His solar idea is pretty dumb, but this does raise the interesting question of "is running my own pumped-storage facility on my property profitable." I think probably not, but hey, let's do a quick Fermi estimate.
Let's assume you have a big property up in the mountains with a cliff that overlooks a lake. The lake provides a handy reservoir to draw water from and dump water into.
Say at the top of your property you have a flat section of land that you're not using - call it one acre. Let's say that you can "easily" excavate this acre 1.5 meters deep to make a storage pond with a regular excavator. Now let's say the vertical distance from the bottom of your storage pond to the top of the lake is pretty big - 10 meters.
The maximum volume of water you can store is 1 acre * 1.5 meters = 6070 m^3.
Potential energy = mass*gravity*height = volume of water * density of water * gravity * height = 6070 m^3 * 1000 kg/m^2 * 9.8 m/s^2 * 10m = 596 MJ = 166 kWh.
Okay, now here comes the really variable part - price of electricity. I'm going to make the assumption that Ontario is fairly representative when it comes to time-of-use pricing.
From the Hydro One website, we can see that the on-peak price is 17.5 cents per kWh, the mid-peak price is 12.8 cents, and the off-peak price is 8.3 cents. Since Ontario has some weird rules about what is off peak and what isn't (changes due to day of week and season), let's just make a blanket statement that for our hypothetical mountain utility we can assume we're "on-peak" for eight hours a day at 16 cents and "off-peak" for 16 hours a day at 8 cents, a nice 2:1 ratio.
The strategy here is to pump water into our reservoir during off-peak and drain it through a turbine during on-peak. Let's say that our pump-turbine has an efficiency of 87% for both pumping and generation (this is a pretty reasonable number for a real-world turbine once all the hydraulic, mechanical, and electrical losses are considered). Round trip efficiency is then (0.87)^2 = 76%.
That means that to put 166 kWh into our reservoir, we actually need 166/0.87 = 191 kWh, which we purchase off-peak for $14.48. When peak rolls around, that 166 kWh can be converted to 166*0.87 = 144 kWh back into the grid, which we sell for $23.04. Each day, we profit $23.04 - $14.48 = $8.96.
So, we make a little less than ten bucks a day, or around $3500 a year. Not a whole lot of cash, but hey, this thing's automated. What kind of pump-turbine do we need to accomplish that?
The hydro equation says that power = eta*Q*rho*g*h, where eta is the machine efficiency, Q is the flow rate, rho is the density of water, g is gravity, and h is the hydraulic head. We need to be able to output 144 kWh in 8 hours, which means the machine's power peak output is 144/8 = 18 kW. Substituting and solving for flow rate gives Q = 0.21 m^3/s, or 210 litres/sec.
That's a pretty big pump-turbine for a small operation. I don't know how much a unit like that would go for, or even if they're available on the open market - you'd likely have to convert a Francis turbine to a pump-turbine.
Considering how small your return is, however, I think that between the cost of the pump-turbine, civil works, and engineering, this thing is a huge money-loser. There's a reason pumped storage facilities tend to be really big and use existing reservoirs - it's the only way to make them economically feasible.
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