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Homework answers / question archive / A university spent $1
A university spent $1.7 million to install solar panels atop a parking garage. 'These panels will have a capacity of 400 kilowatts (kW) and have a life expectancy of 20 years. Suppose that the discount rate is 30%, that electricity can be purchased at $0.10 per kilowatt-hour (kWh), and that the marginal cost of electricity production using the solar panels is zero.
Hint: It may be easier to think of the present value of operating the solar panels for 1 hour per year first. Approximately how many hours per year will the solar panels need to operate to enable this project to break even?
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If the solar panels can operate only for 11,536 hours a year at maximum, the project 'V break even. Continue to assume that the solar panels can operate only for 11,536 hours a year at maximum. In order for the project to be worthwhile (i.e., at least break even), the university would need a grant of at least •ir
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Computation of Number of Hours per Year Solar Panel needs to Operate to enable the project to breakeven:
Initial investment = $1.7 million or $1,700,000
Time n = 20 years
R = 10%
Let, uninform annual cost = P
Then,
1700000 = P*(1-1/1.30^20)/0.30
P = 1700000/3.3158
P = $512,697.70
Let, number of hours per year required to achieve the breakeven = N
Then,
N = $512,697.70/(0.10*400)
= $512,697.70/40
N = 12,817.44 hours
So, the correct option is 1st "12,817.61". The difference is due to rounding off figures.
For a break even it would needs to run for 12,817.61 hours per year. From the above calculation the project will not break-even at 11,536 hours.
Computation of Size of Grant:
The gap between the 12,817.61 hours and 11536 hours, will be fulfilled by the grant.
So,
Size of the grant = 1700000 - (1700000/12,817.44 hours)*11536 hours
Size of the grant = $169,960
