Electrical Engineering (EE) Exam > Electrical Engineering (EE) Questions > An industrial consumer has a load of 1500 kW ...
Start Learning for Free
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?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
An industrial consumer has a load of 1500 kW at a 0.8 p.f. lag for 12 ...
Most Upvoted Answer
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.
Attention Electrical Engineering (EE) Students!
To make sure you are not studying endlessly, EduRev has designed Electrical Engineering (EE) study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Electrical Engineering (EE).
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Similar Electrical Engineering (EE) Doubts
Top Courses for Electrical Engineering (EE)View all
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?
Question Description
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) 2024 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) 2024 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?.
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) 2024 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) 2024 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?.
Solutions 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? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free.
Here you can find the meaning of 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? defined & explained in the simplest way possible. Besides giving the explanation of
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?, a detailed solution 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? has been provided alongside types of 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? theory, EduRev gives you an
ample number of questions to practice 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? tests, examples and also practice Electrical Engineering (EE) tests.
|
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup