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...

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

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 4800 × 1600 = 7, 680, 000 km 2 Area of rights of way after n years 16, 000 × (1. + 0.07) n km 2 Equating area of USA with area of rights of way after n years

7, 680, 000 = 480 16, 000 log10 480 = ∴n = 91.25 years log10 1.07 n (1 + 0.07) =

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 = 5 × π × 0.52 = 3.927 m3 Power of natural gas 3.927 × 60 × 40 = ×106 9.4 GJ/sec = 9.4 GW Power of hydrogen = 9.4 × 1.3

13 = 3.1 GW 40

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 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: 25

Generator 3 operates 20

Generator 2 operates

15

Power (MW) 10

1000

2000

3000

4000 5000 Time (hrs)

6000

7000

8000

8760

0

5714

3077 0

7963

Generator 1 operates

5

754

1.4

9000

The annual load is the integral under the power curve

Load energy over a year = t =8760

∫

t =0

2

3

8760 8760 4000 + 8t − 0.00091t 2 dt = 4000 × 8760 + 8 × 2 − 0.00091× 3 = 138083.25 MWh

Peak power occurs when

dP =0 dt

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 t=

8 ± 64 − 4 × 0.00091× 6000 2 × 0.00091

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 t=

8 ± 64 − 4 × 0.00091× 16000 2 × 0.00091

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 2 3 4 5 6

Cost of energy from storage 2/0.64= 3.125

Profit or (loss) (1.25 (0.125) 0.875 1.875 2.875

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 𝑉 1 ∗ = 239.6 A 𝐈𝒂𝒏 = √3 𝑅𝑎𝑛 𝑉𝑒 −𝑗120 1 𝐈𝒃𝒏 = ∗ = (−59.9 − 𝑗103.75 ) A 𝑅𝑏𝑛 √3 𝑉𝑒 +𝑗120 1 𝐈𝒄𝒏 = ∗ = (−59.9 + 𝑗103.75 ) A 𝑅𝑐𝑛 √3 𝑉𝑒 +𝑗30 𝐈𝒂𝒃 = = (359.4 + 𝑗207.5 ) A 𝑅𝑎𝑏 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 ∗ 𝑒 𝑗(120−𝜙) = (−59.9 + 𝑗203.75 ) A

𝐒 = 𝐈𝒂∗ 𝐕𝒂 + 𝐈𝒃∗ 𝐕𝒃 + 𝐈𝒄∗ 𝐕𝒄 = 𝐈𝒂∗ 2.2

V

√3

+ 𝐈𝐛∗

𝑉𝑒 −𝑗120 √3

+ 𝐈𝐜∗

𝑉𝑒 +𝑗120 √3

= (349.29 + 𝑗35.94 ) kVA

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 1

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

1Ω 1Ω

2Ω

b c

Figure 2.29 Circuit for problem 2.2 Answer a

Ia

Ra

Rb b c

Ib

n

Rc

Ic

𝐈𝐚 R a − 𝐈𝐛 R b = 400 𝐈𝐛 R b − 𝐈𝐜 R c = 400𝑒 −𝑗120 𝐈𝐚 + 𝐈𝐛 + 𝐈𝐜 = 0

𝐈𝐚 = (200 − 𝑗69.28 ) A 𝐈𝐛 = (−200 − 𝑗69.28 ) A 𝐈𝐜 = 𝑗138.56 A 400 𝐕𝐧𝐠 = 𝐕𝐜𝐠 − 𝐈𝐜 R c = 𝑗 V − 2Ω ∗ 𝑗138.56 A = −𝑗46.18 V √3 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.

2

a b c

Ia

Ib

Ic

N

R

S

Figure 2.30 Circuit for problem 2.3 Answer 𝑋𝑐 =

1 = 318.31 Ω 2𝜋𝑓𝐶

Assume V is phase voltage Switch closed: 𝐒 = 𝐈𝒂∗ 𝐕𝒂 + 𝐈𝒃∗ 𝐕𝒃 + 𝐈𝒄∗ 𝐕𝒄 =

P𝑠𝑤𝑖𝑡𝑐ℎ 𝑐𝑙𝑜𝑠𝑒𝑑 =

Switch open:

V2 𝑅

𝐕𝒂∗ 𝐕𝒂 𝐕𝒃∗ 𝐕𝒃 𝐕𝒄∗ 𝐕𝒄 1 2 2 � ∗ + ∗ + ∗ =𝑉 � −𝑗 𝐙𝒂 𝐙𝒃 𝐙𝒄 𝑅 𝑋𝑐

Ia

a

i1 Ib

b

Ic

c

i2

N

R

S

�

−𝑗2𝑋𝑐 𝑗𝑋𝑐

𝐢 𝑗𝑋𝑐 𝐕 � � 𝟏 � = � 𝐚𝐛 � −𝑗𝑋𝑐 + 𝑅 𝐢𝟐 𝐕𝐛𝐜

(𝐕𝒂𝒃 + 2𝐕𝒃𝒄 ) (2𝑅 − 𝑗𝑋𝑐 ) (if 𝐕𝒂 = 𝑉, then 𝐕𝒂𝒃 = √3𝑉𝑒 +𝑗30 and 𝐕𝒃𝒄 = −𝑗√3𝑉) 𝐈𝒄 = −𝐢𝟐 = −

𝑃𝑠𝑤𝑖𝑡𝑐ℎ 𝑜𝑝𝑒𝑛 = 𝐈𝒄 𝐈𝒄∗ 𝑅 =

9𝑅𝑉 2 4𝑅 2 + 𝑋𝑐2

𝑃𝑠𝑤𝑖𝑡𝑐ℎ 𝑐𝑙𝑜𝑠𝑒𝑑 = 𝑃𝑠𝑤𝑖𝑡𝑐ℎ 𝑜𝑝𝑒𝑛 ⇒ 5𝑅 2 = 𝑋𝑐2

𝑅 = 142.35 Ω

3

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

From which

−𝐈𝐜 (−j53) + 𝐈𝐛 (50 + j40) = 𝐕𝐛𝐜 = 400e−j120 𝐈𝐜 = (−6.536 − j3.7736) A 𝐈𝐚 = (11.414 − j0.129) A

2.5

Power in wattmeter = real part of (−𝐕𝒃𝒄 𝐈𝒂∗ ) = 2.33 kW

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: 𝐈𝐑𝐘 𝐙𝐑𝐘 = 𝐕𝐑𝐘 = 400 V ⇒ 𝐈𝐑𝐘 = 4 A

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

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

The line currents are as follows:

4

𝐈𝐑 = 𝐈𝐑𝐘 − 𝐈𝐁𝐑 = (7.464 + j2) A;

|𝐈𝐑 | = 7.727 A

𝐈𝐁 = 𝐈𝐁𝐑 − 𝐈𝐘𝐁 = −j4 A;

|𝐈𝐁 | = 4 A

𝐈𝐘 = 𝐈𝐘𝐁 − 𝐈𝐑𝐘 = (−7.464 + j2) A; (b) The line voltages are as follows:

|𝐈𝐘 | = 7.727 A

𝐈𝐑𝐘 𝐙𝐑𝐘 = 𝐕𝐑𝐘 = 400 V ⇒ 𝐈𝐑𝐘 = 4 A

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

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

The line currents are as follows:

𝐈𝐑 = 𝐈𝐑𝐘 − 𝐈𝐁𝐑 = (0.536 + j2) A;

|𝐈𝐑 | = 2.07 A

𝐈𝐁 = 𝐈𝐁𝐑 − 𝐈𝐘𝐁 = −j4 A;

|𝐈𝐁 | = 4 A

𝐈𝐘 = 𝐈𝐘𝐁 − 𝐈𝐑𝐘 = (−𝟎. 𝟓𝟑𝟔 + 𝐣𝟐) 𝐀;

2.6

|𝐈𝐘 | = 𝟐. 𝟎𝟕 𝐀

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 = 𝐈=

𝐄 𝑗𝑋2 𝑅 ) (𝑗𝑋1 + 𝑗𝑋2 + 𝑅

𝑗𝑋2 𝑅 𝑗𝑋2 +𝑅

𝐄𝐄∗ E 2 (𝑅𝑋22 + 𝑗(𝑅 2 (𝑋1 + 𝑋2 ) + 𝑋1 𝑋22 ) 𝐒 = 𝐄𝐈 = = 𝑗𝑋2 𝑅 ∗ 𝑋12 𝑋22 + 𝑅 2 (𝑋1 + 𝑋2 )2 (𝑗𝑋1 + ) 𝑗𝑋2 + 𝑅 ∗

Therefore P = Re[𝐒] =

E 2 𝑅𝑋22 𝑋12 𝑋22 + 𝑅 2 (𝑋1 + 𝑋2 )2

Using the first derivative test:

dP E 2 𝑋22 (𝑋12 𝑋22 − 𝑅 2 (𝑋1 + 𝑋2 )2 ) = =𝟎 (𝑋12 𝑋22 + 𝑅 2 (𝑋1 + 𝑋2 )2 )2 dR ∴ 𝑋12 𝑋22 − 𝑅 2 (𝑋1 + 𝑋2 )2 = 0

⟹

5

𝑅=

2.7

𝑋1 𝑋2 𝑋1 + 𝑋2

𝑋 𝑋

From 𝑅 = 𝑋 1+𝑋2 it follows: 1 2 1 1 1 = + 𝑅 𝑋1 𝑋2

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 ∗ = 𝐕

𝐕∗ 𝐙∗ ∗

𝟐 𝐕𝒓𝒂𝒕𝒆𝒅 (100)𝟐 𝐙=� = (500 + 𝑗500) Ω � = (10 + 𝑗10)∗ 𝐒

𝐈=

100 500+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

6

2.8

Show that p.u. impedance (obtained from a short circuit test) of a star-delta threephase 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: 𝑉2 𝐺

Z𝑌𝑏𝑎𝑠𝑒 =

Z𝑌𝑝.𝑢. =

Delta side:

Z1

Z𝑏𝑎𝑠𝑒

Z𝐷𝑏𝑎𝑠𝑒 == 𝑉1𝐿𝑁 1 = 𝑉2𝐿𝐿 𝑁

Z𝐷𝑏𝑎𝑠𝑒 =

=

Z𝐺 𝑉2

2 V2𝐿𝐿,𝑏𝑎𝑠𝑒 S𝑏𝑎𝑠𝑒

⟹

𝑉1𝐿𝐿

√3𝑉2𝐿𝐿

𝑁2𝑉 2 3𝐺

=

1 𝑁

⟹

𝑉2𝐿𝐿 =

𝑁𝑉 √3

⟹

𝑍2 in delta = 𝑍𝑁 2

Per phase impedance can be found by delta to star transformation and is given by 𝑍𝑁 2 ⁄3 2.9

Z𝐷𝑝.𝑢. =

Z𝑁 2 3𝐺 3 𝑁2𝑉 2

=

𝑍𝐺 𝑉2

= Z𝑌𝑝.𝑢.

