = 0.6722 MJ-sec/elect-deg

11.2 Synchronous speed2

5022

××= π ω rads/sec = 314.16 rads/sec

KE = 1/2*75000*(314.1593)^2*10^(-6) = 3.7011e+003 MJ

H = (3.7011e+03)/250 = 14.8044 MJ/MVA

M = 250*14.8044/(180*50) = 0.4112 MJ-sec/elec. deg

11.3


2/19

» M=175/(180*50) = 0.0194 MJ-sec/rad

At the time of fault, power input is given by

» Pi=50*0.8 = 40 MW

Power output during fault

» Pu=0.4*40 = 16 MW

Accelerating power is as follows

» Acclpower=Pi-Pu = 24 MW

Acceleration is given by

» Accl=Acclpower/M = 1.2343e+003 elec deg/sec

2

11.4 Pre-fault current

» Ig=Pi*10^6/(sqrt(3)*11*10^3*0.8) = 874.7731 A

Ig=Ig*(0.8-i*0.6) = 6.9982e+002 -5.2486e+002i A

Generator voltage per phase

» Vgpp=(11*10^3/sqrt(3)+i*Ig*2)/(10^3) = 7.4006 + 1.3996i = 7.5318 kV

Line to line generator voltage

» Vgll=abs(Vgpp)*sqrt(3) = 13.0454 kV

Pre-fault rotor angle

» delta=asin(Pi/(Vgll*11))*180/pi = 16.18540

» Accl=Accl*pi/180 = 21.5423 elec. rads/sec2

» t=5/50 = 0.1000 sec

Rotor velocity at the time of occurrence of fault

» ddeltadt=sqrt(2*Accl*(delta*pi/180)) = 3.4887 elec. rads/sec = 66.6290 rpm

Rotor velocity at the end of acceleration period =2

50120× + 66.6290


3/19

= 1566.6290 rpm

Rotor angle at the end of the acceleration period

» deltan=Accl*t^2+delta*pi/180 = 0. 0.4979 rads = 28.52830

11.5 The swing equations of the two machines are

ui

ui

PPdt

d

f

H

PPdt

d

f

H

222

22

2

112

12

1

−=

−=

δ

π

δ

π

( )

( )21

2112

21

21122112

21

21

2

122

21

2112

21

2112

2

212

21

21

21

2112

21

2112

2

22

1

11

2

22

2

12

,,,

1

1

H H

P H P H andP

H H

P H P H P

H H

H H H where

PPdt

d H

f

H H

P H P H

H H

P H P H

dt

d

H H

H H

f

H H

P H P H

H H

P H P H f

H

PP f

H

PP f

dt

d

dt

d

uuequ

iieqieq

equeqieq

uuii

uuii

uiui

+

−=

+

−=−=

+=

−=

+

−−

+

−=⎟

⎟

⎠

⎞⎜⎜

⎝

⎛ −

+

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −−

−=

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ −−⎟⎟

⎠

⎞⎜⎜

⎝

⎛ −=−

δ δ δ

δ

π

δ δ

π

π

π π δ δ

MVA MJ H H

H H H eq /5714.2

5.10

0.65.4

21

21 =×

=+

=

pu H H

P H P H P iieqi 2429.0

5.10

1.15.425.10.6

21

2112 =×−×

=+

−=

Max power =( )

pu9714.115.04.015.0

15.12.1=

++

×

11.6 Input data:

» Pl=0.8;E=1.0;pf=0.85;H=4.0;

» phi=-acos(pf) = -0.5548 rads. = -31.78830

Load current:


4/19

» I=Pl/(E*pf) = 0.9412 pu = 0.8000 - 0.4958i pu

Pre-fault transfer reactance between the internal generator bus and infinite bus

» Xl=0.25+0.15 = 0.4000 pu

Generator voltage behind transient reactance

» Edash=E+i*I*Xl = 1.1983 + 0.3200i pu = 1.2403 pu

Power angle

» delta=angle(Edash) = 0.2610 rads. = 14.95140

During fault transfer reactance between generator internal bus and infinite bus

» Xlf=1.15;

Power output during fault

» Poutf=(abs(Edash)*E/Xlf)*sin(delta) = 0.2783 pu

Accelerating power at the time of fault

» Paccl=0.8-Poutf = 0.5217 pu

Inertia constant

» M=H/(pi*50) = 0.0255 MJ-sec/elec. rad.

