Question

A university spent $1.3 million to install solar panels atop a parking garage. These panels will have a capacity of 200 kilowatts (kW) and have a life expectancy of 20 years. Suppose that the discount rate is 30%, that electricity can be purchased at $0.30 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?

Answer 1

Answer:

It will take 6,534.31 hours per year for the solar panels to operate to enable this project to break even

Explanation:

Discount rate = 30% = 0.3

Looking at one hour of operation in each year = 200 kW x $0.30 Kw/hr

= $60 value of electricity per year

Compound interest factor for a discount rate of 30% = 3.3158

(taken from compound interest factor table or computed using formula ∑1/(1+r)^t , where r = 30%, and t = 1 to 30)

Present value of operating the solar panels for 1 hour per year = 60 × 3.3158 = $ 198.95

For break even it would need to run = 1.3 million ÷ 198.95

= 6,534.31 hours per year