---
title: "Annual saving by pf improvement for 800 kW load"
subject: "Utilization of Electrical Energy"
chapter: 6
key: "semester_7/utilization_of_electrical_energy:installation-800kw-capacitor-annual-saving"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/installation-800kw-capacitor-annual-saving
site: BCT Engineering
---

# Annual saving by pf improvement for 800 kW load

Utilization of Electrical Energy · Chapter 6: Demand Side Management

- Asked in: 2070 Asar
- 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/installation-800kw-capacitor-annual-saving
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-6

## Question

The load on an installation is 800 KW, 0.8 lagging which works for 3000 hours per annum. The tariff is Rs 100 per KVA plus 20 paisa per kwh. If the power factor is improved to 0.9 lagging by means of loss free capacitors costing Rs 60 per KVAR, calculate the annual saving. Allow 10% per annum for interest and depreciation on capacitors.

## Short answer

Since capacitors are loss-free, kWh cost is unchanged. Improving 800 kW from 0.8 to 0.9 pf cuts kVA demand from 1000 to 888.89, saving Rs 11,111 in demand charge. Against 212.54 kVAR capacitors costing Rs 1,275.30 annually, net saving is about Rs 9,836.

## Answer

Capacitors are loss-free, so the energy (kWh) bill does not change; the saving is in the kVA demand charge, minus the annual charges on the capacitors.

#### Demand before and after
$$
\begin{aligned}
S_1 &= \frac{800}{0.8} = 1000\ \text{kVA} \\
S_2 &= \frac{800}{0.9} = 888.89\ \text{kVA} \\
\text{Saving in demand charge} &= 100 \times (1000 - 888.89) = \text{Rs } 11{,}111
\end{aligned}
$$

#### Capacitor rating and its annual cost
$$
\begin{aligned}
\tan\phi_1 &= \tan(\cos^{-1}0.8) = 0.75,\quad \tan\phi_2 = \tan(\cos^{-1}0.9) = 0.4843 \\
Q_C &= 800\,(0.75 - 0.4843) = 212.54\ \text{kVAR} \\
\text{Capital cost} &= 60 \times 212.54 = \text{Rs } 12{,}753 \\
\text{Interest and depreciation} &= 0.10 \times 12{,}753 = \text{Rs } 1275.3
\end{aligned}
$$

#### Net annual saving
$$
\text{Saving} = 11{,}111 - 1275.3 = \text{Rs } 9835.9
$$

The energy charge (0.20 × 800 × 3000 = Rs 480,000) is the same before and after, so it does not affect the saving.

**Answer: Annual saving ≈ Rs 9,836 (capacitors of 212.54 kVAR).**

## Check yourself

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

1. Why does the energy charge not affect the saving?
   A. Capacitors are loss free, so kWh used is unchanged
   B. Energy is free
   C. Power factor changes kWh
   D. Capacitors reduce kWh
   Answer: A. The kW load is the same before and after, so the 480,000 energy bill cancels out. (Key point: Loss-free capacitors)

2. What is the capacitor rating needed?
   A. 800(0.75 - 0.4843) = 212.54 kVAR
   B. 800 x 0.9 = 720 kVAR
   C. 1000 - 888.89 = 111.11 kVAR
   D. 800 x 0.1
   Answer: A. Q_C = P(tan phi1 - tan phi2). The kVA difference is not the kVAR needed. (Key point: Capacitor rating)

3. What is the net annual saving?
   A. Rs 9,836 = 11,111 - 1275.3
   B. Rs 11,111
   C. Rs 12,753
   D. Rs 1275
   Answer: A. Demand saving 100 x 111.11 less 10% of Rs 12,753 capacitor cost. (Key point: Net saving)

## Related in Utilization of Electrical Energy

- [Annual charge and pf improvement (Rs 1300/kVA tariff)](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Apf-improvement-1200kva-1300-tariff) — Utilization of Electrical Energy, chapter 6, asked 3×
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- [Annual cost and saving with pf correction (400 kVA)](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Aindustry-50000-units-pf-saving) — Utilization of Electrical Energy, chapter 6

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