Question

Suppose you own some land next to a 50 MW wind farm located on Colorado’s Front Range. Your land has a fantastic cliff/ridgeline with an elevation difference of 250 m. Evaluate whether or not you can make money by building a pumped hydroelectric energy storage system on your land that interfaces with the wind farm. Your plan is to capture excess energy from the wind farm or buy energy from the wind farm when electricity prices are low and sell that energy back to the grid when electricity prices are high (assume that the utility that owns the grid will grant you a fair contract). You find that a good power rating for your storage system is 50% of the rated wind farm output, or 25 MW. Also your storage system should be able to generate electricity at rated power for 8 h.

a. Determine the flow rate and reservoir size needed to accomplish the required power output and energy capacity. Pumped hydroelectric plants generally run at 80%–90% generating efficiency, depending on the size of the machinery. Suggest a reasonable surface area and depth for your two reservoirs.
b. Assume it costs $500 per kW to install your turbomachinery and penstocks, plus $2 per cubic yard to build reservoirs. Calculate the initial capital cost of your pumped hydroelectric system.
c. Suppose you can buy energy from the wind farm at $0.035/kWh between the hours of 10:00 pm and 8:00 am to charge your storage and sell energy between 1:00 pm and 9:00 pm back to the grid at $0.1/kWh. Select a reasonable simple payback period that would motivate you to invest in energy storage. Calculate the maximum capital cost expenditure on your energy storage system that would allow this payback period.
d. Would you decide to build this energy storage system? Why or why not?

Answer 1

a.) Power output of hydroelectric plant = , where is efficiency, is density of water, F the flow rate. So, for power output of 25MW, we need F=25 x 10⁶ / (90% x 1000 x 9.8 x 250) = 11.33 m³/s.

Also, flow rate = cross sectional area x velocity of water. velocity of water at the base of 250m cliff is = 70m/s

So area of reservoir is 11.33/70 = 0.162m². To contain water flow of cross section 0.162m² at velocity 70m/s for 8 hrs, we require a depth of 40816.3 m.

b.) For 25MW of power, it will require, $500 x 25 x 1000 = $12.5 million. For a reservoir of area 0.162m² and depth 40816.3 m, it requires, $2 x 0.162 x 40816.3 / 0.764555 (conversion factor) = $17293. So two such reservoir would cost $34587. So total capital cost is $12.534587 million.

c.) Energy that can be produced = 25MW x 8h = 200MWh, energy that can be taken = 50MW x 10h = 500MWh

Cost of energy = $0.035/kWh x 500MWh = $17500. Price of energy = $0.1/kWh x 200MWh = $20000, which gives a profit of $2500 per day. So it takes approximately, 12.5million/2500 = 5014 days or 13 years for payback. The maximum capital cost expenditure is less than a million dollars.

d.) No, since within 13 years, the current price of electricity would change, so it will be a loss to buy and sell energy at the current rate. In 13 years the technology is bound to change that would give energy at lower costs than our selling price.