mock exam questions & solutions

8/2/2019 Mock Exam Questions & Solutions

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Practice Exam Paper Electrical Power Systems Engineering

Please note that this paper is out of 88. The use of past exam questions whichhave been judged to be too long has meant that some components of thequestion have been removed.


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Two hours

Mathematical formulae tables supplied by the Examinations Office

UNIVERSITY OF MANCHESTER

Faculty of Engineering and Physical SciencesSchool of Electrical and Electronic Engineering

Practice Examination

2nd December 201109:45 11:45

Answer All questions.

Electronic calculators may be used, provided that they cannot store text.


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Question 1

(a) Describe the formulation of the fast decoupled power load flowalgorithm stating its main features and key attributes.

The Jacobian matrix which is central to Newton Raphson algorithm has theform shown below:

V

QQ

V

PP

J

Inversion of the full Jacobian is time consuming. The decoupled versionexploits the weak coupling between real power flow and voltage magnitudeand between reactive power flow and voltage angles in AC power systems

and thus setting allVP

and

Q elements to zero. This leads to the form

of Jacobian shown below:

V

Q

P

J

0

0

This allows the two remaining smaller sub-matrices to be invertedseparately. This speeds up each iteration hence the problem convergesfaster although the number of iterations increases. In the fast decoupledload flow, further speed gains can be made by not recalculating the blocks

V

Qand

P

.

For a flat start V=1 and 0 , and for fast decoupled load flow we ignore

line charging giving:

B

B

V

Q

P

''

'

Solution is given by:

k

k

k

k

Q

P

B

B

V1''

1'

1

1

[5 marks]


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(b) The node specification for a load flow study, and branch dataof a three-bus power system are given in Table Q1.1 and TableQ1.2 respectively.

Table Q1.1 Node specification

Node PG pu QG pu PL pu QL pu

1 (Slack) - - 0 0

2 (PV bus) 0.5 - 0.3 0

3 (PQ Bus) 0 0 1.0 0.5

Table Q1.2 Branch data

BranchSendingend node

Receivingend node

Xpu

Bpu

1 1 2 0.20833 0.04

2 2 3 0.20833 0.04

(i) Establish the bus admittance matrix for this systemThe network takes the following form:

Node

1

Node

2

Node

3

LoadP = 1.0 pu

Q = 0.5 pu

pujX 20833.012

pujX 20833.023

puP

puV

5.0

0.12

rad

puV

0

0.1

1

1

10Y

LoadP = 0.3 pu

Q = 0.0 pu

1

20Y 2

20Y

30Y

pujBYYYY

Y

pujjj

YY

02.02

013

8.420833.0

1

20833.0

12312

30

2

20

1

2010

Hence by inspection the Y matrix is:

78.48.40

8.456.98.4

08.478.4

jY

[4 marks]


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(ii) Derive the mismatch equations taking node 1 as the slack.The mismatch equations are derived from the active and reactive powerbalance equations at the where P and Q are specified; hence we have:

P2 4.8V2V1 sin21 10 V2 V2 sin22 4.8V2 V3 sin23

P3 0V3V1 sin31 4.8V3 V2 sin32 5 V3 V3 sin 33

Q3 0V3V1 cos 314.8 V3V2 cos32 5 V3 V3 cos 33

Given the following and replacing for known variables radpuVV 0,0.1121

sin11 sin22 sin33 sin0 0

cos11 cos22 cos33 cos0 1

The mismatch equations:

-0.2 4.8sin 2 4.8V3 sin 2

1.0 4.8V3sin

0.5 4.8V3cos 5V3

3-( )

2

2 3- )(

2 3

-( )

P2

P 3

Q3

[4 marks]

(iii)Establish elements of the Jacobian matrix for flat startconditions of a load flow calculation.

The general Jacobian matrix structure is as follows:

3

3

3

3

2

3

3

3

3

3

2

3

3

2

3

2

2

2

V

QQQ

V

PPP

V

PPP

J

By taking partial derivatives of the mismatch equations, we obtain:

)(4.789.56)(4.78)(4.78

)(4.784.78)(4.78

)(4.78)(78.478.4

2333323

233323

3233232

CosVSinVSinV

SinCosVCosV

SinCosVCosVCos

J

4.78

Therefore, to find elements of the Jacobian for with flat start we set

00.1323 andpuV in the above matrix, to give:

78.400

078.478.4

078.456.9

J

[5 marks]


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(iv)Apply the fast decoupled Newton-Raphson technique todetermine the voltage magnitude and angle at the loadbus and voltage angle at the generator bus after 1iteration.

