Microsoft Word - EEPB383 Final Sem 2 1112 With Solution (1)

EEPB383,Semester22011/2012 EEPB383,Semester2201 1/2012

COLLEGEOFENGINEERING PUTRAJAYACAMPUS FINALEXAMINATION SEMESTER22011/2012 PROGRAMME

: Bachelor of Electrical Power Engineering (Honours)

SUBJECT CODE

: EEPB 383

SUBJECT

: Electrical Power System II

DATE

: December 2011

TIME

: 2 ½ hours

INSTRUCTIONSTOCANDIDATES:

1.

ThispapercontainsFIVE(5)questionsinSEVEN(7)pages.

2.

AnswerALLquestions.

THISQUESTIONPAPERCONSISTSOF7PRINTEDPAGESINCLUDINGTHIS COVERPAGE. COVERPAGE.

Page1of17

EEPB383,Semester22011/2012

QUESTION1[15MARKS]

The9–buspowersystemnetworkofanelectricutilitycompanyisshownintheFigure1 below. Plant 3 and 5 are shut-down for maintenance leaving plant 1, 2 and 7 generating powertofeedtheloadsateachofthebusesshownintheFigure1.Theproductioncostsfor plant1,2and7in$/hisgivenasfollows: C 1

=

240 + 6.7 P1 + 0.009 P12

C 2 = 220 + 6.1P2 + 0.005 P22 C 7 = 240 + 6.5 P7 + 0.008P72 

whereP1,P2andP7 are in MW.

Figure1 Withoutconsideringpowerlossesinnetwork,answerthefollowingquestions. a) Findtheincrementalcostsforpowerplant1,2and7intermsofitspowergenerated. [3marks]

b) Assumingallthreepowerplants1,2and7generatewithinitspowergenerationlimits; determine the optimum dispatch of generation.

[8marks]

c) Ifthepowergenerationlimitforplantno.1isbetween80MWand300MW,determine the new optimum dispatch of generation. Page2of17

[4marks]


QUESTION2[15MARKS]

a) Drawtheequivalentcircuitforsub-transientperiodofasynchronousgenerator.Labelall components of the circuit.

[5marks]

b) A100-MVA,20-kVsynchronousgeneratorisconnectedthroughatransmissionlinetoa 100-MVA,20-kVsynchronousmotor.Theperunittransientreactancesofthegenerator andmotorare0.25and0.20,respectively.Thelinereactanceonthebaseof100MVAis 0.1perunit.Themotoristaking50MWat0.8powerfactorleadingataterminalvoltage of 20kV. A three-phaseshortcircuit occurs at the generator terminals. Determine the transientcurrentsineachofthetwomachinesandintheshortcircuit. [10marks]


G1

G2 T1

T2

1

2 CB1

CB2

3

4

Figure2

Page3of17


A 4-bus power system network aboveis representedas a Thevenin equivalent network in termsofitsimpedancevalues.Detailedvaluesareasfollows(allvaluesareinpu): Item

Base MVA

Voltage

X0

Rating

X1

X2

G1

100

20kV

j0.08

j0.25

j0.25

G2

100

20kV

j0.07

j0.2

j0.2

T1

100

20/275kV

j0.09

j0.1

j0.1

T2

100

20/275kV

j0.09

j0.1

j0.1

L12

100

275kV

j0.4

j0.12

j0.12

L13

100

275kV

j0.6

j0.14

j0.14

L23

100

275kV

j0.7

j0.16

j0.16

L34

100

275kV

j0.8

j0.18

j0.18

Theneutralforeachgeneratorisconnectedtogroundviaacurrentlimitingreactorofj0.3per unit. The Generators are operating at no-load, producing voltages at rated values at a frequencyof50Hz.Thesystemvoltagesareatratedvalues.Allvaluesareexpressedinper unitonacommon100MVAbase.CircuitBreaker1(CB1)andCircuitBreaker2(CB2)are in the OPEN position.A bolted3 phase to ground fault occurred at Bus 4. Please dothe following:

a)

Determine the Admittance Matrix

[2marks]

b)

Calculate the fault current at Bus 4 in per unit

[4marks]

c)

DeterminethevoltagesatBus1,Bus2,Bus3andBus4duringfault [10marks]

d)

Calculatethefaultcurrentflowinginline2–3,1–3and3-4

e)

Atreefalls downandshortsall 3phases ina transmissionline toground. Isthisa bolted or non-bolted fault? Explain why?

