---
title: "Annual working cost of power station (a, b, c)"
subject: "Utilization of Electrical Energy"
chapter: 6
key: "semester_7/utilization_of_electrical_energy:power-station-annual-cost-abc"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/power-station-annual-cost-abc
site: BCT Engineering
---

# Annual working cost of power station (a, b, c)

Utilization of Electrical Energy · Chapter 6: Demand Side Management

- Asked in: 2069 Chaitra
- Video: [(2) Calculate Cost of Generation for plant  | Thermal Power Plant | PPE](https://www.youtube.com/watch?v=fKioHqx-4Zo) — Agravat Vishal
- 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/power-station-annual-cost-abc
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-6

## Question

The annual working cost of a power station is represented by the formula Rs (a + b×kW + c×kWh) where the various terms have their usual meaning. Determine the values of a, b and c for a 60 MW station operating at annual load factor of 50% from the following data:
i) Capital cost of building and equipment is Rs. 5×10⁶
ii) The annual cost of fuel, oil, taxation and wages of operating staff is Rs. 9,00,000.
iii) The interest and depreciation on building and equipments are 10% per annum
iv) Annual cost of organization and interest on cost of site etc is Rs.5,00,000.

## Short answer

a = Rs 5,00,000 (annual overheads). b = (0.10×50,00,000)/60,000 = Rs 8.33/kW (interest and depreciation per kW of maximum demand). c = 9,00,000/(60,000×0.5×8760) ≈ Rs 0.003425/kWh (fuel, oil, wages per unit generated).

## Answer

In the expression Rs $(a + b \cdot \text{kW} + c \cdot \text{kWh})$:
- $a$ = fixed cost that does not depend on demand or energy (organisation, site),
- $b$ = cost per kW of maximum demand (interest and depreciation on plant),
- $c$ = cost per kWh generated (fuel, oil, taxation, operating wages).

#### Energy generated per year
$$
\begin{aligned}
\text{Max demand} &= 60\ \text{MW} = 60{,}000\ \text{kW} \\
\text{Units per year} &= 60{,}000 \times 0.5 \times 8760 = 262.8 \times 10^{6}\ \text{kWh}
\end{aligned}
$$

#### Constant a
$$
a = \text{Rs } 5{,}00{,}000 \ \text{(organisation and interest on site cost)}
$$

#### Constant b
$$
\begin{aligned}
\text{Interest and depreciation} &= 0.10 \times 5 \times 10^{6} = \text{Rs } 5{,}00{,}000 \\
b &= \frac{5{,}00{,}000}{60{,}000} = \text{Rs } 8.33\ \text{per kW}
\end{aligned}
$$

#### Constant c
$$
c = \frac{9{,}00{,}000}{262.8 \times 10^{6}} = \text{Rs } 0.003425\ \text{per kWh} \ (\approx 0.34\ \text{paisa/kWh})
$$

So the annual cost is Rs $(5{,}00{,}000 + 8.33\,\text{kW} + 0.003425\,\text{kWh})$. At 60 MW and 50% load factor, the total annual cost is 5 + 5 + 9 = Rs 19 lakh.

**Answer: a = Rs 5,00,000; b = Rs 8.33/kW; c = Rs 0.003425/kWh.**

## Check yourself

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

1. Which cost belongs to constant b?
   A. Interest and depreciation on plant, per kW of maximum demand
   B. Fuel and wages, per kWh
   C. Organisation and site cost, fixed
   D. Taxation only
   Answer: A. a is fixed, b is per kW (plant interest and depreciation), c is per kWh (fuel, oil, taxes, wages). (Key point: Meaning of a, b, c)

2. How are the units generated per year found?
   A. 60,000 kW x 0.5 x 8760 = 262.8 x 10^6 kWh
   B. 60,000 x 8760
   C. 60,000 x 0.5
   D. 60,000 x 0.5 x 365
   Answer: A. Energy = max demand x load factor x hours in a year. (Key point: Annual units)

3. What is b?
   A. Rs 8.33 per kW = 500,000/60,000
   B. Rs 83.3 per kW
   C. Rs 0.003425 per kW
   D. Rs 5,00,000 per kW
   Answer: A. Interest and depreciation = 10% of 5 x 10^6 = Rs 500,000, divided by 60,000 kW. (Key point: Value of b)

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