Hydrostatic problem - Impossible integral!!


by jaumzaum
Tags: hydrostatic, impossible, integral
jaumzaum
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#1
Dec26-12, 01:43 PM
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Just studying hidrostatic over the internet and I saw the following problem:

A U-Tube filled with water, initially at rest in a horizontal table, has A1=40cm2, A2 = 20cm2, A3=30cm2, h1 = 80cm, h2 = 20cm and L = 100cm (below piture). It is pressed in A1 by a constant vertical force F of 4N. Find the position x of the surface A1 in function of time (initial position is x=0). Given: g = 10m/s2, density of water is 1kg/L



How can I solve this? I tried some stuff but it didn't worked.

A1.dx1 = A2.dx2 = A3.dx3 (where dx is the infinitesimal variation of position for a infinitesimal variation of time).

By energy conservation:

F.dx1 = Δ(mechanic energy)

Potential Energy -> Now imagine that the dx1.A1 volume of water has just raised and become the dx2.A2 volume. It raised dx1.(A1+A2)/(2A2)
Ep = dx1.(A1+A2)/(2A2).dm.g
But dm = A1.dx1
Ep = dx12.(A1+A2)/(2A2).A1.ρ.g

Kinetic Enegrgy -> Ek = ∫(dm.v.dv)
dm = A1.v1.dt/ρ (v1 is the instantaneum velocity for the surface A1)
v = A1.v1/A (A is the area of the surface considered)
dv = A1.dv1/A

Ek = ∫[(A13v12.dt.dv1)/(ρ.A2)]

But I don't know how to solve this. I don't know even if it is right.
Can anyone help me?
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SteamKing
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Dec26-12, 02:08 PM
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Remember, water is incompressible.

Apply Pascal's law to the two columns after application of the force.
haruspex
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Dec26-12, 03:48 PM
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Quote Quote by SteamKing View Post
Apply Pascal's law to the two columns after application of the force.
Not sure that helps. The fluid is accelerating, so it's hydrodynamics, not hydrostatics. jaumzaum, I didn't understand your KE calculation. Can you not just add up the KEs of the three sections (ignoring complications at the corners)? At time t the force has advanced x, the volumes will be (h1-x)A1, LA3 and (h1+xA1/A2))A2; the speeds v (=dx/dt), vA1/A3, and vA1/A2.
But the corners bother me. Not sure how to use h2.

jaumzaum
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Dec26-12, 04:34 PM
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Hydrostatic problem - Impossible integral!!


I considered the process to be very slow (as the force is 4N). So I think the column of water tend to move orderly. I don't know how to use the h2 too! Actually this problem was from this university: http://www.ita.br/

It was a test applied on the school. You could use computers and you had to plot the graphic position xtime. The other exercises in the same test resulted in integrals that had to be series-aproximated, so I think you have to do some assumptions. But I don't know what!

[]'s
Joao
jaumzaum
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Dec26-12, 04:41 PM
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Not sure that helps. The fluid is accelerating, so it's hydrodynamics, not hydrostatics. jaumzaum, I didn't understand your KE calculation. Can you not just add up the KEs of the three sections (ignoring complications at the corners)?
I think you can ignore the corners. I calculated KE for a random "surface" (it is actually a infinitesimal cylinder) of area A. Then I just added the 3 integrals for the 3 surfaces. But I still can't solve it! I mean, inside the integral is there dx and dt. To take the integral signal off I have to derivate. But derivate in relation to what? Time? Velocity? Position? And I also don't know how to derivate F.dx, for example! Another thing I noticed is that dx12.(A1+A2)/(2A2).A1.ρ.g is very smaller than F.dx, can we ignore it?

How can I solve this?
haruspex
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Dec26-12, 09:25 PM
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Working in terms of energy, you don't need dx's in the PE expression. Just consider that the total advancement of the force at time t is x. Abbreviating h1 to h:
PE = x2.(A1+A2)/(2A2).A1.ρ.g
2*KE = (h-x)A1ρv2 + Lρv2A12/A3 + (h+xA1/A2)ρv2A12/A2
PE+KE = Fx
v2{h(1+A1/A2)+LA1/A3+x((A1/A2)2-1)} = 2Fx/(ρA1) - x2.g(A1+A2)/A2
Chestermiller
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Dec27-12, 10:38 PM
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Quote Quote by haruspex View Post
Working in terms of energy, you don't need dx's in the PE expression. Just consider that the total advancement of the force at time t is x. Abbreviating h1 to h:
PE = x2.(A1+A2)/(2A2).A1.ρ.g
2*KE = (h-x)A1ρv2 + Lρv2A12/A3 + (h+xA1/A2)ρv2A12/A2
PE+KE = Fx
v2{h(1+A1/A2)+LA1/A3+x((A1/A2)2-1)} = 2Fx/(ρA1) - x2.g(A1+A2)/A2
This looks like a very nice development. The OP should also know that

