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Question about Pascal and pistons

  1. Dec 26, 2008 #1
    i know the usual examples of Pascal's principle (PP) and applications to pistons involves applying a force on the smaller area piston resulting in the amplified force of the large area piston. My question is kind of backwards. The numbers are a little big but just humour me...

    data:

    area of small piston is 1 square meter
    area of large piston is 10000 square meter
    the fluid medium is water
    the downward force of the larger piston is 10 million newtons
    the depth of the bottom of the large piston is 100 meters.
    gravity is present

    My question is 2 fold:

    1. What is the the maximum theoritical distance the smaller piston can be pushed upward towards the sky?

    2. Isn't the column of water in the small piston with the increasing height going to cause a large pressure (because of gravity) so as to limit the downward pressure from the larger piston? In other words, isn't the weight of the fluid taken into account when you answer this question?
     
  2. jcsd
  3. Dec 26, 2008 #2
    First, assume that before the 10 millon N appeared on the large piston, the system was in steady state. The water level on both the thin tube and the fat tube MUST be equal- Stevins law - pressure at the bottom of both thin and fat sides are equal.

    Now the 10 million N suddenly appears on the large side. That tends to push the water down
    on the fat side and it therefore causing it to rise on the thin side. How much does it rise, is what you have asked...
    It rises only so much (x) as to sustain the excess pressure on the fat side:
    xdg = F/A = 10*10^6/10^4; d = 10^3 kg/cu-m
    Therefore x = 1/g meters (where g = acc. due to gravity).

    Whatever water level is in the fat side, after the 10 million N force is applied, the corresponding water column up to the same level on the thin side will balance the pressure due to water on the fat side.
    So anything above that level in the thin column is enough to sustain the 10 million N over
    a 10000 sq-meter area..on the fat side!!
    All it takes is just a 1/g m rise in the level = a little over 10 cm.

    This is regardless of if the thin side is 1 sq-m in area or 100 sq-m.



    The weight is of the fluid is what results in the pressure at the bottom of the columns...
     
  4. Dec 26, 2008 #3

    stewartcs

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    Give this a read, it should answer all of your questions.

    http://hyperphysics.phy-astr.gsu.edu/hbase/pasc.html#hpress

    CS
     
  5. Dec 27, 2008 #4
    yes, very good sight... Ok so this is exactly what i am wondering about.

    W1= work of larger piston
    W2=work of smaller piston

    where W=f * d then:

    f1 *d1 = f2 * d2 (because of the law of conservation of energy)

    and if (given earlier) f1 = 10 million newtons
    and assume d1 = 30 meters

    we derive f2 from f2/a2 = f1/a1 (knowing a2 = 1; a1 = 10,000)

    f2 = 1000

    now plugging in f1 = 10 million newtons
    d1 = 30 meters
    f2 = 1000 newtons

    then d2= (f1 *d1)/f2

    d2 =300,000 meters

    I am not sure i understand how to incorporate this in practice b/c i assume gravity works against 300,000 meters going straight up against gravity. I am having trouble understanding
    the derivation of how Sai got just 10 centimeter. Is that correct? Can someone just tell me a number if gravity prevents d1 from moving too far up, then what exactly is it in the real world with gravity?
     
    Last edited: Dec 27, 2008
  6. Dec 27, 2008 #5

    stewartcs

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    Using the given the numbers, and the equation:

    [tex]d_1 = \frac{A_2}{A_1}d_2 [/tex]

    then yes, you would theoretically get a fluid column that was 300,000 meters high. Obviously, this is not even remotely practical. In order to create a fluid column of that height in a hydraulic piston/cylinder arrangement such as the one we are discussing, you would have to apply an astronomical force to the larger piston. The hydrostatic pressure at the bottom of the fluid column that you are trying to support (using water as the fluid) would be 1000(kg/m^3) x 9.81(m/s^2) x 300,000 m = 2.943 x 10^6 kPa or about 426,846 psi. This pressure would act on the surface area of the larger piston and generate an upward force that would have to be countered with an equal and opposite downward force. The downward force that would need to be applied to the larger piston would then be 2.943 x 10^6 kPa x 10,000 m^2 = 29,430,000,000 kN.

    CS
     
  7. Dec 27, 2008 #6
    Just wait! how can you just assume d1 = 30m? There is water on the large piston side,
    not free space that piston can fall in as much as it wants!!!

    OK here look at it another way...

    Suppose, before the 10 mil N force appeared on the fat side, the water level was
    h on both sides.
    Once the f1 = 10 mil N force appeared, the fat side sunk by y m and
    the thin side rose by x m. Thin side area = a, fat side area = A

    Now the volume of water displaced is constant: so xa = yA (ignore the compression
    of water for now).

    In the new steady state, pressure at the bottom on both sides are equal again:

    (h+x)dg = f1/A + (h-y)dg, where y = x*a/A

    so x = f1/[(A + a)*dg]

    Put f1 = 10^7 N, A = 10^4, a = 1, d = 10^3.

    Now A >> a, so ignore a in the denominator.

    You get x = 1/g meters..!!! Large piston sinks ONLY by
    y = (1/g)*(a/A) = (1/g) / 10000 m.

    Now you apply the energy conservation.

    f2 = f1a/A = 1000 N

    f2 * x = 1000/g J
    F * y = 10^7 / g / 10000 = 1000/g J

    or f1 * y = f2 * x.

    Happy now??

    No matter how you slice it or dice it, it should be the same....
     
  8. Dec 27, 2008 #7

    stewartcs

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    The assumptions being used are that the piston/cylinder assembly has the correct dimensions to enable one to press the larger piston down 30 meters, thereby lifting the smaller one 300,000 meters, and that they both start at the same position and are full of water.

    So, if one pushes the larger piston down into the cylinder 30 meters, the smaller piston will rise 300,000 meters (given the really large area of the large piston and thus large volume of water being displaced).

    CS
     
  9. Dec 27, 2008 #8
    With 10 Mega N on the large piston, you can only push it down 1/(g*10000) m, because the water column rise by 1/g m in the small piston will cause and equal counter force of 10 Mega N upwards on the large piston side. That is my point. To push any further down on the large piston, you need to increase the force beyond 10 Mega N. Of course if you do manage to push it down 30 m (by increasingly applying larger and larger force) on the small piston side you will reach 30 * 10000 meters... but with not just 10 Mega N....no?
     
  10. Dec 27, 2008 #9

    stewartcs

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    Correct, if you are limited to 10 MN of downward force on the larger piston, then you cannot reach the 300,000 meters (I actually didn't see that applied force constraint in the orginal post - my mistake). You would need the 29,430,000,000 kN of downward force like I posted earlier to reach the 300,000 meter mark.

    So if you are limited to just 10 MN of downward force on the larger piston the smaller one would only go about 0.1 meters up.

    CS
     
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