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

[Conductivity]

I have a problem here. I don't get why the stagnation pressure equal the height of the water in the pitot tube.

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 * h

However I fail to analyse the forces on the curved section of the pitot tube. How can I prove that P2 = density* g H
so I can use P2 - P1 = density *g (H-h)

 

[Chestermiller]

Have you learned about Bernoulli's equation yet?

 

[Conductivity]

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 +0.5 density V^2 = P2

What I want you to clarify please is how can I relate P2 to the weight of the fluid in the tube using free body diagram as we did in the simple liquid manometer. What I can't analyse is the pressure at the curve. Also wouldn't there be a pressure due to the weight of the water above the stagnation point (Hydrostatic pressure)?

 

[Chestermiller]

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 +0.5 density V^2 = P2

What I want you to clarify please is how can I relate P2 to the weight of the fluid in the tube using free body diagram as we did in the simple liquid manometer. What I can't analyse is the pressure at the curve. Also wouldn't there be a pressure due to the weight of the water above the stagnation point (Hydrostatic pressure)?

Well, if it were above P1 location, height would be $P_1/\rho g$. The additional height $\Delta h$ over the stagnation tube would be $\rho g \Delta h =\rho v^2/{2}$.

 

[Conductivity]

I think I have a problem in understanding something in the derivation of brenouli's equation

In the derivation of brenouli's equation we assume that we have a constant pressure P that is pushing the fluid and it changes according to the changes in kinetic and potential energy, and I know the derivation and understand it

But when I try for example find the pressure at a point in the fluid, I assumed it will be P + density * g * h where h is the height of the fluid above it. I thought I could perhaps not use that the change in potential energy is mg delta h and say that gravity will cause pressure on each slice of the fluid and will do work as same way that P does. So I integrated the pressure resulting from gravity over the area to find the net force then added it to PA and continued to derive brenouli's equation as $W_{ext} = dKE$ but I was not able to arrive to the same equation using this method.

What am I doing wrong? If it is right and I have done something wrong with the calculation, Could you post a proof of that because it takes a quite long time between replies and this problem caused me quite a mess in my schedule :C

 

[Chestermiller]

I think I have a problem in understanding something in the derivation of brenouli's equation

In the derivation of brenouli's equation we assume that we have a constant pressure P that is pushing the fluid and it changes according to the changes in kinetic and potential energy, and I know the derivation and understand it

But when I try for example find the pressure at a point in the fluid, I assumed it will be P + density * g * h where h is the height of the fluid above it. I thought I could perhaps not use that the change in potential energy is mg delta h and say that gravity will cause pressure on each slice of the fluid and will do work as same way that P does. So I integrated the pressure resulting from gravity over the area to find the net force then added it to PA and continued to derive brenouli's equation as $W_{ext} = dKE$ but I was not able to arrive to the same equation using this method.

What am I doing wrong? If it is right and I have done something wrong with the calculation, Could you post a proof of that because it takes a quite long time between replies and this problem caused me quite a mess in my schedule :C

Sorry. I don't understand what you are saying in your discussion.

 

[Conductivity]

Sorry. I don't understand what you are saying in your discussion.

Oh sorry, let me make it in the form of questions.

First we assume in a tube we have a static pressure as though we are pushing it with a piston from the other side is that right? Then if I want to find a pressure at a certain point, Shouldnt the total pressure be P + density * g * h? where pgh is the pressure resulting from the weight of the fluid above it? If not, Why please?

 

[Chestermiller]

Oh sorry, let me make it in the form of questions.

First we assume in a tube we have a static pressure as though we are pushing it with a piston from the other side is that right? Then if I want to find a pressure at a certain point, Shouldnt the total pressure be P + density * g * h? where pgh is the pressure resulting from the weight of the fluid above it? If not, Why please?

if it is in one of the tubes rising above the pipe, then it is just $\rho gh$ at the bottom of the tube.

