Electric Power Systems - wiley.com · PDF filePeak power occurs when . 0 dP dt = 8-0.00182t=0,...

Electric Power Systems

B.M. WEEDY | B.J. CORY N. JE NKINS | J.B. EKANAYAKE | G. STRBAC

F I F T H E D I T I O N

Solutions

Chapter 1

1.1 In the U.S.A. in 1971 the total area of right of ways for H.V. overhead lines was 16,000 km2. Assuming a growth rate for the supply of electricity of 7 per cent per annum calculate what year the whole of the U.S.A. will be covered with transmission systems (assume area to approximate 4800 x 1600 km). Justify any assumptions made and discuss critically why the result is meaningless.

Answer

Area of USA 24800 1600 7,680,000 km× = Area of rights of way after n years 216,000 (1. 0.07)n km× + Equating area of USA with area of rights of way after n years

7,680,000(1 0.07) 48016,000

n+ = =

10

10

log 480 91.25log 1.07

n years∴ = =

1.2 The calorific value of natural gas at atmospheric pressure and temperature is 40

MJ/m3. Calculate the power transfer in a pipe of lm diameter with gas at 60 atm (gauge) flowing at 5 m/s. If hydrogen is transferred at the same velocity and pressure, calculate the power transfer. The calorific value of hydrogen is 13 MJ/m3 at atmospheric temperature and pressure.

Answer

Volume of gas flowing in 1 second 2 35 0.5 3.927 mπ= × × = Power of natural gas 63.927 60 40 10 9.4 GJ/sec 9.4GW× × × = =

Power of hydrogen = 139.4 3.1GW40

× =

1.3 a) An electric car has a steady output of 10 kW over its range of 100 km when

running at a steady 40 km/h. The efficiency of the car (including batteries) is 65%. At the end of the car's range the batteries are recharged over a period of 10 h. Calculate the average charging power if the efficiency of the battery charger is 90%.

b) The calorific value of gasoline (petrol) is roughly 16,500 kJ/gallon. By assuming an average filling rate at a pump of 10 gallon/minute, estimate the rate of energy transfer on filling a gasoline-driven car. What range and what cost/km would the same car as (a) above produce if driven by gasoline with a 7 gallon tank? (Assume internal combustion engine efficiency is 60% and gasoline costs £3 per gallon)

Answer a) Time of journey at the end of the car's range= 100/40 = 2.5 hours

Power from battery = 10 kW/0.65 = 15.38 kW

Energy from battery = 15.38x2.5 = 38.45 kWh Battery recharged over 10 h, charging power = 38.45/0.9x10=4.27 kW

b) In one minute 165 MJ of petrol energy is added to the tank, That is at 1/60 hrs, 165 MJ or (106/3600)x165 = 45.870 kWh/min is added. This is equal to a rate of 2.75 MW A 7 gallon tank contains 115.5 MJ of energy which at 60% efficiency delivers 115.5x0.6 = 69.3 MJ of useful energy. As 69.3 MJ is equal to 19.25 kWh, the length of journey at 10 kW is 1.925 hours or 40x1.925 = 77 km. A 7 gallon tank costs £21 giving a cost of 27.2 p/km

1.4 The variation of load (P) with time (t) in a power supply system is given by the expression,

P(kW) = 4000 + 8t - 0.00091t2 where t is in hours over a total period of 1 year.

This load is supplied by three 10 MW generators and it is advantageous to fully load a machine before connecting the others. Determine: a) the load factor on the system as a whole; b) the total magnitude of installed load if the diversity factor is equal to 3; c) the minimum number of hours each machine is in operation; d) the approximate peak magnitude of installed load capacity to be cut off to

enable only two generators to be used.

Answer (a) The variation of load with time is drawn as:

0 1000 2000 3000 4000 5000 6000 7000 8000 90000

5

10

15

20

25

Time (hrs)

Pow

er (M

W)

754

7963

3077

5714

Generator 2 operates


8760


The annual load is the integral under the power curve

Load energy over a year = 8760 2 3

2

0

8760 87604000 8 0.00091 4000 8760 8 0.00091 138083.25MWh2 3

t

t

t t dt=

=

+ − = × + × − × = ∫

Peak power occurs when 0dPdt

=

8-0.00182t=0, t=4396 hrs Peak power = 21.58 MW

Therefore the load factor = 138083.25/(21.58x8760) =0.73

(b) If the diversity factor is 3, the installed load = 3 x 21.58 = 64.74 MW (c) 1st machine operates 8760 hrs per year

To find the time that the 2nd machine is in operation, solve the following equation

4000 + 8t - 0.00091t2 = 10000 0.00091t2 - 8t + 6000 = 0

8 64 4 0.00091 60002 0.00091

t ± − × ×=

×

t=754 hrs and 7963 hrs

Therefore generator 2 will operates from 754 - 7963 hrs So total number of hours per day generator 2 is operating = 7209 hrs per year. To find the time that the 3rd machine is in operation, solve the following equation

4000 + 8t - 0.00091t2 = 20000 0.00091t2 - 8t + 16000 = 0

8 64 4 0.00091 160002 0.00091

t ± − × ×=

×

t=3077 hrs and 5714 hrs

Therefore generator 3 will operate from 3077 - 5714 hrs So the total number of hours per day generator 3 operates = 2637 hrs per year.

(d) Peak load is 21.4 MW, giving an excess over 2 generators of 1.4 MW of load or

4.2 MW of installed load.

1.5 a) Explain why economic storage of electrical energy would be of great benefit to power systems.

b) List the technologies for the storage of electrical energy which are available now and discuss, briefly, their disadvantages.

c) Why is hydro power a very useful component in a power system?

d) Explain the action of pumped storage and describe its limitations.

e) A pumped storage unit has an efficiency of 78% when pumping and 82% when generating. If pumping can be scheduled using energy costing 2.0 p/kWh, plot the gross loss/profit in p/kWh when it generates into the system with a marginal cost between 2p and 6p/kWh.

f) Explain why out-of-merit generation is sometimes scheduled.

(From Engineering Council Examination, 1996)

Answer

Overall efficiency = 0.78x0.82 = 0.64

Marginal Cost of electricity

Cost of energy from storage

Profit or (loss)

2 2/0.64= 3.125 (1.25 3 (0.125) 4 0.875 5 1.875 6 2.875

1

Chapter 2

Problems

2.1 The star-connected secondary winding of a three-phase transformer supplies 415V (line to line) at a load point through a four-conductor cable. The neutral conductor is connected to the winding star point which is earthed. The load consists of the following components:

Between a and b conductors a 1Ω resistor Between a and neutral conductors a 1Ω resistor Between b and neutral conductors a 2Ω resistor Between c and neutral conductors a 2Ω resistor Connected to the a, b, and c conductors is an inductor motor taking a balanced current of 100 A at 0.866 power factor (p.f.) lagging. Calculate the current in the four conductors and the total power supplied. Take the ‘a’ to neutral voltage as the reference phasor. The phase sequence is a-b-c.

Answer

𝐕𝒂 = 𝑉√3

= 415√3

V , For induction motor

𝐼𝑚 = 100 𝐴 𝜙 = cos−1(0.866) = 30𝜊

For the resistive load

𝐈𝒂𝒏 =𝑉√3

∗1𝑅𝑎𝑛

= 239.6 A

𝐈𝒃𝒏 =𝑉𝑒−𝑗120

√3∗

1𝑅𝑏𝑛

= (−59.9 − 𝑗103.75 ) A

𝐈𝒄𝒏 =𝑉𝑒+𝑗120

√3∗

1𝑅𝑐𝑛

= (−59.9 + 𝑗103.75 ) A

𝐈𝒂𝒃 =𝑉𝑒+𝑗30

𝑅𝑎𝑏= (359.4 + 𝑗207.5 ) A

Therefore line currents

𝐈𝐚 = 𝐈𝐚𝐛 + 𝐈𝐚𝐧 + 𝐼𝑚 ∗ 𝑒−𝑗𝜙 = (685.6 + 𝑗157.5 ) A

𝐈𝐛 = −𝐈𝐚𝐛 + 𝐈𝐛𝐧 + 𝐼𝑚 ∗ 𝑒−𝑗(120+𝜙) = (−505.9 + 𝑗361.25 ) A

𝐈𝐜 = 𝐈𝐜𝐧 + 𝐼𝑚 ∗ 𝑒𝑗(120−𝜙) = (−59.9 + 𝑗203.75 ) A

𝐒 = 𝐈𝒂∗𝐕𝒂 + 𝐈𝒃∗𝐕𝒃 + 𝐈𝒄∗𝐕𝒄 = 𝐈𝒂∗V√3

+ 𝐈𝐛∗𝑉𝑒−𝑗120

√3+ 𝐈𝐜∗

𝑉𝑒+𝑗120

√3= (349.29 + 𝑗35.94 ) kVA

2.2 The wye-connected load shown in Figure 2.29 is supplied from a transformer whose secondary-winding star point is solidly earthed. The line voltage supplied to the load

2

is 400V. Determine (a) the line currents, and (b) the voltage of the load star point with respect to ground. Take the ‘a’ to ‘b’ phase voltage as the reference phasor. The phase sequence is a-b-c.

a

b

c

1 Ω

2 Ω1 Ω

Figure 2.29 Circuit for problem 2.2

Answer

a

b

c

Ra

RcRb

Ia

Ib

Ic

n

𝐈𝐚Ra − 𝐈𝐛Rb = 400 𝐈𝐛Rb − 𝐈𝐜Rc = 400𝑒−𝑗120 𝐈𝐚 + 𝐈𝐛 + 𝐈𝐜 = 0 𝐈𝐚 = (200 − 𝑗69.28 ) A 𝐈𝐛 = (−200− 𝑗69.28 ) A 𝐈𝐜 = 𝑗138.56 A

𝐕𝐧𝐠 = 𝐕𝐜𝐠 − 𝐈𝐜Rc = 𝑗400√3

V − 2Ω ∗ 𝑗138.56 A = −𝑗46.18 V

2.3 Two capacitors, each of 10μF, and a resistor R, are connected to a 50 Hz three phase supply, as shown in Figure 2.30. The power drawn from the supply is the same whether the switch S is open or closed. Find the resistance of R.

3

a

b

c

Ia

Ib

Ic R

N S

Figure 2.30 Circuit for problem 2.3

Answer

𝑋𝑐 =1

2𝜋𝑓𝐶= 318.31 Ω

Assume V is phase voltage Switch closed:

𝐒 = 𝐈𝒂∗𝐕𝒂 + 𝐈𝒃∗𝐕𝒃 + 𝐈𝒄∗𝐕𝒄 =𝐕𝒂∗𝐕𝒂𝐙𝒂∗

+𝐕𝒃∗𝐕𝒃𝐙𝒃∗

+𝐕𝒄∗𝐕𝒄𝐙𝒄∗

= 𝑉2 1𝑅− 𝑗

2𝑋𝑐

P𝑠𝑤𝑖𝑡𝑐ℎ 𝑐𝑙𝑜𝑠𝑒𝑑 =V2

𝑅

Switch open:

a

b

c

Ia

Ib

Ic R

N S

i1

i2

−𝑗2𝑋𝑐 𝑗𝑋𝑐𝑗𝑋𝑐 −𝑗𝑋𝑐 + 𝑅

𝐢𝟏𝐢𝟐 = 𝐕𝐚𝐛𝐕𝐛𝐜

𝐈𝒄 = −𝐢𝟐 = −(𝐕𝒂𝒃 + 2𝐕𝒃𝒄)(2𝑅 − 𝑗𝑋𝑐)

(if 𝐕𝒂 = 𝑉, then 𝐕𝒂𝒃 = √3𝑉𝑒+𝑗30 and 𝐕𝒃𝒄 = −𝑗√3𝑉)

𝑃𝑠𝑤𝑖𝑡𝑐ℎ 𝑜𝑝𝑒𝑛 = 𝐈𝒄𝐈𝒄∗𝑅 =9𝑅𝑉2

4𝑅2 + 𝑋𝑐2

𝑃𝑠𝑤𝑖𝑡𝑐ℎ 𝑐𝑙𝑜𝑠𝑒𝑑 = 𝑃𝑠𝑤𝑖𝑡𝑐ℎ 𝑜𝑝𝑒𝑛 ⇒ 5𝑅2 = 𝑋𝑐2

𝑅 = 142.35 Ω

4

2.4 The network of Figure 2.31 is connected to a 400 V three-phase supply, with phase sequence a-b-c. Calculate the reading of the wattmeter W.

Figure 2.31 Circuit for Problem 2.4

Answer

𝐈𝐚 + 𝐈𝐛 + 𝐈𝐜 = 0

−𝐈𝐛(50 + j40) = 400

−𝐈𝐜(−j53) + 𝐈𝐛(50 + j40) = 𝐕𝐛𝐜 = 400e−j120 From which

𝐈𝐜 = (−6.536− j3.7736) A

𝐈𝐚 = (11.414− j0.129) A Power in wattmeter = real part of (−𝐕𝒃𝒄𝐈𝒂∗) = 2.33 kW

2.5 A 400 V three-phase supply feeds a delta-connected load with the following branch impedances:

𝐙𝐑𝐘 = 100 Ω 𝐙𝐘𝐁 = j100 Ω 𝐙𝐁𝐑 = −j100 Ω

Calculate the line currents for phase sequences (a) RYB; (b) RBY.

Answer (a) The line voltages are as follows:

𝐈𝐑𝐘𝐙𝐑𝐘 = 𝐕𝐑𝐘 = 400 V ⇒ 𝐈𝐑𝐘 = 4 A

𝐈𝐘𝐁𝐙𝐘𝐁 = 𝐕𝐘𝐁 = 400e−j120 V ⇒ 𝐈𝐘𝐁 = −𝑗4e−j120 A

𝐈𝐁𝐑𝐙𝐁𝐑 = 𝐕𝐁𝐑 = 400ej120 V ⇒ 𝐈𝐁𝐑 = 𝑗4ej120 A

The line currents are as follows:

5

𝐈𝐑 = 𝐈𝐑𝐘 − 𝐈𝐁𝐑 = (7.464 + j2) A; |𝐈𝐑| = 7.727 A

𝐈𝐘 = 𝐈𝐘𝐁 − 𝐈𝐑𝐘 = (−7.464 + j2) A; |𝐈𝐘| = 7.727 A

𝐈𝐁 = 𝐈𝐁𝐑 − 𝐈𝐘𝐁 = −j4 A; |𝐈𝐁| = 4 A

(b) The line voltages are as follows:

𝐈𝐑𝐘𝐙𝐑𝐘 = 𝐕𝐑𝐘 = 400 V ⇒ 𝐈𝐑𝐘 = 4 A

𝐈𝐁𝐑𝐙𝐁𝐑 = 𝐕𝐁𝐑 = 400e−j120 V ⇒ 𝐈𝐁𝐑 = 𝑗4e−j120 A

𝐈𝐘𝐁𝐙𝐘𝐁 = 𝐕𝐘𝐁 = 400ej120 V ⇒ 𝐈𝐘𝐁 = −𝑗4ej120 A

The line currents are as follows:

𝐈𝐑 = 𝐈𝐑𝐘 − 𝐈𝐁𝐑 = (0.536 + j2) A; |𝐈𝐑| = 2.07 A

𝐈𝐘 = 𝐈𝐘𝐁 − 𝐈𝐑𝐘 = (−𝟎.𝟓𝟑𝟔 + 𝐣𝟐) 𝐀; |𝐈𝐘| = 𝟐.𝟎𝟕 𝐀

𝐈𝐁 = 𝐈𝐁𝐑 − 𝐈𝐘𝐁 = −j4 A; |𝐈𝐁| = 4 A

2.6 A synchronous generator, represented by a voltage source in series with an inductive

reactance X1, is connected to a load consisting of a fixed reactance X2 and a variable resistance R in parallel. Show that the generator power output is a maximum when

1/R=1/X1+1/X2

Answer

Load impedance 𝐙 = 𝑗𝑋2||𝑅 = 𝑗𝑋2𝑅𝑗𝑋2+𝑅

𝐈 =𝐄

(𝑗𝑋1 + 𝑗𝑋2𝑅𝑗𝑋2 + 𝑅)

𝐒 = 𝐄𝐈∗ =𝐄𝐄∗

(𝑗𝑋1 + 𝑗𝑋2𝑅𝑗𝑋2 + 𝑅)∗

=E2(𝑅𝑋22 + 𝑗(𝑅2(𝑋1 + 𝑋2) + 𝑋1𝑋22)

𝑋12𝑋22 + 𝑅2(𝑋1 + 𝑋2)2

Therefore

P = Re[𝐒] =E2𝑅𝑋22

𝑋12𝑋22 + 𝑅2(𝑋1 + 𝑋2)2

Using the first derivative test:

dPdR

=E2𝑋22(𝑋12𝑋22 − 𝑅2(𝑋1 + 𝑋2)2)

(𝑋12𝑋22 + 𝑅2(𝑋1 + 𝑋2)2)2= 𝟎

∴ 𝑋12𝑋22 − 𝑅2(𝑋1 + 𝑋2)2 = 0 ⟹

http://en.wikipedia.org/wiki/First_derivative_test

6

𝑅 =𝑋1𝑋2𝑋1 + 𝑋2

From 𝑅 = 𝑋1𝑋2

𝑋1+𝑋2 it follows:

1𝑅

=1𝑋1

+1𝑋2

2.7 A single-phase voltage source of 100 kV supplies a load through an impedance j100

Ω. The load may be represented in either of the following ways as far as voltage changes are concerned:

a) by constant impedance representing a consumption of 10 MW, 10 MVAr at 100kV; or

b) by constant current representing a consumption of 10 MW, 10 MVAr at 100kV.

Calculate the voltage across the load using each of these representations.

Answer (a) constant impedance:

𝐒 = 𝐕𝐈∗ = 𝐕𝐕∗

𝐙∗

𝐙 = 𝐕𝒓𝒂𝒕𝒆𝒅𝟐

𝐒 ∗

=(100)𝟐

(10 + 𝑗10)∗ = (500 + 𝑗500) Ω

𝐈 = 100500+𝑗500+𝑗100

kA

𝐕 = 𝐈𝐙 = (90.164− 𝑗 8.197) kV

(b) constant current:

𝐈 =𝐒∗

𝐕𝒓𝒂𝒕𝒆𝒅∗=

(10 + 𝑗10)∗

100

𝐕 = 𝐄 − 𝑗100 ∗ 𝐈 = 100 − 𝑗10 − 10 = (90 − 𝑗 10) kV

7

2.8 Show that p.u. impedance (obtained from a short circuit test) of a star-delta three-phase transformer is the same whether computed from the star side parameters or from the delta side. Assume a rating of G (volt-amperes), a line-to-line input voltage to the star-winding terminals of V volts, a turns ratio of 1:N ( star to delta), and a short circuit impedance of Z (ohms) per phase referred to the star side.

Answer Star side:

Z𝑌𝑏𝑎𝑠𝑒 =𝑉2

𝐺

Z𝑌𝑝.𝑢. =Z1

Z𝑏𝑎𝑠𝑒=

Z𝐺𝑉2

Delta side:

Z𝐷𝑏𝑎𝑠𝑒 ==V2𝐿𝐿,𝑏𝑎𝑠𝑒2

S𝑏𝑎𝑠𝑒

𝑉1𝐿𝑁𝑉2𝐿𝐿

=1𝑁

⟹ 𝑉1𝐿𝐿√3𝑉2𝐿𝐿

=1𝑁

⟹ 𝑉2𝐿𝐿 =𝑁𝑉√3

⟹

Z𝐷𝑏𝑎𝑠𝑒 =𝑁2𝑉2

3𝐺

𝑍2in delta = 𝑍𝑁2

Per phase impedance can be found by delta to star transformation and is given by 𝑍𝑁2 3⁄

Z𝐷𝑝.𝑢. = Z𝑁2

33𝐺

𝑁2𝑉2= 𝑍𝐺

𝑉2= Z𝑌𝑝.𝑢.

2.9 An 11 kV/132 kV, 50 MVA, three-phase transformer has an inductive reactance of j0.5 Ω referred to the primary (11 kV). Calculate the p.u. value of reactance based on the rating. Neglect resistance.

Answer

𝑍𝐵 =(11 ∗ 103)2

50 ∗ 106= 2.42 Ω

𝑋𝑝.𝑢. =𝑗0.52.42

= 0.207 p. u.

2.10 Express in p.u. all the quantities shown in the line diagram of the three-phase transmission system in Figure 2.32. Construct the single-phase equivalent circuit. Use a base of 100 MVA. The line is 80 km in length with resistance and reactance per km of 0.1 and 0.5 Ω, respectively, and a capacitive susceptance of 10 µS per km (split equally between two ends.)

8

Figure 2.32. Circuit for Problem 2.10

Answer 𝑆𝐵 = 100 MVA

Zone 1 (see diagram below):

𝑉𝐵 = 22 kV

𝐒𝐩.𝐮. =500100

= 5 p. u.

On 500 MVA base, the generator reactance is 2 pu

Reactance on 100 MVA base = 2 ×100500

= 0.4 p. u.

Zone 2:

𝑉𝐵 = 400 kV

𝑍𝐵 =(400 ∗ 103)2

100 ∗ 106= 1600 Ω

𝒁𝒍𝒊𝒏𝒆,𝒑.𝒖. =𝐙𝐥𝐢𝐧𝐞𝑍𝐵

=(0.1 + 𝑗0.5) ∗ 80

1600= (0.005 + 𝑗0.025) p. u.

𝐵𝑝.𝑢. =𝐵1𝑍𝐵

=10 × 10−6 ∗ 80

11600

= 1.28 p. u.

Zone 3:

𝑉𝐵 = 11 kV

For the load

𝐒𝐩.𝐮. =500(0.85 + 𝑗 sin(cos−1 0.85))

100= 4.25 + j2.634 p. u.

