TUTORIAL - 1



Shantilal Shah Engineering College, Bhavnagar

Electrical Engineering Department

B.E. Semester VI (Electrical)

Subject: Electrical Power System – II (2160908)

Assignments

|SR.NO. |TITLE |DATE |SIGN |REMARK |

| | | | | |

| |Trans mission Line | | | |

| |Power Circle Diagrams | | | |

| |Symmetrical Fault Analysis | | | |

| |Symmetrical Components. | | | |

| |Unsymmetrical Fault Analysis | | | |

| |Corona | | | |

Assignment 1: Trans mission Line

1. A three-phase voltage of 11 kV is applied to a line having R = 10 ohm and X = 12 ohm per conductor. At the end of the line is a balanced load of P kW at a leading power factor. At what value of P is the voltage regulation zero when the power factor of the load is (a) 0.707, (b) 0.85?

2. A three-phase 50 Hz transmission line is 400 km long. The voltage at the sending end is 220kV. The line parameters are r = 0.125 ohm/km, x = 0.4 ohm/km and y = 2.8 x 10-6 S/km. Determine sending end current and receiving end voltage when there is no load on the line.

Assignment 2: POWER CIRCLE DIAGRAMS

|1 |The generalized circuit constants fo a transmission line are as follows: |

| | |

| |A=D= 0.895(1.4(, B=182.5(78.6( |

| | |

| |If the line supplies a load of 50 MW at 0.9 p.f. and 215 kV, find the sending end voltage and hence the regulation of line. |

| |For a load of 80 MW at 0.9 p.f. lag, 215 kV, determine the reactive power supplied by the line and by the synchronous capacitor|

| |if the sending end voltage is 236 kV. Also determine the power factor of the line at the receiving end. |

| |Determine the maximum power that can be transmitted if the sending and receiving end voltages are as in (b). |

| | |

| |Ans.[(a) sending end voltage=251 kV, regulation=16.74% (b) Reactive power supplied |

| | |

| |by synchronous capacitor=12.5 MVAr (c) Max. power=227.5 MW |

|2 |3 phase transmission line, 160 km long, transmits a load of 90 MW at 0.8 p.f. lagging. The line voltage at the receiving end is|

| |230 kV. The constants of line are as follows. |

| | |

| |A=D= 0.9785(0.3(, B=85.2(77.44(, C=0.000503(90.1( |

| | |

| |Construct the receiving end and sending end circle diagrams for the transmission line and calculate |

| |sending end voltage, current, power factor, regulation and efficiency of the transmission line |

| |load in kW at 0.8 power factor lagging that could be carried at 8% regulation. |

| | |

| |Ans.[(a)-(i) sending end voltage = 256 kV (ii) sending end current 244 A (iii)sending |

| |end power factor 0.89 (iv) regulation 11.3% (v) 93.8 %. (b) 75000 kW |

Assignment 3: Symmetrical Fault Analysis

|1 |For the radial network shown in figure, a 3-phase fault occurs at F. Determine the fault current and the line voltage at 11 kV |

| |bus under fault conditions. |

| |G1 : 10 MVA, 11 kV, X = 15% |

| |G2 : 10 MVA, 11 kV, X = 12.5% |

| |T11 : 10 MVA, 11/33 kV, X = 10% |

| |TL11 : 30 km, Z = (0.27 + j 0.36) (/km |

| |T22 : 5 MVA, 33/6.6 kV, X = 8%, |

| |TL22 : 3 km, Z = (0.135 + j 0.08) (/km |

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| |Ans. 1715 A, 9.68 kV |

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|2 |Three 6.6 kV generators A, B and C, each of 10% leakage reactance and MVA rating 40, 50 and 25, respectively are interconnected|

| |electrically, as shown in figure, by a tie bar through current limiting reactors, each of 12% reactance based upon the rating |

| |of the machine to which it is connected. A 3-phase feeder is supplied from the bus bar of generator A at a line voltage of 6.6 |

| |kV. The feeder has a resistance of 0.06 (/phase and an inductive reactance of 0.12 (/phase. Estimate the maximum MVA that can |

| |be fed into a symmetrical short circuit at the far end of feeder. |

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| |Ans. 212 MVA |

|3 |A 3-phase transmission line operating at 10 kV and having a resistance of 1 ohm and reactance of 4 ohm is connected to the |