 

[Conductivity]

https://i.imgur.com/oVVX7Tk.png

I mean like this, Just a tube and you have a fluid flowing in it. Do we have a change in pressure in the direction perpendicular to the direction of the flow due to the weight of the fluid above it? So for example at the red point, The total pressure Pt = P +pgh

 

[Chestermiller]

https://i.imgur.com/oVVX7Tk.png

I mean like this, Just a tube and you have a fluid flowing in it. Do we have a change in pressure in the direction perpendicular to the direction of the flow due to the weight of the fluid above it? So for example at the red point, The total pressure Pt = P +pgh

I don't understand your diagram. But, I do understand your original pitot tube diagram. What would you say the fluid velocity is at the horizontal entrance to the pitot tube? What would you say the pressure is at this location in terms of P1, rho and the upstream velocity v? What would you say the pressure is at this location in terms of rho, g, and h, where h is the height the fluid rises in the vertical section of the pitot tube?

 

[Conductivity]

I don't understand your diagram. But, I do understand your original pitot tube diagram. What would you say the fluid velocity is at the horizontal entrance to the pitot tube? What would you say the pressure is at this location in terms of P1, rho and the upstream velocity v? What would you say the pressure is at this location in terms of rho, g, and h, where h is the height the fluid rises in the vertical section of the pitot tube?

1)zero
2)P1 +0.5 p V^2 = P2
3)P2 = pgh

Let's forget a bit about brenoulli's equation and pitot tube. Maybe this will make you understand what I am thinking.

If we have a cup of water, the pressure at the surface is atmospheric pressure and The pressure at a point underneath the surface with distance h is Patm + pgh
So what we see here is as if the atmospheric presssure is there all over the water but the changes in the vertical direction is caused by the weight of water. While in the horizontal we have the same pressure.
https://i.imgur.com/9g6MR3u.png

Now rotate that cup 90 degress, Break the base of it and we have a tube. Patm would be a static pressure that is let's say applied by a piston and we will still have variation in pressure on different stream lines because of the weight of the water above each point
https://i.imgur.com/obLjyq7.png

In brenoulli's equation, We say the force is equal to PA as if the pressure is constant on every point on the surface but shouldn't the pressure change according to pgh as in the static water cup? (Pressure variation normal to stream lines)

 

[Nidum]

You are saying that you think that pressure should vary with height in the pipe carrying the main flow ?

In reality it does but with smaller pipes the variation is small and usually ignored in practical work .

In any case if the Pitot tube is positioned along the central axis of the pipe then errors of measurement are going to be very small since the variations of pressure above and below the central axis cancel out .

Just for interest there is an experiment often done in labs where a small diameter Pitot tube is traversed across a flow of water in a duct and the dynamic pressure profile for the flow recorded .

Later on I had a project which required me to scan the boundary layer profile of a similar water flow in detail . I used a piece of hypodermic syringe with the end flattened down as the Pitot tube and traversed it with a DIY micrometer screw .

 

[Conductivity]

You are saying that you think that pressure should vary with height in the pipe carrying the main flow ?

In the real world it certainly might but for the purpose of simple calculations this pressure variation is usually ignored .

In any case if the Pitot tube is positioned along the central axis of the pipe then the error in any calculations or measurements is usually quite small .

Just for interest there is an experiment often done in labs where a small diameter Pitot tube is traversed across a flow of water in a duct and the pressure profile for the flow recorded . We did our experiments traversing horizontally but it might have been interesting to do the vertical traverse as well . Over the 300 mm square test section of the duct there should have been some small but noticeable difference .

Later on I had a project which required me to scan the boundary layer profile of a similar water flow in detail . I used a piece of hypodermic syringe with the end flattened down as the Pitot tube and traversed it with a DIY micrometer screw .

