## Calculating the suction force produced by a Venturi

Dear all,

As I understand the venturi effect, if a flow of water passes through a constriction, its velocity increases and its pressure decreases at the restriction. The decrease in pressure allows a suction force to be produced.

I am trying to use this principle in order to power a vacuum cup inside a water pipe, but my initial calculations for estimating the suction force are producing seemingly meaningless results.

According to WolframAlpha, the formula describing the venturi effect is:

$$Q=1/4\,\pi\,{{\it D1}}^{2}\sqrt {2}\sqrt {{\frac {{\it P1}-{\it P2}}{ \rho}}}{\frac {1}{\sqrt {{\frac {{{\it D1}}^{4}}{{{\it D2}}^{4}}}-1}}}$$

From my understanding, the vacuum pressure generated by the venturi is $$P1-P2$$. So, rearranging the above equation to make that the subject gives:

$${\it P1}-{\it P2}=8\,{Q}^{2} \left( {\frac {{{\it D1}}^{4}}{{{\it D2}} ^{4}}}-1 \right) \rho{\pi }^{-2}{{\it D1}}^{-4}$$

In my situation, I have the following known information:
• Diameter of pipe: 8"
• Water speed: 0.9m/s
• Water Pressure: 2.8bar

From the diameter and water speed, I calculate that the flow rate is:

$$0.02919\,{\frac {{m}^{3}}{s}}$$

Other information that is required:
• Venturi upstream diameter D1: 10e-3 m
• Venturi downstream diameter D2: 5e-3 m
• Density of water rho: 1000kg/m^3

Substituting all of that information into the formula gives us a pressure difference of 1.035716981*10^9 Pa (which looks huge).

Assuming that my suction cup has a 15mm radius, in order to work out the suction force I use:

$$F=PA$$

This gives me a suction force of 732105N!

This seems extremely high to me, which makes me think that I have misunderstood the principle. Can anyone shed some light on this?

Thanks

--Amr

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 Tags cup, pressure, suction, vacuum, venturi
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