# Power Systems Test 2

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• ###### Question 1
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Increased fuel cost in Rs/Mwh for a plant consisting of units are given by

$$\frac{{d{F_1}}}{{d{P_1}}} = 0.4{P_1} + 400$$

$$\frac{{d{F_2}}}{{d{P_2}}}= 0.48{P_2} + 320$$

The allocation of load P1 and P2 between the unit 1 and unit 2, respectively, for the minimum cost of generation for a total load of 900 MW is

###### Solution
given total load P1 + P2 = 900 MW

$$\frac{{d{F_1}}}{{d{P_2}}} = \frac{{d{F_2}}}{{d{P_2}}}$$

0.41P1 + 400 = 0.48 P­ + 320

-0.4P1 + 0.48P2 = 80

-P1 + 1.2 P2 = 200

P2 = 500 MWP1 = 400 MW

• ###### Question 2
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A short three phase short transmission line in phase co-ordinates is represented by a series impedance matrix as $$\left[ {\begin{array}{*{20}{c}}{{Z_s}}&{{Z_m}}&{{Z_m}}\\{{Z_m}}&{{Z_s}}&{{Z_m}}\\{{Z_m}}&{{Z_m}}&{{Z_m}}\end{array}} \right]$$

If the negative sequence impedance is (2 + 9) Ω and zero sequence impedance is (8 + j42) Ω, which of the following is/are true?

###### Solution
Positive sequence impedance is same as negative sequence impedance. Positive sequence impedance = (2 + j9) Ω Zs – Zm = 2 + j9 Zs + 2 Zm = 8 + j42 By solving above two equations, Zm = 2 + j11 ⇒ Zs = 4 + j20
• ###### Question 3
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A 100 × 100 bus admittance matrix for an electric power system has 600 non-zero elements. Which of the following is/are true?
###### Solution
Size of YBUS matrix = 100 × 100

Number of non-zero elements = 600

Number of diagonal elements = 100

All the diagonal elements are non-zero in a Y-bus matrix.

Number of non-zero diagonal elements = 100

Number of non-zero off diagonal elements – 600 – 100 = 500

Minimum number of branches in the system are

$$= \frac{{500}}{2} = 250$$

• ###### Question 4
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The initial value of DC offset in a three-phase fault depends on:
###### Solution
• AC circuit transient current is composed of symmetrical short circuit current (AC fault current) and DC offset.
• The value of DC offset (magnitude of the initial value of DC offset) can be between zero to the maximum value of AC fault current depending upon different values of (voltage phase angle)
• As the value of α is decided by the instantaneous source voltage at the instant of short circuit the value of DC offset is decided by the instantaneous source voltage at the instant of short circuit.
• ###### Question 5
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Which of the following has a problem in convergence for a system with long radial lines?
###### Solution
Comparison of two load flow methods is given below.

 Type of problem Gauss-Seidel Method Newton-Raphson Method Heavily loaded system Usually cannot solve systems with more than 70° phase shift Solve systems with shifts up to 90° Systems containing negative reactance such as three-winding transformers or series capacitors Unable to solve Solves with ease Systems with slack bus at a desired location Often requires trail and error to find a slack bus location that will yield a solution More tolerant of slack bus location Long and short lines terminating the same bus Usually cannot solve if long to short ratio is above 1000 Can solve a system with a long to short ratio at any bus of 1,000,000 Long radial type of system Difficulty in solving Solves a wide range of such problems Acceleration factor Number of iterations depends on choice of factor Not required
• ###### Question 6
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The positive, negative and zero sequence impedances of 3-phase synchronous generator are j 0.5 pu, j 0.3 pu and j 0.2 pu respectively. When the symmetrical fault occurs on the machine terminals. Find the fault current. The generator neutral is grounded through reactance of j0.1 pu
###### Solution
For three-phase symmetrical fault, the fault current is given by

$${I_f} = {I_{R1}} = \frac{{{E_{R1}}}}{{{Z_{1eq}}}}$$

ER1 is pre fault voltage = 1 pu

Z1eq is positive sequence impedance

Z1eq = j 0.5 pu

$${I_f} = \frac{1}{{j0.5}} = - j2.0\;pu$$

Note:

• The positive sequence currents involved in all type of systems
• Negative sequence currents involved in only unbalanced systems
• Zero sequence currents involved in systems involving earth
• Thus, in a balanced three-phase system both negative and zero sequence currents are zero
• ###### Question 7
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A 100 MVA synchronous generator operates on full load at a frequency of 50 Hz. Inertia constant is 8 MJ/MVA. The load is suddenly reduced 100 MW. Due to time lag in governor system, the steam valve begins to close after 0.4 seconds. The change in frequency that occurs in this time is_____
###### Solution
The rating of the machine (S) = 100 MVA

