---
title: "Optimum pf under two-part tariff (200 kW)"
subject: "Industrial Power Distribution and Illumination"
chapter: 10
key: "semester_6/industrial_power_distribution_and_illumination:two-part-tariff-optimum-pf"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/two-part-tariff-optimum-pf
site: BCT Engineering
---

# Optimum pf under two-part tariff (200 kW)

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

- Asked in: 2069 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/two-part-tariff-optimum-pf
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-10

## Question

An industrial consumer takes steady load of 200 kW at p.f. of 0.707 lagging. He is charged according to two part tariff Rs.10/- per kWh and Rs. 200/- per kVA/month. The cost of power factor correcting equipment with switching device is Rs. 1200/- per kVAr.
a) What will be the optimum p.f. at which consumer should take power for minimum annual over all expenditure
b) What will be the capacity of the p.f. correcting equipment for maximum economy
c) The annual energy bill if he takes power for 16 hours daily for 350 days in a year.

## Short answer

Optimum pf = 0.866 lagging (sin φ = B/A = 0.5). PF correcting equipment capacity = 84.6 kVAR. Annual energy bill = Rs 11,754,256 (approx Rs 1.175 crore).

## Answer

**Data:** P = 200 kW at cos φ1 = 0.707; energy charge Rs 10/kWh; demand charge Rs 200/kVA/month; correcting equipment Rs 1200/kVAR.

**Annual kVA charge:**
```text
A = 200 × 12 = Rs 2400 /kVA/year
B = Rs 1200 /kVAR
```
(No interest/depreciation rate is given, so the full equipment cost is set against one year's kVA charge.)

#### a) Optimum power factor
Total annual cost = A·P/cos φ2 + B·P(tan φ1 − tan φ2) is minimum when
```text
sin φ2 = B / A = 1200 / 2400 = 0.5
cos φ2 = √(1 − 0.5²) = 0.866 lagging
```

#### b) Capacity of pf correcting equipment
```text
Q1 = 200 × tan(cos⁻¹0.707) = 200 × 1.0003 = 200.06 kVAR
Q2 = 200 × tan(cos⁻¹0.866) = 200 × 0.5774 = 115.47 kVAR
Qc = 200.06 − 115.47 = 84.59 kVAR
Cost = 84.59 × 1200 ≈ Rs 101 508
```

#### c) Annual bill (16 h/day, 350 days)
```text
Energy = 200 × 16 × 350        = 1 120 000 kWh
Energy charge = 1 120 000 × 10 = Rs 11 200 000
kVA demand = 200 / 0.866       = 230.94 kVA
Demand charge = 230.94 × 2400  = Rs 554 256
Annual bill = 11 200 000 + 554 256 = Rs 11 754 256
```

For comparison, without correction the demand is 200/0.707 = 282.89 kVA and the demand charge is Rs 678 925, so the bill would be Rs 11 878 925; correction saves Rs 124 669 per year in demand charge.

Note: if a 10 % annual interest + depreciation on the equipment is assumed instead (B = Rs 120/kVAR/year), sin φ2 = 0.05 and the optimum pf becomes about 0.999.

**Answer: (a) optimum pf = 0.866 lagging, (b) ≈ 84.6 kVAR, (c) annual bill ≈ Rs 1.175 crore (Rs 11 754 256).**

## Check yourself

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

1. With A = Rs 2400/kVA/year and B = Rs 1200/kVAR, what gives the minimum annual overall expenditure?
   A. sin phi2 = B/A = 0.5, so pf = 0.866
   B. cos phi2 = B/A = 0.5
   C. tan phi2 = 0.5
   D. sin phi2 = A/B = 2
   Answer: A. Total annual cost is minimum when sin phi2 = B/A = 0.5, so cos phi2 = 0.866 lagging. (Key point: Optimum pf condition)

2. How is the pf correcting equipment capacity found?
   A. Q = P / cos phi2
   B. Q = P(tan phi1 - tan phi2) = 200.06 - 115.47 = 84.6 kVAR
   C. Q = P x sin phi2
   D. Q = P(cos phi1 - cos phi2)
   Answer: B. The leading kVAR needed is the drop in lagging kVAR from 0.707 pf to 0.866 pf, about 84.6 kVAR. (Key point: Capacity of equipment)

3. What is the annual energy charge for 200 kW, 16 h/day, 350 days at Rs 10/kWh?
   A. Rs 11 200 000
   B. Rs 554 256
   C. Rs 1 120 000
   D. Rs 678 925
   Answer: A. Energy = 200 x 16 x 350 = 1 120 000 kWh, giving 1 120 000 x 10 = Rs 11 200 000. (Key point: Annual energy bill)

4. What kVA demand is used for the demand charge after correction?
   A. 200 / 0.707 = 282.9 kVA
   B. 200 kVA
   C. 200 / 0.866 = 230.9 kVA
   D. 84.6 kVA
   Answer: C. After correction the pf is 0.866, so kVA = 230.94 and demand charge = 230.94 x 2400 = Rs 554 256 per year. (Key point: Demand charge after correction)

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