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Thread: Work needed to pump water

  1. #1
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    Work needed to pump water

    Work needed to pump water-capture.jpg
    ok, somebody please tell me where I am going wrong.

    Basically, I am integrating along the side with length C, so my largest recangle would be 6 times 9 (or a times b). I place the x-axis along side with length c of the triangle that's facing us. so each of the layer i am working with is vertical. This should work, becuase if you basically rotate the image, you can still visualize the water as being pumped towards the faucet right?

    i used similar triangles, so using the triangle that faces us, we get (y/(13-x) = 6/13. thus 13y = 78-6x so y = 6-(6x/13).

    volume of each prism would equal (6-(6x/13)) * 9 * delta x.

    Force = 62.5 * 9 * (6-(6x/13)) * deltax
    work = 562.5 * (6-(6x/13)) * x deltax

    i integrated the work function from 0 to 13, and i get 95062.5 but the my online homework checker says its wrong. it doesnt tell me the correct answer, but can somebbody please tell me where I am going wrong?
    Last edited by toesockshoe; Feb 7th 2015 at 10:25 PM.
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    Re: Work needed to pump water

    Hey toesockshoe.

    Using some basic high school algebra, the area of a triangle is half the base times the height (perpendicular height). Although integration is recommended, for this simple example we can use a simpler approach.

    We have the area of the triangle = 1/2*c*a = 1/2 * 13 * 6 = 39 square feet.

    The volume is given by 39 * 9 = 351 cubic feet.

    Now with regard to the physics I have to say that this is a little ambiguous because the way you pump it in terms of direction and orientation makes a massive difference. If you pump it against the direction of gravity then you can use the fact that you need to cancel the force of the weight of the water (not its mass but weight due to gravity) against that of what is sucking up the water.

    Since everything has the same density across the surface of the volume we use the fact that since the water weighs 62.5 pounds per cubic feet then we have a total mass (in pounds) of 351*62.5 = 21937.5 pounds.

    Now you have to convert to Newtons by first converting to kilograms and using a = 9.8 metres per second^2. In kilograms we get 21937.5*0.453592 (thanks to google) which is 9950.682617 kg.

    In Newtons we have 9950.682617 * 9.8 = 97516.6896466 Newtons.

    Now this assumes that you are sucking the water in the opposite direction to the forces of gravity and that no other forces exist to counteract the water which means that once the forces are counter-acted and the water is flowing then it should keep flowing.
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    Re: Work needed to pump water

    @chiro
    I am using rectangles (vertical rectangles) , not triangles . also, our teacher will expect us to use integrals on the exam because that is what the current chapter is about
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    Re: Work needed to pump water

    also, i need an exact answer. the teacher (nor the online homework) allows an approximation. how would i covert from pounds to km without apporximating. also, is it approriate to convert from pounds to kg in this case? pounds is a force while kg is a mass. wouldnt that mean we are going backwards? we already have force... why would we need to go back to mass?
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    Re: Work needed to pump water

    Pounds is a unit of mass - not a force.

    A force is in Newtons which is kg*m/s^2.

    Pound (mass) - Wikipedia, the free encyclopedia

    I looked at the Force aspect but not the distance aspect (since Work requires Force to be dissipated over some path - which is what Work is).

    Work (physics) - Wikipedia, the free encyclopedia

    Force is always in Newtons and Newtons is always in kilograms.

    Also with regard to angles and rectangles, that just depends on how you do the integration. Integrals just add up tiny little squares anyway but whatever floats your boat.

    I guess you need to integrate though because the displacement will depend on how the level of the water after a certain time.

    If this is the case then you will need to make your displacement as a function of the water already drained and that requires a more complex calculation than you are doing. The reason is because the displacement you have to move the water depends on how much water has left and how much water remains to figure out what level the water is at.

    So you will need to find a way to get the levels as a function of how much water is left.

    Picture the tank having x kilograms of water and finding a function d(x) where d(x) represents the distance between the top of the tank where the spout starts and where the level of the water actually is.

    If you do this then we can further guide you through the process.
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    Re: Work needed to pump water

    . wait. then what is this: http://en.wikipedia.org/wiki/Pound_(force). this is getting confusing. when is pound used as a force, and when is it used as mass?
    Last edited by toesockshoe; Feb 7th 2015 at 11:49 PM.
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    Re: Work needed to pump water

    Pounds has always been a unit of force. The unit of mass in the English system is the slug.
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    Re: Work needed to pump water

    Work = \int WALT

    Where ...
    W = weight density of the liquid, 62.5 lbs/ft^3
    A = cross-sectional area of a representative horizontal slice, 9 \cdot \frac{13y}{6}
    L = lift distance for that slice, 6-y
    T = slice thickness, dy

    Work = \int_0^6 62.5 \cdot 9 \cdot \frac{13y}{6} \cdot (6-y) \, dy = 43875 \, ft \cdot lbs
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    Re: Work needed to pump water

    thanks. but is there a reason why I can not use vertical cross sections? vertical cross sections would still fill up the tank. would it take more or less work to pump the water sideways and out of the tank than it does from the bottom to top? the problem didnt specify which way the water is being pumped out so how do we know whether to use vertical or horizontal cross sections?
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    Re: Work needed to pump water

    Pumping problems are normally set up to calculate work required to overcome the force of gravity in the vertical direction, hence the use of a horizontal cross-section to make calculations easier.

    I suppose one could use vertical cross-sections, but that would require a center of mass calculation with unique distances for lift of each cross-section ... I've never tried it, so I really can't vouch for such a method.
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    Re: Work needed to pump water

    oh alright. also this might be a stupid question, but wouldnt the work required to overcome the force of of gravity need to be greater than 9.8? if gravity is -9.8, then applying a force of 9.8 would make the water just stay as it is right? applying a force of something above 9.8 would allow the water to move up. again, can you tell me where I am going wrong conceptually?
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  12. #12
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    Re: Work needed to pump water

    Initially the force would have to be greater than the force of gravity, but once the mass is moving all that is required is a constant force equal to the force of gravity to keep it moving. The mass would be moving upward with zero net force, i.e. not accelerating.

    That initial force greater than the force of gravity is usually ignored as insignificant compared to the overall work required.
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