eeen60301 solutions

8/2/2019 EEEN60301 Solutions

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EEEN60301

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Number of hours

Mathematical formulae tables supplied by the Examinations Office

UNIVERSITY OF MANCHESTER

Faculty of Engineering and Physical Sciences

School of Electrical and Electronic Engineering

Paper title: Power System Modelling

DateTime

Answer All questions.

Insert any special instructions here, for example:

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


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Solutions

Question 1: Solution

(a) Describe what is meant by the dielectric permittivity of a material. In

particular discuss why it is necessary to use a complex number todefine relative permittivity and what tan refers to.

1 mark each for polarisation

1 mark for ratio of capacitance/charge storage with and without the material

1 mark for discussing in-phase current

1 mark for defining tan delta in terms of power storage or permittivity

1 mark for coherent description or mathematical represenation of real and imaginary currentsetc.

[5 marks]

(b) Explain the difference between DC conductivity and relaxation

processes.polarisation arises from a finite displacement of charges in a steady electric field and

conduction arises from a finite average velocity of motion of charges in a steady electric

field.

[2 marks]

(c) Considering a parallel-plate capacitor in which dielectric loss isdominated by DC conduction

(i) show that tan= /0r

where and 0rare the materials conductivity and permittivityrespectively and is the applied signal angular frequency.

The power loss in the dielectric is given by: C V2

tan = C0 V2r tan

If losses are only due to conduction: C0 V2r tan = I

2Rdc = V

2/Rdc

and in particular for a parallel-plate capacitor of areaA and plate separation d

[0A/d] V2r tan =V

2A/d

and so tan = /0r

1 mark for each line of this proof[4 marks]

(ii) how does the plot oftanvary with low frequencies?

Tan delta againast log(f) has a slope of -1.

[1 marks]

(d) The system of Figure Q1.1 requires a path from X to Y to operatesuccessfully. If the components A,B,C,D, or E fail they go open circuit.Each component has a reliability of 0.9, and is independent of theothers.

(i) What is the probability of all the elements working successfully atany given time?

Answer: 0.95=0.59049

[1 mark]


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(ii) What is the probability that at least 3 of the elements worksuccessfully at any given time?

Answer =0.95+50.9

40.1

1+10x0.9

30.1

2or 1-(10x0.9

20.1

3+5x0.9

10.1

4+0.1

5) =0.99144

Note the 10 comes from 5C3.

3 marks for correct answer, 2 marks if arithmatic wrong,

1 mark for only knowing this meant adding likelihood of 3, 4 or 5 working.[3 marks]

By considering separately the two possible states of element B, useconditional probability to derive an expression for system reliability interms of the reliabilities of each component RA, RB, RC, RD, and RE.

Probability of system success = (Prob of success if B works x Prob of B working)

+ (Prob of success if B fails x Prob of B failing)

If B works the system looks like:

or

and the reliability of this system is given by R(system if B works) = 1 (1-RD)(1-RE)

If B fails:

the system looks like

and the reliability of the system is given by R(system if B failed) = 1 (1-RARD)(1-RCRE)

and so the system reliability RS = RB{1 (1-RD)(1-RE)} + (1-RB){ 1 (1-RARD)(1-RCRE)}

4 marks for the condition probability statement and 3 marks for the maths[7 marks]

(iii) If each component has a reliability of 0.9, show the reliability ofthe system is XXX.

I will accept any method of getting the right answer.

Substituting the above gives: 0.89100+0.009639 = 0.98739

[2]

A

D E

C

ED

A

D E

C


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Question 2: Solution

(a) Overhead lines anddistributed circuitconsisting of two r

(L).

(i) Draw and cleaseveral repeat

Two marks for all correct: 1 mar

(ii) Referring to thdescribe the qvalues of comoverhead lines

R1: Conductor resistance essenti

inductance and because is a geo

R2: leakage current/loss through

from insulator surface currents, i

L: self inductance of conductorsand because is a geometric magn

conductor and sheath and OHL

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Figure Q1.1

cables can each be modelled by luodels. The model consists of repeasistors (R1, R2) one capacitor (C) an

rly label the arrangement of the mounits

k if one component wrongly placed, or repeats n

R 1R2 CL

e nature of the physical origins of tuantitative differences and similaritionents used in the models for tranand cables.

ally similar in each case because of current carry

etric magnetic effect.

insulation are different because leakage on over

n cables we are talking dielectric loss.

essentially similar in each case because they areetic effect: mutual indcutance is different becaus

etween all conductors an so is a bit higher.

