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Thread: Work Required To Pump Water

  1. #1
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    Work Required To Pump Water

    A reservoir has the shape of a vertical cylinder with height 3m and radius 4m and it is filled with water.

    i) Let x be the height in meters measured from the bottom of the reservoir. The weight in Newtons of a thin layer of water between x and x+delta x is approximately P(x)(delta x). What is P(x)?

    ii) The work in Joules required to pump the thin water layer to 3 meters above the reservoir is approximately w(x)(delta x). What is w(x)?

    iii) Using previous results, find, in Joules, the amount of work required to pump all the water 3 meters above the reservoir.

    I am unable to solve part (i), so if someone could shed a little light on this question, I may be able to solve the rest, thanks!
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  2. #2
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    Re: Work Required To Pump Water

    Since when is weight measured in Newtons? Oh wait, force due to mass and gravity, yep a force, ok Newtons works

    Do you have a formula to find weight from a volume?
    Thanks from GHunder
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  3. #3
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    Re: Work Required To Pump Water

    Quote Originally Posted by Prove It View Post
    Since when is weight measured in Newtons? Oh wait, force due to mass and gravity, yep a force, ok Newtons works

    Do you have a formula to find weight from a volume?
    The only things we are given are density = 1000 kg/m^3 and g = m/s^2. What I did was find the volume of the cylinder, and then multiply that by 1000 and 9.8 in an attempt to find the weight, but I kept getting the wrong formula. :s
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  4. #4
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    Re: Work Required To Pump Water

    I don't know calculus but here is my attempt
    mass of water in tank (kg)*9.8* integral from 0 to 3 of (6-x) dx = W in joules
    Thanks from GHunder
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  5. #5
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    Re: Work Required To Pump Water

    This is easier than you think.

    1. The weight of water in a layer dx thick is its mass times g, and the mass is the volume times density. So:  dw = \rho A g dx, where A is the area of the cylinder. Here I'm using lower case w for weight.
    2. The work to raise water of weight dw to by height h is dW = hdw. Here I'm using upper case W for work. So for a slice of water at height x, the distance it must be raised is 6-x, and the work required is dW=(6-x)dw.
    3. The total amount of work to pump all the water to a height of 6 meters is  W = \int _0 ^3 (6-x)dw = \int _0 ^3 (6-x)\rho A g dx
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    Re: Work Required To Pump Water

    Quote Originally Posted by ebaines View Post
    This is easier than you think.

    1. The weight of water in a layer dx thick is its mass times g, and the mass is the volume times density. So:  dw = \rho A g dx, where A is the area of the cylinder. Here I'm using lower case w for weight.
    2. The work to raise water of weight dw to by height h is dW = hdw. Here I'm using upper case W for work. So for a slice of water at height x, the distance it must be raised is 6-x, and the work required is dW=(6-x)dw.
    3. The total amount of work to pump all the water to a height of 6 meters is  W = \int _0 ^3 (6-x)dw = \int _0 ^3 (6-x)\rho A g dx
    If you don't mind me asking, why are you including the area in the formula then, if I need to use volume x density?

    I have tried inputting the following:
    P(x) = pg(pi 4^2 x)
    P(x) = pg[2pi(4)(x)+2pi(4^2)]
    obviously substituting the density and gravity for their respective values... And we are not supposed to include the delta x at the same.

    I am ripping my hair out because of this question haha.
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  7. #7
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    Re: Work Required To Pump Water

    I have stumbled upon the solution on my own finally.. It turns out, I was not expected to include the height variable in my equation as it was already supplied outside of the answer box. The answer for part 1 was in fact 156800pi; volume of cylinder (without the x variable for height) x density of water x gravitational force.
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