Energy Generation through Gravity (mining operation monorail)

  1. 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. jcsd
  3. mfb

    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.
  4. 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.
  5. mfb

    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.
  6. 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
  7. mfb

    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.

  8. 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.
  9. 13.6 kW is enough to heat an average house in winter. It is about 20 hp.
  10. sophiecentaur

    sophiecentaur 13,292
    Science Advisor
    Gold Member

    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.
  11. 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 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.
  12. mfb

    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.
  13. 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

    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?
  14. sophiecentaur

    sophiecentaur 13,292
    Science Advisor
    Gold Member

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

    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.
  16. Very Rough Design.jpg
  17. 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.
  18. sophiecentaur

    sophiecentaur 13,292
    Science Advisor
    Gold Member

    Pictures like those always put me in mind of PM machines. GPE is independent of path, so the shape is irrelevant.
  19. 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.
  20. Use a low RPM multipole generator. As for the torque it depends on the electrical load and the power disappation there-in.
  21. 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.
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