Yes that is what I am suggesting. Difference in pressure in the vertical direction due to the weight of the fluid. Also for a tube, If I assume pressure is isotropic, equal in every direction then I can never get steady flow in one direction only because in a circular tube if we take a horizontal cross section we will find that on this cross section points will have different pressure due to the different heights of fluid. I tried a lot and didnt find a way to make it laminar. Also if we assume gravity as a external force and then we integrate to find the pressure of gravity on one side of the tube and the other then find the net work done, I couldn't make it equal to the change in potential energy. How do we neglect gravity but it will also do work.

If I try to make the approximation of of constant presssure, In venturi tube, They take into account that across the tube there is a pressure difference equal to pg(D/2) if we take the center line.

 

[Nidum]

I've edited my post slightly to make it a little clearer .

Let's forget about the Pitot tube for the present .

Draw me a diagram showing all the pressures acting on a slice of fluid taken from the flow in a large diameter tube .

 

[Chestermiller]

I agree with Nidum. If you are asking whether the static pressure varies with depth within the horizontal pipe, then, yes it does. But this does not mean that pressure can't vary with horizontal position within the horizontal pipe. Is that what you are asking about?

As you correctly indicated, the fluid pressure at the centerline of the horizontal pipe is higher than at the top of the pipe and lower than at the bottom of the pipe.

 

[Conductivity]

I've edited my post slightly to make it a little clearer .

Let's forget about the Pitot tube for the present .

Draw me a diagram showing all the pressures acting on a slice of fluid taken from the flow in a large diameter tube .

I think I missed up :/

https://imgur.com/a/6MCh7

In the center point, We have the height of the fluid above it is (D/2) so the pressure is rho (D/2) where D is the diameter.
on any other point on the slice, the height of the fluid above it is variable h. So each point has a different pressure.

The fluid above and the bottom exerts a static pressure on the slice ( I thought of static pressure as pressure that exists everywhere and is equal in all directions)
Lastly perhaps there would be a pressure on the bottom of the slice to counter act the imbalance of forces

 

[Chestermiller]

I think I missed up :/

https://imgur.com/a/6MCh7

In the center point, We have the height of the fluid above it is (D/2) so the pressure is rho (D/2) where D is the diameter.
on any other point on the slice, the height of the fluid above it is variable h. So each point has a different pressure.

The fluid above and the bottom exerts a static pressure on the slice ( I thought of static pressure as pressure that exists everywhere and is equal in all directions)

At a given spatial point, it is the same in all directions. But, it varies with spatial location.

Also, in an enclosed tube, unless we know that pressure at some particular location, we can only talk about relative pressure changes. So the pressure at the centerline is $(D/2)\rho g$ higher than at the top and $(D/2)\rho g$ lower than at the bottom.

 

[Nidum]

Re: Your pm .

You say ' I would really like if you could show me how the pressure changes with elevation (force analysis)in the tube and then use the approximation to get the brenoulli's equation '

Happy to help if I can but I'm not quite sure what you are asking about . Can you explain a little more clearly please ?

 

[russ_watters]

Oh sorry, let me make it in the form of questions.

First we assume in a tube we have a static pressure as though we are pushing it with a piston from the other side is that right? Then if I want to find a pressure at a certain point, Shouldnt the total pressure be P + density * g * h? where pgh is the pressure resulting from the weight of the fluid above it? If not, Why please?

What's throwing me is the weird inverted manometers in the diagrams. Can we get rid of them and just do pressure gauges? Because you aren't talking about the manometer fluid depth, you are talking about the height of the air in your vessel, right?

The height of the air in a system (like a fan, air pump or wind tunnel) is usually ignored, but if you want to include it...

...the midpoint of the system has the average pressure (say, by force on a piston) and you can add ρgh from there to get variation with depth.

...er...you aren't trying to use water and air simultaneously in the same equation, are you?

 

[Conductivity]

I don't really understand why my question isn't clear :c. Sorry.

Can we forget about the pitot tube for a sec because if I solve this complication I can answer the original question..

Assume there is no gravity and that we have a tube with water flowing in it. On one side we have a piston that is pushing the water with Presssure P. If the flow of water is laminar then the pressure at every point must be equal to P or I would have turbulent flow. Is this point correct?

