Flow of water in a venturimeter

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SUMMARY

The discussion centers on calculating the maximum throughput of water in a horizontal venturimeter with a diameter of 100 mm and a throat of 50 mm, given an entrance pressure of 0.65 bar and a throat pressure limit of 0.3 bar. The Bernoulli equation is applied to relate the pressures and velocities at different points in the venturimeter. The calculations yield a negative velocity, indicating an error in the assumptions or inputs, particularly concerning the specific weight (γ) of water. The participants emphasize the importance of verifying the value of γ to resolve the negative velocity issue.

PREREQUISITES
  • Understanding of fluid dynamics principles, specifically Bernoulli's equation.
  • Familiarity with venturimeter design and operation.
  • Knowledge of pressure units and conversions, particularly bar to pascal.
  • Basic algebra for solving equations involving fluid flow.
NEXT STEPS
  • Review the application of Bernoulli's equation in fluid mechanics.
  • Learn about the specific weight (γ) of fluids and its significance in calculations.
  • Explore the design and functionality of venturimeters in various engineering applications.
  • Investigate common errors in fluid dynamics calculations and how to troubleshoot them.
USEFUL FOR

Engineering students, fluid mechanics practitioners, and professionals involved in hydraulic system design and analysis will benefit from this discussion.

Karol
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Homework Statement


Water flows in a horizontal venturimeter. the diameter is 100 mm, and the throat is 50 mm.
The pressure at the entrance is 0.65[bar].
What is the maximum throughput that is allowed so that the absolute pressure in the throat will not drop under 0.3[bar]

Homework Equations


Bernoully equation:
[tex]H_1+\frac{V_1^2}{2g}+\frac{P_1}{\gamma}+H_P=H_2+ \frac{V_2^2}{2g} +\frac{P_2}{\gamma}[/tex]

The Attempt at a Solution


The velocity V2 in the throat:
[tex]V_2=V_1 \frac{0.1^2}{0.005^2}=4V_1[/tex]
Bernoully equation:
[tex]\frac{0.65E5}{1E5}+\frac{V_1^2}{20}=\frac{0.3E5}{1E5}+\frac{16V_1^2}{20}[/tex]
Which produces negative velocity.
Even if i use absolute pressures, the same. i use 1.65 instead of 0.65.
 
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Karol said:
[tex]\frac{0.65E5}{1E5}+\frac{V_1^2}{20}=\frac{0.3E5}{1E5}+\frac{16V_1^2}{20}[/tex]
Which produces negative velocity.

Hi, Karol.

I don't see how you're getting a negative velocity here.

Also, are you sure that your value of [itex]\gamma[/itex] is correct?
 
you are right!
Thanks...
 

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