Sample/practice exam 13 June 2019, questions PDF

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REVISION ELECTRICAL POWER SYSTEMS REVISION PROBLEMS THE REVISION PROBLEMS BELOW ARE TAKEN FROM TUTORIALS AND ARE ONLY A REFLECTION OF THE SCOPES OF THE UNIT. Problem 1 All the examples in Topics Lecture Notes and assignment problems. Problem 2 Model the transmission lines as inductors, and denote as...

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REVISION

ELECTRICAL POWER SYSTEMS REVISION PROBLEMS  THE REVISION PROBLEMS BELOW ARE TAKEN FROM TUTORIALS 2-10 AND ARE ONLY A REFLECTION OF THE SCOPES OF THE UNIT.

Problem 1 All the examples in Topics 2-6 Lecture Notes and assignment problems. Problem 2 Model the transmission lines as inductors, and denote 𝑆𝑖𝑗 as the complex power flowing between bus i and bus j, with 𝑆𝑗𝑖 = −𝑆𝑗𝑖∗ . Compute S13, S31, S23 and SG3 in the figure below, using power conservation law (KCL).

[Answer]

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REVISION Problem 3 Given the impedance diagram of a simple system as shown in the Figure, draw the admittance diagram for the system and develop the 4 × 4 admittance matrix by inspection.

1

[Answer]

Y33  y13  y 23  y 34

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REVISION Problem 4 A 250 MVA, 24 kV, 60 Hz, three-phase alternator (synchronous machine) is operating at rated voltage, rated frequency, and rated apparent power with a terminal power factor of 0.8 lagging. Armature (stator) coil resistance is negligible. The internally generated voltage E f  20 kV . a) Draw the per phase equivalent circuit. b) Determine the power angle δ, and c) the value of synchronous reactance X s . [Answer] a) Per phase equivalent circuit

b) Use the terminal (line to neutral) voltage V t as the reference. The armature current is Ia 

S 3VL

  cos 1  PF  

250  106 3  24,000 

  cos 1  0.8  6014.2  36.87 A

Since the electric power converted must equal output power, 3E f I a cos      3Vt I a cos

where φ is the phase of I a and the power angle Vt cos    E f 

  cos1 

 24,000  0.8    3   cos 1  0.8  19.47   cos 1  20,000    

c)

E f  Vt  j I a X s

From

Xs 

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E f V t jIa

24,000  0 3   1.38  1 90 6014.2  36.87  20,00019.47 

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REVISION Problem 5

Shown above is a power system with three loads operating at 480 V. Load 1 is an induction motor consuming 100 kW at 0.78 PF lagging, load 2 is an induction motor consuming 200 kW at 0.8 PF lagging, and load 3 is a synchronous motor with real power consumption 150 kW. a) If the synchronous motor is adjusted to 0.85 PF lagging, what is the line current? b) If the synchronous motor is adjusted to 0.85 PF leading, what is the line current? c) Assuming that the line losses are PLL = 3IL2RL, how do these losses compare in the two cases? [Answer] a) The real power of load 1 is 100 kW, and the reactive power of load 1 is

Q1  P1 tan 1  100 tan  cos1 0.78  80.2kVAR The real power of load 2 is 200 kW, and the reactive power of load 2 is

Q2  P2 tan  2  200 tan  cos1 0.8  150 kVAR The real power of load 3 is 150 kW, and the reactive power of load 3 is

Q3  P3 tan 3  150 tan  cos1 0.85  93 kVAR The total real load is Ptot  P1  P2  P3  100  200  150450kW

The total reactive load is Qtot  Q1  Q2  Q3  80.2  150  93kVAR

The equivalent system PF is

   1Q 1 323.2   0.812 lagging PF  cos   cos  tan  tot   cos  tan  Ptot  450    Jingxin Zhang, 2019

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REVISION The line current is

IL 

Ptot  3VL cos

450000  667 A 3  480  0.812

b) The real and reactive powers of loads 1 and 2 are the same. The reactive power of load 3 is

Q3  P3 tan 3  150 tan   cos1 0.85   93 kVAR The total real load is Ptot  P1  P2  P3  100  200  150450kW

The total reactive load is Qtot  Q1  Q2  Q3  80.2  150  93kVAR

The equivalent system PF is

   1Q 1 137.2   0.957lagging PF  cos   cos  tan  tot   cos  tan  Ptot  450    The line current is

IL 

Ptot  3VL cos 

450000  566 A 3  480  0.957

c) The transmission line losses in the first case are PLL  3 I L2 RL  1334667RL

The transmission line losses in the second case are

PLL  3 I L2 RL  96168 RL Notice that the transmission power losses are 28% less in the second case, while the real power supplied to the loads remains the same. Problem 6 A synchronous generator (SG) with synchronous reactance XS = 0.32 per unit is connected to an infinite bus (power grid) through a transmission line with reactance XL = 0.2 per unit. The bus voltage Vt = 10° per unit and the SG delivers real power P = 1 per unit at 0.95 PF lagging. a) Draw the per phase equivalent circuit. Determine b) the current Ia following from the SG to the infinite bus, c) the internal voltage Ef of the SG, d) the equation for the electric power delivered by the SG versus its torque angle δ, and e) the torque angle δ at which the SG delivers the electric power. Note: Per unit values are normalized physical quantities and are dimensionless. For details, see Topic 3 lecture notes. Jingxin Zhang, 2019

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REVISION [Answer] a) Per phase equivalent circuit

b) From the circuit diagram, the current Ia from SG to the infinite bus is

P  cos (PF) 1  cos 0.95 Ia    1.05263  18.195 A per unit Vt (PF) 1 0.95 1

1

c) The SG’s internal voltage Ef is

E f  Vt  j(X S  X L ) I a  10  j0.52(1.05263  18.195 )  1 0  0.54737 71.805  1.1709  j 0.5200  1.2812 23.946 V per unit c) Using P 

Vt E f sin  XS  XL P

we have

Vt Ef sin  (1.0)(1.2812) sin    2.4638sin  XS  XL 0.52

d) Since SG delivers P = 1 W per unit to the infinite bus, we have

P  2.4638sin   1 1   sin 1  23.946 2.4638 which is the angle between Ef and Vt . Problem 7 For the system shown, draw an impedance diagram in per unit, by choosing 100 kVA as the base kVA and 2400 V as the base voltage for the generators. Draw the per phase equivalent circuit.

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REVISION

[Answer]

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REVISION Problem 8 The single-line diagram and equipment ratings of a three phase electrical system are given below. Draw the network circuit diagram for the system using a 1000MVA base, and a 765kV base in the zone of line 1-2. Neglect the effects of -Y transformer phase shifts.

Transformers:  T1 : 1000MVA,  T2 : 1000MVA,  T3 : 500MVA,  T4 : 750MVA, Transmission Lines :

15 kV  / 765 kV Y , X = 0.1 p.u. 15 kV  / 765 kV Y , X = 0.1 p.u. 15 kV  / 765 kV Y , X = 0.12 p.u. 15 kV  / 765 kV Y , X = 0.11 p.u.

 1-2 : 765 kV, X = 50 ,  1-3 : 765 kV, X = 40 ,  2-3 : 765 kV, X = 40 , Synchronous Generators :    

G1 : 1000MVA, G2 : 1000MVA, G3 : 500MVA, G4 : 750MVA,

15 kV, 15 kV, 13.8 kV, 13.8 kV,

X = 0.18 p.u. X = 0.20 p.u. X = 0.15 p.u. X = 0.30 p.u.

[Answer] The networks circuit diagram for the system is shown below. The per unit impedance values are calculated as follows: Sbase  1000 MVA VbaseHV  765kV

Zone of Transmission Lines.

