Power systems miscellaneous

Power systems miscellaneous

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Power Systems

Power systems miscellaneous
  1. The A, B, C, D constants of a 220 kV line are:
    A = D = 0.94 ∠1°,
    B = 130 ∠73°,
    C = 0.001 ∠90°
    If sending end voltage of the line for a given load delivered at nominal voltage is 240 kV, then % voltage regulation of the line is








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    Voltage at sending-end,
    Vs = AVR + BIR
    at no-load,
    IR = 0
    and

    VRO =
    Vs
    		 =
    240
    A0.94∠1°

    = 255.32 Volts
    % voltage regulation =
    VRO - VRL
    		 × 100
    VRL

    Where VRL is the full-load voltage at receiving-end
    =
    255032 - 220
    		 × 100 = 16%
    220

    Correct Option: C

    Voltage at sending-end,
    Vs = AVR + BIR
    at no-load,
    IR = 0
    and

    VRO =
    Vs
    		 =
    240
    A0.94∠1°

    = 255.32 Volts
    % voltage regulation =
    VRO - VRL
    		 × 100
    VRL

    Where VRL is the full-load voltage at receiving-end
    =
    255032 - 220
    		 × 100 = 16%
    220

  1. A generator is connected t hr ough a 20 MVA, 13.8/138 kV step down transformer, to a transmission line. At the receiving end of the line a load is supplied through a step down transformer of 10 M VA, 138/69 kV rating. A 0.72 puload, evaluated on load side transformer ratings as base values, is supplied from the above system. For system base values of 10 MVA and 69 kV in load circuit, the value of the load (in per unit) in generator circuit will be








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    Base impedance on load side, ZBL

    =
    (69 kV)²
    		 = 476 Ω
    (10 MVA)

    Then, value of load,
    ZL = ZL (p.v.) × ZBL
    = 0.72 × 476 = 342.8 Ω
    Base impedance on generator side,
    ZBG =
    (13.8 kV)²
    		 = 9.522 Ω
    (20 MVA)

    Then, in p.u. impedenace
    Zp.u = =
    ZL
    		 =
    342.8
    		 = 36p.u
    ZBG9.522

    Correct Option: A


    Base impedance on load side, ZBL

    =
    (69 kV)²
    		 = 476 Ω
    (10 MVA)

    Then, value of load,
    ZL = ZL (p.v.) × ZBL
    = 0.72 × 476 = 342.8 Ω
    Base impedance on generator side,
    ZBG =
    (13.8 kV)²
    		 = 9.522 Ω
    (20 MVA)

    Then, in p.u. impedenace
    Zp.u = =
    ZL
    		 =
    342.8
    		 = 36p.u
    ZBG9.522

  1. Three identical star connected resist or s of 1.0 pu are connected to an unbalanced 3 phase supply. The load neutral is isolated. The symmetrical components of the line voltages in pu are:
    Vab1 = X ∠1, Vab2 = Y∠2.
    If all the pu calculations are with the respective base values, the phase to neutr al sequence voltages are








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    Assuming Va as reference phasor,
    Van1 = V∠0°
    Vbn1 = V∠– 120°
    Vab1 = Van1 - Vbn1
    = V∠0° – V∠– 120°
    = √3Van1 ∠30°
    Given, Vab1 = X ∠θ2
    Then, Y ∠θ2 = √3 Van2 ∠ - 30°
    ⇒ Van2 = Y√3 ∠θ2 + 30°

    Correct Option: C


    Assuming Va as reference phasor,
    Van1 = V∠0°
    Vbn1 = V∠– 120°
    Vab1 = Van1 - Vbn1
    = V∠0° – V∠– 120°
    = √3Van1 ∠30°
    Given, Vab1 = X ∠θ2
    Then, Y ∠θ2 = √3 Van2 ∠ - 30°
    ⇒ Van2 = Y√3 ∠θ2 + 30°

  1. A single phase transmission line and a telephone line are both symmetrically strung one below the other, in horizontal configurations, on a common t ower. The shor t est and l ongest distances between the phase and telephone conductors are 2.5 m and 3 m respectively. The volt age (volt /km) induced in the telephone circuit, due to 50 Hz current of 100 amps in the power circuit is








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    Mutual inductance between lower line and telephone line,

    M = 4 × 10-7 In
    DP1T2
    		 H/m
    DP1T1

    = 4 × 10-7 In
    3
    		 H/m
    2.5

    = 0.73 × 10–7 H/m
    Hence voltage induced by power line into the telephone line,
    |VT| = ωMI
    = 314 × (0.73 × 10–7) × 100
    = 2.29 H/km

    Correct Option: C


    Mutual inductance between lower line and telephone line,

    M = 4 × 10-7 In
    DP1T2
    		 H/m
    DP1T1

    = 4 × 10-7 In
    3
    		 H/m
    2.5

    = 0.73 × 10–7 H/m
    Hence voltage induced by power line into the telephone line,
    |VT| = ωMI
    = 314 × (0.73 × 10–7) × 100
    = 2.29 H/km

  1. The Gauss Seidel load flow method has following disadvantages. Tick the incorrect statement.








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    The time required for a single it eration in G-S method is smaller with respect to NewtonRaphson method, but no. of iteration is quite large as it is proportional to the size of system. Also, convergence property also gets affected by the selection of slack bus.
    1.

    =
    dC1
    		 = 2 × 0.01P1 + 30 = 0.02 P1 + 30
    dP2

    =
    dC2
    		 = 2 × 0.05P2 + 10 = 0.01 P2 + 10
    dP2

    =
    dC1
    		 =
    dC2
    dP1dP2

    = 0.02P1 + 30 = 30 = 30 = 0.1 P2 + 10
    2P1 + 3000 = 10P2 + 1000
    2P1 + 2000 = 10P2
    P1 + P2 = 200
    P2 = 200
    P1 = 0
    =
    dC1
    		 = 20 Rs/Mwh
    dP1

    Correct Option: D

    The time required for a single it eration in G-S method is smaller with respect to NewtonRaphson method, but no. of iteration is quite large as it is proportional to the size of system. Also, convergence property also gets affected by the selection of slack bus.
    1.

    =
    dC1
    		 = 2 × 0.01P1 + 30 = 0.02 P1 + 30
    dP2

    =
    dC2
    		 = 2 × 0.05P2 + 10 = 0.01 P2 + 10
    dP2

    =
    dC1
    		 =
    dC2
    dP1dP2

    = 0.02P1 + 30 = 30 = 30 = 0.1 P2 + 10
    2P1 + 3000 = 10P2 + 1000
    2P1 + 2000 = 10P2
    P1 + P2 = 200
    P2 = 200
    P1 = 0
    =
    dC1
    		 = 20 Rs/Mwh
    dP1