# Energy Generation through Gravity (mining operation monorail)

1. Jan 20, 2014

### danny.mcshane

Hello, I am looking for some help in determining a potential energy generation project.

Our open cast mine produces roughly 250 metric tonnes of raw rock per hour which is transported to the process area by an industrial monorail. The interesting thing is this monorail needs no power as the weight of the full carts travelling down is enough to carry the empty carts back to the top. The monorail is 80m long (40m down and 40m up) at a severe angle of approx. 30 degrees.

I am thinking that with a geared drive shaft and chain I might be able to generate electricity as well through this process which would be amazing as there is a far from constant supply to the mine. Having read as many posts as I can and barely understanding them am I right that 550lbs = 1HP so my 250 tonnes would equal 1000HP? I understand there may be some loss here but even so that sounds brilliant.

For design I am thinking installing a chain around the monorail that runs over a large gear that turns a drive shaft to power a motor. I think this would work but I would be delighted if you could say otherwise.

I appreciate any help/advice that can be offered.

EDIT. I thought it might be important. It takes between 55-65 seconds for one cycle. Amazingly I timed it a number of times and it always seems to be the same.

Last edited: Jan 20, 2014
2. Jan 20, 2014

### Staff: Mentor

Those units don't match. HP is a power, tonnes and lbs are masses. They cannot be equal.

Does your monorail has some braking mechanism that slows it currently? If not, gravity is just strong enough to counter friction.

A track length of 40m with a slope of 30° corresponds to a height difference of 20m. The kinetic energy of 250 tons is then $E=mgh = 250,\!000kg * 9.81m/s^2 * 20m = 49,\!050,\!000\ kg\ m^2/s^2 \approx 13.6 kWh$ - and you get this per hour. A perfect machine could produce 13.6kW (on average). A more realistic value would have to take friction (and other losses) into account.

3. Jan 20, 2014

### danny.mcshane

Thank you very much for the reply. Yes I believe it does have some form of inbuilt braking to co-ordinate the dropping of the ore into the carts. The conveyors run stop/start so they always drop their compartment into a cart. I am afraid I don't know how this is implemented currently though.

13.6kWh?????? Assuming friction losses I would still be producing more power than we could hope to use. I know this will be obvious to most people here but would the use of appropriate gears be able to reduce the amount generated by, well, lots. This was a silly discussion between supervisors and I wasn't expecting anything like that.

Note, I am sorry for my stupidity with the 550lbs statement. I read that 550lbs falling one foot per second could generate 1HP and oversimplified.

4. Jan 20, 2014

### Staff: Mentor

13.6kW, not kWh.
This is equal to 13.6 kWh per hour.
You want to reduce the power? That is simple, just use the brakes in addition to the generator.

I have no idea about the efficiency of the monorail, so I cannot give a lower estimate. Concerning the upper estimate: more than 5kW would be surprising, I think.

5. Jan 20, 2014

### danny.mcshane

Sorry, I saw kWh but thought MWh. I am not making a good first impression.

OK so 5kWh is about two days of our residential use (we think) so it would be still be wonderful. I will look into methods of storing electricity but from what I have read it seems quite tricky or expensive. Is there a clever way to store it without investing in industrial batteries? I couldn't find one but I thought I would ask.

I will check with the engineers tomorrow and try to get the full site usage details. We have a thermal drying process which we have to use generators for so I am guessing it will be quite high. We could use all our additional power here if needed I suppose.

Taking a punt at looking like a complete fool again, could a generator similar to that in a wind turbine be used by a chain driven drive shaft? I could provide the same amount of motion with any desired RPM by using the correct gears couldn't I? Looking at those they seem to produce 6MWh from a revolving shaft, I doubt we would use that much but it would at least prove we could power the operation. If we could generate power for the bosses as well they might even pay for it and save us some pennies.

Sorry for all the questions but we are a few Europeans in the middle of nowhere trying to come up with a way to enjoy a few home comforts. I said I would look into it while I was home and I would hate to go back with nothing so all help is greatly appreciated.

Last edited: Jan 20, 2014
6. Jan 20, 2014

### Staff: Mentor

kW, not kWh[/red]. That is an important difference.

Can you store rocks on the top? That would be an easy way to store some energy. Batteries are expensive, and all other methods have some disadvantages as well.

Probably.

7. Jan 20, 2014

### danny.mcshane

Many thanks for the help, I really appreciate it. I'm going to look into it and I will let you know how I get on.