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 =

2.10

(11 ∗ 103 )2 = 2.42 Ω 50 ∗ 106

𝑋𝑝.𝑢. =

𝑗0.5 = 0.207 p. u. 2.42

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.) 7

Figure 2.32. Circuit for Problem 2.10 Answer 𝑆𝐵 = 100 MVA

Zone 1 (see diagram below): 𝑉𝐵 = 22 kV 𝐒𝐩.𝐮. =

500 = 5 p. u. 100

On 500 MVA base, the generator reactance is 2 pu Reactance on 100 MVA base = 2 × Zone 2:

100 = 0.4 p. u. 500

𝑉𝐵 = 400 kV 𝑍𝐵 =

(400 ∗ 103 )2 = 1600 Ω 100 ∗ 106

𝒁𝒍𝒊𝒏𝒆,𝒑.𝒖. = 𝐵𝑝.𝑢. = Zone 3:

𝐙𝐥𝐢𝐧𝐞 (0.1 + 𝑗0.5) ∗ 80 = = (0.005 + 𝑗0.025) p. u. 𝑍𝐵 1600

𝐵 10 × 10−6 ∗ 80 = = 1.28 p. u. 1 1 𝑍𝐵 1600

𝑉𝐵 = 11 kV

For the load

𝐒𝐩.𝐮. =

500(0.85 + 𝑗 sin(cos−1 0.85)) = 4.25 + j2.634 p. u. 100 8

(0.05+j0.025) p.u. 5 p.u.

0.1 p.u.

0.4 p.u. 0.1 p.u.

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

0.64 p.u. 0.64 p.u. Load

(0.05+j0.025) p.u.

0.64 p.u.

Zone 1 2.11

0.64 p.u.

Zone 2

Zone 3

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 220 =1 220

𝑉𝑝.𝑢. = 𝑍𝐵 =

(220)2 = 4.84 Ω 10000

𝑅𝑝.𝑢. = 𝐈𝐩.𝐮. =

2.12

20 = 4.1232 4.84

𝑉𝑝.𝑢. = 0.242 𝑅𝑝.𝑢.

𝐒𝒑.𝒖. = 𝐕𝐈 ∗ = 0.242

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

9

S

Z I

V1

V2

𝐒 = (18 + 𝑗6) kVA

(a)

𝐒

Load

Ss

∗

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

(b)

𝐈=�

(c)

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

√3𝑉2

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

2.13

pf = cos �tan−1 �

6.93 �� = 0.9345 18.186

Two transmission circuits are defined by the following ABCD constants: 1, 50, 0, 1, -4 and 0.9 ∠∠2º, 91º, 150∠ 0.9∠ 79º, 2º, 9X10 Determ ine 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 2.14

D0=C1B2 +D1D2 = 0.9 ∠2º

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 = 139.392 Ω 125 ∗ 106

𝐕𝐩.𝐮. =

132 =1 132

𝐒 = 125(0.9 + 𝑗 sin(cos −1 0.9)) = (112.5 + 𝑗54.486) MVA 10

𝐒𝒑.𝒖. =

𝐒 = 0.9 + 𝑗0.4359 𝑆𝐵

Line length 16 km 𝐙𝒑.𝒖. =

Accurate:

𝐙 16(0.156 + 𝑗0.4125) = = 0.0179 + 𝑗0.0473 𝑍𝐵 139.392

𝑅𝑃 𝑋𝑄 2 𝑋𝑃 𝑅𝑄 2 + � +� − � 𝑉 𝑉 𝑉 𝑉 0.0179 ∗ 0.9 0.0473 ∗ 0.4359 2 + � 𝐸 2 = �1 + 1 1 0.0473 ∗ 0.9 0.0179 ∗ 0.4359 2 +� − � 1 1 𝐸 2 = �𝑉 +

𝐸 2 = (1.03673)2 + (0.03477)2 = 1.076

𝐸 = 1.0373 p. u. = 1.0373 ∗ 132 V = 136.92 V

Approximate:

𝐸=𝑉+

𝑅𝑃 𝑋𝑄 + = 1.0367 p. u. 𝑉 𝑉

𝐸 = 𝐸 ∗ 𝑉𝐵 = 1.03673 ∗ 132 V = 136.85 V Line length 80 km 𝐙𝒑.𝒖. =

Accurate:

𝐙 80(0.156 + 𝑗0.4125) = = 0.0895 + 𝑗0.2367 𝑍𝐵 139.392

𝑅𝑃 𝑋𝑄 2 𝑋𝑃 𝑅𝑄 2 + � +� − � 𝑉 𝑉 𝑉 𝑉 0.0895 ∗ 0.9 0.2367 ∗ 0.4359 2 + � 𝐸 2 = �1 + 1 1 0.2367 ∗ 0.9 0.0895 ∗ 0.4359 2 +� − � 1 1 𝐸 2 = �𝑉 +

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

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

Approximate:

𝐸=𝑉+

𝑅𝑃 𝑋𝑄 + = 1.18373 p. u. 𝑉 𝑉

𝐸 = 𝐸 ∗ 𝑉𝐵 = 1.18373 ∗ 132 V = 156.25 V 11

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 𝑋 = 10 𝑅

𝑅 2 + 𝑋 2 = 22

⟹

4 = 0.199 𝑅=� 101 𝐙 = 0.199 + 𝑗1.99 𝐕 = 1∠ − 30° = 0.866 − 𝑗0.5 𝐄−𝐕 𝐈= = 0.2903 − 𝑗0.3901 𝐙 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

Z= 2∠ tan −1 10 = 2∠84.29 E = 1.7 p.u V = 1.0 p.u

V2 VV PG =G cos θ − G L cos(θ + δ ) Z Z 2 1.7 1.7 × 1 cos(84.29 + 30 ) = cos84.29 − 2 2 = 0.494 p.u.

VV V2 PL =L cos θ − L G cos(θ − δ ) Z Z 2 1.0 1.7 ×1 = cos84.29 − cos(84.29 − 30 ) 2 2 = 0.446 p.u. 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; 12

(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)

(b)

𝑉𝑙𝑖𝑛𝑒 = √3𝑉 = 415.6922 V

𝐕 = 12.7892 − 𝑗19.1852 = 23.0573∠ − 56.3119° A 𝐙 = 𝐈𝒂𝒃 − 𝐈𝒄𝒂 = 𝐈𝒍𝒐𝒂𝒅 − 𝐈𝒍𝒐𝒂𝒅 ej120 = 39.9364∠ − 86.3119° A

𝐈𝒍𝒐𝒂𝒅 = (c)

(i)

𝐈𝒍𝒊𝒏𝒆

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

Star-connected capacitor

𝑄 = 𝑄𝑐 = 3

(ii)

𝑉𝑐 2 = 3 ∗ 2402 ∗ 2𝜋𝑓𝐶 𝑋𝑐

23.925 × 103 = 440.72 𝜇F 𝐶= 3 ∗ 2402 ∗ 2𝜋 ∗ 50

Delta-connected capacitor

𝑉𝑐 2 = 3 ∗ (√3 ∗ 240)2 ∗ 2𝜋𝑓𝐶 𝑄 = 𝑄𝑐 = 3 𝑋𝑐

𝐶=

23.925 × 103 = 146.91 𝜇F 3 ∗ 3 ∗ 2402 ∗ 2𝜋 ∗ 50

13

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

′ ′ = 𝛿𝑚𝑒𝑐ℎ ∙ 𝑝 = 4° ∙ 𝛿𝑒𝑙

No load: 𝐸=

6.6 √3

kV

𝑃𝑠𝑦𝑛𝑐ℎ = 3 × 𝑛𝑠 =

60 × 𝑓 𝑝

𝐸2 ′ × 𝛿𝑒𝑙 2𝑋𝑠

𝑇𝑠𝑦𝑛𝑐ℎ =

3.2

𝜋 [𝑟𝑎𝑑] 180

2 𝑃𝑠𝑦𝑛𝑐ℎ 6.6 1 𝜋 1 3 = 3 × � × 10 � × × 4° × × 𝑛𝑠 2 × 10 180 2𝜋 × 60 × 50 √3 2𝜋 ∗ 60 2 × 60

= 968 𝑁𝑚

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∙𝑉 ∙ 𝑠𝑖𝑛𝛿 𝑋

1

𝑃𝑚𝑎𝑥 =

1.1 × 1 = 0.52381 p. u. 2.1

𝑠𝑖𝑛𝛿 = 1 ⟹ 𝛿 = 90°

𝐈 = 3.3

𝑗1.1 − 1 𝐄−𝐕 = = 0.5215 + 𝑗0.4787 = 0.7079 ∠42.5497° 0.01 + 𝑗2.1 𝐙

𝑐𝑜𝑠𝜑 = 0.7367 leading

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 = 𝑃=3

6.6 √3

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

𝐸𝑉 3 × 𝑉 × 1.25𝐸 𝑠𝑖𝑛𝛿 = ∙ 𝑠𝑖𝑛𝛿, 𝑋𝑠 𝑋𝑠

𝑃𝑚𝑎𝑥 = 31.655 𝑀𝑊 𝑄=

3 ∙ 𝑉 ∙ (1.25𝐸 ∙ 𝑐𝑜𝑠𝛿 − 𝑉) , 𝑋𝑠

𝑄 = −10.89 𝑀𝑉𝐴𝑟

pf = cos �tan−1 � 3.4

𝑄

𝑃𝑚𝑎𝑥

maximum power, angle 𝛿 = 90°

angle 𝛿 = 90°

�� = 0.9456

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.

Procentage negulation =

From Figure 3.4, 𝑋𝑠 =

⇒

𝑆 ∗

𝐈 = �� � � = 1 𝑝. 𝑢. 𝑉

𝑉𝑜𝑐 − 𝑉𝑙𝑜𝑎𝑑 𝐸−𝑉 ∗ 100 = ∗ 100 𝑉𝑙𝑜𝑎𝑑 𝑉

𝐾𝐹 1.95 𝑝. 𝑢. = = 1.95 𝐾𝐶 1 𝑝. 𝑢.

2

cos φ= 0.8 lagging: 𝐈 = 𝐼�cosφ − 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 =

cos φ = 1:

𝐸−𝑉 2.6725 − 1 × 100 = × 100 = 167.25% 𝑉 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 = 𝐼�cosφ + 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°

3.5

Procentage regulation = 76.14%

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.,

𝐈 ∙ 𝐙 = 𝐈 × 0.3 = 0.3𝐼𝑑 + 𝑗0.3𝐼𝑞

𝛿 = 30°,

𝐸 = 1.4𝑉

𝑋𝑑 = 𝑋𝑑′ + 0.3 = 1.8 𝑝. 𝑢. 𝑋𝑞 = 𝑋𝑞′ + 0.3 = 1.3 𝑝. 𝑢.

𝑃=

3.6

𝑉 2 �𝑋𝑑 − 𝑋𝑞 � 𝑉𝐸 𝑠𝑖𝑛𝛿 + 𝑠𝑖𝑛2𝛿 = 0.4814 𝑝. 𝑢. 𝑋𝑑 �2𝑋𝑑 𝑋𝑞 �

𝑃𝑠𝑦𝑛𝑐𝑜𝑒𝑓

𝑉 2 �𝑋𝑑 − 𝑋𝑞 � 𝑑𝑃 𝑉𝐸 = = 𝑐𝑜𝑠𝛿 + 𝑐𝑜𝑠2𝛿 = 0.7804 𝑝. 𝑢. 𝑑𝛿 𝑋𝑑 𝑋𝑑 𝑋𝑞

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 3

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.

