Power systems miscellaneous


  1. What is the maximum value of a load which consume 500 kWh per day at a load factor of 0.40, if the consumer increases the load factor of 0.50 without increasing the maximum demand?









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    Load factor =
    Energy consumed in 24 hours
    Maximum demand in kW × 24

    or 0.40 =
    500
    Maximum demand × 24

    Maximum demand =
    500
    = 52.08kW
    0.40 × 24

    Correct Option: A

    Load factor =
    Energy consumed in 24 hours
    Maximum demand in kW × 24

    or 0.40 =
    500
    Maximum demand × 24

    Maximum demand =
    500
    = 52.08kW
    0.40 × 24


  1. The yearly load duration curve of a power plant is a straight line. The maximum load is 750 MW and the minimum load is 600 MW. The capacity of the plant is 900 MW. What is the capacity factor and utilization factor?









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    Average Annual load =
    750 + 600
    = 675 MW
    2

    Capacity factor =
    Average Annual load
    =
    675
    = 0.75
    Capacity of the Plant900

    Utilization factor =
    Maximum demand
    =
    750
    = 0.83
    Capacity of the plant900

    Correct Option: B

    Average Annual load =
    750 + 600
    = 675 MW
    2

    Capacity factor =
    Average Annual load
    =
    675
    = 0.75
    Capacity of the Plant900

    Utilization factor =
    Maximum demand
    =
    750
    = 0.83
    Capacity of the plant900



  1. What is the utilization factor of a power station which supplies the following loads ?
    Load A: Motor load of 200 kW between 10 AM to 7 PM
    Load B: Lighting load of 100 k W bet ween 7 PM to 11 PM
    Load C: Pumping load of 110 kW between 3 PM to 10 AM









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    Sum of individual maximum demands
    = (200 + 100 + 110) kW = 410 kW.

    TimeLoad
    10 AM to 3 PM200 kW
    3 PM to 7 PM(200 + 110) kW = 310 kW
    7 PM to 11 PM(100 + 110) kW = 210 kW
    11 PM to 10 AM   110 kW

    Maximum demand of the whole system is 310 kW. Utiligation factor
    =
    Sum of individual maximum demand
    =
    410
    = 1.32
    Maximum demand of whole system132

    Correct Option: B

    Sum of individual maximum demands
    = (200 + 100 + 110) kW = 410 kW.

    TimeLoad
    10 AM to 3 PM200 kW
    3 PM to 7 PM(200 + 110) kW = 310 kW
    7 PM to 11 PM(100 + 110) kW = 210 kW
    11 PM to 10 AM   110 kW

    Maximum demand of the whole system is 310 kW. Utiligation factor
    =
    Sum of individual maximum demand
    =
    410
    = 1.32
    Maximum demand of whole system132


  1. A power station supplies the peak load of 50 MW, 40 MW and 70 MW to three localities. The annual load factor is 0.50 p.u. and the diversity factor of the load at the station is 1.55. The maximum demand on the station and average load respectively will be









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    Diversity factor =
    Sum of individual maximum demand
    Maximum demand on the station

    or 1.55 =
    (50 + 40 + 70) MW
    maximum demand on the station

    ∴ Maximum demand on the station =
    160 MW
    = 103.22 MW
    1.55

    Average load = load factor × maximum demand
    = (0.50 × 103.22) MW = 51.61 MW

    Correct Option: C

    Diversity factor =
    Sum of individual maximum demand
    Maximum demand on the station

    or 1.55 =
    (50 + 40 + 70) MW
    maximum demand on the station

    ∴ Maximum demand on the station =
    160 MW
    = 103.22 MW
    1.55

    Average load = load factor × maximum demand
    = (0.50 × 103.22) MW = 51.61 MW



  1. In a DC transmission line









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    NA

    Correct Option: C

    NA