8. Jan 20, 2014

### pikpobedy

13.6 kW is enough to heat an average house in winter. It is about 20 hp.

9. Jan 21, 2014

### sophiecentaur

What you are proposing is virtual the same a Hydroelectric Power generation - except that the Sun's energy provides you with an endless supply of water into the dam at the top. The rock supply will run out, when the hill is flattened and, once you start to dig near or below the level of the processing plant, you need to put energy into the system.

10. Jan 21, 2014

### danny.mcshane

I am stuck again, although I think this might be too complicated to work out.

The monorail itself is an ultra low friction unit without the electric drive rope. I don't think there will be any resistance from this. The timing of the carts is regulated by a hydraulic pressure system close to the top that is in sync with the conveyor. This can be taking out though as it is an add on.

Assuming very little friction the weight of the rocks is clearly enough to lift the empty carts back up again, but how much torque (I think this is right) will I be able to apply to a gear? Is there a formula to work this out?

There are 36 mining carts each weighing approx. 220kg. At any one time there will be 1 at the apex of the egg shaped loop being loaded, 3 at the flatter bit at the bottom being dragged round and then 16 on the down side and 16 on the up side. I hope that makes sense. Assuming 250 tonnes of rocks pushing on the downside I guess what I am trying to figure out is:

250 tonnes plus weight of down carts MINUS weight of up and bottom carts = force left to turn the drive shaft.

This may be totally wrong!! Is there a formula I can apply that could work this out or am I looking for Mathematics and Physics to do the impossible. Even if it is multiple formulas I don't mind spending time on it. I have looked at numerous complicated looking formulas but I don't know if they are what I need or not.

Ideally I want to know the maximum force I could exert on a gear to drive a generator something like this http://www.theswitch.com/2013/12/02...tor-passes-initial-no-load-tests-with-succes/. I don't know how removing the turbine blades would affect power output but assuming I could gear up to the same torque (force?) that is a lot of power.

They are massively expensive but if I could power the operation it would be worth it.

Any help would be greatly appreciated.

11. Jan 21, 2014

### Staff: Mentor

There will, it's just open how much.

Depends on the rotation speed. torque*angular velocity = power.
How much rock is lowered each time? Probably not 250 tons. This value (together with the time it needs) allows to calculate the available power while the monorail is operating.

That is certainly too large for a few kW, I guess it will be inefficient (and still way too expensive) in this power range.

12. Jan 21, 2014

### danny.mcshane

So 250 tonnes of rock drops 20m every hour and the whole thing keeps turning with no power required at all. In a vacuum (I think) this would generate 13.6kW (kind thanks to mfb for this). I will lose some from friction by adding a gear driven shaft, also then I will lose some from the monorail of course. Lets say I lose 0.6kw (so simplifying I know) and I am left with a system that has 13kW of potential energy.

torque*angular velocity = power so........ my velocity is 20meters per hour my power is 13.6kW

Is it torque = velocity over power so torque = 20/13.6 = 1.47 ?

Also everything I have read talks about angular vectors etc like here http://en.wikipedia.org/wiki/Torque

My set up is more like an egg shape with the narrow bit at the top and the slightly fatter bit at the bottom. I will be putting the gear at the top with the chain running across the top 220 degrees (8 to 4 on a clock face) so everything on the left hand side pulls downwards keeping the gear constantly turning counter clockwise. Does this make angular vectors and such irrelevant?

13. Jan 21, 2014

### sophiecentaur

Since when did adding gears automatically improve efficiency of a machine?

14. Jan 21, 2014

### Staff: Mentor

20m per hour is the vertical height (assuming your rocks really need 1hour to go along those 40m of monorail). The monorail moves by 40m during this time.

Don't forget units. And you have to convert between linear velocity v and angular velocity $\omega$ first: $\omega r = v$ where r is the radius of the wheel.

I don't think you need "angular vectors" (this expression does not appear at all in the wikipedia page), or any other vectors.

15. Jan 21, 2014

### danny.mcshane

16. Jan 21, 2014

### danny.mcshane

This is a very rough sketch but hopefully it gives an idea.

The monorail is continually moving and around 250 tonnes of ore is transported per hour. No power is required as gravity powers the monorail.

My idea was to fix a chain around the carts that runs over a gear I would install at the top(if anything else would be better I would be very appreciative). Then this directly turns a shaft or if it would be better is geared up to a higher ratio. This could then power a generator of some sort.

