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

EEPB383, Semester2 2011/2012

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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.


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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:

2111 009076240 PPC .. ++=

2222 005016220 PPC .. ++=

2777 008056240 PPC .. ++=

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. [4 marks]


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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]


12

3

G1G2

T1T2

4

CB1 CB2

Figure 2


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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

Rating X0 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 [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? [3 marks]


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

Figure 3

Generators G1 and G2: 1.021 jXX == p.u; 05.00jX = p.u; Voltage: 11 kV

Transformers T1 and T2: 15.0021 jXXX === p.u; Voltage: 11/275 kV

Line: 05.021jZZ == p.u, 1.00

jZ = p.u ; 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. [5 marks]


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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. [15marks]

b) Find the initial operating angle. [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


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Formula Sheet

∑=

−=

n

i i

iDP

1 2γ

βλ;

∑

∑

=

=

+

=n

i i

n

i i

iDP

1

1

2

1

2

γ

γ

β

λ

02

2

0

=∆+∆

+∆

δδδ

π SPdt

dD

dt

d

f

H

em PPdt

d

f

H−=

2

2

0

δ

π

( )θωξ

δδδ ξω +

−

∆+= −

te dtn sin

1 2

00

te dtn n ω

ξ

δωωω ξω sin

1 2

00

−

−

∆−=

SHP

fD 0

2

πξ =

Sn PH

f0πω =

21 ξωω −= nd ξθ 1cos−=

f

aaaa

ZZZZ

EIII

3021

210

+++=== : Single line to ground fault

f

aaa

ZZZ

EII

++=−=

21

21 : Line-Line fault

f

aaa

ZZ

IZEI

30

110

+

−−= : Double Line to Ground Fault

2

112

Z

IZEI aa

a

−−= : Double Line to Ground Fault


of 17

f

f

ZZZ

ZZZ

aa

Z

EI

3

31

1

02

02

++

+

+

= : Double Line to Ground Fault

=

2

1

0

2

2

1

1

111

a

a

a

c

b

a

V

V

V

aa

aa

V

V

V

( ) add t

dd

t

dd

t

dd

asy eX

Et

Xe

XXe

XXEti

τττ δδω −−− ++

+

−+

−= sin2sin

111112)(

''0

''''0

'''

B

puS

SS = ,

B

puV

VV = ,

B

puI

II = ,

B

puZ

ZZ = ,

( )

B

BB

S

VZ

2

= , B

BB

V

SI

3=

kk

B

X

SSCC =

2

=

newB

oldB

oldB

newBold

punewpu

V

V

S

SZZ


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Solution

Question 1[15 marks] a)

11

11 018.07.6 P

dP

dC+==λ $/MWh

22

22 010.01.6 P

dP

dC+==λ $/MWh

77

77 016.05.6 P

dP

dC+==λ $/MWh

b) Neglecting losses, total power demand,

MW

PPPPPD

345

100801060401025200

...... 9321

=

++++++++=

++++=

∑

∑

=

=

+

=

7,2,1

7,2,1

2

1

2

i i

i i

iDP

γ

γ

β

λ =

)008.0(2

1

)005.0(2

1

)009.0(2

1

)008.0(2

5.6

)005.0(2

1.6

)009.0(2

7.6345

++

+++

= 7.95 $/MWh

Solve for P based on optimal dispatch,

95.7018.07.6 1 =+ P

44.691 =P MW

95.7010.01.6 2 =+ P

1852 =P MW

95.7010.05.6 7 =+ P

625.907 =P MW

c)

Set 801 =P MW, optimise 2P and 7P


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)008.0(2

1

)005.0(2

1

)008.0(2

5.6

)005.0(2

1.680345

+

++−

=λ = 7.885 $/MWh

885.7010.01.6 2 =+ P

5.1782 =P MW

885.7010.05.6 7 =+ P

56.867 =P 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)

50 36.870.625 36.87 .

0.8 100mS p u

∠ − °= = ∠ − °

×

[1 mark]

201 0 .

