Viewing a single comment thread. View all comments

The_Ombudsman t1_j0m3na6 wrote

Neat!

Now I have a question: for a given volume of water, how much power is produced coming down through the turbines, and how much power would it cost to pump that same volume of water back up to the upper reservoir?

I really doubt they can get more power out than they use; otherwise it'd just be running all the time, free electricity! But I'm curious what the figures would be anyway.

−1

FriendlyDespot t1_j0m55fq wrote

Between inefficiencies in the pumping, and inefficiencies in the generating turbines, pumped hydro storage is around 75-85% efficient. It's pretty great.

14

Deathbeddit t1_j0mb39x wrote

The Ludington plant in MI is about 77% efficient iirc, not including energy for maintenance etc. It’s an open system pulling water in from Lake Mi and kills TONS of fish. Like bonkers amounts of fish, even though there are nets to try to exclude them. Also filling is mostly powered by fossil fuels today. An increasing fraction of energy to fill it comes from Renewables but that’s a work in progress. It is a steep design in a lot of ways and I am expecting new technologies will continue to be more appealing. I don’t know what the failure of the embankments would look like but I wouldn’t want to be near it myself.

3

ckach t1_j0mbgn8 wrote

It's probably somewhat common knowledge, but the physics are pretty straightforward.

Energy = (mass of the water) * (acceleration of gravity) * (height it goes down)

Acceleration of gravity is ~10m/s/s

Height is in meters, usually several hundred for these projects

Mass of the water is in kg. 1 cubic meter of water is 1,000kg, so you can get the mass straight from the size of the reservoir.

The energy will be in Joules. For there are 3.6 million joules in a kwh, so divide to get it in kwh.

So, if you have 50k cubic meters of water going down 500m, that would be

(50,000 * 1000)kg * 10m/s/s * 500m / 3,600,000 ~ 70,000 kwh

50,000 cubic meters of water would be 10m x 50m x 100m, for example.

1

Weareallgoo t1_j0mg5y9 wrote

This is not quite correct. Your formula is for potential energy. To calculate the power generated by a hydro power turbine, you need to use flow rate (l/s) or mass rate (kg/s), and include head loss and turbine efficiency.

Also, 1 cubic meter of water has a mass of 1000kg, not 50,000kg

1

ckach t1_j0mi9yu wrote

Yeah, I just wanted to go over energy rather than power since it's what is usually more important to the discussion. Another comment mentioned efficiency, so I didn't bother with it.

The 50k kg was definitely a mistake on my part. I tried editing it from 1k to 50k cubic meters in the example, but must have changed the wrong 1,000.

1

schwangeronis t1_j0mhf9i wrote

The main reason these batteries are attractive is that they only pump when there’s excess power in the system, which is a hurdle for wind and solar power. So if the sun isn’t shining and the wind isn’t blowing the battery is discharging water.

1

The_Ombudsman t1_j0mxhxq wrote

That's all well and good, and was covered in the article.

Doesn't really answer my question though.

1

schwangeronis t1_j0n87ig wrote

Oh right, there’s loss in the system. I can’t answer how it compares to a conventional battery, but I can say it’s cheaper

1