I saw the generators wind turbines used and though it looked similar, obviously 250 tonnes pulls down with a lot of force so I was hoping it would be similar to wind blowing blades. They are super expensive but given our power issues and reliance on diesel generators it could be a big help.

Would this even design even work? Is there a much simpler way? I am open to all suggestions.

17. Jan 21, 2014

### sophiecentaur

Pictures like those always put me in mind of PM machines. GPE is independent of path, so the shape is irrelevant.

18. Jan 21, 2014

### danny.mcshane

I found those while I was searching for ideas. I guess as long as the mine is operating it kind of is a PM machine. Well forgetting the power used in the electric powered conveyor carrying the rock up out of the mine :) I just think we could capture some of this potential energy but how much?

If my 13.6kW potential energy could turn a shaft in a PMG it would work. Problem is I am struggling to figure out what size generator would match the GPE figure. I know torque is involved so I am going to keep at it and see if the penny drops.

19. Jan 21, 2014

### pikpobedy

Use a low RPM multipole generator. As for the torque it depends on the electrical load and the power disappation there-in.

20. Jan 22, 2014

### danny.mcshane

OK so I am confused.

So potential energy = mgh which in this instance equals 250,000 x 9.81 x 20 = 49.05MW of potential energy.

An earlier post said 13.6kW though and I tend to not trust my maths so which is right?

Also, I have been working out some torque equations and moments of inertia etc so I think I know how much force is required to turn different drive shafts in generators.

The thing I really need to know is how much force I can generate. Assuming a vacuum and 250 tonnes comes down turning my shaft every hour, what is the maximum resistance I can overcome? I just cant figure that bit out.

21. Jan 23, 2014

### sophiecentaur

It's difficult to take a thread seriously when, after 20 posts, I read "49.05MW of potential energy"

Work out the available Power (rate of work done per second - Watts) and see how that relates to the Power needed to move your train at a given speed up a slope or against some estimated resistance. Knock off 3/4 of the power for practical considerations and that will give you a clue about viability.

The 'force' needed for a drive shaft has nothing to do with the energy required for continuously operation. MI is largely irrelevant once a wheel or shaft is turning.

22. Jan 23, 2014

### danny.mcshane

The train is already moving completing a full revolution in a minute although this is slowed with a hydraulic system for timing reasons. It also is moving free of any electrical energy. The weight of the objects dropped into the carts is sufficient to keep the whole loop moving. It just goes round and round as if it were powered but it isn't.

I will find out how much the breaking slows it so I can get an accurate work per second figure in watts. Given the breaking used there is clearly more power going in than is required to keep it moving. That means there will be some free energy I can take out right?

I know I keep getting things wrong and I am sorry guys, I am trying to understand things as best I can. If it is a dead duck I will leave it but something tells me there is a good idea somewhere in here.

The 49MW thing came from working out the potential energy of the ore being dropped into the carts. I just did E=mgh to get the potential energy. I know the 13.6kW figure someone else got will be right but I don't know why.....I just followed the formula.

Also, once the wheel and/or shaft is turning is there a way to figure out how much torque 250 tonnes will exert on a rotating cylinder? From what I understand I need to match the power being put in to the power I will get out to spec the generator. I have found equations to work it out on levers, as a downward force but what formula would need to be applied to my operation.

23. Jan 23, 2014

### sophiecentaur

That's just the problem. You won't even use the correct units for Work or Power. This has already been pointed out to you.

24. Jan 23, 2014

### danny.mcshane

Sorry, I will give it a go.

The work being done is 16 carts being raised 20 meters. The weight of 16 carts is approx. 3520kg. I don't know enough to apply friction forces yet but I will learn soon.

So here Work (j) = Weight (N) x Height (m) = 34519.4081008128 x 20 = Work = 690388.162016256 joules

Time is tough to work out. It currently takes 30 seconds to lift the weight but breaking is being applied. Is there a way to work out how long a 40m loop would take excluding breaking as none of us know the exact amount of breaking applied. If I stick with 30 seconds though for now.

Power (W) = Work (j) / Time (s) = 690388.162016256/30 = 23012.9387338752 Watts

So I am currently using 23012.9387338752 Watts of power to keep the loop turning.

To work out the power the rocks could be generating I need to state what is happening. I have 250 tonnes of ore falling 20m every hour. Before I get it wrong again my thinking is E = mgh. Is that wrong. Argh, physics is tough.

25. Jan 23, 2014

### danny.mcshane

Oh hang on, times. Keep the times the same