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Force used to lift water

  1. May 11, 2012 #1

    Finding it hard to find exact information on the maths and forces behind this on the internet, (maybe my search terms to blade)

    If I have a 10cm diamter cylinder thats 10cm deep and filled with water and a piston (weightless piston for the example) at the top and a 1 cm diameter pipe running from the base, how much weight would I have to place on the piston to push all of the water 20cm high into another recepticle?

    Thanks in advance for anyone that can answer this, Also the formulas required to calculate it would be great

  2. jcsd
  3. May 11, 2012 #2
    Assuming you know some classical mechanics, first calculate the mass of the water from its density and volume, then figure out how much force is needed to "beat" gravity in this case.
  4. May 11, 2012 #3
    very basic mechanics so here goes

    (rounded for ease)

    785 cubic centimetres
    0.785 kilograms of mass

    Finding hard to find gravitys specific force online

    Also doesn't pascals law and the surface area of piston come in somewhere?
  5. May 11, 2012 #4


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    There seems to be important information missing. What shape is the second receptacle and how exactly is it positioned in relation to the first? E.g. is it the same shape, with its base 20cm higher? Or a very much wider vessel?
    The reason you need to know is that a weight sufficient to start the transfer might not be enough to complete it. This gets very complicated if we have to take into account the momentum in the water.
    So I'll answer a simpler question for now...
    Suppose the second vessel is like the first, 20cm higher, and take the pipe to be very narrow (so no momentum to worry about).
    To get the flow started, you just need the pressure at the bottom of the first cylinder to exceed that needed to support a 20cm column. You have 10cm of pressure coming from the water in the first vessel, so you only need an extra 10cm worth from the weight. I.e. the weight would need to equal the weight of the water.
    But as the flow continues, you're getting less help from the weight of water in the first vessel and more push back from that in the second. At the end, you're having to push water up 30cm with no help at all, so you now need 3 times the weight.
    Thinking about it some more, I suspect that the second vessel is irrelevant: the pipe rises 20cm and all you are intended to do is push all of the water into the pipe (with it overflowing from the top).
    Last edited: May 11, 2012
  6. May 12, 2012 #5
    Thanks for the reply

    Yeah, sorry I didn't make that clear, i've attached a diagram

    Thanks again

    Attached Files:

    Last edited: May 12, 2012
  7. May 12, 2012 #6


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    OK, so the details of the second vessel are irrelevant. It will never exert any pushback on the water. All that matters is pushing the water to the top of the pipe.
    As I said, the key thing to consider is the pressure at the bottom of the pipe. For the water to be reaching the top of the pipe this must be at least the equivalent of 20cm of water (the height of the pipe). At first, the water in the first vessel supplies 10cm of pressure, so you need another 10cm pressure from the weight. This means the weight will be the same as the weight of the water, π/4 kg.
    But as the water level in the first vessel goes down, the extra weight needed will increase. Strictly speaking, we'd have to take into account the momentum of the water flowing in the pipe, which all gets pretty complicated; but if we assume it flows arbitrarily slowly then we can just consider the static position when the first vessel is almost empty. At this point the weight on top is getting no help from the weight of water and has to supply the entire 20cm of pressure. Hence the weight needed will be twice as much, π/2 kg.
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