Power Systems Test 2

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 MW

Positive sequence impedance is same as negative sequence impedance.

Positive sequence impedance = (2 + j9) Ω

Z_s – Z_m = 2 + j9

Z_s + 2 Z_m = 8 + j42

By solving above two equations,

Z_m = 2 + j11

⇒ Z_s = 4 + j20 

Size of Y_BUS 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

\(= \frac{{500}}{2} = 250\) 

• 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.

Comparison of two load flow methods is given below.

For three-phase symmetrical fault, the fault current is given by

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

E_R1 is pre fault voltage = 1 pu

Z_1eq is positive sequence impedance

Z_1eq = 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

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

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

\(\Rightarrow {f_2} = 51.23\;Hz\)

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 (X­_L) = 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  

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

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

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, 

P_L = 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

\({P_{f1}} = \frac{1}{{1 - 0.74}} = 3.846\)

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.

V_R = 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°

Given that, terminal voltage (V_t) = 1 pu

Internal emf (Eg) = (1 + j0.5) pu

Steady state synchronous reactance (X_d) = 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

E_f = V_t + (jX"_d) I_f

= 1 + (j 0.2) (0.625)

= 1 + j 0.125

Magnitude = \(\sqrt {1 + {{\left( {0.125} \right)}^2}\;} \)

= 1.0078 pu

ε₁ = 4, ε₂ = 3, ε₃ = 2

E_max = 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

Length = 320 km

L = 0.98 × 320 = 313.6 mH

C = 9.93 × 10 × 320 = 3.177 × 10^-6 F

V_R = 400 kV

V_s = 400 kV

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

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

= 0.313 rad

= 17.93°

Value of shunt reacted

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

\( = \frac{{314.15\sin 17.93^\circ }}{{1 - \cos 17.93^\circ }} = 1991.33\)

When the fault occurs

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

P_e = P_max sin δ

P_max = 400 MW

P_e = 80 MW = P_m

At steady state = P_e = P_m

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

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

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

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

= 69.8° = 1.218 rad

Since P_e = 80 MW and before the fault P_e = P_m also in fault moment P_e = 0 the P_m = 80

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

The clearing time t_c is:

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

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