𝐸 = 0.7857 𝑝. 𝑢. 0𝑐 √2

𝑋𝑠 =

𝑋′ = 𝑋 ′′ =

𝐸 = 0.2481 𝑝. 𝑢. 0𝑎 √2

𝐸 = 0.1768 𝑝. 𝑢. 0𝑏 √2

On load: 𝑉2 = 1,

𝑝𝑓2 = 0.8,

𝐼=1

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

𝐄 = 𝐕 + 𝑗𝑋𝑠 𝐈 = 1 + 𝑗0.7857(0.8 − 𝑗0.6) = 1.604∠23.2° As per Figure 3.6(b), 𝐼" = 𝐼′ =

𝐼" = 3.7

1.604 = 9.07 p. u. 0.1768

1.604 = 6.47 p. u. 0.2481

1.604 = 2.04 p. u. 0.7857

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

Answer 4

φ = 𝑐𝑜𝑠 −1 0.9 = 25.842

𝑃 = 0.9 × 500 = 450 MW 𝑋𝑠 = 𝑉=

1 Ω = 1.818 𝑆𝐶𝑅 𝑝ℎ𝑎𝑠𝑒

22

√3

𝑘𝑉

𝐸=0 𝑄=

⇒

𝐈=

22

1 ∠90° = 𝑗6.986 √3 𝑋𝑠 ⋅

3𝑉 2 = 266.23 MVAr Leading 𝑋𝑠

With respect to the performance chart shown in 3.12

MW

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.

E=2.5 p.u

0.9 lagging 500

e

f g

450

a b

h

E=V=1 p.u

c

j d

3.8

-200

200

400

500

MVAr

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

5

𝑉 = 275 kV, 𝑙 = 96 𝑘𝑚, 𝐼 = 800𝐴, 𝑟 = 0.078 Ω/km, phase

𝑝. 𝑓. = 0.9 lagging

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 = − 𝑌𝑐 =

1 1 =− = 952.7954 Ω 𝜔𝑐 2𝜋𝑓𝑐

1 = 0.00105 𝑠 𝑋𝑐

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 + 𝐕𝐫

𝐈𝐬 = 𝐈 + 𝐕𝐬

𝑌 2 𝑌 2

𝐀=𝐃=1+

𝐙𝐘 = 0.9799 + 𝑗0.0039 2

𝐁 = 𝐙 = 7.483 + 𝑗38.2179 𝐂 = �1 +

𝐙𝐘 −0.0021 + 𝑗1.039 �𝐘 = � � 4 1000

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

6

Medium line representation, T method of solution: 𝐈𝐫 ⋅ 𝐙 275 ⋅ 103 (720 − 𝑗384.71)(7.488 + 38.2179) = + 𝐕𝐜 = 𝐕𝐫 + 𝟐 2 √3 = (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 (7.488 + 𝑗38.2179)(706.9 × 172.3) 𝐙 ⋅ 𝐈𝐬 𝐕𝐬 = 𝐕𝐜 + = (168.13 + 𝑗12.453) + 𝟐 2

3.9

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

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 = 300�cos𝜑 − 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

𝐙 𝐁 = � sinh √𝐙𝐘 = 11.5553 + 𝑗95.3144 𝐘

𝐘 𝐂 = � sinh √𝐙𝐘 = ( − 0.0048 + 𝑗1.5056) × 10−3 𝐙

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−𝟏 𝐁

= (0.0125 + 𝑗7.8193) × 10−5 7

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 × 10−6 S × 83 = 𝑗7.9099 × 10−4 S

This is a medium line; From π network representation 𝐀=𝐃=1+

𝐙𝐘 = 0.98917 + j0.00256 2

𝐁 = 𝐙 = 6.474 + j27.39 𝐂 = �1 +

𝐙𝐘 � 𝐘 = −1.0126 × 10−6 + j7.8671 × 10−4 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 8

∗ 𝐒 𝐈 = � 𝐕𝐫 � = (377.9 − 𝑗283.4)A √𝟑

𝐕𝐫 1 1.25 11 × 103 1 𝐈circular = 𝑛𝑡 ⋅ 𝑣𝑡 ⋅ � � ⋅ =5× × × = −𝑗99.2321 A 100 𝑗2 × 2 √3 𝑗2𝑋𝑙 √3 𝐈T1 = 𝐈T2 =

𝐈 + 𝐈circular = 189 − 𝑗 240.9 = 306.1979∠ − 51.8962° A 2 𝐈 − 𝐈circular = 189 − 𝑗 42.5 = 193.6676∠ − 12.6709° A 2

𝑄 = 3 ⋅ 𝐼𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 2 ⋅ 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. 𝑋 2 𝑠 2 − 𝑉 2 𝑅2 𝑠 + 𝑅22 = 0 ∴𝑠=

𝑉 2 𝑅2 − �𝑉 4 𝑅22 − 4𝑋 2 𝑅22 𝑋2

From equivalent circuit 𝐼=

𝑉

�𝑋 2 + (𝑅2 ⁄𝑠)2

𝑉2 𝑄= + 𝐼2 𝑋 𝑋𝑚

With 𝑋𝑚 = 3 p. u., 𝑋 = 0.2 p. u. and 𝑅2 = 0.04 p. u., the above equations were used 9

to obtained Q-V characteristic 1.07 1.06 1.05

Q p.u.

1.04 1.03 1.02 1.01 1 0.99 0.98 0.8

0.85

0.9

0.95 V p.u.

1

1.05

1.1

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

𝑆𝑐𝑟 =

𝑟2 0.04 = = 0.2 p. u. 𝑋𝑠 0.2

10

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.

XL=0.3 p.u.

X=0.1 p.u.

XL=0.3 p.u.

X=1.5 p.u.

X=2 p.u. Load

𝐙𝐞 = 𝑗 �1.5 + 0.1 + Busbar L:

0.3 + 0.1� = 𝑗1.85 2

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

𝐒𝐋,(𝐩.𝐮.) =

𝐒𝐋 = 0.5 + 𝑗0.4197 100 MVA

Generator 1:

∗

𝐒𝐋,(𝐩.𝐮.) 𝐈=� � = 0.5 − 𝑗0.4197 𝐕

𝐄𝟏 = 𝐈 ∙ 𝐙𝐞 + 𝑉𝐿 = (0.5 − 𝑗0.4197) × 𝑗1.85 + 1 = 1.7765 + 𝑗0.925 = 2.0029 ∠ 27.505𝑜

Generator 2: 𝐈𝟐 = �

∗

𝐒𝐠𝐞𝐧𝟐 50 + 𝑗20 ∗ � = 0.5 − 𝑗0.2 � =� 𝐕 100

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

11

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

12

Answer Substation A generators Substation A transformers 48 km, 258 mm2 line

Vbase=16 kV; Sbase=100 MVA Sbase=100 MVA

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

Zbase = 1322/100 = 174.24

48 km, 113 mm2 line

Zbase = 174.24

80 km, 258 mm2 line

Zbase = 174.24

Substation B Synch condensator Star equivalent of the 132/66/11 kV transformer

Vbase=11 kV; Sbase=100 MVA

From Table 3.2(a) Z=(0.068+j0.404)x48/174.24 = 0.0187+j0.1113 From Table 3.2(a) Z=(0.155+j0.41)x48/174.24 = 0.0427+j0.113 From Table 3.2(a) Z=(0.068+j0.404)x80/174.24 = 0.03122+j0.1855 X=1.5x100/30=5 p.u.

Substation C transformer Substation C load

Sbase=100 MVA

Substation D generators Substation D transformer Substation D load Cable

X=0.1x100/125=0.08 p.u.

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 X=0.15x100/150=0.1 p.u.

Vbase=66 kV; Sbase=100 MVA Vbase=11 kV; Sbase=100 MVA Sbase=100 MVA

S = I = 3 – j1.86 p.u Z = 1/I = 0.2408 + j0.1493 X=1.0x100/75=1.333 p.u.

Vbase=66 kV; Sbase=100 MVA Zbase = 662/100 = 43.56

S = I = 0.5 – j0.375 p.u Z = 1/I = 1.28 + j0.96 Z=(2+j4)/43.56 = 0.0459+j0.0918

X = 0.1 p.u.

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. 13

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 − 𝑗𝑋𝑙 𝐈 = 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. =

𝑆 = 0.8 + 𝑗0.6 𝑆𝐵

𝑺𝒑.𝒖. ∗ 𝐄 = 𝑉𝑇 + 𝑗𝑋𝑠 ∙ 𝐈 = 1 + 𝑗1.5 × � � 𝑉𝑇 0.8−𝑗0.6 � 1

𝐄 = 1 + 𝑗1.5 × �

b) 𝑄 =

𝑉𝑇 (𝐸 𝑋𝑠

∙ cos 𝛿 − 𝑉𝑇 )

= 1.9 + 𝑗1.2 = 2.2472 ∠ 32.2756𝑜

𝑉𝑇2 1 =− = −0.6667, cos 𝛿 = 0 𝑋𝑠 1.5 = 𝑄𝑀𝐴𝑋,𝑝.𝑢. ∙ 𝑆𝐵 = −0.6667 × 500 MVA = 333.33 MVAr

𝑄𝑀𝐴𝑋,𝑝.𝑢. = −

c)

𝑄𝑀𝐴𝑋

𝛿 = 0 ⟹ 𝑄𝑀𝐴𝑋,𝑑𝑒𝑙𝑖𝑣.

14

𝑉𝑇 1 (𝐸 ∙ cos 𝛿 − 𝑉𝑇 ) = (2.25 − 1) = 0.8333 𝑋𝑠 1.5 = 𝑄𝑀𝐴𝑋,𝑝.𝑢. ∙ 𝑆𝐵 = 0.8333 × 500 MVA = 416.67 MVAr

𝑄𝑀𝐴𝑋,𝑑𝑒𝑙𝑖𝑣.𝑝.𝑢. = d)

MW

𝑄𝑀𝐴𝑋,𝑑𝑒𝑙𝑖𝑣.

E=2.25 p.u

0.8 lagging 500

e

a b

MVA limit

f 400

Prime mover limit

g

h

E=V=1 p.u

c Stability limit

d

Excitation limit

j 500

MVAr

15

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

4 500 ×10 = 2000 MWs = 2000 MJ Initial stored energy =× 6

Additional energy added to spinning mass = 400 × 10 × 0.4 = 160 MJ 6

2160 × 3000 = 3117.9 r.p.m or 52 Hz 2000

Increase in speed = 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 fullload 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.

4%

3% α

x

100- x 60 MW

60 MW Assume base MVA as 100 MVA: From the figure

α

α

= 0.03 = 0.04 and x 60 (100 − x ) 60 0.04 x= 3 − 0.03x 0.07 x = 3 x = 42.9 MW 100 - x = 57.1 MW 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.

3% 4%

α’

α f

50 MW

50 MW

If the operating frequency is f p.u. No load speed of 3% generator = f + 0.03 ×

50 =f + 0.025 p.u. 60

The no load speed of 4% machine before change in its no load speed is equal to

f + 0.015 p.u.

