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Streamline, stream-tube

  1. Jan 1, 2008 #1
    1. The problem statement, all variables and given/known data

    q1.

    a)Explain the terms stream-tube.

    b)State the equation of continutity along a stream-tube for an incomperssible fluid.

    c)A hosepipe with an internal cross sectional area A1 has at its end a nozzle with a hole whose cross-sectional area is A2 (A2<A1). If the speed of the water as it emerges from the nozzle is vo, determine its speed in the hosepipe. Also, determine the excess pressure inside the hosepipe, which is horizontal.



    2. Relevant equations

    p + pgh + 1/2 pv^2

    3. The attempt at a solution

    a) A thin bundle of adjacent streamlines forming a stream tube.

    b) equation of continutity :

    v1*d*A1 = v2*d*A2

    for an incompressible fluid.

    c)

    p + Pgh + 1/2 Pv^2

    v1*d*A1 = v2*d*A2

    V1= vo

    need help on this
     
  2. jcsd
  3. Jan 1, 2008 #2

    Doc Al

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    Staff: Mentor

    Use the continuity equation.
    Use Bernoulli's equation.
     
  4. Jan 1, 2008 #3
    v1*d*A1 = v2*d*A2

    v2= (vo *A1 ) / A2

    ----------------------------

    p + Pgh + 1/2 Pv^2= constant

    v2= (vo *A1 ) / A2

    p + Pgh + 1/2 P ((vo *A1 ) / A2) ^2
     
  5. Jan 1, 2008 #4
    v1*d*A1 = v2*d*A2

    v2= (vo *A1 ) / A2

    ----------------------------

    p + Pgh + 1/2 Pv^2= constant

    v2= (vo *A1 ) / A2

    p + Pgh + 1/2 P ((vo *A1 ) / A2) ^2
     
  6. Jan 1, 2008 #5

    Doc Al

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    Staff: Mentor

    Careful with your notation. The nozzle area is A2 and the nozzle speed is v0. Redo this.
    Redo this with the correct values for the speeds. Take advantage of the fact that the pipe is horizontal.

    Set it up like this:
    (p + Pgh + 1/2 Pv^2) for hosepipe = (p + Pgh + 1/2 Pv^2) for nozzle

    Find the change in pressure.
     
  7. Jan 2, 2008 #6
    v1*d*A1 = v2*d*A2

    v1*d*A1 = vo*d*A2

    v0 = ( v1 * A1 ) / A2

    is this correct for the speed?
     
  8. Jan 2, 2008 #7

    Doc Al

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    Staff: Mentor

    It would be, except that you should be solving for the speed in the hosepipe, which is v1, not v0. :wink:
     
  9. Jan 2, 2008 #8
    v1*d*A1 = v2*d*A2

    v1*d*A1 = vo*d*A2

    v1 = (vo * A2 ) / A1
     
  10. Jan 2, 2008 #9
    (p + Pgh + 1/2 Pv^2) for hosepipe = (p + Pgh + 1/2 Pv^2) for nozzle

    if v1 = (vo * A2 ) / A1

    now to find the change in the pressure-

    (p + Pgh + 1/2 Pv^2) for hosepipe - (p + Pgh + 1/2 Pv^2) for nozzle = 0
     
  11. Jan 2, 2008 #10

    Doc Al

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    Staff: Mentor

    Right.

    Plug in the values for hosepipe (p1, v1) and nozzle (p2, v0) and find the difference in pressure (p1 - p2); since the pipe is horizontal, the Pgh terms drop out.
     
  12. Jan 5, 2008 #11
    c) determine the excess pressure inside the hosepipe, which is horizontal.

    (p1 + Pgh + 1/2 Pv1^2) for hosepipe - (p2 + Pgh + 1/2 Pv2^2) for nozzle = 0

    since pipe is horizontal Pgh drops out.

    (p1 + 1/2 Pv1^2) for hosepipe - (p2 + 1/2 Pv2^2) for nozzle = 0

    if v1 = (vo * A2 ) / A1

    then

    (p1 + 1/2 P *v1^2) - (p2 + 1/2 Pv2^2) = 0

    (p1 + 1/2 P*[(vo * A2 ) / A1]^2) - (p2 + 1/2 Pv2^2) = 0
     
  13. Jan 5, 2008 #12

    Doc Al

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    Staff: Mentor

    OK, now find p1 - p2.

    Note: The speed in the nozzle is given as v0, so replace v2 with v0.
     
  14. Jan 5, 2008 #13
    (p1 + 1/2 P*[(vo * A2 ) / A1]^2) = (p2 + 1/2 Pv0^2)

    [vo * A2 ) / A1]^2 = (vo * A2 ) / A1 * (vo * A2 ) / A1 = A1 ^-2 (vo^2 + 2A2 + A2^2)

    p1 + 1/2 P*[(A1 ^-2 (vo^2 + 2A2 + A2^2)] = (p2 + 1/2 Pv0^2)

    p1-p2= 1/2 Pv0^2 - 1/2 P*[(A1 ^-2 (vo^2 + 2A2 + A2^2)]
     
  15. Jan 5, 2008 #14

    Doc Al

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    Staff: Mentor

    That last step is incorrect. Instead:

    [tex](\frac{v_0 A_2}{A_1})^2 = \frac{v_0^2 A_2^2}{A_1^2}[/tex]
     
  16. Jan 5, 2008 #15
    Hi doc, what programme did you use to get your text with that type?
     
  17. Jan 5, 2008 #16

    Doc Al

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    Staff: Mentor

    Equations are written using Latex. Read about it here, or click on the [itex]\Sigma[/itex] in the post editing menu.
     
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