Electrical Engineering (EE) Exam > Electrical Engineering (EE) Questions > A load of 460 kVA draws power from 480 V, 3-...
Start Learning for Free
A load of 460 kVA draws power from 480 V, 3-Ф network at a power factor of 0.87 lag. Power factor has been improved to 0.97 lag by installing a capacitor at terminals of the load. Charges for power are 4.75 Rs/kVAhr, then the savings per annum after installing the capacitor is ______ Rs. (rounded upto two decimal places)
Correct answer is '1981884.3'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
A load of 460 kVA draws power from 480 V, 3-Ф network at a power fact...
Understanding the Load and Power Factor
The load of 460 kVA operates at a voltage of 480 V with a power factor (PF) of 0.87 lagging. The real power (kW) can be calculated using the formula:
- **Real Power (kW) = kVA × PF**
- **Real Power (kW) = 460 kVA × 0.87 = 400.2 kW**
Calculating Apparent Power and Reactive Power
At the initial power factor of 0.87, we can also find the reactive power (kVAR):
- **kVAR = √(kVA² - kW²)**
- **kVAR = √(460² - 400.2²) ≈ 189.98 kVAR**
After installing the capacitor, the power factor improves to 0.97. The new real power remains the same (400.2 kW), but the new apparent power can be calculated as:
- **New Apparent Power (kVA) = kW / PF**
- **New Apparent Power (kVA) = 400.2 kW / 0.97 ≈ 412.37 kVA**
Now, the new reactive power can be found:
- **New kVAR = √(kVA² - kW²)**
- **New kVAR = √(412.37² - 400.2²) ≈ 90.48 kVAR**
Determining the Capacitor Size
The required capacitor size to offset the reactive power can be calculated as:
- **Capacitor Size (kVAR) = Initial kVAR - New kVAR**
- **Capacitor Size = 189.98 kVAR - 90.48 kVAR = 99.5 kVAR**
Cost Savings Calculation
The charge for power is 4.75 Rs/kVAhr. The annual savings can be calculated based on the difference in kVA usage:
- **Annual kVAhr Savings = (Initial kVA - New kVA) × Hours/year**
- **Annual kVAhr Savings = (460 - 412.37) × 8760 ≈ 412.63 kVAhr**
Now, the annual savings in Rs:
- **Annual Savings = Annual kVAhr Savings × Charge per kVAhr**
- **Annual Savings = 412.63 kVAhr × 4.75 Rs/kVAhr ≈ 1981884.3 Rs**
Thus, the savings per annum after installing the capacitor is approximately **1981884.3 Rs**.
The load of 460 kVA operates at a voltage of 480 V with a power factor (PF) of 0.87 lagging. The real power (kW) can be calculated using the formula:
- **Real Power (kW) = kVA × PF**
- **Real Power (kW) = 460 kVA × 0.87 = 400.2 kW**
Calculating Apparent Power and Reactive Power
At the initial power factor of 0.87, we can also find the reactive power (kVAR):
- **kVAR = √(kVA² - kW²)**
- **kVAR = √(460² - 400.2²) ≈ 189.98 kVAR**
After installing the capacitor, the power factor improves to 0.97. The new real power remains the same (400.2 kW), but the new apparent power can be calculated as:
- **New Apparent Power (kVA) = kW / PF**
- **New Apparent Power (kVA) = 400.2 kW / 0.97 ≈ 412.37 kVA**
Now, the new reactive power can be found:
- **New kVAR = √(kVA² - kW²)**
- **New kVAR = √(412.37² - 400.2²) ≈ 90.48 kVAR**
Determining the Capacitor Size
The required capacitor size to offset the reactive power can be calculated as:
- **Capacitor Size (kVAR) = Initial kVAR - New kVAR**
- **Capacitor Size = 189.98 kVAR - 90.48 kVAR = 99.5 kVAR**
Cost Savings Calculation
The charge for power is 4.75 Rs/kVAhr. The annual savings can be calculated based on the difference in kVA usage:
- **Annual kVAhr Savings = (Initial kVA - New kVA) × Hours/year**
- **Annual kVAhr Savings = (460 - 412.37) × 8760 ≈ 412.63 kVAhr**
Now, the annual savings in Rs:
- **Annual Savings = Annual kVAhr Savings × Charge per kVAhr**
- **Annual Savings = 412.63 kVAhr × 4.75 Rs/kVAhr ≈ 1981884.3 Rs**
Thus, the savings per annum after installing the capacitor is approximately **1981884.3 Rs**.
Free Test
FREE
| Start Free Test |
Community Answer
A load of 460 kVA draws power from 480 V, 3-Ф network at a power fact...