VbaseLV  15kV

Zone Generators.

ZbaseHV 

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VbaseHV 2  765kV 2 Sbase

1000MVA

 585.23

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REVISION

S base 1000MVA   0.7547 kA 3VbaseHV 3 765kV  The per unit impedances of the generators are then given by: I baseHV 

X G 1  0.18 p.u.

X G 2  0.20 p .u . 2

13.8   1000  X G 3  0.15       0.2539 p.u.  15   500  2

13.8   1000  X G 4  0.3       0.3386 p.u.  15   750  The per unit impedances of the transformers are then given by:

X T1  0.1 p. u. X T 2  0.1 p.u. 2

 1000     0.24 p.u.  500 

2

 1000     0.1467 p .u .  750 

 15  X T 3  0.12    15 

 15 X T 4  0.11   15

The per unit sequence impedances of the transmission lines are then given by:

X 12 

50  0.08544 p .u . 585.23

X 13 

40  0.06835 p .u . 585.23

X 23 

40  0.06835 p.u. 585.23

Bus 1 j0.18pu

EG1

XG1

j0.06835pu

j0.1 pu

XT1

j0.24pu j0.254pu

X13_1 j0.08544 pu

j0.06835pu

X12

X23 Bus 2

j0.1pu

XT2

Bus 3

XT3 XT4

j0.147pu

XG3_

XG4_ j0.339pu

EG4 j0.20pu

XG2

EG3

Network circuit diagram

EG2 Jingxin Zhang, 2019

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REVISION Problem 9 Consider a three-phase generator rated 300 MVA, 23 kV, supplying a system load of 240 MVA and 0.9 power factor lagging at 230 kV through a 330 MVA, 23 / 230 Y kV step-up transformer with a leakage reactance of 0.11 per unit. (a) Neglecting the exciting current and choosing base values at the load of 100 MVA and 230 kV, find the phasor currents IA, IB and IC supplied to the load in per unit. (b) By choosing the load terminal voltage VA as reference, specify the proper base for the generator circuit and determine the generator voltage V as well as the phasor currents Ia, Ib, and Ic, from the generator. (Note: Take into account, the phase shift of the transformer.) (c) Find the generator terminal voltage in kV and the real power supplied by the generator in MW. (d) By omitting the transformer phase shift altogether check to see whether you get the same magnitude of generator terminal voltage and real power delivered by the generator. [Answer]

−

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REVISION

I a  2.4  55.84 ;I b  2.4  175.84 ;I c  2.4 64.16

Problem 10 In the figure below, the three-phase transformer bank is made up of three identical single-phase transformers, each specified by Xl = 0.24 Ω (on the low-voltage side), negligible resistance and magnetizing current, and turns ratio N2/N1 = 10. The transformer bank is delivering 100 MW at 0.8 PF lagging to a substation bus whose voltage is 230 kV. a) Determine the primary current magnitude, primary voltage (line-to-line) magnitude, and the threephase complex power supplied by the generator. Choose the line-to-neutral voltage at the bus as the reference. Account for the phase shift. b) Find the phase shift between the primary and secondary voltages. To rest of the system Generator

Y

Step-up transformer bank

230 kV substation bus

[Answer] a) The reactance of the transformer (at each phase) at low voltage side is given as Xl= 0.24 Ω. This low-voltage (primary) side is connected as ∆ and the reactance is converted to Y value as X 0.24 X l   l   0.08 Ω. 3 3 Jingxin Zhang, 2019

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REVISION As in ∆, the line-to-line and phase voltages are same but in Y, these are not same. Therefore, the transformer turns ratio will be modified as

Van 3

N2 V  10  an  10 3 N1 Van



Van N  2  10 3 N1

Finally, the per phase diagram can be drawn as follows jXg

Ia

j0.08

I a

+

+

+

Ea

Van

V

_

_

_

+

ej30

Van _

1 : 10 3

Transformer

Generator

Bus

The current flowing through the line can be calculated as (as we did in tutorial 3)

