Derive the Venturi Meter eqn from the Bernoulli eqn

AI Thread Summary
The discussion revolves around deriving the Venturi meter equation from the Bernoulli equation while addressing issues with LaTeX formatting. The user identifies a contradiction in their equations, noting that pressure at point 2 (p2) must be less than at point 1 (p1) due to increased velocity, but their substitutions suggest otherwise. They seek clarification on expressing p1 in terms of other variables and relating p2 to another pressure term. Acknowledgment of the contradiction leads to a clearer understanding of the physical principles involved. The conversation highlights the importance of consistent variable definitions in fluid dynamics equations.
GreyNoise
Gold Member
Messages
32
Reaction score
6
Homework Statement
By applying Bernoulli's equation and the equation of continuity to points 1 and 2 of Fig. 16-14 [see attached file], show that the speed of the flow at the entrance is
v1 = a*sqrt{(2(dens' - dens)gh)/(dens(A^2-a^2))}
Relevant Equations
0.5*dens*v_1^2 + p_1 = 0.5*dens*v_2^2 + p_2 Bernoulii eqn
A*v_1 = a*v_2 continuity eqn
Advanced apologies for this format; I am posting my question as an the image b/c the Latex is being very buggy with me, and I lost a kind of lengthy post to it. Can anyone show me what I am doing wrong? I have attached a pdf version for easier reading if need be.
pr-43-p-290-h-r-text-ed.jpg
 

Attachments

Physics news on Phys.org
GreyNoise said:
1657147269864.png


From equation (1) you can see that ##p_2## must be less than ##p_1## because ##v_2 > v_1## (from the continuity equation). So, ##p_2 < p_1##.

However, in equation (2) you let ##p_1 = \rho g h## and ##p_2 =\rho' \, gh##.
But ##\rho' \, > \rho##. So, these substitutions would imply that ##p_2 > p_1##, which contradicts ##p_2 < p_1##. So, letting ##p_1 = \rho g h## and ##p_2 =\rho' \, gh## can't be correct.

Assume we can take points 1 and 2 to be at the same horizontal level:
1657148480656.png


Introduce the height ##H## as shown. Can you express ##p_1## in terms of ##p_c## , ##\rho##, ##g##, ##H##, and ##h##? Likewise, can you relate ##p_2## and ##p_d##?
 
Thnx so much for the response TNsy. Pointing out my contradiction between lines (1) and (2) was the big aha moment for me, and including the ##\rho gH## term makes the physical sense clear now.
 
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A. I know that the field depends only on z since there is translational invariance in x and y directions because the planes are...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top