Real power requirement = 460x0.87 = 400kW
kVA of load at 0.97 lag pf = 400/0.97 = 412.37 kVA
Annual bill paid without capacitor = 460x24x365x4.75 = 19140600 Rs
Annual bill with capacitor = 412.37x24x365x4.75 = 17158715.7 Rs
Savings = 19140600 - 17158715.7 = 1981884.3 Rs
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
|
|
Top Courses for Electrical Engineering (EE)View all
A load of 460 kVA draws power from 480 V, 3-Ф network at a power factor of 0.87 lag. Power factor has been improved to 0.97 lag by installing a capacitor at terminals of the load. Charges for power are 4.75 Rs/kVAhr, then the savings per annum after installing the capacitor is ______ Rs. (rounded upto two decimal places)Correct answer is '1981884.3'. Can you explain this answer?
Question Description
A load of 460 kVA draws power from 480 V, 3-Ф network at a power factor of 0.87 lag. Power factor has been improved to 0.97 lag by installing a capacitor at terminals of the load. Charges for power are 4.75 Rs/kVAhr, then the savings per annum after installing the capacitor is ______ Rs. (rounded upto two decimal places)Correct answer is '1981884.3'. 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 A load of 460 kVA draws power from 480 V, 3-Ф network at a power factor of 0.87 lag. Power factor has been improved to 0.97 lag by installing a capacitor at terminals of the load. Charges for power are 4.75 Rs/kVAhr, then the savings per annum after installing the capacitor is ______ Rs. (rounded upto two decimal places)Correct answer is '1981884.3'. 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 A load of 460 kVA draws power from 480 V, 3-Ф network at a power factor of 0.87 lag. Power factor has been improved to 0.97 lag by installing a capacitor at terminals of the load. Charges for power are 4.75 Rs/kVAhr, then the savings per annum after installing the capacitor is ______ Rs. (rounded upto two decimal places)Correct answer is '1981884.3'. Can you explain this answer?.
A load of 460 kVA draws power from 480 V, 3-Ф network at a power factor of 0.87 lag. Power factor has been improved to 0.97 lag by installing a capacitor at terminals of the load. Charges for power are 4.75 Rs/kVAhr, then the savings per annum after installing the capacitor is ______ Rs. (rounded upto two decimal places)Correct answer is '1981884.3'. 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 A load of 460 kVA draws power from 480 V, 3-Ф network at a power factor of 0.87 lag. Power factor has been improved to 0.97 lag by installing a capacitor at terminals of the load. Charges for power are 4.75 Rs/kVAhr, then the savings per annum after installing the capacitor is ______ Rs. (rounded upto two decimal places)Correct answer is '1981884.3'. 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 A load of 460 kVA draws power from 480 V, 3-Ф network at a power factor of 0.87 lag. Power factor has been improved to 0.97 lag by installing a capacitor at terminals of the load. Charges for power are 4.75 Rs/kVAhr, then the savings per annum after installing the capacitor is ______ Rs. (rounded upto two decimal places)Correct answer is '1981884.3'. Can you explain this answer?.
Solutions for A load of 460 kVA draws power from 480 V, 3-Ф network at a power factor of 0.87 lag. Power factor has been improved to 0.97 lag by installing a capacitor at terminals of the load. Charges for power are 4.75 Rs/kVAhr, then the savings per annum after installing the capacitor is ______ Rs. (rounded upto two decimal places)Correct answer is '1981884.3'. 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 A load of 460 kVA draws power from 480 V, 3-Ф network at a power factor of 0.87 lag. Power factor has been improved to 0.97 lag by installing a capacitor at terminals of the load. Charges for power are 4.75 Rs/kVAhr, then the savings per annum after installing the capacitor is ______ Rs. (rounded upto two decimal places)Correct answer is '1981884.3'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A load of 460 kVA draws power from 480 V, 3-Ф network at a power factor of 0.87 lag. Power factor has been improved to 0.97 lag by installing a capacitor at terminals of the load. Charges for power are 4.75 Rs/kVAhr, then the savings per annum after installing the capacitor is ______ Rs. (rounded upto two decimal places)Correct answer is '1981884.3'. Can you explain this answer?, a detailed solution for A load of 460 kVA draws power from 480 V, 3-Ф network at a power factor of 0.87 lag. Power factor has been improved to 0.97 lag by installing a capacitor at terminals of the load. Charges for power are 4.75 Rs/kVAhr, then the savings per annum after installing the capacitor is ______ Rs. (rounded upto two decimal places)Correct answer is '1981884.3'. Can you explain this answer? has been provided alongside types of A load of 460 kVA draws power from 480 V, 3-Ф network at a power factor of 0.87 lag. Power factor has been improved to 0.97 lag by installing a capacitor at terminals of the load. Charges for power are 4.75 Rs/kVAhr, then the savings per annum after installing the capacitor is ______ Rs. (rounded upto two decimal places)Correct answer is '1981884.3'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A load of 460 kVA draws power from 480 V, 3-Ф network at a power factor of 0.87 lag. Power factor has been improved to 0.97 lag by installing a capacitor at terminals of the load. Charges for power are 4.75 Rs/kVAhr, then the savings per annum after installing the capacitor is ______ Rs. (rounded upto two decimal places)Correct answer is '1981884.3'. 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
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup