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Pascal's Principle with Hydraulic Pumps

  1. Jan 9, 2012 #1
    1. The problem statement, all variables and given/known data

    A hydraulic can crusher is shown in the figure. The large piston has an area of 8m2 and exerts a force F2 of magnitude 2x106 N on the cans. Calculate the magnitude of the force F1 exerted by the small piston (area 10cm2) on the fluid. Do not ignore that the large pistons is 1 m higher than the small piston.


    2. Relevant equations

    P1 = P2
    Thus
    F1/A1 = F2/A2

    Figure:
    http://bit.ly/yuDaZ7 [Broken]


    3. The attempt at a solution

    What is the significance of the height of the piston ? I cannot solve it without knowing that information is it very important?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 9, 2012 #2

    gneill

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    The height difference will create a pressure difference due to the action of gravity and the weight of the working fluid. See Pascal's Principle.

    The problem doesn't state what the hydraulic working fluid is. You'll need to know its density in order to calculate the pressure difference due to height.
     
    Last edited by a moderator: May 5, 2017
  4. Jan 9, 2012 #3
    I don't know what you mean. Our professor never really elaborated in Pascal's Principle all she said was P1 = P2 :/
     
  5. Jan 9, 2012 #4
    ρgh = F2 / A2

    ρ(9.8m/s^2)(1m) = 2x10^6 N / 8m^2


    Unknown Density? Can we assume that it is water?
     
  6. Jan 9, 2012 #5

    gneill

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    Pascal's Principle implies that any pressure induced into a (confined) fluid at one location will be felt equally at every other location in the fluid. This is independent of any pressure differences that may exist due to other reasons like gravitational potential differences (height differences). When the hydraulic system is assumed to be all at the same height then the pressure is the same everywhere, hence your P1 = P2.

    In this problem you are told that there is a height difference, so the pressure at the large piston will be less than the pressure at the small piston by an amount fixed by the height difference and the density of the working fluid: ρgh.

    You should be able to write an equation that incorporates this fixed pressure difference into a relation between the forces at the piston faces.
     
  7. Jan 9, 2012 #6

    gneill

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    If you are not told what the working fluid is then you can either make the assumption that it is water (state this assumption in the solution that you hand in!), or you can leave the density in symbolic form and present a symbolic result.
     
  8. Jan 9, 2012 #7
    So what you are saying is that P1<P2 when there is height? So must I change the equation to inequalities then? or its still the same as P1 = P2 but this time it will be --> ρgh = F/A?
     
  9. Jan 9, 2012 #8

    gneill

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    First let's make it clear which piston is which. The problem places piston 2 higher than piston 1. That means that the pressure P2 at piston 2 will be less than the pressure P1 at piston 1 by a fixed amount.

    The equation will be an equality, not an inequality.
     
  10. Jan 9, 2012 #9

    I don't need to consider the pressure of the Larger Piston because it is in a different elevation.

    Thus:

    P1 = ρgh

    F/A = ρgh

    F = A(ρgh)

    F = 1x10^-3 m^3 (ρ)(9.8 m/s^2)(1m)

    F = ρ( 9.8 x 10 ^-3 m/s^2 )

    Correct?

    *EDIT* I think this is highly wrong.. Im really confused.

    Can you just walk me through ?
     
    Last edited: Jan 9, 2012
  11. Jan 9, 2012 #10

    gneill

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    :confused: How do you justify that??? :confused:

    Of course you must consider the pressure at the large piston! It's the pressure at the large piston that is creating the force to crush the cans. You are given a value for that force, and you are given the area of the piston...
    I cannot just present the solution to you. I can continue to give hints and suggestions.

    Since you have the can-crushing force applied by piston 2, and you have the area of piston 2, what is the pressure of the hydraulic fluid at piston 2?
     
  12. Jan 9, 2012 #11
    Pressure = 2x10^6 N / 8 m^2

    Pressure = 250000 Pa

    P1 = P2 + ρgh ? Can this be? so that two pressures will be equal?

    F/A = 250,000Pa + ρ(9.8m/s^2)(1m)
    F = 1x10^-3 m^2 [ 250,000Pa + ρ(9.8m^2/s^2) ]

    F = 250 N + (9.8 x10^-3 m^4/s^2) ρ

    ASSUMING FLUID IS WATER

    F = 250 N + ((9.8 x10^-3 m^2/s^2) 1000 kg/m^3

    F = 250 N + 9.8 N
    F = 259.8 N

    Correct?

    What if the machine has the efficiency of 90%? what does this mean?
     
  13. Jan 9, 2012 #12
    Pressure = 2x10^6 N / 8 m^2

    Pressure = 250000 Pa

    P1 = P2 + ρgh ? Can this be? so that two pressures will be equal?

    F/A = 250,000Pa + ρ(9.8m/s^2)(1m)
    F = 1x10^-3 m^2 [ 250,000Pa + ρ(9.8m^2/s^2) ]

    F = 250 N + (9.8 x10^-3 m^4/s^2) ρ

    ASSUMING FLUID IS WATER

    F = 250 N + ((9.8 x10^-3 m^2/s^2) 1000 kg/m^3

    F = 250 N + 9.8 N
    F = 259.8 N

    Correct?

    What if the machine has the efficiency of 90%? what does this mean?
     
  14. Jan 9, 2012 #13

    gneill

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    Yes! That result looks good.
    Generally such a statement means that only 90% of the work done at the input end of a machine gets delivered to the output. Typically this is due to frictional or other losses in the mechanism.

    The result will be that a force higher than you calculated without taking such losses into consideration will be required at piston 1 in order to deliver the desired can-crushing force at piston 2.
     
  15. Jan 9, 2012 #14
    Yay! Thanks! I also found out in the web that the formula is called "Pressure Vs Depth of Static Fluid."
    P1 = P2 + ρgh.


    So in terms of computation of Efficiency, if the machine has only 90% efficiency it means it only delivers 0.9(259.8N) or 233.82 N to the output?

    EDIT:

    Or do we still need to take account that

    Efficiency = AMA / TMA x100% ?
     
    Last edited: Jan 9, 2012
  16. Jan 9, 2012 #15

    gneill

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    It is stated that the machine delivers a particular force at the output, so presumably you will want to calculate the required force at the input that will result in that force being delivered at the output despite the losses due to efficiency.
     
  17. Jan 9, 2012 #16
    There is no given input force.. so is this even possible?

    TMA = requires distance moved and is also not given
     
  18. Jan 9, 2012 #17

    gneill

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    You calculated the force required at piston 1 given that the machine is 100% efficient. You need to modify this value to account for the stated efficiency.
    I think you'll find that the work done (being force x distance) is proportional to the force. The distances involved can be assumed to remain the same despite efficiency. Thus you need to apply the efficiency consideration to the force at the input.
     
  19. Jan 9, 2012 #18
    259.8N is the force of required of the piston 1 at 100% efficiency.
    at 90% efficiency it only gives about 233.82 N of Force.

    Correct?
     
  20. Jan 9, 2012 #19

    gneill

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    No, it should require MORE force at the input to achieve the same 2 x 106N at the output.
     
  21. Jan 10, 2012 #20

    So, 90% of it is only 233.82

    We need the force to be 110% so it can match the output force which will be

    285.78 N of force in the input ? Correct? (
     
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