EEEN60301

otal [25 marks]

pedt unitsd one inductor

del: include

t shown.

he modeles between themission

ing requirement

ead lines results

all long and straighte cable is between


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C: Conductor capacitance to ground, higher for cables because of simple geometry, also for relative

dielectric permittivity increases cable capacitance beyond geometric capacitance.

One mark each for sensible comment on each: two marks extra for additional comments understanding.

[2,6 marks]

(b) A simpler model can be used for a shorter line as shown in FigureQ2.1.

VrVs

IrjXLR

-jXC/2jXC/2

Is

Figure Q2.1

(i) Show that this arrangement four terminal circuit can berepresented by:

Lc

rrs

rrs

jXRZandjXYand

ZYYCZB

ZYDA

where

DICVI

BIAVV

+==

+==

+==

+=

+=

/1

411

2

rrs

rrs

Lr

rrs

crs

rs

I

ZYZY

YVI

VZY

ZIV

jXRZZYV

IVV

jXYYVYV

II

++

+=

++=

+=

++=

=++=

1241

12

or

)where(2

)/1where(22

(ii) Under what circumstances might the capacitive reactance benegligible?

For short overhead lines, say < 50 km. (one mark for each of OHL and < 50 km. (The word short in itself is not

enough for a mark)

[3,2 marks]

(c)

(i) For a full-bridge AC-DC converter, the DC terminal voltage VDC_t


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==

++

)cos(

2)sin(2

1_ t

VtdtVV sstDC

.cos22

)]cos()[cos(2

ssVV

=+= (2)

DCtDC ERIV +=_ V168180206.0 == (2)

(ii)

Since VV

V stDC 168cos22

_ ==

, (2)

778.022402

)168(cos =

=

(2)

o141= (2)

(iii): Assuming converter is lossless,

hence AC power output = DC machine power output

The DC machine power output = 180x20 202 x0.6 = 3.36kW

(2)

Total [25 marks]

Question 3: Solutions

(a)

(i) Since: 3810V3

6600V == , Xd = 5.2/ph, P=5 x10

6 , cos =0.871 (1)

Hence = 29.42 , and sin =0.49, (1)

Since Pout=3VIcos ,871.038103

1056

=I = 502A (1)

Since += IjXVE d0 = 42.29)5022.5(

3

6600+ j (1)

= 3810+j 2610(cos -j sin)=3810+j2273+ j(-j2610*0.49)


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E= 3810+ j 2273 + 1278.9 = 5089 + j 2273 = 5573 24o (1)

Hence E = 5573 V/ph (1)

(ii)

(2)

Where V=3810V, E = 5573V, I= 502A, Xd = 5.2 , =29o , and =24o (2)

(b)

(i) Since the synchronous speedp

f2S = = (2*3.14*50)/3 104.7 rad/s

or ns=60*f /p= 60*50/3=1000 rpm

The motor rotor speed nr = (1-s) ns = (1- 0.025)*ns = 975 rpm

(2)

(ii) Refer to the steady-state equivalent circuit of the induction motorbelow:

V1

jX1R1 jX2

I2I1

R2

S

The total impedance of the circuit is

Z = R1+R2/S +j(X1+X2) = 0.5+ 0.35/0.025 + j(1.2+1.2) = 14.5+ j 2.4 =14.7 9.398o.


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The terminal voltage per-phase

V1 = 415/3 = 240 0o (line-to-neutral)

So that the stator current is found as follows:

I1 = V1/Z = (240 0o) / (14.7 9.40o) = 16.33 -9.40o

(4)

(iii) The motor torque at slip of 2.5% is

mNsR

sstart =

+++

=

+++= 56.106

)2.12.1()025.0/35.05.0(

025.0/35.0)240(3

6.104

1

)X(X)/(R

)/(RV3

1

22

2

221

221

22

1

s

(3)

(iv) Since the motor power factor is cos () = cos ( 9.40o ) = 0.987lagging

The input power is given by

Pin = 3V1I1 cos(9.40o) = 3*240*16.33*0.987 11604.75 W

The stator loss is given by:

stator

losses

P = 3I12R1= 3 * (16.33)

2*0.5 400W

As the rotor current I2 = I1, hence the rotor losses

rotor

lossesP = 3I22R2= 3 * (16.33)

2*0.35 280W

The output power is found as follows.

