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An industrial consumer has a load of 1500 kW at a 0.8 p.f. lag for 12 hours and 1000 KW upf for 12 hours during the day. The daily load factor is
- a)0.833
- b)0.666
- c)1.5
- d)0.8
Correct answer is option 'A'. Can you explain this answer?
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An industrial consumer has a load of 1500 kW at a 0.8 p.f. lag for 12 ...
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An industrial consumer has a load of 1500 kW at a 0.8 p.f. lag for 12 ...
Daily Load Factor Calculation:
The daily load factor can be calculated by dividing the total energy consumed in a day by the maximum energy that could have been consumed if the load was constant at the maximum demand level.
Step 1: Calculate Energy Consumed
To calculate the energy consumed, we need to calculate the total kilowatt-hours (kWh) for each load and add them together.
Load 1: 1500 kW at a power factor of 0.8 lag for 12 hours
Energy consumed for Load 1 = Power (kW) x Time (hours) = 1500 kW x 12 hours = 18000 kWh
Load 2: 1000 kW at unity power factor (upf) for 12 hours
Energy consumed for Load 2 = Power (kW) x Time (hours) = 1000 kW x 12 hours = 12000 kWh
Total energy consumed = Energy consumed for Load 1 + Energy consumed for Load 2 = 18000 kWh + 12000 kWh = 30000 kWh
Step 2: Calculate Maximum Energy
The maximum energy that could have been consumed if the load was constant at the maximum demand level is calculated by multiplying the maximum demand level by the total time period.
Maximum demand level = 1500 kW (maximum load during Load 1)
Total time period = 24 hours (12 hours for each load)
Maximum energy = Maximum demand level x Total time period = 1500 kW x 24 hours = 36000 kWh
Step 3: Calculate Load Factor
Load factor = Total energy consumed / Maximum energy
Load factor = 30000 kWh / 36000 kWh = 0.8333 (rounded to 4 decimal places)
Conclusion:
The daily load factor is 0.833, which is option 'A'. This means that the industrial consumer is utilizing about 83.33% of the maximum energy that could have been consumed if the load was constant at the maximum demand level throughout the day.
The daily load factor can be calculated by dividing the total energy consumed in a day by the maximum energy that could have been consumed if the load was constant at the maximum demand level.
Step 1: Calculate Energy Consumed
To calculate the energy consumed, we need to calculate the total kilowatt-hours (kWh) for each load and add them together.
Load 1: 1500 kW at a power factor of 0.8 lag for 12 hours
Energy consumed for Load 1 = Power (kW) x Time (hours) = 1500 kW x 12 hours = 18000 kWh
Load 2: 1000 kW at unity power factor (upf) for 12 hours
Energy consumed for Load 2 = Power (kW) x Time (hours) = 1000 kW x 12 hours = 12000 kWh
Total energy consumed = Energy consumed for Load 1 + Energy consumed for Load 2 = 18000 kWh + 12000 kWh = 30000 kWh
Step 2: Calculate Maximum Energy
The maximum energy that could have been consumed if the load was constant at the maximum demand level is calculated by multiplying the maximum demand level by the total time period.
Maximum demand level = 1500 kW (maximum load during Load 1)
Total time period = 24 hours (12 hours for each load)
Maximum energy = Maximum demand level x Total time period = 1500 kW x 24 hours = 36000 kWh
Step 3: Calculate Load Factor
Load factor = Total energy consumed / Maximum energy
Load factor = 30000 kWh / 36000 kWh = 0.8333 (rounded to 4 decimal places)
Conclusion:
The daily load factor is 0.833, which is option 'A'. This means that the industrial consumer is utilizing about 83.33% of the maximum energy that could have been consumed if the load was constant at the maximum demand level throughout the day.
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An industrial consumer has a load of 1500 kW at a 0.8 p.f. lag for 12 hours and 1000 KW upf for 12 hours during the day. The daily load factor isa)0.833b)0.666c)1.5d)0.8Correct answer is option 'A'. Can you explain this answer? for Electrical Engineering (EE) 2026 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about An industrial consumer has a load of 1500 kW at a 0.8 p.f. lag for 12 hours and 1000 KW upf for 12 hours during the day. The daily load factor isa)0.833b)0.666c)1.5d)0.8Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An industrial consumer has a load of 1500 kW at a 0.8 p.f. lag for 12 hours and 1000 KW upf for 12 hours during the day. The daily load factor isa)0.833b)0.666c)1.5d)0.8Correct answer is option 'A'. Can you explain this answer?.
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