Acceleration at the time of fault

» Accl=Paccl/M = 20.4886 elec. rads./sec2

δ δ δ

sin2941.423725.31sin15.1

0.1*2403.18.0

0255.0

12

2

−=⎟ ⎠

⎞⎜⎝

⎛ −=

dt

d


5/19

11.7 From the previous problem'

E =1.2403 pu, M = 0.0255 MJ-sec/elec. rad.,


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δ0 = 0.2610 rads. = 14.95140. Power input 0iP = 0.8 pu

Fault duration t = 10 cycles = 0.2 secs.

Rotor angle δ at the time of fault clearance is given by

020

2δ δ += t

M

Pi = 0.8885 rads.

Accelerating area A1 = Pi0 (δ – δ0) = 0.5020

Decelerating area A2 = ( )[ ] ( )δ δ δ δ −−−×

mm 8.0coscos4.0

0.12403.1

= ( )[ ] ( )δ δ δ δ −−− mm 8.0coscos1007.3

For stability, A1 = A2 , that is,

( )[ ] ( )δ δ δ δ −−− mm 8.0coscos1007.3 = 0.5020

f (δm) = 3.1007 cos δm + 0.8 δm – 2.1641

( )m

m

m

d

df δ

δ

δ sin1007.3=

» tolr=0.0001;

» [deltam] = P1207(tolr);

initial estimate of deltam60*pi/180

deltam = 66.52200

Maximum swing of the rotor angle is δm = 66.52200.

Since δm is less than (π – δ0) the system will remain stable.

(ii) Assume that the critical clearing angle is δc.

Accelerating area A1 = Pi0 (δc – δ0) = 0.8 (δc – 0.2610 ) = 0.8 δc – 0.2088

Decelerating area A2 = ( )cmiPd m

c

δ δ δ δ δ

δ −−∫ 0sin1007.3


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[ ] )(8.0coscos1007.3 002 cc A δ δ π δ δ −−−+=

= 3.1007 cos δc + 0.8 δc +0.6912

For stability A1 = A2, that is,

0.8 δc – 0.2088 = 3.1007 cos δc + 0.8 δc +0.6912

( ) 01 8733.106.8653.11007.3

6912.02088.0cos ==⎟

⎠

⎞⎜⎝

⎛ +−= − radscδ

( ) ( )3198.0

8.0

2610.08653.10255.022

0

0 =−×

=−

=i

cc

P

M t

δ δ secs. = 16 cycles

11.8 Input data:

» Pi0=0.8;Einfint=1.0;Xlprefault=0.4;Xlfault=1.05;Xlpfault=0.55;pf=1.0;

Load current

» I=0.8/(Einfint*pf) = 0.8000 pu

Voltage behind generator transient reactance

» Edash=Einfint+i*Xlprefault*I = 1.0000 + 0.3200i pu = 1.05 pu

Initial power angle

» delta0=angle(Edash) = 0.3097 rads. = 17.74470

Pre fault maximum power

» PmA=Edash*Einfint/Xlprefault = 2.6249 pu

Post fault maximum power

» PmB=Edash*Einfint/Xlpfault = 1.9090 pu

During fault maximum power

» PmC=Edash*Einfint/Xlfault = 1.0000 pu

Maximum power angle

» deltam=(180-(asin(Pi0/PmB)*180/pi)) = 155.22430 = 2.7092 rads


8/19

Cosine of critical clearing power angle

» cosdeltac=(Pi0*(deltam-delta0)+PmB*cos(deltam)-PmC*cos(delta0))/(PmB-

PmC) = -0.8427

Critical clearing power angle

» deltac=acos(cosdeltac)*180/pi = 147.43070

Critical clearing time computation:

» tc=sqrt(2*M*(147.4307-17.7447)/0.8) = 0.3796 secs. = 18.9800 cycles.

Plotting the power output versus power angle

» delta=linspace(0,pi,500);

» P1=2.6249*sin(delta);

» P2=1.0*sin(delta);

» P3=1.9090*sin(delta);

» plot(delta*180/pi,P1,'k-')

» grid

» hold on

» plot(delta*180/pi,P2,'k-.')

