Question

A university spent $1.7 million to install solar panels atop a parking garage. These panels will have a capacity of 300 kilowatts (kW) and have a life expectancy of 20 years. Suppose that the discount rate is 20%, 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? 10,472.99 17,454.99 5,818.33 11,636.66 If the solar panels can operate only for 10,473 hours a year at maximum, the project break even. Continue to assume that the solar panels can operate only for 10,473 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

Answer 1

Answer:

the solar panels need to operate 11,636.66 hours to brak even

at  10,473 hours per year the project don't break-even It will need a grant for 170.026,74 dollars to do so.

Explanation:

300 kilowatts x 0.1 = 30 dollars of saving per hour:

we need to find the couta of a present value of 1,700,000 for 20 years at 20% discount rate:

$PV \div \frac{1-(1+r)^{-time} }{rate} = C$

PV  $1,700,000.00

time 20

rate 0.2

$1700000 \div \frac{1-(1+0.2)^{-20} }{0.2} = C$

C  $ 349,106.102

Each hour saves 30 dollars

so 349,106.10/30 = 11.636,87‬ hours

IF the solar panels operate 10,473 hours per year then:

$C \times \frac{1-(1+r)^{-time} }{rate} = PV$

C 314,190

time 20

rate 0.2

$314190 \times \frac{1-(1+0.2)^{-20} }{0.2} = PV$

PV $1,529,973.2565

the present value of the panel is 1,529,973.26

The university would need a grant for 170.026,74