The above Jacobian matrix for the decoupled load flow is as follows:

''

'

B

B

J

78.400

078.478.4

078.456.9

The matrix above can be inverted easily by partitioning into two submatrices one 2 by 2 and one 1 by 1, which can then be inverted to give:

k

k

k

k

Q

P

B

B

V1''

1'

1

1

The voltage at node 3 is found by the following formula Newton Raphsontechnique:

The mismatches for flat start are: 5.0;0.1;2.0 03

0

3

0

2 QPP

5.0

0.1

2.0

2092.000

04184.02092.0

02092.02092.0

*

3

0

3

0

2

1

3

3

2

Q

P

P

J

V

1046.0

3766.0

1674.0

3

3

2

V

There

8954.0

3766.0

1674.0

1046.0

3766.0

1674.0

0.1

0

0

3

3

2

V

Therefore the voltage at node 3 after iteration 1 = 0.8954 pu anglenegative 0.1674 radians and the voltage angle at node 2 is negative0.3766 radians

[7 marks]

Total [25 marks]


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Question 2

(a) Two switches are in parallel. As long as both of them can beopened simultaneously, the rest of the network can operatesuccessfully. For example if one or both is stuck in the open

position, the system can still be operational, but if one is fusedclosed the system fails. The switches are identical. Each hasthe probability of correct operation of 97%. The probability ofeach failing open circuit is 1% and failing closed circuit is 2%.

(i) List the potential states of the system identifying thosewhich work and those which do not

By state enumeration: Let us call the situations W-working, O-stuck open,Sstuck Shut.

We can have WW, WO, OW, WS, SW, OS, SO, SS, OO.

These give G G G B B B B B G where G is works, B is fails

Successful operation if WW or WO or OW or OOFails if WS or SW or SS or SO or OS

[4 marks]

(ii) What is the likelihood of the system being operational?

Successful operation if WW or WO or OW or OO =.97*.97+.97*.01+.01*.97+.01*.01 = 0.9604

[4 marks]

(iii)What is the main assumption in your calculation?Independence of failure likelihood

[2 marks]

(b) A set of 40 switches is tested to failure in the laboratory.

After only 30 have failed the test was terminated. Had thetest run until all the samples failed they would have fit aWeibull distribution with a shape parameter of 4.

(i) Why is it not appropriate to determine the Weibullparameters treating the 30 failures as a complete dataset?

You systematically have missed the longest surviving data. If you considerthe prob density function, you have missed the right-hand end (rightcensored data). Your data set is not a representative sample.

[3 marks]


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(ii) If Weibull parameters are determined from the 30 sampleswhat is the implication for the shape parameter obtained(explaining why)?

Since you have missed the longer life samples, the pdf is shifted to

the right. The consequence is a peakier distribution than should bederived and so the shape parameter determined is larger than 4.

[3 marks]

(c) A manufacturer wishes to design a new single core HV powercable to be used as part of a three-phase circuit. The cable willuse an insulation thickness of 15 mm. Using equations 2.1 orotherwise and given a working electric field of 10 MV/m and arelative permittivity of 2.5 for the cable insulation:

)2.2()/ln(

2)1.2(

)/ln(

1

rRC

rRx

VE

ro

x

(i) Calculate the maximum voltage at which this cable shouldoperate. State this voltage as the RMS line voltage.

The working voltage will depend on the cable size. Take a typical size of185mm2 and a packing factor of 0.9.

r=sqrt(185mm2/0.9/)=8.1mm

V=Exrxln(R/r) = 10 x 106 x 8.1mm x ln(23.1/8.1) x 10-3 = 85.9kV.

This would be the peak phase to earth voltage.(4 marks)

(ii) Calculate what the capacitance of this cable will be andthe expected dielectric losses for a frequency of 50Hz andtaking the tan of the insulation to be 0.0004.

Using capacitance equation:

C=2 x 3.14159 x 8.854 x 10-12 x 2.5 x 1/ln(23.1/8.1)=146pF/m

Losses=V2wCtandelta=(48.3kV/2)2x2xpix50x146pFx0.0005=85mW/m

(3 marks)

iii. Qualitatively state the effect of an increase in insulationthickness on the current carrying capability of the cable.

An increased insulation thickness would increase T1 the thermalresistance between the conductor and the sheath. This would mean that


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the conductor temperature would be increased for a given value ofcurrent, i.e. the current carrying capacity would be reduced.

