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Weight to electrical energy.

  1. Sep 2, 2013 #1
    I am trying to calculate the output of a weight imposed on a electrical generator via a gearbox.

    Can anyone please help me with this?

    Let say the weight is 500 000 metric tonnes. How much electricity could this produce if putting a constant force on the gearbox?

    This is a theoretical question.

    Like a truck full of load rolling down a hill. Using the gearbox to brake the truck to a constant decent. How much electricity would the dynamo produce?

    Best regards,

  2. jcsd
  3. Sep 2, 2013 #2

    Andrew Mason

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    Welcome to PF!

    How high is the hill?

  4. Sep 2, 2013 #3


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    Staff: Mentor

    To expand: foce isn't energy. You need the force and the distance traveled.
  5. Sep 2, 2013 #4
    Thank you!

    Ok to be more precise.

    Ill give a better description:

    A oil tanker is ankered to the bottom of the sea with a contraption connected to a pinon( that keeps constant pressure on the gears except when the tide turns) which is connected to a gearbox. The ship weights 500 000 Metric tonnes. The tide makes the water level lower itself 1 meter in 6 hours before it pushes the boat back up for the next 6 hours. the cycle repeats itself off course until the oil companies torpedoes the ship to oblivion. An electro-generator is powered by the gears. How do i calculate the electricity produced?

    thank you again and thanx for a great forum!

    Last edited: Sep 2, 2013
  6. Sep 2, 2013 #5


    Staff: Mentor

    The change in gravitational PE is mgh. For m = 500E6 kg, g = 10 m/s^2, h = 1 m we get PE = 5 GJ. That is about 1400 kWhr which is worth about $160.

    I doubt the oil company will sweat too much.
  7. Sep 2, 2013 #6
    so a 500 000 tonnes ship produces as much as a small diesel generator?
  8. Sep 2, 2013 #7


    Staff: Mentor

    1400 kWhr over 6 hr is a little more than 200 kW, so yes. Plus, you can run the generator all day instead of only 50% of the time.
  9. Sep 2, 2013 #8
    You can still run the ship generator 24 hours except the 4 moments where the tide is at its absolute bottom or top. am i right? Gravity will take it down an boyancy will lift it up. 500 000 tonnes cranked into a gearbox must be able to do a lot more than a 0,5 Hp diesel generator...am i missing something?
  10. Sep 2, 2013 #9
    and the distance the tide is moving up and down is irrelevant. the degree of change on the pinon on the crown wheel is not hovever. it will be the same. up and down is the same distance.
  11. Sep 2, 2013 #10


    Staff: Mentor

    I wouldn't see how. The anchor chain can only exert tension so you will only get power during a rising tide.

    200 kW is more like 250 Hp.
    Last edited: Sep 2, 2013
  12. Sep 3, 2013 #11
    The pinon drives the crown wheel like the wheels of a truck. Whether the truck rolls backwards or forwards is irellevant. The boyancy will push the ship up and gravity will pull it down. Each putting force on the crown wheel.
  13. Sep 3, 2013 #12


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    Assume that the distance spanned by the tide in meters is given by h. [We have been given that h = 1]

    How much additional sea water does the tanker displace if one pushes it down by this much. Call this displaced weight D. A moment's thought should convince you that for a 500,000 metric ton tanker with a keel depth of more than one meter and a reasonable hull shape, D < 500,000 metric tons.

    How much less sea water does the tanker displace if one pushes it up by this much? Let us assume a hull profile so that this quantity approximately equal to D.

    Assume that the tanker is afloat.

    On the downstroke, the force that will be applied on the generator mechanism can be no more than D. On the upstroke, the force that can be applied to the generator mechanism can be no more than D. Worse, if the generator is actually resisting with this much force, the tanker will neither rise nor fall at all and the power generated will be exactly zero.

    It seems obvious that the optimum strategy for power generation would be resist the downstroke until tide is at rock bottom and then harvest energy equal to D*h/2.

    Then on the upstroke, the optimum strategy would be to resist the upstroke until tide is at its peak and then harvest energy equal to D*h/2.

    The factor of two assumes that the upward or downward force varies linearly through the stroke.
  14. Sep 3, 2013 #13


    Staff: Mentor

    Regardless, it is still just 200 kW or so. Even at 100% duty cycle it is not a whole lot.
  15. Sep 4, 2013 #14


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    Certainly not a lot. In addition, the achievable power is lower than this by a large factor.

    500,000 metric tons divided by 28 meters draft gives an available force of about 18,000 metric tons(force) per meter that the water rises or falls. This costs a factor of 28.

    18,000 metric tons = 18,000,000 kg. Multiply by gravity and that's 180,000,000 Newtons maximum. As explained previously, this maximum force is halved over the full tidal stroke. This costs a factor of 2. Multiply 180,000,000 Newtons by a 1 meter stroke and divide by 2. That's 90,000,000 Joules per stroke.

    Divide 90,000,000 Joules by six hours to get about 4000 Joules per second. That's 4 KW.

    [As a sanity check, 200 KW divided by 28 divided by 2 is about 4 KW, so the numbers agree].
  16. Sep 4, 2013 #15


    Staff: Mentor

    Hmm, that is a good point that I didn't think about. The 500 kTonne is not necessarily relevant. It displaces that much water and becomes neutrally buoyant. Neutral buoyancy means 0 force and therefore 0 power extracted. As the tide drops you take in the chain so that it is not slack but no tension either. Then, you lock the chain as the tide rises generating some positive buoyancy and tension in the chain. That positive buoyancy reaches 18 kTonne * g. Power is extracted as you let the chain play out under tension, but that tension is always less than 18 kTonne * g.

    Thanks for correcting me.
  17. Sep 4, 2013 #16
    Please explain this. Im dont get it. sorry.

    what chain are we talking about here? im lost
  18. Sep 4, 2013 #17


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    Let me try to say it slightly differently then.

    Suppose that the tide dropped so far that the tanker was completely out of the water. Then you would get the entire 500 kilotonnes of weight pushing downward. Any lesser drop in water depth would result in a lesser downforce. The draft of a supertanker of this size is in the neighborhood of 28 meters. If the tide only goes out by one meter than one could estimate that only 1/28th of the weight of the tanker would be available as downforce.

    Similarly on the way back up. If the water rose so much that the draft of the supertanker doubled to 56 meters (without sinking the tanker or staving in its sides) then one could get 500 kilotonnes of net upward force -- the first 28 meters of draft having been taken up just to get to neutral buoyancy and zero net force. If the water rose by only one meter then one could estimate that only 1/28 of the weight of the tanker would be available as upforce.

    Either way, instead of a one meter stroke multipled by 500 kilotonnes of force you have a one meter stroke multiplied by 18 kilotonnes of net force.

    Does that part make sense?
  19. Sep 4, 2013 #18


    Staff: Mentor

    The anchor chain. You mentioned that it was "ankered" to the ocean floor in post 4.
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