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ECE 422/SPRING 2015 – FINAL EXAM – IN-CLASS (40%)

NAME:

BE SURE TO SHOW YOUR WORK CLEARLY AND FULLY. SHOWING YOUR THINKIING HELPS

YOU GET MORE POINTS.

1. Short Answer Questions (40 points).

a. (8 points) Consider the right figure about the limits of

real and reactive power outputs with a hydro unit

(1) If its real power output is fixed at 50MW, what

are the secure ranges of its reactive power outputs

respectively at 100% and 115% of the rated

terminal voltage?

100%: -88 ~ 50 (2pts)

115%: -102~60 (2pts)

The numbers don’t have to be accurate

(2) When the generator is at the operating point (60MW, 50Mvar) in the figure with 100% rated terminal

voltage, which one is more likely violated?

(a) field current heating limit (4pts)

(b) armature current heating limit

b. (3 points) Like AGC, which can maintain frequency everywhere in the system at 60Hz, AVRs with

generators are able to maintain voltages everywhere at the rated voltage level. True or false, and why?

False (2pts)

Reason: Voltage cannot be effectively controlled remotely (1pt)

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c. (4 points) Give at least one example for each of these two types of voltage control equipment

(1) It maintains the voltage level at a specific bus but not a var source.

ULTC transformer (2pts)

(2) It is able to either supply or absorb var automatically but has no rotating part

SVC (2pts)

d. (3 points) The lower the frequency of power oscillation, the less damaging it can be to the power system.

True of false, and why?

False (2pts)

Reason: Slow oscillation is more dangerous to power system operations since it is more difficult to damp

than fast oscillation. (1pt)

e. (5 points) Which of the following may improve the transient stability of a generator connected to the grid,

and why?

(1) Reduce its loading level (Y, 1pt)

(2) Increase its inertia (Y, 1pt)

(3) Reduce its leakage reactance Ll (Y, 1pt)

(4) Reduce its damping coefficient KD (N, 1pt)

(5) Add a shunt capacitor on its power transmission path (Y, 1pt)

f. (4 points) Why is the Trapezoidal Rule method called an implicit method for transient stability simulation?

What is its main advantage compared to explicit methods like the R-K methods?

There is no explicit equation to calculate Xn+1 from Xn (2pts)

Advantage: no numerical instability (2pts)

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g. (3 points) List at least 3 disadvantages/limitations of using the “simplified model” (transient stability - slide

27) for transient stability simulation of multi-machine systems

3pts if given any three:

Classical model (optimistic)

Constant impedance load model (optimistic)

Ignoring damping (conservative)

Governor control ignored (conservative)

Excitation control ignored (conservative)

h. (10 points) About the generator connected with an infinite bus

through a transformer and two parallel lines. How does each of the

following change influence the oscillation frequency and damping

ratio of the generator?

(1) Reduce its inertia

(2) Reduce the system synchronous frequency from 60Hz to 50Hz

(3) Reduce the reactance of one of the two lines

(4) Reduce its loading, i.e. steady-state value of power angle

(5) Reduce its E’ by adjusting field current ifd.

1pt each

Damping Freq

(1) Increase Increase

(2) Decrease Decrease

(3) Decrease Increase

(4) Decrease Increase

(5) Increase Decrease

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ECE 422/SPRING 2015 – FINAL EXAM – TAKE-HOME (60%)

NAME:

Due by Wednesday 8AM. Please either hand your solution directly to GTA Denis at MK205 (please don’t slide

it into my office) or email the scan to me and cc Denis (dosipov@vols.utk.edu)

2. (10 points) A generating unit has a simplified

linearized AVR system as shown in the figure. If

A=0.1 sec, E=0.5 sec, G=1 sec, R=0.05 sec,

KR=1, KE=1 and KG=1,

a. Use the Routh-Hurwitz array to find the range of KA for control system stability

b. If KA is set to 1/2 of the upper limit determined from a, then estimate the steady-state step response

5 pts each

a.

b.