9

Zone 1 Zone 2 Zone 3

5 p.u. 0.4 p.u. 0.1 p.u.0.64 p.u.0.64 p.u.

0.64 p.u.0.64 p.u.

(0.05+j0.025) p.u.

(0.05+j0.025) p.u.

0.1 p.u. S=(4.25+j2.634) p.u.

Load

2.11 A wye-connected load is supplied from three-phase 220 V mains. Each branch of a load is a resistor of 20 Ω. Using 220 V and 10kVA base, calculate the p.u. values of the current and power taken by the load.

Answer

𝑉𝑝.𝑢. =220220

= 1

𝑍𝐵 =(220)2

10000= 4.84 Ω

𝑅𝑝.𝑢. =20

4.84= 4.1232

𝐈𝐩.𝐮. =𝑉𝑝.𝑢.

𝑅𝑝.𝑢.= 0.242

𝐒𝒑.𝒖. = 𝐕𝐈∗ = 0.242

2.12 A three-phase supply is connected to three star-connected loads in parallel, through a feeder of impedance (0.1+j0.5) Ω per phase. The loads are as follows: 5kW, 4 kVAr; 3kW, 0 kVAr; 10 kW, 2 kVAr. Assuming voltage at the load end is 440 V, determine:

a) line current; b) power and reactive power losses in the feeder per phase; c) power and reactive power from the supply and the supply power factor.

Answer We assume that V2=440 V. If V1= 440 V, an iterative method would be required (converging

10

Z

V1 V2

S

I

Ss

Load

𝐒 = (18 + 𝑗6) kVA

(a)

𝐈 = 𝐒

√3𝑉2∗

= (23.619− 𝑗7.873) A = 24.8965∠− 18.435 A

(b)

𝚫𝐒𝐩𝐞𝐫 𝐩𝐡𝐚𝐬𝐞 = 𝑍 ∗ |𝐈|2 = (61.983 + 𝑗309.92) VA

(c) 𝐒𝐬𝐮𝐩𝐩𝐥𝐲 = 𝐒 + 3𝚫𝐒 = (18.186 + 𝑗6.93) kVA

pf = cos tan−1 6.93

18.186 = 0.9345

2.13 Two transmission circuits are defined by the following ABCD constants: 1, 50, 0, 1, and 0.9 ∠ 2º, 150∠ 79º, 9X10-4∠ 91º, 0.9∠ 2º, Determine the ABCD constants of the

circuit comprising these two circuits in series.

Answer A0=A1A2 +B1C2 =1 x 0.9∠2º+50 x 9X10-4∠91º = 0.9019 ∠4.8595º

B0=A1B2 +B1D2 = 166.0176∠63.686º

C0=C1A2 +D1C2 = 0.0009 ∠91º

D0=C1B2 +D1D2 = 0.9 ∠2º

2.14 A 132 kV overhead line has a series resistance and inductive reactance per phase per kilometre of 0.156 and 0.4125 Ω, respectively. Calculate the magnitude of the sending-end voltage when transmitting the full line capability of 125 MVA when the power factor is 0.9 lagging and the received voltage is 132 kV, for 16 km and 80km lengths of line. Use both accurate and approximate methods.

Answer 𝑉𝐵 = 132 kV , 𝑆𝐵 = 125 MVA

𝑍𝐵 =(132 ∗ 103)2

125 ∗ 106= 139.392 Ω

𝐕𝐩.𝐮. =132132

= 1

𝐒 = 125(0.9 + 𝑗 sin(cos−1 0.9)) = (112.5 + 𝑗54.486) MVA

11

𝐒𝒑.𝒖. =𝐒𝑆𝐵

= 0.9 + 𝑗0.4359

Line length 16 km

𝐙𝒑.𝒖. =𝐙𝑍𝐵

=16(0.156 + 𝑗0.4125)

139.392= 0.0179 + 𝑗0.0473

Accurate:

𝐸2 = 𝑉 +𝑅𝑃𝑉

+𝑋𝑄𝑉2

+ 𝑋𝑃𝑉−𝑅𝑄𝑉2

𝐸2 = 1 +0.0179 ∗ 0.9

1+

0.0473 ∗ 0.43591

2

+ 0.0473 ∗ 0.9

1−

0.0179 ∗ 0.43591

2

𝐸2 = (1.03673)2 + (0.03477)2 = 1.076 𝐸 = 1.0373 p. u. = 1.0373 ∗ 132 V = 136.92 V

Approximate:

𝐸 = 𝑉 +𝑅𝑃𝑉

+𝑋𝑄𝑉

= 1.0367 p. u.

𝐸 = 𝐸 ∗ 𝑉𝐵 = 1.03673 ∗ 132 V = 136.85 V

Line length 80 km

𝐙𝒑.𝒖. =𝐙𝑍𝐵

=80(0.156 + 𝑗0.4125)

139.392= 0.0895 + 𝑗0.2367

Accurate:

𝐸2 = 𝑉 +𝑅𝑃𝑉

+𝑋𝑄𝑉2

+ 𝑋𝑃𝑉−𝑅𝑄𝑉2

𝐸2 = 1 +0.0895 ∗ 0.9

1+

0.2367 ∗ 0.43591

2

+ 0.2367 ∗ 0.9

1−

0.0895 ∗ 0.43591

2

𝐸2 = ( 1.18373)2 + (0.17402)2 = 1.4315

𝐸 = 1.1965 p. u. = 1.1965 ∗ 132 V = 157.938 V

Approximate:

𝐸 = 𝑉 +𝑅𝑃𝑉

+𝑋𝑄𝑉

= 1.18373 p. u.

𝐸 = 𝐸 ∗ 𝑉𝐵 = 1.18373 ∗ 132 V = 156.25 V

12

2.15 A synchronous generator may be represented by a voltage source of magnitude 1.7 p.u. in series with an impedance of 2 p.u. It is connected to a zero-impedance voltage source of 1 p.u. The ratio of X/R of the impedance is 10. Calculate the power generated and the power delivered to the voltage source if the angle between the voltage sources is 30º.

Answer 𝑋𝑅

= 10 𝑅2 + 𝑋2 = 22 ⟹

𝑅 = 4101

= 0.199

𝐙 = 0.199 + 𝑗1.99 𝐕 = 1∠ − 30° = 0.866 − 𝑗0.5

𝐈 =𝐄 − 𝐕𝐙

= 0.2903− 𝑗0.3901 𝐒𝐠𝐞𝐧𝐞𝐫𝐚𝐭𝐞𝐝 = 𝐄𝐈∗ = 0.4934 + 𝑗0.6631 𝐒𝐝𝐞𝐥𝐢𝐯𝐞𝐫𝐞𝐝 = 𝐕𝐈∗ = 0.4464 + 𝑗0.1927

Therefore power generated is 0.49 pu and power delivered is 0.44 pu.

Alternative answer would be

12 tan 10 2 84.29Z −= ∠ = ∠

E = 1.7 p.u

V = 1.0 p.u 2

2

cos cos( )

1.7 1.7 1cos84.29 cos(84.29 30 )2 2

0.494 p.u.

G G LG

V V VPZ Z

θ θ δ= − +

×= − +

=

2

2

cos cos( )

1.0 1.7 1cos84.29 cos(84.29 30 )2 2

0.446 p.u.

L GLL

V VVPZ Z

θ θ δ= − −

×= − −

=

2.16 A three-phase star-connected 50 Hz generator generates 240 V per phase and supplies three delta-connected load coils each having a resistance of 10 Ω and an inductance of 47.75 mH.

Determine:

(a) the line voltage;

13

(b) the load and line currents; (c) the total real power and reactive power dissipated by the load.

Determine also the values of the three capacitors required to correct the overall power factor to unity when the capacitors are (i) star- and (ii) delta-connected across the load.

State an advantage and a disadvantage of using the star connection for power factor correction.

Answer 𝑋 = 2𝜋𝑓𝐿 = 2𝜋 ∗ 50 ∗ 47.75 ∗ 10−6 = 15.0011 Ω 𝐙 = (10 + 𝑗15.0011) Ω (a)

𝑉𝑙𝑖𝑛𝑒 = √3𝑉 = 415.6922 V

(b)

𝐈𝒍𝒐𝒂𝒅 =𝐕𝐙

= 12.7892− 𝑗19.1852 = 23.0573∠ − 56.3119° A

𝐈𝒍𝒊𝒏𝒆 = 𝐈𝒂𝒃 − 𝐈𝒄𝒂 = 𝐈𝒍𝒐𝒂𝒅 − 𝐈𝒍𝒐𝒂𝒅ej120 = 39.9364∠ − 86.3119° A (c)

𝐒 = 3𝐕𝐈𝐥𝐨𝐚𝐝∗ = 3𝐙|𝐈𝐥𝐨𝐚𝐝|𝟐 = (15.949 + 𝑗23.925) kVA = 28.754∠ 56.3119° kVA (i) Star-connected capacitor

𝑄 = 𝑄𝑐 = 3𝑉𝑐2

𝑋𝑐= 3 ∗ 2402 ∗ 2𝜋𝑓𝐶

𝐶 =23.925 × 103

3 ∗ 2402 ∗ 2𝜋 ∗ 50= 440.72 𝜇F

(ii) Delta-connected capacitor

𝑄 = 𝑄𝑐 = 3𝑉𝑐2

𝑋𝑐= 3 ∗ (√3 ∗ 240)2 ∗ 2𝜋𝑓𝐶

𝐶 =23.925 × 103

3 ∗ 3 ∗ 2402 ∗ 2𝜋 ∗ 50= 146.91 𝜇F

1

Chapter 3

Problems

3.1 When two four pole, 50 Hz synchronous generators are paralleled their phase displacement is 2 mechanical. The synchronous reactance of each machine is 10Ω/phase and the common busbar voltage is 6.6 kV. Calculate the synchronizing torque.

Answer 𝑝 = 2 𝑝𝑎𝑖𝑟𝑠 𝑜𝑓 𝑝𝑜𝑙𝑒𝑠

𝑓 = 50 𝐻𝑧

𝑋𝑠 = 10 Ω/𝑝ℎ𝑎𝑠𝑒

𝑉𝑙 = 6.6 kV

𝛿𝑚𝑒𝑐ℎ′ = 2° 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡

𝛿𝑒𝑙′ = 𝛿𝑚𝑒𝑐ℎ′ ∙ 𝑝 = 4° ∙

𝜋180

[𝑟𝑎𝑑]

No load:

𝐸 =6.6√3

kV

𝑃𝑠𝑦𝑛𝑐ℎ = 3 × 𝐸2

2𝑋𝑠× 𝛿𝑒𝑙′

𝑛𝑠 =60 × 𝑓𝑝

𝑇𝑠𝑦𝑛𝑐ℎ = 𝑃𝑠𝑦𝑛𝑐ℎ

2𝜋 ∗ 𝑛𝑠60= 3 ×

6.6√3

× 1032

×1

2 × 10× 4° ×

𝜋180

×1

2𝜋 × 60 × 502 × 60

= 968 𝑁𝑚

3.2 A synchronous generator has a synchronous impedance of 2 p.u. and a resistance of 0.01 p.u. It is connected to an infinite busbar of voltage 1 p.u. through a transformer of reactance j0.1 p.u. If the generated (no-load) e.m.f. is 1.1 p.u. calculate the current and power factor for maximum output.

Answer 𝐙 = 0.01 + 𝑗0.1 + 𝑗2 = 0.01 + 𝑗2.1

𝑃 =𝐸 ∙ 𝑉𝑋

∙ 𝑠𝑖𝑛𝛿

2

𝑃𝑚𝑎𝑥 =1.1 × 1

2.1= 0.52381 p. u.

𝑠𝑖𝑛𝛿 = 1 ⟹ 𝛿 = 90°

𝐈 =𝐄 − 𝐕𝐙

=𝑗1.1 − 1

0.01 + 𝑗2.1= 0.5215 + 𝑗0.4787 = 0.7079 ∠42.5497°

𝑐𝑜𝑠𝜑 = 0.7367 leading

3.3 A 6.6 kV synchronous generator has negligible resistance and synchronous reactance of 4Ω/phase. It is connected to an infinite busbar and gives 2000A at unity power factor. If the excitation is increased by 25% find the maximum power output and the corresponding power factor. State any assumptions made.

Answer 𝐼2 = 2000𝐴 , 𝑐𝑜𝑠𝜑 = 1

𝑉 =6.6√3

× 103 𝑉

𝐄 = 𝐕 + 𝑗𝑋𝑠 ∙ 𝐈 =6.6√3

× 103 + 𝑗4 × 2000 = (3.8105 + 𝑗8) 𝑘𝑉 = 8.8612∠64.5309 𝑘𝑉

𝑃 = 3𝐸𝑉𝑋𝑠

𝑠𝑖𝑛𝛿 =3 × 𝑉 × 1.25𝐸

𝑋𝑠∙ 𝑠𝑖𝑛𝛿, maximum power, angle 𝛿 = 90°

𝑃𝑚𝑎𝑥 = 31.655 𝑀𝑊

𝑄 =3 ∙ 𝑉 ∙ (1.25𝐸 ∙ 𝑐𝑜𝑠𝛿 − 𝑉)

𝑋𝑠, angle 𝛿 = 90°

𝑄 = −10.89 𝑀𝑉𝐴𝑟

pf = cos tan−1 𝑄

𝑃𝑚𝑎𝑥 = 0.9456

3.4 A synchronous generator whose characteristic curves are given in Figure 3.4 delivers full load at the following power factors: 0.8 lagging, 1.0 and 0.9 leading. Determine the percentage regulation at these loads.

Answer

𝑉 = 1𝑝.𝑢., 𝑆 = 1 𝑝.𝑢. ⇒ 𝐈 = 𝑆𝑉∗ = 1 𝑝.𝑢.

Procentage negulation =𝑉𝑜𝑐 − 𝑉𝑙𝑜𝑎𝑑

𝑉𝑙𝑜𝑎𝑑∗ 100 =

𝐸 − 𝑉𝑉

∗ 100

From Figure 3.4,𝑋𝑠 =𝐾𝐹𝐾𝐶

=1.95 𝑝.𝑢.

1 𝑝.𝑢.= 1.95

3

cos φ= 0.8 lagging: 𝐈 = 𝐼cosφ− 𝑗sin(cos−1(𝑐𝑜𝑠φ)) = 0.8 − 𝑗0.6

𝐄 = 𝐕 + 𝑗𝑋𝑠 ∗ 𝐈 = 1 + 𝑗1.95(0.8− 𝑗0.6) = 2.17 + 𝑗1.56 = 2.6725∠35.7121°

Procentage negulation =𝐸 − 𝑉𝑉

× 100 =2.6725− 1

1× 100 = 167.25%

cos φ = 1: 𝐈 = 1 𝑝.𝑢.

𝐄 = 𝐕 + 𝑗𝑋𝑠 ∗ 𝐈 = 1 + 𝑗1.95 = 2.1915∠62.8503°

𝑃𝑟𝑜𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑟𝑒𝑔𝑢𝑙𝑎𝑡𝑖𝑜𝑛 = 119.15%

cos φ = 0.9 leading: 𝐈 = 𝐼cosφ + 𝑗sin(cos−1(𝑐𝑜𝑠φ)) = 0.9 + 𝑗0.4359

𝐄 = V + 𝑗𝑋𝑠 ∗ 𝐈 = 1 + 𝑗1.95(0.9 + 𝑗0.4359) = 0.15 + 𝑗1.755 = 1.7614∠85.1143°

Procentage regulation = 76.14%

3.5 A salient-pole, 75 MVA, 11kV synchronous generator is connected to an infinite busbar through a link of reactance 0.3 p.u. and has Xd = 1.5 p.u. and Xq = 1 p.u., and negligible resistance. Determine the power output for a load angle 30º if the excitation e.m.f. is 1.4 times the rated terminal voltage. Calculate the synchronizing coefficient in this operating condition. All p.u. values are on a 75 MVA base.

Answer 𝑋𝑑′ = 1.5 𝑝.𝑢., 𝑋𝑞′ = 1 𝑝.𝑢., 𝛿 = 30°, 𝐸 = 1.4𝑉

𝐈 ∙ 𝐙 = 𝐈 × 0.3 = 0.3𝐼𝑑 + 𝑗0.3𝐼𝑞

𝑋𝑑 = 𝑋𝑑′ + 0.3 = 1.8 𝑝.𝑢.

𝑋𝑞 = 𝑋𝑞′ + 0.3 = 1.3 𝑝.𝑢.

𝑃 =𝑉𝐸𝑋𝑑

𝑠𝑖𝑛𝛿 +𝑉2𝑋𝑑 − 𝑋𝑞2𝑋𝑑𝑋𝑞

𝑠𝑖𝑛2𝛿 = 0.4814 𝑝.𝑢.

𝑃𝑠𝑦𝑛𝑐𝑜𝑒𝑓 =𝑑𝑃𝑑𝛿

=𝑉𝐸𝑋𝑑

𝑐𝑜𝑠𝛿 +𝑉2𝑋𝑑 − 𝑋𝑞

𝑋𝑑𝑋𝑞𝑐𝑜𝑠2𝛿 = 0.7804 𝑝.𝑢.

3.6 A synchronous generator of open-circuit terminal voltage 1 p.u. is on no load and then suddenly short-circuited; the trace of current against time is shown in Figure 3.6(b). In Figure 3.6(b) the current 0c= 1.8 p.u., 0a=5.7 p.u., and 0b= 8 p.u. Calculate the values of Xs, X’ and X’’. Resistance may be neglected. If the machine

4

is delivering 1 p.u, current at 0.8 power factor lagging at the rated terminal voltage before the short circuit occurs, sketch the new envelope of the 50 Hz current waveform.

Answer No load: 𝐸 = 𝑉 = 1

0𝑐 = 1.8 𝑝.𝑢.

0𝑎 = 5.7 𝑝.𝑢.

0𝑏 = 8 𝑝.𝑢.

𝑋𝑠 =𝐸0𝑐√2

= 0.7857 𝑝.𝑢.

𝑋′ =𝐸

0𝑎√2

= 0.2481 𝑝.𝑢.

𝑋′′ =𝐸

0𝑏√2

= 0.1768 𝑝.𝑢.

On load: 𝑉2 = 1, 𝑝𝑓2 = 0.8, 𝐼 = 1

𝐈 = 𝐼𝑝𝑓2 − 𝑗sin(cos−1(𝑝𝑓2)) = 0.8 − 𝑗0.6

𝐄 = 𝐕 + 𝑗𝑋𝑠𝐈 = 1 + 𝑗0.7857(0.8− 𝑗0.6) = 1.604∠23.2°

As per Figure 3.6(b),

𝐼" =1.604

0.1768= 9.07 p. u.

𝐼′ =1.604

0.2481= 6.47 p. u.

𝐼" =1.604

0.7857= 2.04 p. u.

3.7 Construct a performance chart for a 22 kV, 500 MVA , 0.9 p.f. generator having a short-circuit ratio of 0.55.

Answer

5

φ = 𝑐𝑜𝑠−10.9 = 25.842

𝑃 = 0.9 × 500 = 450 MW

𝑋𝑠 =1𝑆𝐶𝑅

= 1.818Ω

𝑝ℎ𝑎𝑠𝑒

𝑉 =22√3

𝑘𝑉

𝐸 = 0 ⇒ 𝐈 =22√3

⋅1𝑋𝑠

∠90° = 𝑗6.986

𝑄 =3𝑉2

𝑋𝑠= 266.23 MVAr Leading

With respect to the performance chart shown in 3.12

At point ‘g’: P=450 MW and 𝑄 = √5002 − 4502=217.9 MVAr ‘efg’ at 450 MW ‘gh’ is a circular part with radius 500 MVA Assume maximum excitation current of 2.5 pu to construct ‘gj’ Point ‘b’ was found by reducing the power at E=V=1 p.u. by 10%, i.e. by 50 MW; then ‘ecd’ was constructed.

-200 200 400 500

ab c

d

e fg h

j

E=V=1 p.u

E=2.5 p.u

500

450

MVAr

MW

0.9 lagging

3.8 A 275 kV three-phase transmission line of length 96 km is rated 800A. The values of resistance inductance and capacitance per phase per kilometre are 0.078Ω, 1.056mH, and 0.029 μF, respectively. The receiving-end voltage is 275 kV when full load is transmitted at 0.9 power factor lagging. Calculate the sending-end voltage and current, and the transmission efficiency, and compare with the answer obtained by the short-line representation. Use nominal π and T methods of solution. The frequency is 60 Hz.

