---
title: "Most economical power factor for 50 kW factory"
subject: "Industrial Power Distribution and Illumination"
chapter: 10
key: "semester_6/industrial_power_distribution_and_illumination:optimum-pf-50kw-factory"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/optimum-pf-50kw-factory
site: BCT Engineering
---

# Most economical power factor for 50 kW factory

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

- Asked in: 2070 Magh, 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/optimum-pf-50kw-factory
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-10

## Question

A factory has an average annual demand of 50 KW and an annual load factor of 0.5. The power factor is 0.75 lagging, the tariff is Rs. 190/KVA/month and energy charge is Rs.7.50 Kwh. If loss free capacitors costing Rs. 600/KVAR are to be utilized. Find the value of power factor at which maximum saving will result. Also determine the saving after improving p.f. and payback period.

## Short answer

Most economical pf ≈ 0.965 lagging (sin φ2 = B/A, A=Rs 2280/kVA/yr, B=Rs 600/kVAR). Capacitor: 60.9 kVAR. Annual kVA charge saving: Rs 67,670, payback 0.54 year.

## Answer

**Maximum demand:** load factor = average demand / maximum demand, so
```text
Pmax = 50 / 0.5 = 100 kW
```
The kVA charge applies to this maximum demand.

**Most economical pf** (loss-free capacitors, kWh unchanged):
Let A = annual kVA charge per kVA, B = cost per kVAR of capacitor. The net saving is maximum when
```text
sin φ2 = B / A
```
No interest/depreciation rate is given, so the full capacitor cost is compared with one year's kVA charge (i.e. capital recovered in one year):
```text
A = 190 × 12 = Rs 2280 /kVA/year
B = Rs 600 /kVAR
sin φ2 = 600 / 2280 = 0.2632
cos φ2 = √(1 − 0.2632²) = 0.9648 lagging
```

**Capacitor size:**
```text
Q1 = 100 × tan(cos⁻¹0.75)   = 100 × 0.8819 = 88.19 kVAR
Q2 = 100 × tan(cos⁻¹0.9648) = 100 × 0.2728 = 27.28 kVAR
Qc = 88.19 − 27.28          = 60.91 kVAR
```

**Saving:**
```text
kVA1 = 100 / 0.75   = 133.33 kVA
kVA2 = 100 / 0.9648 = 103.65 kVA
ΔkVA = 29.68 kVA
Annual saving in kVA charge = 29.68 × 2280 = Rs 67 670
Capacitor cost = 60.91 × 600 = Rs 36 549
Net saving in first year = 67 670 − 36 549 = Rs 31 121
```
The energy charge (Rs 7.50/kWh) is the same before and after, since capacitors are loss-free and kW is unchanged.

**Payback period:**
```text
= 36 549 / 67 670 = 0.54 year ≈ 6.5 months
```

Note: if instead an annual interest + depreciation of 10 % on the capacitor were assumed (B = Rs 60/kVAR/year), sin φ2 = 60/2280 and the economic pf would be about 0.9997, i.e. practically unity.

**Answer: Most economical pf ≈ 0.965 lagging; capacitor ≈ 60.9 kVAR; annual saving ≈ Rs 67 670 (net Rs 31 121 in the first year); payback ≈ 0.54 year.**

## Check yourself

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

1. How is the maximum demand obtained from the average demand and load factor?
   A. Pmax = 50 x 0.5 = 25 kW
   B. Pmax = average / load factor = 100 kW
   C. Pmax = 50 / 0.75 kW
   D. Pmax = 50 + 0.5 kW
   Answer: B. Load factor = average / maximum demand, so Pmax = 50/0.5 = 100 kW, on which the kVA charge applies. (Key point: Maximum demand)

2. With A = annual kVA charge per kVA and B = capacitor cost per kVAR, at what condition is saving maximum?
   A. cos phi2 = B/A
   B. tan phi2 = B/A
   C. sin phi2 = B/A
   D. sin phi2 = A/B
   Answer: C. Net saving is maximum when sin phi2 = B/A = 600/2280 = 0.2632, giving cos phi2 = 0.965. (Key point: Most economical pf condition)

3. What capacitor rating is needed to go from pf 0.75 to 0.9648 for 100 kW?
   A. About 88 kVAR
   B. About 27 kVAR
   C. About 115 kVAR
   D. About 61 kVAR
   Answer: D. Qc = 88.19 - 27.28 = 60.91 kVAR. (Key point: Capacitor size)

4. Why does the Rs 7.50/kWh energy charge not affect the saving?
   A. Loss-free capacitors leave kW and kWh unchanged
   B. Energy charge is always zero
   C. It is included in kVA charge
   D. Capacitors reduce kWh by half
   Answer: A. Capacitors only change reactive power, so the kWh billed is the same before and after. (Key point: Role of energy charge)

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