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Generating electricity

  1. Jun 9, 2013 #1
    What possible rate of electricity would be generated if a 10,000 lb weight goes down a 30 degree inclined track, attached by a cable that spins the rotor in a generator mounted at the top of the track? Please assume that friction on the track is minimal and that there is a large supply of these weights already at the top ready to go down so that the process continues uninterrupted. Thanks!
     
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  3. Jun 9, 2013 #2
    Do you know what units "rate of electricity" are in? (I assume you mean rate of electrical energy production)
    I think that how fast the 10,000 lb weight goes down will be directly proportional to how much power is produced. And I think how fast the 10,000 lb weight goes down is determined by the mechanical resistance or "load" it feels from pulling on the rotor which is generating the electrical power.
     
  4. Jun 9, 2013 #3

    russ_watters

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    Welcome to PF! Do you know the equation for gravitational potential energy? Can you try to apply it here?
     
  5. Jun 9, 2013 #4
    You have not provided any information which would limit the rate at which electricity could be generated.
     
  6. Jun 10, 2013 #5
    Thanks for the advice about grav potential energy! The equation gave me an answer of 10.2 million joules. I took that to mean the system would produce 10.2 million watts every second, but since it's on a 30 degree incline rather than 90 degrees, only 1/3, or 3.4 million watts, would be generated. Is this a reasonable ballpark number or is it way off in left field? Thanks.
     
  7. Jun 10, 2013 #6

    russ_watters

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    That's the total energy, not the energy per second (power). To find the power(in watts, not watts per second), you need to divide by the number of seconds it takes to lower the weight.

    Also, the angle is irrelevant: only the height matters.
     
  8. Jun 10, 2013 #7
    So if the descent takes 4 minutes, then 10.2 million joules of energy divided by 240 seconds equals 42,500 watts of power? Thanks a ton (in this case 5 tons) for all the advice!
     
  9. Jun 10, 2013 #8

    russ_watters

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    Is the height 230 meters? If so, then you've calculated correctly.
     
  10. Jun 10, 2013 #9

    berkeman

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    So dismantling the pyramids will generate electricity for a while :smile:

    But in the end, you just end up with a buch of stone slabs on the desert floor...
     
  11. Jun 12, 2013 #10
    Thanks for the chuckle! The weight coming down to generate electricity is in the realm of physics, but getting that weight back up to the top also has a lot to do with economics. A simple, sustainable and profitable method will generate electricity for much longer than the pyramids have stood. Watching the climate change in front of my eyes has got me thinking outside the sarcophagus for a method like that. I might have one, since I got help from you fine people at PF! Does $2 worth of energy every hour to get that 10,000 lbs back up the hill sound like a climate-healthy way to generate that 42,500 watts of electricity, up to the very day we run out of gravity to bring it back down? The ancient Egyptians might even be impressed with such a cheap way to get weight up to the top.
     
  12. Jun 12, 2013 #11

    russ_watters

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    Unless you have slaves like the Egyptians did, you won't be able to get the weight back up cheaply (and even then, you have to feed them). However, this could make for a viable energy storage device.
     
  13. Jun 12, 2013 #12

    turbo

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    You don't need slaves if you have the Sun. Solar energy can evaporate water, instill it into clouds in the sky, and make it rain when the moist air encounters cooler air. If the rain falls in drainages leading to river-valleys with hydro-dams, then you have essentially free power.

    Got to pay for the construction of the dams and the transmission system, but after that the maintenance costs are quite modest.
     
  14. Jun 12, 2013 #13

    jbriggs444

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    If that $2 pays for 10,000 pounds to get back to the top of the hill then at 15 trips per hour that's $30/hour. $30 for 42.5 kwh is $0.70/kwh. That's pretty high.

    If that $2 can lift 150,000 pounds back to the top of the hill than that's $0.05/kwh. That's competitive. Though various other costs are still likely to kill you.

    Why be coy -- what's your scheme?
     
  15. Jun 15, 2013 #14
    The equation for gravitational potential energy gives a higher number of KW's the less time it takes a weight to descend, but it seems like the more time a weight spends descending the higher the KW number would be because the rotor would have a higher resistance for a longer time. Am I missing something, or does this maybe fall into the category of turning power into energy ( KW's into kwh's)? How do I find out how many kwh's of energy can actually be produced by 42.5 KW? How do I make the theory jive with what I think should happen? Thanks!
     
  16. Jun 15, 2013 #15

    jbriggs444

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    If the weight is descending at any steady rate then the resistance of the rotor is a fixed function of the size of the weight [and gearing]. It does not depend on the rate of fall at all.

    That is Newton's second law in action. F = ma. If the acceleration of the weight is zero then the net force on the weight must be zero. If the force of gravity is fixed then the force between the weight and the generator must be fixed. [And gearing doesn't buy you anything].

    The faster the weight is descending, the faster energy is produced. This is a consequence of the definition of work. Work = force times distance. The faster an object moves, the more distance it covers in a unit of time. The rate of production of energy is called power.

    How many KWh of energy can be produced by 42.5 KW of power? 42.5 every hour. That's by definition.
     
  17. Jun 15, 2013 #16

    Redbelly98

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    It's a pretty basic calculation. Multiply 42.5 kW times the number of hours that the system is generating power at 42.5 kW.

    Not to be disrespectful, but people might have a hard time trusting your numbers and claims if you were not aware of the relationship between kW-h and kw.
     
  18. Jun 15, 2013 #17

    russ_watters

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    You're missing something basic you already calculated: the kWh is fixed, it is just a converted unit from joules(3,600,000 joules = 1 kWh). The kW changes based on how fast you use your available kWh. So lowering the weight slower means less kW for more time, for the same kWh.
     
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