[2 marks]

Total [25 marks]


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Question 3

(a) State the main objectives of reactive compensation devices in:(i) Electric power transmission systems and(ii) Electric power distribution systems.Transmission networks

Maintain a uniform voltage profile in the network Compensation of reactive losses Enhancement of stability Reactive power reserve

[2 marks]Distribution networks

Power-factor correction

The reduction of active and reactive power losses Feeder voltage control Balancing of non-symmetric loads

[2 marks]

(b) Name two reactive compensation devices applied intransmission systems stating the advantages anddisadvantages of each device.

Marks given for any 2 devices complete with an advantage and

disadvantage.

Series capacitorsAdvantages:

Used for improving transient stability and increasing active power flows of longtransmission linesSeries capacitor is self-regulating

Disadvantage: Can cause sub-synchronous resonanceShunt capacitors

Advantages:

Supply reactive power and boost local voltagesTheir principal advantages are low cost and flexibility of installation andoperation

Disadvantage:Reactive power output is proportional to the square of the voltage.Consequently, the reactive power output is reduced at low voltages when it islikely to be needed most

Static Var Systems (SVS)SVS are shunt-connected static generators and/or absorbers whose output arevaried so as to control specific parameters of the electric power system

Advantage:


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SVSs application include enhancement of stability, prevention of voltagecollapse, control of overvoltages.dynamic voltage control, where shunt reactors or capacitors are steady statevoltage control facilities

Disadvantage:

possibility of instability when the SVS is pushed to its limitSynchronous condenser

Advantages:during power swings there is an exchange of kinetic energy between asynchronous condenser and the power system and a synchronous condensercan supply a large amount of reactive powerit has about 10% to 20% overload capability for up to 30 minutescopes much better with low system voltagestheir reactive power production is not affected by the system voltagecan generate and absorb MVars

Disadvantageis that they are high capital and operating costs

[4 marks]

(c) A distribution system planner is considering installing a shuntcapacitor C at the end of the 33kV feeder shown in Figure Q4. Node 1is the bulk supply point while a three-phase load SL equal to 25 MVAat 0.85 power factor lagging is connected at node 2. The systemparameters are given in Table Q4.

.Figure Q4 Radial distribution system

Table Q4 System parameters

Voltage at node 2(line voltage)

FeederLength

R(per phase)

X(per phase)

33kV 20km 0.2 per km 0.3 per km

After installation of a three-phase 10MVar shunt capacitor at node 2in the system in Figure Q4 compute:


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(i) The voltage magnitude and angle at node 1(ii) Active power, PBSP and reactive power QPSP node 1

Base values MVA base 100 MVA

V base 33 kV

Current base = 1.7495 kA =

Z base 10.89 ohms

Load at node 2

25 MVA at PF = 0.85

P 21.25 MW

Q 13.17 Mvar

P per unit 0.2125

Q per unit 0.1317

V2 (kV) 33 kV

V2 per unit 1.00 per unitImpedance

R

ohms per km

X

ohms per km

line length

(km)

0.2 0.3 20

R X

actual values ==> 4 6

per unit ==> 0.36731 0.55096

LLbase

base

V

MVA

*3


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Total [13 marks]

With shunt capacitorLoad

Equivalent SL

25 MVA at PF 0.85 21.49 0.989

P 21.250 MW

New Q w ith shunt capacitor 3.170 Mvar Shunt capacitor = 10 Mvar

P per unit = 0.213 Q per unit = 0.032

Current, I = (P - jQ) / V

Ireal Iim Magnitude ngle (degrees)

per unit ==> 0.213 j -0.032 0.21485 -8.5

actual ==> 0.372 j -0.055 0.37589 -8.5

Voltage drop = I*(R+jX)

delta V (per unit) = 0.09552 j 0.10544 0.14227 47.8

delta V (actual) = 1.81983 j 2.00886 2.71059 47.8

V1 per unit = 1.09552 j 0.10544 1.10058 5.5

V1 actual = 20.872 j 2.009 20.969 5.5 [3 marks]

Losses

Real power loss 0.0170 1.6955

Reactive power loss 0.0254 2.5433

per unit actual

P Sending end = 0.2295 22.946 23.65 14.56698

Q sending end = 0.0571 5.713 [2 marks]

Sending end power = P & Q load + P & Q power loss

New PF at node 2

Sending end voltage V1 = V2 + voltage drop

RI2

XI2


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Question 4

The power system shown in Figure Q4.1 has the following ratedparameters for the generator, transmission lines and transformers:

Generator G1: XG1= 0.03 pu VG1= 12.1 kV SG1 = 150 MVA H=4.5 sTransformer T1: XT1 = 0.08 pu 11 kV/110 kV ST1 = 160 MVATransformer T2: XT2 = 11% 115 kV/400 kV ST3 = 150 MVALine L1: Rated voltage 110kV XL1 = 48.40

Line L2: Rated voltage 110kV XL2 = 72.6

All per-unit (pu) and % values above are based on the individual ratings ofequipment. The system power base is SB= 100MVA.

