1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Volume of water flowing out a cylindrical pipe

  1. Sep 9, 2006 #1
    Water is flowing down a cylindrical pipe of radius r.

    (a) Write a formula for the volume, V, of water that emerges from the end of the pipe in one second if the water is flowing at a rate of

    (i) 3 cm/sec (ii) k cm/sec

    (b) Graph your answer to part (a)(ii) as a function of
    (i) r, assuming k is constant
    (ii) k, assuming r is constant



    All right, I just don't know where to start at all... I'm thinking that I'll probably need the formula for the volume of a cylinder. Besides that, I don't know what I have to do with the formula. I'm completely clueless on how to implement the rates of water flow into the volume of a cylinder formula.. if that's even what I have to do... I'm sure that if I can figure out part (a), part (b) won't be much of a problem. (just solving for different variables and graphing)
     
  2. jcsd
  3. Sep 9, 2006 #2

    radou

    User Avatar
    Homework Helper

    The cross area of a cylinder equals A = r^2*PI, so the volume that flows through the cylinder in one second equals V = v*A [m^3/sec], where v is the speed of water.
     
  4. Sep 9, 2006 #3
    you said the volume that flows through is in m^3/sec... does that mean I have to convert to m/sec, or will the rates work if i keep them as cm/sec?

    Should work as cm/sec, right?

    After you respond to this.. I'm gonna give it a shot, and you can let me know how i'm doing.

    Thanks a lot.
     
  5. Sep 9, 2006 #4

    radou

    User Avatar
    Homework Helper

    Nope, you don't have to convert anything, of course it works perfectly well with cm^3/sec, too. I just forgot that the speed was given in cm/sec.
     
  6. Sep 9, 2006 #5
    ok, let's see if i did it correctly

    (a) (i) V = 3pi*r2 (where V is in cm^3/sec)
    (ii) V = kpi*r2 (where V is in cm^3/sec)

    (b) (i) V on the vertical axis, r on the horizontal axis. It's just a parabola .. it'll look like x2.
    (ii) V is on the vertical axis, k is on the horizontal axis. It's just a linear function with a positive slope.

    How's this? (and is there any way to be more specific with the graphs, like to plot out any points, even if the points are just defined as variables?)
     
  7. Sep 9, 2006 #6

    radou

    User Avatar
    Homework Helper

    Looks great.

    The graphs are perfectly ok, and I don't see a way to be more specific than you already are, except maybe to mention that for V(r) it makes sense if r > 0, since the cylinder radius can't be negative. As for V(k), k should be greater or equal than zero, since a negative volume doesn't make any sense.
     
  8. Sep 9, 2006 #7
    thanks a lot
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Volume of water flowing out a cylindrical pipe
Loading...