v=dx/dt

This results in a first order ordinary differential equation for calculating x as a function of t. After taking the square root of both sides of the equation, it probably would be best to substitute
x=y2 and solve for y.
jaumzaum
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Dec28-12, 09:19 PM
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Quote Quote by haruspex View Post
Working in terms of energy, you don't need dx's in the PE expression. Just consider that the total advancement of the force at time t is x. Abbreviating h1 to h:
PE = x2.(A1+A2)/(2A2).A1.ρ.g
2*KE = (h-x)A1ρv2 + Lρv2A12/A3 + (h+xA1/A2)ρv2A12/A2
PE+KE = Fx
v2{h(1+A1/A2)+LA1/A3+x((A1/A2)2-1)} = 2Fx/(ρA1) - x2.g(A1+A2)/A2
When I try to solve this (using Wolfram Mathematica of couse), the function does not accept the contour value of 0 for initial value of position.

v2 (3.73333 + 3 x) = 2 x - 30 x2

http://www.wolframalpha.com/input/?i...+30+x[t]^2

What this means?
haruspex
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Dec28-12, 11:52 PM
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As you can see, plugging in x=0 gives v=0, which is true. You would get the same trying to determine a trajectory under gravity using the energy equation y'2 = 2gy. This stops any numerical attempt to produce a plot from there - things never move. You can validly get around it by plugging in a very small nonzero starting value.
Chestermiller
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Dec29-12, 12:00 AM
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Quote Quote by jaumzaum View Post
When I try to solve this (using Wolfram Mathematica of couse), the function does not accept the contour value of 0 for initial value of position.

v2 (3.73333 + 3 x) = 2 x - 30 x2

http://www.wolframalpha.com/input/?i...+30+x[t]^2

What this means?
What it means is that you need to follow the advice I gave you in my previous response. Substitute v = dx/dt, divide both sides of the equation by (3.733 + 3x), take the square root of both sides of the equation, substitute x = y2, and factor y out from both sides of the equation. Then solve the resulting first order ordinary differential equation for y(t). At short times, the solution should be of the approximate form y ~ kt, or x ~ at2/2, where a is the initial acceleration, and where a can be determined analytically in terms of the constants in your equation.

Wolframalpha won't be able to solve the equation until you work it into the form that I indicated.
Chestermiller
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Dec29-12, 01:04 AM
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Quote Quote by haruspex View Post
As you can see, plugging in x=0 gives v=0, which is true. You would get the same trying to determine a trajectory under gravity using the energy equation y'2 = 2gy. This stops any numerical attempt to produce a plot from there - things never move. You can validly get around it by plugging in a very small nonzero starting value.
As I said in my previous post, plugging in a small starting value isn't necessary. If

y'2 = 2gy

If you take the square root of both sides, you get:

[tex]y'=\sqrt{2gy}[/tex]

Next, substitute y = z2 to get

[tex]2z\frac{dz}{dt}=z\sqrt{2g}[/tex]
or
[tex]\frac{dz}{dt}=\frac{\sqrt{2g}}{2}[/tex]

This can now easily be integrated with respect to t. The same type of approach applies to the problem at hand.
haruspex
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Dec29-12, 02:44 PM
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Quote Quote by Chestermiller View Post
As I said in my previous post, plugging in a small starting value isn't necessary.
Sure - an analytic solution gets past the problem. Just wanted to explain that numerical methods can still be made to work when necessary.
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Dec29-12, 03:48 PM
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Quote Quote by haruspex View Post
Sure - an analytic solution gets past the problem. Just wanted to explain that numerical methods can still be made to work when necessary.
Actually, I was alluding to using this approach (substitution of x = y2) to solve the problem numerically, starting out at y = x = 0 at t = 0. I've done this trick many times in obtaining numerical solutions to problems of this type of form.
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Dec29-12, 06:04 PM
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Quote Quote by haruspex View Post
PE+KE = Fx
v2{h(1+A1/A2)+LA1/A3+x((A1/A2)2-1)} = 2Fx/(ρA1) - x2.g(A1+A2)/A2
I believe the factor shown in blue should be 1 + (A1/A2)^2. [EDIT: Nevermind, my mistake!!!]

Anyway, for the numbers given in the problem, I think it's safe to neglect the term in red. Then to a good approximation, the solution is SHM with a small amplitude. [Edit: I'm getting an amplitude of 10/3 cm and a period of about 2.2 s]


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