| |generating station bus bars through 5 MVA step up transformer having a reactance of 5%. The bus bars are supplied by a 10 MVA |

| |alternator having 10% reactance. Calculate the short circuit kVA fed to symmetrical fault between phases if it occurs |

| |At the load end of transmission line |

| |At the high voltage terminal of the transformer. |

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| |Ans. (i) 16440 KVA, (ii) 50,000 KVA |

|4 |The system shown in figure is initially at no load. Calculate the sub transient fault current that results when a three phase |

| |fault occurs at F at 66 kV. Take 69 kV as base voltage and 100 MVA as base MVA. |

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| |Ans. –j 2.586 pu |

|5 |For the system shown in the figure, the ratings of various components are: |

| |Generator : 25 MVA, 12.4 kV, Xd” = 10% |

| |Motor: 20 MVA, 3.8 kV, a sub-transient reactance = 15% |

| |Transformer T1: 25 MVA, 11/33 kV, reactance = 0.08 p.u. |

| |Transformer T2: 20 MVA, 33/3.3 kV, reactance = 0.10 p.u. |

| |Transmission Line Reactance = 20 ohms/phase. |

| |The system is loaded so that the motor is drawing 15 MW, at 0.9 leading p.f., the motor terminal voltage being 3.1 kV. |

| |Find the sub-transient current in generator and motor for a fault at generator bus. |

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| |Ans. 8.89 KA, 4.93 KA |

|6 |A station contains 4 bus bars sections to each of which is connected a generating unit of 30 MVA having 12 V and leakage |

| |reactance of 12%, the bus bar reactor having a reactance of 10%. Calculate the maximum MVA fed into the fault on any bus bar |

| |section and also the maximum MVA if the number of similar bus bar sections were increased to infinity. |

| |Ans. 5499 MVA |

Assignment 4: Symmetrical components

|1. |An unbalanced, wye-connected load consisting off the phase resistacnes Ra = 60(, Rb = 40 ( and Rc = 80 ( is connected to a |

| |440 V, three-phase, balanced supply. Calculate the line currents by the method of symmetrical components. |

| | |

| |Ans. 4.473(100.67( A, 5.138(-34.71( A, 3.696(156.63( A |

|2. |The sequence voltages in the red phase are as under: |

| |ER0 = 100 ; ER1 = 200 – j 100 ; ER2 = -100 |

| |Find the phase voltages ER, EY and EB. |

| | |

| |Ans. ER = 223.6 ( -26.56(, EY = 213( 99.89( ; EB = 338.57( 66.2( |

|3. |The zero and positive sequence components of red phase are as under: |

| |ER0 = 0.5 – j 0.866 ; ER1 = 2 ( 0( Find ER2 if ER = 3 ( 0( |

| | |

| |Ans. 1 (60( |

|4. |In a 3-phase 4-wire-system |

| |IR1 = 200( 0( A ; IR2 = 100( 60( Find IR0 if the total current flowing back into the supply in the neutral conductor is 300 ( |

| |300(. |

| | |

| |Ans. 100 (300( |

|5. |A balanced star connected load takes 90 A from a balanced 3-phase, 4-wire supply. If the fuses in the Y and B phases are removed,|

| |find the symmetrical components of the line currents |

| |(i) before the fuses are removed (ii) after the fuses are removed. |

| | |

| |Ans. (i) IR1 = 90 ( 0( ; IY1 = 90 ( -120( ; IB1 = 90 ( -240( |

| |IR0 = IR2 = IY0 = IY2 = IB0 = IB2 = 0 |

| |(ii) IR0 = 30 ( 0(; IR1 = 30 ( 0(; IR2 = 30 ( 0( |

| |IY0 =30( 0( ; IY1 30( -120( ; IY2 =30( -240( |

| |IB0 =30( 0( ; IB1 =30( -240(; IB2 =30( -120( |

Assignment 5: UNSYMMETRICAL FAULT ANALYSIS-I

1: A 25 MVA 13.2 kV, alternator with solidly grounded neutral has sub transient reactance of 0.25 pu. The negative and zero sequence reactances are 0.35 and 0.1 pu respectively. A single line to ground fault occurs at the terminals of an unloaded alternator. Determine the fault current and line to line voltage.