New no load speed of 4% generator = f + 0.04 ×

50 =f + 0.0333 p.u. 60

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

α1 = 0.05 100 200 0.025 ∴α1 =

For Generator B

=

α2 = 0.08 227.5 350

0.052 ∴α 2 =

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

α2 8% 5% α1 50 Hz

200 MW

100 MW

227.5 MW

350 MW

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 = = ∆f KA KA

300 = ∆f KB

Therefore, K B = 1.5K A As f A= f −

300 300 and f B= f + KA KB

fB − f A =

300 300 + K A KB

1 1 50 = − 49 300 + 1.5 K KA A 500 1= KA K A = 500 MW/Hz

= K B 1.5 = K A 750 MW/Hz

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 frequencybias-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 0.01 ∆f = 200 + 2000 × ∆f 0.1 0.1 100 + 300∆f= 200 + 200∆f 100 + 3000 × ∆f = 1Hz 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, ∆PA =1000 MW and ∆PB =-1000 MW f A= f +

1000 KA

= 50 + 50.5

f B= f −

1000 KA

and

K A = 2000 MW/Hz

b) If change of flow in line is ∆P For system A:

−∆P = ∆f KA

For system B:

−600 + ∆P = ∆f KB

Therefore:

1000 KB

= 50 − 49

1000 KB

K B = 1000 MW/Hz

−∆P −600 + ∆P = KA KB −∆P −600 + ∆P = 2000 1000 ∆P = 400 MW

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 + j 0.01 × (0.8 − j 0.39) (0.014 + j 0.04) + (0.03 + j 0.01)

= 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 + j 0.04 × (0.8 − j 0.39) (0.014 + j 0.04) + (0.03 + j 0.01)

= 0.5650 - j0.0388 Power through the line 5.2

= 0.5650 + j0.0388 p.u = 56.5 + j3.89 MVA

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 When = ∆V

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

RP + XQ = 0 V

RP 5 × 25 − = − = −6.25 MVAr Q= X 20

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 0.15/t2

1 p.u I=

1 / t −1 0.15 / t 2

1/t

I 1 p.u

Therefore VAr flow =

1 / t −1 0.15 / t 2

When t=0.85 VAr flow =

1/ 0.85 − 1 0.15 / 0.852

= 0.85 p.u = 425 MVAr

When t=1.1 VAr flow = 5.5

1 / 1.1 − 1 0.15 / 1.12

= −0.733 p.u = -367 MVAr

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 ∆V=

90Q = 7.51× 103 76.21× 103

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 Q 25 × 16.67 × 106 + 90 × L − 6.36 × 106 3 = 7.51× 103 ∆V= 3 76.21× 10

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 =

= I BASE = Z

332 = 10.89Ω 100

100 ×106 = 1749.5 A 3 × 33 ×103

5 + j 20 = 0.46 + j1.836 10.89

Fault current (ignoring resistance)

1 I FAULT = ×1749.5 =0.953 kA 1.836 ∂Q = 953 × 3 = 1.65 MVAr / kV ∂V Volt drop due to transport of real power

∆V = PR =

25 5 × × 33= 3.78 kV 100 10.89

Q = 3.78 ×1.65 = 6.25 MVAr 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 ts2VsVr − Vr2 = ( PR + QX )ts2

[0.14 ×1 + 0.38 × 0.48] ts2

2 0.9ts2 − 0.9 =

= 0.3224ts2

ts2 = 1.4 ts = 1.184 t= r

5.8

1 = 0.844 ts

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 BC AB DA TA

Rated voltage kV 0.415 11 33 33/11

Rating MVA

TB

11/0.415

2.5

10

Nominal tap ratio

Impedance (Ω)

0.0217+j0.00909 1.475+j2.75 1.475+j2.75 30.69/11 1.09+j9.8 Referred to 33 kV 10.45/0.415 0.24+j1.95 Referred to 11 kV

Impedance p.u.

10 MVABASE

Impedance (Ω)

0.415

Nominal tap ratio

Rated voltage kV

BC

Rating MVA

Line or transformer

Answer

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

TB

33/11

10

11/0.415 2.5

30.69/11

1.09+j9.8

ZBASE=108.9Ω

Referred to 33 kV

0.01+j0.09

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

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

ZBASE=12.1Ω 0.02+j0.16

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 P 50Q = 8.07 V 3V P P 275 × 103 − 8.07 = 158771.3 − 8.07 V= V V 3 ∆V=

V 2 − 158771.3V + 8.07 P = 0

V=

158771.3 ± 2.52 × 1010 − 32.28 P 2

Value closer to 275 kV was selected, 158771.3 + 2.52 × 1010 − 32.28 P VL = 3 2

275

270

Line Voltage (kV)

265

260

255

250

245

0

50

100

150 Power (MW)

200

250

300

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 + XQ ∆V = V 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

= ∆V

RP + XQ 0.156 × 50 × 30 + 0.4125 × 50 ×14.52 = = 7 kV V 132 3

Sending end voltage = 132 + 3 × 7 = 144.1 kV (b) = ∆δ

tan δ =

XP − RQ 0.4125 × 50 × 30 − 0.156 × 50 ×14.52 3 = = ×10 6633 V 132 3

6633 0.0797 = 144.1×103 3

δ = 4.56 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).

RP XQ + 2 VS t 2 t V= − R t VR

Figure 5.29 Circuits for Problem 5.11

Answer = VS

400 = 230.94 kV 3

P per phase = 333.33 MW Q per phase = (133.3-Qcompensator) MW If VR needs to be maintained at 10%

230.94 × 0.9 V= = 207.85 kV R 6 230.94 ×103 7 × 333 ×10 + 70(133.33 − Qcompensator ) = ×10 − 207.85 0.9 207.85 ×103 × 0.92 3

Compensator reactive power per phase, Qcompensator = 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 = 0.8325V 2 Taking the voltage drop from generator terminal to load bus; from approximate equation: 2 XQ 0.143 × (1.314 − 0.8325V ) 1 − V= = V V

0.8809V 2 − V + 0.1879 = 0 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 Assuming that receiving end voltage is V Q supplied by the line = 0.8325V 2 Taking the voltage drop from generator terminal to load bus; from approximate equation: 2 XQ 0.083 × (1.314 − 0.8325V ) 1 − V= = V V

0.9309V 2 − V + 0.1091 = 0

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.

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. Assuming that receiving end voltage is V Q supplied by the line = 0.259V 2 Q supplied by the shunt compensator =

120V 2 = 0.24V 2 500

From approximate equation: 2 2 XQ 0.266 × (1.314 − 0.259V − 0.24V ) 1 − V= = V V

→

(A)

0.8672V 2 − V + 0.3495 = 0 There is no feasible answer. So 345 kV design will not be satisfactory. Note that from equation (A) (in the form aV 2 − bV + c = 0: If shunt compensation becomes zero, still 4ac>b2 so no solution If series compensation becomes zero, still 4ac>b2 so no solution

Assume voltage = 500 kV Select Vb=500 kV and Sb = 500 MVA Zb = 500 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 = 0.127 p.u. Susceptance of the line Y = 2.27 x 10-6 x 480 = 0.00109 S = 0.00109 x 500 p.u. = 0.545 p.u.

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. Assuming that receiving end voltage is V Q supplied by the line = 0.545V 2 Q supplied by the shunt compensator =

120V 2 = 0.24V 2 500

From approximate equation:

1 − V=

2 2 XQ 0.127 × (1.314 − 0.545V − 0.24V ) = V V

→

(A)

0.9V 2 − V + 0.167 = 0

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

42 + 0.532 Approximate line current = 1

0.5

= 4.04 p.u.

Receiving end voltage:

= VS AVR + BI R VR= (1 − j 0.127 × 4.04∠ − 18.19 ) / 0.9655 = 0.87 − j 0.505 =1.006∠ − 30.13 Evan though voltage is within the required limit, angle is not. So not a satisfactory design.

Assume voltage = 765 kV Select Vb=765 kV and Sb = 500 MVA

Zb = 1170.45 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 = 0.054 p.u. Susceptance of the line Y = 0.00109 x 1170.45 p.u. = 1.276 p.u. 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.

Assuming that receiving end voltage is V Q supplied by the line = 1.276V 2 Q supplied by the shunt compensator =

120V 2 = 0.24V 2 500

From approximate equation: 2 2 XQ 0.054 × (1.314 − 1.276V − 0.24V ) 1 − V= = V V

→

(A)

0.9181V 2 − V + 0.071 = 0

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

42 + 0.2022 Approximate line current = 1 Receiving end voltage:

0.5

= 4.005 p.u.

= VS AVR + BI R VR= (1 − j 0.054 × 4.005∠ − 18.19 ) / 0.9655 = 0.966 − j 0.213 =0.989∠ − 12.4 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 V= tVS − XQt 2 R t 0.3 × 0.15t 2 1 =− 0.045t 2 − t + 1 =0

Therefore 1 ± 1 − 4 × 0.045 1 ± 0.9055 = 2 × 0.045 2 × 0.045 t = 1.05

= 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

1 ∆V1 = − 1 =−0.13 p.u. 1.15 Voltage drop across the reactance of 1:0.85 transformer

∆V= 2

1 −= 1 0.176 p.u. 0.85

Reactive power absorbed =

∆V12 + ∆V22 = 0.266 p.u = 31.9 MVAr X

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 PL − jQL 600 − j 210 = ×103= 0.866 − j 0.303 kA * VL 3 × 400 I = 0.917 kA

(i) I=

Active losses = 3I 2 R =× 3 0.917 2 × 2.7 = 6.82 MW Reactive losses = 3I 2 X = 3 × 0.917 2 × 30.4 = 76.7 MVAr (ii) Busbar 1 voltage =

400 ×103 + (0.866 − j 0.303) ×103 × (2.7 + j 30.4) 3 = 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)

G1 Figure 6.15 System for example 6.2 (i) (ii)

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

Answer 1 − 1 0.4 j 0.4 j −2.5 j 2.5 j (i) Ybus = = 1 2.5 j −2.5 j 1 − 0.4 j 0.4 j (ii) Assume V2(0) =1.0 p.u

P2 − jQ2 V (0)* − Y21V1 2 1 −1 + j 0.5 − j 0.4 − 0.2 + 1 = − j 2.5 ×1 = − j 2.5 1

V2(1) =

1 Y22

= 0.8 − j 0.4 = 0.894∠ − 26.57

1 P2 − jQ2 − Y21V1 (1)* Y22 V2 1 −1 + j 0.5 = − j 2.5 ×1 − j 2.5 0.894∠26.57 = 0.6 − j 0.3 p.u.

V2(2) =

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.

Figure 6.16 System for Problem 6.3

Answer

Ybus

j2 0 − j 4 j 2 j 2 − j 3 j1 0 = 0 j1 − j 2 j1 j1 − j 3 0 j2

Assume V2(0) =1.0 p.u, V3(0) =1.0 p.u and V4(0) =1.0 p.u

1 P2 − jQ2 − Y21V1 − Y23 V3(0) (0)* Y22 V2 1 −0.1 = − j 2 ×1.05 − j1×1 − j 3 1 = 1.0333 − j 0.0333 V2(1) =

P3 − jQ3 (1) (0) V (0)* − Y32 V2 − Y34 V4 3 1 0.5 − j 0.2 = − j1× (1.0333 − j 0.0333) − j1× 1 − j2 1 = 1.1167 + j 0.2333 V3(1) =

1 Y33

1 P4 − jQ4 − Y41V1 − Y43 V3(1) (0)* Y44 V4 1 −0.4 + j 0.05 = − j 2 ×1.05 − j1× (1.1167 + j 0.2333) − j3 1 = 1.0556 − j 0.0556 V4(1) =

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.

Figure 6.17 System for Problem 6.4 Determine the flow of power in line AC using two iterations of voltages at each bus.

