Question

Answer following earth science questoins

1.The nearby town of Bellefonte, PA features the Gamble Mill, an old mill originally used to grind grain into flour - 1800s hydropower. The mill dam, pond, millrace and tail race are still in place and functional. The mill building is now a restaurant (currently for sale), but all the water-works are owned by the borough. The municipal manager wants to put the system back to work as a small hydroelectric system to help supply the borough's electricity. Does this idea make sense? How much power could this system produce?

The height difference between the mill race and the tailrace is ten feet (3 meters), and the stream can supply, on average, 100 cubic feet of water per second (2.8 m^3 / sec). The planned turbine would be 61% efficient under those conditions. Power (watts) = density of water * height * flow rate * acceleration of gravity * coefficient of efficiency

NOTE: the coefficient of gravity is 9.81 for this problem

Dont round answer

2.

If the system produces 44,874 watts, what is the energy worth per day? Assume electricity costs of $0.06 per kWh.

3.

If the system costs $90,587, how many years would it take to pay off the investment? Assume the energy is worth $81 per day. Assume there are 306 operational days per year (the rest are maintenance days).

4.

In Arizona, photovoltaic (PV) panels can generate an average annual power of 1 MW/acre. If there are 322 sunny days per year in Arizona, but only 162 sunny days per year in Pennsylvania, how much average power (in MW/acre) could the same PV panels generate in Pennsylvania? Please include two decimal places in your answer. Note: assume that a cloudy day delivers 0 watts/acre of solar power. This is not true, but it simplifies the problem for us.

5.

In Arizona, where there are 332 sunny days per year, it requires 335 acres of land to generate 300 MW of average annual power. If there are only 152 sunny days per year in Pennsylvania, how many acres of land would it require to generate the same average annual power? Please round to the nearest acre. Note: assume that a cloudy day delivers 0 watts/acre of solar power. This is not true, but it simplifies the problem for us.

Answer 1

Calculating the power of the hydroelectric system in Bellefonte, PA gives an output of 52,529.88 watts, which can be economically viable for electricity generation. Photovoltaic panels generate about 0.50 MW/acre in Pennsylvania due to fewer sunny days than Arizona. To generate the same power as in Arizona, approximately 705 acres would be needed in Pennsylvania.

Step-by-step explanation:

Understanding the Potential of a Hydroelectric System at Bellefonte, PA

When considering the hydroelectric system's potential at Bellefonte, PA, we can start by calculating the power this system could generate. Using the formula Power (watts) = density of water * height * flow rate * acceleration of gravity * coefficient of efficiency, where the density of water is 1000 kg/m³, the height difference is 3 meters, the flow rate is 2.8 m³/s, the acceleration of gravity is 9.81 m/s², and the coefficient of efficiency is 61% (or 0.61), we can determine the power output.

Thus, Power = 1000 kg/m³ * 3 m * 2.8 m³/s * 9.81 m/s² * 0.61 = 52,529.88 watts. This calculation suggests that the mill's existing water-works could be viable for generating electricity, assuming that the current flow and other conditions remain constant.

Photovoltaic Power Generation in Different States

For the photovoltaic (PV) panels, since the PV panels generate an average annual power of 1 MW/acre in Arizona with 322 sunny days, we need to adjust this for the number of sunny days in Pennsylvania. With only 162 sunny days in Pennsylvania and assuming a cloudy day provides 0 watts/acre, the average power generated would be proportionally less.

The average power in Pennsylvania can be calculated as (1 MW/acre) * (162 sunny days / 322 sunny days) = 0.50 MW/acre.

To replace the same amount of power that 335 acres in Arizona would provide, we would need more land in Pennsylvania due to fewer sunny days. If Arizona requires 335 acres for 300 MW, then Pennsylvania would need (335 acres) * (322 sunny days / 152 sunny days) which approximately equals 705 acres (rounded).

Answer 2

Answering the student's questions, the potential power output for the hydroelectric system is 52,252.68 W. The daily value of this energy is $64.62, with a payback period of 3.67 years. Photovoltaic panels would generate 0.50 MW/acre in Pennsylvania and require 640 acres to match Arizona's power generation.

Step-by-step explanation:

Earth Science and Physics Calculations

1. To assess whether the small hydroelectric system makes sense for Bellefonte, PA, we first calculate its potential power output. Using the provided formula, the power (in watts) can be calculated as:

Power (W) = (density of water) * (height difference) * (flow rate) * (acceleration due to gravity) * (efficiency coefficient)

Inserting the values given:

Power (W) = 1000 kg/m³ * 3 m * 2.8 m³/s * 9.81 m/s² * 0.61 = 52,252.68 W (or watts).

2. The energy produced per day can be calculated by multiplying the power output by the hours in a day and then converting it to kilowatt-hours:

44,874 W * 24 hours/day = 1,076,976 Wh/day = 1,076.976 kWh/day

The value of this energy per day at $0.06 per kWh:

1,076.976 kWh/day * $0.06/kWh = $64.62/day

3. To calculate the payback period of the system that costs $90,587:

Payback Period (years) = Total Cost / (Energy Value per Day * Operational Days per Year)

Payback Period = $90,587 / ($81/day * 306 days/year) = 3.67 years

4. Since Pennsylvania gets fewer sunny days compared to Arizona, the average power generated would be proportionally less:

Average Power (MW/acre) = (Sunny Days in PA / Sunny Days in AZ) * Power in AZ

Average Power = (162 days/year) / (322 days/year) * 1 MW/acre = 0.50 MW/acre

5. To generate the same average annual power in Pennsylvania with fewer sunny days:

Average Power (PA) = Average Power (AZ) * (Acres in AZ) / (Sunny Days in PA) * (Sunny Days in AZ)

Acres Required in PA = 300 MW / (0.50 MW/acre * 152/322) = 640 acres