Power Systems Test 1

Size of Jacobian matrix = (2n - 2 - m) × (2n - 2 - m)

n = number of buses

m = number of generator bus, reactive power support bus, 

One bus is treated as slack bus.

where as shunt capacitor buses are considered as load buses.

Size = 2(300) - 2 - (19 + 10 )

Line voltage, V_L = 11 kV

Phase voltage, V_ph = (11/√3) kV

Real power (P) = 200 kW

Power factor = cos ϕ = 0.8

kVAR demand of load \(Q = P\tan \phi = 200 \times \frac{{0.6}}{{0.8}} = 150\;kVAR\)

kVAR demand of load at unity power factor = 0

kVAR rating of Δ connected capacitor bank = 150 kVAR

⇒ X_Cph = 806.67 Ω

(1 / 2π f C) = 806.67 Ω 

C = 3.948 μF 

Z = (3.25 + j 7.55) ohms

R = 3.25 Ω

X = 7.55 Ω

Receiving end voltage, V_r = 11 kV

\({V_r}\left( {phase} \right) = \frac{{11}}{{\sqrt 3 }}kV = 6.35\;kV\) 

Load power factor = cos ϕ_r = 0.8 lag

Load current, \({I_L} = \frac{{2395 \times {{10}^3}}}{{3 \times 6.35 \times {{10}^3}}}\)

= 125.72 A

Sending end voltage (V_s) per phase

= V_r + IR cos ϕ_r + IX sin ϕ_r

= 6.35 + 10³ + (125.72) (3.25) (0.8) + (125.72) (7.55) (0.6)

= 7.25 kV

Sending end power factor,

\(\cos {\phi _s} = \frac{{{V_r}\cos {\phi _r} + IR}}{{{V_s}}}\) 

\(= \frac{{\left( {6.35 \times {{10}^3} \times 0.8} \right) + \left( {125.72 \times 3.25} \right)}}{{7.25 \times {{10}^3}}}\) 

Voltage (V_R) = 6.6 kV

MVA rating = 20 MVA

X^’’_d = 10% = 0.1

X^’_d = 20% = 0.2

X_d = 100% = 1

Rated current \({I_R} = \frac{{\left( {VA} \right)}}{{\sqrt 3 \times V}}\)

\(= \frac{{20 \times {{10}^6}}}{{\sqrt 3 \times 6.6 \times {{10}^3}}}\)

= 1749.5 A

Let I₀ be the initial symmetrical rms short circuit current.

At initial state, reactance = X^’’_d = 0.1 pu

\({I_0} = \frac{{{V_{pu}}}}{{X_{{d_{pu}}}^{''}}} = \frac{1}{{0.1}} = 10pu\)

\({I_{0\left( {actual} \right)}} = {I_{0\left( {pu} \right)}} \times {I_R}\)

Voltage (E) = 11 kV

Phase voltage (E_ph) \(= \frac{{11}}{{\sqrt 3 }}kV = 6.35\;kV\)

Three phase fault current = 5610 A

\({I_{f\left( {3 - \phi } \right)}} = \frac{{{E_{ph}}}}{{{X_1}}}\) 

\(\Rightarrow 5610 = \frac{{6.35 \times {{10}^3}}}{{{X_1}}}\) 

⇒ X₁ = 1.132 Ω

L-L fault current = 6760 A

\({I_{f\left( {L - L} \right)}} = \frac{{\sqrt 3 {E_{ph}}}}{{{X_1} + {X_2}}}\) 

\(\Rightarrow 6760 = \frac{{\sqrt 3 \times 6.35 \times {{10}^3}}}{{1.132 + {X_2}}}\) 

⇒ X₂ = 0.495 Ω

L-G fault current = 8630 A

\({I_{f\left( {L - G} \right)}} = \frac{{3{E_{ph}}}}{{{X_1} + {X_2} + {X_0}}}\) 

\(\Rightarrow 8630 = \frac{{3 \times 6.35 \times {{10}^3}}}{{1.132 + 0.495 + {X_0}}}\) 

⇒ X₀ = 0.58 Ω

\({X_{base}} = \frac{{\left( {kV} \right)_{base}^2}}{{{{\left( {MVA} \right)}_{base}}}} = \frac{{{{\left( {11} \right)}^2}}}{{25}} = 4.84\;{\rm{\Omega }}\) 

L = 12 × 10^-3 C = .03 × 10^-6

Voltage across the breaker poles

\(= i\sqrt {\frac{L}{C}} = \sqrt[8]{{\frac{{12 \times {{10}^{ - 3}}}}{{.03 \times {{10}^{ - 6}}}}}} = 5059.64\;V\)