Answer Assuming Sbase = 1 GW

4002 Therefore Z base = = 160 Ω 1000

Ybus

j 40 j16 j 40 − j 96 j 40 − j 66.67 j 26.67 0 = j16 j 26.67 − j 74.67 j 32 − j 72 0 j 32 j 40

1 PB − jQB − YBA VA − YBC VC(0) (0)* YBB VB 1 −1 = − j 40 ×1 − j 26.67 ×1 − j 66.67 1 = 1.0 − j 0.015

VB(1) =

1 PC − jQC − YCA VA − YCB VB(1) − YCD VD(0) (0)* YCC VC 1 −3 = − j16 ×1 − j 26.67 × (1.0 − j 0.015) − j 32 ×1 − j 74.67 1 = 1.0 − j 0.0455

VC(1) =

PB − jQB (1) V (1)* − YBA VA − YBC VC B −1 1 − j 40 ×1 − j 26.67 × (1.0 − j 0.0455) = − j 66.67 1.0 + j 0.015

VB(2) =

1 YBB

= 0.9998 − j 0.0332

1 PC − jQC − YCA VA − YCB VB(2) − YCD VD(1) (1)* YCC VC 1 −3 = − j16 ×1 − j 26.67 × (0.9998 − j 0.0332) − j 32 × (1.0 − j 0.0455) − j 74.67 1 = 0.9981 − j 0.0725 VC(2) =

Power flow in line AC = Re VA y AC ( VA − VC ) = Re 1× j16 × (1 − 0.9981 + j 0.0725 ) = 1.16 p.u. = 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.

Figure 6.18 System for Problem 6.5

Answer

j5 − j 25 j 20 = Ybus j 20 − j 60 j 40 j 5 j 40 − j 45 1 P2 − jQ2 − Y21V1 − Y23 V3(0) (0)* Y22 V2 1 −0.8 + j 0.6 = − j 20 ×1 − j 40 ×1 − j 60 1 = 0.99 − j 0.0133

V2(1) =

Q3 = − Im V3(0)* (Y31V1 + Y32 V2(1) + Y33 V3(0)

= − Im [1× ( j 5 ×1 + j 40 × (0.99 − j 0.0133) − j 45 × 1] = 0.4

1 P3 − jQ3 − Y31V1 − Y33 V2(1) (0)* Y33 V3 1 0.6 − j 0.4 = − j 5 ×1 − j 40 × (0.99 − j 0.0133) − j 45 1 = 1.0 + j 0.0015

V3(1) =

V3(1) = 1.0 + j 0.0015 = 1 V3(1) = V3(1) / abs (V3(1) ) = 1.0 + j 0.0015

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. G1

G2

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

5j −7.5 j 2.5 j = Ybus 2.5 j −6.5 j 4 j 5 j 4j −9 j Take busbar 2 as the slack bus

V1(1) =

1 P1 − jQ1 − Y12 V2 − Y13 V3(0) (0)* Y11 V1

1 0.9 − j 2.5 ×1 − j 5.0 ×1 − j 7.5 1 = 1.0 + j 0.12 =

Q3 = − Im V3(0)* (Y31V1(1) + Y32 V2 + Y33 V3(0) )

= − Im [1× ( j 5 × (1.0 + j 0.12) + j 4 ×1 − j 9 ×1) ] =0

1 P3 − jQ3 − Y31V1(1) − Y33 V2 (0)* Y33 V3 1 −1.5 + j 0 = − j 5 × (1.0 + j 0.12) − j 4 ×1 − j9 1 = 1.0 − j 0.1

V3(1) =

V3(1) =1.0 − j 0.1 =1.005 V3(1) = V3(1) / abs (V3(1) ) = 0.995 − j 0.0995 V1(2) = =

1 P1 − jQ1 − Y12 V2 − Y13 V3(1) (1)* Y11 V1 0.9 1 − j 2.5 ×1 − j 5.0 × (0.995 + j 0.0995) − j 7.5 1.0 − j 0.12

= 0.9825 + j 0.052

Q3 = − Im V3(1)* (Y31V1(2) + Y32 V2 + Y33 V3(1) )

= − Im [1× ( j 5 × (0.9825 + j 0.052) + j 4 × 1 − j 9 × (0.995 + j 0.0995)) ] = 0.1576 1 P3 − jQ3 − Y31V1(2) − Y33 V2 (1)* Y33 V3 1 −1.5 + j 0.1576 − j 5 × (0.9825 + j 0.052) − j 4 × 1 = − j 9 0.995 − j 0.0995

V3(2) =

= 0.9911 − j 0.1387

V3(1) =0.9911 − j 0.1387 = 1.0008 V3(1) = V3(1) / abs (V3(1) ) = 0.9903 − j 0.1386 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.

Figure 6.20 System for Problem 6.8 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

Ybus

j5 j 2.5 j10 − j17.5 j5 − j15 j10 0 = j 2.5 j10 − j17.5 j5 0 j5 − j15 j10

Assume V2(0) =1.01 p.u, V3(0) =1.01 p.u and V4(0) =1.01 p.u

P2 − jQ2 (0) V (0)* − Y21V1 − Y23 V3 2 1 −0.5 + j 0.2 = − j 5 ×1.01 − j10 ×1.01 − j15 1.01 = 0.9968 − j 0.033 V2(1) =

1 Y22

1 P3 − jQ3 − Y31V1 − Y32 V2(1) − Y34 V4(0) (0)* Y33 V3 1 −0.4 + j 0.1 = − j 2.5 ×1.01 − j10 × (0.9968 − j 0.033) − j 5 × 1.01 − j17.5 1.01 = 0.9968 − j 0.0415 V3(1) =

1 P4 − jQ4 − Y41V1 − Y43 V3(1) (0)* Y44 V4 1 −0.2 = − j10 ×1.01 − j 5 × (0.9968 − j 0.0415) − j15 1.01 = 1.0056 − j 0.027 V4(1) =

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

0.2

0.4

0.2

D

C

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 0.1

0.1

0.1

0.1

0.1

0.1

Z eq = 0.0533 p.u.

0.2

Fault level =

50 = 937.5 MVA 0.0533

937.5 ×106 Fault current = = 49, 206 A 3 ×11×103 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 MVA 0.2 p.u.

100 MVA 0.1 p.u. 0.035+j0.14 Ω/km

200 MVA 0.1 p.u.

On 100 MVA base

Z= base

1322 = 174.24 Ω 100

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

= Z eq

0.2 + 0.1 100 + 0.032 + 0.1× = 0.232 p.u. 2 200

Fault level =

7.3

100 = 431 MVA 0.232

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

B

X

60 MVA 0.15

30 MVA 0.1

P

Select Sbase = 60 MVA Fault level at P should be limited to 240 / 60 = 4 p.u Therefore Z eq should be= 1= 0.25 p.u.

4

However from the circuit,

0.15 0.15 60 = + X + 0.1× // Z eq 30 2 2 0.075(0.075 + X ) = + 0.2 0.075 + 0.075 + X 0.0056 + 0.075 X = + 0.2 0.15 + X By equating Z eq ,

= Z eq

0.0056 + 0.075 X = + 0.2 0.25 0.15 + X

∴X = 0.076 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

Z= base

662 = 43.56 Ω 100

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

Generator

Positive

Negative

Zero

18/43.56=0.413

14.5/43.56=0.333

150/43.56 = 3.444

p.u.

p.u.

p.u.

Transformer

10.9/43.56=0.25 p.u.

Line

44/43.56=1.01 p.u.

Total

1.673 p.u.

From equation 7.6

If =

= I base

3 =0.447 p.u. 1.673 + 1.593 + 3.444

100 ×106 874.77 A = 3 × 66 ×103

Fault current = 0.447 x 874.77 = 391 A

1.593 p.u.

3.444 p.u.

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

I1 =

1 1.673 + [1.593 × 3.444 / (1.593 + 3.444) ]

= 0.362 p.u. From (7.9) and (7.10)

1.593 (1.593 + 3.444) = −0.114 p.u.

I0 = −0.362 ×

3.444 (1.593 + 3.444) = −0.248 p.u.

I2 = −0.362 ×

Therefore

IY = a 2 × 0.362 − a × 0.248 − 0.114 (−0.5 − j 0.866) × 0.362 − (−0.5 + j 0.866) × 0.248 − 0.114 = = −0.171 − j 0.528 p.u. 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 + j 0.6) + (0.3 + j 0.55) + (1 + j 0.78) + 3 × 0.66 = 0.649 - j 0.35

If =

= 0.737∠ − 28.3

= V f (0.649 - j 0.35) × 0.66 = 0.43 − j 0.23 p.u. 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

VB =

− j 3E A [ Z 2 − aZ 0 ] Z1 + Z 2 + Z 0

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:

To positive-sequence currents

Generator (p.u.) j0.2

Feeder (Ω/p.h.) 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):

I= I= I= 1 2 0

EA Z1 + Z 2 + Z 0

VB = a 2 E A − I 0 Z 0 − a 2 I1Z1 − aI 2 Z 2 = a 2 E A − (Z 0 − a 2 Z1 − aZ 2 )I1 (Z 0 − a 2 Z1 − aZ 2 )E A = a EA − Z1 + Z 2 + Z 0 2

=

(a 2 − a)Z 2 + (a 2 − 1)Z 0 EA Z1 + Z 2 + Z 0

(a 2 − a) = −0.5 − j

3 3 + 0.5 − j = −j 3 2 2

(a 2 − 1) =−0.5 − j

3 3 3 − 1 =− − j 2 2 2

3 3 3 a× j 3 = − −j = (a 2 − 1) −0.5 + j × j 3 = 2 2 2

VB =

− j 3E A [ Z 2 − aZ 0 ] Z1 + Z 2 + Z 0

Taking Sbase = 30 MVA, then on 6.6 kV side

Z= base

6.62 = 1.452 Ω 30 Positive

Negative

Zero

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 Total

0.4/1.452=0.275 p.u. j0.513 p.u.

j0.493 p.u.

3x0.275+j0.275+j0.06 =0.825+j0.335

From equation 7.6

If =

3 j 0.513 + j 0.493 + 0.825 + j 0.335

= 1.904∠ − 58.45

= I base

30 ×106 2624.3A = 3 × 6.6 ×103

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

75 500

150 800

250 1500

350 2000

Figure 7.30 System for Problem 7.7 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.

CB3

11 kV

X

M

X

M

X

M

CB1

X

Grid infeed 166.6 MVA

X 30 MVA 15%

132 kV

CB2

CB6

X

M

X A

CB7

5 MVA

On 100 MVA base Grid infeed = 1.666 p.u.

X s = 0.6 p.u. Transformer reactance = 0.15 x 100/30 = 0.5 p.u. For a fault at A,

X eq = 0.6 +

0.5 0.4 + =1.05 p.u. 2 2

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 =

1 (6 × 4 + 5) ×106 × = 761.05 A 2 3 ×11×103

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

Rated current through CB3 to CB6 =

6 ×106 = 315 A 3 ×11×103

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 X = X= X 0 . The initial fault contribution of the 1 2 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 X s = 1/40 = 0.025 p.u.

Positive 132 kV busbar

Zero No contribution as

0.025 p.u.

0.025 p.u.

132 kV side is delta connected

Each

0.1x100/15

transformer

=0.667 p.u. 0.025+0.667/3 =

Total

Negative

0.247 p.u.

0.667 p.u.

0.667 p.u. 0.667/3 = 0.222

0.247 p.u.

p.u.

From equation 7.6

3 0.247 + 0.247 + 0.222 = 4.19 p.u.

If =

= I base

100 ×106 5248.64 A = 3 ×11×103

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

′′ X= ′′ Generators G1 and G2: X = 1 2

j 0.1 p.u.; 11 kV

Transformers T1 and T2: X = X= X= 1 2 0

j 0.1 p.u.; 11/275 kV

(Earthed star-delta) Line: Z= Z= j 0.05 p.u., Z 0 = j 0.1 p.u.; 275 kV 1 2 (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) I base =

100 ×106 = 209.94 A 3 × 275 ×103 0.1

0.1

= Z eq

0.1

0.1

0.05

0.2 × 0.25 = 0.111 p.u. 0.2 + 0.25

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 + 0.066 = 10.41 p.u.

