Solving Pressure Equations: P=pgh & Bernoulli's

[Physicist1011]

I need help understanding P=pgh and Bernoulli's equation.

 

[Chestermiller]

I need help understanding P=pgh and Bernoulli's equation.

Which part don't you understand?

 

[Physicist1011]

I don;t understand how these equations make sense - how are they derived??

 

[Chestermiller]

I don;t understand how these equations make sense - how are they derived??

The derivations of these equations are presented in every freshman physics book. Which part of the derivations don't you understand?

 

[Physicist1011]

Well I don't really understand a problem I am trying to solve which relates to Bernoulli's equation:
For the diagram below: Why is the pressure in the fountain the difference in the hydrostatic pressures in B and C?

 

[Physicist1011]

The hydrostatic pressure at C is P2=pgh2 and at B is P1=pgh1 but how is the difference equal to the hydrostatic pressure of the fountain?

 

[Chestermiller]

Thanks. This is very helpful. Now, to be sure we are on the same page, can you please provide a complete and exact statement of the problem?

 

[Physicist1011]

It is: 'Can you use Bernoulli's equation to calculate the height of the water fountain?'
And to do this I want to calculate the pressure of the water in the fountain, but I do not understand why the difference in hydrostatic pressures of C and B is equal to the pressure in the water fountain.

 

[Chestermiller]

Let's work the problem together using Bernoulli. Take as the datum for elevation z the top of the liquid surface in C; so this is z = 0. Taking as our (first) two points for applying the Bernoulli equation the top of the liquid surface in C and the top of the liquid surface in the reservoir in A, what is the pressure at C?

 

[Physicist1011]

P=pgh = 1000kg/m3 * 9.8m/s * 0 = 0 Pa
Wait is this correct?

 

[Chestermiller]

P=pgh = 1000kg/m3 * 9.8m/s * 0 = 0 Pa
Wait is this correct?

No. $$p_A+\rho g h_2=p_C+0$$How does this look?

 

[Physicist1011]

Ok so Pc = 101325 Pa + 1000 kg/m3 * 9.8 m/s2 * 1.5m = 116025 Pa (h2=1.5m)

 

[Chestermiller]

Ok so Pc = 101325 Pa + 1000 kg/m3 * 9.8 m/s2 * 1.5m = 116025 Pa (h2=1.5m)

Let's do it algebraically, if that's OK with you. There is gas above the surface of B that connects to the gas above the surface of C. How does the pressure at the surface of B, $p_B$, compare with the pressure at the surface of C, $p_C$?

 

[Physicist1011]

Is the same

 

[Chestermiller]

Is the same

OK. So we have: $$p_B=p_A+\rho g h_2$$Now we will take our next two Bernoulli points as follows:
1. The surface of liquid in B, where the pressure is $p_B$, the elevation is $z_B$ and the fluid velocity is essentially zero
2. The outlet of the fountain, where the pressure is $p_A$, the elevation is $z_B+h_1$, and the fluid velocity is v

So, what is your Bernoulli equation involving these two points?

 

[Physicist1011]

How is where the pressure is Pa the elevation is zB+h2 isn't it zB+h1?

Edit: also isn't water at A moving with velocity through the tube ( you have canceled out 1/2pv2)

 

[Chestermiller]

How is where the pressure is Pa the elevation is zB+h2 isn't it zB+h1?

EDIT: Yes. I've corrected that now. The labels on the diagram were difficult to read.

We are looking at the left column of liquid now. $z_B$ is the elevation of the liquid surface at B above the liquid surface at C. And the outlet of the left tube is $h_2$ above the liquid surface at B. The pressure at the outlet of the left tube (where water is squirting out) is atmospheric, $p_A$. So,
$$p_A+\rho g (z_B+h_1)+\frac{1}{2}\rho v^2=p_B+\rho g z_B$$ or, equivalently, $$p_A+\rho g h_1+\frac{1}{2}\rho v^2=p_B$$

 

[Physicist1011]

Ok I got that too, but how does PC-PB=PA (I don't understand how this occurs).