Answer

6

𝑉 = 275 kV, 𝑙 = 96 𝑘𝑚, 𝐼 = 800𝐴, 𝑝.𝑓. = 0.9 lagging

𝑟 = 0.078 Ω/km, phase

𝐿 = 1.056 mH/km, phase

𝐶 = 0.029𝜇F/km, phase

𝑓 = 60 𝐻𝑧

𝑅 = 𝑟 ⋅ 𝑙 = 7.488 Ω

𝑋𝑙 = 𝜔𝐿 = 2𝜋𝑓𝐿 = 38.2179 Ω

𝑋𝑐 = −1𝜔𝑐

= −1

2𝜋𝑓𝑐= 952.7954 Ω

𝑌𝑐 =1𝑋𝑐

= 0.00105 𝑠

𝜑 = 𝑎𝑟𝑐𝑐𝑜𝑠0.9 = 25.8419°

𝑉𝑟 =275√3

kV

Short line representation 𝐈𝐫 = 𝐼𝑟∠ − 𝜑 = 800∠ − 25.8419°𝐴 = 720 − 𝑗348.71 𝐴

𝐙 = 𝑅 + 𝑗𝑋𝑙

𝐕𝐬 = 𝐕𝐫 + 𝐈 ⋅ 𝑍 = (177.49 + 𝑗24.906)𝑉 = 179.23 ∠ 7.9877° 𝑘𝑉

Medium line representation, π method of solution: 𝐕𝐬 = 𝐕𝐫 + 𝐈 ⋅ 𝑍

𝐈 = 𝐈𝐫 + 𝐕𝐫𝑌2

𝐈𝐬 = 𝐈 + 𝐕𝐬𝑌2

𝐀 = 𝐃 = 1 +𝐙𝐘2

= 0.9799 + 𝑗0.0039

𝐁 = 𝐙 = 7.483 + 𝑗38.2179

𝐂 = 1 +𝐙𝐘4𝐘 =

−0.0021 + 𝑗1.0391000

𝐈𝐬 = 𝐂𝐕𝐫 + 𝐃𝐈𝐫 = (706.6 − 𝑗173.92)A = 727.693∠ − 13.827° A

𝐕𝐬 = 𝐀𝐕𝐫 + 𝐁𝐈𝐫 = (174.31 + 𝑗25.53)kV = 176.17∠ 8.3326° kV

7

Medium line representation, T method of solution:

𝐕𝐜 = 𝐕𝐫 +𝐈𝐫 ⋅ 𝐙𝟐

= 275 ⋅ 103

√3+

(720 − 𝑗384.71)(7.488 + 38.2179)2

= (168.13 + 𝑗12.453)kV

𝐈𝐜 = 𝐕𝐜 ∗ 𝐘𝐜 = (168.13 + 𝑗12.453) ⋅ 𝑗0.001 = (−13.07 + 𝑗176.5)A

𝐈𝐬 = 𝐈𝐫 + 𝐈𝐜 = (706.9− 𝑗172.25)A

𝐕𝐬 = 𝐕𝐜 +𝐙 ⋅ 𝐈𝐬𝟐

= (168.13 + 𝑗12.453) +(7.488 + 𝑗38.2179)(706.9 × 172.3)

2

𝐕𝐬 = (174.07 + 𝑗25.317)𝑘𝑉 = 175.9025∠ 8.2751° kV

3.9 A 220 kV, 60 Hz three-phase transmission line is 320km long and has the following constants per phase per km:

Inductance 0.81 mH Capacitance 12.8 nF Resistance 0.038 Ω

Ignore leakage conductance. If the line delivers a load of 300A, 0.8 power factor lagging, at a voltage of 220 kV, calculate the sending-end voltage. Determine the π circuit which will represent the line.

Answer

𝐈𝐫 = 300cos𝜑 − 𝑗sin(cos−1(cos𝜑)) = (240 − j180)𝐴

𝐙 = 𝑙 ⋅ (𝑟 + 𝑗2πf𝐿) = (17.6 + j15.683)Ω

𝐘 = 𝑙 ⋅ 𝑦 = 𝑙 ⋅ (𝑗2πf𝐶) = 𝑗0.02654 S

Equivalent four-terminal network for long line: 𝐀 = 𝐃 = cosh√𝐙𝐘 = 0.9255 + 𝑗0.0092

𝐁 = 𝐙𝐘

sinh√𝐙𝐘 = 11.5553 + 𝑗95.3144

𝐂 = 𝐘𝐙

sinh√𝐙𝐘 = ( − 0.0048 + 𝑗1.5056) × 10−3

𝐕𝐬 = √𝟑(𝐀𝐕𝐫 + 𝐁 𝐈𝐫) = (238.13 + j38.03)kV = 241.14∠ 9.07° kV

π representation parameters:

Series element: 𝐁 = 11.5553 + 𝑗95.3144

Shunt element: 𝐀−𝟏𝐁

= (0.0125 + 𝑗7.8193) × 10−5

8

3.10 Calculate the A B C D constants for a 275 kV overhead line of length 83 km. The parameters per kilometre are as follows:

Resistance 0.078Ω Reactance 0.33 Ω Admittance (shunt capacitative) 9.53 x 10-6 S

The shunt conductance is zero.

Answer

𝐙 = 𝑙 ⋅ (𝑟 + 𝑗x) = 83 × (0.078 + j0.33)Ω = (6.474 + j27.39)Ω

𝐘 = 𝑗 ⋅ 𝑦 ⋅ 𝑙 = 𝑗 9.53 × 10−6S × 83 = 𝑗7.9099 × 10−4 S

This is a medium line; From π network representation

𝐀 = 𝐃 = 1 +𝐙𝐘2

= 0.98917 + j0.00256

𝐁 = 𝐙 = 6.474 + j27.39

𝐂 = 1 +𝐙𝐘4𝐘 = −1.0126 × 10−6 + j7.8671 × 10−4

3.11 A 132 kV, 60 Hz transmission line has the following generalized constants: A = 0.9696 ∠0.49˚ B = 52.88 ∠74.79˚Ω C = 0.001177 ∠90.15˚S

If the receiving-end voltage is to be 132kV when supplying a load of 125MVA at 0.9 p.f. lagging, calculate the sending-end voltage and current.

Answer

√3 × 132 × 103|𝐈𝐫| = 125 × 106 ; Therefore |𝐈𝐫| = 546.8 A 𝐈𝐫 = 546.8∠−𝑐𝑜𝑠−1(0.9) = 546.8∠ − 25.8419° A 𝐕𝐬 = √𝟑(𝐀𝐕𝐫 + 𝐁 𝐈𝐫) = (160.87 + j38.857)kV = 165.5∠ 13.58° kV

𝐈𝐬 = 𝐂𝐕𝐫 + 𝐃𝐈𝐫 = (478.83− j137.28) A = 498.12∠ − 16° A

3.12 Two identical transformers each have a nominal or no-load ratio of 33/11 kV and a leakage reactance of 2Ω referred to the 11kV side; resistance may be neglected. The transformers operate in parallel and supply load of 9 MVA, 0.8 p.f. lagging. Calculate the current taken by each transformer when they operate five tap steps apart (each step is 1.25% of the nominal voltage). Also calculate the kVAr absorbed by this tap setting.

Answer

S = 9 × (0.8 − 𝑗 sin(cos−1(0.8))) = (7.2 − 𝑗5.4) MVA

9

𝐈 = 𝐒√𝟑

𝐕𝐫∗

= (377.9− 𝑗283.4)A

𝐈circular = 𝑛𝑡 ⋅ 𝑣𝑡 ⋅ 𝐕𝐫√3 ⋅

1𝑗2𝑋𝑙

= 5 ×1.25100

×11 × 103

√3×

1𝑗2 × 2

= −𝑗99.2321 A

𝐈T1 =𝐈2

+ 𝐈circular = 189− 𝑗 240.9 = 306.1979∠ − 51.8962° A

𝐈T2 =𝐈2− 𝐈circular = 189− 𝑗 42.5 = 193.6676∠− 12.6709° A

𝑄 = 3 ⋅ 𝐼𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟2 ⋅ 2𝑋𝑙 = 118.2 kVAr

3.13 An induction motor, the equivalent circuit of which is shown in Figure 3.40 is connected to supply busbar which may be considered as possessing a voltage and frequency which is independent of the load. Determine the reactive power consumed for various busbar voltages and construct the Q-V characteristic. Calculate the critical voltage at which motor stalls and the critical slip, assuming that the mechanical load is constant.

Figure 3.40 Equivalent circuit of 500 kW, 6.6 kV induction motor in Problem 3.14. All p.u. values refer to rated voltage and power (P = 1 p.u. and V = 1 p.u.)

Answer

From equation 3.8 with P = 1 p.u.

𝑋2𝑠2 − 𝑉2𝑅2𝑠 + 𝑅22 = 0

∴ 𝑠 =𝑉2𝑅2 − 𝑉4𝑅22 − 4𝑋2𝑅22

𝑋2

From equivalent circuit

𝐼 =𝑉

𝑋2 + (𝑅2 𝑠⁄ )2

𝑄 =𝑉2

𝑋𝑚+ 𝐼2𝑋

With 𝑋𝑚 = 3 p. u., 𝑋 = 0.2 p. u. and 𝑅2 = 0.04 p. u., the above equations were used

10

to obtained Q-V characteristic

𝑉𝑐 = 2 ⋅ 𝑃 ⋅ 𝑋𝑠 = √2 × 1 × 0.2 = 0.6325p. u.

𝑆𝑐𝑟 =𝑟2𝑋𝑠

=0.040.2

= 0.2 p. u.

0.8 0.85 0.9 0.95 1 1.05 1.10.98

0.99

1

1.01

1.02

1.03

1.04

1.05

1.06

1.07

V p.u.

Q p

.u.

11

3.14 A 100 MVA round-rotor generator of synchronous reactance 1.5 p.u. supplies a substation (L) through a step up transformer of reactance 0.1 p.u., two lines each of reactance 0.3 p.u. in parallel and a step down transformer of reactance 0.1 p.u. The load taken at L is 100 MW at 0.85 lagging. L is connected to a local generating station which is represented by an equivalent generator of 75 MVA and synchronous reactance of 2 p.u. All reactances are expressed on a base of 100 MVA.

Draw the equivalent single-phase network. If the voltage at L is to be 1p.u. and the 75MVA machine is to deliver 50 MW, 20 MVAr, calculate the internal voltages of the machines.

Answer

X=0.1 p.u. X=0.1 p.u.

X=2 p.u.

XL=0.3 p.u.

XL=0.3 p.u.X=1.5 p.u.

Load

𝐙𝐞 = 𝑗 1.5 + 0.1 +0.32

+ 0.1 = 𝑗1.85

Busbar L:

𝐒𝐋 = 𝐒𝐥𝐨𝐚𝐝 − 𝐒𝐠𝐞𝐧𝟐 = (100(1 + 𝑗tan(cos−1 (0.85)) − (50 + 𝑗20)= (50 + 𝑗41.9744) 𝑀𝑉𝐴

𝐒𝐋,(𝐩.𝐮.) =𝐒𝐋

100 MVA= 0.5 + 𝑗0.4197

Generator 1:

𝐈 = 𝐒𝐋,(𝐩.𝐮.)

𝐕 ∗

= 0.5 − 𝑗0.4197

𝐄𝟏 = 𝐈 ∙ 𝐙𝐞 + 𝑉𝐿 = (0.5 − 𝑗0.4197) × 𝑗1.85 + 1 = 1.7765 + 𝑗0.925

= 2.0029 ∠ 27.505𝑜

Generator 2:

𝐈𝟐 = 𝐒𝐠𝐞𝐧𝟐𝐕

∗

= 50 + 𝑗20

100∗

= 0.5 − 𝑗0.2

𝐄𝟐 = 𝐈𝟐 ∙ 𝑗𝑋𝑠 + 𝑉𝐿 = (0.5− 𝑗0.2) × 𝑗2 + 1 = 1.4 + 𝑗1 = 1.7205 ∠ 35.5377o

12

3.15 The following data applies to the power system shown in Figure 3.41.

Generating station A: Four identical turboalternators, each rated at 16kV, 125 MVA, and of synchronous reactance 1.5 p.u. Each machine supplies a 125 MVA, 0.1 p.u. transformer connected to a busbar sectioned as shown.

Substation B: Two identical 150 MVA, three-winding transformers each having the following reactances between windings: 132/66 kV windings 10%; 66/11 KV windings 20%; 132/11kV windings 20%; all on 150MVA base.

The secondaries supply a common load of 200MW at 0.95 p.f. lagging. To each tertiary winding is connected a 30 MVA synchronous compensator of synchronous reactance 1.5 p.u.

Substation C: Two identical 150 MVA, transformers each of 0.15 p.u. reactance, supply a common load of 300 MW at 0.85 p.f. lagging.

Generating station D: Three identical 11kV, 75 MVA generators, each of 1p.u. synchronous reactance, supply a common busbar which is connected to an outgoing 66kV cable trough two identical 100 MVA transformers each of 0.1 p.u. reactance. Load 50 MW, 0.8 p.f. lagging.

Determine the equivalent circuits for balanced operation giving component values on a base of 100 MVA. Treat the loads as impedances.

Figure 3.41 Line diagram of system in Problem 3.15

13

Answer

Substation A generators

Vbase=16 kV; Sbase=100 MVA

X=1.5x100/125=1.2 p.u.

Substation A transformers

Sbase=100 MVA X=0.1x100/125=0.08 p.u.

48 km, 258 mm2 line Zbase = 1322/100 = 174.24

From Table 3.2(a) Z=(0.068+j0.404)x48/174.24 = 0.0187+j0.1113

48 km, 113 mm2 line Zbase = 174.24 From Table 3.2(a) Z=(0.155+j0.41)x48/174.24 = 0.0427+j0.113

80 km, 258 mm2 line Zbase = 174.24 From Table 3.2(a) Z=(0.068+j0.404)x80/174.24 = 0.03122+j0.1855

Substation B Synch condensator


X=1.5x100/30=5 p.u.

Star equivalent of the 132/66/11 kV transformer

X132+X11=0.2x100/150 =0.133 p.u. X132+X66 =0.1x100/150 =0.067 p.u. X66+X11 =0.2x100/150 =0.133 p.u. 132 side = 0.5(0.133+0.067-0.133) = 0.033 p.u. 66 kV side = 0.5(0.067+0.133-0.133) = 0.033 p.u 11 kV side =0.5(0.133+0.133-0.067) = 0.1 pu

Substation C transformer

Sbase=100 MVA X=0.15x100/150=0.1 p.u.

Substation C load Vbase=66 kV; Sbase=100 MVA

S = I = 3 – j1.86 p.u Z = 1/I = 0.2408 + j0.1493

Substation D generators


X=1.0x100/75=1.333 p.u.

Substation D transformer

Sbase=100 MVA X = 0.1 p.u.

Substation D load Vbase=66 kV; Sbase=100 MVA

S = I = 0.5 – j0.375 p.u Z = 1/I = 1.28 + j0.96

Cable Zbase = 662/100 = 43.56 Z=(2+j4)/43.56 = 0.0459+j0.0918

3.16 Distinguish between kW, kVA and kVAr. Explain why

(a) generators in large power systems usually run overexcited, generating VAr. (b) remote hydro generators need an underexcited rating so that they can absorb

VAr. (c) loss of an overexcited generator in a power system will normally cause a drop

in voltage at its busbar.

14

A load of 0.8 p.u. power and 0.4 p.u. VAr lagging is supplied from a busbar of 1.0 p.u. voltage through an inductive line of reactance 0.15p.u.. Determine the load terminal voltage assuming that p.u. current has the same value as p.u. VA.

(From E.C. Examination, 1996)

Answer

If .p.u. current has the same value as p.u. VA:

𝐈 = 0.8− 𝑗0.4

𝐕𝐭 = 𝐕𝐬 − 𝑗𝑋𝑙𝐈 = 1 − 𝑗0.15 × (0.8− 𝑗0.4) = 0.94− 𝑗0.12 = 0.9476∠ −7.275𝑜

3.17 Sketch the performance chart of a synchronous generator indicating the main

limits. Consider a generator with the following nameplate data 500MVA, 20 kV, 0.8p.f.( power factor), Xs = 1.5p.u.

(a) Calculate the internal voltage and power angle of the generator operating at 400 MW with cosφ=0.8 (lagging) with 1p.u. terminal voltage.

(b) What is the maximum reactive power this generator can absorb from the system?

(c) What is the maximum reactive power this generator can deliver to the system, assuming a maximum internal voltage of 2.25p.u.?

(d) Place the numerical values calculated on the performance chart. A graphical solution is acceptable.

(From E.C. Examination, 1996) Answer

a) 𝑆𝐵 = 500MVA

𝑆𝑝.𝑢. =𝑆𝑆𝐵

= 0.8 + 𝑗0.6

𝐄 = 𝑉𝑇 + 𝑗𝑋𝑠 ∙ 𝐈 = 1 + 𝑗1.5 × 𝑺𝒑.𝒖.

𝑉𝑇∗

𝐄 = 1 + 𝑗1.5 × 0.8−𝑗0.61

= 1.9 + 𝑗1.2 = 2.2472 ∠ 32.2756𝑜

b) 𝑄 = 𝑉𝑇𝑋𝑠

(𝐸 ∙ cos𝛿 − 𝑉𝑇)

𝑄𝑀𝐴𝑋,𝑝.𝑢. = −𝑉𝑇2

𝑋𝑠= −

11.5

= −0.6667, cos𝛿 = 0

𝑄𝑀𝐴𝑋 = 𝑄𝑀𝐴𝑋,𝑝.𝑢. ∙ 𝑆𝐵 = −0.6667 × 500 MVA = 333.33 MVAr

c) 𝛿 = 0 ⟹ 𝑄𝑀𝐴𝑋,𝑑𝑒𝑙𝑖𝑣.

15

𝑄𝑀𝐴𝑋,𝑑𝑒𝑙𝑖𝑣.𝑝.𝑢. =𝑉𝑇𝑋𝑠

(𝐸 ∙ cos𝛿 − 𝑉𝑇) =1

1.5(2.25 − 1) = 0.8333

𝑄𝑀𝐴𝑋,𝑑𝑒𝑙𝑖𝑣. = 𝑄𝑀𝐴𝑋,𝑝.𝑢. ∙ 𝑆𝐵 = 0.8333 × 500 MVA = 416.67 MVAr

d)

500

ab c

d

e fg h

j

E=V=1 p.u

E=2.25 p.u

500

400

MVAr

MW

0.8 lagging

MVA limit

Excitation limit

Stabilitylimit

Prime mover limit

Chapter 4

Problems

4.1 A 500 MVA, 2 pole, turbo-alternator delivers 400 MW to a 50 Hz system. If the generator circuit breaker is suddenly opened and the main steam valves take 400 ms to operate what will be the over-speed of the generator? The stored energy of the machine (generator and turbine) at synchronous speed is 4 kWs/kVA.

Answer

Initial stored energy 64 500 10 2000MWs 2000 MJ= × × = = Additional energy added to spinning mass 6400 10 0.4 160MJ= × × =

Increase in speed = 2160 3000 3117.9 r.p.m or 52 Hz2000

× =

4.2 Two identical 60 MW synchronous generators operate in parallel. The governor

settings on the machines are such that they have 4 and 3% droops (no-load to full-load percentage speed drop). Determine (a) the load taken by each machine for a total of 100MW; (b) the percentage adjustment in the no-load speed to be made by the speeder motor if the machines are to share the load equally.

Answer As two generators are identical it is reasonable to assume that their open circuit frequencies are identical.

60 MW 60 MW

4% 3%

α

x 100- x

Assume base MVA as 100 MVA: From the figure

0.0460xα

= and ( )0.03

100 60xα

=−

0.04 3 0.030.07 3

42.9 MW100 - 57.1 MW

x xxxx

= −===

So power delivered by generators are 42.9 MW and 57.1 MW

For equal sharing of power, the no-load speed of 4% droop machine should be increased as shown in the figure.

4%3%

α

50 MW 50 MW

α’

f

If the operating frequency is f p.u.

No load speed of 3% generator = 500.03 0.025 p.u.60

f f+ × = +

The no load speed of 4% machine before change in its no load speed is equal to 0.015 p.u.f +

New no load speed of 4% generator = 500.04 0.0333 p.u.60

f f+ × = +

% increase required = ( 0.0333) -( 0.0250) 100%=0.83%f f+ + ×

4.3 a) Explain how the output power of a turbine-generator operating in a constant frequency system is controlled by adjusting the setting of the governor. Show the effect on the generator power-frequency curve.

b) Generator A of rating 200 MW and generator B of rating 350 MW have governor droops of 5 and 8%, respectively, from no-load to full-load. They are the only supply to an isolated system whose nominal frequency is 50 Hz. The corresponding generator speed is 3000 r.p.m. Initially, generator A is at 0.5 p.u. load and generator B is at 0.65 p.u. load, both running at 50 Hz. Find the noload speed of each generator if it is disconnected from the system.

c) Also determine the total output when the first generator reaches its rating.

Answer b) For Generator A: For Generator B

1 0.05100 200

1 0.025

α

α

=

∴ =

2 0.08227.5 350

2 0.052

α

α

= =

∴ =

Therefore open circuit speed of generator A = 1.025 x 3000 = 3075 rev/min Open circuit speed of generator B = 1.052 x 3000 = 3156 rev/min

200 MW350 MW

5%

8%

α1

100 MW 227.5 MW

α2

50 Hz

c) At the rating of generator A, speed = 1.025-0.05 = 0.975 p.u

Open circuit speed of generator B = 1.052 p.u. So power generated by generator B at a speed of 0.975 p.u.

=1.052 0.975 350 337 MW

0.08−

× =

Therefore total power = 537 MW

4.4 Two power systems A and B are interconnected by a tie-line and have power frequency constants KA and KB MW/Hz. An increase in load of 500 MW on system A causes a power transfer of 300 MW from B to A. When the tie-line is open the frequency of system A is 49 Hz and of system B 50 Hz. Determine the values of KA and KB, deriving any formulae used.

Answer For system A: For system B:

500 300 200

A Af

K K−

= = ∆ 300

Bf

K= ∆

Therefore, 1.5B AK K=

As 300A

Af f

K= − and 300

BB

f fK

= +

300 300

1 150 49 3001.5

5001

500 MW/Hz

B AA B

A A

A

A

f fK K

K K

KK

− = +

− = +

=

=

1.5 750 MW/HzB AK K= =

4.5 Two power systems, A and B, having capacities of 3000 and 2000 MW, respectively, are interconnected through a tie-line and both operate with frequency-bias-tieline control. The frequency bias for each area is 1% of the system capacity per 0.1 Hz frequency deviation. If the tie-line interchange for A is set at 100MW and for B is set (incorrectly) at 200 MW, calculate the steady-state change in frequency.

