I How can I prove that P2 = density* g H

[Chestermiller]

What about the first part? Work done by gravity? It is as if pressure increases as I go up the tube and not decrease.

You are calling h depth, not height. h is not elevation. It is the total depth of fluid. The correct equation for hydrostatics should read $$\frac{dp}{dz}=-\rho g$$where z is the local elevation.

 

[Conductivity]

You are calling h depth, not height. h is not elevation. It is the total depth of fluid. The correct equation for hydrostatics should read $$\frac{dp}{dz}=-\rho g$$where z is the local elevation.

I didnt? Sorry If I didn't catch what you said.
"In the lower tube I assumed the pressure is equal to rho g (H2 -y) where y is the elevation from the point 0. In the upper one rho g(H2 -y)"
derivative is -pg.

I am saying that when I did this and calculated the work due to gravity done I found that it is positive (Higher pressure down) while it is supposed to be negative.

 

[Chestermiller]

Sorry. I don't follow. Do you not understand the force balance involved? Why are you looking at the work due to gravity? Maybe you can show your analysis in detail?

 

[Conductivity]

Sorry. I don't follow. Do you not understand the force balance involved? Why are you looking at the work due to gravity? Maybe you can show your analysis in detail?

Well, we have the value of pressure at a specific depth so I thought I could try and integrate to find the force and calculate the work done by gravity to see if I get the same thing if I used change in potential energy

What I did with like this, Assume a rectangular tube for simplicity. Make the reference point as following.
https://i.imgur.com/TwInjDb.png
Make an expression for the pressure of the lower tube in terms of H2 ($\rho g (H2-y)$) and for the upper tube ($\rho g (H2-y)$), I will neglect static pressure as I only want to find the work done by gravity. Since there is no change in energy for the middle part where the fluid goes up, it must be on the sides and we can calculate the external work done.

Integrate for the lower tube From y = 0 to y = H1 if the width of the tube is L to get the total force. I get

$F_T = \int_0^{H_1} \rho g (H_2 - y) L \, dy = L(\rho g H_2 H_1 - \frac{\rho g H^2_1}{2} )$​

integrate for the lower tube From y = h2 to y = H2

$F_T= \int_{h_2}^{H_2} \rho g (H_2 - y) L \, dy = L \rho g \frac{(H_2 -h_2)^2}{2}$​

Net external work, ($d_1$ is the distance the fluid moves in the lower tube, $d_2$ is the distance the fluid moves in the upper tube)

$W_{ext} = L(\rho g H_2 H_1 - \frac{\rho g H^2_1}{2} ) d_1 - L \rho g \frac{(H_2 -h_2)^2}{2} d_2$​

If we divide by V to get the work done per unit volume, we get:

$\rho g (H_2 -\frac{H_1}{2} ) - \rho g (\frac{H_2 -h_2}{2})$​

rearranging gives,

$\rho g \frac{H_2+h_2}{2} - \rho g \frac{H_1}{2}$​

Which is the exact value of the change in potential energy but it is supposed to be negative because it is the work done. I honestly don't know if this is supposed to work. So it seems this method will only work if only the pressure increases as we go up so we can get negative work. For example, in a vertical tube (where the fluid is going down) to get the same thing as using the change in potential energy we have to say that the pressure decreases with pgh which is a bit counter intuitive...

Sorry if I didnt write all the steps, Hopefully you get what I was trying to do since the start of the thread. Many thanks in advance.

 

[Chestermiller]

Well, we have the value of pressure at a specific depth so I thought I could try and integrate to find the force and calculate the work done by gravity to see if I get the same thing if I used change in potential energy

What I did with like this, Assume a rectangular tube for simplicity. Make the reference point as following.
https://i.imgur.com/TwInjDb.png
Make an expression for the pressure of the lower tube in terms of H2 ($\rho g (H2-y)$) and for the upper tube ($\rho g (H2-y)$), I will neglect static pressure as I only want to find the work done by gravity. Since there is no change in energy for the middle part where the fluid goes up, it must be on the sides and we can calculate the external work done.

Integrate for the lower tube From y = 0 to y = H1 if the width of the tube is L to get the total force. I get

$F_T = \int_0^{H_1} \rho g (H_2 - y) L \, dy = L(\rho g H_2 H_1 - \frac{\rho g H^2_1}{2} )$​

integrate for the lower tube From y = h2 to y = H2

$F_T= \int_{h_2}^{H_2} \rho g (H_2 - y) L \, dy = L \rho g \frac{(H_2 -h_2)^2}{2}$​

Net external work, ($d_1$ is the distance the fluid moves in the lower tube, $d_2$ is the distance the fluid moves in the upper tube)

$W_{ext} = L(\rho g H_2 H_1 - \frac{\rho g H^2_1}{2} ) d_1 - L \rho g \frac{(H_2 -h_2)^2}{2} d_2$​

If we divide by V to get the work done per unit volume, we get:

$\rho g (H_2 -\frac{H_1}{2} ) - \rho g (\frac{H_2 -h_2}{2})$​

rearranging gives,

$\rho g \frac{H_2+h_2}{2} - \rho g \frac{H_1}{2}$​

Which is the exact value of the change in potential energy but it is supposed to be negative because it is the work done. I honestly don't know if this is supposed to work. So it seems this method will only work if only the pressure increases as we go up so we can get negative work. For example, in a vertical tube (where the fluid is going down) to get the same thing as using the change in potential energy we have to say that the pressure decreases with pgh which is a bit counter intuitive...