Pout = Pin PSLP

RL -PL = 11604.75 400280 - 500 = 10424.75 W

Therefore the efficient is:

= Pout/Pin = 10424.75 / 11604.75 = 89.8%

(6)

Total [25 marks]


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Question 4: Solutions

Sync. generator G1: 75MVA, 13.2kV, sync. reactance Xd = j2.0 p.u.Generator transformer T1: 100MVA, 13.2kV/132kV, X = j0.10 p.u.Generator transformer T2: 50MVA, 13.2kV/132kV, X = j0.10 p.u.132kV transmission line L1: Z = (0.20 + j0.40)/km, length = 10km.Step-down transformer T3: 60MVA, 145kV/11.8kV, X = j0.15 p.u.Equivalent load ZL at bus 4: 55MVA at 11kV, cos = 0.95 (lagging).

Assume in your answers to the following questions, generator G1 isoperating at a terminal voltage of 13.9kV (i.e. actual Bus 1 voltage is

13.9kV), the base voltage on L1 is 132kV and the base power is 100MVA.a) If T1 & T2 are operating in parallel and using per-unit quantities

calculate the load current in per-unit.

Base V = 13.2kV @ bus 1, 132kV @ buses 2 & 3, 10.74 @ bus 4.

Base MVA = 100

V1 = 13.9/13.2 = 1.053 pu

XT1 = j0.1 pu, XT2 = j0.1 x 100/50 = j0.2 pu. XT1//XT2 = 0.0667 pu.

ZL1 = (0.2 +j0.4) x 100 / (1322/100) = 0.0115 + j 0.0230 pu.

XT3 = j0.15 x 100/60 x 1452/1322 = j0.3017 pu.

ZL = 55MVA at 11kV = 53.7MVA at 10.74kV. power factor load = 0.95 lag

ZL = (11.02/55) / (10.742/100) = 1.907 18.2 = 1.812 +j 0.596 pu

1.053 0 = IL ((1.812+0.0115) + j(0.0667+0.0230+0.03017+0.596))

IL = 1.053 0 / (1.824 + j 0.9874) = 0.5077 -28 pu

[7 marks]

(i) voltage in kV at the load (bus 4)

VL = (0.5077 -28) x (1.907 18.2) = 0.968 -10 pu = 10.39kV.[4 marks]

G1

T2

132kV

L1

T3

Bus 4

ZL

T1

11kV

Bus 3

Bus 2Bus 1

132kV

13.2kV


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(ii) real and reactive power in kW & kVAr received by the load.

P = 0.968 x 0.5077 cos (18.2) = 0.467 pu = 46.7MW.

Q = 0.968 x 0.5077 sin (18.2) = 0.153 pu = 15.3MVAr

[4 marks]

b) If transformer T1 has been disconnected from the network (i.e. only T2supplies the load), calculate voltage in kV at the load (bus 4). Finally,discuss any problems that might be experienced if the network isoperated according to this scenario.

IL = 1.053 0 / (1.824 + j 1.121) = 0.492 -32 pu

VL = (0.492 -32) x (1.907 18.2) = 0.938 -14 pu = 10.07kV.

[5 marks]

Discussion:

Voltage at load is 10.07kV or 0.938pu. The nominal voltage is 11kV, hencevoltage is 9% below nominal value. This is unacceptable, hence iftransformer T1 is out of service, load should be disconnected.

Alternative approach is to increase terminal voltage of generator, problemneed to increase voltage by about 4% to bring terminal voltage toacceptable value (hence require approx 14.4kV). This might be acceptable

if no loads are connected to bus 1.The load at 10.07kV is 46.1MVA. This has to be supplied by transformerT2. However if voltage at bus 1 is 1.053pu, T2 needs to supply 0.492 x1.053 = 0.518 pu = 51.8 MVA.

The transformer is rated at 50 MVA. Hence load is 3.6% above rating. Thisis acceptable for a short time.

However, if voltage is increased to 1.095pu, load is about 7% aboverating. During an emergency condition , this might be acceptable for avery short time, but longer term will damage the transformer.

I expect the student to discuss some of the issues.

[5 marks]

Total [25 marks]

END OF EXAMINATION PAPER

eeen60301 solutions

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