Venturi meter pressure difference

• pressurised
In summary, the conversation discusses solving for the pressure difference in a Bernoulli equation using the equations p1+½ρU12+ρgz1=p2+½ρU22+ρgz2 and Q=A1U1=A2U2. The attempt at a solution involves rearranging the equations and eliminating A1 to find the solution. The summarizer also points out an error in the rearranging process and suggests expressing A1 in terms of A2 and the diameters for easier solving.
pressurised

Homework Equations

p=ρgh
Q=A1U1=A2U2
Bernoulli:
p1+½ρU12+ρgz1=p2+½ρU22+ρgz2

The Attempt at a Solution

[/B]
So I got that the pressure difference is gh(ρm-ρ). So I rearranged Bernoulli for p1-p2 to get (assuming the potential is the same I eliminated it),

p1-p2=½ρU22-½ρU12
I carried on rearranging to get:
√(2gh(ρm-ρ))/ρ = Q/A2-Q/A1

I am unsure how to eliminate A1 so I can proceed with the question

pressurised said:
p1-p2=½ρU22-½ρU12
I carried on rearranging to get:
√(2gh(ρm-ρ))/ρ = Q/A2-Q/A1
Looks like you made an error in the rearranging. I'm not sure, but it appears that you assumed that ##\sqrt{U_2^2 - U_1^2} = U_2 - U_1##

I am unsure how to eliminate A1 so I can proceed with the question
How can you express A1 in terms of A2 and the diameters D1 and D2?

TSny said:
Looks like you made an error in the rearranging. I'm not sure, but it appears that you assumed that ##\sqrt{U_2^2 - U_1^2} = U_2 - U_1##

Thank you it seems I did.

How can you express A1 in terms of A2 and the diameters D1 and D2?

Oh I see, if I use A1U1=A2U2 And rearrange for U2, and use d2/D2 for the area ratio?

pressurised said:
Oh I see, if I use A1U1=A2U2 And rearrange for U2, and use d2/D2 for the area ratio?
Yes. But you might want to rearrange for U1. Depends on how you are doing the algebra to get to the result.

TSny said:
Yes. But you might want to rearrange for U1. Depends on how you are doing the algebra to get to the result.

Thank you! I got it now! :)

1. What is a Venturi meter and how does it measure pressure difference?

A Venturi meter is a device used to measure the pressure difference between two points in a fluid flow. It consists of a constriction in the fluid flow path, which causes the fluid to speed up and the pressure to decrease. This pressure difference can then be measured using a manometer or pressure gauge.

2. What are the applications of a Venturi meter?

Venturi meters are commonly used in fluid mechanics to measure the flow rate and pressure difference in pipes and channels. They are also used in industries such as water treatment, oil and gas, and air conditioning systems.

3. How accurate is a Venturi meter in measuring pressure difference?

The accuracy of a Venturi meter depends on various factors such as the design, size, and fluid properties. Generally, they have a high degree of accuracy, with an error margin of less than 1%.

4. Can a Venturi meter be used for all types of fluids?

No, a Venturi meter is designed to measure the pressure difference in a specific type of fluid. It is important to consider the fluid properties, such as density and viscosity, before using a Venturi meter.

5. How do you calculate the pressure difference using a Venturi meter?

The pressure difference can be calculated using the Bernoulli's equation, which states that the sum of pressure, kinetic energy, and potential energy at any point in a fluid flow is constant. By measuring the fluid velocity at the constriction and the inlet, the pressure difference can be calculated using this equation.

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