Answer 0.01 0.01100 3000 200 20000.1 0.1

100 300 200 2001Hz

f f

f ff

+ × ∆ = + × ∆

+ ∆ = + ∆∆ =

4.6 a) (i) Why do power systems operate in an interconnected arrangement? (ii) How is the frequency controlled in a power system? (iii) What is meant by the stiffness of a power system?

b) Two 50 Hz power systems are interconnected by a tie-line, which carries

1000MW from system A to system B, as shown in Figure 4.16. After the outage of the line shown in the figure, the frequency in system A increases to 50.5 Hz, while the frequency in system B decreases to 49 Hz.

Figure 4.16 Interconnected systems of Problem 4.7 (b)

(i) Calculate the stiffness of each system. (ii) If the systems operate interconnected with 1000 MW being transferred from A to B, calculate the flow in the line after outage of a 600 MW generator in system B.

Answer a) When a change occurs, AP∆ =1000 MW and BP∆ =-1000 MW

1000

100050.5 50

2000 MW/Hz

AA

A

A

f fK

KK

= +

= +

=

and

1000

100049 50

1000 MW/Hz

BB

B

B

f fK

KK

= −

= −

=

b) If change of flow in line is P∆

For system A: A

P fK−∆

= ∆

For system B: 600

B

P fK

− + ∆= ∆

Therefore:

600

6002000 1000

400 MW

A B

P PK K

P P

P

−∆ − + ∆=

−∆ − + ∆=

∆ =

So flow in the line is 1000 + 400 = 1400 MW

Chapter 5

Problems

5.1 An 11kV supply busbar is connected to an 11/132 kV, 100 MVA, 10% reactance transformer. The transformer feeds a 132 kV transmission link consisting of an overhead line of impedance (0.014+j0.04) p.u. and a cable of impedance (0.03+j0.01) p.u. in parallel. If the receiving end is to be maintained at 132 kV when delivering 80 MW, 0.9 p.f. lagging, calculate the power and reactive power carried by the cable and the line. All p.u. values relate to 100 MVA and 132 kV bases.

Answer

At the receiving end

P = 80/100 = 0.8 p.u

Q = 0.8xtan(cos-10.9) = 0.39 p.u

I = 0.8 – j0.39

Current through the line = 0.03 0.01 (0.8 0.39)(0.014 0.04) (0.03 0.01)

j jj j

+× −

+ + +

= 0.2350 - j0.3512

Power through the line = 0.2350 + j0.3512 p.u = 23.5 + j35.1 MVA

Current through the cable = 0.014 0.04 (0.8 0.39)(0.014 0.04) (0.03 0.01)

j jj j

+× −

+ + +

= 0.5650 - j0.0388

Power through the line = 0.5650 + j0.0388 p.u = 56.5 + j3.89 MVA

5.2 A three-phase induction motor delivers 500 hp at an efficiency of 0.91, the operating power factor being 0.76 lagging. A loaded synchronous motor with a power consumption of 100 kW is connected in parallel with the induction motor. Calculate the necessary kVA and the operating power factor of the synchronous motor if the overall power factor is to be unity.

Answer

Assume 100 kVA base

Active power drawn by the motor = 500 x 746/0.91 = 409.9 kW = 4.099 p.u

Reactive power = 4.1tan(cos-10.76) = 3.505 p.u

Current drawn by the motor = 4.099 – j3.505

If the overall power factor is to be unity,

Current drawn by the synchronous motor = 1 + j3.505 = 3.645 74.08∠

MVA of the synchronous motor = 364.5 MVA

Power factor = cos 74.08o=0.274

5.3 The load at the receiving end of a three-phase overhead line is 25 MW, power factor 0.8 lagging, at a line voltage of 33 kV. A synchronous compensator is situated at the receiving end and the voltage at both ends of the line is maintained at 33 kV. Calculate the MVAr of the compensator. The line has resistance of 5 Ω per phase and inductive reactance of 20 Ω per phase.

Answer

For the load PL = 25 MW and QL = 25tan(cos-10.8) = 18.75 MVAr

When 0RP XQVV+

∆ = =

5 25 6.25 MVAr20

RPQX

×= − = − = −

Therefore Q that should be supplied by the synchronous compensator

= 18.75 +6.25 = 25 MVAr

5.4 A transformer connects two infinite busbars of equal voltage. The transformer is rated at 500 MVA and has a reactance of 0.15 p.u. Calculate the VAr flow for a tap setting of (a) 0.85:1; (b) 1.1:1.

Answer

1 p.u 1 p.u1/t

0.15/t2I

21/ 1

0.15 /tI

t−

=

Therefore VAr flow = 21/ 1

0.15 /t

t−

When t=0.85

VAr flow = 21/ 0.85 1 0.85

0.15 / 0.85−

= p.u = 425 MVAr

When t=1.1

VAr flow = 21/1.1 1 0.733

0.15 /1.1−

= − p.u = -367 MVAr

5.5 A three-phase transmission line has resistance and inductive reactance of 25 Ω and 90 Ω, respectively. With no load at the receiving end, a synchronous compensator there takes a current lagging by 90°; the voltage is 145 kV at the sending end and 132 kV at the receiving end. Calculate the value of the current taken by the compensator.

When the load at the receiving end is 50 MW, it is found that the line can operate with unchanged voltages at the sending and receiving ends, provided that the compensator takes the same current as before, but now leading by 90°.

Calculate the reactive power of the load.

Answer

Sending end phase voltage = 83.72 kV

Receiving end phase voltage = 76.21 kV

∆V = 7.51 kV

When there is no load on the receiving end

33

90 7.51 1076.21 10

QV∆ = = ××

Q = 6.36 MVAr

VI = 6.36 x 106

I = 83.45 A

When 50 MW load is connected, PL = 50/3 = 16.67 MW

If load reactive power is QL and as compensator takes the same current in the

reverse direction,

Q = QL/3- 6.36

6 6

33

25 16.67 10 90 6.36 1037.51 10

76.21 10

LQ

V

× × + × − × ∆ = = ××

QL = 24.27 MVAr

5.6 Repeat Problem 5.3 making use of Q V∂ ∂ at the receiving end.

Answer

Use Bases of 100 MVA and 33 kV.

ZBASE

233 10.89100

= = Ω

6

BASE 3

100 10I 1749.5A3 33 10

×= =

× ×

5 20 0.46 1.83610.89

j j+= = +Z

Fault current (ignoring resistance)

1 1749.5 0.953 kA1.836FAULTI = × =

953 3 1.65 /Q MVAr kVV∂

= × =∂

Volt drop due to transport of real power

25 5 33 3.78 kV100 10.89

3.78 1.65 6.25 MVAr

V PR

Q

∆ = = × × =

= × =

Total Q required is 6.25+18.75=25 MVAr.

5.7 In Example 5.3, determine the tap ratios if the receiving-end voltage is to be maintained at 0.9 p.u. of the sending-end voltage.

Answer

Sending end voltage = 1.0 p.u

Receiving end voltage = 0.9 p.u

[ ]

2 2 2

2 2 2

2

( )

0.9 0.9 0.14 1 0.38 0.48

0.3224

s s r r s

s s

s

t V V V PR QX t

t t

t

− = +

− = × + ×

=

2 1.41.1841 0.844

s

s

rs

tt

tt

=

=

= =

5.8 In the system shown in Figure 5.28, determine the supply voltage necessary at D to maintain a phase voltage of 240 V at the consumer's terminals at C. The data in Table 5.2 apply.

Figure 5.28 Line diagram for system in Problem 5.8

Table 5.2 Data for Problem 5.8

Line or transformer

Rated voltage kV

Rating MVA Nominal tap ratio

Impedance (Ω)

BC 0.415 0.0217+j0.00909 AB 11 1.475+j2.75 DA 33 1.475+j2.75 TA 33/11 10 30.69/11 1.09+j9.8

Referred to 33 kV TB 11/0.415 2.5 10.45/0.415 0.24+j1.95

Referred to 11 kV

Answer

Line

or

trans

form

er

Rat

ed v

olta

ge

kV

Rat

ing

MVA

Nom

inal

tap

ratio

Impe

danc

e (Ω

)

Impe

danc

e p.

u.

10 M

VAB

AS

E

BC 0.415 0.0217+j0.00909 ZBASE=0.0172Ω

1.26+j0.528

AB 11 1.475+j2.75 ZBASE=12.1Ω

0.122+j0.227

DA 33 1.475+j2.75 ZBASE=108.9Ω

0.0135+j0.0252

TA 33/11 10 30.69/11 1.09+j9.8

Referred to 33 kV

ZBASE=108.9Ω

0.01+j0.09

TB 11/0.415 2.5 10.45/0.415 0.24+j1.95

Referred to 11 kV

ZBASE=12.1Ω

0.02+j0.16

Ignore losses in circuit. Use 10 MVA base.

At C:

P=0.12MW, Q=0.09MVAr, P=0.012 p.u., Q=0.009p.u.

ΔV=0.012x1.26+0.009x0.528=0.0198 p.u.

At B:

P=1.08MW, Q=0.81MVAr, P=0.108p.u.,Q=0.081

ΔV=0.108x0.142+0.081x0.387=0.047 p.u.

At A:

P=5.88MW, Q=4.41MVAr, P=0.588p.u.,Q=0.441

ΔV=0.588x0.0235+0.441x0.1152=0.065 p.u.

Therefore total volt drop is 1.98+4.7+6.5=13.2%

Boost through transformers = 1/(0.93x0.95)-1=0.1319 = 13.2%

So Busbar A should be 33 kV

5.9 A load is supplied through a 275 kV link of total reactance 50 Ω from an infinite busbar at 275 kV. Plot the receiving-end voltage against power graph for a constant load power factor of 0.9 lagging. The system resistance may be neglected.

Answer

QL = P.tan(cos-10.9)=0.484P

3

2

50 8.073

275 10 8.07 158771.3 8.073

158771.3 8.07 0

Q PVV V

P PVV V

V V P

∆ = =

×= − = −

− + =

10158771.3 2.52 10 32.28

2PV ± × −

=

Value closer to 275 kV was selected,

10158771.3 2.52 10 32.283

2LPV

+ × − =

0 50 100 150 200 250 300245

250

255

260

265

270

275

Power (MW)

Line

Vol

tage

(kV

)

5.10 Describe two methods of controlling voltage in a power system.

(a) Show how the scalar voltage difference between two nodes in a network is given approximately by:

RP XQVV+

∆ =

Each phase of a 50 km, 132 km overhead line has a series resistance of 0.156 Ω/km and an inductive reactance of 0.4125 Ω/km. At the receiving end the voltage is 132 kV with a load of 100 MVA at a power factor of 0.9 lagging. Calculate the magnitude of the sending-end voltage.

(b) Calculate also the approximate angular difference between the sending-end and receiving-end voltages.

Answer

(a) P per phase = 100 x 0.9/3 = 30 MW

Q per phase = 30 tan(cos-10.9) = 14.52 MVAr

0.156 50 30 0.4125 50 14.52 7 kV132 3

RP XQVV+ × × + × ×

∆ = = =

Sending end voltage = 132 3 7 144.1 kV+ × =

(b) 30.4125 50 30 0.156 50 14.52 10 6633132 3

XP RQV

δ − × × − × ×

∆ = = × =

3

6633tan 0.0797144.1 10 3

4.56

δ

δ

= =×

=

5.11 Explain the limitations of tap-changing transformers. A transmission link (Figure 5.29(a) connects an infinite busbar supply of 400 kV to a load busbar supplying 1000 MW, 400 MVAr. The link consists of lines of effective impedance (7+j70)Ω feeding the load busbar via a transformer with a maximum tap ratio of 0.9:1. Connected to the load busbar is a compensator. If the maximum overall voltage drop is to be 10% with the transformer taps fully utilized, calculate the reactive power requirement from the compensator.

Note: Refer voltage and line Z to load side of transformer in Figure 5.29(b).

2 2S

RR

RP XQV t tVt V

+ = −

Figure 5.29 Circuits for Problem 5.11

Answer

400 230.94 kV3SV = =

P per phase = 333.33 MW

Q per phase = (133.3-Qcompensator) MW

If RV needs to be maintained at 10%

230.94 0.9 207.85 kVRV = × =

63

33 2

7 333 10 70(133.33 )230.94 10207.85 100.9 207.85 10 0.9

compensatorQ × × + −×× = − × ×

Compensator reactive power per phase, compensatorQ = 49.35 MVAr

Therefore compensator should supply 148 MVArs.

5.12 A generating station consists of four 500 MW, 20 kV, 0.95 p.f. (generating VArs) generators, each feeding through a 525 MVA, 0.1 p.u. reactance transformer onto a common busbar. It is required to transmit 2000 MW at 0.95 p.f. lagging to a substation maintained at 500 kV in a power system at a distance of 500 km from the generating station. Design a suitable transmission link of nominal voltage 500 kV to achieve this, allowing for a reasonable margin of stability and a maximum voltage drop of 10%. Each generator has synchronous and transient reactances of 2 p.u. and 0.3 p.u. respectively, and incorporates a fast-acting automatic voltage regulator. The 500 kV transmission lines have an inductive reactance per phase of 0.4Ω/km and a shunt capacitive susceptance per phase of 3.3 x 10-6 S/km. Both series and shunt capacitors may be used if desired and the number of three-phase lines used should be not more than three-fewer if feasible. Use approximate methods of calculation, ignore resistance, and state clearly any assumption made. Assume shunt capacitance to be lumped at the receiving end only.

Answer

Assume one line:

On 500 MVA, 500 kV base

Zbase = 5002/500=500 Ω

Xline = 0.4 x 500 = 200 Ω

Assume that there is 70% compensation so X = 60 Ω = 0.12 pu

Transformer reactance = (0.1x500/525)/4 = 0.023 pu

For the load P = 2000 MW = 4 p.u.

cos-10.95 = 18.19o

Q = 2000 x tan(18.19o) = 657.37 MVAr = 1.314 p.u.

Line suceptance in p.u. = 3.33x10-6x500x500 = 0.8325

Assuming that receiving end voltage is V

Q supplied by the line = 20.8325V

Taking the voltage drop from generator terminal to load bus; from approximate equation:

( )20.143 1.314 0.83251

VXQVV V

× −− = =

20.8809 0.1879 0V V− + =

Solving the above quadratic equation: V =0.898 p.u.

As the voltage is less than the required voltage this is not a satisfactory design.

Assume two lines in parallel:

In this case with a 70% compensation, line reactance is = 30 Ω = 0.06 pu

Transformer reactance = (0.1x500/525)/4 = 0.023 pu



Taking the voltage drop from generator terminal to load bus; from approximate equation:

( )20.083 1.314 0.83251

VXQVV V

× −− = =

20.9309 0.1091 0V V− + =

Solving the above quadratic equation: V =0.951 p.u.

This design is suitable.

5.13 It is required to transmit power from a hydroelectric station to a load centre 480 km away using two lines in parallel for security reasons.

Assume sufficient bundle conductors are used such that there are no thermal limitations, and the effective reactance per phase per km is 0.44 Ω and that the resistance is negligible. The shunt capactive susceptance of each line is 2.27x10-6 S per phase per km, and each line may be represented by the nominal π-circuit with half the capacitance at each end. The load is 2000 MW at 0.95 lagging and is independent of voltage over the permissible range.

Investigate, from the point of view of stability and voltage drop, the feasibility and performance of the link if the sending-end voltage is 345, 500, and 765 kV assuming the transmission angle is not to exceed 30°.

The lines may be compensated up to 70 per cent by series capacitors and at the load-end compensators of 120 MVAr capacity are available. The maximum permissible voltage drop is 10%. As two lines are provided for security reasons, your studies should include the worst-operating case of only one line in use.

Answer

Assume voltage = 345 kV

Select Vb=345 kV and Sb = 500 MVA

Zb = 238.05 Ω

Under worst case scenario, it was assumed that only one line is in operation.

Zline = 480 x 0.44 = 211.2 Ω

If 70% of series compensation provided

Z = 0.3 x 211.2 = 63.36 Ω = 0.266 p.u.

Susceptance of the line

Y = 2.27 x 10-6 x 480 = 0.00109 S = 0.00109 x 238.05 p.u. = 0.259 p.u.


cos-10.95 = 18.19o

Q = 2000 x tan(18.19o) = 657.37 MVAr = 1.314 p.u.



Q supplied by the shunt compensator = 2

2120 0.24500

V V=

From approximate equation:

( )2 20.266 1.314 0.259 0.241 (A)

V VXQVV V

× − −− = = →

20.8672 0.3495 0V V− + =

There is no feasible answer. So 345 kV design will not be satisfactory.

Note that from equation (A) (in the form 2 0aV bV c− + = :

If shunt compensation becomes zero, still 4ac>b2 so no solution

If series compensation becomes zero, still 4ac>b2 so no solution



Zb = 500 Ω


Zline = 480 x 0.44 = 211.2 Ω


Z = 0.3 x 211.2 = 63.36 Ω = 0.127 p.u.


Y = 2.27 x 10-6 x 480 = 0.00109 S = 0.00109 x 500 p.u. = 0.545 p.u.


cos-10.95 = 18.19o

Q = 2000 x tan(18.19o) = 657.37 MVAr = 1.314 p.u.




2120 0.24500

V V=


( )2 20.127 1.314 0.545 0.241 (A)

V VXQVV V

× − −− = = →

20.9 0.167 0V V− + =

V =0.986 p.u.

However as line is a nominal π-circuit, the above approximation is not true.

ZY/2 = -0.0346

A = D = 1 + ZY/2 = 0.9654

B = Z = 0.127

Approximate line current =

0.52 24 0.534.04

1

+ = p.u.

Receiving end voltage:

(1 0.127 4.04 18.19 ) / 0.96550.87 0.505

=1.006 30.13

jj

= +

= − × ∠−= −

∠−

S R R

R

V AV BI

V

Evan though voltage is within the required limit, angle is not. So not a satisfactory design.



Zb = 1170.45 Ω


Zline = 480 x 0.44 = 211.2 Ω


Z = 0.3 x 211.2 = 63.36 Ω = 0.054 p.u.


Y = 0.00109 x 1170.45 p.u. = 1.276 p.u.


cos-10.95 = 18.19o

Q = 2000 x tan(18.19o) = 657.37 MVAr = 1.314 p.u.




2120 0.24500

V V=


( )2 20.054 1.314 1.276 0.241 (A)

V VXQVV V

× − −− = = →

20.9181 0.071 0V V− + =

V =1.012 p.u.

However as line is a nominal π-circuit, the above approximation is not true.

ZY/2 = -0.0345

A = D = 1 + ZY/2 = 0.9655

B = Z = 0.054

Approximate line current =

0.52 24 0.2024.005

1

+ = p.u.

Receiving end voltage:

(1 0.054 4.005 18.19 ) / 0.96550.966 0.213

=0.989 12.4

jj

= +

= − × ∠−= −

∠−

S R R

R

V AV BI

V

Satisfactory design

5.14 Explain the action of a variable-tap transformer, showing, with a phasor diagram, how reactive power may be despatched from a generator down a mainly reactive line by use of the taps. How is the level of real power despatch controlled?

Power flows down an H.V. line of impedance 0 + j0.15 p.u. from a generator whose output passes through a variable-ratio transformer to a large power system. The voltage of the generator and the distant large system are both kept at 1.0 p.u. Determine the tap setting if 0.8 p.u. power and 0.3 p.u. VAr are delivered to a lagging load at the power system busbar. Assume the reactance of the transformer is negligible.

Answer 2

2

2

1 0.3 0.15

0.045 1 0

R SV tV XQt

t t

t t

= −

= − ×

− + =

Therefore

1 1 4 0.045 1 0.90552 0.045 2 0.045

1.05

t

t

± − × ±= =

× ×=

5.15 Two substations are connected by two lines in parallel, of negligible impedance, each containing a transformer of reactance 0.18 p.u. and rated at 120 MVA. Calculate the net absorption of reactive power when the transformer taps are set to 1:1.15 and 1:0.85, respectively (i.e. tap-stagger is used). The p.u. voltages are equal at the two ends and are constant in magnitude.

Answer

Voltage drop across the reactance of 1:1.15 transformer

11 1 0.13

1.15V∆ = − = − p.u.

Voltage drop across the reactance of 1:0.85 transformer

21 1 0.176

0.85V∆ = − = p.u.

Reactive power absorbed = 2 2

1 2 0.266V VX

∆ + ∆= p.u = 31.9 MVAr

Chapter 6

6.1 Consider a simple power system composed of a generator, a 400kV overhead transmission line and a load, as shown in Figure 6.14.

Figure 6.14 400 kV system for Example 6.1

If the desired voltage at the consumer busbar is 400 kV, calculate the:

(i) Active and reactive losses in the line (ii) Voltage magnitude and phase angle at busbar 1. (iii) Active and reactive output of the generator at busbar 1

Answer

(i) 3*

600 210 10 0.866 0.303 kA3 400

L L

L

P jQ j j− −= = × = −

×I

V

0.917 kAI =

Active losses = 2 23 3 0.917 2.7 6.82I R = × × = MW

Reactive losses = 2 23 3 0.917 30.4 76.7I X = × × = MVAr

(ii) Busbar 1 voltage 3

3400 10 (0.866 0.303) 10 (2.7 30.4)3

j j×= + − × × +

= 243.84 6∠

Line to line voltage = 422.34 kV and angle = 6

(iii) Active power at the generator = 600 + 6.8 = 606.6 MW Reactive power at the generator = 210 + 76.7 = 286.7 MW

6.2 A load of 1+j0.5 p.u. is supplied through a transmission line by a generator G1 that maintains its terminal voltage at 1 p.u. (Figure 6.15)

Figure 6.15 System for example 6.2

(i) Form the YBUS matrix for this system. (ii) Perform two iterations of the Gauss – Seidel load flow.

