--- title: "PF improvement of 490 kW load at 6.6 kV" subject: "Industrial Power Distribution and Illumination" chapter: 10 key: "semester_6/industrial_power_distribution_and_illumination:pf-490kw-6-6kv-payback" source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/pf-490kw-6-6kv-payback site: BCT Engineering --- # PF improvement of 490 kW load at 6.6 kV Industrial Power Distribution and Illumination · Chapter 10: Electrical Energy Audit in Industry - Asked in: 2071 Bhadra, 5 marks - Video: [Capacitor Bank Capacity for improve Power Factor | power factor correction capacitor formula](https://www.youtube.com/watch?v=om-O0Km7rTg) — Electrical Dost - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/pf-490kw-6-6kv-payback - Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-10 ## Question The p.f. of an industrial three-phase load of 490 kW is to be improved from 0.7 lagging to 0.97 lagging by connecting loss free delta connected capacitor across 6.6 kV, 50 Hz supply. The cost of kVAr controlling device with switching is NRs. 1200.00 per kVAr. Calculate: (i) the total kVAr required, (ii) the required value of capacitance and (iii) payback period. ## Short answer Total kVAr required: 377.1 kVAR (P tan φ1 – P tan φ2). Capacitance per phase in delta: 9.19 μF. Payback ≈ 0.97 year (≈ 11.6 months) assuming Rs 200/kVA/month demand charge. ## Answer Capacitors connected in delta across the 6.6 kV supply provide the leading kVAR needed to raise the pf from 0.7 to 0.97. #### (i) Total kVAR required ```text Q1 = P tan φ1 = 490 × tan(cos⁻¹0.7) = 490 × 1.0202 = 499.90 kVAR Q2 = P tan φ2 = 490 × tan(cos⁻¹0.97) = 490 × 0.2506 = 122.81 kVAR Qc = Q1 − Q2 = 377.09 kVAR ``` #### (ii) Capacitance (delta connection) In delta, each capacitor sees the line voltage of 6.6 kV and supplies one-third of Qc. ```text Qc per phase = 377.09 / 3 = 125.70 kVAR Qph = V² ω C → C = Qph / (2π f V²) C = 125 698 / (2π × 50 × 6600²) = 125 698 / (314.16 × 43 560 000) = 9.19 × 10⁻⁶ F = 9.19 µF per phase ``` #### (iii) Payback period The question does not give the demand tariff, so assume a kVA demand charge of **Rs 200/kVA/month** (as given in the companion version of this problem). Since the capacitors are loss-free, the kWh consumption does not change; the saving is only in kVA demand. ```text kVA before = 490 / 0.7 = 700.00 kVA kVA after = 490 / 0.97 = 505.15 kVA Reduction = 194.85 kVA Annual saving = 194.85 × 200 × 12 = Rs 467 629 Capital cost = 377.09 × 1200 = Rs 452 513 Payback = 452 513 / 467 629 = 0.968 year ≈ 11.6 months ``` **Answer: Qc ≈ 377.1 kVAR; C ≈ 9.19 µF per phase (delta); payback ≈ 0.97 year (about 11.6 months) with an assumed demand charge of Rs 200/kVA/month.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the total kVAR required for pf correction found? A. Qc = P(tan phi1 - tan phi2) B. Qc = P(cos phi2 - cos phi1) C. Qc = P(tan phi1 + tan phi2) D. Qc = P / tan phi2 Answer: A. The capacitors supply the difference between initial and final lagging kVAR: 499.90 - 122.81 = 377.09 kVAR. (Key point: Capacitor kVAR required) 2. For the delta-connected bank, what voltage does each capacitor see and what kVAR does it carry? A. Phase voltage 3.81 kV, full 377 kVAR B. Line voltage 6.6 kV, one-third of Qc C. Line voltage 6.6 kV, full Qc D. Phase voltage 3.81 kV, one-third of Qc Answer: B. In delta each capacitor is across the line voltage and carries Qc/3 = 125.7 kVAR, so C = Qph/(2 pi f V^2) = 9.19 uF. (Key point: Delta capacitance) 3. Why does the saving come only from kVA demand charge in the payback calculation? A. Capacitors reduce the kW of the load B. Capacitors reduce the energy tariff rate C. Loss-free capacitors leave kW and kWh unchanged but lower kVA D. Capacitors increase the voltage Answer: C. Real power and kWh stay the same, while kVA falls from 700 to 505.15, so only the kVA demand charge is saved. (Key point: Saving basis for payback) 4. With capital Rs 452 513 and annual saving Rs 467 629, what is the payback? A. About 2.1 years B. About 0.5 year C. About 1.5 years D. About 0.97 year (11.6 months) Answer: D. Payback = capital / annual saving = 452 513 / 467 629 = 0.968 year. (Key point: Payback result) ## Close variants set in other exams - [PF improvement of 490 kW load with demand charge](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Apf-490kw-6-6kv-demand-charge) — Industrial Power Distribution and Illumination, chapter 10, asked 2× ## Related in Industrial Power Distribution and Illumination - [PF improvement of 490 kW load at 11 kV](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Apf-490kw-11kv) — Industrial Power Distribution and Illumination, chapter 10 - [PF improvement of 1020 kVA factory load](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Afactory-1020kva-pf-improvement) — Industrial Power Distribution and Illumination, chapter 10, asked 2× ## Related in other subjects - [Delta capacitor per phase for 50 kW motor](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Acapacitor-delta-415v-50kw-0-72) — Basic Electrical Engineering, chapter 8 - [Capacitor for 6 kW load pf improvement](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Ainductive-load-6kw-pf-capacitor) — Basic Electrical Engineering, chapter 7 - [Annual saving by pf improvement for 800 kW load](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Ainstallation-800kw-capacitor-annual-saving) — Utilization of Electrical Energy, chapter 6 - [Capacitor bank for 3-phase motor pf](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Acapacitor-bank-60kw-induction-motor) — Basic Electrical Engineering, chapter 8