If =

Fault current = I f = 10.41× 209.94 =2.18 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 + 0.2 = 4.29 p.u.

If =

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

X 1 = 0.2

X 2 = 0.16

X 0 = 0.14

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 X= X= X0 . 1 2 Using a base of 500 MVA calculate the base current and impedance of each voltage level. Calculate the fault current in amperes for: (a) (b) (c) (d)

a 275kV busbar three-line fault; a 275 kV single-line-to earth fault on the busbar; a 20 kV three-line fault on the generator terminals; a 20 kV single-line-to-earth fault on the generator terminals.

Answer On 500 MVA, 275 kV base

= I base

500 ×106 1049.7 A = 3 × 275 ×103

On 500 MVA, 20 kV base

= I base

Z base =

500 ×106 14.43kA = 3 × 20 ×103 202 = 0.8 Ω 500

Generator MVA = 400/0.8 = 500 MVA So on 500 MVA, for the generator: X 1 = 0.2 , X 2 = 0.16 , X 0 = 0.14 For transformer: X = X= X 0 = 0.15x500/550 = 0.136 p.u. 1 2 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 + 0.136 2.98 ×1049.7 = 3.124 kA =

If =

(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 + 0.136

= If

(c) Three phase fault at 20 kV barbar

1 = 5 p.u. 0.2 5 ×14.43 = 72.17 kA =

= If

(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.

If =

3 j 0.2 + j 0.16 + 3.75 + j 0.14

I f = 0.79 p.u.=11.44 kA

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:

j1.1

j1.3

j1.0

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

1

3

j0.1

0.8+j0.3

1.53 -31.60

2

1 p.u

Using Equation 2.8 and working from the infinite busbar voltage, the voltage at point 3 is given by 2 2 QX PX VA = V + + V V

0.3 × 0.1 2 0.8 × 0.1 2 =+ 1 + 1 1 =

1.032 + 0.082 = 1.033p.u.

−1 0.08 = = δ tan 4.44 1.03

i.e. VA 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

P2 + Q2 0.82 + 0.32 = I R2 X = X 0.1 2 12 V = 0.073p.u. The actual load taken by 3 (if represented by an impedance) is given by

VA2 1.0332 = = 0.594 + j 0.365p.u. Z 1.53∠ − 31.60 The total load supplied by link from generator to 3

=( 0.8 + 0.594 ) + j ( 0.3 + 0.073 + 0.365 ) = 1.394 + j 0.738 p.u. Internal voltage of generator E1

2 2 0.738 × 0.71 1.394 × 0.71 = 1.033 + + 1.033 1.033

= 1.542 + 0.9582 = 1.814 −1 0.958 = = δ tan 31.88 1.54

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.

j0.45

j0.4

(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

j0.4

j0.45 j0.2

j0.2 Fault

From delta to star transformation

j0.1

j0.1

j0.45 j0.05 Fault Again from star to delta transformation

Z12 Z11

Z22 Fault

Z12 = j0.55+j0.1+j 0.55*0.1/0.05 = j1.75 p.u Maximum power transfer = 1.05 x 1.0 / 1.75 = 0.6 p.u. (c) Post-fault reactance = j0.45+j0.4 = j0.85 p.u. Maximum power transfer = 1.05 x 1.0 / 0.85 = 1.24 p.u. H on 100 MVA base = 3 x (60/0.9)/100 p.u = 2 p.u

M =

2 = 2.2 ×10−4 p.u 180 f

A time interval of 0.05 s will be used. Hence

(∆t ) 2 = 11.36 M

The initial operating angle, −1 1 = = δ 0 sin 38.1 1.62

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

∆P =1 − 0.6sin δ 0 = 0.63 p.u. The value of ∆P for the commencement of the period is 0.63/2 = 0.315 p.u

(∆t ) 2 ∆Pn −1 M

δn

∆t (s)

∆P (p.u.)

∆δ n = ∆δ n −1 +

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)

800 700

Angle (deg)

600 500 400 300 200 100 0 0

0.2

0.4

0.6 0.8 Time(s)

1

1.2

1.4

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. Rating 30 MVA; Inertia constant l0 kWs/kVA.

Generator:

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

= M

8.5

3.4 = 0.00038 p.u 180 f

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

1.05 1.03 1.09

1.025 1.017 1.045

1 1 1

0.95 0.97 0.93

0.9 0.94 0.885

0.85 0.92 0.86

0.8 0.98 0.84

0.75 0.87 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 2 2 QX PX From E = 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.

2.04

2.02

2

E(p.u)

1.98

1.96

1.94

1.92

1.9

1.88

1.86 0.75

8.6

0.8

0.85

0.9

0.95 V(p.u)

1

1.05

1.1

1.15

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

Vt =

(1 + 0.6 × 0.6 )

= 1.44 p.u.

2

+ (0.8 × 0.6) 2

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 V = − dV V X Since Q is a constant,

V 2 1.442 = = 3.45 p.u. Q 0.6

= X

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

(V − 0.8)

2

= 0.2 ( Q − 0.8 )

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 V = − dV V X

(A)

Since Q =5 (V − 0.8 ) + 0.8 2

dQ = 10(V − 0.8) dV Further,

QX V Q 1−V ∴ = V X 1= V +

Substituting in (A):

1 − V V 1 − 2V − = X X X 1 − 2V ∴X = 10(V − 0.8) 10(V − 0.8)=

5 (V − 0.8 ) + 0.8 2

V

=

10(1 − V )(V − 0.8) (1 − 2V )

3V 2 − 8V + 4 = 0 V = 0.67 p.u X = 0.25 p.u. Above solution is only true if

PX QX << V + V V P = 1 p.u.

PX 0.25 QX = = 0.373 and V + = 1 ; therefore the assumption is correct. V 0.67 V 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

1= (1 + 0.35Q ) + (0.8 × 0.35)2 2

Q = −0.114 p.u. 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: δ1

δ2

0 ∫ P dδ + δ∫ ( P − P sin δ )dδ = 0

δ0

0

2

1

P0 (δ 2 − δ 0 ) + P2 cos δ 2 − P2 cos δ1 = 0 0.8 × (143.10 − 26.10 ) × π /1800 + 1.32 cos143.10 − 1.32 cos δ1 = 0 cos δ1 =

1.634 − 1.056 1.32

δ1 = 64.030 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

1= (1 + 0.3Q ) + (0.9 × 0.3)2 2

Q = −0.123 p.u. = 62 MVAr (b) Therefore, taking infinite busbar voltage as the reference

E =(1 − 0.5 × 0.123) + j (0.9 × 0.5) = 0.938 + j 0.45 E = 1.04 p.u. (c) From equal area criteria: δ1

δ2

0 ∫ P dδ + δ∫ ( P − P sin δ )dδ = 0

δ0

0

1

1

P0 (δ 2 − δ 0 ) + P1 cos δ 2 − P1 cos δ1 = 0

= P1

1 = 2 p.u. 0.2 + 0.3

δ0 = sin-1(0.9/2) = 26.74o δ2 = 180o-26.74o = 153.26o

0.9 × (153.260 − 26.740 ) × π /1800 + 2 cos153.260 − 2 cos δ1 = 0

δ1 = 84.20

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):

Vd = (a) When

α = 0° and γ = 15°:

Vd = (b) When

1.35 ×120 cos 30 + = cos 45 127.4 kV 2

α = 60° and γ = 15°:

Vd = 9.2

1.35 ×120 cos 0 + cos15 159.2 kV = 2

α = 30° and γ = 15°:

Vd = (c) When

1.35VL cos α + cos (α + γ ) 2

1.35 ×120 cos 60 + = cos 75 61.5 kV 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):

Vd = = Vd

1.35VL cos α + cos (α + γ ) 2 1.35 × VL cos 30 + cos = 45 100 kV 2

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)

= Id

π V0 6X

.[ cos α − cos(α + γ ) ]

π ×1.35 × 94.2 ×103

I d 800 = =

6X

cos 30 − cos 45

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°

δ 0 = 10°; Allow 5° margin on δ 0 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)

V cos α − V0i cos δ ∴ I d =0 r RL + Rcr − Rci

V0 r = V0i = 1.35 × VL = 1.35 ×120 = 162 kV R= R= cr ci

3 X 3 ×15 = = 14.3 Ω

π

π

When α = 10° and δ = 15°

Id

162 ×103 = [cos10 − cos15 ] 611.7 A ( 5 + 14.3 − 14.3)

At maximum transmittable power α = 0°

= Id 9.5

162 ×103 = [cos 0 − cos15 ] 1104 A ( 5 + 14.3 − 14.3)

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):

V cos α − V0i cos β ∴ I d =0 r RL + Rcr + Rci

β = γ + δ = 30 Id 9.6

162 ×103 = [cos15 − cos 30 ] 481.7 A ( 5 + 14.3 + 14.3)

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 constantδ 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):

V cos α − V0i cos δ ∴ I d =0 r RL + Rcr − Rci

V0 r = 1.35 × VL = 1.35 ×100 = 135 kV V0i = 1.35 × VL = 1.35 × 90 = 121.5 kV R= R= cr ci

3 X 3 ×15 = = 14.3 Ω

π

π

When α = 10° and δ = 15°

= Id

[135cos10 − 121.5cos15 ] 1.55 kA =

(10 + 14.3 − 14.3)

Vd = 1.35 × VL cos α = 1.35 ×100 cos10 = 133 kV P = 1.55 x 133 = 206 MW For I d = 800 A

800 =

135 ×103 cos10 − V0i cos15

(10 + 14.3 − 14.3)

V0i = 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:

Va.c.VVSC sin [ψ − θ ] X 120 ×125 × sin 5 = = 261.5 MW 5 P=

Q=

Va.c. {VVSC cos [ψ − θ ] − Va.c.}

X 120 × (125cos 5 − 120) = = 108.6 MVar 5 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:

Vs

(a) Sbase = 150 MVA

Z= base

332 = 7.26 Ω 150

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 VS = 1.2 pu and

ψ −θ = 0

VSVR 1.2 ×1 = = 0.3 pu X 4

= Pmax

Maximum power is 0.3 x 150 = 45 MW

(b) At 10 MW:

P=

VSVR sin δ X

10 (1.2)(1.0) sin δ = 150 4 0.0666 = 0.3sin δ δ 12.83° =

(c)

Maximum reactive power generation occurs when VSC output is inphase with the infinite bus voltage and it is 1.2 pu

I=

(1.2 − 1) = − j 0.05 pu j4

Voltage at the 33 kV busbar:

V33 = 1.0 − j 0.05 × j1 = 1.05 pu

Qmax =1.05 × 0.05 = 0.0525 pu = 7.875 MVar

(d) Maximum

reactive power absorption occurs when VSC output is inphase with

the infinite bus voltage and it is 0.7 pu

= I

(0.7 − 1) = j 0.075 pu j4

Voltage at the 33 kV busbar:

V33 = 1.0 + j 0.075 × j1 = 0.925 pu

Qmax = 0.925 × 0.075 = 0.069 pu = 10.4 MVar

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 =

345 ×103 / 3 = 4.98 Ω 40 ×103

4.98 = 0.0132 H 120π

∴= L

Surge impedance

L == C

0.0132 = 726.9 Ω 25 ×10−9

1 1 Frequency = = = 8760 Hz 2π LC 2π 0.0132 × 25 ×10−9 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

