Question

A company spends $5 million to install solar panels with a 1,000-kw capacity and a life expectancy of 20 years. The discount rate is 8%. a) If electricity can be purchased for costs of $0.25 per kw hour, how many hours each year will the solar panels need to operate for this project to break-even? b) If efficient systems operate for 2000 kw-hours per year, would the project break-even?

Answer 1

a) Let annual working hours = h

then savings = 1000 * h * 0.25 = 250 * h

For breakeven

5000000 = 250*h * (P/A,8%,20)

h * 9.818147 = 20000

h = 20000 / 9.818147 = 2037.04

Therefore minimum hour to operate for breakeven is 2037

b) As 2000 is less than 2037, system will not breakeven