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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?
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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**.
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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
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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?
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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?.
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