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Homework answers / question archive / The plant engineer of a major food processing corporation is evaluating alternatives to supply electricity to the plant
The plant engineer of a major food processing corporation is evaluating alternatives to supply
electricity to the plant. He desires to know if he should build a 4000-kilowatt power plant. His
operating costs (other than fuel) for such a power plant are estimated to be $130,000 per year. He
is considering two alternative fuels:
WOOD. Installed cost of the power plant is $1200/kilowatt. Fuel consumption is 30,000 tons per
year. Fuel cost for the first year is $30 per ton and is estimated to increase at a rate of $5 per ton
for each year after the first. OIL. Installed cost is $1000/kw. Fuel consumption is 46,000 barrels per year. Fuel cost is $54 per
barrel for the first year and is estimated to increase at $5/barrel per year for each year after the first.
If MARR is 12%, and the analysis period is 10 years, which fuel alternative should the engineer
choose?
We have the following information
Wood Option
Installation Cost = $1200/kilowatt. The Power plant is 4000 kilowatt
So, Installation Cost = $1200 × 4000 = $4,800,000
Annual operating Cost = $130,000
Wood Consumption Cost table
|
Year |
Wood Consumption in Tons |
Wood Cost per ton ($) |
Wood Consumption Cost ($) |
|
1 |
30,000 |
30 |
9,00,000 |
|
2 |
30,000 |
35 |
1,050,000 |
|
3 |
30,000 |
40 |
1,200,000 |
|
4 |
30,000 |
45 |
1,350,000 |
|
5 |
30,000 |
50 |
1,500,000 |
|
6 |
30,000 |
55 |
1,650,000 |
|
7 |
30,000 |
60 |
1,800,000 |
|
8 |
30,000 |
65 |
1,950,000 |
|
9 |
30,000 |
70 |
2,100,000 |
|
10 |
30,000 |
75 |
2,250,000 |
Life (n) = 10 years
MARR = 12%
PW(12%) = 4,800,000 + 130,000(P/A, 12%, 10) + 9,00,000(P/F, 12%, 1) + 1,050,000(P/F, 12%, 2) + 1,200,000(P/F, 12%, 3) + 1,350,000(P/F, 12%, 4) + 1,500,000(P/F, 12%, 5) + 1,650,000(P/F, 12%, 6) + 1,800,000(P/F, 12%, 7) + 1,950,000(P/F, 12%, 8) + 2,100,000(P/F, 12%, 9) + 2,250,000(P/F, 12%, 10)
PW(12%) = 4,800,000 + 130,000[((1+0.12)10 – 1)/0.12 (1+0.12)10] + 9,00,000/(1 + 0.12)1 + 1,050,000/(1 + 0.12)2 + 1,200,000/(1 + 0.12)3 + 1,350,000/(1 + 0.12)4 + 1,500,000/(1 + 0.12)5 + 1,650,000/(1 + 0.12)6 + 1,800,000/(1 + 0.12)7 + 1,950,000/(1 + 0.12)8 + 2,100,000/(1 + 0.12)9 + ,2,250,000/(1 + 0.12)10
PW(12%) = 4,800,000 + 734,529 + 803,571 + 837,054 + 854,136 + 857,949 + 851,140 + 835,941 + 814,229 + 787,572 + 757,281 + 724,440
PW(12%) of Wood Option = $13,657,843
Oil Option
Installation Cost = $1000/kilowatt. The Power plant is 4000 kilowatt
So, Installation Cost = $1000 × 4000 = $4,000,000
Annual operating Cost = $130,000
Oil Consumption Cost table
|
Year |
Oil Consumption in Tons |
Oil Cost per barrel ($) |
Oil Consumption Cost ($) |
|
1 |
46,000 |
54 |
2,484,000 |
|
2 |
46,000 |
59 |
2,714,000 |
|
3 |
46,000 |
64 |
2,944,000 |
|
4 |
46,000 |
69 |
3,174,000 |
|
5 |
46,000 |
74 |
3,404,000 |
|
6 |
46,000 |
79 |
3,634,000 |
|
7 |
46,000 |
84 |
3,864,000 |
|
8 |
46,000 |
89 |
4,094,000 |
|
9 |
46,000 |
94 |
4,324,000 |
|
10 |
46,000 |
99 |
4,554,000 |
Life (n) = 10 years
MARR = 12%
PW(12%) = 4,000,000 + 130,000(P/A, 12%, 10) + 2,484,000 (P/F, 12%, 1) + 2,714,000 (P/F, 12%, 2) + 2,944,000 (P/F, 12%, 3) + 3,174,000 (P/F, 12%, 4) + 3,404,000 (P/F, 12%, 5) + 3,634,000 (P/F, 12%, 6) + 3,864,000 (P/F, 12%, 7) + 4,094,000 (P/F, 12%, 8) + 4,324,000 (P/F, 12%, 9) + 4,554,000 (P/F, 12%, 10)
PW(12%) = 4,000,000 + 130,000[((1+0.12)10 – 1)/0.12 (1+0.12)10] + 2,484,000/(1 + 0.12)1 + 2,714,000/(1 + 0.12)2 + 2,944,000/(1 + 0.12)3 + 3,174,000/(1 + 0.12)4 + 3,404,000/(1 + 0.12)5 + 3,634,000/(1 + 0.12)6 + 3,864,000/(1 + 0.12)7 + 4,094,000/(1 + 0.12)8 + 4,324,000/(1 + 0.12)9 + 4,554,000/(1 + 0.12)10
PW(12%) = 4,000,000 + 734,529 + 2,217,857 + 2,163,584 + 2,095,481 + 2,017,134 + 1,931,521 + 1,841,097 + 1,747,877 + 1,653,498 + 1,559,278 + 1,466,266
PW(12%) of Wood Option = $23,428,123
Since, the Present Worth (PW) of Cost of Wood option is lower, so Wood Alternative should be used.
