---
title: "PF improvement of 490 kW load with demand charge"
subject: "Industrial Power Distribution and Illumination"
chapter: 10
key: "semester_6/industrial_power_distribution_and_illumination:pf-490kw-6-6kv-demand-charge"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/pf-490kw-6-6kv-demand-charge
site: BCT Engineering
---

# PF improvement of 490 kW load with demand charge

Industrial Power Distribution and Illumination · Chapter 10: Electrical Energy Audit in Industry

- Asked in: 2071 Magh, 8 marks
- Video: [Power Factor Explained – Your Electricity Bill Money Drain (Reactive Power)](https://www.youtube.com/watch?v=NIrKOVZrqnU) — The Engineering Mindset
- 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-demand-charge
- 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. Assume kVA demand charge is Rs. 200/kVA/Month. Calculate: a) the total kVAr required b) the required value of capacitance c) pay back period

## Short answer

a) 377.1 kVAR, b) 9.19 μF per phase (delta), c) payback 0.97 year (≈ 11.6 months).

## Answer

Loss-free capacitors connected in delta across the 6.6 kV supply provide leading kVAR. The real power (490 kW) and kWh stay the same; only the kVA demand falls, so the saving is in the kVA demand charge.

#### a) Total kVAR required
```text
tan(cos⁻¹0.7)  = 1.0202
tan(cos⁻¹0.97) = 0.2506
Q1 = 490 × 1.0202 = 499.90 kVAR
Q2 = 490 × 0.2506 = 122.81 kVAR
Qc = Q1 − Q2      = 377.09 kVAR
```

#### b) Capacitance required
In a delta connection, each phase capacitor is across the line voltage of 6.6 kV.
```text
Qc/phase = 377.09 / 3 = 125.70 kVAR
C = Qph / (ω V²)
  = 125 698 / (2π × 50 × 6600²)
  = 125 698 / (1.3685 × 10¹⁰)
  = 9.19 µF per phase
```
(Line current of the capacitor bank = 377 090 / (√3 × 6600) = 33.0 A.)

#### c) Payback period
```text
kVA1 = 490 / 0.70 = 700.00 kVA
kVA2 = 490 / 0.97 = 505.15 kVA
ΔkVA = 194.85 kVA

Annual saving in demand charge
  = 194.85 × 200 × 12 = Rs 467 629 /year

Cost of capacitor bank with switching
  = 377.09 × 1200 = Rs 452 513

Payback = 452 513 / 467 629
        = 0.968 year ≈ 11.6 months
```

| Quantity | Before | After |
|---|---|---|
| pf | 0.70 | 0.97 |
| kVAR | 499.9 | 122.8 |
| kVA | 700.0 | 505.2 |
| Demand charge/month | Rs 140 000 | Rs 101 031 |

**Answer: (a) 377.1 kVAR, (b) 9.19 µF per phase in delta, (c) payback ≈ 0.97 year (≈ 11.6 months).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which pair of values is used to find Qc for 490 kW from pf 0.7 to 0.97?
   A. tan = 0.7 and 0.97
   B. sin = 0.7 and 0.97
   C. tan = 1.0202 and 0.2506
   D. cos = 1.0202 and 0.2506
   Answer: C. Q = P tan(cos^-1 pf), giving 499.90 and 122.81 kVAR, so Qc = 377.09 kVAR. (Key point: Reactive power calculation)

2. Which formula gives the delta capacitance per phase?
   A. C = Qph x omega x V^2
   B. C = Qph / (omega x V)
   C. C = Qph / (omega x V^2) with V = 6.6 kV
   D. C = Qph / (omega x V^2) with V = 6.6/sqrt3 kV
   Answer: C. Qph = omega C V^2 and in delta V is the full line voltage 6.6 kV, so C = 125 698/(314.16 x 6600^2) = 9.19 uF. (Key point: Capacitance formula)

3. What is the reduction in kVA demand after correction?
   A. About 195 kVA (700 to 505)
   B. About 377 kVA
   C. About 490 kVA
   D. About 100 kVA
   Answer: A. kVA = kW/pf: 490/0.7 = 700 and 490/0.97 = 505.15, a reduction of 194.85 kVA. (Key point: kVA reduction)

4. How is the payback computed?
   A. Annual saving divided by capital cost
   B. Capital cost divided by annual demand-charge saving
   C. Capital cost divided by monthly saving times 12 energy charge
   D. kVAR divided by kVA
   Answer: B. Payback = Rs 452 513 / Rs 467 629 per year = 0.968 year, with annual saving = 194.85 x 200 x 12. (Key point: Payback calculation)

## Close variants set in other exams

- [PF improvement of 490 kW load at 6.6 kV](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Apf-490kw-6-6kv-payback) — Industrial Power Distribution and Illumination, chapter 10, asked 2×

## Related in Industrial Power Distribution and Illumination

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