T s( )K

0.5 s 1( ) s 1( ) 0.05 s 1( ) 0.1 s 1( )

ch 0.5 s 1( ) s 1( ) 0.05 s 1( ) 0.1 s 1( ) K 11

chexpand

collect s0.0025s

4 0.0825s

3 0.73 s

2 1.65 s K 1

n 4 a

400 K 400

660.0

292.0

33.0

1.0

400

b2

an 1

an 4

an

0

an 1

aaa

b1

an 1

an 2

an

an 3

an 1

a

c2

b1

0 an 1

0

b1

b

c1

b1

an 3

an 1

b2

b1

b

d1

c1

b2

an 1

c2

c1

c

c1

0 solve K 12.6 d1

0 solve K 1.0

K 12.6

K12.6

2 Vtss

K

1 K0.863

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3. (25 points) The circuit model below represents the steady-state operating condition of a single-machine-

infinite-bus system. Et=EB=1.0pu. The generator is represented by the classical model with H=10 (MWs/MVA)

and KD=3 (pu torque/pu speed deviation). Consider a small disturbance ended at t=0s.

a. Write the linearized state equations of the system. Determine the undamped natural frequency (n) of

oscillation in both rad/s and Hz, damping ratio (), and damped frequency (d) in both rad/s and Hz.

b. Consider a small disturbance of =10o and =0 rad/s at t=0s. Determine the equations of and on

the zero-input response of the generator, and plot (degree) and (rad/s) for t=0~10s.

c. If the input power of the generator is

increased by 10% at t=0s with =0o

and =0 rad/s, determine the

equations of and on the zero-

state response of the generator, and

plot (degree) and (rad/s) for

t=0~10s.

a. Xt=Xtr+X1//X2=j0.8

It=(Et-EB)/Xtr=0.625+j0.1675 1’

P+jQ=EtIt*=0.625+j0.1675 1’

E’=Et+X’dIt=0.8325+j0.625=1.04136.90o 1’

0=36.90o

XT=X’d+Xt= j1

KS=E’EBcos0/XT=0.8325 2’

A=0

/ 2 / 2 0.15 0.0416

0 377 0

D SK H K H

2’

12= -0.075 ± j3.9608

ωd=3.9608 rad/s=0.6304Hz 2

ωn=√12=3.9615 rad/s=0.6305Hz 2

=02 2

D

S

K

K H =0.0189 2

b.

2’

rad

010

1800.175 acos ( ) 1.552

t( ) 00

1 2

e n t

sin d t 0.644 0.175e0.075 t

sin 3.961 t 1.552( )

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2’

2’

c.

2’

2’

2’

rad/s

rad

rad/s

t( ) 0

n 0

1 2

e n t

sin d t 376.991 0.692e0.075 t

sin 3.961 t( )

S It

Et 0.625 0.167i Pm Re S( ) 0.625 P 0.1 Pm 0.063

t( ) 0

f0 P

H n2

11

1 2

e0.075 n t

sin d t

= 0.644 0.075 1 1.0002e0.075 t

sin 3.961 t 1.552( ) =

t( ) 0

f0 P

H n 1 2

e n t

sin d t 376.991 0.297e0.075 t

sin 3.961 t( )

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4. (25 points) Continue with the system in Question 3 with the same steady-state operating condition, but

neglect KD. Consider a contingency with this sequence of events:

at t=0s, a three-phase short circuit fault happens at one end of a transmission line as indicated;

at t1, the breakers on both ends of that line are opened to clear the fault, and (t1) reaches 1;

at t2, when (t2) reaches 2, those two breakers succeed in reclosing the line since the fault disappears

during t1~t2.

If 1=60o and 2=80o, apply the Equal-Area Criterion to judge the system’s transient stability with respect to the

contingency described above. Sketch the P- curve of the generator under that contingency, and indicate the

path of the system state and the accelerating and decelerating areas. If the system is transiently stable, what is

the maximum rotor angle m?

a.

From Question 1,

Xeq0=j1

Pm=0.625, 0=36.9o

Pmax0= |E’EB/Xeq0|=1.041 2’

t<t1:

Xeq1=inf 1’

Pmax1=|E’EB/Xeq1|=0 2’

t=t1~t2

Xeq2=j0.25+j0.2+j0.6=j1.6 1’

Pmax2=|E’EB/Xeq2|=0.6506 2’

t>t2

Xeq3=Xeq0=j1 1’

Pmax3= Pmax0= |E’EB/Xeq0|=1.041 2’

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1

0max1( sin )

acc mA P P d = Pm(1-0)+Pmax1(cos1-cos0)=0.252 3’

2

11 max 2 m max 2 2 1 m 2 1( sin ) (cos cos ) ( )

decA P P d P P =-0.00583 3’

22 max 3 m max 3 2 m 2( sin ) (cos cos ) ( )

m

dec m mA P P d P P =0.325

The maximum value may reach under this case, i.e. the angle of the returning point:

m=118.5o 3’

5’