Page4of17

[6marks]

[3marks]



RefertothepowersystemshowninFigure3andtheparametersgiven.

Figure3

GeneratorsG1andG2: X 1

= X

TransformersT1andT2: X 1 Line: Z 1

= Z

2

=

2

= 2

0

j 0.1 p.u; X = j 0.05 p.u; 0

= X = X =

j 0.15 p.u;

0

j 0.05 p.u, Z = j 0.1 p.u ;

Voltage:11kV Voltage: 11/275 kV Voltage: 275 kV

Allperunitvaluesareonabaseof100MVA,andpre-faultvoltageis1.0perunit.

a) Drawthepositive,negativeandzerosequenceimpedancenetworksforthepowersystem. Stateinyourdrawingthenumericalvaluesofalltheimpedance.

[2+2+2marks]

b) Determinethefaultcurrentinperunitforaboltedsingle–line-groundfaultatbusB. [4marks]   c) Determinethecurrentinperunitflowinginthefaultedphaseoftheoverheadlinefora bolted single-line-ground fault at busbar B.

[10marks]

d)DeterminethecurrentflowingintheneutralofT1andT2forasingle-line-groundfaultat bus B given the fault impedance is j0.5 per unit.

Page5of17

[5marks]



Figure3 A50Hz,H=5MJ/MVAgeneratorisconnectedthroughparalleltransmissionlinestoa grid system. The machine is delivering 1.0pu power and both the terminal voltage and the infinite-busare1.0pu.Thevaluesofthereactanceonacommonsystembasearegivenasin thediagram.Thetransientreactanceofthegeneratoris0.20puasindicated. Determine: a) Thepower-angleequationduringsteadystatecondition. b) Find the initial operating angle.

[15marks] [2marks]

c) A temporary fault occurs at a distance of 20% of the line length away from the sending end terminal of the line. When the fault is cleared by an auto-reclose protection,bothlinesareintact.Determinethecriticalclearingangleandthecritical fault clearing time.

[3marks]

ENDOFQUESTIONPAPER

Page6of17


FormulaSheet n

P D +

n

P D =

λ − β i ; 2 γ i =1 i

∑

λ =

i =1

n

+ D

π f 0 dt 2

d ∆δ

+

dt

i

1

∑ 2γ i =1

H d 2 ∆δ

β i

∑ 2γ i

PS ∆δ = 0

2

H d δ

=

π f 0 dt 2 δ = δ 0

∆δ 0

+

2

ω n ∆δ 0

−

2

−ξω nt

e

D

π f 0

2

HPS

π f 0

=

ω d

= ω n

H

sin (ω d t + θ )

−ξω nt

1 − ξ

ω n

sin ω d t

PS

1 − ξ 2

I a0 = I a1 = I a2 =

I 1a = − I a2 =

θ = cos −1 ξ

E a 1

2

Z + Z + Z 0 + 3 Z f E a

1

2

Z + Z + Z f

:

:

Singlelinetogroundfault

Line-Line fault

1 1

0

I a = −

I a2

e

1 − ξ

ω = ω 0

ξ =

Pm − Pe

=−

E a − Z I a 0

Z + 3 Z f

E a

:

Double Line to Ground Fault

:


1 1

− Z I a

Z 2

Page7of17


I a1 =

E a   1   Z + 2 0 Z + Z +3 Z f Z 2  Z 0 +3 Z f 

V a  1 1  b  2 V  = 1 a V c  1 a    iasy (t ) =

S pu =

S S B

SCC =

:


1  V a0 









a V a1   2  2 a  V a

 1 E 0 1  −t τ d ''  1 1  −t τ d ' 1  e e ( t ω − + − + + δ ) + 2 sin sin δ e −t τ a  ''  X '' X '   X ' X  X d  X d  d d  d   d