Sorry if I didnt write all the steps, Hopefully you get what I was trying to do since the start of the thread. Many thanks in advance.

I'm sorry. I'm just not able to follow what you are doing here. Maybe someone else can help.

 

[boneh3ad]

So, I see what you did, but I don't really understand why you did it. It seems you found a force acting on a cross section, then converted that into some kind of work, (while neglecting any mention of the fluid having to travel up through the inclined portion, during which work would be done), and called that work done by gravity. However, I don't really see how you can can this work done by gravity. All the work you calculated was perpendicular to the direction of gravity.

You've basically calculated the work done by pressure, except that there would be no work done by pressure because in this example, you are assuming the fluid isn't moving. If the fluid was moving, you'd need to include the effects of acceleration on the local pressure.

So what is it you are actually trying to prove or show here? I am not entirely certain what your goal is. It doesn't seem to relate to your original question about a Pitot tube. With that question, the answer is relatively easy: the flow stagnates against the tip of the probe, so it reaches stagnation pressure, there is a hole that transmits that pressure into the tube, and then the hydrostatic pressure from the column of liquid in the tube has to exactly equal the stagnation pressure at the tip in order for the force balance to work.

 

[Conductivity]

So, I see what you did, but I don't really understand why you did it. It seems you found a force acting on a cross section, then converted that into some kind of work, (while neglecting any mention of the fluid having to travel up through the inclined portion, during which work would be done), and called that work done by gravity. However, I don't really see how you can can this work done by gravity. All the work you calculated was perpendicular to the direction of gravity.

Sorry the thread drifted apart from the original question. I was trying to find the work done by hydrostatic pressure to see if it will equal negative the change in potential energy. Of course what is right though is gravity will do work in portion where the fluid goes up but it seemed that if I have hydrostatic pressure it will also do work so I tried to calculate that.
Also how can gravity do work other than by pressure?

"(while neglecting any mention of the fluid having to travel up through the inclined portion, during which work would be done)"

Well, I took the same approach as the derivation for brenoulli's equation where you calculate the net external work using this method. That was my justification for not calculating the work done in the portion where the fluid goes up because these are internal forces

you are assuming the fluid isn't moving. If the fluid was moving, you'd need to include the effects of acceleration on the local pressure.

I am sorry I didnt get that, How is the fluid not moving?Thanks for answering

 

[boneh3ad]

Sorry the thread drifted apart from the original question. I was trying to find the work done by hydrostatic pressure to see if it will equal negative the change in potential energy. Of course what is right though is gravity will do work in portion where the fluid goes up but it seemed that if I have hydrostatic pressure it will also do work so I tried to calculate that.

Gravity acts vertically, though, so there's no reason horizontal motion should factor into work done by gravity.

Well, I took the same approach as the derivation for brenoulli's equation where you calculate the net external work using this method because the part in the middle is replaced by another so there isn't a net change in energy, However a small part from the lower tube at the end becomes at the top. That was my justification for not calculating the work done in the portion where the fluid goes up

Gravity is vertical, so it doesn't contribute to work in the horizontal direction, which is all you considered in your analysis.

I am sorry I didnt get that, How is the fluid not moving?

If the fluid is moving, then the hydrostatic pressure is not the only contributor to pressure, as you have assumed. That's the point of behind Bernoulli's equation.

Typically, when analyzing flow through a pipe or duct between two points, 1 and 2, we use a modified version of Bernoulli's equation, namely
\dfrac{p_1}{\rho g} + \dfrac{v_1^2}{2g} + h_1 = \dfrac{p_2}{\rho g} + \dfrac{v_2^2}{2g} + h_2 + h_L,
where $h_L$ is called head loss and is a combination of terms used to include the various effects of viscosity and pumps and such. So, you haven to account for $v^2$ if you want a good answer.

 

[Conductivity]

Gravity acts vertically, though, so there's no reason horizontal motion should factor into work done by gravity.
Gravity is vertical, so it doesn't contribute to work in the horizontal direction, which is all you considered in your analysis.
If the fluid is moving, then the hydrostatic pressure is not the only contributor to pressure, as you have assumed. That's the point of behind Bernoulli's equation.

Typically, when analyzing flow through a pipe or duct between two points, 1 and 2, we use a modified version of Bernoulli's equation, namely
\dfrac{p_1}{\rho g} + \dfrac{v_1^2}{2g} + h_1 = \dfrac{p_2}{\rho g} + \dfrac{v_2^2}{2g} + h_2 + h_L,
where $h_L$ is called head loss and is a combination of terms used to include the various effects of viscosity and pumps and such. So, you haven to account for $v^2$ if you want a good answer.

Oh, I haven't assume that is not moving nor that it is the only pressure component. I just removed all the other factors for the being to do the calculation because it wouldn't affect it.

Okay, I definitely know that gravity acts vertically. But what did I calculate here? Why don't we calculate the work done by hydrostatic pressure in the derivation of brenoulli's equation ( They assume that p is constant through out the height of the pipe on both sides) (Or is it inside the change in potential energy)?

That should be it, Thank you again. Sorry if I seem persistent, I don't mean to.