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!

Fluid problem

  1. Sep 16, 2008 #1

    tony873004

    User Avatar
    Science Advisor
    Gold Member

    1. The problem statement, all variables and given/known data
    Water flows into a nozzle at 3.00 m/s at a pressure of 1.31*105Pa. What should the ratio of input to output diameter be if the flow is to remain steady? What is the flow speed at the exit?


    2. Relevant equations
    [tex]
    P_{{\rm{input}}} + \frac{{\rho v_{{\rm{input}}}^2 }}{2} = P_{{\rm{atmosphere}}} + \frac{{\rho v_{{\rm{output}}}^{\rm{2}} }}{2}
    [/tex]

    [tex]A_1 v_1 = A_2 v_2 [/tex]

    3. The attempt at a solution
    My first guess is to simply say that a 1:1 input:eek:utput ratio with the water flowing out at the same speed as it flows in should do the trick. But that would be too easy. I guess that I'm to assume that the pressure outside is air pressure. If so, am I doing it right?
    [tex]
    \begin{array}{l}
    P_{{\rm{input}}} + \frac{{\rho v_{{\rm{input}}}^2 }}{2} = P_{{\rm{atmosphere}}} + \frac{{\rho v_{{\rm{output}}}^{\rm{2}} }}{2} \\
    \\
    P_{{\rm{input}}} + \frac{{\rho v_{{\rm{input}}}^2 }}{2} - P_{{\rm{atmosphere}}} = \frac{{\rho v_{{\rm{output}}}^{\rm{2}} }}{2} \\
    \\
    \frac{{2\left( {P_{{\rm{input}}} + \frac{{\rho v_{{\rm{input}}}^2 }}{2} - P_{{\rm{atmosphere}}} } \right)}}{\rho } = v_{{\rm{output}}}^{\rm{2}} \\
    \\
    v_{{\rm{output}}}^{} = \sqrt {\frac{{2\left( {P_{{\rm{input}}} + \frac{{\rho v_{{\rm{input}}}^2 }}{2} - P_{{\rm{atmosphere}}} } \right)}}{\rho }} \\
    \\
    v_{{\rm{output}}}^{} = \sqrt {\frac{{2\left( {1.31 \times 10^5 \,{\rm{Pa}} + \frac{{\left( {1000\frac{{{\rm{kg}}}}{{{\rm{m}}^{\rm{3}} }}} \right)\left( {3.00\,{\rm{m/s}}^{\rm{2}} } \right)}}{2} - 101.325 \times 10^3 \,{\rm{Pa}}} \right)}}{{\left( {1000\frac{{{\rm{kg}}}}{{{\rm{m}}^{\rm{3}} }}} \right)}}} = 5.58{\rm{ m/s}} \\
    \end{array}
    [/tex]

    [tex]\begin{array}{l}
    A_1 v_1 = A_2 v_2 \\
    \\
    \frac{{A_1 }}{{A_2 }} = \frac{{v_2 }}{{v_1 }} = \frac{{3.00\;{\rm{m/s}}}}{{{\rm{5}}{\rm{.58}}\,{\rm{m/s}}}} = 0.5376 \\
    \end{array}[/tex]
     
  2. jcsd
  3. Sep 17, 2008 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi tony! :smile:

    Yes … air pressure …

    but it would be simpler just to say

    [tex] \Delta(v^2) = \left(\frac{2}{\rho}\right)\,\Delta P[/tex]

    [tex] v_1 = \sqrt{v_0^2\ +\ \left(\frac{2}{\rho}\right)\,\Delta P\ }[/tex] :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Fluid problem
  1. Fluid problem. (Replies: 3)

  2. Fluid problem (Replies: 2)

  3. Fluids Problem (Replies: 2)

  4. Fluid problem (Replies: 1)

  5. Fluids Problem (Replies: 4)

Loading...