A university spent 1.8 million to install solar panels atop a parking garage. Th

Question

A university spent 1.8 million to install solar panels atop a parking garage. These panels will have a capacity of 500kw, with a life expectancy of 20yrs and suppose the discount rate is 10% a. If electricity can be purchased for cost of $0.10 per kwh, how many hour per yr. will the solar panels have to operate to make this project break even. b. If efficient systems operate for 2,400 hrs pr year, would the project break evens? c. The univ. is seeking a grant to cover capital costs. How big of a grant would make this project worthwhile (to the university)?

Explanation / Answer

ANSWER:

1) If there is 1 hour of generation per year , the electricity generated will be = capacity * purchasing cost = 500 kw * 0.10 kwh = $50

i = 10% and n = 20 years

present value = value of electricity generated per year(p/a,i,n)

pv = 50(p/a,10%,20)

pv = 50 * 8.514 = $425.7

2) money spent on solar panels = $1,800,000

pv = $425.7

break even = money spent on solar panels / pv = $1,800,000 / 425.7 = 4,228.32

if systems operate for 2,400 hours then the efficient system is not productive.

3) if we take consideration for an efficient system of 2400 hours the present value will be 2400 * 425.7 = $1,021,680

grant required = money spent on installing solar panels - pv = $1,800,000 - $1,021,680 = $778,320

so they should require grant of $778,320.

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