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During June-2019, the plant has recorded a maximum demand of 600 kVA and
average PF is observed to be 0.82 lag, the minimum average PF to be maintained
is 0.92 lag as per the independent utility supplier and every one % dip in PF
attracts a penalty of Rs 10,000/in each month. Calculate new kVA and the
improvement in PF for July-2019 by installing 100 kVAr capacitors.​?
Most Upvoted Answer
During June-2019, the plant has recorded a maximum demand of 600 kVA a...
Calculation of New kVA:
1. Given data:
- Maximum demand in June 2019: 600 kVA
- Average power factor observed: 0.82 lag
- Minimum average power factor required: 0.92 lag
- Penalty for each 1% dip in power factor: Rs 10,000 per month

2. To calculate the new kVA, we need to consider the power factor improvement and the penalty cost.
- The power factor improvement will be calculated by finding the difference between the observed power factor and the required power factor.
- The penalty cost will be calculated by multiplying the difference in power factor by the penalty rate.

3. Calculation of power factor improvement:
- Power factor improvement = Required power factor - Observed power factor
- Power factor improvement = 0.92 lag - 0.82 lag
- Power factor improvement = 0.1 lag

4. Calculation of penalty cost:
- Penalty cost = Power factor improvement * Penalty rate
- Penalty cost = 0.1 lag * Rs 10,000
- Penalty cost = Rs 1,000

5. Calculation of new kVA:
- New kVA = Maximum demand / Power factor improvement
- New kVA = 600 kVA / 0.1 lag
- New kVA = 6000 kVA

Installation of 100 kVAr Capacitors:
1. Capacitors are used to improve the power factor of the system. By installing capacitors, the reactive power can be compensated, leading to an improvement in the power factor.

2. The capacitors will provide a leading reactive power, which will reduce the lagging reactive power in the system and improve the power factor.

3. The installation of 100 kVAr capacitors will provide a leading reactive power of 100 kVAr.

4. The new power factor can be calculated by considering the improved reactive power provided by the capacitors.

5. Calculation of new power factor:
- New power factor = (Real power) / (Apparent power)
- Apparent power = kVA
- Real power = Apparent power * Power factor
- Real power = 6000 kVA * 0.92 lag
- Real power = 5520 kW
- New power factor = 5520 kW / 6000 kVA
- New power factor = 0.92 lag

Conclusion:
By installing 100 kVAr capacitors, the power factor can be improved from the observed value of 0.82 lag to the required minimum value of 0.92 lag. The new kVA required for July 2019 will be 6000 kVA. The installation of capacitors will help in reducing the lagging reactive power and improving the power factor, thereby avoiding penalties imposed by the utility supplier.
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During June-2019, the plant has recorded a maximum demand of 600 kVA andaverage PF is observed to be 0.82 lag, the minimum average PF to be maintainedis 0.92 lag as per the independent utility supplier and every one % dip in PFattracts a penalty of Rs 10,000/in each month. Calculate new kVA and theimprovement in PF for July-2019 by installing 100 kVAr capacitors.​?
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During June-2019, the plant has recorded a maximum demand of 600 kVA andaverage PF is observed to be 0.82 lag, the minimum average PF to be maintainedis 0.92 lag as per the independent utility supplier and every one % dip in PFattracts a penalty of Rs 10,000/in each month. Calculate new kVA and theimprovement in PF for July-2019 by installing 100 kVAr capacitors.​? 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 During June-2019, the plant has recorded a maximum demand of 600 kVA andaverage PF is observed to be 0.82 lag, the minimum average PF to be maintainedis 0.92 lag as per the independent utility supplier and every one % dip in PFattracts a penalty of Rs 10,000/in each month. Calculate new kVA and theimprovement in PF for July-2019 by installing 100 kVAr capacitors.​? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for During June-2019, the plant has recorded a maximum demand of 600 kVA andaverage PF is observed to be 0.82 lag, the minimum average PF to be maintainedis 0.92 lag as per the independent utility supplier and every one % dip in PFattracts a penalty of Rs 10,000/in each month. Calculate new kVA and theimprovement in PF for July-2019 by installing 100 kVAr capacitors.​?.
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