Answer

(i) 1 1

2.5 2.50.4 0.4Y

1 1 2.5 2.50.4 0.4

bus

j jj jj j

j j

− − = = − −

(ii) Assume (0)2V =1.0 p.u

2

(1) 2 221 1(0)*

222

1

1 1 0.5 0.4 0.2 12.5 12.5 1

0.8 0.4 0.894 26.57

P jQ

j jjj

j

− −=

− + = = − − +− × −

= − = ∠−

Y VVVY

2

(2) 2 221 1(1)*

222

1

1 1 0.5 2.5 12.5 0.894 26.57

0.6 0.3

P jQ

j jj

j

− −=

− + = − × − ∠ = −

Y VVVY

p.u.

6.3 List the information which may be obtained from a load-flow study. Part of a power system is shown in Figure 6.16. The circuit reactances and values of real and reactive power (in the form P jQ± ) at the various busbars are expressed as per unit values on a common MVA base. Resistance may be neglected. By the use of an iterative method, calculate the voltages at the stations after the first iteration.

G1

Figure 6.16 System for Problem 6.3

Answer

4 2 0 22 3 1 0

0 1 2 12 0 1 3

bus

j j jj j j

j j jj j j

− − = − −

Y

Assume (0)2V =1.0 p.u, (0)

3V =1.0 p.u and (0)4V =1.0 p.u

2

(0)(1) 2 221 1 23 3(0)*

222

1

1 0.1 2 1.05 1 13 1

1.0333 0.0333

P jQ

j jj

j

− − −= − = − × − × −

= −

Y V Y VVVY

3

(1) (0)3 3(1)32 2 34 4(0)*

333

1

1 0.5 0.2 1 (1.0333 0.0333) 1 12 1

1.1167 0.2333

P jQ

j j j jj

j

− − −=

− = − × − − × − = +

Y V Y VVVY

4

(1)(1) 4 441 1 43 3(0)*

444

1

1 0.4 0.05 2 1.05 1 (1.1167 0.2333)3 1

1.0556 0.0556

P jQ

j j j jj

j

− − −= − + = − × − × + −

= −

Y V Y VVVY

6.4 A 400 kV interconnected system is supplied from busbar A, which may be considered to be an infinite busbar. The loads and line reactances are as indicated in Figure 6.17.


Determine the flow of power in line AC using two iterations of voltages at each bus.

Answer

Assuming Sbase = 1 GW

Therefore 2400 160

1000baseZ = = Ω

96 40 16 4040 66.67 26.67 016 26.67 74.67 3240 0 32 72

bus

j j j jj j jj j j jj j j

− − = − −

Y

(0)(1)(0)*

1

1 1 40 1 26.67 166.67 1

1.0 0.015

B BBA A BC CB

BBB

P jQ

j jj

j

− − −=

− = − × − × − = −

Y V Y VVVY

(1) (0)(1)(0)*

1

1 3 16 1 26.67 (1.0 0.015) 32 174.67 1

1.0 0.0455

C CCA A CB B CD DC

CCC

P jQ

j j j jj

j

− − − −=

− = − × − × − − × − = −

Y V Y V Y VVVY

(1)(2)(1)*

1

11 40 1 26.67 (1.0 0.0455)1.0 0.01566.67

0.9998 0.0332

B BBA A BC CB

BBB

P jQ

j j jjj

j

− − −=

− − × − × −= +− = −

Y V Y VVVY

(2) (1)(2)(1)*

1

1 3 16 1 26.67 (0.9998 0.0332) 32 (1.0 0.0455)74.67 1

0.9981 0.0725

C CCA A CB B CD DC

CCC

P jQ

j j j j jj

j

− − − −=

− = − × − × − − × − − = −

Y V Y V Y VVVY

Power flow in line AC = ( )Re A AC A Cy − V V V

= ( )Re 1 16 1 0.9981 0.0725 1.16 p.u.j j× × − + =

= 1.16 GW

6.5 Determine the voltage at busbar (2), voltage at busbar (3) and the reactive power at bus (3) as shown in Figure 6.18, after the first iteration of a Gauss-Seidel load flow method. Assume the initial voltage to be 1 0∠ p.u. All the quantities are in per unit on a common base.


Answer

25 20 520 60 405 40 45

bus

j j jj j jj j j

− = − −

Y

(0)(1) 2 221 1 23 32 (0)*

222

1

1 0.8 0.6 20 1 40 160 1

0.99 0.0133

P jQ

j j jj

j

− − −=

− + = − × − × − = −

Y V Y VVVY

[ ]

(0)* (1) (0)3 3 31 1 32 2 33 3Im (

Im 1 ( 5 1 40 (0.99 0.0133) 45 10.4

Q

j j j j

= − + + = − × × + × − − ×

=

V Y V Y V Y V

(1)3 3(1)31 1 33 23 (0)*

333

1

1 0.6 0.4 5 1 40 (0.99 0.0133)45 1

1.0 0.0015

P jQ

j j j jj

j

− − −=

− = − × − × − − = +

Y V Y VVVY

(1)3 1.0 0.0015 1V j= + =

(1) (1) (1)3 3 3/ ( )

1.0 0.0015absj

== +

V V V

6.6 Active power demand of the 132 kV system shown in Figure 6.19 is supplied by two generators G1 and G2. System voltage is supported by generator G2 and a large synchronous compensator SC (see Figure 6.19), which both maintain the voltage at 1 pu at their respective nodes. Generator G1, connected at node 1, has no reactive power capacity available for voltage control.

Figure 6.19 132 kV system for problem 6.6

(i) Form the Ybus matrix for this system. (ii) Perform two iterations of the Gauss – Seidel load flow.

Answer

7.5 2.5 52.5 6.5 45 4 9

bus

j j jj j j

j j j

− = − −

Y

Take busbar 2 as the slack bus

(0)(1) 1 112 2 13 31 (0)*

111

1

1 0.9 2.5 1 5.0 17.5 1

1.0 0.12

P jQ

j jj

j

− − −= = − × − × −

= +

Y V Y VVVY

[ ]

(0)* (1) (0)3 3 31 1 32 2 33 3Im ( )

Im 1 ( 5 (1.0 0.12) 4 1 9 1)0

Q

j j j j

= − + + = − × × + + × − ×

=

V Y V Y V Y V

G1 G2

(1)3 3(1)31 1 33 23 (0)*

333

1

1 1.5 0 5 (1.0 0.12) 4 19 1

1.0 0.1

P jQ

j j j jj

j

− − −= − + = − × + − × −

= −

Y V Y VVVY

(1)3 1.0 0.1 1.005V j= − =

(1) (1) (1)3 3 3/ ( )

0.995 0.0995abs

j== −

V V V

(1)(2) 1 112 2 13 31 (1)*

111

1

0.91 2.5 1 5.0 (0.995 0.0995)1.0 0.127.5

0.9825 0.052

P jQ

j j jjj

j

− − −= − × − × += −−

= +

Y V Y VVVY

[ ]

(1)* (2) (1)3 3 31 1 32 2 33 3Im ( )

Im 1 ( 5 (0.9825 0.052) 4 1 9 (0.995 0.0995))0.1576

Q

j j j j j

= − + + = − × × + + × − × +

=

V Y V Y V Y V

(2)3 3(2)31 1 33 23 (1)*

333

1

1.5 0.15761 5 (0.9825 0.052) 4 10.995 0.09959

0.9911 0.1387

P jQ

j j j jjj

j

− − −= − + − × + − ×= −−

= −

Y V Y VVVY

(1)3 0.9911 0.1387 1.0008V j= − =

(1) (1) (1)3 3 3/ ( )

0.9903 0.1386abs

j== −

V V V

6.7 In Figure 6.20 the branch reactances and busbar loads are given in per unit on a common base. Branch resistance is neglected.

Explain briefly why an iterative method is required to determine the busbar voltages of this network.


Form the YBUS admittance matrix for this network. Using busbar 1 as the slack (reference) busbar, carry out the first iteration of a Gauss-Seidel load-flow algorithm to determine the voltage at all busbars. Assume the initial voltages of all busbars to be 1.01 p.u.

Answer

17.5 5 2.5 105 15 10 0

2.5 10 17.5 510 0 5 15

bus

j j j jj j j

j j j jj j j

− − = − −

Y

Assume (0)2V =1.01 p.u, (0)

3V =1.01 p.u and (0)4V =1.01 p.u

2

(0)(1) 2 221 1 23 3(0)*

222

1

1 0.5 0.2 5 1.01 10 1.0115 1.01

0.9968 0.033

P jQ

j j jj

j

− − −=

− + = − × − × − = −

Y V Y VVVY

3

(1) (0)3 3(1)31 1 32 2 34 4(0)*

333

1

1 0.4 0.1 2.5 1.01 10 (0.9968 0.033) 5 1.0117.5 1.01

0.9968 0.0415

P jQ

j j j j jj

j

− − − −=

− + = − × − × − − × − = −

Y V Y V Y VVVY

4

(1)(1) 4 441 1 43 3(0)*

444

1

1 0.2 10 1.01 5 (0.9968 0.0415)15 1.01

1.0056 0.027

P jQ

j j jj

j

− − −=

− = − × − × − − = −

Y V Y VVVY

Chapter 7

7.1 Four 11 kV generators designated A, B, C, and D each have a subtransient reactance of 0.1 p.u. and a rating of 50MVA. They are connected in parallel by means of three 100 MVA reactors which join A to B, B to C, and C to D; these reactors have per unit reactances of 0.2, 0.4, and 0.2, respectively. Calculate the volt-amperes and the current flowing into a three-phase symmetrical fault on the terminals of machine B. Use a 50 MVA base.

Answer

A B

CD

0.2

0.4

0.2

On 50 MVA base, the reactances of line AB, BC and CD are 0.1 p.u., 0.2 p.u. and

0.1 p.u..

Then the p.u. equivalent circuit

0.2

0.1

0.10.1

0.10.1

0.1

0.0533eqZ = p.u.

Fault level = 50 937.5

0.0533= MVA

Fault current =6

3

937.5 10 49,2063 11 10

×=

× × A

7.2 Two 100 MVA, 20 kV turbogenerators (each of transient reactance 0.2 p.u.) are connected, each through its own 100 MVA, 0.1 p.u. reactance transformer, to a common 132kV busbar. From this busbar, a 132kV feeder, 40 km in length, supplies an 11 kV load through a 132/11 kV transformer of 200 MVA rating and reactance 0.1 p.u. If a balanced three-phase short circuit occurs on the low-voltage terminals of the load transformer, determine, using a 100 MVA base, the fault current in the feeder and the rating of a suitable circuit breaker at the load end of the feeder. The feeder impedance per phase is (0.035 +j0.14) Ω/km.

Answer 100 MVA0.2 p.u.

100 MVA0.1 p.u. 200 MVA

0.1 p.u.0.035+j0.14 Ω/km

On 100 MVA base

2132 174.24

100baseZ = = Ω

Line impedance in p.u. = 0.008 + j0.032 p.u.

0.2 0.1 1000.032 0.1 0.2322 200eqZ +

= + + × = p.u.

Fault level = 100 431

0.232= MVA

7.3 Two 60 MVA generators of transient reactance 0.15 p.u. are connected to a busbar designated A. Two identical machines are connected to another busbar B. Busbar A is connected to busbar B through a reactor, X. A feeder is supplied from A through a step-up transformer rated at 30 MVA with 10% reactance.

Calculate the reactance X, if the fault level due to a three-phase fault on the feeder side of the 30 MVA transformer is to be limited to 240 MVA. Calculate also the voltage on A under this condition if the generator voltage is 13kV (line).

Answer

A BX

30 MVA0.1

60 MVA0.15

P

Select Sbase = 60 MVA

Fault level at P should be limited to 240 / 60 = 4 p.u

Therefore eqZ should be 1 0.254= = p.u.

However from the circuit,

0.15 0.15 60/ / 0.12 2 30

0.075(0.075 ) 0.20.075 0.0750.0056 0.075 0.2

0.15

eqZ X

XX

XX

= + + × +

= ++ ++

= ++

By equating eqZ ,

0.0056 0.075 0.2 0.25

0.15eqXZ

X+

= + =+

0.076X∴ = p.u.

Voltage at A = 4 x 0.2 = 0.8 p.u.

= 13 x 0.8 = 10.4 kV

7.4 A single line-to-earth fault occurs on the red phase at the load end of a 66 kV transmission line. The line is fed via a transformer by 11 kV generators connected to a common busbar. The line side of the transformer is connected in star with the star point earthed and the generator side is in delta. The positive-sequence reactances of the transformer and line are j10.9 Ω and j44 Ω, respectively, and the equivalent positive and negative-sequence reactances of the generators, referred to the line voltage, are j18 Ω and j14.5 Ω, respectively. Measured up to the fault the total effective zero sequence reactance is j150 Ω. Calculate the fault current in the lines if resistance may be neglected. If a two-line-to-eafth fault occurs between the blue and yellow lines, calculate the current in the yellow phase.

Answer

Assume Sbase = 100 MVA, then on 66 kV side

266 43.56

100baseZ = = Ω

p.u. reactances of the generator and transformers + line are:

Positive Negative Zero

Generator 18/43.56=0.413

p.u.

14.5/43.56=0.333

p.u.

150/43.56 = 3.444

p.u.

Transformer 10.9/43.56=0.25 p.u.

Line 44/43.56=1.01 p.u.

Total 1.673 p.u. 1.593 p.u. 3.444 p.u.

From equation 7.6

3

1.673 1.593 3.444f = + +I =0.447 p.u.

6

3

100 10 874.77 A3 66 10baseI ×

= =× ×

Fault current = 0.447 x 874.77 = 391 A

From equation (7.8), for a two line to earth fault

[ ]11

1.673 1.593 3.444 / (1.593 3.444)0.362 p.u.

=+ × +

=

I

From (7.9) and (7.10)

01.5930.362

(1.593 3.444)0.114 p.u.

= − ×+

= −

I

23.4440.362

(1.593 3.444)0.248 p.u.

= − ×+

= −

I

Therefore

2 0.362 0.248 0.114( 0.5 0.866) 0.362 ( 0.5 0.866) 0.248 0.114

0.171 0.528 p.u.

Y

j jj

= × − × −= − − × − − + × −= − −

I a a

Magnitude of Y phase current = 0.555 x 874.77 = 485.5 A

7.5 A single-line-to-earth fault occurs in a radial transmission system. The following sequence impedances exist between the source of supply (an infinite busbar) of voltage 1 p.u. to the point of the fault: Z1 = 0.3 +j0.6 p.u., Z2 = 0.3 + j0.55 p.u., Z0 = 1 +j0.78 p.u. The fault path to earth has a resistance of 0.66 p.u. Determine the fault current and the voltage at the point of the fault.

Answer

From equation 7.6 by considering fault resistance

3(0.3 0.6) (0.3 0.55) (1 0.78) 3 0.660.649 - 0.350.737 28.3

f j j jj

=+ + + + + + ×

=

= ∠−

I

(0.649 - 0.35) 0.66

0.43 0.23 p.u.f j

j= ×

= −

V

7.6 Develop an expression, in terms of the generated e.m.f. and the sequence impedances, for the fault current when an earth fault occurs on phase (A) of a three-phase generator, with an earthed star point. Show also that the voltage to earth of the sound phase (B) at the point of fault is given by

[ ]2 0

1 2 0

3 AB

j E Z ZV

Z Z Z− −

=+ +

a

Two 30 MVA, 6.6 kV synchronous generators are connected in parallel and supply a 6.6 kV feeder. One generator has its star point earthed through a resistor of 0.4 Ω and the other has its star point isolated. Determine: (a) the fault current and the power dissipated in the earthing resistor when an earth fault occurs at the far end of the feeder on phase (A); and (b) the voltage to earth of phase (B). The generator phase sequence is ABC and the impedances are as follows:

Generator (p.u.) Feeder (Ω/p.h.) To positive-sequence currents j0.2

j0.6

To negative-sequence currents j0.16

j0.6

To zero-sequence currents j0.06 j0.4

Use a base of 30 MVA.

Answer

From equation (7.5):

1 2 01 2 0

A= = =+ +

EI I IZ Z Z

2 20 0 1 1 2 2

2 20 1 2 1

22 0 1 2

1 2 02 2

2 0

1 2 0

( )

( )

( ) ( 1)

B A

A

AA

A

= − − −

= − − −

− −= −

+ +

− + −=

+ +

V a E I Z a I Z aI Za E Z a Z aZ I

Z a Z aZ Ea EZ Z Z

a a Z a Z EZ Z Z

2 3 3( ) 0.5 0.5 32 2

j j j− = − − + − = −a a

2 3 3 3( 1) 0.5 12 2 2

j j− = − − − = − −a

23 3 33 0.5 3 ( 1)2 2 2

j j j j

× = − + × = − − = −

a a

[ ]2 0

1 2 0

3 AB

j− −=

+ +E Z aZ

VZ Z Z

Taking Sbase = 30 MVA, then on 6.6 kV side

26.6 1.452

30baseZ = = Ω


Generator j0.2 p.u. j0.16 p.u. j0.06 p.u.

Feeder j0.6/1.452=j0.413 p.u. j0.413 p.u. j0.4/1.452=j0.275 p.u.

Earth resistor 0.4/1.452=0.275 p.u.

Total j0.513 p.u. j0.493 p.u. 3x0.275+j0.275+j0.06

=0.825+j0.335

From equation 7.6

3

0.513 0.493 0.825 0.3351.904 58.45

f j j j=

+ + +

= ∠−

I

6

3

30 10 2624.3A3 6.6 10baseI ×

= =× ×

Fault current = 1.904 58.45 2624.3 5000 58.45f = ∠− × = ∠−I A

Power dissipated in the resistor = 50002 x 0.4 = 10 MW

From the equation proved before

VB = -0.6894 - j0.3545 p.u.

Phase voltage = 3810.5 V

Therefore VB = -2627 - j1351 V

7.7 An industrial distribution system is shown schematically in Figure 7.30. Each line has a reactance of j0.4 p.u. calculated on a 100 MVA base; other system parameters are given in the diagram. Choose suitable short-circuit ratings for the oil circuit breakers, situated at substation A, from those commercially available, which are given in the table below.

Short circuit (MVA) 75 150 250 350 Rated current (A) 500 800 1500 2000


The industrial load consists of a static component of 5 MVA and four large induction motors each rated at 6 MVA. Show that only three motors can be started simultaneously given that, at starting, each motor takes five times full-load current at rated voltage, but at 0.3 p.f.

Answer As discussed in Chapter 7 when selecting CB1 and CB2, the contribution from

induction motors is neglected.

Grid infeed166.6 MVA

30 MVA15%

M

M

M

M

5 MVA

X

X

X

X

X

X

X

CB1

CB2

CB3

CB6

CB7

132 kV

11 kV

A

On 100 MVA base

Grid infeed = 1.666 p.u.

sX = 0.6 p.u.

Transformer reactance = 0.15 x 100/30 = 0.5 p.u.

For a fault at A,

0.5 0.40.6 1.05 p.u.2 2eqX = + + =

Fault level at A = 1/1.05 = 0.952 p.u. = 95.2 MVA

Since the CB is oil it was assumed that the breaker opening time is 8-cycle

Therefore the fault level interrupted by each CB = 95.2/2 = 47.6 MVA

Current through each line = 6

3

1 (6 4 5) 10 761.052 3 11 10

× + ×× =

× × A

Due to the required current rating a 150 MVA CB should be selected.

Rated current through CB3 to CB6 = 6

3

6 10 315 A3 11 10×

=× ×

As starting current is 5 rimes the rated current, the corresponding MVA at starting

is 30 MVA.

Assuming the fault current is as in the same order as the starting current CB3 to

CB6 should have an interrupting capability of 30 MVA. Select 75 MVA, 500 A CBs

for CB3 to CB6.

When all motors are starting at the same time, the starting current creates an MVA

level of 150 MVA at the busbar A. With 5 MVA static load this is greater than short

term rating of the CB1 and 2. So only 3 motors could be started simultaneously.

7.8 Explain how the Method of Symmetrical Components may be used to represent any 3 p.h. current phasors by an equivalent set of balanced phasors.

A chemical plant is fed from a 132kV system which has a 3 p.h. symmetrical fault level of 4000 MVA. Three 15 MVA transformers, connected in parallel, are used to step down to an 11 kV busbar from which six 5 MVA, 11 kV motors are supplied. The transformers are delta-star connected with the star point of each 11 kV winding, solidly earthed. The transformers each have a reactance of 10% on rating and it may be assumed that 1 2 0X X X= = . The initial fault contribution of the motors is equal to five times rated current with 1.0 p.u. terminal voltage.

Using a base of 100 MVA, (a) calculate the fault current (in A) for a line-to-earth short circuit on the 11 kV

busbar with no motors connected; (b) calculate the 3 p.h. symmetrical fault level (in MVA) at the 11 kV busbar if

all the motors are operating and the 11 kV busbar voltage is 1.0 p.u.

Answer (a) On 100 MVA base, fault level at 132 kV busbar = 4000/100 = 40 p.u.

Therefore sX = 1/40 = 0.025 p.u.


132 kV

busbar 0.025 p.u. 0.025 p.u.

No contribution as

132 kV side is delta

connected

Each

transformer

0.1x100/15

=0.667 p.u. 0.667 p.u. 0.667 p.u.

Total 0.025+0.667/3 =

0.247 p.u. 0.247 p.u.

0.667/3 = 0.222

p.u.

From equation 7.6

3

0.247 0.247 0.2224.19 p.u.

f = + +=

I

6

3

100 10 5248.64A3 11 10baseI ×

= =× ×

Fault current = 4.19 5248.64 22f = × =I kA

(b) If all the motors are connected

Fault level provided by each motor = 5 x 5 MVA = 25 MVA = 0.25 p.u.

Reactance = 1/0.25 = 4 p.u.

Since 6 motors are connected in parallel, equivalent reactance

= 4/6 = 0.667 p.u.

