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Energy Generation through Gravity (mining operation monorail)

by danny.mcshane
Tags: energy, generation, gravity, mining, monorail, operation
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pikpobedy
#19
Jan21-14, 08:17 PM
P: 39
Use a low RPM multipole generator. As for the torque it depends on the electrical load and the power disappation there-in.
danny.mcshane
#20
Jan22-14, 06:16 PM
P: 31
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.
sophiecentaur
#21
Jan23-14, 03:58 AM
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Quote Quote by danny.mcshane View Post
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.
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.
danny.mcshane
#22
Jan23-14, 04:16 AM
P: 31
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.
sophiecentaur
#23
Jan23-14, 04:33 AM
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Quote Quote by danny.mcshane View Post
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.
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.
danny.mcshane
#24
Jan23-14, 05:02 AM
P: 31
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.
danny.mcshane
#25
Jan23-14, 05:03 AM
P: 31
Oh hang on, times. Keep the times the same
danny.mcshane
#26
Jan23-14, 05:28 AM
P: 31
So 250 tonnes of rock an hour is 69.44kg of rock per second.

E = 69.44 x 9.81 x 20 = 13624.128 Watts

Oh bugger, that doesn't even work. It has to be more than the power to make it go round.

EDIT: Ok, does the fact that there are 16 x full carts moving down mean I multiply the above by 16?

If that was the case it would be each cart I providing 13kw so 16 = 217986.048 Watts of power. Is that the energy generated on the side going down. And then the total energy in the system is:

Power in 217986.048 - Power Used (Excluding friction) 23012.9387338752 = 194973.1092661248 Watts of spare energy. And this is with the breaking system in place.

OK so if I can match the resistance of whatever generator I use to get the exact right RPMs to time with the conveyor I can extract energy, keep everything in time and maybe even be right for once :)

So my generator needs to be no more than 194KW Output to not waste anything and I need 1RPM. Am I on the right track?
danny.mcshane
#27
Jan23-14, 10:54 AM
P: 31
I cant find anything like that but.....would a 50kW Generator and 4RPM be the same? and 25kW with 8RMP and so on. Would they generate the same power overall or is the power rating its production total regardless of anything?
mfb
#28
Jan23-14, 02:58 PM
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Quote Quote by danny.mcshane View Post
So 250 tonnes of rock an hour is 69.44kg of rock per second.

E = 69.44 x 9.81 x 20 = 13624.128 Watts

Oh bugger, that doesn't even work. It has to be more than the power to make it go round.
What has to be more? You worked out the 13.6kW we had all the time.

EDIT: Ok, does the fact that there are 16 x full carts moving down mean I multiply the above by 16?
No, each cart is responsible for 1/16 of that value. You don't have to care about the number of carts if the amount of mass lowered at the same time is constant.

So my generator needs to be no more than 194KW Output to not waste anything and I need 1RPM. Am I on the right track?
I have no idea where this 1RPM comes from, but it is wrong. Your generator needs something like 15kW as design power (a bit more is fine, a bit less is also fine if you can use the brakes), the design rotation speed depends on the mechanism between monorail and generator.

You should roughly match the designed power (or at least don't be above it) and be in the right RPM range (not above the specifications) at the same time.
sophiecentaur
#29
Jan23-14, 03:34 PM
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This is all very interesting stuff but open cast mining is big business. Why are you not employing a professional consultant for this project? Whatever you may learn from this thread, can you risk making any decisions based on it when there is so much money at stake?

This is the 'department of free advice', of course. But you have no idea how much of this advice is good and absolutely no comeback on us. It makes me wonder.
danny.mcshane
#30
Jan24-14, 03:08 AM
P: 31
I know, but we are backed by a Chinese company and to be honest they couldn't care less if we have power or not on an evening. We asked for a generator for the residence block and were refused point blank. I would love to hire a consultant but I am one up from a grunt (I used to be a grunt in the UK) and I cant authorise purchasing teacups.

There isn't a lot to do on an evening so after countless nights moaning we were thinking about what we could do. This seemed obvious but it is all 's to the top brass. I want to present it to them but they will rip me to pieces if I am wrong so I really want to get it right.

If I could tell them it will cost X but they wont need the generators anymore then they might listen. Sadly I was the 'smartest' one and got lumped with the task of investigating it. I really appreciate all the advice and I am trying to take as much in as I can. All I need to do is get the confidence to request the meeting.

Ultimately I will present it and they will run with it so it will get the proper attention it deserves I am sure.
danny.mcshane
#31
Jan24-14, 03:29 AM
P: 31
Quote Quote by mfb View Post

No, each cart is responsible for 1/16 of that value. You don't have to care about the number of carts if the amount of mass lowered at the same time is constant.
But there will be 16 carts full of 69.44kg of ore all moving down at any 1 time. So the amount of mass being lowered will be 1111.04kg per meter per second.
sophiecentaur
#32
Jan24-14, 05:18 AM
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Quote Quote by mfb View Post
What has to be more? You worked out the 13.6kW we had all the time.

No, each cart is responsible for 1/16 of that value. You don't have to care about the number of carts if the amount of mass lowered at the same time is constant.

I have no idea where this 1RPM comes from, but it is wrong. Your generator needs something like 15kW as design power (a bit more is fine, a bit less is also fine if you can use the brakes), the design rotation speed depends on the mechanism between monorail and generator.

You should roughly match the designed power (or at least don't be above it) and be in the right RPM range (not above the specifications) at the same time.
You will need the brakes for when there is no load on the generator, of course. There may well be a problem with speed regulation as the alternator will need to be run within fairly tight limits and it will not be 'convenient' to regulate the speed of the rock carrying to fit in with the electrical demands.

I am rapidly coming to the conclusion that a diesel generating set is the right answer for this application. Off the shelf, a known cost and can be sold afterwards. A number of small petrol generators could be even more practical. A bit of lighting and a TV would easily be fed from a 1kW set from a DIY / tool store.
danny.mcshane
#33
Jan24-14, 06:03 AM
P: 31
Sorry I should have said, the residence block has 1500 people living in it. The site is supplied by mains electricity although lately this has been off more than it has been on. Typically we see cuts regularly through the day but they are short. On a night it is possible to have no mains supply at all.

When the power stops to site we stay running for about 30 minutes through the UPS type thing they have.

Essential processes are powered by generators because that makes money. Workers are left in the dark as we don't matter. We had a generator of our own, probably like the one you are mentioning. It got very difficult though when the supervisors had power and the workers didn't. We took the decision to use it as a sort of emergency light system.

We are now in a kind of all or nothing scenario where either everyone has power or no-one does.
mfb
#34
Jan25-14, 11:18 AM
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Quote Quote by danny.mcshane View Post
But there will be 16 carts full of 69.44kg of ore all moving down at any 1 time. So the amount of mass being lowered will be 1111.04kg per meter per second.
No, and I have no idea how you combined your numbers to get that value.
250 tons/hour over 20m is 5000 tons*meters/hour, or 1389kg*meter/second.


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