Finding Input to Output Ratio for Steady Water Flow

[tony873004]

Homework Statement

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?

Homework 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]

The Attempt at a Solution

My first guess is to simply say that a 1:1 inpututput 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]

 

[tiny-tim]

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?
…
My first guess is to simply say that a 1:1 inpututput 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?

Hi tony!

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]

 

[baba89]

I would first check the equations and units used in the solution. The equations used seem to be appropriate for solving this problem, but I would double check the units to make sure they are consistent.

Next, I would consider the assumptions made in the solution. The solution assumes that the pressure outside the nozzle is equal to atmospheric pressure, which may not always be the case. I would also consider the assumptions made about the flow being steady and the water flowing in and out at the same speed.

I would also consider the limitations of the solution. This solution assumes a steady flow and does not take into account any turbulence or other factors that may affect the flow. It also assumes a perfect nozzle with no losses or obstructions.

Finally, I would suggest testing the solution through experimentation or using computational fluid dynamics to verify the results. This would help to ensure that the solution is accurate and applicable in real-world situations.