Therefore positive sequence reactance = 0.247//0.667 = 0.18 p.u.

Three phase short circuit level = 100/0.18 = 554.8 MVA

7.9 Describe the effect on the output current of a synchronous generator following a solid three-phase fault on its terminals.

For the system shown in Figure 7.31 calculate (using symmetrical components):

a) the current flowing in the fault for a three-phase fault at busbar A; b) the current flowing in the fault for a one-phase-to-earth fault at busbar B; c) the current flowing in the faulted phase of the overhead line for a one-phase-to

earth fault at busbar B.

Generators G1 and G2: 1 2 0.1X X j′′ ′′= = p.u.; 11 kV

Transformers T1 and T2: 1 2 0 0.1X X X j= = = p.u.; 11/275 kV (Earthed star-delta) Line: 1 2 0.05Z Z j= = p.u., 0 0.1Z j= p.u.; 275 kV (All p.u. values are quoted on a base of 100 MVA) Figure 7.31 Circuit for Problem 7.9 Assume the pre-fault voltage of each generator is 1 p.u. and calculate the symmetrical fault currents (in amps) immediately after each fault occurs.

Answer

(a) 6

3

100 10 209.94A3 275 10baseI ×

= =× ×

0.1 0.1

0.1 0.1 0.05

0.2 0.25 0.1110.2 0.25eqZ ×

= =+

p.u.

Fault current = 1/0.111 = 9 p.u. = 9 x 209.94 = 1.89 kA

(b) Positive and negative sequence equivalent reactance = 0.111 p.u.

For zero sequence

0.1 0.1

0.1

Equivalent impedance = 0.2//0.1 = 0.066 p.u.

From equation 7.6

3

0.111 0.111 0.06610.41 p.u.

f = + +=

I

Fault current = 10.41 209.94 2.18f = × =I kA

(c) When only consider current in overhead line

Positive and negative sequence equivalent reactance = 0.25 p.u.

Zero sequence impedance = 0.2 p.u.

3

0.25 0.25 0.24.29 p.u.

f = + +=

I

Fault current = 4.29 209.94 0.89f = × =I kA

7.10 Why is it necessary to calculate short-circuit currents in large electrical systems?

A generator rated at 400 MW, 0.8 power factor, 20 kV has a star-connected stator winding which is earthed at its star point through a resistor of 1 Ω. The generator reactances, in per unit on rating, are:

1 0.2X = 2 0.16X = 0 0.14X =

The generator feeds a delta-star-connected generator transformer rated at 550 MVA which steps the voltage up to a 275 kV busbar. The transformer star-point is solidly earthed and the transformer reactance is 0.15 p.u. on its rating. The 275 kV busbar is connected only to the transformer. Assume that for the transformer

1 2 0X X X= = .

Using a base of 500 MVA calculate the base current and impedance of each voltage level.

Calculate the fault current in amperes for:

(a) a 275kV busbar three-line fault; (b) a 275 kV single-line-to earth fault on the busbar; (c) a 20 kV three-line fault on the generator terminals; (d) a 20 kV single-line-to-earth fault on the generator terminals.

Answer On 500 MVA, 275 kV base

6

3

500 10 1049.7 A3 275 10baseI ×

= =× ×

On 500 MVA, 20 kV base

6

3

500 10 14.43kA3 20 10baseI ×

= =× ×

220 0.8

500baseZ = = Ω

Generator MVA = 400/0.8 = 500 MVA

So on 500 MVA, for the generator: 1 0.2X = , 2 0.16X = , 0 0.14X =

For transformer: 1 2 0X X X= = = 0.15x500/550 = 0.136 p.u.

Earthing resistor = 1/0.8 = 1.25 p.u.

(a) For a three phase fault on 275 kV busbar

1 2.98 p.u.

0.2 0.1362.98 1049.7 3.124 kA

f = =+

= × =

I

(b) Single phase to earth fault on 275 kV basbar

Positive sequence reactance = 0.2+0.136=0.336 p.u.

Negative sequence reactance = 0.16+0.136 = 0.296 p.u.

Zero sequence reactance = 0.136 p.u

3 3.906 p.u.=4.1 kA

0.336 0.296 0.136f = =+ +

I

(c) Three phase fault at 20 kV barbar

1 5 p.u.

0.25 14.43 72.17 kA

f = =

= × =

I

(d) Single phase to earth fault on 275 kV basbar

Positive sequence reactance = 0.2 p.u.

Negative sequence reactance = 0.16 p.u.

Zero sequence reactance = 3x1.25+j0.14 p.u = 3.75+j0.14 p.u.

30.2 0.16 3.75 0.14

0.79 p.u.=11.44 kA

f

f

j j j=

+ + +

=

I

I

Chapter 8

Problems

8.1 A round-rotor generator of synchronous reactance 1 p.u. is connected to a transformer of 0.1 p.u. reactance. The transformer feeds a line of reactance 0.2 p.u. which terminates in a transformer (0.1 p.u. reactance) to the LV side of which a synchronous motor is connected. The motor is of the round-rotor type and of 1 p.u. reactance. On the line side of the generator transformer a three-phase static reactor of 1 p.u. reactance per phase is connected via a switch. Calculate the steady-state power limit with and without the reactor connected. All per unit reactances are expressed on a 10 MVA base and resistance may be neglected. The internal voltage of the generator is 1.2 p.u. and of the motor 1 p.u.

Answer

Total reactance without the shunt reactor = 1 + 0.1 + 0.2 + 0.1 + 1 = 2.4 p.u.

Steady state power limit = 1.2x1/2.4 = 0.5 p.u. = 5 MW

With the reactor:

j1.0

j1.1 j1.3

Converting star to delta (see chapter 2)

Reactance between two sources = 1.1 + 1.3 + 1.1*1.3/1 = 3.83

Steady state power limit = 1.2x1/3.83 = 0.313 p.u. = 3.13 MW

8.2 In the system shown in Figure 8.20, investigate the steady state stability. All per unit values are expressed on the same base and the resistance of the system (apart from the load) may be neglected. Assume that the infinite busbar voltage is 1 p.u.

Figure 8.20 Line diagram of system in Problem 8.2

Answer

j0.71 j0.1

1.53 -31.60

0.8+j0.3

1 p.u

1 23

Using Equation 2.8 and working from the infinite busbar voltage, the voltage at point 3 is given by

2 2

2 2

2 2

0.3 0.1 0.8 0.111 1

1.03 0.08 1.033p.u.

AQX PXV VV V

= + +

× × = + +

= + =

1 0.08tan 4.441.03

δ − = =

i.e. AV has an angle of 4.440 to the infinite busbar

The reactive power absorbed by the line from point 3 to point 2 (the infinite busbar) is

2 2 2 22

2 2

0.8 0.3 0.11

0.073p.u.

RP QI X X

V + +

= =

=

The actual load taken by 3 (if represented by an impedance) is given by

2 2

0

1.033 0.594 0.365p.u.1.53 31.6

AV jZ

= = +∠−

The total load supplied by link from generator to 3

( ) ( )0.8 0.594 0.3 0.073 0.3651.394 0.738p.u.

jj

= + + + +

= +

Internal voltage of generator E1

2 2

2 2

0.738 0.71 1.394 0.711.0331.033 1.033

1.54 0.958 1.814

× × = + +

= + =

1 0.958tan 31.881.54

δ − = =

Hence, the angle between E1 and V is: 31.88+4.44 =36.320

Since this angle is much less than 90°, the system is stable.

8.3 A generator, which is connected to an infinite busbar through two 132 kV lines in parallel, each having a reactance of 70 Ω/phase, is delivering 1 pu to the infinite busbar. Determine the parameters of an equivalent circuit, consisting of a single machine connected to an infinite busbar through a reactance, which represents the above system

a) Pre-fault b) When a three-phase symmetrical fault occurs halfway along one line. c) After the fault is cleared and one line isolated If the generator internal voltage is 1.05 p.u. and the infinite busbar voltage is 1.0 p.u, what is the maximum power transfer pre-fault, during the fault and post-fault?

Determine the swing curve for a fault clearance time of 125 ms.

The generator data are as follows: Rating 60 MW at power factor 0.9 lagging; Transient reactance 0.3 p.u. Inertia constant 3 kWs/kVA

Answer

On 100 MVA base, Zbase = 1322/100 = 174.24

Line reactance = 70/174.24 = 0.4 p.u.

For generator

Transient reactance on 100 MVA base = 0.3*100/(60/0.9) = 0.45 p.u.

j0.45j0.4

(a) Pre-fault reactance = j0.45+j0.4/2 = j0.65 p.u.

Maximum power transfer = 1.05 x 1.0 / 0.65 = 1.62 p.u.

(b) When a three-phase symmetrical fault occurs halfway along one line

j0.45j0.4

Fault

j0.2 j0.2

From delta to star transformation

j0.45

j0.05

Fault

j0.1 j0.1

Again from star to delta transformation

Z12

Fault

Z11 Z22

Z12 = j0.55+j0.1+j 0.55*0.1/0.05 = j1.75 p.u


(c) Post-fault reactance = j0.45+j0.4 = j0.85 p.u.


H on 100 MVA base = 3 x (60/0.9)/100 p.u = 2 p.u

42 2.2 10 p.u180

Mf

−= = ×

A time interval of 0.05 s will be used. Hence

2( ) 11.36t

M∆

=

The initial operating angle,

10

1sin 38.11.62

δ − = =

Just before the fault accelerating power 0P∆ = . Immediately after the fault

01 0.6sin 0.63 p.u.P δ∆ = − =

The value of P∆ for the commencement of the period is 0.63/2 = 0.315 p.u

t∆ (s) P∆ (p.u.) 2

1 1( )

n n nt P

Mδ δ − −

∆∆ = ∆ + ∆ nδ

0.05 0.315 11.36 x 0.315 = 3.58o 38.1+3.58 = 41.68o

0.1 1-0.6sin41.68o = 0.6 3.58o+11.36 x 0.6 = 10.4o 41.68o+10.4o = 52.08o

0.125 Fault is cleared

0.15 1-0.6sin52.08o = 0.527

10.4o +11.36 x 0.527 = 16.39o 52.08o+16.39o = 68.47o

As fault was cleared between 0.1 - 0.15 s period from here onwards post fault values applied.

0.2 1-1.24sin68.47o = -0.15

16.39o-11.36 x 0.15 = 14.69o 68.47o+14.69o = 83.16o

0.25 1-1.24sin83.16o = -0.23

14.69o -11.36 x 0.23 = 12.08o 83.16o+12.08o = 95.24o

The following matlab code was developed:

clear Dd(1)=0; d(1)=38.1; Pi=1.0; for i=2:25 t(i)=0.05*i; if i==2, DP=(Pi-0.6*sin(pi*d(1)/180))/2; elseif i<5 DP=Pi-0.6*sin(pi*d(i-1)/180); else DP=Pi-1.24*sin(pi*d(i-1)/180); end Dd(i)=Dd(i-1)+11.36*DP; d(i)=d(i-1)+Dd(i); end plot(t,d)

0 0.2 0.4 0.6 0.8 1 1.2 1.40

100

200

300

400

500

600

700

800

Time(s)

Ang

le (d

eg)

Pre-fault power is 1 p.u.

8.4 An induction motor and a generator are connected to an infinite busbar. What is the equivalent inertia constant of the machines on 100 MVA base. Also calculate the equivalent angular momentum.

Data for the machines are: Induction motor Rating 40 MVA;

Inertia constant 1 kWs/kVA. Generator: Rating 30 MVA;

Inertia constant l0 kWs/kVA. Answer

Heq on 100 MVA base = 1x40/100 + 10x30/100 p.u = 3.4 p.u

3.4 0.00038 p.u

180M

f= =

8.5 The P-V, Q-V characteristics of a substation load are as follows:

V 1.05 1.025 1 0.95 0.9 0.85 0.8 0.75 P 1.03 1.017 1 0.97 0.94 0.92 0.98 0.87 Q 1.09 1.045 1 0.93 0.885 0.86 0.84 0.85

The substation is supplied through a link of total reactance 0.8 p.u. and negligible resistance. With nominal load voltage, P = 1 and Q = 1 p.u. By determining the supply voltage-received voltage characteristic, examine the stability of the system by the use of dE/dV. All quantities are per unit.

Answer

From 2 2QX PXE V

V V = + + ,

V vs E was obtained. It can be seen that at 1 p.u voltage dE/dV is positive indicating

a stable system.

0.75 0.8 0.85 0.9 0.95 1 1.05 1.1 1.151.86

1.88

1.9

1.92

1.94

1.96

1.98

2

2.02

2.04

V(p.u)

E(p

.u)

8.6 A large synchronous generator, of synchronous reactance 1.2 p.u., supplies a load through a link comprising a transformer of 0.1 p.u. reactance and an overhead line of initially 0.5 p.u. reactance; resistance is negligible. Initially, the voltage at the load busbar is 1 p.u. and the load P+jQ is (0.8 +j0.6) p.u. regardless of the voltage. Assuming the internal voltage of the generators is to remain unchanged, determine the value of line reactance at which voltage instability occurs.

Answer

Under initial impedance

Generator terminal voltage

( )2 21 0.6 0.6 (0.8 0.6)

1.44 p.u. tV = + × + ×

=

If the internal voltage of the generator remains unchanged and as P and Q are constants the generator terminal voltage remains unchanged.

Under the limit of stability

dQ Q VdV V X

= −

Since Q is a constant,

2 21.44 3.45

0.6VXQ

= = = p.u.

Here X is the total impedance; i.e. 1.2+0.1+Xline. Therefore Xline = 2.15 p.u.

8.7 A load is supplied from an infinite busbar of voltage 1p.u. through a link of series reactance 1p.u. and of negligible resistance and shunt admittance. The load consists of a constant power component of 1 p.u. at 1 p.u. voltage and a per unit reactive power component (Q) which varies with the received voltage (V) according to the law

( ) ( )20.8 0.2 0.8V Q− = −

All per unit values are to common voltage and MVA bases. Determine the value of X at which the received voltage has a unique value and the corresponding magnitude of the received voltage. Explain the significance of this result in the system described. Use approximate voltage-drop equations.

Answer

A receiving end voltage becomes unique at the knee point of X-V curve. At this point:

dQ Q VdV V X

= − (A)

Since ( )25 0.8 0.8Q V= − +

10( 0.8)dQ VdV

= −

Further,

1

1

QXVV

Q VV X

= +

−∴ =

Substituting in (A):

1 1 210( 0.8)

1 210( 0.8)

V V VVX X X

VXV

− −− = − =

−∴ =

−

( )25 0.8 0.8 10(1 )( 0.8)

(1 2 )V V V

V V− + − −

=−

23 8 4 00.67 p.u

V VV

− + ==

X = 0.25 p.u.

Above solution is only true if

PX QXVV V

<< +

P = 1 p.u.

0.25 0.3730.67

PXV

= = and 1QXVV

+ = ; therefore the assumption is correct.

8.8 Explain the criterion of stability based on the equal-area diagram.

A synchronous generator is connected to an infinite busbar via a generator transformer and a double-circuit overhead line. The transformer has a reactance of 0.15 p.u. and the line an impedance of 0+j0.4p.u. per circuit. The generator is supplying 0.8 p.u. power at a terminal voltage of 1 p.u. The generator has a transient reactance of 0.2 p.u. All impedance values are based on the generator rating and the voltage of the infinite busbar is 1 p.u.

a) Calculate the internal transient voltage of the generator.

b) Determine the critical clearing angle if a three-phase solid fault occurs on the sending (generator) end of one of the transmission line circuits and is cleared by disconnecting the faulted line.

Answer

(a) Since generator terminal voltage and infinite bus voltage are 1 p.u

( )2 21 1 0.35 (0.8 0.35)0.114 p.u.

QQ= + + ×

= −

Hence, using the infinite busbar as the reference bus

E = (1-0.114x0.55)+j0.8x0.55

= 0.937+j0.44 p.u.

|E|=1.035 p.u

(b) Prior to the fault

X = 0.2 + 0.15 + 0.4/2 = 0.55 p.u.

P2 ≈ 1/0.55 = 1.818 p.u.

δ0 = sin-1(0.8/1.818) = 26.1o

During the fault P = 0

After the fault

X = 0.2 + 0.15 + 0.4 = 0.75 p.u.

P2 ≈ 1/0.75 = 1.333 p.u.

δ2 = 180o-sin-1(0.8/1.333) = 143.1o

From equal area criteria:

1 2

0 1

0 0 2

0 2 0 2 2 2 10 0 0 0

1

1

01

( sin ) 0

( ) cos cos 0

0.8 (143.1 26.1 ) /180 1.32cos143.1 1.32cos 01.634 1.056cos

1.3264.03

P d P P d

P P P

δ δ

δ δ

δ δ δ

δ δ δ δ

π δ

δ

δ

+ − =

− + − =

× − × + − =−

=

=

∫ ∫

8.9 A 500 MVA generator with 0.2 p.u. reactance is connected to a large power system via a transformer and overhead line which have a combined reactance of 0.3 p.u. All p.u. values are on a base of 500 MVA. The amplitude of the voltage at both the generator terminals and at the large power system is 1.0 p.u. The generator delivers 450 MW to the power system.

Calculate a) the reactive power in MVAr supplied by the generator at the transformer input

terminals; b) the generator internal voltage; c) the critical clearing angle for a 3 p.h. short circuit at the generator terminals.

Answer

(a) Since generator terminal voltage and infinite bus voltage are 1 p.u

( )2 21 1 0.3 (0.9 0.3)0.123 p.u. = 62 MVAr

QQ= + + ×

= −

(b) Therefore, taking infinite busbar voltage as the reference

( )1 0.5 0.123 (0.9 0.5)0.938 0.451.04 p.u.

E jj

E

= − × + ×

= +

=

(c) From equal area criteria:

1 2

0 1

0 0 1

0 2 0 1 2 1 1

( sin ) 0

( ) cos cos 0

P d P P d

P P P

δ δ

δ δ

δ δ δ

δ δ δ δ

+ − =

− + − =

∫ ∫

11 2 p.u.

0.2 0.3P = =

+

δ0 = sin-1(0.9/2) = 26.74o

δ2 = 180o-26.74o = 153.26o

0 0 0 01

01

0.9 (153.26 26.74 ) /180 2cos153.26 2cos 0

84.2

π δ

δ

× − × + − =

=

Chapter 9

Problems

9.1 A bridge-connected rectifier is fed from a 230 kV/120 kV transformer from the 230 kV supply. Calculate the direct voltage output when the commutation angle is 15° and the delay angle is (a) 0°; (b) 30°; (c) 60°.

Answer

From equations (9.4) and (9.6):

( )1.35 cos cos2

Ld

VV α α γ = + +

(a) When α = 0° and γ = 15°:

1.35 120 cos 0 cos15 159.2 kV2dV × = + =

(b) When α = 30° and γ = 15°:

1.35 120 cos30 cos 45 127.4 kV2dV × = + =

(c) When α = 60° and γ = 15°:

1.35 120 cos 60 cos 75 61.5 kV2dV × = + =

9.2 It is required to obtain a direct voltage of 100 kV from a bridge-connected rectifier operating with α = 30° and γ = 15°. Calculate the necessary line secondary voltage of the rectifier transformer, which is nominally rated at 345 kV/150 kV; calculate the tap ratio required.

Answer

From equations (9.4) and (9.6):

( )1.35 cos cos2

Ld

VV α α γ = + +

1.35 cos30 cos 45 100 kV2

Ld

VV × = + =

VL = 94.2 kV

Tap ratio = 150/94.2 = 1.6

9.3 If the rectifier in Problem 9.2 delivers 800 A d.c, calculate the effective reactance X (Ω) per phase.

Answer

From equation (9.5)

[ ]0 . cos cos( )6dVIX

π α α γ= − +

31.35 94.2 10800 cos30 cos 456dI

Xπ × × × = = −

X = 13.2 Ω

9.4 A d.c. link comprises a line of loop resistance 5 Ω and is connected to transformers giving secondary voltage of 120 kV at each end. The bridge-connected converters operate as follows:

Rectifier: Inverter:

α = 10° 0δ = 10°; Allow 5° margin on 0δ for δ

X = 15 Ω X = 15 Ω

Calculate the direct current delivered if the inverter operates on constant δ control. If all parameters remain constant, except α , calculate the maximum direct current transmittable.

Answer

From equation (9.17)

0 0cos cosr id

L cr ci

V VIR R R

α δ−∴ =

+ −

0 0 1.35 1.35 120 162 kVr i LV V V= = × = × =

3 3 15 14.3cr ciXR Rπ π

×= = = = Ω

When α = 10° and δ = 15°

( )[ ]

3162 10 611.7 Acos10 cos155 14.3 14.3dI ×

= =−+ −

At maximum transmittable power α = 0°

( )[ ]

3162 10 1104 Acos 0 cos155 14.3 14.3dI ×

= =−+ −

9.5 The system in Problem 9.4 is operated with α = 15° and on constant β control. Calculate the direct current for γ = 15°.

Answer

For constant β from equation (9.16):

0 0cos cosr id

L cr ci

V VIR R R

α β−∴ =

+ +

30β γ δ= + =

( )[ ]

3162 10 481.7 Acos15 cos305 14.3 14.3dI ×

= =−+ +

9.6 For a bridge arrangement, sketch the current waveforms in the valves and in the transformer windings and relate them, in time, to the anode voltages. Neglect delay and commutation times. Comment on the waveforms from the viewpoint of harmonics.

Answer See Figure 9.12

9.7 A direct current transmission link connects two a.c. systems via converters, the line voltages at the transformer-converter junctions being l00 kV and 90 kV. At the l00 kV end the converter operates with a delay angle of 10°, and at the 90 kV end the converter operates with a δ of 15°. The effective reactance per phase of each converter is 15 Ω and the loop resistance of the link is 10 Ω. Determine the magnitude and direction of the power delivered if the inverter operates on constant-δ control. Both converters consist of six valves in bridge connection. Calculate the percentage change required in the voltage of the transformer, which was originally at 90 kV, to produce a transmitted current of 800 A, other controls being unchanged. Comment on the reactive power requirements of the converters.

