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Equation of Continuity

  1. Jan 7, 2009 #1
    1. The problem statement, all variables and given/known data
    A water line with an internal radius of 6.1*10^-3 m is connected to a shower head that has 24 holes. The speed of the water in the line is 1.2 m/s.

    (b) At what speed does the water leave one of the holes (effective radius = 4.6*10^-4 m) in the head?

    2. Relevant equations
    Av = Av

    3. The attempt at a solution
    6.1*10^-3(1.2)= 4.6*10^-4(v)
    V = 175.851

    Obviously this is wrong because I didn't use the 24 holes information, but where does that fit in?
  2. jcsd
  3. Jan 7, 2009 #2


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    Isn't the area the total of the area of the 24 holes and not just the area of 1?
  4. Jan 7, 2009 #3
    I tried that and got 4220.42 m/s, which it says is also wrong.
  5. Jan 7, 2009 #4
    Those are radiuses. And when you calculate the area you have to square them, since the area of the cross section of a cylinder is a circle, and the area of the circle is: [tex]R^2\pi[/tex].

    And you have to do what LowlyPion said, that you consider all the 24 holes, which are also circles ;)
  6. Jan 7, 2009 #5
    I actually found the areas correctly when I did my work earlier. I just typed it in wrong.

    What do you mean by consider the 24 holes?
  7. Jan 7, 2009 #6
    You considered only one hole. Whereas there are 24 holes, so you have to divide your answer by 24, and you will get: [tex]v\approx 8.79 m/s[/tex]
  8. Jan 7, 2009 #7
    How did you do that? I got 175/24 = 7.3
  9. Jan 7, 2009 #8


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    The equation of continuity that you are using is really the conservation of mass since the water is an incompressible fluid.

    The area of the input pipe times flow velocity through it will equal the total area of all the output holes times the flow velocity. Discarding pi ...

    Ri² * Vi = 24 * Ro² * Vo
  10. Jan 7, 2009 #9
    I got 6.63. How did you get that?

    6.1*10^-6 (1.2) = 24 * (4.6*10^-8) (v)

    I solved for v and got 6.63.
  11. Jan 7, 2009 #10
    you have to square 6.1 too i.e. [tex]6.1^2[/tex] and [tex] 4.6^2[/tex]
  12. Jan 7, 2009 #11
    Thank you for the help. I get it now.
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