2 E 0 

,

S B X kk

V pu =

V V B

,

new old Z pu = Z pu

I pu =

I I B

,

S Bnew  V Bold 



Z pu =

Z Z B

2



S Bold  V Bnew 

Page8of17

,

Z B =

(V B )2 S B

, I B

=

S B

3V B


Solution

Question1[15marks] a) dC 1

λ 1 =

dP1

λ 2

=

λ 7

=

dC 2 dP2 dC 7 dP7

= 6.7 + 0.018 P1 $/MWh

= 6.1 + 0.010 P2 $/MWh

= 6.5 + 0.016 P7 $/MWh

b) Neglectinglosses,totalpowerdemand, P D = P1 + P2 + P3 + ...... + P9

= 0 + 20 + 25 + 10 + 40 + 60 + 10 + 80 + 100 = 345 MW P D +

λ =

β i i =1, 2, 7 2γ i

∑

1 i =1, 2 ,7 2γ i

6 .7 6.1 6.5 + + 2(0.009) 2(0.005) 2(0.008) =7.95$/MWh 1 1 1 + + 2(0.009) 2(0.005) 2(0.008)

345 + =

∑

SolveforPbasedonoptimaldispatch, 6.7 + 0.018P1 P1

= 7.95

= 69.44 MW

6.1 + 0.010 P2 = 7.95 P2 = 185 MW 6.5 + 0.010 P7

= 7.95

P7 = 90.625 MW

c) Set P1

= 80 MW,optimise P2 and P7

Page9of17


6 .1 6.5 + 2(0.005) 2(0.008) =7.885$/MWh 1 1 + 2(0.005) 2(0.008)

345 − 80 +

λ =

6.1 + 0.010 P2 = 7.885 P2 = 178.5 MW 6.5 + 0.010 P7

= 7.885

P7 = 86.56 MW

Total=80+178.5+86.56=345MW

Questions2[15marks] a)

Xl:leakagereactance Xad:reactanceduetoarmaturereaction Xf :reactanceduetofiledwinding b)

Sm

=

50∠ − 36.87° 0.8 ×100

= 0.625∠ − 36.87° p.u

[1 mark]

Vm

=

20 20

= 1∠0° p.u

[1 mark] Before the fault, *

I m

=

Sm V m

=

0.625∠36.870 1

=

0.625∠36.87 0 p.u [1 mark]

Page10of17


Generator emf behind transient reactance is,

E g '

= Vm +

' j ( X dg

+

X l )I m

= 1∠ 0 +

j (0.25 + 0.1) *0.625∠ 36.87° = 0.8688 + j 0.175 = 0.8862∠11.39° p.u [2 marks]

Motor emf behind transient reactance is, ' Em ' = Vm − j ( X dm ) I m = 1∠0 − j (0.2) *0.625∠36.87 ° = 1.075 − j 0.10 = 1.0796 ∠ − 5.32 ° p.u

[2 marks]

During fault, the generator short circuit current is, '

I g

=

'

0.8862∠11.390 = ' j ( X dg j 0.25 ) E g

= 3.545∠ − 78.61° p.u

[1 mark] During fault, the motor short circuit current is,

I

' m

=

E m '

'

=

1.0796∠ − 5.320

j ( X dm + X l )

= 3.5988∠ − 95.32° p.u

j 0.30

[1 mark] The short circuit current is

I f' = I g' + Im' = 3.545∠ − 78.61° + 3.5988∠ − 95.32° = 7.0679∠ − 87.03° p.u

Questions3[25marks] a) Determine the Admittance Matrix

Converting to Admittance values Item

[2 marks]

X1

Y1

G1

j0.25

-j4

G2

j0.2

-j5

T1

j0.1

-j10

T2

j0.1

-j10

L12

j0.12

-j8.333

L13

j0.14

-j7.143

L23

j0.16

-j6.25

L34

j0.18

-j5.556

Page 11of17


b) Calculate the fault current at Bus 4 in per unit

[4 marks]

Calculating Z 44: Item

X1

G1

j0.25

G2

j0.2

T1

j0.1

T2

j0.1

L12

j0.12

L13

j0.14

L23

j0.16

L34

j0.18

Calculating Fault Current at Bus 4

c) DeterminethevoltagesatBus1,Bus2,Bus3andBus4duringfault marks]


[10


d) Calculatethe faultcurrentflowinginline2–3,1–3and3-4

[6marks]

e) Atreefallsdownandshortsall3phasesinatransmissionlinetoground.Isthisa bolted or non-bolted fault?. Explain why?

