Fuel Cost of Non-Solar Energy System Examples

The first example is about the fuel cost of a non-solar or conventional energy system. It examines the time value of an inflating fuel cost.

Example 12.5

Calculate the cost of fuel of a conventional (non-solar) energy system for 15 years if the total annual load is 114.9 GJ and the fuel rate is $17.2/GJ, the mar­ket discount rate is 7%, and the fuel inflation rate is 4% per year.

Solution

The first-year fuel cost is obtained from Eq. (12.4) as

t

CL = CFL JLdt = 17.2 X 114.9 = $1,976.30

0

(because the total annual load is given, the integral is equal to 114.9 GJ).

The fuel costs in various years are shown in Table 12.4. Each year’s cost is estimated with Eq. (12.13) or from the previous year’s cost multiplied by

Table 12.4 Fuel Costs in Various Years for Example 12.5

Year

Fuel cost ($)

PW of fuel cost ($)

1

1,976.30

1,847.01

2

2,055.35

1,795.22

3

2,137.57

1,744.89

4

2,223.07

1,695.97

5

2,311.99

1,648.42

6

2,404.47

1,602.20

7

2,500.65

1,557.28

8

2,600.68

1,513.62

9

2,704.70

1,471.18

10

2,812.89

1,429.93

11

2,925.41

1,389.84

12

3,042.42

1,350.87

13

3,164.12

1,313.00

14

3,290.68

1,276.18

15

3,422.31

1,240.40

Total PW of fuel cost

$22,876

(1 + i). Each value for the present worth is estimated by the corresponding value of the fuel cost using Eq. (12.11).

Подпись: or image1649 Подпись: 1 Подпись: [1.04 15 1.07, Подпись: 11.5752

An alternative method is to estimate PWF(n, i, d) from Eq. (12.18) or Appendix 8 and multiply the value with the first year’s fuel cost as follows. From Eq. (12.18),

Present worth of fuel cost = 1,976.3 X 11.5752 = $22,876

As can be seen, this is a much quicker method, especially if the calcula­tions are done manually, and the same result is obtained but the intermediate values cannot be seen.

Although in the previous example, a fixed fuel inflation rate is used for all years, this may vary with time. In the case of a spreadsheet calculation, this can be easily accommodated by having a separate column representing the fuel infla­tion rate for each year and using this rate in each annual estimation accordingly. So, in this case, either the same value for all years or different values for each year can be used without difficulty. These estimations can also be performed with the help of the PWF, as shown in Example 12.5, but as the number of dif­ferent rates considered increases, the complexity of the estimation increases, because the PWF needs to be calculated for every time period the rate changes, as shown in the following example.

Example 12.6

Calculate the present worth of a fuel cost over 10 years if the first year’s fuel cost is €1,400 and inflates at 8% for 4 years and 6% for the rest of the years. The market discount rate is 7% per year.

Solution

Подпись: PWF(n, i, d) image1653 Подпись: 1

The problem can be solved by considering two sets of payments at the two inflation rates. The first set of five payments has a first payment that is €1,400 and inflates at 8%. Therefore, from Eq. (12.18),

Подпись: or
Подпись: '1.08Ї5 1.07,
image1656
Подпись: PWF(5, 0.08, 0.07) Подпись: 4.7611

Thus, the present worth of the first set is 1,400 X 4.7611 = €6,665.54.

The second set starts at the beginning of the sixth year and, for this period, i = 6% per year. To find the initial payment for this set, €1,400 is inflated four times by 8% and one time by 6%. Therefore,

Initial payment for the second set is 1,400(1.08)4(1.06) = €2,018.97.

Подпись: or image1660 Подпись: 1.06 )5 1.07, Подпись: 4.5864

As before, for the second series of payments,

Подпись: PW Подпись: 4.5864 X 2,018.97 (1.07)5 Подпись: €6,602.11

The second set of payments needs to be discounted to the present worth by

So the answer is the sum of the present worth of the two sets of payments: 6,665.54 + 6,602.11 = €13,267.65.

Updated: August 29, 2015 — 4:52 pm