Bernoulli's Principle (Possibly involving Venturi's Effect)

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SUMMARY

The discussion revolves around applying Bernoulli's Principle and the Venturi Effect to solve a fluid dynamics problem involving two large tanks and a horizontal pipe with a constriction. The key equation used is Bernoulli's equation: p1 + dgy1 + (1/2)(dv1)² = p2 + dgy2 + (1/2)d(v2)². The challenge lies in determining the velocity at the constriction and how it relates to the height of the liquid in the vertical pipe. Participants emphasize the importance of correctly interpreting kinetic and potential energy in the context of fluid flow.

PREREQUISITES
  • Understanding of Bernoulli's equation and its components
  • Familiarity with the concept of fluid density and flow rate
  • Knowledge of the Venturi Effect and its implications in fluid dynamics
  • Basic principles of pressure energy, kinetic energy, and potential energy in fluids
NEXT STEPS
  • Study the derivation and applications of Bernoulli's equation in various fluid systems
  • Learn how to calculate flow rates and velocities in pipes with varying cross-sectional areas
  • Explore the principles of the Venturi Effect and its practical applications in engineering
  • Investigate the relationship between pressure, velocity, and height in fluid dynamics
USEFUL FOR

Students of fluid mechanics, engineers working with hydraulic systems, and anyone interested in understanding the principles of fluid dynamics and their applications.

JPOconnell
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Homework Statement


Two very large open tanks A and F (the figure (Figure 1) ) contain the same liquid. A horizontal pipe BCD, having a constriction at C and open to the air at D, leads out of the bottom of tank A, and a vertical pipe E opens into the constriction at C and dips into the liquid in tank F. Assume streamline flow and no viscosity.

If the cross-sectional area at C is one-half the area at D and if D is a distance below the level of the liquid in A, to what height will liquid rise in pipe E? (in ration of (h2/h1), or just an equation for h1 in h2 is fine)

Link to picture:

Homework Equations


d here is density of the fluid, in this case it is constant.

Bernoulli's equation: p1 + dgy1 + (1/2)(dv1)2 = p2 + dgy2 + (1/2)d(v2)2

Density = Av (A is the area of flow surface, while v is the velocity)

The Attempt at a Solution



I tried to set up a Bernoulli's equation for this, but there are 3 sections of the pipe, and I'm not sure how to add them together. I was looking for a solution and I saw Venturi's Effect, which just basically said that the pressure in the choked pipe is lower, and I don't know if it is even relevant.

But I know for sure that the height h2 got to do with the fluid kinetic energy in pipe C, but I don't know how to find the velocity in pipe C... Well, I tried to mix it around and ended up cancelling out and nothing remained. Really weird problem, given it just today that I learned about this guy's equation.

Thanks in advance
 
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Wil try to be of some assistance.
First off
"Density = Av (A is the area of flow surface, while v is the velocity)" this is incorrect, what is the units of Area and velocity? m^2 and m/s so multiplying this, what is the end unit? Is this the units of density, check what you get.

Bernoulli's principle:
P/Rho is the pressure energy, V^2/2 is the kinetic energy and gh is the potential energy. So to say that h_2 is a result of the kinetic energy in C is somewhat correct but in itself it is potential energy.

If there was no constriction at C, what would you say would be the pressure in the pipe at D? First simplify the pipeline... Answer these questions and let's work from there.
 

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