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Fluid Dynamics Flow Rate

  1. May 2, 2008 #1
    1. The problem statement, all variables and given/known data
    a 4mm hole is 1m below the surface of 2m diameter water tank.
    a.what is the volume flow rate through the hole in L/min?
    b.what is the rate in mm/min at which the water level in the tank will drop if the water is not replenished.
    [​IMG][/IMG]

    2. Relevant equations
    Q=vA
    v1A1=v2A2
    V1=A1[tex]\Delta[/tex](x1)=v1A1[tex]\Delta[/tex](t)
    V2=A2[tex]\Delta[/tex](x2)=v2A2[tex]\Delta[/tex](t)
    p+.5[tex]\rho[/tex]v1^2+[tex]\rho[/tex]gy=constant


    3. The attempt at a solution
    a. A1=[tex]\pi[/tex]r^2=[tex]\pi[/tex]*1^2=3.14m^2, A2=[tex]\pi[/tex]r^2=[tex]\pi[/tex]*.002^2=1.25x10^-5m^2
    im sorry i really dont understand fluid dynamics, i have no idea how to find v, to get Q... ans part b, i dont even undestand the question.... plz someone help me,
    can i find v by, [tex]\rho[/tex]+[tex]\rho[/tex]gy=[tex]\rho[/tex]+.5[tex]\rho[/tex]v^2
    and all the densities cancel so v=[tex]\sqrt{}2gy[/tex]???
     
    Last edited: May 2, 2008
  2. jcsd
  3. May 2, 2008 #2
    i think V1=A1*x=3.14*1=3.14m^3??
     
  4. May 2, 2008 #3
    btw the answers are 3.3L/min and 1.06mm/min
     
  5. May 2, 2008 #4
    can anyone help??
     
  6. May 2, 2008 #5

    alphysicist

    User Avatar
    Homework Helper

    Hi fredrick08,

    That's the right idea, but this isn't written quite correctly. It needs to be:

    [tex]
    P_1 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2
    [/tex]

    after the simplifications you have already made (v_1=0 and h_2=0). What do you get?
     
  7. May 2, 2008 #6
    huh??? sorry i dont understand??? if v1=0 then doesnt v2=0 aswell?????
    coz v2=[tex]\sqrt{}v1[/tex]??

    and how does v1 and h2 both equal 0????? plz explain??
     
  8. May 2, 2008 #7

    alphysicist

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    Homework Helper

    Didn't you already set v1=0? Bernoulli's equation applied to two points is:

    [tex]
    P_1 + \rho g h_1+\frac{1}{2}\rho v_1^2 = P_2 + \rho g h_2 + \frac{1}{2} \rho v_2^2
    [/tex]

    but your equation only had one velocity and one height.

    But it does make sense. Point 1 is at the top of the tank, and point 2 is where the water comes out. So v1 is the speed at which the water level is dropping at the top of the tank, and v2 is the speed at which the water is leaving. Since v1 is very small (the overall water level drops very slowly since the hole is so small), it can be neglected here.

    (Since they give you the diameter of the entire water tank, you don't need to neglect v1 in Bernoulli's equation; but it should not make a practical difference in v2 whether you include it or not. If you neglect v1, you can get v2 directly from Bernoulli's equation; if you include v1, you'll have to solve Bernoulli's and the continuity equation together.)

    As for h2, it acts just like the regular potential energy mgh where we had the freedom to set any height we wished equal to zero. You can either set the height h2 to be zero at the hole (and then h1 would be +1) or set h1 to be zero (and then h2 would be -1).
     
  9. May 3, 2008 #8
    oh yes, ty = ) that makes perfect sense... duh... thankyou, i was getting confused coz i needed to draw and properly label a diagram, now i can see wat i trying to find lol...
     
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