Power Systems Test 1

A = D = 0.9 ∠0. B = 200 ∠90° Ω and C = 0.95 × 10^-3 ∠90° S

\(\left[ {\begin{array}{*{20}{c}}{A'}&{B'}\\{C'}&{D'}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0\\Y&1\end{array}} \right]\)

A’ = 1

⇒ A + BY = 1

\(\Rightarrow 0.9 + 200\angle 90\left( Y \right) = 1\)

H₁ = 0.8 pu, S₁ = 500 MVA, S_new = 200 MVA

H₂ = 0.5 pu, S₂ = 1000 MVA, S_new = 200 MVA

Inertia constant \(H \propto \frac{1}{S}\)

\({H_{1new}} = {H_1} \times \frac{{{S_1}}}{{{S_{new}}}} = 0.8 \times \frac{{500}}{{200}} = 2\;pu\)

\({H_{2new}} = {H_2} \times \frac{{{S_2}}}{{{S_{new}}}} = 0.5 \times \frac{{1000}}{{200}}\)

H_{2 new} = 2.5 pu

As the pair is a coherent pair:

H_eq = H_{1 new} + H_2new

= 2 + 2.5 = 4.5 pu

In a delta connected system:

• The zero sequence component of the line current is always zero, whereas the zero sequence component of phase current may or may not be zero.
• The positive sequence component of the line current is √3 times of positive sequence component of phase current and lags the phase current by 30^o.
• Negative sequence component of line current is √3 times of negative sequence component of phase current and leads the phase current by 30^o 

In a bus-impedance matrix, Z₄₄ = j0.5

If a capacitor having a reactance of -j3.5 pu is now added to the network between node-4 and the reference node.

Now, the Z₄₄ = j0.5 – j3.5 = -j 3 = 3∠-90° pu

The voltage at node 4, V₄ = 1.5∠2° pu

Rated secondary current of current transformer = 5 A

Pickup current(I_pickup) = Current setting(A) \(= \frac{{current\;setting\left( \% \right)}}{{100}} \times {I_{relay}}\)

Current setting(A) = 1.5 × 5 = 7.5 A

Plug-setting multiplier\( = \frac{{{\rm{Fault\;current}}}}{{{\rm{C.T. ratio\times current\;setting(A)}}}}\)

\(= \frac{{5000}}{{\frac{{400}}{{5}}\;\times7.5}} = 8.33\)

Steady-state stability limit = \(\frac{{{V_1}\;{V_2}}}{X}\)

More steady-state stability represents more transient stability of the system.

Steady-state stability limit of the system can be increased by the following methods:

• Upgrading voltage on the existing transmission system or opting for higher voltages on the new transmission system.
• The use of an additional parallel transmission line reduces transfer reactance X, thereby increases it.
• The use of a transformer with lower leakage reactance improves a steady-state stability limit.
• Series capacitive compensation of transmission line reduces transfer reactance X and hence steady-state stability limit increases.
• The use of bundled phase conductors reduces the reactance thereby increasing the stability limit.

Total no of elements in Y_bus = 100 × 100 = 10000

Given that 80% is sparse bus = number of zero element in Y_BUS = 10000 × 0.8 = 8000

Total number of non-zero element in Y_bus = 10000 – 8000 = 2000

Total number of non-zero non diagonal element in Y_bus = 2000 – 100 = 1900

Two non-zero non diagonal element in bus matrix form one transmission line

Hence total number of transmission line = \(\frac{1900}{2}=950\)

X_L = 8 Ω

⇒ 2πfL = 8

\(\Rightarrow L = \frac{8}{{2\pi \times 50}} = 25.46\;mH\)

C = 0.025 μF

Natural frequency of oscillations

\({f_n} = \frac{1}{{2\pi \sqrt {LC} }}\)

\(= \frac{1}{{2\pi \sqrt {25.46 \times {{10}^{ - 3}} \times 0.025 \times {{10}^{ - 6}}} }}\)

= 6.3 kHz

P_D + j Q_D = 2.0 + j2 tan cos^-1 (0.85)