S   cos 1 ( pf ) 3V L

I a 

In this problem, P  100 106 W, pf  cos  0.8 , V L  230 10 3 V and we know P  S cos  100 10 6 W

S  I a 

100 10 6  125 10 6 VA 0.8

125 10 6   cos 1 (0.8)  313.8  36.87 A 3 3  230 10

By considering the phase shift between primary and secondary, we can write the primary current as  j30  j 30   I a  10 3e I a  10 3  e  313.8  36.87  5435  66.87 A

Hence, the magnitude of primary current is 5435 A. The primary phase voltage V an can be written as Van  V  j 0.08I a

230  103

Here, V 

Van 10 3

e j 30 

3 10 3





  30  7666.7  30

Therefore, Jingxin Zhang, 2019

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REVISION Van  7666.7  30  j 0.08  5435  66.8  7666.7  30  434.8 23.13  7935.7 -27.4905 V

Therefore, the magnitude of the line-to-line primary voltage  3  7935.7  13.75 kV The three-phase complex power supplied by the generator is

S3   3Van Ia *  3(7935.7 -27.4905 )(543566.87 )  129.4 39.3795 MVA b) The secondary phase leads the primary phase by 27.4905 and this phase shift applies line-toneutral (phase) as well as line-to-line voltages. Problem 11 A 200 km, 230 kV, 60 Hz three-phase line has a series impedance z = 0.08 + j0.48 Ω/km and a shunt admittance y = j3.33 × 10-6 S/km. At full load, the line delivers 250 MW at 0.99 p.f. lagging and at 220 kV. Using the nominal π circuit, calculate (a) the ABCD parameters, (b) the sending-end voltage and current, and (c) the percent voltage regulation.

[Answer]

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REVISION Problem 12 A 400 km, 500 kV, 60-Hz uncompensated three-phase line has a series impedance z = 0.03 + j0.35 Ω/km and a shunt admittance y = j4.4 × 10-6 S/km. Calculate (a) Zc, (b) γl, and (c) the exact ABCD parameters for the line. [Answer]

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REVISION Problem 13 At full load, the line in Problem 12 delivers 1000 MW at unity power factor and at 475 kV. Calculate (a) (b) (c) (d) (e)

the sending-end voltage, the sending-end current, the sending-end power factor, the full load line losses, and the percent voltage regulation.

[Answer]

Using the results of Problem 12

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REVISION Problem 14 A 230 kV, 100 km, 60 Hz, three-phase overhead transmission line with a rated current of 900 A/phase has a series impedance z = 0.088 + j0.465 Ω/km and a shunt admittance y = j3.524 μS/km. (a) Obtain the nominal π circuit in normal units and in per unit on a base of 100 MVA (three phase) and 230 kV (line-to-line). (b) Determine the three-phase rated MVA of the line. (c) Compute the ABCD parameters. (d) Calculate the SIL. [Answer]

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REVISION Problem 15 A 400 km, 500 kV, 60-Hz uncompensated three-phase line has the following parameters: A = 0.87940.66° per unit, B = 134.885.3° Ω, C = 1.688 × 10-3∠90.2 S. Determine, (a) the practical line loadability in MW, assuming Vs = 1.0 per unit, V R ≈ 0.95 per unit, and δmax = 35°; (b) the full-load current at 0.99 p.f. leading, based on the above practical line loadability; (c) the exact receiving-end voltage for the full-load current in (b); (d) the percent voltage regulation. [Answer] (a) Using (p.66 Topic 5 Lecture Notes) with δ = 35°

500 35   0.8794 0.66  V RFL 0   134.8 85.3 1.213 8.109  3 288.6835  0.8794V RFL 0.66  163.593.41 Solving for V RFL (288 .68 35  163.593.41 )  280.0419 0.0076 0.8794 0.66  280. 0419 kVL N