Inertia constant = H = 8 MJ/MVA

Kinetic energy stored in the rotating parts of the generator = SH = 800 MJ

Energy transferred in 0.4 sec = 100 × 0.4 = 40 MJ

$$\begin{array}{l} \frac{{K{E_1}}}{{K{E_2}}} = {\left( {\frac{{{f_1}}}{{{f_2}}}} \right)^2}\\ \Rightarrow \frac{{800}}{{840}} = {\left( {\frac{{50}}{{{f_2}}}} \right)^2} \end{array}$$

$$\Rightarrow {f_2} = 51.23\;Hz$$Change in frequency = 1.23 Hz

• ###### Question 8
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In a 132 kV, 50 Hz system, reactance and capacitance up to the location of the circuit breaker is 5 Ω and 0.02 μF respectively. A resistance of 500 Ω is connected across the breaker of the circuit breaker. The damped frequency of oscillation is ______ (in kHz)
###### Solution
Concept:

The frequency of damped oscillations is given by

$$f = \frac{1}{{2\pi }}\sqrt {\frac{1}{{LC}} - \frac{1}{{4{C^2}{R^2}}}}$$

Calculation:

Given that, inductive reactance (X­L) = 5Ω

2πfL = 5

2π × 50 × L = 5

L = 15.91 mH

Capacitance (C) = 0.02 μF

Resistance (R) = 500 Ω

$$f = \frac{1}{{2\pi }}\sqrt {\frac{1}{{15.91\; \times \;{{10}^{ - 3}}\; \times\; 0.02\; \times \;{{10}^{ - 6}}}} - \frac{1}{{4\; \times\; {{\left( {0.02\; \times \;{{10}^{ - 6}}} \right)}^2}\; \times\; {{\left( {500} \right)}^2}}}}$$

= 4.03 kHz

• ###### Question 9
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A circuit breaker is employed to quench the magnetizing current of a 80 MVA transformer at 230 kV. The magnetizing current is 4% of full load current. Now, the circuit breaker is opened when the magnetizing current is as its peak value. The stray capacitance is 4 pF and inductance is 25 mH. The maximum voltage appears across the circuit is ______(in kV)
###### Solution
Full load current $$\left( {{I_L}} \right) = \frac{{80 \times {{10}^6}}}{{\sqrt 3 \times 230 \times {{10}^3}}} = 200.8\;A$$

Magnetizing current = (0.04) (200.8) = 8 A

Maximum value of magnetizing current $$= 8\sqrt 2\;A$$

Voltage that appears across circuit breaker

$$V = i\sqrt {\frac{L}{C}} = 8\sqrt 2 \times \sqrt {\frac{{25 \times {{10}^{ - 3}}}}{{4 \times {{10}^{ - 12}}}}} = 897.43\;kV$$

• ###### Question 10
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The power system has two generating plants and power being dispatched economically with P1 = 150 MW and P2 = 200 MW. The loss co-efficient are B11 = 0.003 MW-2, B22 = 0.001 MW-2 and B12 = - 0.0004 MW-2. The penalty factor for the plant 1 will be _________ (up to two decimal places)
###### Solution
Concept:

The transmission loss is given as

$${P_L} = \mathop \sum \limits_m \mathop \sum \limits_n \;{P_m}{B_{mn}}{P_n}\;$$

For two plant system, m = 1, 2 and n = 1, 2

Then,

PL = P1 B11 P1 + P1 B12 P2 + P2 B21 P1 + P2 B22 P2

B12 = B21

$$\Rightarrow {P_L} = P_1^2{B_{11}} + 2{P_1}{P_2}{B_{12}} + P_2^2{B_{22}}$$

Penalty factor for plant 1 is,

$${P_{f1}} = \frac{1}{{1 - \frac{{\partial {P_L}}}{{\partial {P_1}}}}}$$

$$\frac{{\partial {P_L}}}{{\partial {P_1}}} = 2\;{P_1}\;{B_{11}} + 2\;{P_2}\;{B_{12}}$$

Penalty factor for plant 2 is

$${P_{f2}} = \frac{1}{{1 - \frac{{\partial {P_L}}}{{\partial {P_2}}}}}\;$$

$$\frac{{\partial {P_L}}}{{\partial {P_2}}} = 2{P_1}\;{B_{12}} + 2\;{P_2}\;{B_{22}}$$