= 2 + j 1.24

- j Q_c = - j 2.1

P_R + j Q_R = 2 + j 1.24 – j 2.1 = 2 – j 0.86

= 2.18 MVA, 23.3° leading

Power factor = 0.918

Z = 3 + j10 = 10.44 ∠73.3°

\({{I}_{R}}=\frac{2.18~\angle 23.3}{\sqrt{3}\times 11}=0.114~\angle 23.3~kA\)

V_S = V_R + z I_R

\(=\frac{11}{\sqrt{3}}+10.44~\angle 73.3\cdot 0.114~\angle 23.3=6.33~\angle 10.8{}^\circ ~kV\)

The voltage drop across the reactance of the transformer

When taps are set to 1 : 11

\({\rm{\Delta }}{V_1} = \frac{1}{{11}} - 1 = - 0.909\;pu\)

The voltage drop across the reactance of the transformer

When taps are set to 1 : 0.9

\({\rm{\Delta }}{V_2} = \frac{1}{{0.9}} - 1 = 0.111\;pu\)

Reactive power absorbed \(= \frac{{{\rm{\Delta }}V_1^2 + {\rm{\Delta }}V_2^2}}{X}\)

\(= \frac{{{{\left( { - 0.909} \right)}^2} + {{\left( {0.111} \right)}^2}}}{{0.18}} = 4.6589pu\)

Reactive power in MVAr = 4.6589 × 200

Voltage rating (V) = 220 kV

VA rating = 100 MVA

Inductance (L) = 30 H

Capacitance (C) = 2500 μF

Full load current \(\left( {{I_L}} \right) = \frac{{100 × {{10}^6}}}{{√ 3 × 220 × {{10}^3}}}\)

= 262.43 A

Magnetizing current \(\left( {{I_\mu }} \right) = \frac{5}{{100}} × {I_L}\)

\({I_\mu } = \frac{5}{{100}} × 262.43 = 13.12\;A\)

Magnetizing current is interrupted at 53% of its peak value.

Magnetizing current at the time of interruption

= i = 0.53 × I_μ = 0.53 × √2 × 13.12 = 9.83 A

The maximum voltage which may appear across the circuit breaker contacts.

\(V = i√ {\frac{L}{C}} \)

Concept:

Accelerating torque = (Accelerating power, P_a)/(Synchronous speed, ω_s)

P_a is the accelerating power = fault mechanical input – fault mechanical output

Calculation:

Given that,

P = 4, f = 50 Hz, pf = 0.8

At fault, electrical output = 60%

MVA rating of a turbo generator is 500 MVA

Before fault:

Mechanical input (P_s1) = electrical output (P_e1) = 500 × 0.6 × 10⁶ = 300 MW

During fault:

Electrical output, P_e2 = 0.6 × 400 × 10⁶ = 240 MW

Mechanical input, P_s2 = 300 MW

Accelerating power, P_a = P_s2 – P_e2 = 300 – 240 = 60 MW

Now, speed \({N_s} = \frac{{120\; \times \;f}}{P} = 1500\;rpm\) 

⇒ ω_s = 157.07 rad/sec

Accelerating torque, \({T_a} = \frac{{{P_a}}}{{{\omega _s}}} = \frac{{60\; \times \;{{10}^6}}}{{157.07}} = 0.3819\;MN - m\) 

Voltage (V) = 11 kV

VA rating = 10000 kVA

Full load current \(\left( {{I_L}} \right) = \frac{{10 \times {{10}^6}}}{{\sqrt 3 \times 11 \times {{10}^3}}} = 524.86\;A\)

Out of balance current = 20% of full load current

= 0.2 × 524.86

= 104.97 A

Let x% of the winding remain unprotected.

The fault current \(= \frac{{11 \times {{10}^3}}}{{\sqrt 3 }} \times \frac{1}{7} \times \frac{x}{{100}}\)

\(\Rightarrow \frac{{11 \times {{10}^3} \times x}}{{\sqrt 3 \times 7 \times 100}} > 104.97\) 

⇒ x > 11.57

The percentage of unprotected winding = 11.57%

The percentage of protected winding = 100 – 11.57