Undergrad What is the energy output of a Gravity Battery System?

[ScottyP99]

TL;DR

Gravity Battery Actual Energy Generated

I have been seeing quite a bit of news on Gravity Batteries, and am curious about the actual energy generated from the stored GPE ?

I know GPE = mass x height x gravity

h = 50 ft
m = 6000 lbs
g = 32.2 ft/sec^2

GPE = 9,660,660 joules = 2.68 KWh

But what is the actual energy output from this ?

Thanks in advance !

Scott

 

[Bystander]

Depends on the conversion efficiency.

 

[hmmm27]

and a correct calculation of PE : Imperial units don't convert directly to Joules.

 

[ScottyP99]

OK thanks. What are real world efficiency conversion factors in a generic system ?

Just lookin for some real world data to go with the system potential of something like http://www.EnergyVault.com

 

[berkeman]

OK thanks. What are real world efficiency conversion factors in a generic system ?

Just lookin for some real world data to go with the system potential of something like http://www.EnergyVault.com

Can you just please convert your calculation into SI units so we can chek your work?

 

[ScottyP99]

hmmm27 s this formula not correct:

 

[ScottyP99]

SI Unit calc:

h = 15.24 m
w = 2722 kg
g = 9.8 m/s^2

GPE = 406536 Joules

 

[berkeman]

SI Unit calc:

h = 15.24 m
w mass = 2722 kg
g = 9.8 m/s^2

GPE = 406536 Joules

Okay with correction.

 

[ScottyP99]

OK so can I use the joule to kwh conversion on the SI values ?

 

[berkeman]

OK so can I use the joule to kwh conversion on the SI values ?

I would think so. Just be sure to use the correct conversion factor(s).

 

[berkeman]

BTW, if you're not sure how to do that conversion, we can show you how to multiply by "1" (with the right units in the numerator and denominator) to do that conversion.

 

[ScottyP99]

Would this be correct using the SI values and the joule --> kwh formula:

 

[hmmm27]

a Joule is a Watt second ; 3600J in a Wh : 3.6MJ in a kWh ;

406,536J = 0.1129 kWh,

ie: humping 3 tons up 50 feet takes and gets you enough energy to run a 100W light-bulb for a little over an hour.

[edit: yup, what you said]

 

[Baluncore]

h = 50 ft
m = 6000 lbs
g = 32.2 ft/sec^2

m = 2720 kg
g = 9.8 m/s2
h = 15.24 m

PE = m·g·h
PE = 2720 * 9.8 * 15.24 = 406237.4 joule.

12 hours overnight = 12 * 60 * 60 = 43200 sec.
watts = joules / seconds = 406237 / 43200 = 9.4 W.
9.4 W for 12 hours = 112.8 watt·hours.

The efficiency of the process would depend on the method used to convert the fall to usable energy.

 

[russ_watters]

12 hours overnight = 12 * 60 * 60 = 43200 sec.

watts = joules / seconds = 406237 / 43200 = 9.4 W.
9.4 W for 12 hours = 112.8 watt·hours.

I get 122 W-h. Also, no need to multiply and then divide by 12. Just use one hour. Anyway, not a huge difference, so...

But what is the actual energy output from this ?

I'd expect upwards of 90% efficiency of such a storage system based on the efficiency of motors/generators. But as you can see, it takes a lot of mass or a lot of height to store a significant amount of energy this way.

 

[jrmichler]

Industrial servo motors are available with regenerative drives that pump regenerated power back into the power lines. Here's a link to one such family of drives: https://literature.rockwellautomation.com/idc/groups/literature/documents/in/2198-in014_-en-p.pdf. You can find more information by searching regenerative drives. That 90% or higher efficiency is right for modern industrial servo motors and drives.

These are used in elevators and overhead cranes. Regenerative drives have a large advantage over previous technology in overhead cranes. The previous technology used a mechanical brake in the crane to lower the load. The lift motor would turn a cam to release the brake, and the brake converted the energy from lowering into heat. That heat went into the crane motor/reducer/brake assembly and was the main limiting factor for crane usage. That's why those cranes all have specific limits of allowable usage in foot-pounds per hour.

 

[Baluncore]

Also, no need to multiply and then divide by 12. Just use one hour. Anyway, not a huge difference, so...

That was because, at night it is usually dark for 12 hours, and an LED light needs about 10 watt.

The hill behind my house is 30 m. I need 300 watt for 12 hours = 43,200 sec, to run my base load through the night. I have solar to re-energise the system during the day.
How much mass do I need? 300 watt * 43200 sec = 12.96 MJ.
Mass = 12.96e6 / ( 9.8 * 30 ) = 44.1 tonne.

If I use a massive 50 tonne trolley on rails, the rope drum and gearbox will be inefficient, and it will be dangerous.

But it could be done with 45 m³ of water. I need at least another 5 m³ to cover the inefficiency. If I use water, then pumps and turbines have poor efficiency, but I can select the flow per revolution that drives the alternator.

Is it worth the effort? 300 W for 12 hours is 3.6 kW⋅hr. My buy price per night is about AU$1.20 or AU$440 PA. It would take 10 years to just pay off the tanks. The pumped storage system is probably "not worth the candle".

The real pity is that, over the last 40 years I have paid the State to build a hydro-electric system. I could store my solar energy during the day by delaying the fall of water in the State's system, but the sale price of my energy during the day is 25% of my buy price which makes that storage system less than 25% efficient. For each dollar I save, I must give them three.

 

[ScottyP99]

Thank you !

I am trying to understand the metrics behind Energy Vault or similar, where they use excess power to lift large concrete blocks with cranes to store the GPE, and then release and generate energy when needed by dropping the weights.

Any ideas on how they can effectively generate enough energy from the GPE to make it worthwhile and effective ?

 

[Baluncore]

Any ideas on how they can effectively generate enough energy from the GPE to make it worthwhile and effective ?

Such a mechanical system would be used only to stabilise the short term fluctuations in a solar or wind energy system. It would give standby energy generation plants time to “fire up” when they are needed.