It is a non-Bolted Fault because the tree has impedance.

Question4[25marks] a)

Positivesequencenetwork

Negativesequencenetwork


[3 marks]


Zerosequencenetwork

b) 0

I a

1

E a

2

= I a = I a =

1

2

0

Z B + Z B + Z B

+ 3 Z f

1

Z B = ( j 0.1 + j 0.15 + j 0.05) //( j 0.15 + j 0.1) =j0.1364 2

Z B

=

1

Z B =j0.1364

0

Z B = ( j 0.15 + j 0.1) //( j 0.15) =j0.09375 0

I a

1

2

= I a = I a =

0

I B ( F ) = 3 I a

1.0 =-j2.728.pu j 0.1364 + j 0.1364 + j 0.09375 + 0

= − j8.1844 p.u

c) PositivesequencevoltageatbusesAandB,

1

1

∆V A = ( j 0.1 + j 0.15) * I B ( F )(

1

1

0.25 ) = j 0.25 * j 2.728 * 0.4545 =-0.31 0.55

1

V A ( F ) = V A (0) + ∆V A =1.0–0.31=0.69p.u 1

1

∆V B = ( j 0.1 + j 0.15 + j 0.05) * I B ( F )(

0.25 ) =-0.372 0.55 Page 14of17


1

1

1

V B ( F ) = V B (0) + ∆V B =1.0–0.372=0.628p.u

NegativesequencevoltagebusesAandB, V A2 ( F ) = V A2 (0) + ∆V A2 =0.0–0.31=-0.31 2

2

2

V B ( F ) = V B (0) + ∆V B =0.0–0.372=-0.372

ZerosequencevoltagebusesAandB,

0

0

∆V A = ( j 0.15) * I B ( F )(

0

0

0.25 ) =-0.186 0.55

0

V A ( F ) = V A (0) + ∆V A =0.0–0.186=-0.186p.u

0.25 ) =-0.31p.u 0.55 0 0 0 V B ( F ) = V B (0) + ∆V B =0.0–0.31=-0.31p.u 0

0

∆V B = ( j 0.15 + j 0.1) * I B ( F )(

012

I AB

 V A0 ( F ) − V B0 ( F )   − 0.186 − (−0.31)     0  z AB j 0.1     − j1.24  V A1 ( F ) − V B1 ( F )   0.69 − 0.628    = − j1.24 p.u =  =  1    j 0.05 z AB      1 . 24 − j  V A2 ( F ) − V B2 ( F )   − 0.31 − (−0.372)      2 j 0.05  z AB  

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1 1  abc 012 I AB ( F ) = AI AB ( F ) = 1 a 2  1 a

1  − j1.24   a − j1.24

  a  − j1.24 2

a I AB = 3( − j1.24) =-j3.72p.u b I AB = 1.24∠ − 90° + 1.24∠ − 90 + 240° + 1.24∠ − 90° + 120° =0.0p.u c I AB = 1.24∠ − 90° + 1.24∠ − 90 + 120° + 1.24∠ − 90° + 240° =0.0p.u

Question5[20marks]

AtinfinitenodeH=α Thereactancebetween1and3,X=j(0.1+0.5/2)=j0.35pu

1mar

Outputpowerofgenerator Powerangleequation: 2mar

, which is equivalent to Vtanglerelativetotheinfinitebus 2mar Theterminalvoltage:Vt=1∠20.5°pu=0.94+j0.35pu

1mar

Theoutputcurrentofthegenerator:

2mar

Thetransientinternalvoltage: Page16of17


2mar

Thetotalseriesresistance: 1mar

ThepowerangleequationrelatingE’andV: 2mar

b)&c) Theinitialoperatingangleisgivenby: 2mar

Sincebothlinesareintactwhenthefaultiscleared,thepower-angleequationbeforeandafter thefaultremainsthesame: 2mar

1mar

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Microsoft Word - EEPB383 Final Sem 2 1112 With Solution (1)

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