Power Systems Test 1

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Power Systems Test 1
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  • Question 1
    2 / -0.33
    The ABCD parameters of a 3-phase overhead transmission line are A = D = 0.9 ∠0. B = 200 ∠90° Ω and C = 0.95 × 10-3 ∠90° S. At no-load condition a shunt inductive, reaction is connected at the receiving end of the line to limit the receiving-end voltage to be equal to the sending-end voltage. The ohmic value of the reactor is
    Solution
    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

    \(\Rightarrow 0.9 + 200\angle 90\left( Y \right) = 1\)\(\Rightarrow Y = \frac{{0.1}}{{200\angle 90^\circ }} \Rightarrow Z = 2000\;{\rm{\Omega }}\)

  • Question 2
    2 / -0.33
    A power station consists of two synchronous generators of rating 500 MVA and 1000 MVA with inertia 0.8 pu and 0.5 pu respectively on their own buses MVA rating. The equivalent pu inertia constant on 200 MVA common base is:
    Solution
    H1 = 0.8 pu, S1 = 500 MVA, Snew = 200 MVA

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

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

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

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

    H2 new = 2.5 pu

    As the pair is a coherent pair:

    Heq = H1 new + H2new

    = 2 + 2.5 = 4.5 pu

  • Question 3
    2 / -0.33
    For a delta connected system which of the following statements is true?
    Solution
    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 30o.
    • Negative sequence component of line current is √3 times of negative sequence component of phase current and leads the phase current by 30
  • Question 4
    2 / -0.33
    For a power system network with 10 nodes, Z44 of its bus impedance matrix is j0.5 pu. The voltage at node-4 is 1.5∠2° pu. If a capacitor having a reactance of -j3.5 pu is now added to the network between node-4 and the reference node, the current drawn by the capacitor in pu is _______.
    Solution
    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° puThe current drawn by the capacitor, \({I_C} = \frac{{{V_4}}}{{{Z_{44}}}} = \frac{{1.5\angle 2^\circ }}{{3\angle - 90^\circ }} = 0.5\angle 92^\circ pu\)

  • Question 5
    2 / -0.33

    A 10-ampere, over current relay having a current setting of 150% and a time setting multiplier of a 0.8 connected to supply circuit through a 400/5 current transformer when the circuit carries a fault current of 5000 A. Find the plug-setting multiplier.

    Solution
    Rated secondary current of current transformer = 5 A

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

    Current setting(A) = 1.5 × 5 = 7.5 A

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

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

  • Question 6
    2 / -0.33

    Consider the methods given below:

    A. Use of additional parallel transmission lines.

    B. Upgrading voltage on the existing transmission system.

    C. Use of bundled conductors

    D. Use of a transformer with lower leakage reactance.

    Which of the above methods can be used to improve the steady-state stability limit of a system?

    Solution
    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.
  • Question 7
    2 / -0.33
    The Ybus matrix of a 100-bus interconnected system is 80% sparse. Then, the number of transmission lines in the system must be
    Solution
    Total no of elements in Ybus = 100 × 100 = 10000

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

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

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

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

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

  • Question 8
    2 / -0.33
    In a 220 kV, 50 c/s power system, the reactance and capacitance up to the circuit breaker are 8 Ω and 0.025 μF respectively. A resistance of 600 Ω is connected across the circuit breaker contacts. The natural frequency of oscillation is _______ (in kHz)
    Solution
    XL = 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

  • Question 9
    2 / -0.33
    ​A three-phase feeder having a resistance of 3 Ω and a reactance of 10 Ω supplies a load of 2 MW at 0.85 lagging power factor. The receiving end voltage is maintained at 11 kV by means of a static condenser drawing 2.1 MVAR from the line. The sending end voltage is
    Solution
    PD + j QD = 2.0 + j2 tan cos-1 (0.85)

    = 2 + j 1.24

    - j Qc = - j 2.1

    PR + j QR = 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°

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

    VS = VR + z IR

    \(=\frac{11}{\sqrt{3}}+10.44~\angle 73.3\cdot 0.114~\angle 23.3=6.33~\angle 10.8{}^\circ ~kV\)|VS| (line) = √3 × 6.33 = 10.97 kV

  • Question 10
    2 / -0.33
    Two substations are connected by two lines in parallel with negligible impedance, but each containing a tap changing transformer of reactance 0.18 pu on the basis of its rating of 200 MVA. Find the net absorption of reactive power in MVAr when the transformer taps are set to (1 : 11) and (1 : 0.9) respectively. Assume per unit voltages to be equal at two ends of the substation.
    Solution
    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= 931.78 MVAr

  • Question 11
    2 / -0.33
    A circuit breaker interrupts the magnetizing current of 100 MVA transformer at 220 kV. The magnetizing current of the transformer is 5% of the full load current. The maximum voltage (in volts) which may appear across the gap of the breaker when the magnetizing current is interrupted at 53% of its peak value. The stray capacitance 2500 μF and the inductance is 30 H.
    Solution
    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 × Iμ = 0.53 × √2 × 13.12 = 9.83 A

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

    \(V = i√ {\frac{L}{C}} \)\(= 9.83 × \sqrt {\frac{{30}}{{2500 × {{10}^{ - 6}}}}} = 1077.24\;V\)

  • Question 12
    2 / -0.33
    A 50 Hz, 4 pole, 500 MVA, 22 kV turbo generator is delivering rated MVA at 0.6 pf. Suddenly a 3.-ph fault occurs at the generator terminal, which will reduce the electrical output by 40% and assume constant power input to the shaft. The accelerating torque of the generator in M(mega) N-m at the fault will be ______ (up to three decimal places)
    Solution
    Concept:

    Accelerating torque = (Accelerating power, Pa)/(Synchronous speed, ωs)

    Pa 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 (Ps1) = electrical output (Pe1) = 500 × 0.6 × 106 = 300 MW

    During fault:

    Electrical output, Pe2 = 0.6 × 400 × 106 = 240 MW

    Mechanical input, Ps2 = 300 MW

    Accelerating power, Pa = Ps2 – Pe2 = 300 – 240 = 60 MW

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

    ⇒ ωs = 157.07 rad/sec

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

  • Question 13
    2 / -0.33

    A three phase, 2 pole, 11 kV 10000 kVA alternator has neutral earthed through a resistance of 7 Ω. The machine has current balance protection which operates upon out of balance current that exceed 20% of full load. The % of winding protected against earth fault is _______

    Neglect the % reactance of generator winding.
    Solution
    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= 88.43%

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