# Economics of Power Generation - Part 2

Power plants generate electrical energy in hundreds (and sometimes even thousands) of Mega Watts. Extensive usage of the equipment in power plants take its toll over time. They undergo wear and tear and hence require maintenance, repair and replacement. Hence, over time, the value of the power plant equipment as well as the building decreases. This decrease in the value is termed as 'depreciation'.
Ideally, the power plant equipment work at their rated capacities and at the efficiencies they are designed for. However, practically, it is not possible. Over time, wear and tear affects the efficiency as well as the life of a power plant.
We have to keep in mind that any plant will have a large capital (installation) cost. Such a huge sum of money is generally loaned. Interest is also charged upon this amount.
Combining the above mentioned points, the suppliers have to pay for:
1. Interest on the amount.
2. Depreciation in value of the plant
The suppliers, of course, charge the consumers for these and such charges are included in the fixed and semi - fixed costs of electrical energy.
The planning and management section of the plant sets aside some amount of money every year. As a result, once the plant's useful lifespan ends, these amounts set aside are used for the replacement of the plant. The amount set aside is termed as 'annual depreciation charges' and it has to be calculated for proper management and control. It is calculated by the following methods:

## Methods to Calculate Annual Depreciation Cost of a Power Plant

### Straight Line Method

In this method, provision is made from setting aside a certain and fixed amount of money every year. This value remains fixed for every year and depends upon the useful lifespan of the plant. It can be given as, total depreciation value divided by the useful life of the plant. The total depreciation value is calculated by subtracting the 'salvage (scrap) value after the lifespan' from the 'initial cost'.

Annual Depreciation = (Initial cost - Scrap value) / Useful life of the plant

For example, if a plant initially costs ₹ 500,000 and its scrap (salvage) value is ₹ 20,000 after 40 years of useful life, then,
Annual depreciation charges = (500,000 - 20,000) / 40 = 12,000 ₹

• Simple method
• Easy to apply
• In actual practice, depreciation of equipment is not constant every year.
• It does not consider the amount of interest earned by the annual depreciation amount set aside annually.
Graphically, it can be expressed as follows:

### Diminishing Value Method

In this method, a fixed rate of depreciation value is set. This rate is first applied to the capital cost (P) and then to the diminishing value. The rate is decided according to the useful lifespan of the plant. Yearly depreciation value can be calculated as follows:
Let, P = Capital cost of the equipment,
x = Annual unit depreciation (if annual depreciation is 10%, then annual unit depreciation is 0.1)
After 1 year, the value of equipment = P - Px = P (1 - x)
The annual depreciation after 1st year is equal to = (P - Px)x = Px - Px2
Hence, value of the equipment after 2 years = Diminishing value - Annual depreciation
= P - Px - Px - Px2
= P(1 - x)2
After n years (useful lifespan), value of the equipment = P(1 - x)n
Graphically,
We can see that depreciation charges are higher in the early years and reduce with time.

• Better distribution of charges: In the early years, depreciation charges are more while maintenance and repair charges are less. In the later years, depreciation charges are less while maintenance and repair charges are higher.
• In the early years, the plant is supposed to collect money and then collect interest on it as time passes. But in this method, the amount of interest is not taken into account.

### Sinking Fund Method

In this method, the arrangement is made such that a fixed amount is set aside annually and then invested at a certain interest rate which is compounded yearly. This fixed depreciation charges will be such that sum of these charges and the interest collected must be equal to the cost of replacement of the equipment.

Let, P = initial value of plant
n = useful life span
S = Salvage value after n years of useful life
r = Annual rate of interest
After n years, the salvage value will be received.
Therefore, Cost of equipment replacement = P - S
If the depreciation amount set aside every year is q, then interest will be received on this amount till n years are completed. Also, after n years, the total amount must be equal to the cost of replacement (i.e. P - S)

If we deposit amount q for the first year, then,
Interest after 1 year = rq
Amount after 1 year = q + rq = q(1 + r)
Similarly, after n years, amount = q(1 + r)n
But, amount q is added at the end of 1st year. Hence, it will collect interest only for (n-1) years.
Therefore, amount q (deposited at the end of 1st year) = q(1+r)n-1
Similarly, amount q (deposited at the end of 2nd year) = q(1+r)n-2
Amount q (deposited at the end of (n-1)th year) = q(1+r)n-(n-1) = q(1+r)
Therefore, the total sum (after n years) = q[(1+r)n-1 + (1+r)n-2 + ... + (1+r)]

The terms in the brackets constitute a Geometric progression with a ratio (1+r)-1
Therefore, Total funds collected = Sum of G.P = q[(1+r)n - 1]/r
But, total funds must be equal to the cost of replacement, i.e. P-S. Therefore,

This method is not widely used in practice. However, it does find application in economic assessments.