Answer

For constant δ from equation (9.17):

0 0cos cosr id

L cr ci

V VIR R R

α δ−∴ =

+ −

0 1.35 1.35 100 135 kVr LV V= × = × =

0 1.35 1.35 90 121.5 kVi LV V= × = × =

3 3 15 14.3cr ciXR Rπ π

×= = = = Ω

When α = 10° and δ = 15°

[ ]( )

135cos10 121.5cos15 1.55 kA10 14.3 14.3dI −= =

+ −

1.35 cos 1.35 100cos10 133 kVd LV V α= × = × =

P = 1.55 x 133 = 206 MW

For dI = 800 A

( )

30135 10 cos10 cos15

80010 14.3 14.3

iV × − =+ −

0iV = 129.4 kV

Percentage change = (129.4-121.5)*100/121.5 = 6.5%

9.8 Draw a schematic diagram showing the main components of an h.v.d.c. link connecting two a.c. systems. Explain briefly the role of each component and how inversion into the receiving end a.c. system is achieved.

Discuss two of the main technical reasons for using h.v.d.c. in preference to a.c. transmission and list any disadvantages.

(From E.C. Examination, 1996)

9.9 A VSC is fed from a 230kV/120kV transformer from the 230 kV supply. Total series reactance between the VSC and the a.c. supply is 5 Ω (including the transformer leakage reactance). The output voltage of the VSC is 125 kV and leads the a.c. supply voltage by 5°. What is the active and reactive power delivered to the a.c. system?

Answer

From equation (9.20) and (9.21), per phase active and reactive power are:

[ ]. . sin

120 125 sin 5 261.5MW5

a c VSCV VP

Xψ θ−

=

× ×= =

[ ] . . . .cos

120 (125cos5 120) 108.6 MVar5

a c VSC a cV V VQ

Xψ θ −−

=

× −= =

9.10 A VSC based h.v.d.c. link is connected into a 33 kV a.c. system with a short circuit level of 150 MVA. The VSC can operate between 0.7-1.2 pu voltage and has a 21.8 Ω coupling reactor. Assume a voltage of 1 pu on the infinite busbar, and using a base of 150 MVA calculate:

(a) The maximum active power that the d.c. link can inject into the a.c. system. (b) The angle of the VSC when operating at 10 MW exporting power. (c) The maximum reactive power (in MVar) that the d.c. link can inject into the a.c.

system (d) The maximum reactive power (in MVar) that the d.c. link can absorb from the

a.c. system

Answer

Per unit equivalent of the VSC is shown below:

(a) 150baseS = MVA

233 7.26150baseZ = = Ω

Impedance of the line = 21.8/7.26 = 3 pu

System impedance on 150 MVA is 1 pu

From equation (9.20) maximum power transfer occurs when 1.2SV = pu and

0ψ θ− =

Vs

max

1.2 1 0.3 pu4

S RV VPX

×= = =

Maximum power is 0.3 x 150 = 45 MW

(b) At 10 MW:

sinS RV VP

Xδ=

10 (1.2)(1.0) sin150 40.0666 0.3sin

12.83

δ

δδ

=

== °

(c) Maximum reactive power generation occurs when VSC output is inphase with

the infinite bus voltage and it is 1.2 pu

(1.2 1) 0.05 pu4

jj−

= = −I

Voltage at the 33 kV busbar:

33 1.0 0.05 1 1.05 puV j j= − × =

max 1.05 0.05 0.0525pu7.875 MVar

Q = × =

=

(d) Maximum reactive power absorption occurs when VSC output is inphase with

the infinite bus voltage and it is 0.7 pu

(0.7 1) 0.075 pu4

jj−

= =I

Voltage at the 33 kV busbar:

33 1.0 0.075 1 0.925 puV j j= + × =

max 0.925 0.075 0.069pu10.4 MVar

Q = × =

=

Chapter 10

10.1 A 345 kV, 60 Hz system has a fault current of 40kA. The capacitance of a busbar to which a circuit breaker is connected is 25 000 pF. Calculate the surge impedance of the busbar and the frequency of the restriking (recovery) voltage on opening.

Answer

Since the fault current is 40 kA, reactance of the line 3

3

345 10 / 3 4.9840 10×

= = Ω×

4.98 0.0132

120L

π∴ = = H

Surge impedance = 9

0.0132 726.925 10

LC −= = Ω

×

Frequency = 9

1 1 87602 2 0.0132 25 10LCπ π −

= =× ×

Hz

10.2 A highly capacitive circuit of capacitance per phase 100 µF is disconnected by

circuit breaker, the source inductance being 1 mH. The breaker gap breaks down when the voltage across it reaches twice the system peak line-to-neutral voltage of 38 kV. Calculate the current flowing with the breakdown, and its frequency, and compare it with the normal charging current of the circuit.

Answer

Surge impedance = 3

0 6

1 10 3.16100 10

LZC

−

−

×= = = Ω

×

Frequency = 3 6

1 1 5032 2 1 10 100 10LCπ π − −

= =× × ×

Hz

Peak current = 38 2 243.16×

= = kA

10.3 A l0 kV, 64.5 mm2 cable has a fault 9.6 km from a circuit breaker on the supply side of it. Calculate the frequency of the restriking voltage and the maximum voltage of the surge after two cycles of the transient. The cable parameters are (per km), capacitance per phase = 1.14µF, resistance = 5.37Ω, inductance per phase = 1.72mH. The fault resistance is 6 Ω.

Answer Since capacitance per km is 1.14 µF and inductance per km is 1.72 mH and line length is 9.6 km,

Surge impedance = 3

0 6

1.72 10 38.81.14 10

L lZC c

−

−

×= = = = Ω

×

Frequency = 3 6

1 1 3742 2 1.72 10 9.6 1.14 10 9.6LCπ π − −

= =× × × × ×

Hz

Time for 2 cycles = 2/374 = 0.0053 s

3

1 1 5.37 9.6 6 1742.742 2 1.72 10 9.6

RL

α −

× + = = = × ×

1742.74 0.00531 1 0.99te eα− − ×− = − ≈

Maximum terminal voltage set-up with the CB opening on a short circuit

310 102 2 16.33

3×

= × × = kV

Voltage after 2 cycles = ( )16.33 1 16.32te α−− = kV

10.4 A 132kV circuit breaker interrupts the fault current flowing into a symmetrical three-phase-to-earth fault at current zero. The fault infeed is 2500 MVA and the shunt capacitance, C, on the source side is 0.03 µF. The system frequency is 50 Hz. Calculate the maximum voltage across the circuit breaker and the restriking-voltage frequency. If the fault current is prematurely chopped at 50 A, estimate the maximum voltage across the circuit breaker on the first current chop.

Answer

Since the fault level is 2500 MVA,

Reactance of the line 3 2

6

(132 10 ) 6.972500 10

×= = Ω

×

6.97 0.022100

Lπ

∴ = = H

Surge impedance = 6

0.022 8560.03 10

LC −= = Ω

×

Frequency = 6

1 1 6.172 2 0.022 0.03 10LCπ π −

= =× ×

kHz

Maximum voltage across the CB 3132 102 2 215.5

3×

= × × = kV

Maximum voltage across the CB on first current chop 50 856 42.8= × = kV

10.5 An overhead line of surge impedance 500 Ω is connected to a cable of surge impedance 50 Ω. Determine the energy reflected back to the line as a percentage of incident energy.

Answer

Reflection coefficient 1 0

1 0

50 500 0.81850 500

Z ZZ Z

α − −= = = −

+ +

Reflected voltage 0.818 iv= −

Reflected current 0.818 ii=

Reflected energy 20.818 0.67i iv i= − = − × incident surge energy

10.6 A cable of inductance 0.188 mH per phase and capacitance per phase of 0.4 µF is connected to a line of inductance of 0.94 mH per phase and capacitance 0.0075 µF per phase. All quantities are per km. A surge of 1 p.u. magnitude travels along the cable towards the line. Determine the voltage set up at the junction of the line and cable.

Answer

For the cable:

3

0 6

0.188 10 21.680.4 10

Z−

−

×= = Ω

×

For the transmission line:

3

1 6

0.94 10 3540.0075 10

Z−

−

×= = Ω

×

Voltage at the junction = 1

0 1

2 1.88ZZ Z

=+

p.u.

10.7 A long overhead line has a surge impedance of 500 Ω and an effective resistance at the frequency of the surge of 7 Ω/km. If a surge of magnitude 500 kV enters the line at a certain point, calculate the magnitude of this surge after it has traversed 100 km and calculate the resistive power loss of the wave over this distance. The wave velocity is 3 x 105 km/s.

Answer

From equation 10.21, with G=0,

0

1exp2

1 7500exp 100 248.3 kV2 500

iRv v xZ

= −

= − =

From equation 10.22

Power at x =0

700exp exp 0.25500i i i i i i

Rv i x v i v iZ

− = − =

Since input power is i iv i , power loss = 0.75 i iv i

0

500 1500

ii

viZ

= = = kA

Power loss = 0.75 500 1 375× × = MW

10.8 A rectangular surge of 1 p.u. magnitude strikes an earth (ground) wire at the centre of its span between two towers of effective resistance to ground of 200 Ω and 50 Ω. The ground wire has a surge impedance of 500 Ω. Determine the voltages transmitted beyond the towers to the earth wires outside the span.

Answer

Voltage transmitted to the 200 Ω tower side =

200 / /500 142.862 2 0.44500 200 / /500 500 142.86i i iv v v = = + +

Voltage transmitted to the 50 Ω tower side =

50 / /500 45.452 2 0.17500 50 / /500 500 45.45i i iv v v = = + +

10.9 A system consists of the following elements in series: a long line of surge impedance 500 Ω, a cable (Z0 of 50 Ω), a short line (Z0 of 500 Ω), a cable (Z0 of 50 Ω), a long line (Z0 of 500 Ω). A surge takes 1 µs to traverse each cable (they are of equal length) and 0.5 µs to traverse the short line connecting the cables. The short line is half the length of each cable. Determine, by means of a lattice diagram, the p.u. voltage of the junction of the cable and the long line if the surge originates in the remote long line.

Answer

α β

Line to cable 50 500 0.81850 500

−= −

+ 2 50 0.182

50 500×

=+

Cable to line 500 50 0.81850 500

−=

+ 2 500 1.818

50 500×

=+

Line to cable 50 500 0.81850 500

−= −

+ 2 50 0.182

50 500×

=+

Cable to line 500 50 0.81850 500

−=

+ 2 500 1.818

50 500×

=+

Figure 10.41 Solution of Problem 10.9

10.10 A 3 p.h., 50 Hz, 11 kV star-connected generator, with its star point earthed, is connected via a circuit breaker to a busbar. There is no load connected to the busbar. The capacitance to earth on the generator side terminals of the circuit breaker is 0.007 µF per phase. A three-phase-to-earth short circuit occurs at the busbar with a symmetrical subtransient fault current of 5000 A. The fault is then

cleared by the circuit breaker. Assume interruption at current zero. (a) Sketch the voltage across the circuit breaker terminals of the first phase to

clear. (b) Neglecting damping, calculate the peak value of the transient recovery voltage

of this phase. (c) Determine the time to this peak voltage and hence the average rate of rise of

recovery voltage.

Answer

From equation (10.6)

0ˆ( ) 1 costv t V e tα ω−= −

So peak value of the transient recovery voltage = 311 102 2 17.96

3×

× × = kV

If line reactance is X,

311 10 50003

1.27X

X

×=

∴ = Ω

Inductance = 0.004 H

Time to peak = 60.004 0.007 10 16.7LCπ π −= × × = µs

Rate of rise of recovery voltage = 17.96/16.7=1.075 kV/µs

10.11 A very long transmission line AB is joined to an underground cable BC of length 5 km. At end C, the cable is connected to a transmission line CD of 15 km length. The transmission line is open-circuit at D. The cable has a surge impedance of 50 Ω and the velocity of wave propagation in the cable is 150 x 106 m/s. The transmission lines each have a surge impedance of 500 Ω. A voltage step of magnitude 500 kV is applied at A and travels along AB to the junction B with the cable. Use a lattice diagram to determine the voltage at: (a) D shortly after the surge has reached D; (b) D at a time 210 µs after the surge first reaches B; (c) B at a time 210 µs after the surge first reaches B. Sketch the voltage at B over these 210 µs.

Answer

Answer is very similar to the answer for 10.13 except travelling times are

different.

Time taken to travel along line BC = 5000/150x106 = 33 µs

Time taken to travel along line CD = 15000/3x108 = 50 µs

Line to cable: 1 0.818α = − and 1 0.182β =

Line to cable: 2 0.818α = and 2 1.818β =

33

500 Ω 500 Ω50 Ω

133

166

200

233

Tim

e (µ

s)

A B C D

66

100

1

α1

β1

1 2β α

21 2β α

31 2β α

41 2β α

21 2 2β β α 1 2β β

1 2β β

21 2β β

1 2 1β β α

21 2 2β β α

51 2β α

2 21 2 2β β α6

1 2β α

21 2 2β β α

1 2 1β β α

(a) Voltage at D shortly after the surge has reached D

= 1 22 2 0.182 1.818 500 330.9kVivβ β = × × × =

(b) Voltage at D at a time 210 µs after the surge first reaches B

21 2 2 2

2

2 2 2

2 0.182 1.818 500 1 0.818 0.818

823kV

ivβ β α α = + + = × × × + +

=

(c) Voltage at B at a time 210 µs after the surge first reaches B

[ ]

[ ]

2 3 4 5 61 2 2 2 2 2 2

21 2 2

2 3

2

1

1

0.182 500 1 0.818 0.818 0.818 ....

0.182 1.181 500 1 0.818413kV

i

i

v

v

β α α α α α α

β β α

= + + + + + + + +

= × + + + + + × × +

=

Chapter 11 11.1 A 132 kV supply feeds a line of reactance 13 Ω which is connected to a 100 MVA

132/33 kV transformer of 0.1 p.u. reactance. The transformer feeds a 33 kV line of reactance 6 Ω which, in turn, is connected to an 80 MVA, 33/11 kV transformer of 0.1 p.u. reactance. This transformer supplies an 11 kV substation from which a local 11 kV feeder of 3 Ω reactance is supplied. This feeder energizes a protective overcurrent relay through 100/1 A current transformers. The relay has a true inverse-time characteristic and operates in 10 s with a coil current of 10 A.

Figure 11.36 Circuit for Question 11.1 If a three-phase fault occurs at the load end of the 11 kV feeder, calculate the fault current and time of operation of the relay.

Answer

T1

L1

T2

L2 L3

132 kV 33 kV 11 kV

Use a system base of 100 MVA and voltage bases of 132 kV, 33 kV and 11 kV. The reactance of the circuit to the fault is; L1= 0.075 p.u. T1= 0.1 p.u. L2= 0.551 p.u. T2= 0.125 p.u. L3= 2.479 p.u. Therefore total reactance to the fault is 3.33 p.u..

At 11 kV: 6

base 3

100 10I 524911 10 3

A×= =

× ×

Ifault = 5249/3.33 = 1575 A, Irelay = 15.75 A

If the relay inverse-time characteristic is given by ktI

= and when I = 10 A, t =

10 sec; k =100

Time of operation = 100 6.345

15.76s=

11.2 A ring-main system consists of a number of substations designated A, B, C, D, and E, connected by transmission lines having the following impedances per phase (Ω): AB (1.5 +j2); BC (1.5 +j2); CD (1 + j1.5); DE (3 + j4); EA (1 +j1).

The system is fed at A at 33 kV from a source of negligible impedance. At each substation, except A, the circuit breakers are controlled by relays fed from 1500/5 A current transformers. At A, the current transformer ratio is 4000/5. The characteristics of the relays are as follows: Current (A) 7 9 11 15 20 Operating times of relays at A, D and C 3.1 1.95 1.37 0.97 0.78

Operating times at relays at B and E 4 2.55 1.8 1.27 1.01

Examine the sequence of operation of the protective gear for a three-phase symmetrical fault at the midpoint of line CD.

Assume that the primary current of the current transformer at A is the total fault current to the ring and that each circuit breaker opens 0.3 s after the closing of the trip-coil circuit. Comment on the disadvantages of this system.

Answer

ZLH = 4.5 +j5.75 Ω, ZRH = 3.5 +j4.75 Ω

7.3

5.9LH

RH

Z

Z

= Ω

= Ω

33 1. 2.61 kA7.33

33 1. 3.23 kA5.93

LH

RH

I

I

= =

= =

A

B

CD

E

1+j1 1.5+j2

1.5+j23+j4

0.5+

j0.7

5

0.5+

j0.7

5

Relay Position

Fault current (kA)

Relay Current (A)

Time of relay operation (s)

Time of CB and relay operation (s)

A 5.84 7.3 3 3.3 B 3.23 10.77 1.95 2.25 C 3.23 10.77 1.45 1.75 D 2.61 8.7 2.15 2.45 E 2.61 8.7 2.8 3.25

The main difficulty with this scheme is the long clearing times.

11.3 The following currents were recorded under fault conditions in a three-phase system:

AI 1500 45 A= ∠ , BI 2500 150 A= ∠ , CI 1000 300 A= ∠

If the phase sequence is A-B-C, calculate the values of the positive, negative, and zero phase-sequence components for each line.

Answer

IA = 1061+j1061 A IB = -2165 +j1250 A IC = 500-j866 A

A0 A2

A1 B2

A2 C

I 1 1 1 I1I 1 . I3

I 1 Ia aa a

=

a = 0.5-j0.866 IA0 = -201+j482 A IA1 = 20-j480 IA2 = 1242+j1059 A

11.4 Determine the time of operation of 1A standard IDMT over-current relay having a Plug Setting (PS) of 125% and a Time Multiplier setting (TMS) of 0.6. The CT ratio is 400:1 and the fault current 4000A.

Answer

Plug setting of the relay = 1.25 * 400 = 500 A

Plug setting multiplier = 4000/500 = 8

0.02

0.14 0.6 2sec8 1

t ×= =

−

11.5 The radial circuit shown in Figure 11. 37 employs two IDMT relays of 5A. The plug setting of the relays are 125% and time multiplier of relay A is 0.05 sec. Find the time multiplier setting of relay B to coordinate two relays for a fault of 1400 A. Assume a grading margin of 0.4 sec.

Figure 11.37 Circuit for Question 11.5

Answer

Plug setting of the relay A = 1.25 * 100 = 125 A

Plug setting multiplier = 1400/125 = 11.2

Operating time of relay A = 0.02

0.14 0.0511.2 1

t ×=

− = 0.14 sec

Since the grading margin is 0.4 sec, the operating time of relay B should be set to 0.4 + 0.14 = 0.54 sec

Plug setting of the relay B = 1.25 * 200 = 250 A

Plug setting multiplier = 1400/250 = 5.6

0.02

0.14 TMS 0.545.6 1

TMS 0.135

t ×= =

−=

Select 0.15

11.6 A 66 kV busbar having a short circuit level of 800 MVA is connected to a 15 MVA 66/11kV transformer having a leakage reactance of 10% on its rating as shown in Figure 11.38.

(a) Write down the three-phase short circuit current (in kA) for a fault on the 66 kV terminals of the transformer

(b) Calculate the three-phase short circuit current (in kA) for a fault on the 11 kV feeder.

(c) The 11 kV relay has a Current Setting (Plug Setting) of 100% and Time Multiplier of 0.5. Use the IDMT characteristic to calculate the operating time for a 3-phase fault on the 11 kV feeder.

(d) The 66 kV relay has a Current Setting (Plug Setting) of 125%. Choose a Time Multiplier to give a grading margin of 0.4 seconds for a fault on the 11 kV feeder.

(e) What is the operating time for a three-phase fault on the 66 kV winding of the transformer?

(f) In practice, what form of transformer protection would operate first for a fault on the 66 kV winding?

Figure 11.38 Circuit for Question 11.6 Answer

a) If = 6

3

800 10 69883 66 10fI A×

= =× ×

, 7 kA

b) Using a 15 MVA base

6

3

15 0.1 0.11875800

1 8.421 p.u.0.11875

15 10 787 A3 11 10

6.63kA

f

base

f

X

I

I

I

= + =

= =

×= =

× ×=

c) Relay current = 6630/600 = 11.05 A, From Figure 11.16, this gives a time of

2.8 sec reduced to 1.4 sec with a TM of 0.5.

Applying the characteristic formula

0.02

0.14 0.5 1.425sec11 1

t ×= =

−

d) Operating time of 66 kV protection should be 1.825 sec. Fault current is 8.5

kA at 11 kV and so 1.416 kA at 66 kV and a relay current of 7.1 A. Therefore current as a multiple of PS is 7.1/1.25 = 5.7. To give an operating time of 1.825 sec:

0.02

0.14 TM 1.825sec5.7 1

t ×= =

−

TM = 0.46 Select TM as 0.5.

e) Fault current for a 66 kV fault is 7 kA, relay current is 35 A, current as a multiple of setting is 28 giving an operating time of around 2 secs reduced by the TM to 1 sec.

f) In practice differential protection (or the Bucholtz relay) would operate first for

faults inside the transformer.

11.7 A three-phase, 200 kVA, 33/11 kV transformer is connected as delta-star. The CTs on the 11 kV side have turns ratio of 800/5. What should be the CT ratio on the HV side?

