Power systems miscellaneous
- 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
- 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 Plant 900 Utilization factor = Maximum demand = 750 = 0.83 Capacity of the plant 900 Correct Option: B
Average Annual load = 750 + 600 = 675 MW 2 Capacity factor = Average Annual load = 675 = 0.75 Capacity of the Plant 900 Utilization factor = Maximum demand = 750 = 0.83 Capacity of the plant 900
- 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.Time Load 10 AM to 3 PM 200 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 system 132 Correct Option: B
Sum of individual maximum demands
= (200 + 100 + 110) kW = 410 kW.Time Load 10 AM to 3 PM 200 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 system 132
- 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 MWCorrect 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
- In a DC transmission line
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NA
Correct Option: C
NA