--- title: "Annual cost and saving with pf correction (400 kVA)" subject: "Utilization of Electrical Energy" chapter: 6 key: "semester_7/utilization_of_electrical_energy:industry-50000-units-pf-saving" source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/industry-50000-units-pf-saving site: BCT Engineering --- # Annual cost and saving with pf correction (400 kVA) Utilization of Electrical Energy · Chapter 6: Demand Side Management - Asked in: 2081 Baisakh, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/industry-50000-units-pf-saving - Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-6 ## Question An industry consumes 50000 units of electricity in a year at average power factor of 0.7 lag. The recorded max. demand is 400 kVA. The tariff is Rs. 120/kVA plus Rs. 5.50 per kWhr. Calculate the annual cost of supply and find out annual saving in cost by installing power factor improving device costing Rs. 1000 per kVAR which raises the power factor from 0.7 to 0.9 lagging. Allow 10% per year on the cost of the improving device to cover all additional cost. ## Short answer Annual cost at 0.7 pf: Rs 323,000. Required capacitor Qc = 280(1.0202−0.4843) = 150.05 kVAR. With yearly demand charge, MD saving Rs 10,667, corrector annual cost Rs 15,005, net loss Rs 4,338; not economical. ## Answer **Given:** energy 50,000 kWh/year, pf 0.7 lag, maximum demand 400 kVA, tariff Rs 120/kVA + Rs 5.50/kWh, corrector Rs 1000/kVAR, annual charges 10%, pf raised to 0.9. The demand charge is taken **per kVA per year**, as printed. #### Annual cost of supply at pf 0.7 $$ \begin{aligned} \text{MD charge} &= 400 \times 120 = \text{Rs } 48\,000 \\ \text{Energy charge} &= 50\,000 \times 5.50 = \text{Rs } 275\,000 \\ \text{Annual cost} &= \text{Rs } 323\,000 \end{aligned} $$ #### Corrector rating Maximum-demand kW $= 400 \times 0.7 = 280$ kW. $$ \begin{aligned} Q_C &= 280(\tan\phi_1 - \tan\phi_2) = 280(1.0202 - 0.4843) = 150.05\ \text{kVAR} \\ \text{Cost} &= 150.05 \times 1000 = \text{Rs } 150\,047 \\ \text{Annual charge} &= 0.10 \times 150\,047 = \text{Rs } 15\,005 \end{aligned} $$ #### New demand and saving $$ \begin{aligned} S_2 &= \frac{280}{0.9} = 311.11\ \text{kVA} \\ \text{Reduction in MD charge} &= (400 - 311.11) \times 120 = \text{Rs } 10\,667 \\ \text{Net annual saving} &= 10\,667 - 15\,005 = -\text{Rs } 4\,338 \end{aligned} $$ The energy charge (kWh) does not change. With the demand charge per year, the device costs more than it saves (net loss of Rs 4,338), so it is **not economical**. #### If the demand charge is Rs 120/kVA per month This is the usual way demand is billed (e.g. NEA bills demand monthly): $$ \begin{aligned} \text{Annual cost} &= 400 \times 120 \times 12 + 275\,000 = \text{Rs } 851\,000 \\ \text{MD saving} &= 88.89 \times 1440 = \text{Rs } 128\,000 \\ \text{Net saving} &= 128\,000 - 15\,005 = \text{Rs } 112\,995 \end{aligned} $$ **Answer:** Annual cost Rs 323,000; corrector 150.05 kVAR costing Rs 150,047. On a yearly demand charge the net saving is −Rs 4,338 (a loss). On a monthly demand charge the annual cost is Rs 851,000 and the net saving is about Rs 112,995 per year. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the annual cost at pf 0.7 found? A. Only 50,000 x 5.50 B. 400 x 0.7 x 120 plus energy C. Energy charge multiplied by pf D. MD charge 400 x 120 = Rs 48,000 plus energy 50,000 x 5.50 = Rs 275,000, total Rs 323,000 Answer: D. The tariff is kVA charge plus kWh charge. (Key point: Annual cost) 2. What corrector rating is needed? A. Q_C = 280(1.0202 - 0.4843) = 150.05 kVAR using MD kW = 400 x 0.7 B. 400(1.0202 - 0.4843) C. 280(0.9 - 0.7) D. 311.11 - 280 Answer: A. Use the maximum demand kW (280) and the change in tan phi. (Key point: Corrector rating) 3. Why is the device not economical when demand charge is per kVA per year? A. Energy charge rises B. MD saving (400 - 311.11) x 120 = Rs 10,667 is less than annual charge Rs 15,005 C. kVA does not change D. The device costs nothing Answer: B. Net saving is -Rs 4,338, a loss. (Key point: Saving on yearly demand charge) 4. What changes if the demand charge is per month? A. Energy charge increases 12 times B. Corrector cost becomes monthly C. MD saving becomes 88.89 x 1440 = Rs 128,000, so net saving about Rs 112,995 D. Net saving remains a loss Answer: C. Demand charge applies 12 times a year, making the improvement worthwhile. 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