The generator delivers power P=120 MW at 0.8 lagging power factor to the

grid during the pre-fault condition. The grid voltage is Vinf= 400 kV.

G1 11 kV T1

AB 110kV

L1 L2

C 110kVT2

Vinf(400kV)S=P+jQ

Grid

Figure Q4.1


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(a) Draw phasor diagrams of the whole system clearly indicating,voltage and current phasors and all relevant angles. Commenton the flows of real and reactive power and the sign of rotorangle. Is the generator over-excited or under-excited?

[4 marks]

(b) Write the formula for the quasi steady state approximation offault current and define each of the variables (parameters)involved.

''')'''()'( dd

T

t

T

t

sssseIIeIIII

[2 marks]Iss rms value of steady state current

I rms value of transient state currentI rms value of sub-transient state currentt - timeTd transient short circuit time constantTd sub-transient short circuit time constant

[1 marks]

cos -laggingQ>0, production of QP>0Generator overexcited>0


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(c) Write a classical (constant flux linkage) model of asynchronous generator used for power system dynamicstudies and define all parameters involved.

[2 marks] rotor speed rotor angle

0 synchronous speedH - inertia constantTmmechanical torqueTeelectrical torque

KDdamping constant[2 marks]

(d) State the difference between rotor angle stability and voltagestability of a power system.

Rotor angle stability: The ability of interconnected synchronous machines of a powersystem to remain in synchronism. (Study of electromechanical oscillations inherent topower system is required.)

Voltage stability: The ability of a power system to maintain steady acceptable

voltages at all buses in the system under normal operating conditions and after beingsubjected to a disturbance. (The main factor is inability of power system to meet thedemand for reactive power.)

[4 marks]

(e) Following a three-phase fault on line L2 close to bus C (see FigQ4.1), the faulted line is disconnected and the systemcontinues to operate with only one line (L1) in service. Usingthe equal area criterion determine the critical fault clearingtime to ensure that the system retains stability following the

fault.

XG= 0.03 (100/150)(12.1/11)2= 0.0242 pu

XT1 = 0.08(100/160) = 0.05 puXT2 = 0.11(100/150)(115/110)

2 = 0.08 puZB = (110*10

3)2/(100*106) = 121 ohmXL1 = 48.4/121 = 0.4 puXL2 = 72.6/121 = 0.6 pu

VB1 = 11kV

VB2 = 110kVVB3 = 110(400/115) = 382.6kV

)1(

))1((2

1

0

p

KTTHp Dem


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Vinf= 400/382.6 = 1.045 puS = P+j*P*tg = (1.2 + j*0.9)puPm = 1.2 pu

[3 marks]Pre fault:

Xe = (0.0242+0.05)+0.4||0.6+0.08=0.3942 puEg= V+j* Xe*(S

*/ Vinf*) = 1.456618.1o

0 = 18.1o = 0.316 rad

Pe0 = [(1.4566*1.045)/0.3942]sin = 3.86 sin[2 marks]

During fault:Pe1 = 0

Post fault:Xe2 = (0.0242+0.05)+0.4+0.08=0.5542 pu

Pe2 = [(1.4566*1.045)/0.5542]sin = Pe2m sin = 2.7466 sinPe2m = 2.746620 = arcsin(1.2/2.7466)=25.9

o2 = 180

o 25.9o = 154.1 o = 2.689 rad[2 marks]

From equal area criterion:Pm(2 - 0)= Pe2m(coscr - cos2)

coscr =[1.2(2.689 - 0.316) + 2.7466*(-0.9)]/2.7466 = 0.1368cr = 82.14

o

=cr-o= 82.14o

-18.1o

= 64.04o

t2=(*4*H)/[18000*(1.2-0)]= (64.04o*4*4.5)/(18000*1.2)

t=231ms [3 marks][10 Marks]

Total [25 Marks]

END OF EXAMINATION PAPER

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