== Z0

L = C

1×10−3 = 3.16 Ω 100 ×10−6

1 1 Frequency = = = 503 Hz −3 2π LC 2π 1× 10 × 100 × 10−6

Peak current = =

10.3

38 × 2 = 24 kA 3.16

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

= = Z0

L = C

1.72 ×10−3 = 38.8 Ω 1.14 ×10−6

l = c

1 1 Frequency = = = 374 Hz 2π LC 2π 1.72 ×10−3 × 9.6 ×1.14 ×10−6 × 9.6 Time for 2 cycles = 2/374 = 0.0053 s = α

1 R 1 5.37 × 9.6 + 6 = = 1742.74 2 L 2 1.72 ×10−3 × 9.6

1 − e −α t = 1 − e −1742.74×0.0053 ≈ 0.99

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

=× 2

10 ×103 × 2= 16.33 kV 3

Voltage after 2 cycles = 16.33 (1 − e −α t ) = 16.32 kV 10.4

A 132kV circuit breaker interrupts the fault current flowing into a symmetrical threephase-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 restrikingvoltage 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 = L ∴=

Surge impedance

(132 ×103 ) 2 = 6.97 Ω 2500 × 106

6.97 = 0.022 H 100π =

L = C

0.022 = 856 Ω 0.03 ×10−6

1 1 Frequency = = = 6.17 kHz 2π LC 2π 0.022 × 0.03 ×10−6

Maximum voltage across the CB =× 2

132 ×103 × 2= 215.5 kV 3

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 Z1 − Z 0 50 − 500 = = −0.818 Z1 + Z 0 50 + 500

Reflection coefficient α =

Reflected voltage = −0.818vi

Reflected current = 0.818ii Reflected energy = −0.8182 vi ii = −0.67 × 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: = Z0

0.188 ×10−3 = 21.68 Ω 0.4 ×10−6

For the transmission line: = Z1

0.94 ×10−3 = 354 Ω 0.0075 ×10−6

Voltage at the junction = 10.7

2 Z1 = 1.88 p.u. Z 0 + Z1

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,

1 R v vi exp − x = 2 Z0 1 7 500 exp − 100 = 248.3 kV = 2 500 From equation 10.22 R Power at x = vi ii exp − Z0

700 x = vi ii exp − = 0.25vi ii 500

Since input power is vi ii , power loss = 0.75vi ii = ii

vi 500 = = 1 kA Z 0 500

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 / /500 142.86 2vi = vi 2= 0.44vi 500 + 200 / /500 500 + 142.86

Voltage transmitted to the 50 Ω tower side = 50 / /500 45.45 2vi = vi 2= 0.17vi 500 + 50 / /500 500 + 45.45 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 = −0.818 50 + 500

2 × 50 = 0.182 50 + 500

Cable to line

500 − 50 = 0.818 50 + 500

2 × 500 = 1.818 50 + 500

Line to cable

50 − 500 = −0.818 50 + 500

2 × 50 = 0.182 50 + 500

Cable to line

500 − 50 = 0.818 50 + 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) v(t ) = Vˆ 1 − e −α t cos ω0t So peak value of the transient recovery voltage =

11×103 ×2 = 2× 17.96 kV 3

If line reactance is X, 11× 103 = 5000 3X ∴ X = 1.27Ω

Inductance = 0.004 H −6 Time to peak = π LC = π 0.004 × 0.007 ×10= 16.7 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: α1 = −0.818 and β1 = 0.182 Line to cable: α 2 = 0.818 and β 2 = 1.818

50 Ω

500 Ω A

33

1

α1

B

C

Time (µs)

133 166 200

D

β1 β1α 2

β1β 2

66 100

500 Ω

β1α 22

β1α

3 2

2 β1α 24 β1 β 2

β12 β 2α 2

β1β 2α 22 β1β 2α1

β1α 25

2 2 β1α 26 β1 β 2α 2

233

(a) Voltage at D shortly after the surge has reached D = 2 β1β 2 vi =× 2 0.182 ×1.818 × 500 = 330.9 kV

β1β 2

β1β 2α 22 β1β 2α1

(b) Voltage at D at a time 210 µs after the surge first reaches B = β1β 2 vi 2 + 2α 2 + 2α 22 2 × 0.182 ×1.818 × 500 1 + 0.818 + 0.8182 = = 823kV (c) Voltage at B at a time 210 µs after the surge first reaches B = β1vi 1 + α 2 + α 22 + α 23 + α 24 + α 25 + α 26 + β12 β 2 vi [1 + α 2 ]

= 0.182 × 500 1 + 0.818 + 0.8182 + 0.8183 + .... + 0.1822 ×1.181× 500 [1 + 0.818]

= 413kV

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

132 kV L1

T1

33 kV

T2

L2

11 kV L3

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: I base

100 ×106 = 5249 A 11×103 × 3

Ifault = 5249/3.33 = 1575 A, Irelay = 15.75 A If the relay inverse-time characteristic is given by t =

k and when I = 10 A, t = I

10 sec; k =100 Time of operation =

11.2

100 = 6.345 s 15.76

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) Operating times of relays at A, D and C Operating times at relays at B and E

7

9

11

15

20

3.1

1.95

1.37

0.97

0.78

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

Z LH = 7.3 Ω Z RH = 5.9 Ω

33 1 . 2.61 kA = 3 7.3 33 1 . 3.23 kA = I RH = 3 5.9

= I LH

A 1+j1

1.5+j2

E 0.5+j0.75

0.5+j0.75

B

3+j4 D

1.5+j2 C

Relay Position

Fault current (kA)

Relay Current (A)

Time of relay operation (s)

A B C D

5.84 3.23 3.23 2.61

7.3 10.77 10.77 8.7

3 1.95 1.45 2.15

Time of CB and relay operation (s) 3.3 2.25 1.75 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:

= I A 1500∠45 A ,

= I B 2500∠150 A , = IC 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

I A0 1 1 I = 1 1 a A1 3 I A2 1 a 2

1 IA a 2 . I B a IC

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

= t

11.5

0.14 × 0.6 = 2sec 80.02 − 1

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 = t =

0.14 × 0.05 = 0.14 sec 11.20.02 − 1

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

0.14 × TMS = 0.54 5.60.02 − 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 = I f =

800 ×106 = 6988 A , 7 kA 3 × 66 ×103

b) Using a 15 MVA base

15 + 0.1 = 0.11875 800 1 = If = 8.421 p.u. 0.11875 15 ×106 787 A = I base = 3 × 11× 103 I f = 6.63kA X=

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

= t

0.14 × 0.5 = 1.425sec 110.02 − 1

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:

= t

0.14 × TM = 1.825sec 5.7 0.02 − 1

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)

11.7

In practice differential protection (or the Bucholtz relay) would operate first for faults inside the transformer.

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 =

I × 33 11

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

=

I × 33 5 × × 3 11 800

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 =

I × 33 5 × × 3 11 800

=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

V VT ratio V (132000 / 220) = I / CT ratio I / (1000 /1) = 1.67 × Actual Impedance

=

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: • PMAX = 550 [MW] •

Ccoal = 1.26 [£ / MJ]

•

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

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

C= ( P ) H ( P ) × Ccoal

C ( P ) =158.76 + 11.214P + 0.003654P 2 [£ / h] The marginal cost is given by:

dC ( P ) = 11.214 + 0.007308P [£ / MWh] dP

a)

dC ( P ) dP

= 13

11.214 + 0.007308P = 13 P = 244.4 [MW] b)

dC ( P ) dP

= 22

11.214 + 0.007308P = 22 P → = 1, 475.9168 > PMAX P = 550 [MW]

12.2

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

dF1 = 0.003P1 + 0.7 dP1 dF2 = 0.004 P2 + 0.5 dP2 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: •

dF1 £ → £ 0.003P1 + 0.7 F 0.7P1 + 0.0015P12 = 1 ( P1 ) = MWh h dP1

•

dF2 £ → 0.004P2 + 0.5 F 0.5P2 + 0.002P22 £ = 2 ( P2 ) = MWh h dP2

•

D = 150 MW [ ]

•

Pmin = 20 [MW]

•

Pmax = 125 [MW]

P) min {F1 ( P1 ) + F2 (P2 )} P1 ,P2

Subject to:

P1 + P2 = D Pmin ≤ P1 ≤ Pmax Pmin ≤ P2 ≤ Pmax P1 , P2 ≥ 0 The Lagrange function of this optimization problem is given by:

= L F1 ( P1 ) + F2 ( P2 ) + λ[D − P1 − P2 ] dL dF1 (P1 ) = −= λ 0 dP1 dP1 dL dF2 (P2 ) = = − λ 0 dP2 dP2

Eq (1) Eq (2)

dL = D − P1 + P2 = 0 dλ ∴ P1 + P2 = D = 150

= P1 150 − P2

Eq (3)

Equating equations (1) and (2):

dF1 (P1 ) dF2 (P2 ) = dP1 dP2

0.003P1 + 0.7 = 0.004P2 + 0.5 0.004P2 − 0.003P1 = 0.2 Substituting for P1 from equation (3) into the previous expression:

0.004P2 − 0.003 × (150 − P2 ) = 0.2 P2 =

0.2 + 150 × 0.003 0.004 + 0.003

P2 = 92.86 [MW] P1 = 57.14 [MW] The total cost of generation is given by:

CT ( P= F1 ( P1 ) + F2 ( P2 ) 1 , P2 ) The saving is then given by:

Saving = CT ( P1 = 75, P2 = 75 ) − CT ( P1 = 57.14, P2 = 92.86 ) = 109.6875 − 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:

C1 ( P= 0.04P12 + 2P1 + 250 1) C2 ( P2= ) 0.02P22 + 3P2 + 450 C3 ( P3= ) 0.01P32 + 5P3 + 250 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: •

) £ → dC1 ( P1= 0.08 P1 + 2 £ C1 ( P1= ) 0.04 P12 + 2 P1 + 250 h MWh dP1

•

) £ → dC2 ( P2= 0.04 P2 + 3 £ C2 ( P2= ) 0.02 P22 + 3P2 + 450 h MWh dP2

•

dC3 ( P3 ) £ → C3 ( P3= = 0.02 P3 + 5 £ ) 0.01P32 + 5P3 + 250 h MWh dP3

•

D = 525 [ MW ]

dC ( P ) dP1

1 1 b) =

(1) (2)

dC2 ( P2 ) dC3 ( P3 ) = dP2 dP3

dC1 ( P1 ) dC2 ( P2 ) = 0.08 → P1 = + 2 0.04 P2 + 3 2 → P1 −= P2 25 dP1 dP2 dC1 ( P1 ) dC3 ( P3 ) = 0.08 → P1 = + 2 0.02 P3 + 5 4 → P1 −= P3 150 dP1 dP3

(3) P1 + P2 + P3 = 525

Rearranging the equations: (1) = P2 2P1 − 25

(2) = P3 4P1 − 150 (3) P1 + P2 + P3 = 525

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

P1 + 2P1 − 25 + 4P1 − 150 = 525 7P → = 700 1

P1 = 100 MW [ ]

P2 = 175 [MW] P3 = 250 [MW]

a)

dC1 (P1 ) = 0.08P1 + 2 dP1 dC1 (P1 ) = 10 £ MWh dP1

c) CTotal = C1 ( P1 ) + C2 ( P2 ) + C3 (P3 )

CTotal = 4,562.5 £ h 12.4

A power system is supplied from three generating units that have the following cost functions: Unit A: 13 + 1.3PA + 0.037PA2 [$/h] Unit B: 23 + 1.7PB + 0.061PB2 [$/h] Unit C: 19 + 1.87PC + 0.01PC2 [$/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: PAMAX = 110 MW PBMAX = 85 MW PCMAX = 266 MW