Answer

If the line current on the high voltage side is I, then line current on the low voltage side

=33

11I ×

The CTs on the low voltage side are connected in delta. Therefore the line current in the CT secondary

33 5 3

11 800I ×

= × ×

The CTs on the high voltage side are connected in star and the line current on that side of the CT secondary = I CT ratio× For differential protection to work, line currents on the secondary sides of both set

of CTs should be equal, i.e I CT ratio× 33 5 3

11 800I ×

= × ×

=0.032 Select 150:5 CT ratio.

11.8 A three zone distance protection relay is located at busbar A as shown in Figure 11.39 (square). The VT ratio is 132 kV/220 V and CT ratio is 1000/1. Find the impedance setting for zone 1 and zone 2 protection of the relay assuming that zone 1 covers 80% of line section AB and zone 2 covers 100% of line section AB plus 30% of the shortest adjacent line.

Figure 11.39 Circuit for Question 11.8

All quantities are in p.u. on 100 MVA, 132 kV base)

Answer

Impedance seen by the relay = Voltage/Current seen by the relay

(132000 / 220)

/ / (1000 /1)1.67 Actual Impedance

V VT ratio VI CT ratio I

= =

= ×

Zone 1 protection 80% of line AB = j 0.1 x 0.8 x 1.67 = j 0.134 pu Zone 2 protection 100% of line AB and 30% of shortest adjacent line (which is line BC) = (j 0.1 x 1.0 + j 0.2 x 0.3) x 1.67 = j 0.27 pu

Chapter 12 12.1 The input-output curve of a coal-fired generating unit (with a maximum output of

550MW) is given by the following expression: H(P) = 126+8.9P+0.0029P2 [MJ/h] If the cost of coal is 1.26 £/MJ, calculate the output of the unit when the system marginal cost is

(a) 13 [£/MWh] and (b) 22 [£/MWh].

Answer

Data given: • MAXP 550 [MW]=

• coalC 1.26 [£ / MJ]=

• ( ) 2H P 126 8.9P 0.0029P [ MJ / h]= + +

The generation cost of the coal-fired plant is given by:

( ) ( ) coalC P H P C= ×

( ) 2C P 158.76 11.214P 0.003654P [£ / h]= + +

The marginal cost is given by:

( )dC P11.214 0.007308P [£ / MWh]

dP= +

a) ( )13

dC PdP

=

11.214 0.007308P 13+ =

P 244.4 [MW]=

b) ( )22

dC PdP

=

MAX11.214 0.007308P 22 P 1,475.9168 P+ = → = >

P 550 [MW]=

12.2 The incremental fuel costs of two units in a generating station are as follows:

11

1

22

2

0.003 0.7

0.004 0.5

dF PdPdF PdP

= +

= +

marginal cost are in £/MWh and unit outputs are in MW. Assuming continuous running with a total load of 150MW calculate the saving per hour obtained by using the most economical division of load between the units as compared with loading each equally. The maximum and minimum operational loadings are the same for each unit and are 125MW and 20MW.

Answer

Data given:

• ( ) 211 1 1 1 1

1

dF £ £0.003P 0.7 F P 0.7P 0.0015P MWh hdP = + → = +

• ( ) 222 2 2 2 2

2

dF £ £0.004P 0.5 F P 0.5P 0.002P MWh hdP = + → = +

• [ ]D 150 MW=

• minP 20 [MW]=

• maxP 125 [MW]=

( ) 1 2

1 1 2 2P ,PP) min F P F (P )+

Subject to:

1 2P P D+ =

min 1 maxP P P≤ ≤

min 2 maxP P P≤ ≤

1 2P ,P 0≥ The Lagrange function of this optimization problem is given by:

( ) ( )1 1 2 2 1 2L F P F P λ[D P P ]= + + − −

1 1

1 1

dF (P )dL λ 0dP dP

= − = Eq (1)

2 2

2 2

dF (P )dL λ 0dP dP

= − = Eq (2)

1 2dL D P P 0dλ

= − + =

1 2P P D 150 ∴ + = =

1 2P 150 P= − Eq (3) Equating equations (1) and (2):

1 1 2 2

1 2

dF (P ) dF (P )dP dP

=

1 20.003P 0.7 0.004P 0.5+ = +

2 10.004P 0.003P 0.2− =

Substituting for P1 from equation (3) into the previous expression:

( )2 20.004P 0.003 150 P 0.2− × − =

20.2 150 0.003P

0.004 0.003+ ×

=+

2P 92.86 [MW]=

1P 57.14 [MW]= The total cost of generation is given by:

( ) ( ) ( )T 1 2 1 1 2 2C P ,P F P F P= + The saving is then given by:

( ) ( )T 1 2 T 1 2Saving C P 75,P 75 C P 57.14,P 92.86109.6875 108.5714

£1.12 MW

= = = − = =

= −

=

12.3 What is the merit order used for when applied to generator scheduling?

A power system is supplied by three generators. The functions relating the cost (in £/h) to active power output (in MW) when operating each of these units are:

( )( )( )

21 1 1 1

22 2 2 2

23 3 3 3

P 0.04P 2P 250

P 0.02P 3P 450

P 0.01P 5P 250

C

C

C

= + +

= + +

= + +

The system load is 525 MW. Assuming that all generators operate at the same marginal cost, calculate:

(a) the marginal cost;

(b) optimum output of each generator;

(c) the total hourly cost of this dispatch.

Answer

Data given:

• ( ) 2 1 11 1 1 1 1

1

( )£ £0.04 2 250 0.08 2 h MWhdC PC P P P P

dP = + + → = +

• ( ) 2 2 22 2 2 2 2

2

( )£ £0.02 3 450 0.04 3 h MWhdC PC P P P P

dP = + + → = +

• ( ) ( )3 323 3 3 3 3

3

£ £0.01 5 250 0.02 5 h MWhdC P

C P P P PdP

= + + → = +

• 525 [ ]D MW=

b) ( )3 31 1 2 2

1 2 3

( ) ( ) dC PdC P dC PdP dP dP

= =

(1) ( ) ( )1 1 2 2

1 2 1 21 2

0.08 2 0.04 3 2 25dC P dC P

P P P PdP dP

= → + = + → − =

(2) ( )3 31 1

1 3 1 31 3

( ) 0.08 2 0.02 5 4 150dC PdC P P P P P

dP dP= → + = + → − =

(3) 1 2 3 525P P P+ + =

Rearranging the equations:

(1) 2 1P 2P 25= −

(2) 3 1P 4P 150= −

(3) 1 2 3P P P 525+ + =

Substituting equations (1) and (2) in (3):

1 1 1 1P 2P 25 4P 150 525 7P 700+ − + − = → =

[ ]1P 100 MW=

2P 175 [MW]=

3P 250 [MW]=

a) 1 11

1

dC (P ) 0.08P 2dP

= +

1 1

1

dC (P ) £10 MWhdP =

c) ( ) ( )Total 1 1 2 2 3 3C C P C P C (P )= + +

Total£C 4,562.5 h =

12.4 A power system is supplied from three generating units that have the following

cost functions:

Unit A: 2A A13 1.3P 0.037P+ + [$/h]

Unit B: 2B B23 1.7P 0.061P+ + [$/h]

Unit C: 2C C19 1.87P 0.01P+ + [$/h]

a) How should these units be dispatched if a load of 380 MW is to be supplied

at minimum cost? b) If, in addition to supplying a 380MW load, the system can export (sell)

energy on the neighbouring country in which the system marginal costs is 10.65 $/MWh, what is the optimal amount of power that it should be exported?

c) Repeat problem (b) if the outputs of the generating units are limited as follows:

PA

MAX = 110 MW PB

MAX = 85 MW PC

MAX = 266 MW

Answer

Data given:

• ( ) 2 A AA A A A A

A

dC (P )$ $C P 0.037P 1.3P 13 0.074P 1.3 h MWhdP = + + → = +

( ) 2 B BB B B B B

B

dC (P )$ $C P 0.061P 1.7P 23 0.122P 1.7 h MWhdP = + + → = +

( ) ( )C C2C C C C C

C

dC P$ $C P 0.01P 1.87P 19 0.02P 1.87 h MWhdP = + + → = +

a) D 380 [MW]=

Equations are given by:

( )C CA A B B

A B CA B C

dC PdC (P ) dC (P ) & P P P 380dP dP dP

= = + + =

Then:

(1) A A B BA B

A B

dC (P ) dC (P ) 0.074P 1.3 0.122P 1.7dP dP

= → + = +

A B0.074P 0.122P 0.4− =

(2) ( )C CA A

A CA C

dC PdC (P ) 0.074P 1.3 0.02P 1.87dP dP

= → + = +

A C0.074P 0.02P 0.57− =

(3) A B CP P P 380+ + =

Then, writing the equations in matrix form:

A

B

C

0.074 0.122 0 P 0.40.074 0 0.02 P 0.57

1 1 1 P 380

− − =

Solving this matrix:

[ ]AP 77.60 MW=

[ ]BP 43.79 MW=

CP 258.61 [MW]=

b) ( )C CA A B B

A B C

dC PdC (P ) dC (P ) $10.65 MWhdP dP dP = = =

Then:

(1) A AA A

A

dC (P ) 0.074P 1.3 10.65 P 126.35[ MW]dP

= + = → =

(2) B BB B

B

dC (P ) 0.122P 1.7 10.65 P 73.36[ MW]dP

= + = → =

(3) ( )C C

C CC

dC P0.02P 1.87 10.65 P 439 [MW]

dP= + = → =

exp A B CP P P P 380= + + −

expP 258.71 [MW]= c)

(1) [ ]MaxA AP 110 MW P 110[ MW]= → =

(2) [ ]MaxB BP 85 MW P 73.36[ MW]= → =

(3) [ ]MaxC CP 266 MW P 266[ MW]= → =

exp A B CP P P P 380= + + −

expP 69.36 [MW]=

12.5 For the system with the load duration curve given in Figure 12.19 below, design a generation system to minimize the total investment and operating costs. The cost characteristics of different generation technologies are given in Table below. Value of lost load is 8,000£/MWh.

Figure 12.19 19 Load Duration Curve of Problem 12.5

Technology Investment cost [£/kW/y]

Operating costs [£/MWh]

Base Load 250 5 Mid Merit 80 50

Peak Load 30 90 Calculate the amount of energy produced by each technology and the energy not served.

Answer

Data given: • £VOLL 8,000 MWh

=

Let’s calculate 1t , 2t and 3t :

(1) B M1 1

M B

I I 250,000 80,000t t 3,777.78[ h]O O 50 5

− −= = → =

− −

(2) M P2 2

P M

I I 80,000 30,000t t 1, 250[ h]O O 90 50

− −= = → =

− −

(3) P P3 3

P

I It t 3.75[ h]VOLL O VOLL

= ≈ → =−

Using these values in combination with the LDC: Baseload

( )B 124 7.5K 7.5 t 8,760

900 8,760−

= + × −−

BK 17.96 [GW]=

Then, the energy served by this type of generation technology is:

( ) ( )BB B 1 1

K 7.5ES K t 8,760 t

2+

= × + × −

BES 131.27 [TWh]=

Mid-merit

( )M 2 B24 7.5K 7.5 t 8,760 K

900 8,760−

= + × − −−

MK 5.31 [GW]= Then, the energy served by this type of generation technology is:

( )1 2M M

t tES K

2+

= ×

MES 13.34 [TWh]= Peaking

( )P 3 B M30 24K 24 t 900 K K0 900

−= + × − − −

−

PK 6.71 [GW]= Then, the energy served by this type of generation technology is:

( ) ( ) ( ) ( )3 2P B M P B M

t 900 900 tES K K K 24 24 K K

2 2+ +

= × + + − + × − −

PES 3.49 [TWh]=

Finally, the Energy Not Served (ENS) is given by:

( )B M P3

30 K K KENS t

2− − −

= ×

ENS 46.9 [MWh]=

12.6 A power system contains 5 identical generators of capacity of 120MW and

availability of 96% that supply demand with the Load Duration Curve presented in Figure 12.20.

Figure 12.20 Load Duration Curve of Problem 12.6 Calculate the LOLP at peak and LOLE. Answer

Data given:

• 5 identical generators • [ ]P 120 MW=

• vA 96%= Let’s first calculate the system state probabilities. The probability of each state is given by a binomial distribution as follows:

( ) ( )n kkv v

nPr K k A 1 A

k−

= = × × −

where:

• k is the number of available generators. • n is the total amount of generators. • vA is the availability of each generator.

Available

generators k

Available generation

[MW]

State probability

Cumulative probability*

5 600 0.815373 0.815373 4 480 0.169869 0.985242 3 360 0.014156 0.999398 2 240 0.000590 0.999988 1 120 0.000012 1.000000 0 0 0.000000 1.000000

(*) Probability of available generation being greater than or equal to generation

stated in column 2 Then, the LOLP of each demand level is:

Demand block

i

Demand level [MW]

Duration [h] LOLP

1 470 180 0.014758 2 315 2,320 0.000602 3 240 2,900 0.000012 4 115 3,360 0.000000

The LOLP at peak load is:

470LOLP 1.48%= Finally, the LOLE is given by:

4

i ii 1

LOLE Duration LOLP=

= ×∑

LOLE 4.0894 [h]=

12.7 A power system consists of two areas that are currently not connected and the national transmission operator is considering building and interconnector.

Generation costs in the two areas are:

with the demand in the two areas being DA=600MW and DB=1000MW.

Calculate the benefits in enhancing the efficiency of the overall generation system operation that an interconnector of

a) 200MW and b) 400MW

would create.

Answer

Data given:

• ( ) 2 ( )£ £0.025 6 0.05 6 h MWhA A

A A A A AA

dC PC P P P PdP

= + → = +

• ( ) 2 ( )£ £0.11 10 0.22 10 h MWhB B

B B B B BB

dC PC P P P PdP

= + → = +

• [ ]600 AD MW= 1,000 [ ]BD MW= The total generation cost is given by:

( ) ( ) ( ) 2 2T A B A A B B A A B BC P ,P C P C P 6P 0.025P 10P 0.11P= + = + + +

( )originalT A B

£C P ,P 132,600 h =

a) A A ABP D F= +

B B ABP D F= −

Then:

original newT T TC C C∆ = −

T T A AB B ABC 132.600 C (D F , D F )∆ = − + −

T TC 132.600 C (800, 800)∆ = −

T£C 33,400 h ∆ =

b) 132.600 (1000, 600)T TC C∆ = −

£56,000 h ∆ = TC

12.8 A power system of a small country is composed of two regions that are not connected. Generators 1 and 2 are located in the Northern Region while generators 3 and 4 are located in the Southern Region. The load in the Northern Region is 100 MW and the load in the Southern Region is 420 MW. Marginal cost of these generators are:

Northern Region

Southern Region

a) Calculate the marginal costs in both regions and the corresponding generation dispatches.

b) The transmission company is considering building a 450km long transmission link between the two regions. Show that the optimal transmission capacity required to connect the two regions should be about 100 MW, when the annuitsied investment cost of transmission is 37£/MW.km.year.

Answer

Data given:

• [ ]100 MWND =

• 1 1£3 0.02 MWhMC P = +

• 2 2£4 0.04 MWhMC P = +

• [ ]420 MWSD =

• 3 3£3.6 0.025 MWhMC P = +

MC P MWh1 13 0 02= + . [£ / ]

MC P MWh2 24 0 04= + . [£ / ]

MC P MWh3 336 0 025= +. . [£ / ]

MC P MWh4 44 2 0 025= +. . [£ / ]

• 4 4£4.2 0.025 MWhMC P = +

a) North

1 2 1 2MC MC 0.02P 0.04P 1= → − = Eq (1)

1 2 2 1P P 100 P 100 P+ = → = − Eq (2)

Substituting equation (2) in (1):

( )1 10.02P 0.04 100 P 1− × − =

11 4P0.06+

=

[ ]1P 83.33 MW=

2P 16.67 [MW]=

N£MC 4.67 MWh =

South

3 4 3 4MC MC 0.025P 0.025P 0.6 = → − =

3 4P P 24∴ − = Eq (3)

3 4 4 3P P 420 P 420 P+ = → = − Eq (4)


3 3P 420 P 24− + =

3420 24P

2+

=

[ ]3P 222 MW=

4P 198 [MW]=

N£MC 9.15 MWh =

b) l 450 [km]=

T0

kl 37 450 £c 1.9 MWhτ 8760× = = ≈

Then:

T N S 1 3π MC MC 3 0.02P 3.6 0.025P = − = + − −

T 1 3π 0.02P 0.025P 0.6∴ = − − Eq (5)

1 2 1 2MC MC P 2P 50= → − = Eq (6)

1 2 NS 2 NS 1P P 100 F P 100 F P+ = + → = + − Eq (7)


NS1 NS 1 1

250 2FP 200 2F 2P 50 P3+

− − + = → =

Then:

3 4 3 4MC MC P P 24= → − = Eq (8)

3 4 NS 4 NS 3P P F 420 P 420 F P+ + = → = − − Eq (9)


NS3 NS 3 3

444 FP 420 F P 24 P2−

− + + = → =

Substituting P1 and P3 in equation (5):

( ) NS NST NS

250 2F 444 Fπ F 0.02 0.025 0.63 2+ − = × − × −

The optimal transmission capacity is such that supply and demand for transmission are in equilibrium. Therefore:

T Tπ c=

NS NS250 2F 444 F0.02 0.025 0.6 1.93 2+ − × − × − =

NSF 247.097 [MW]=

12.9 Two areas, A and B, of a power system are linked by transmission link, with a

secure capacity of 900 MW. System load is concentrated in area B. In winter, the load is 3,500MW while summer load is 2,000 MW. The cost of generation in the areas can be modelled by the following expressions, where P is in MW:

[£/h] for area A, and

[£/h] for area B.

a) Determine the optimal levels of generation in areas A and B for each of the winter and summer seasons neglecting the constraints of the transmission system. Calculate the marginal cost of production in each season.

b) If necessary, modify the levels of generation computed in (a) to take into

2001.070)( AAA PPPC ++=

2002.0250)( BBB PPPC ++=

consideration the capacity of the existing transmission link. What are the marginal cost prices in areas A and B in each of the seasons?

c) Assuming that the duration of the winter period is 2,500h, calculate the generation cost in each season. What are the total annual generation costs?

d) The transmission company is considering doubling the transmission capacity between the two areas in order to reduce generation cost. Assuming that the annuitised investment cost of the reinforcement is 1,000,000 £/year, determine if this proposed investment is justified.

Answer

Data given:

• [ ]WD 3,500 MW=

• SD 2,000 [MW]=

• TK 900 [MW]=

• ( ) 2 A AA A A A A

A

dC (P )£ £C P 0.001P P 70 0.002P 1 h MWhdP = + + → = +

• ( ) 2 B BB B B B B

B

dC (P )£ £C P 0.002P 2P 50 0.004P 2 h MWhdP = + + → = +

a)

(1) A B A B A BMC MC 0.002P 0.004P 1 P 2P 500= → − = → − =

(2) A ABP F=

(3) B AB B ABP F D P D F+ = → = −

Substituting equations (2) and (3) in (1):

( )AB ABF 2 D F 500− × − =

AB500 2DF

3+

=

Then:

Winter Summer Units AP 2,500 1,500 MW BP 1,000 500 MW ABF 2,500 1,500 MW MC 6.00 4.00 £/MWh

Winter Summer Units AC 8,820 3,820 £/h BC 4,050 1,550 £/h

TotalC 12,870 5,370 £/h

Then:

WC 12,870 2,500 32,175,000[ £]= × =

SC 6,450 6,260 33,616,200[ £]= × =

Finally:

uncostrainedTOTAL W SC C C 65,791,200 [£]= + =

b)

Winter Summer Units AP 900 900 MW BP 2,600 1,100 MW ABF 900 900 MW AMC 2.80 2.80 £/MWh BMC 12.40 6.40 £/MWh

c) [ ]Wt 2,500 h=

St 6, 260 [h]=

Winter Summer Units AC 1,780 1,780 £/h BC 18,770 4,670 £/h

TotalC 20,550 6,450 £/h

Then:

WC 20,550 2,500 51,375,000[ £]= × =

SC 6,450 6,260 40,377,000[ £]= × =

Finally:

900TOTAL W SC C C 91,752,000 [£]= + =

d) [ ]TK 1,800 MW=

Annuitized investment cost of 1,000,000[ £ / yr]

Winter Summer Units AP 1,800 1,500 MW BP 1,700 500 MW ABF 1,800 1,500 MW AMC 4.60 4.00 £/MWh BMC 8.80 4.00 £/MWh

Winter Summer Units AC 5,110 3,820 £/h BC 9,230 1,550 £/h TotalC 14,340 5,370 £/h

Then:

WC 14,340 2,500 35,850,000[ £]= × =

SC 5,370 6,260 33,616,200[ £]= × = 1800TOTALC 69,466,200 [£]=

Finally:

900 1800TOTAL TOTAL TOTALC C C∆ = −

TOTALC 22,285,800 1,000,000∆ = >

Investment is justified.

12.10 A distribution substation will supply 160 households of two types. Large

detached houses (45) with expected peak demand of 12kW and smaller terraced houses (115) with peak demand of 8kW. Coincidence factors for an infinite number of detached and terraced houses are 0.12 and 0.18 respectively. Assuming that peaks of both types of houses coincide, calculate the peak demand of the proposed 11kV/0.4kV substation.

Answer

Data given:

• 160 households: o 45 Big (B) o 115 Small (S)

• [ ]1 12 kWBP =

• [ ]1 8 kWSP =

• 0.12Bnj =

• 0.18Snj =

We know that:

1n nP j n P= × × Then:

0.12 45 12 0.18 115 8T B Sn n nP P P= + = × × + × ×

230.4 [kW]T

nP =

Electric Power Systems - wiley.com · PDF filePeak power occurs when . 0 dP dt = 8-0.00182t=0,...

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Transcript of Electric Power Systems - wiley.com · PDF filePeak power occurs when . 0 dP dt = 8-0.00182t=0,...