Calculation:

Given that,

P1 = 150 MW, P2 = 200 MW

B11 = 0.003 MW-2

B22 = 0.001 MW-2

B12 = -0.0004 MW-2

$$\frac{{\partial {P_L}}}{{\partial {P_1}}} = 2\;{P_1}\;{B_{11}} + 2\;{P_2}\;{B_{12}}$$

= 2(150) (0.003) + 2(200) (-0.0004)

= 0.74

$${P_{f1}} = \frac{1}{{1 - 0.74}} = 3.846$$

• ###### Question 11
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The sending and voltage of a three phase 50 Hz lossless transmission line is maintained at 400 kV. The inductance and capacitance of the line are 8 mH/km/phase and 200 nF/ km/ Phase respectively.

To maintain a voltage of 400 kV at receiving end, when the line is delivering 600 MW load, which of the following is/are true?

###### Solution
Inductance (L) = 8 mH/km/phase

Capacitance (C) = 200 nF/km/phase

Surge impedance, $$Z = \sqrt {\frac{L}{c}}$$

$$= \sqrt {\frac{{8\; \times \;{{10}^{ - 3}}}}{{200 \;\times \;{{10}^{ - 9}}}}} \;$$

= 200 Ω

Surge impedance loading = $$\frac{{{V^2}}}{{{Z_c}}}$$

$$= \frac{{{{\left( {400\; \times \;{{10}^3}} \right)}^2}}}{{200}} = 800\;MW$$

Given load is 600 MW, and it is lesser than the surge impedance loading.To maintain rated voltage at receiving end, the compensation must be inductive.

• ###### Question 12
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A balanced 3ϕ load of 30 MW is supplied at 132 kV, 50 Hz & 0.85 pf lagging by means of transmission line. The series impedance of single conductor is (20 + j52) Ω and total phase-neutral admittance is 315 × 10-6 mho. For a nominal T-method, which of the following is / are true regarding the A, B, C, D constants of the line.

###### Solution
VR = 132 × 103 V Phase voltage $$= \frac{{132}}{{\sqrt 3 }}kV = 76.21\;kV$$ z = 20 + j52 Ω Y = j 315 × 10-6$$A = D = 1 + \frac{{YZ}}{2} = 1 + \frac{{\left( {315 \times {{10}^{ - 6}}j} \right)\left( {20 + j52} \right)}}{2}$$ = 0.99 ∠0.182 $$B = z\left( {1 + \frac{{YZ}}{4}} \right)$$ $$= \left( {20 + 52j} \right)\left( {1 + \frac{{\left( {315 \times {{10}^{ - 6}}j} \right)\left( {20 + 52j} \right)}}{4}} \right)$$ = 55.48 ∠69.05 C = Y = j315 × 10-6 = 0.315 × 10-3 ∠90°
• ###### Question 13
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A cylindrical rotor synchronous generator is operating at a terminal voltage of 1 pu with an internal emf of (1 + j0.5) pu. It has a steady state synchronous reactance of 0.8 pu and sub transient reactance of 0.2 pu.

If there is a three-phase solid short circuit fault at the terminal of the generator, which of the following is/are true?

###### Solution
Given that, terminal voltage (Vt) = 1 pu Internal emf (Eg) = (1 + j0.5) pu Steady state synchronous reactance (Xd) = 0.8 pu Sub transient synchronous reactance (X"d) = 0.2 pu Fault current, $${I_f} = \frac{{{E_g} - {V_o}}}{{{X_d}}}$$ $$= \frac{{1 + j0.5 - 1}}{{j0.8\;}}$$  = 0.625 pu Sub transient internal emf Ef = Vt + (jX"d) If = 1 + (j 0.2) (0.625) = 1 + j 0.125 Magnitude = $$\sqrt {1 + {{\left( {0.125} \right)}^2}\;}$$ = 1.0078 pu
• ###### Question 14
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A cable is graded with three dielectrics of permittivities 4, 3 and 2. The maximum permissible potential gradient for all dielectrics is same and equal to 30 kV / cm. The core diameter is 1.5 cm and sheath diameter is 5.5 cm the rms value of working voltage is ______ (in kV)
###### Solution
ε1 = 4, ε2 = 3, ε3 = 2

