Power systems miscellaneous

Power systems miscellaneous

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Power Systems

Power systems miscellaneous
  1. A single line to ground fault occurs on an unloaded generator in phase a. If xd = x2 = 0.25 p.u., x0 = 0.15 p.u., reactance connected in the neutral, xn = 0.05 p.u. and the initial prefault voltage is 1.0 p.u., then the magnitude of the fault current will be ________ p.u.








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    NA

    Correct Option: C

    NA

  1. A consumer has a maximum demand of 60 kW at 0.50 load factor. If tariff is ₹ 750 per kW of maximum demand plus ₹ 1.20 per kWh, then overall cost per kWh, will be _________








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    Annual energy consumption
    = 60 × 0.50 × 8760 = 262800 kWh.
    Cost of annual energy consumption, CE = (1.20) × 262800 = Rs. 315360
    Fixed charge per year, CF = Rs. 750 × 67 = Rs. 45000
    Total annual charge = CE + CF = Rs. (315360 + 45000) = Rs. 360360

    Overall cost per kWh =
    360360
    		 = 1.37
    262800

    Correct Option: A

    Annual energy consumption
    = 60 × 0.50 × 8760 = 262800 kWh.
    Cost of annual energy consumption, CE = (1.20) × 262800 = Rs. 315360
    Fixed charge per year, CF = Rs. 750 × 67 = Rs. 45000
    Total annual charge = CE + CF = Rs. (315360 + 45000) = Rs. 360360

    Overall cost per kWh =
    360360
    		 = 1.37
    262800

  1. A power system has two generating plants and the power is being dispatched economically with P1 = 125 MW and P2 = 250 M W. The loss coefficients are B11 = 0.10 × 102 MW–1 , B12 = – 0.01 × 10–2 MW–1, B22 = 0.13 × 10–2 MW–1to raise the total load on the system by 1 MW will cost an additional ₹ 200 per hour. The penalty factor for plant 1 will be __________








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    PL = P²1 B11 + 2P1 P2 B12 + P²2 B22

    ∂P1
    		 = 2P1B11 + 2P2B22 = 0
    ∂P1

    = 2 × 125 × 0.10 × 10–2 + 2 × 250 × (– 0.01 × 10–2) + 0 = (0.250 – 0.05) = 0.20
    Putting factor for Plant 1,
    L1 =
    1
    		 =
    1
    		 = 1.25
    1 -
    ∂PL
    		(1 - 0.20)
    ∂P1

    Correct Option: C

    PL = P²1 B11 + 2P1 P2 B12 + P²2 B22

    ∂P1
    		 = 2P1B11 + 2P2B22 = 0
    ∂P1

    = 2 × 125 × 0.10 × 10–2 + 2 × 250 × (– 0.01 × 10–2) + 0 = (0.250 – 0.05) = 0.20
    Putting factor for Plant 1,
    L1 =
    1
    		 =
    1
    		 = 1.25
    1 -
    ∂PL
    		(1 - 0.20)
    ∂P1

  1. A 3 – φ supply, 5 kW induction motor has a power fact or 0.75 lagging. A bank of capacitors is connected in delta across the supply terminals and power factor raised to 0.9 lagging. kVAR rating of the capacitors connected in each phase, will be _________ kVAr








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    p.f., cos φ1 = 0.75
    ⇒ φ1 = cos–1 (0.75) = 41°24'
    and tan φ1 = tan 41°24' = 0.8819
    Final power factor, cos φ2 = 0.9 (lagging)
    ⇒ φ2 = cos–1 (0.9) = 25°50'
    and tan φ2 = tan 25°50' = 0.4843
    Motor input, P = 5 k watt.
    Efficiency, η = 100% (assuming)
    Leading kVAr supplied by the condenser bank
    = P (tan φ1 – tan φ2)
    = 5 (0.8819 – 0.4813)
    = 1.99 kVAr.
    Rating of capacitors connected in each phase

    =
    1.99
    		 = 0.663 kVAr
    3

    Correct Option: A

    p.f., cos φ1 = 0.75
    ⇒ φ1 = cos–1 (0.75) = 41°24'
    and tan φ1 = tan 41°24' = 0.8819
    Final power factor, cos φ2 = 0.9 (lagging)
    ⇒ φ2 = cos–1 (0.9) = 25°50'
    and tan φ2 = tan 25°50' = 0.4843
    Motor input, P = 5 k watt.
    Efficiency, η = 100% (assuming)
    Leading kVAr supplied by the condenser bank
    = P (tan φ1 – tan φ2)
    = 5 (0.8819 – 0.4813)
    = 1.99 kVAr.
    Rating of capacitors connected in each phase

    =
    1.99
    		 = 0.663 kVAr
    3

  1. A 10 kVA, 400 V/200 V single phase transformer with 10% impedance draws a steady short circuit line current of ___________ A








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    NA

    Correct Option: C

    NA