Solving Pressure Equations: P=pgh & Bernoulli's

[Chestermiller]

What? How is PA on both sides of the 2nd equation? PB is on the other side.

$$p_B=p_A+\rho g h_2\tag{1}$$
and $$p_A+\rho g h_1+\frac{1}{2}\rho v^2=p_B\tag{2}$$
Adding Eqns. 1 and 2 together, we get:
$$p_A+p_B+\rho g h_1+\frac{1}{2}\rho v^2=p_A+p_B+\rho g h_2\tag{3}$$or$$\rho g h_1+\frac{1}{2}\rho v^2=\rho g h_2\tag{4}$$

 

[Physicist1011]

But I thought for the 2nd equation PA was the pressure of the water coming out of the fountain and in the first equation it was the pressure of the water in the open container A which would be atmospheric pressure.

 

[Chestermiller]

But I thought for the 2nd equation PA was the pressure of the water coming out of the fountain and in the first equation it was the pressure of the water in the open container A which would be atmospheric pressure.

Both are atmospheric pressure. The pressure at the outlet where the fountain is emerging is atmospheric also.

 

[Physicist1011]

What? how is that?

 

[Chestermiller]

What? how is that?

The atmosphere imposes its own pressure on the fluid at the exit of a tube. What did you think?

 

[Physicist1011]

I thought the pressure would be different since it will have a different pressure in the tube where it is being pushed by air towards the exit of the tube.

 

[Chestermiller]

I thought the pressure would be different since it will have a different pressure in the tube where it is being pushed by air towards the exit of the tube.

Not so. Imagine a jet of fluid coming out of the tube. What is the radial force per unit area exerted by the surrounding atmosphere on the free surface of the jet?

 

[Physicist1011]

The force will be PA so atmospheric pressure times the area of the tube.

 

[Chestermiller]

The force will be PA so atmospheric pressure times the area of the tube.

Yes. The radial force per unit area on the free surface of the jet will be PA. And, this must match the fluid pressure within the jet. And, since, (according to Pascal's law) pressure acts equally in all directions, the pressure throughout the fluid jet at the exit of the tube will be PA.

 

[Physicist1011]

Oh yes. Thank you for your help.

 

[SirCurmudgeon]

Hello All, The arrangement of this apparatus is interesting. Does it have a name? Does it get "set up" with a finger over the
open-topped tube? Just wondering.
Thank you.

 

[Physicist1011]

Yes. The radial force per unit area on the free surface of the jet will be PA. And, this must match the fluid pressure within the jet. And, since, (according to Pascal's law) pressure acts equally in all directions, the pressure throughout the fluid jet at the exit of the tube will be PA.

Hi again I have another question to do with the one we just solved that I cannot figure out. What is the diameter of the tubes is changed, how would I calculate the change in height of the herons fountain using equations then?

 

[Chestermiller]

Hi again I have another question to do with the one we just solved that I cannot figure out. What is the diameter of the tubes is changed, how would I calculate the change in height of the herons fountain using equations then?

What are your thoughts on this? Your thinking should be along the lines of "what are the limitations of the derivation of the Bernoulli equation, and of how the Bernoulli equation was applied to this problem?

 

[Physicist1011]

What are your thoughts on this? Your thinking should be along the lines of "what are the limitations of the derivation of the Bernoulli equation, and of how the Bernoulli equation was applied to this problem?

So the limitations are that it cannot detect a change in velocity due to a different diameter.
For the second part 'How the Bernoulli equation was applied to this problem' I don't think it can help with diameter.

 

[Chestermiller]

The Bernoulli equation is based on fluid viscosity effects being negligible and, in applying the Bernoulli equation to the present system at hand, we assume that the fluid velocities within reservoirs B and C are very small (so they can be neglected). But, what if the diameters of the tubing in the apparatus were much larger so that the fluid velocities in the tanks were not negligible? And what if the diameters of the tubing in the apparatus were much smaller (like hypodermic needles) so that viscous effects are very significant? So large changes in the diameters of the tubes would negate the analysis we have performed. But, as things stand, for small changes in the diameters of the tubes, since the tube diameters don't even appear in our equations, the predicted height of the fountain would not change.

 

[Physicist1011]

The viscosity of the fluid involved is used - p - density in the equation.
Also how would I calculate the height of the fountain then if I was looking at the effects of different diameters?

 

[Chestermiller]

The viscosity of the fluid involved is used - p - density in the equation.
Also how would I calculate the height of the fountain then if I was looking at the effects of different diameters?

You are aware that density is not the same thing as viscosity, right?

 

[Physicist1011]

oh whoops sorry.