Emax = 30 kV / cm

d = 1.5 cm, D = 5.5 cm

r = 0.75 cm, R = 2.75 cm

ε1r = ε2r1 = ε3r2

ε1r = ε2r1

$$\Rightarrow {r_1} = \frac{{{{\rm{\varepsilon }}_1}r}}{{{{\rm{\varepsilon }}_2}}} = \frac{{4 \times 0.75}}{3} = 1\;cm$$

ε1r = ε3r2

$$\Rightarrow {r_2} = \frac{{{{\rm{\varepsilon }}_1}r}}{{{{\rm{\varepsilon }}_3}}} = \frac{{4 \times 0.75}}{2} = 1.5\;cm$$

V = V1 + V2 + V3

$$= {E_{max}}\left[ {r\ln \left( {\frac{{{r_1}}}{r}} \right) + {r_1}\ln \left( {\frac{{{r_2}}}{{{r_1}}}} \right) + {r_2}\ln \left( {\frac{R}{{{r_L}}}} \right)} \right]$$

$$= 30\;\left[ {0.75\ln \left( {\frac{1}{{0.75}}} \right) + \ln \left( {\frac{{1.5}}{1}} \right) + 1.5\ln \left( {\frac{{2.75}}{{1.5}}} \right)} \right]$$

= 45.9 kV$${V_{rm}} = \frac{{{V_m}}}{{\sqrt 2 }} = \frac{{45.9}}{{\sqrt 2 }} = 32.4\;kV$$

• ###### Question 15
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A 3ϕ 50 Hz 320 km long transmission line has inductance of 1.98 mH / km / phase & C = 9.93 × 10-9 F / km. The line supplies a load of 75 MVA, 0.8 pf lagging at 400 kV. The MVAR rating of 3ϕ shunt reactor to be installed at the receiving end to sending end voltage at rated value is
###### Solution
Length = 320 km

L = 0.98 × 320 = 313.6 mH

C = 9.93 × 10 × 320 = 3.177 × 10-6 F

VR = 400 kV

Vs = 400 kV

$$z = \sqrt {\frac{L}{C}} = \sqrt {\frac{{313.6 \times {{10}^{ - 3}}}}{{3.1776 \times {{10}^{ - 6}}}}} = 314.15\;{\rm{\Omega }}$$

$$\beta = 2\pi \sqrt {LC} = 2\pi \times 50 \times \sqrt {\left( {313.6 \times {{10}^{ - 3}}} \right)\left( {3.177 \times {{10}^{ - 6}}} \right)}$$

= 17.93°

Value of shunt reacted

$${X_L} = \frac{{{z_c}\sin {\beta _L}}}{{\frac{{{V_s}}}{{{V_R}}} - \cos {\beta _L}}}$$

$$= \frac{{314.15\sin 17.93^\circ }}{{1 - \cos 17.93^\circ }} = 1991.33$$MVAR rating $$= \frac{{{V^2}}}{{{X_L}}} = \frac{{{{\left( {400} \right)}^2}}}{{1991.33}} = 80.34\;MVAR$$

• ###### Question 16
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A 50 Hz, a synchronous generator capable to supply 400 MW connected to a large power system and delivering 80 MW when a three-phase fault occurs at its terminals. If the maximum power angle is to be 85°, the time in which the fault must be cleared is _____ (in sec)

(Assume the base value is 100 MW  and consider the inertia constant H = 7 MJ / MVA )
###### Solution
When the fault occurs

$$\delta = {\delta _1} = 85^\circ = \frac{{85 \times \pi }}{{180}} = 1.4835\;rad$$

Pe = Pmax sin δ

Pmax = 400 MW

Pe = 80 MW = Pm

At steady state = Pe = Pm

$$\sin {\delta _0} = \frac{{{P_e}}}{{{P_{max}}}} = \frac{{80}}{{400}}$$

$$\Rightarrow {\delta _0} = 11.54^\circ = \frac{{11.54 \times \pi }}{{180}} = 0.2\;rad$$

δc = cos-1 [cos δ1 + (δ1 – δ0) sin δ0]

= cos-1 [cos (1.4835) + (1.4835 – 0.201) sin (0.201)]

Since Pe = 80 MW and before the fault Pe = Pm also in fault moment Pe = 0 the Pm = 80

If we take a base of 100 MW then Pm = 0.8 pu

The clearing time tc is:

$${t_c} = \sqrt {\frac{{2H\left( {{\delta _c} - {\delta _0}} \right)}}{{\pi f{P_m}}}}$$

$$= \sqrt {\frac{{2 \times 7 \times \left( {1.218 - 0.201} \right)}}{{\pi \times 50 \times 0.8}}}$$= 0.336 sec

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