Question

A university spent $1.6 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 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?

8,214.40
9,857.28
5,750.08
11,500.16
If the solar panels can operate only for 7,393 hours a year at maximum, the project break even.
Continue to assume that the solar panels can operate only for 7,393 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

- Approximately how many hours per year will the solar panels need to operate to enable this project to break even? 8,214.40

- In order for the project to be worthwhile (i.e., at least break even), the university would need a grant of at least $159,967.88

Explanation:

We can calculate the break-even point for the solar panels by finding the number of hours they need to operate to cover the initial cost of $1.6 million. Since the marginal cost of electricity production using the solar panels is zero, the university saves $0.10 per kilowatt-hour (kWh) that would otherwise be spent on purchasing electricity.

Given:

- Capacity of solar panels: 400 kW

- Electricity cost: $0.10 per kWh

- Discount rate: 20%

- Life expectancy of solar panels: 20 years

- Maximum operating hours per year: 7,393 hours

First, we'll calculate the present value of operating the solar panels for 1 hour per year:

`\[\text{Savings per hour} = 400 \, \text{kW} * \$0.10 \, \text{per kWh} = \$40\]`

Now, we'll calculate the present value of these savings over 20 years, taking into account the discount rate:

`\[\text{Present value of savings per hour per year} = \sum_(i=1)^(20) (\$40)/((1 + 0.20)^i)\]`

We can use this value to find the number of hours per year the solar panels need to operate to break even:

`\[\text{Break-even hours per year} = \frac{\$1,600,000}{\text{Present value of savings per hour per year}}\]`

Given that the solar panels can operate only for 7,393 hours a year at maximum, we can calculate the grant needed to make the project worthwhile:

`\[\text{Grant needed} = (\text{Break-even hours per year} - 7,393) * \text{Present value of savings per hour per year}\]`

Let's calculate these values.

The calculated values are as follows:

- Present value of savings per hour per year:
`\$194.78`

- Break-even hours per year:
`8,214.26`

- Grant needed:
`\$159,967.88`

So the correct answers are:

- Approximately how many hours per year will the solar panels need to operate to enable this project to break even? 8,214.40

- In order for the project to be worthwhile (i.e., at least break even), the university would need a grant of at least $159,967.88