--- title: "kVAr and techno-economic analysis for 1000 kVA transformer" subject: "Industrial Power Distribution and Illumination" chapter: 10 key: "semester_6/industrial_power_distribution_and_illumination:transformer-1000kva-kvar-techno-economic" source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/transformer-1000kva-kvar-techno-economic site: BCT Engineering --- # kVAr and techno-economic analysis for 1000 kVA transformer Industrial Power Distribution and Illumination · Chapter 10: Electrical Energy Audit in Industry - Asked in: 2076 Bhadra, 8 marks - 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/transformer-1000kva-kvar-techno-economic - Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-10 ## Question In a industry transformer has following details: a) Rating of the transformer = 1000 kVA b) Average loading on transformer = 800 kVA c) Power factor = 0.6 lag d) No load loss = 1600 watt e) Full load loss = 13500 watt Calculate the kVAr require to improve p.f. to 0.95, reduction of kVA demand and techno-economic analysis, if the cost of capacitor bank with controlling switches NRs. 1500/kVAr and demand charge is NRs. 210/kVA/month. ## Short answer Capacitor required is 482.23 kVAr. kVA demand reduces from 800 to 505.26 kVA (saving 294.74 kVA). Investment NRs 7.23 lakh, monthly demand saving NRs 61,895 yields payback of about 11.7 months; technically and economically justified. ## Answer **Given:** transformer 1000 kVA, average load 800 kVA at 0.6 pf (lag), no-load loss 1.6 kW, full-load loss 13.5 kW, capacitor cost NRs 1500/kVAr, demand charge NRs 210/kVA/month. #### kVAr required to improve pf to 0.95 ```text P = 800 × 0.6 = 480 kW Q₁ = 800 × 0.8 = 640 kVAr Q₂ = 480 × tan(cos⁻¹0.95) = 480 × 0.3287 = 157.77 kVAr Qc = 640 − 157.77 = 482.23 kVAr ``` (In practice a standard 500 kVAr APFC panel would be used; the calculation uses 482.23 kVAr.) #### Reduction of kVA demand ```text S₂ = 480 / 0.95 = 505.26 kVA ΔS = 800 − 505.26 = 294.74 kVA ``` #### Reduction in transformer loss No-load loss (1.6 kW) stays the same; copper loss ∝ (load/rating)²: ```text Cu loss before = 13.5 × (0.8)² = 8.640 kW Cu loss after = 13.5 × (0.5053)² = 3.446 kW Reduction = 5.194 kW ``` If the transformer runs all year (8760 h): 5.194 × 8760 = 45 496 kWh saved per year (extra benefit; energy rate not given). #### Techno-economic analysis ```text Cost of capacitor bank = 482.23 × 1500 = NRs 723 347 Demand charge saving = 294.74 × 210 = NRs 61 895 / month = NRs 742 737 / year Simple payback = 723 347 / 61 895 = 11.7 months ≈ 0.97 year ``` | Item | Value | |---|---| | Capacitor needed | 482.23 kVAr | | kVA demand reduced | 294.74 kVA | | Cu loss reduced | 5.19 kW | | Investment | NRs 7.23 lakh | | Annual demand saving | NRs 7.43 lakh | | Payback | ≈ 11.7 months | The investment is recovered in less than one year from demand charge alone; the loss reduction, lower current, better voltage and released transformer capacity (about 295 kVA) are extra gains. So the pf improvement is **technically and economically justified**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For 800 kVA at 0.6 pf, what is the initial reactive power Q1? A. 480 kVAr B. 640 kVAr C. 800 kVAr D. 157.77 kVAr Answer: B. P = 480 kW and Q1 = 800 x 0.8 = 640 kVAr. (Key point: kVAr required) 2. What capacitor rating raises pf to 0.95? A. 157.77 kVAr B. 640 kVAr C. 482.23 kVAr D. 294.74 kVAr Answer: C. Q2 = 480 x tan(cos^-1 0.95) = 157.77, so Qc = 640 - 157.77 = 482.23 kVAr. (Key point: Capacitor rating) 3. What is the kVA demand reduction? A. 505.26 kVA B. 294.74 kVA C. 482.23 kVA D. 800 kVA Answer: B. S2 = 480/0.95 = 505.26 kVA, so reduction = 800 - 505.26 = 294.74 kVA. (Key point: kVA demand reduction) 4. How is simple payback found and what is it here? A. Investment / monthly demand saving, about 11.7 months B. Investment x saving, about 1 month C. Saving / investment, about 1 year D. Investment / loss reduction, about 5 months Answer: A. Investment NRs 723 347 divided by monthly saving NRs 61 895 gives about 11.7 months. 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