Answer Data given: •

$ → dCA (PA )= 0.074P + 1.3 $ CA ( PA = ) 0.037PA2 + 1.3PA + 13 A h MWh dPA $ → dCB (PB )= 0.122P + 1.7 $ CB ( PB )= 0.061PB2 + 1.7PB + 23 B h MWh dPB

$ → dCC ( PC ) = 0.02P + 1.87 $ CC ( PC ) = 0.01PC2 + 1.87PC + 19 C h MWh dPC

a) D = 380 [MW] Equations are given by:

dCA (PA ) dCB (PB ) dCC ( PC ) = = & PA + PB = + PC 380 dPA dPB dPC Then: (1)

dCA (PA ) dCB (PB ) = 0.074P → 1.3 0.122PB + 1.7 A += dPA dPB

0.074PA − 0.122PB = 0.4 (2)

dCA (PA ) dCC ( PC ) = 0.074P → = 0.02PC + 1.87 A + 1.3 dPA dPC 0.074PA − 0.02PC = 0.57

(3) PA + PB + PC = 380 Then, writing the equations in matrix form:

0 PA 0.4 0.074 −0.122 0.074 0.57 0 −0.02 PB = 1 1 1 PC 380 Solving this matrix:

PA = 77.60 MW [ ]

PB = 43.79 MW [ ]

PC = 258.61 [MW]

b)

dCA (PA ) dCB (PB ) dCC ( PC ) = = = 10.65 $ MWh dPA dPB dPC Then: (1)

dCA (PA ) = 0.074PA += 1.3 10.65 P → = 126.35[ MW] A dPA

(2)

dCB (PB ) = 0.122PB + 1.7 = 10.65 P → = 73.36[ MW] B dPB

(3)

dCC ( PC ) = 0.02PC + 1.87= 10.65 P → C= 439 [ MW] dPC

Pexp = PA + PB + PC − 380 Pexp = 258.71 [MW] c)

PAMax 110 [ MW ] P →= 110[ MW] (1) = A PBMax 85 [ MW ] P → = 73.36[ MW] (2) = B PCMax 266 [ MW ] P → = 266[ MW] (3) = C Pexp = PA + PB + PC − 380

Pexp = 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 Base Load Mid Merit Peak Load

Investment cost [£/kW/y] 250 80 30

Operating costs [£/MWh] 5 50 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 t1 , t 2 and t 3 : (1) = t1

IB − IM 250, 000 − 80, 000 = t → = 3, 777.78[ h] 1 OM − OB 50 − 5

(2) = t2

IM − IP 80, 000 − 30, 000 = t → = 1, 250[ h] 2 OP − OM 90 − 50

(3) = t3

IP IP ≈ t →= 3.75[ h] 3 VOLL − O P VOLL

Using these values in combination with the LDC: Baseload

K B = 7.5 +

24 − 7.5 × ( t1 − 8, 760 ) 900 − 8, 760

K B = 17.96 [GW] Then, the energy served by this type of generation technology is:

ESB = K B × t1 +

( K B + 7.5) × 2

(8, 760 − t1 )

ESB = 131.27 [TWh] Mid-merit

24 − 7.5 × ( t 2 − 8, 760 ) − K B 900 − 8, 760 K M = 5.31 [GW] K M = 7.5 +

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

= ESM

( t1 + t 2 ) × K

M

2 ESM = 13.34 [TWh]

Peaking

K P = 24 +

30 − 24 × ( t 3 − 900 ) − K B − K M 0 − 900

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

= ESP

( t 3 + 900 ) ×

2 ESP = 3.49 [TWh]

( K B + K M + K P − 24 ) +

( 900 + t 2 ) × 2

( 24 − K B − K M )

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

= ENS

( 30 − K B − K M − K P ) × t

2 ENS = 46.9 [MWh]

12.6

3

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 [ ]

•

A v = 96%

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

n n −k Pr ( K = k ) = × A kv × (1 − A v ) k

where: • • •

k is the number of available generators. n is the total amount of generators. A v is the availability of each generator. Available generators

k

Available generation [MW]

State probability

Cumulative probability*

5 4 3 2 1 0

600 480 360 240 120 0

0.815373 0.169869 0.014156 0.000590 0.000012 0.000000

0.815373 0.985242 0.999398 0.999988 1.000000 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

Duration [h]

LOLP

i

Demand level [MW]

1 2 3 4

470 315 240 115

180 2,320 2,900 3,360

0.014758 0.000602 0.000012 0.000000

The LOLP at peak load is:

LOLP470 = 1.48% Finally, the LOLE is given by:

LOLE =

4

∑Duration × LOLP i

i =1

LOLE = 4.0894 [h]

i

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: •

£ → dC A ( PA ) = C A ( PA ) = 0.025 PA2 + 6 PA 0.05 PA + 6 £ h MWh dPA

•

£ → dCB ( PB ) = CB ( PB ) = 0.11PB2 + 10 PB 0.22 PB + 10 £ h MWh dPB

•

DA = 600 [ MW ]

DB = 1, 000 [ MW ]

The total generation cost is given by:

CT ( PA , PB ) =CA ( PA ) + CB ( PB ) =6PA + 0.025PA2 + 10PB + 0.11PB2 Coriginal ( PA , PB ) = 132, 600 £ h T a) P= D A + FAB A

P= D B − FAB B Then:

∆C= Coriginal − CTnew T T ∆= CT 132.600 − CT (D A + FAB , D B − FAB ) ∆= CT 132.600 − CT (800, 800)

£ 33, 400 ∆CT = h b) ∆= CT 132.600 − CT (1000, 600)

56, 000 £ ∆CT = h 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

MC1 = 3 + 0.02 P1

[£ / MWh]

MC 2 = 4 + 0.04 P2

[£ / MWh]

Southern Region

MC 3 = 3.6 + 0.025P3

[£ / MWh]

MC 4 = 4.2 + 0.025P4

[£ / MWh]

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: •

DN = 100 MW [ ]

•

MC1= 3 + 0.02 P1 £ MWh

•

MC2= 4 + 0.04 P2 £ MWh

•

DS = 420 MW [ ]

•

MC= 3.6 + 0.025 P3 £ 3 MWh

•

MC= 4.2 + 0.025 P4 £ 4 MWh

a) North

MC1 = MC2 0.02P → 1 − 0.04P2 = 1

Eq (1)

P1 + P2 = 100 P → 2 = 100 − P1

Eq (2)

Substituting equation (2) in (1):

0.02P1 − 0.04 × (100 − P1 ) = 1 P1 =

1+ 4 0.06

P1 = 83.33 MW [ ] P2 = 16.67 [MW] MC N = 4.67 £ MWh South

MC3 = MC4 0.025P → 3 − 0.025P4 = 0.6 ∴ P3 − P4 = 24

Eq (3)

P3 + P4 = 420 P → 4 = 420 − P3

Eq (4)

Substituting equation (4) in (3):

P3 − 420 + P3 = 24 P3 =

420 + 24 2

P3 = 222 MW [ ] P4 = 198 [MW] MC N = 9.15 £ MWh b) l = 450 [km]

c= T

kl 37 × 450 = ≈ 1.9 £ MWh τ0 8760

Then:

π T =MC N − MCS = 3 + 0.02P1 − 3.6 − 0.025P3

∴ π= 0.02P1 − 0.025P3 − 0.6 T

Eq (5)

MC = MC2 P → = 50 1 1 − 2P 2

Eq (6)

P1 + P2 = 100 + FNS P → 2 = 100 + FNS − P1

Eq (7)

Substituting equation (7) in (6):

P1 − 200 − 2FNS + 2P1 = 50 P → 1=

250 + 2FNS 3

Then:

MC = MC4 P → = P4 24 3 3 −

Eq (8)

P3 + P4 + FNS = 420 P → 4 = 420 − FNS − P3

Eq (9)

Substituting equation (9) in (8):

P3 − 420 + FNS + P= 24 P → = 3 3

444 − FNS 2

Substituting P1 and P3 in equation (5):

250 + 2FNS 444 − FNS π T ( FNS ) =0.02 × − 0.025 × − 0.6 3 2 The optimal transmission capacity is such that supply and demand for transmission are in equilibrium. Therefore:

π T = cT 250 + 2FNS 444 − FNS 0.02 × − 0.025 × 1.9 − 0.6 = 3 2

FNS = 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:

C ( PA ) = 70 + PA + 0.001PA2 [£/h] for area A, and

C ( PB ) = 50 + 2 PB + 0.002 PB2

[£/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

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: •

D W = 3,500 MW [ ]

•

DS = 2, 000 [MW]

•

K T = 900 [MW]

•

A) £ → dCA (P CA ( = PA ) 0.001PA2 + PA + 70 0.002PA + 1 £ = h MWh dP A

•

B) £ → dCB (P= CB ( P= 0.002PB2 + 2PB + 50 0.004PB + 2 £ B) h MWh dP B

a) (1) MC A = MCB 0.002P → → A − 0.004PB = 1 P A − 2PB = 500 (2) PA = FAB (3) PB + FAB =D P → D − FAB B = Substituting equations (2) and (3) in (1):

FAB − 2 × ( D − FAB ) = 500 FAB =

500 + 2D 3

Then:

PA PB FAB MC

Winter

Summer

Units

2,500

1,500

MW

1,000

500

MW

2,500 6.00

1,500 4.00

MW £/MWh

Winter

Summer

Units

CA CB

8,820

3,820

£/h

4,050

1,550

£/h

CTotal

12,870

5,370

£/h

Then:

C W = 12,870 × 2,500 = 32,175, 000[ £] CS = 6, 450 × 6, 260 = 33, 616, 200[ £] Finally: uncostrained CTOTAL = C W + CS = 65, 791, 200 [£]

b) Winter

PA PB FAB MCA MCB

Summer

Units

900

900

MW

2,600

1,100

MW

900

900

MW

2.80

2.80

£/MWh

12.40

6.40

£/MWh

c) t W = 2,500 h [ ]

t S = 6, 260 [h] Winter

Summer

Units

CA CB

1,780

1,780

£/h

18,770

4,670

£/h

CTotal

20,550

6,450

£/h

Then:

C W = 20,550 × 2,500 = 51,375, 000[ £] CS = 6, 450 × 6, 260 = 40,377, 000[ £] Finally:

C900 TOTAL = C W + CS = 91, 752, 000 [£] d) K T = 1,800 MW [ ] Annuitized investment cost of 1, 000, 000[ £ / yr] Winter

PA PB FAB MCA MCB

Summer

Units

1,800

1,500

MW

1,700

500

MW

1,800

1,500

MW

4.60

4.00

£/MWh

8.80

4.00

£/MWh

Summer

Units

Winter

CA CB

5,110

3,820

£/h

9,230

1,550

£/h

CTotal

14,340

5,370

£/h

Then:

C W = 14,340 × 2,500 = 35,850, 000[ £] CS = 5,370 × 6, 260 = 33, 616, 200[ £]

C1800 TOTAL = 69, 466, 200 [£] Finally: 1800 ∆CTOTAL = C900 TOTAL − C TOTAL

∆CTOTAL = 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)

•

P1B = 12 kW [ ]

•

P1S = 8 kW [ ]

•

jnB = 0.12

•

jnS = 0.18

We know that:

Pn = jn × n × P1 Then:

PnT = PnB + PnS = 0.12 × 45 ×12 + 0.18 ×115 × 8 PnT = 230.4 [kW]