» plot(delta*180/pi,P3,'k--')

» x=linspace(0,180,500);

» y=0.8;

» plot(x,y,'k-')

» xlabel('Power angle in degrees')

» ylabel('Power output in pu')


9/19

» legend('Pre-fault power output','During fault power output','Post fault power

output')


10/19

0 20 40 60 80 100 120 140 160 1800

0.5

1

1.5

2

2.5

3

Power angle in degrees

P o w e r o u t p u t i n p u

Pre-fault power output

During fault power output

Post fault power output

11.9 Input data:

» Pi0=1.0;Edash=1.20;Einfinit=1.0;Xlprefault=0.6;Xlfault=1.0267;

» tf=15/50;H=3.5;

(i) Rotor angle prior to the fault

» delta0=asin(Pi0*Xlprefault/(Edash*Einfinit)) = 0.5236 rads. = 30.00000

(ii) Computation of generator output, accelerating power, and rotor acceleration

» Poutgen=(Edash*Einfinit/Xlfault)*sin(delta0) = 0.5844 pu

» Paccl=Pi0-Poutgen = 0.4156 pu

» M=H/(pi*50) = 0.0223 MJ-sec./elec. rad.

» Accl=Paccl/M = 18.6522 rads./sec2

Computation of rotor angle and decelerating power at the instance of fault clearance:


11/19

» delta=delta0+(Paccl/(2*M))*tf^2 = 1.3629 rads. = 78.09120

Generator output at the instance of fault clearance

» Poutgenf=(Edash*Einfinit/Xlfault)*sin(delta) = 1.1436 pu

Genrator accelerating power at the instance of fault clearance

» Paccl=Pi0-Poutgenf = -0.1436 pu

11.10

f unct i on ydot = p1210( t , y) ;ydot =t 3̂- y 2̂;

Input to the command window

» tspan=[0 1];

» y0=0.5;

» [t,y]=ode23('p1210',tspan,y0)

Output

t = y =

0 0.5000


12/19

0.1000 0.4762

0.2000 0.4549

0.3000 0.4367

0.4000 0.4226

0.5000 0.4144

0.6000 0.4141

0.7000 0.4242

0.8000 0.4477

0.9000 0.4875

1.0000 0.5469

Plot of y vs i

» plot(t,y,'k--');

» hold on

» grid

» xlabel('t')

» ylabel('y')

» title('Solution of differential eq. of problem 12.10')

» hold off

»


13/19

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.4

0.42

0.44

0.46

0.48

0.5

0.52

0.54

0.56

t

y

Solution of differential eq. of problem 12.10

11.11 The second order differential swing equation is written as two first order

differential equations, that is,

Z 1 = δ and.

2 δ = Z

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=⎥⎥

⎦

⎤

⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=⎥⎥

⎦

⎤

⎢⎢

⎣

⎡=⎥

⎦

⎤⎢⎣

⎡=

.

2.

2

2

.

.

2

1.

.

2

1,

aP

Z

Z

Z

Z

Z Z

Z

Z Z

δ

δ

Function ‘p1211’ comput es t he st ate vari abl es

f unct i on Zdot = p1211( t , Z) ;

% I nput i s t i me t and Z = [ Z( 1) ; Z( 2) ]

% Out put i s Zdot

i f t >= 0. 25;Pa=0. 75- 1. 5*si n( Z( 1) ) ;

el sePa=0. 75- 0. 30*si n( Z( 1) ) ;

endZdot=[ Z( 2) ; Pa] ;


14/19

The f ol l owi ng scr i pt f i l e execut es ode23

» tspan=[0 2.0]; % Sets the time span

» Z0=[30*pi/180;0]; % Assigns the initial rotor angle in radians

» [t,Z]=ode23('P1211',tspan,Z0) % Executes ode23

Output:

t vs δ and t vs.

δ is obtained and plotted

» x=Z(:,1);y=Z(:,2); % Sets first and second columns of Z equal to x and y

respectively

» subplot(2,1,1)

» plot(t,x*180/pi,'k--')

» grid

» xlabel('Time t in sec.')

» ylabel('Rotor angle in radians')

» title('Rotor Angle Versus Time')

» subplot(2,1,2)

» plot(t,y,'k--')

» grid

» xlabel('Time t in sec.')

» ylabel('Rotor velocity in radians/sec.')

