Applying Bernoulli's to this geometry.

In summary: Yes.Look at the given equation r^2 + (R - z)^2 = R^2so if z=0, r=0 as the figure corroborates.It also states dz/dt < 0 which again implies z=0 at the bottom.
  • #1
pyroknife
613
3
I have attached the geometry of interest with some parts of the solution. The geometry is a vessel that is half of a sphere with an orifice at the bottom.

The first expression that they have written, the "A*(2*g*z)^0.5=..." is from conservation of flow rate. 2*g*z is the velocity at the inlet of the orifice.

I don't understand how they got that velocity since there should be a pressure drop from z to the orifice opening.
Basically it just seems like they did Potential energy (@ z) = Kinetic Energy at orifice opening.
So that gives g*z=0.5*v^2=>v=(2*g*z)^0.5

but isn't there a pressure gain as well of density*gravity*z?
The only way they could have gotten that velocity is if the pressure at both locations are the same.
 

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  • #2
This problem is an application of Torricelli's Law, which is a particular form of the more general Bernoulli equation:

http://en.wikipedia.org/wiki/Torricelli's_law

The pressure at the open top of the sphere and at the bottom where the outlet is are approximately equal to each other and to the ambient atmospheric pressure.
 
  • #3
SteamKing said:
This problem is an application of Torricelli's Law, which is a particular form of the more general Bernoulli equation:

http://en.wikipedia.org/wiki/Torricelli's_law

The pressure at the open top of the sphere and at the bottom where the outlet is are approximately equal to each other and to the ambient atmospheric pressure.

Thanks for the reply. What you stated above I agree with, but isn't that different from what the drawing shows? it looks like at height z is at a location that is not exposed to the ambient pressure and same with at z=0 (orifice opening)? That's what has me confused.

If Z is indeed at the top, wouldn't Z=R?

Thus the expression should be 2*g*R, instead of 2*g*z.

It looks like z is some arbitary height, at z=R (P=Patm), but at other points below R, it should be Patm+(density*gravity*(R-z)) from the hydrostatic pressure relation.
 
  • #4
I'm wondering why the velocity on the surface v1 = 0. Seems to me that v1 is finite (= -dz/dt). Maybe it's deliberately ignored. But if not, (v1^2)/2 = (v2^2)/2 - gz > 0.
 
  • #5
pyroknife said:
Thanks for the reply. What you stated above I agree with, but isn't that different from what the drawing shows? it looks like at height z is at a location that is not exposed to the ambient pressure and same with at z=0 (orifice opening)? That's what has me confused.

If Z is indeed at the top, wouldn't Z=R?

Thus the expression should be 2*g*R, instead of 2*g*z.

It looks like z is some arbitary height, at z=R (P=Patm), but at other points below R, it should be Patm+(density*gravity*(R-z)) from the hydrostatic pressure relation.

I think z is the instantaneous level of the fluid.
 
  • #6
rude man said:
I'm wondering why the velocity on the surface v1 = 0. Seems to me that v1 is finite (= -dz/dt). Maybe it's deliberately ignored. But if not, (v1^2)/2 = (v2^2)/2 - gz > 0.

In all the classes I've taken, we assume the velocity at the free space is Zero.
 
  • #7
rude man said:
I think z is the instantaneous level of the fluid.
Sorry, what does that mean?
 
  • #8
rude man said:
I'm wondering why the velocity on the surface v1 = 0. Seems to me that v1 is finite (= -dz/dt). Maybe it's deliberately ignored. But if not, (v1^2)/2 = (v2^2)/2 - gz > 0.
Hi rude man,

The cross sectional area at the top of a huge tank like this is usually assumed to be much larger than the cross sectional area of the oriface, so the kinetic energy at the top is usually neglected in Bernoulli.

chet
 
  • #9
pyroknife said:
Sorry, what does that mean?

It means the fluid surface is at z.
 
  • #10
Chestermiller said:
Hi rude man,

The cross sectional area at the top of a huge tank like this is usually assumed to be much larger than the cross sectional area of the oriface, so the kinetic energy at the top is usually neglected in Bernoulli.

chet

Hi Chet,
I defer to the majority ...

PS congrats for making Sci Advisor! Richly deserved!
 
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  • #11
rude man said:
It means the fluid surface is at z.

Hmmmm I see, that would explain velocity=0 there and P=Patm. What about at z=0. Would you make an assumption that's the bottom surface (lack of a better term)?
 
  • #12
pyroknife said:
Hmmmm I see, that would explain velocity=0 there and P=Patm. What about at z=0. Would you make an assumption that's the bottom surface (lack of a better term)?

Yes.
Look at the given equation r^2 + (R - z)^2 = R^2
so if z=0, r=0 as the figure corroborates.
It also states dz/dt < 0 which again implies z=0 at the bottom.

The figure should have indicated z as going + from the bottom to the surface instead of an ambiguous two-way arrow.
 

Related to Applying Bernoulli's to this geometry.

What is Bernoulli's principle?

Bernoulli's principle is a fundamental law of fluid mechanics that states that as the speed of a fluid (such as air or water) increases, its pressure decreases. This principle is based on the conservation of energy and is commonly used to explain the lift force on an airplane wing or the flow of water through a pipe.

How is Bernoulli's principle applied to geometry?

In geometry, Bernoulli's principle is often used to analyze the flow of fluids through various shapes and structures. For example, it can be applied to calculate the lift force on an airfoil or the pressure distribution on a curved surface. By understanding how the speed and pressure of a fluid change as it moves through a geometry, engineers and scientists can design more efficient and effective structures.

What are some real-world applications of Bernoulli's principle in geometry?

Bernoulli's principle has many practical applications in various fields, such as aviation, architecture, and hydrodynamics. It is used to design efficient airplane wings, wind turbines, and sails. In architecture, it is used to design buildings that can withstand strong winds. In hydrodynamics, it is used to study the flow of water in rivers and pipes, and to design structures such as dams and bridges.

What are the limitations of applying Bernoulli's principle to geometry?

While Bernoulli's principle is a useful tool in analyzing fluid flow in geometry, it has some limitations. It assumes that the fluid is incompressible and inviscid, which may not always be the case in real-world scenarios. Additionally, it does not take into account factors such as turbulence and boundary layer effects, which can significantly impact the accuracy of the results.

How can I use Bernoulli's principle in my own experiments or projects?

Bernoulli's principle can be applied in various experiments and projects, depending on your area of interest. For example, you could use it to design a paper airplane that can fly longer and higher, or to investigate the flow of water in a fountain. It can also be used to analyze the efficiency of different shapes in generating lift, such as airfoils or wind turbine blades. With some creativity and basic understanding of fluid mechanics, you can incorporate Bernoulli's principle into your own experiments and projects.

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