Power systems miscellaneous

Power systems miscellaneous

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Power Systems

Power systems miscellaneous
  1. Bundled conductors are employed to








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    NA

    Correct Option: C

    NA

  1. For a 500 Hz frequency excitation, a 50 km long power line will be modelled as








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    NA

    Correct Option: C

    NA

  1. The insulation level of 400 kV EHV overhead transmission line is decided on the basis of








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    NA

    Correct Option: B

    NA

  1. A 3φ, 50 Hz, 132 kV transmission line consists of conduct or of 1.17 cm diameter and spaced equilaterely at a distance of 3m. The line conductors have smooth surface with value for m = 0.96. The barometric pressure is 72 cm of Hg and temperature of 20°C. The corona loss in foul weather condition will be








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    Diameter =1.17 cm; r = 0.585 cm; d = 3 m

    ∴
    d
    		 =
    300
    		 = 512.8
    r0.585

    or √
    r
    		 = √
    1
    		 = 0.044
    d512.8

    δ =
    3.92 × b
    		 =
    3.92 × 72
    		 = 0.96
    273 + t293

    V =
    132
    		 = 76.21 kV/phase
    √3

    ∴ Vd = 21.1 × 0.96 × 0.96 × 0.585 ln (512.8)
    = 70.98 kV (line to neutral)
    Corona loss under foul weather condition will be when the disruptive voltage is taken as
    Vd = 0.8 × fair weather value
    = 0.8 × 70.98 = 56.78
    PC = 241 × 10–5
    ƒ + 25
    		 × √
    r
    		(V - Vd)
    6d

    = 241 × 10–5 ×
    75
    		 × √
    1
    		(76.21 - 56.78)
    0.96512.8

    = 3.12 k W/k m/phase

    Correct Option: A

    Diameter =1.17 cm; r = 0.585 cm; d = 3 m

    ∴
    d
    		 =
    300
    		 = 512.8
    r0.585

    or √
    r
    		 = √
    1
    		 = 0.044
    d512.8

    δ =
    3.92 × b
    		 =
    3.92 × 72
    		 = 0.96
    273 + t293

    V =
    132
    		 = 76.21 kV/phase
    √3

    ∴ Vd = 21.1 × 0.96 × 0.96 × 0.585 ln (512.8)
    = 70.98 kV (line to neutral)
    Corona loss under foul weather condition will be when the disruptive voltage is taken as
    Vd = 0.8 × fair weather value
    = 0.8 × 70.98 = 56.78
    PC = 241 × 10–5
    ƒ + 25
    		 × √
    r
    		(V - Vd)
    6d

    = 241 × 10–5 ×
    75
    		 × √
    1
    		(76.21 - 56.78)
    0.96512.8

    = 3.12 k W/k m/phase

  1. What is the corona loss of a 3 – φ line, 160 km long, conductor diameter 1.036 cm, 2.44 delta spacing, air temperature 26°, corresponding to a barometric pressure of 72 cm, operating voltage 110 kV at 50 Hz. m = 0.78?








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    Here, L = 160 km; diameter = 1.036;
    d = 2.44 m; r = 0.518 cm = 0.518 × 10-2m

    δ =
    3.92 b
    		 =
    3.92 × 72
    		 = 0.943
    273 + t273 + 26

    ∴ Vd = 21.1 × 0.78 × 0.943 × 0.518
    In
    2.44
    		 × 100 = 8.039 × In
    244
    		= 49.47 (Line to neutral) kV.
    0.5180.518

    Power Loss
    = 241 × 10-5
    ƒ + 25
    		 × √
    r
    		(V - Vd)²kW/phase/km.
    δd

    = 241 × 105
    50 + 25
    		 × √
    0.518
    		 × 100(63.5 - 9.4)²
    0.9432.44

    = (0.1916 × 0.046 × 198.81) = 1.75 kW/phase/km.
    = 5.25 kW/km.
    = 841 kW for three phases.

    Correct Option: D

    Here, L = 160 km; diameter = 1.036;
    d = 2.44 m; r = 0.518 cm = 0.518 × 10-2m

    δ =
    3.92 b
    		 =
    3.92 × 72
    		 = 0.943
    273 + t273 + 26

    ∴ Vd = 21.1 × 0.78 × 0.943 × 0.518
    In
    2.44
    		 × 100 = 8.039 × In
    244
    		= 49.47 (Line to neutral) kV.
    0.5180.518

    Power Loss
    = 241 × 10-5
    ƒ + 25
    		 × √
    r
    		(V - Vd)²kW/phase/km.
    δd

    = 241 × 105
    50 + 25
    		 × √
    0.518
    		 × 100(63.5 - 9.4)²
    0.9432.44

    = (0.1916 × 0.046 × 198.81) = 1.75 kW/phase/km.
    = 5.25 kW/km.
    = 841 kW for three phases.