 

[Chestermiller]

Ok I got that too, but how does PC-PB=PA (I don't understand how this occurs).

Who said that? We already said that pC=pB

 

[Physicist1011]

Hmm... I have a book which says "the pressure of the water in the fountain is the difference of the hydrostatic pressures in C and B"

 

[Chestermiller]

Hmm... I have a book which says "the pressure of the water in the fountain is the difference of the hydrostatic pressures in C and B"

The book is wrong. The pressure at the location where the water is coming out is atmospheric. And the pressure in B is the same as the pressure in C.

 

[Physicist1011]

Ok. I am struggling to get my head around to how I will calculate the height of the water fountain.

 

[Chestermiller]

Ok. I am struggling to get my head around to how I will calculate the height of the water fountain.

Well, from the equations we’ve written down, you can determine the upward velocity at the outlet. Then it’s just the same as a particle thrown upward.

 

[Physicist1011]

In the equation PA+pgh2+1/2pv2=PB the bolded h2 is calculating gravitational potential energy so is the height from the water surface at C (why isn't it distance from the ground?) and in the equation for calculating pressure P=pgh the h is the distance from the surface of the water higher up, right?

CHET MILLER EDIT: The OP spotted an error in this equation. It should be h1, not h2. I misread the diagram.

 

[Chestermiller]

In the equation PA+pgh2+1/2pv2=PB the bolded h2 is calculating gravitational potential energy so is the height from the water surface at C (why isn't it distance from the ground?)

Actually, it's the height from the water surface at B. Our two Bernoulli points are (a) the water surface at B and (b) the outlet of the fountain tube.

CHET MILLER EDIT: Again, the OP is correct about this. I misread h2 from the diagram, rather than h1.

and in the equation for calculating pressure P=pgh the h is the distance from the surface of the water higher up, right?

In Bernoulli form, this equation would read $$P-\rho g h=P_A+0$$ where, here, the datum for elevation is taken as the water surface, our upper Bernoulli point is at the surface and our lower Bernoulli point is at elevation -h (or equivalently depth h) where the pressure is P. Note that it doesn't matter where our datum for elevation is taken as long as we are consistent between the two sides of the equation.

 

[Physicist1011]

ok, but in calculating the velocity from the equation PA+pgh2+1/2pv2=PB PA is the pressure of the water outlet, which I do not know meaning I cannot calculate the velocity from this equation.
Also for the equation P-pgh=PA+0 do I need to use atm pressure in it? I thought I could just use P=pgh to calculate pressure of water in B?

 

[Chestermiller]

ok, but in calculating the velocity from the equation PA+pgh2+1/2pv2=PB PA is the pressure of the water outlet, which I do not know meaning I cannot calculate the velocity from this equation.

We have two equations, one for each of the two columns:
$$p_B=p_A+\rho g h_2$$
and $$p_A+\rho g h_1+\frac{1}{2}\rho v^2=p_B$$

What do you get if you substitute pB from the 1nd equation into the 2nd equation.

Also for the equation P-pgh=PA+0 do I need to use atm pressure in it? I thought I could just use P=pgh to calculate pressure of water in B?

It depends on if you are working with gauge pressures or absolute pressures. If you are working with absolute pressures, you need to include the PA. If you are working with gauge pressures, you can omit the PA. Either way, if you are working with an incompressible fluid, the results of the calculations are the same, except for the pressures, which will all differ by the atmospheric pressure. That is, (absolute pressure) = (gauge pressure) + (atmospheric pressure)

 

[Physicist1011]

(just to check the first column you are talking about is the water fountain and the 2nd is from a to c right?)
Ok if I substitute pB into the 2nd equation I will still have 2 variables to solve for PA and v.

 

[Chestermiller]

(just to check the first column you are talking about is the water fountain and the 2nd is from a to c right?)
Ok if I substitute pB into the 2nd equation I will still have 2 variables to solve for PA and v.

Isn't PA on both sides of the equation?

 

[Physicist1011]

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

 

[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.