--- title: "PF improvement of 1020 kVA factory load" subject: "Industrial Power Distribution and Illumination" chapter: 10 key: "semester_6/industrial_power_distribution_and_illumination:factory-1020kva-pf-improvement" source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/factory-1020kva-pf-improvement site: BCT Engineering --- # PF improvement of 1020 kVA factory load Industrial Power Distribution and Illumination · Chapter 10: Electrical Energy Audit in Industry - Asked in: 2073 Bhadra, 2+2+2+2 marks; 2068 Bhadra (old course), 8 marks (asked 2 times) - Video: [Capacitor Bank Sizing (KVAR) for Power Factor Improvement](https://www.youtube.com/watch?v=qrJDMBR7i1s) — ElectricalEngineeringPlanet - 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/factory-1020kva-pf-improvement - Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-10 ## Question A factory takes an average load of 1020 kVA at 0.64 p.f.(lag) from the 1600 kVA transformer. The no-load loss and full load loss of transformer are 2.4kW and 18.57 KW respectively. If the unit cost of capacitor bank per kVAr is Rs.800, maximum demand charge is Rs.190 per kVA and unit charge of energy is Rs.7.50. Calculate: i) kVAR required to improve p.f. to 0.95 ii) Reduction in loss in transformer if the loading duration is 2920 hours per year. iii) Reduction in maximum demand charge iv) Payback Period ## Short answer Required capacitor 569.18 kVAr. Transformer copper loss drops by 4.12 kW (12 036 kWh/yr, Rs 90 267). Maximum demand reduces by 332.84 kVA saving Rs63 240/month. Payback period 0.54 year (6.4 months). ## Answer **Given:** average load 1020 kVA at 0.64 pf (lag); transformer 1600 kVA; no-load loss 2.4 kW; full-load (copper) loss 18.57 kW; capacitor cost Rs 800/kVAr; maximum demand charge Rs 190/kVA (taken as **per month**); energy charge Rs 7.50/kWh; loading 2920 h/year. #### i) kVAr required to improve pf to 0.95 ```text P = 1020 × 0.64 = 652.8 kW Q₁ = 1020 × sin(cos⁻¹0.64) = 1020 × 0.7684 = 783.74 kVAr Q₂ = P × tan(cos⁻¹0.95) = 652.8 × 0.3287 = 214.56 kVAr Qc = Q₁ − Q₂ = 569.18 kVAr New kVA S₂ = 652.8 / 0.95 = 687.16 kVA ``` #### ii) Reduction in transformer loss No-load loss (2.4 kW) does not change. Copper loss ∝ (S/S_rated)²: ```text Cu loss before = 18.57 × (1020/1600)² = 7.547 kW Cu loss after = 18.57 × (687.16/1600)² = 3.425 kW Reduction = 4.122 kW Energy saved = 4.122 × 2920 = 12 035.6 kWh/yr Cost saved = 12 035.6 × 7.50 = Rs 90 266.8/yr ``` #### iii) Reduction in maximum demand charge ```text kVA reduction = 1020 − 687.16 = 332.84 kVA Saving = 332.84 × 190 = Rs 63 240 per month = Rs 758 880 per year ``` #### iv) Payback period ```text Capacitor cost = 569.18 × 800 = Rs 455 342 Annual saving = 758 880 + 90 266.8 = Rs 849 147 Payback = 455 342 / 849 147 = 0.536 year ≈ 6.4 months ``` **Answer:** (i) 569.18 kVAr; (ii) 4.12 kW less loss, 12 036 kWh/yr = Rs 90 267/yr; (iii) Rs 63 240/month (Rs 758 880/yr); (iv) payback ≈ 0.54 year (about 6.4 months). (If the demand charge is taken as a one-time yearly figure, annual saving = 63 240 + 90 267 = Rs 153 507 and payback ≈ 2.97 years.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the kVAr needed to improve pf from 0.64 to 0.95 for 1020 kVA? A. 783.74 kVAr B. 214.56 kVAr C. 652.8 kVAr D. 569.18 kVAr Answer: D. P = 652.8 kW, Q1 = 783.74, Q2 = 214.56, so Qc = 569.18 kVAr. (Key point: Capacitor kVAr) 2. Which transformer loss changes when pf is improved, and how? A. Copper loss, which varies as the square of kVA loading; no-load loss is unchanged B. No-load loss, which falls with the load C. Both losses change equally D. Neither loss changes Answer: A. Copper loss is proportional to (S/S_rated)^2, while iron loss depends only on voltage. (Key point: Transformer loss dependence) 3. What is the copper loss reduction after correction (1020 kVA to 687.16 kVA)? A. About 4.12 kW B. About 7.55 kW C. About 3.43 kW D. About 12.0 kW Answer: A. Before: 18.57 x (1020/1600)^2 = 7.547 kW; after: 3.425 kW; reduction 4.122 kW. (Key point: Loss reduction result) 4. How is the maximum demand charge saving computed, and what is its value per month? A. kW reduction x Rs 190 = Rs 124 000 B. kVAr reduction x 190 = Rs 108 000 C. kVA reduction (332.84) x Rs 190 = Rs 63 240 D. kWh saved x 7.5 = Rs 90 267 Answer: C. The demand charge is per kVA; the reduction is 1020 - 687.16 = 332.84 kVA, giving Rs 63 240 per month. 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