- #1
amrbekhit
- 11
- 1
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 http://www.wolframalpha.com/input/?i=venturi", the formula describing the venturi effect is:
Q=1/4\,\pi\,{{\it D1}}^{2}\sqrt {2}\sqrt {{\frac {{\it P1}-{\it P2}}{<br /> \rho}}}{\frac {1}{\sqrt {{\frac {{{\it D1}}^{4}}{{{\it D2}}^{4}}}-1}}}<br />
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}}<br /> ^{4}}}-1 \right) \rho{\pi }^{-2}{{\it D1}}^{-4}
In my situation, I have the following known information:
From the diameter and water speed, I calculate that the flow rate is:
0.02919\,{\frac {{m}^{3}}{s}}
Other information that is required:
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
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 http://www.wolframalpha.com/input/?i=venturi", the formula describing the venturi effect is:
Q=1/4\,\pi\,{{\it D1}}^{2}\sqrt {2}\sqrt {{\frac {{\it P1}-{\it P2}}{<br /> \rho}}}{\frac {1}{\sqrt {{\frac {{{\it D1}}^{4}}{{{\it D2}}^{4}}}-1}}}<br />
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}}<br /> ^{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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