20mV p u= = ∠ °

[1 mark]

Before the fault,

* 000.625 36.87

0.625 36.87 .1

mm

m

SI p u

V

∠= = = ∠

[1 mark]


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Generator emf behind transient reactance is,

' '( ) 1 0 (0.25 0.1)*0.625 36.87 0.8688 0.175 0.8862 11.39 .g m dg l m

E V j X X I j j p u= + + = ∠ + + ∠ ° = + = ∠ °

[2 marks]

Motor emf behind transient reactance is,

' '( ) 1 0 (0.2)*0.625 36.87 1.075 0.10 1.0796 5.32 .m m dm m

E V j X I j j p u= − = ∠ − ∠ ° = − = ∠ − °

[2 marks]

During fault, the generator short circuit current is,

' 0'

'

0.8862 11.393.545 78.61 .

( ) 0.25

g

g

dg

EI p u

j X j

∠= = = ∠ − °

[1 mark]

During fault, the motor short circuit current is,

' 0'

'

1.0796 5.323.5988 95.32 .

( ) 0.30m

m

dm l

EI p u

j X X j

∠ −= = = ∠ − °

+

[1 mark]

The short circuit current is

' ' ' 3.545 78.61 3.5988 95.32 7.0679 87.03 .f g m

I I I p u= + = ∠ − ° + ∠ − ° = ∠ − °

Questions 3 [25 marks]

a) Determine the Admittance Matrix [2 marks]

Converting to Admittance values Item 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


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

Calculating Z44: 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 [10

marks]


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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? [3 marks]

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

Question 4[25 marks] a)

Positive sequence network

Negative sequence network


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Zero sequence network

b)

fBBB

aaaa

ZZZZ

EIII

3021

210

+++===

)1.015.0//()05.015.01.0(1jjjjjZB +++= = j0.1364

12BB ZZ = = j0.1364

)15.0//()1.015.0(0jjjZB += = j0.09375

009375.01364.01364.0

0.1210

+++===

jjjIII aaa = -j2.728 .pu

1844.83)( 0jIFI aB −== p.u

c) Positive sequence voltage at buses A and B,

)55.0

25.0)((*)15.01.0( 11

FIjjV BA +=∆ = 4545.0*728.2*25.0 jj = -0.31

111 )0()( AAA VVFV ∆+= = 1.0 – 0.31 = 0.69 p.u

)55.0

25.0)((*)05.015.01.0( 11

FIjjjV BB ++=∆ = -0.372


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111 )0()( BBB VVFV ∆+= = 1.0 – 0.372 = 0.628 p.u

Negative sequence voltage buses A and B,

222 )0()( AAA VVFV ∆+= = 0.0 – 0.31 = - 0.31

222 )0()( BBB VVFV ∆+= = 0.0 – 0.372 = -0.372

Zero sequence voltage buses A and B,

)55.0

25.0)((*)15.0( 00

FIjV BA =∆ = - 0.186

000 )0()( AAA VVFV ∆+= = 0.0 – 0.186 = -0.186 p.u

)55.0

25.0)((*)1.015.0( 00

FIjjV BB +=∆ = - 0.31 p.u

000 )0()( BBB VVFV ∆+= = 0.0 – 0.31 = -0.31 p.u

−

−

−

=

2

22

1

11

0

00

012

)()(

)()(

)()(

AB

BA

AB

BA

AB

BA

AB

z

FVFV

z

FVFV

z

FVFV

I =

−−−

−

−−−

05.0

)372.0(31.005.0

628.069.01.0

)31.0(186.0

j

j

j

=

−

−

−

24.1

24.1

24.1

j

j

j

p.u


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)()( 012FAIFI AB

abcAB = =

−

−

−

24.1

24.1

24.1

1

1

111

2

2

j

j

j

aa

aa

)24.1(3 jIaAB −= = -j3.72 p.u

°+°−∠+°+−∠+°−∠= 1209024.12409024.19024.1bABI = 0.0 p.u

°+°−∠+°+−∠+°−∠= 2409024.11209024.19024.1cABI = 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 Output power of generator Power angle equation:

, which is equivalent to Vt angle relative to the infinite bus

The terminal voltage: Vt = 1∠20.5° pu = 0.94 + j0.35pu The output current of the generator:

The transient internal voltage:

1mar

2mar

1mar

2mar

2mar


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The total series resistance:

The power angle equation relating E’ and V:

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

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

2mar

1mar

2mar

2mar

2mar

1mar

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

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