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Torricelli's Theorem and etc. help

  1. Nov 17, 2008 #1
    1. The problem statement, all variables and given/known data
    A large tank, 25m in height and open at the top, is completely filled with saltwater (density of 1025kg/m^3). A small drain plug witha cross- sectional area of 4.0 x 10^-5 m^2 is located 5 m from the bottom of the tank.

    a. calculate the force exerted by the water on the plug before the plug breaks
    b. calculate the speed of the water as it leaves the hole in the side of the tank.
    c. calculate the volume flow rate of the water from the hole.

    3. The attempt at a solution

    for b i used bernoulli's equation to solve this.
    P1 + pgy1 + (1/2)pv^2 = P2 + pgy2 + (1/2)pv^3
    reduced down to: v = sqr(2gh)
    v= sqr((2)(9.8)(20))
    v is approx 19.8 m/s

    for c, flow rate = av
    (4.0 x 10^-5 m^2)(19.8)
    7.9196 x 10^-4

    for a, f= pa
    f= 101300 x (4.0 x 10^-5 m^2)

    this right?
     
    Last edited: Nov 17, 2008
  2. jcsd
  3. Nov 17, 2008 #2

    alphysicist

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    Hi crhscoog,

    I don't believe this is correct. This would give the force from the air on the plug, but you want to find the force from the water on the plug. Do you see how to find that?
     
  4. Nov 17, 2008 #3
    hmm i see what you mean. ill look into my notes and see if i can find anything on it.

    *from a glance at the notes, would it have to do something with absolute pressure?
     
  5. Nov 17, 2008 #4

    alphysicist

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    You need to find the water pressure at that point, which is 20m below the surface. What would that be?
     
  6. Nov 17, 2008 #5
    pgh 1000 x 9.8 x 20 = 196000 Pa
     
    Last edited: Nov 17, 2008
  7. Nov 17, 2008 #6

    alphysicist

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    That's the right idea; however they give a density (not 1000), and also what you have here is just the increase in pressure; remember that the full formula is:

    [tex]
    P = P_0 + \rho g h
    [/tex]
     
  8. Nov 17, 2008 #7
    ah right forgot about the given density

    P= 101300 + (1025)(9.8)(20)
    P= 200900 Pa

    Using P=f/a, I multiply 200900 by (4.0 x 10^-5) to get:
    f= 8.036 N
     
  9. Nov 17, 2008 #8

    alphysicist

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    I think you forgot to add the first term.
     
  10. Nov 17, 2008 #9
    ah snap. im rushing this... sry =D

    P(total)= 302200 Pa
    F= 302200 x .00004
    F= 12.088 N
     
  11. Nov 17, 2008 #10

    alphysicist

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    That looks right to me for the water force on the plug.
     
  12. Nov 17, 2008 #11
    thank you for your time, patience, and help. i really appreciate it!
     
  13. Nov 17, 2008 #12

    alphysicist

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    You're welcome; I'm glad to help!
     
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