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Homework Help: Fluid problem

  1. Sep 16, 2008 #1

    tony873004

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    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

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    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:
     
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