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Fluid Mechanics: Point like sink

  1. Jan 24, 2007 #1
    Fluid Mechanics: Point like "sink"

    I am badly badly confused by the following problem and its solution. I hope that someone would be kind enough to help me out! Thank you for your help and your time!:smile:

    1) The plunger of a hypodermic needle (a point-like "sink") is being pulled up with speed v_p as shown in the first link/picture. The fluid at the needle is v_n given by the equation of continuity. What is v(r), the speed field, in the surrounding fluid? (Hint: draw a sphere of radius r centered at the tip of the needle)
    [​IMG]

    The model solution is:
    [​IMG]

    NOTE: Q is the volume flow rate.

    And I am completely lost after raeding this solution. I don't understand a single step from the beginning to the end.

    (i) What is the REASON of drawing a sphere around the tip of the needle?
    (ii) Why is it true that dQ=v(r)dA? What I have learnt is that Q=vA, not dQ=v(r)da.
    (iii) What is dA?
    (iv) Why did they take the indefinite integral to both sides?
    (v) How come v(r) is brought OUT of the integral sign? v(r) should be varying and is NOT (vi) a constant, right?
    (vii) How actually did they get the 4pi(r^2) out of nowhere?
    (viii) Is the origin r=0 set to be at the centre of the sphere (i.e. the tip of the needle)? What does v(1) represent, for example?
    (ix) v_p and v_n are not used in any part of the model solution. Does it mean that the speed fluid given by v(r)=Q/(4*pi*r^2) would exist even if the plunger is stationary (i.e. not pulled)? How is this possible? Shouldn't the surrounding fluid be STATIC?

    (x) One more question: when it says that the plunger is being pulled up at speed v_p, but this is not the speed of the FLUID. Then how can it be related to the equation of continuity? Is it true that the fluid below the plunger but above the top of the needle is ALSO moving up at speed v_p? If so, why?


    I apologize for the amount of questions being asked, but any help to any part of the problem is greatly appreciated!
     
  2. jcsd
  3. Jan 24, 2007 #2
    i. Because outside the needle, the motion of fluid flowing into its tips is symmetric
    "spherically". If you draw a sphere... the speed of fluid flow is same every where on the
    surface of the sphere, regardless of its radii....
    ii. Because v(r) is constant by part (i), differential what you have learnt i.e. Q=vA yeild
    dQ = v(r) dA
    iii. An arbitrary infinitely small area on the surface of sphere....
    iv. Becuase we want the total volume Q..... but dQ is only part of it.... therefore, we do an integration
    v. It is a constant by part (i)
    vii. This is the surface area of a sphere... if you integrate over dA, you get the total
    surface area of the sphere, which is 4pi r^2...
    viii Yes, v1 is the volume flow rate over area of the fluid at r = 1.
    ix. No, Q is depend on v_p, the larger the v_p, the more fluid is sucked into the needle...
    your solution just didn't show there relationship
    x. v_p and v_n is related by the cross-section area... obviously, v_p * A_p = v_n * A_n...
    (A_x is thier cross section area respectively)... and by definition, Q is how much fluid
    being sucked by the needly, therefore, Q = v_n * A_n
     
  4. Jan 24, 2007 #3

    andrevdh

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    Homework Helper

    What the integration is describing is the volume flow rate through the surface of spheres centered around the tip of the needle.

    If one starts with a small sphere centered around the tip of the needle the water will flow throught the surface of the sphere towards this region of low pressure. Since the local pressure is the same in all directions the sphere will collapse at the same rate along all directions to fill the void created by the water that entered the needle.

    The volume flow rate throught the surfaces of successively larger spheres will be constant - that is the whole of the liquid is "falling" inwards towards the tip of the needle. This does not mean that v(r) will be constant since as the surface area increases the rate of flow decreases to keep the total volume flow rate constant. That is v(r) will be a decreasing function with increasing r.

    To come back to the integration - what it is determining is the volume flow rate through the surface of a particular sphere. The integration is therefore over the surface of the sphere. The flow rate through a small portion of the sphere will then be per definition given by

    [tex]dQ = v(r) dA[/tex]
     
    Last edited: Jan 24, 2007
  5. Jan 25, 2007 #4
    Thanks for explaining! I have to think about all of these before possibly making another reply!
     
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