» title('Rotor Velocity Versus Time')

»


15/19

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 228

30

32

34

36

38

Time t in sec.

R o t o r a n g l e

i n r a d i a n s

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-0.1

-0.05

0

0.05

0.1

0.15

Time t in sec.

R o t o r v e l o c i t y

i n r a d i a n s / s e c . Rotor Angle Versus Time

Rotor Angle Versus Time

11.12 For derivation please see 7.3.1 Power flow equations in [Y bus] frame of reference.

The power flow equation is given by

( )12121221111121 coscos δ θ θ −+= Y E E Y E P where δ12 = δ1 – δ2 (I)

Eq. (I) modifies as follows for the problem case

( )12121221121221 coscos δ θ θ −+= Y E E Y E P

For maximum power transfer; ( )1212 δ θ − = 0. Hence, δ12 = θ12. θ12 of a transmission line

having resistance is < 900.

If Z

Y 1

12 = and Z

R=12cosθ . Assuming Z = R + j X .

The maximum steady limit is written as

Z

E E

Z

R E P 21

2

21 +−= (II)


16/19

[Note: In deriving (I) in Sec.7.3.1, θ12 was substituted by (180 – θ12)]

When resistance is neglected, θ11 = 900. Therefore power transferred is given by

1221 sinδ

X

E E P =

X

E E P 21max = when δ12 = 90

0 (III)

Comparison of (II) and (III) shows that the steady limit is more when resistance of the

line is neglected.

11.14 Computation of reactance between generator terminal and infinite bus

» X12=0.4/2+0.4 = 0.6000 pu

Pf angle:

» phi=-acos(0.85) = -0.5548 rads. = -31.78830

Load current

» I=0.75/(1.0*0.85) = 0.8824 pu = 0.7500 - 0.4648i pu

Voltage behind generator reactance

» E2=1.0+i*X12*I = 1.2789 + 0.4500i pu

Magnitude of voltage behind generator reactance

» Emag2=abs(E2) = 1.3557 pu

Power angle δ0 prior to the disturbance

» delta0 = angle(E2) = 0.3383 rads. = 19.38540

Synchronizing coefficient

» Ps=(Emag2*1.0/X12)*cos(delta0) = 2.1315 pu

Coefficient of damping power

» Pd=0.15;


17/19

» M=8/(pi*50) = 0.0509 MJ-sec/elec. rad

Damping factor

» Pd/M = 2.9452

» Ps/M = 41.8514

Natural frequency of oscillations 8514.41= M

Ps = 6.4693 rads/sec.

The equation representing the motion of the rotor following a disturbance is, thus,

written as

08514.419452.2

...

=+Δ+Δ δ δ

The second order differential equation is written as two first order differential

equations for employing MATLAB to plot variation of rotor angle δ and rotor

velocity ω against time.

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

Δ

Δ=⎥

⎦

⎤⎢⎣

⎡= .

2

1

δ

δ

x

x x

The two first order differential equations become

122

.

21

.

8514.419452.2 x x x

x x

−−=

=

The MATLAB function ‘p1214’ computes the state variables

f unct i on xdot = p1214( t , x) ;xdot=[ x(2) ; - 2. 9452*x( 2) - 41. 8514*x( 1) ] ;

The following script file executes ode23

» tspan=[0 5.0];

» x0=[12.5*pi/180;0];

» [t,x]=ode23('P1214',tspan,x0)


18/19

» x1=x(:,1);y1=x(:,2);

» plot(t,x1*180/pi,'k-')

» grid

» hold on

» plot(t,y1,'k-.')

» xlabel('Time t in secs.')

» ylabel('Rotor angle in degrees')

» ylabel('Rotor velocity in rads./sec')

» ylabel('Rotor angle in degrees & velocity in rads./sec')

» legend('Rotor angle delta','Rotor velocity')

» title('Plot of rotor angle and velocity versus time')

»

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-6

-4

-2

0

2

4

6

8

10

12

14

Time t in secs.

R o t o r a n g l e i n d e g r e e s &

v e l o c i t y i n r a d s . / s e c

Plot of rotor angle and velocity versus time

Rotor angle delta

Rotor velocity


19/19

279 39 Solutions Instructor Manual Chapter 11 Power System Stability

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Transcript of 279 39 Solutions Instructor Manual Chapter 11 Power System Stability