---
title: "PF improvement of 490 kW load at 11 kV"
subject: "Industrial Power Distribution and Illumination"
chapter: 10
key: "semester_6/industrial_power_distribution_and_illumination:pf-490kw-11kv"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/pf-490kw-11kv
site: BCT Engineering
---

# PF improvement of 490 kW load at 11 kV

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

- Asked in: 2075 Baisakh, 8 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-11kv
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-10

## Question

The power factor on an industrial 3φ load 490 kW is to be improved from 0.7 lagging to 0.97 lagging by connecting loss free delta connected capacitor across 11 kV, 50 Hz supply. The cost of Var controlling device with switching is Nrs.2000 per kVAR.

## Short answer

Capacitor bank rating 377.09 kVAr total (125.7 kVAr/phase in delta), capacitance 3.31 µF per phase. Cost NRs 7.54 lakh; kVA demand falls from 700 to 505.15 kVA.

## Answer

**Given:** 3-phase load P = 490 kW, pf improved from 0.7 lag to 0.97 lag, delta-connected loss-free capacitors on 11 kV, 50 Hz; cost of var control with switching NRs 2000/kVAr.

#### kVAr rating of capacitor bank
```text
tan φ₁ = tan(cos⁻¹0.7)  = 1.0202
tan φ₂ = 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  (total, 3-phase)
```

#### Capacitance per phase (delta)
In delta, each capacitor has full line voltage 11 kV across it.
```text
Qc per phase = 377.09 / 3 = 125.70 kVAr
C = Q_ph / (ω V²)
  = 125 698 / (2π × 50 × 11 000²)
  = 125 698 / (314.16 × 1.21 × 10⁸)
  = 3.307 × 10⁻⁶ F = 3.307 µF
Current per capacitor = 125.70 kVAr / 11 kV = 11.43 A
```

#### Cost
```text
Cost = 377.09 × 2000 = NRs 7,54,189
```

#### Effect
```text
kVA before = 490 / 0.7  = 700.0 kVA
kVA after  = 490 / 0.97 = 505.15 kVA
Reduction  = 194.85 kVA
```

**Answer: capacitor bank of 377.09 kVAr (125.7 kVAr per phase), C ≈ 3.31 µF per phase in delta, cost ≈ NRs 7.54 lakh; kVA demand falls from 700 to 505.2 kVA.**

## Check yourself

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

1. What total kVAr improves 490 kW from 0.7 to 0.97 pf?
   A. 499.90 kVAr
   B. 122.81 kVAr
   C. 377.09 kVAr
   D. 194.85 kVAr
   Answer: C. Qc = 490 x (1.0202 - 0.2506) = 377.09 kVAr. (Key point: Capacitor kVAr)

2. In a delta-connected bank on 11 kV, what voltage is across each capacitor?
   A. 6.35 kV
   B. 11 kV, the full line voltage
   C. 19 kV
   D. 3.67 kV
   Answer: B. In delta, phase voltage equals line voltage. (Key point: Delta connection)

3. How is capacitance per phase calculated?
   A. C = Q_ph / (omega x V^2) with Q_ph = 125.7 kVAr, giving about 3.31 uF
   B. C = Q_total x V, giving 4 uF
   C. C = V / (omega x Q), giving 1 uF
   D. C = Q_ph x omega x V^2, giving 3.31 F
   Answer: A. Q_ph = omega C V^2, so C = 125 698 / (314.16 x 1.21e8) = 3.31 uF. (Key point: Capacitance formula)

4. What is the reduction in kVA demand?
   A. 700 kVA
   B. 505.15 kVA
   C. 377.09 kVA
   D. 194.85 kVA
   Answer: D. 700 - 505.15 = 194.85 kVA. (Key point: kVA reduction)

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