How can I prove that P2 = density* g H

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In summary, the liquid manometer is easy. We analyse the forces on the horizontal contact surface of the tube and the liquid manometer so P1 = Density * g * hHowever I fail to analyse the forces on the curved section of the pitot tube. How can I prove that P2 = density* g Hso I can use P2 - P1 = density *g (H-h)Have you learned about Bernoulli's equation yet?Yes I have.If we choose a stream line that passes through the tube then change in height is equal to zero. Also since the fluid will have zero velocity at the opening of the tube we get that P1 +0.
  • #36
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.
 
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  • #37
boneh3ad said:
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?
boneh3ad said:
"(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

boneh3ad said:
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
 
  • #38
Conductivity said:
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.

Conductivity said:
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.

Conductivity said:
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
[tex]\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,[/tex]
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.
 
  • #39
boneh3ad said:
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
[tex]\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,[/tex]
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.
 
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