VRFL  VRFL

VRFL  280.041 9 3  484.9742 kVL L  0.9699 pu > 0.95 pu Hence voltage drop requirement sati sfied

(d)

VRNL  VS / A  500 / 0.8794  568.6kV L L 568.6  484.97 %VR   100  17.24% 484.97

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REVISION Problem 16 A 200 km, 500 kV, 60-Hz uncompensated three-phase line has the following parameters: A = 0.96940.154° per unit, B = 69.5485.15° Ω. Determine the practical line loadability in MW under VS = 1 per unit, VR = 0.95 per unit, max = 35º, and 0.99 leading power factor at the receiving end. [Answer]

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REVISION

Problem 17 For a 400 km, 500 kV, 60-Hz uncompensated three-phase line, the following parameters have been calculated for its equivalent π circuit: Z′ = R′ + jX′= 11.0 + j134.3 = 134.8∠85.3° Ω Y′ = G′ + jB′ = 3.14×10−6 + j1.796×10−3 S. At full load the receiving end voltage is 475 kV line-to-line. The sending end voltage at full load has been calculated as 526.4 kV line-to-line and the voltage regulation as 26%. To improve the voltage regulation, identical shunt reactors are installed at both ends of the line during light loads, providing 65% total shunt compensation. The reactors are removed at full load. (a) Determine the voltage regulation for the compensated line and comment on the result. (b) Also calculate the impedance of each shunt reactor.

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REVISION [Answer] (a)

where VS = 526.14 kV is given by the problem. VRFL = 475 kVL-L is the same as no compensation case since the shunt reactors are removed

(b)

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REVISION Problem 18 For the uncompensated line in Problem 17, the ABCD parameters have been calculated as A = 0.87940.66° per unit, B = 134.885.3° Ω, C = 1.688 × 10-3∠90.2 S. Now identical series capacitors are installed at both ends of the line, providing 40% total series compensation. (a) Calculate the impedance of each series capacitor. (b) Determine the equivalent ABCD parameters of the line. (c) Determine the theoretical maximum real power that the series-compensated line can deliver when |Vs| = |VR| = 1 per unit. (d) Compare your result with the theoretical maximum power transfer capability of the uncompensated line which was calculated as PRmax = 1702 MW. [Answer] (a) From Problem 17

(b) Using the ABCD parameters given in the problem, and using the ABCD parameters of series networks given on page 13 of Topic 5 Lecture Notes, the equivalent ABCD parameters of the compensated lines are

A  eq C  eq

B  eq   1  j 26.86  0.8794 0.66     D  0 1   1.688 103 90.2  eq   Receiving end series capacitor

Uncompensated line

Compensated line

 0.9248 0.64  1.688 10 3 90.2 



6  

 Sending end series capacitor



(c)

Using Aeq  0.92480.64 p.u., Beq  86.66 82.31  , Beq  Z eq and PRmax

VRVS AVR2   cos( Z   A ) with VR  VS  500 kVL-L , A  Aeq and  Z   Zeq Z Z

we have

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REVISION

which is 46.7% larger than the value PRmax = 1702MW calculated for the uncompensated line. Problem 19 Fig 1 gives a four bus power system, where y12 = −jb12, y13 = −jb13, y23 = −jb23, y34 = −jb34 and bij > 0 for all i and j. Derive the nodal equations that relate the current injected into each bus and the voltage at each bus via the network admittances. Does this set of equations have a unique solution? If not, show how defining one of the buses to be a swing bus leads to a unique set of solutions.

Fig 1: Four Bus Power System [Answer] By the rules (see pp 46-52, Topic 1 Lecture Notes), the admittance matrix is given by: Diagonal elements: Yii = sum of admittances connected to bus i. Off-diagonal elements: = − (sum of admittances connected between buses i and j), i ≠ j. And in matrix form the nodal equations can be written as Ibus = YbusVbus. Using these we get  I1  I ...