Solving Simple Fluid Problems: Bernoulli's Equation

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Homework Help Overview

The problem involves fluid dynamics, specifically applying Bernoulli's equation to determine the velocity of liquid flowing out of a tube from a container filled with liquid at a constant height.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster expresses confusion about the application of Bernoulli's equation and seeks clarification on deriving the velocity of the liquid. Some participants discuss assumptions regarding pressure and surface velocity, while others explore the implications of these assumptions on the equation.

Discussion Status

The discussion is active, with participants providing insights into the assumptions that can be made in applying Bernoulli's equation. There is a recognition of the relationship between the problem and Torricelli's Theorem, and some participants confirm the reasoning behind the derived velocity expression.

Contextual Notes

Participants are navigating assumptions about pressure conditions and the relative sizes of fluid surface area versus outlet area, which are critical to the problem's setup.

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Homework Statement


A container is filled with liquid, density d. The depth of the liquid is h. A narrow, horizontal, tube at the bottom of the container is letting the liquid flow out at a speed v.
The height of the liquid is kept constant.

What is the velocity of the liquid flowing out of the tube?

Homework Equations


Bernoulli's equation:

p + (1/2)dv^2 + dgh = constant

The Attempt at a Solution


I really don't know how to do this! I know the answer is (2gh)^(1/2), but i don't know why that is, as my textbook has no examples. Could someone please explain how to do these kinds of problems?
 
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Ah... this is an example of a special case of the Bernoulli's Equation, known as Torricelli's Theorem.

I will leave the details of the working out of the equation to you, but I'll point out some important aspects you'll need to consider:

1) You may make the assumption that the pressure on top = pressure below = atmospheric pressure, since both the top and the bottom are open surfaces
2) The surface fluid area of the container is very large compared to the area of the bottom. This means you make the approximation that the surface velocity is...
 
Thanks for your reply.
So can we approximate the surface velocity to be zero?

In which case:

[tex] p_{1} + \frac{1}{2} \rho v_{1}^2 + \rho gh = p_{2} + \frac{1}{2} \rho v_{2}^2[/tex]

But since [tex]p_{1} = p_{2}[/tex] and [tex]v_{1}=0[/tex] then we get [tex]v_{2} = \sqrt{2gh}[/tex].
 
Correct, you are absolutely right. Just as an added note, notice that this is also the equation describing the kinematics of the free fall of an object initially at rest.
 

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