[4685A, 7.717 kV, 15.087 kV, 7.717 kV]

2: A 10 MVA, 6.9 kV generator has Z1 = 10%, Z2 = 7%, Z0 = 3% and its neutral is grounded solidly. The generator is open circuited and excited to its rated voltage. Find fault amount and phase voltages for the following two causes (1) L –G fault at generator terminal (2)L – L fault at generator terminal

.

Ans :12525 amp., Va = 0 kV, Vb = 3.08 ( 253( kV, Vc = 3.08 ( 107(kV

Ans : 8517 amp., Va = 3.278 kV, Vb = 1.639 kV, Vc = 1.639 kV.

3. The power system shown in figure, find fault current for the following three cases, when a fault occurs at the mid point of the transmission line.(i) L - G(ii) L-L(iii) L-L-G

Neglect prefault current. The reactances are given in p.u. on the same base.

Equipment Positive Seq. Negative Seq. Zero Seq.

Generator 0.30 0.20 0.05

Motor 0.25 0.15 0.05

Transformer (both) 0.10 0.10 0.10

Line 0.15 0.15 0.40

Ans. (i) –j 5.46, (ii) –j 4.35, (iii) LY = -4.08 + j 2.59, IB = 4.08 + j 2.59, IF = j 5.19

Assignment : Corona

|1 |In a three-phase overhead line the conductors have each a radius of 1.2 cm and are spaced symmetrically 3 m apart. If the |

| |dielectric strength of air is 30 kV per cm (peak), find disruptive critical voltage. |

| |[243.4 kV line-to-line] |

|2 |In a three-phase overhead line the conductors are arranged in the form of an equilateral triangle. The diameter of each |

| |conductor is 30 mm. Assuming fair weather conditions, air density factor of 0.94, irregularity factor 0.93 and breakdown |

| |strength of air 30 kV per cm (peak), find the minimum spacing between conductors, if the disruptive critical voltage is not|

| |to exceed 220 kV between lines. |

| |[1.44 m] |

|3 |A 3-phase, 220 kV, 50 Hz transmission line consists of 30 mm diameter conductor spaced 2.5 metres apart in the form of an |

| |equilateral triangle. In the temperature is 38( and atmospheric pressure is 76 cm of Hg, calculate the corona loss per km |

| |of the line. Irregularity factor is 0.83 to be assumed. |

| |[1941 W] |

|4 |A 132 kV, 50 Hz, three-phase transmission line 190 km long consists of three 1.2 cm diameter stranded copper conductors |

| |spaced in 2.8 metres delta arrangement. Assuming |

| |dielectric strength of air = 21.21 kV per cm (rms) |

| |temperature = 30( |

| |irregularity factor = 0.86 |

| |irregularity factor for local corona = 0.74 |

| |irregularity factor for general corona = 0.84, |

| |Determine the corona characteristics of the transmission line. |

| |[Vc =65.2 kV/phase (rms) |

| |for local corona – Vv = 78.2 kV/phase |

| |for general corona – Vv = 88.8 kV/phase |

| |corona will take place but will not be visual |

| |Corona loss using Peek’s formula – P = 1.0437 kW/km/phase, total corona loss P=59439 kW |

| |Corona loss using Peterson’s formula – P = 4.985 kW/km/phase, total corona loss P= 2841.45 kW] |

|5 |A three phase 220 kV, 50 Hz transmission line consists of 1.5 cm radius conductor spaced 2 metres apart in equilateral |

| |triangle formation. If the temperature is 40 degree centigrade and atmospheric pressure is 76 cm, calculate the corona loss|

| |per km of the line using Peek’s formula. Take mo = 0.85 |

| | |

| |[Total corona loss per km = 0.5998 kW] |

|6 |A 132 kV line with 1.956 cm diameter conductors is built so that corona takes place if the line voltage exceeds 210 kV |

| |(rms). If the value of potential gradient at which ionization occurs can be taken as 30 kV per cm find the spacing |

| |between the conductors. |

| |[D = 341 cm] |

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F

G1

G2

T2

T1

TL1

TL2

F

G

M

F

F

F

G1

G2

50 MVA, 15 kV

Xd”=0.2 pu

25VA, 15 kV

Xd”=0.2 pu

A

66 kV

XT”=0.1 pu

Delta/Star transformer

M

G

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