Now let's add the gravity to the situation, If we say that at each point the pressure is P + rho g h where h is the perpendicular distance from the point to the top of the tube, Then we have different pressure at each point which means it shouldn't be laminar? Can we find an equation that relates P and h to the pressure at a point? Should the pressure be constant at every point to have laminar flow?

Hopefully it is clear now :/

 

[russ_watters]

I think this is clearer:

Can we forget about the pitot tube for a sec because if I solve this complication I can answer the original question..

Assume there is no gravity and that we have a tube with water flowing in it. On one side we have a piston that is pushing the water with Presssure P. If the flow of water is laminar then the pressure at every point must be equal to P or I would have turbulent flow. Is this point correct?

Correct (no, laminar vs turbulent flow has no impact here). The one minor caveat is that somewhere in the system there must be an obstruction for the backpressure to push against, but this is a minor issue.

Now let's add the gravity to the situation, If we say that at each point the pressure is P + rho g h where h is the perpendicular distance from the point to the top of the tube, Then we have different pressure at each point...

Fine.

...which means it shouldn't be laminar?

...Should the pressure be constant at every point to have laminar flow?

No, it doesn't need to be laminar. Can you explain why you think that matters?

Can we find an equation that relates P and h to the pressure at a point?

Isn't that what you've already done?

 

[Conductivity]

I think this is cleare

No, it doesn't need to be laminar. Can you explain why you think that matters?

Oh yea, Sorry it doesn't need to be laminar.
Only non-viscous and steady flow

Isn't that what you've already done?

But wouldn't difference in pressure mean the there is flow in different directions other than the direction along the pipe?

 

[russ_watters]

Oh yea, Sorry it doesn't need to be laminar.
Only non-viscous and steady flow

Right.

But wouldn't difference in pressure mean the there is flow in different directions other than the direction along the pipe?

Oh, I'd gotten a vibe that cross flow might have been part of this...

The fact that the fluid is flowing may be part of the confusion. If you remove the flow from the situation, then all you have is hydrostatic pressure, so named because it is static: all forces are in balance.

To see how it works, start with a stack of books on a table. Each book has a larger force being applied to it from the book below than the book above. Yet it doesn't move. Why? Because it also has its own weight to deal with. Summing the 3 forces yields 0 net force.

In an incompressible fluid like water, you can replace the books with a stack of blocks, full of water. Same math applies.

For a compressible fluid, like air, you have a small caveat that for equal volumes, each block has a little less mass and weight than the block below it. But that doesn't change the force balance: still zero net force.

 

[Conductivity]

Right.

Oh, I'd gotten a vibe that cross flow might have been part of this...

The fact that the fluid is flowing may be part of the confusion. If you remove the flow from the situation, then all you have is hydrostatic pressure, so named because it is static: all forces are in balance.

To see how it works, start with a stack of books on a table. Each book has a larger force being applied to it from the book below than the book above. Yet it doesn't move. Why? Because it also has its own weight to deal with. Summing the 3 forces yields 0 net force.

In an incompressible fluid like water, you can replace the books with a stack of blocks, full of water. Same math applies.

For a compressible fluid, like air, you have a small caveat that for equal volumes, each block has a little less mass and weight than the block below it. But that doesn't change the force balance: still zero net force.

Yes, But it is balanced in the vertical direction but is it on the horizontal? Look at one open of the tube, Cut a horizontal slice, Some points in this slice will have higher pressure than other because the difference in height (Circular cross section). This part is what confusing me.

 

[russ_watters]

Yes, But it is balanced in the vertical direction but is it on the horizontal? Look at one open of the tube, Cut a horizontal slice, Some points in this slice will have higher pressure than other because the difference in height (Circular cross section). This part is what confusing me.

I'm not following: with a horizontal slice, all points on that plane are at the same elevation.

 

[Conductivity]

I'm not following: with a horizontal slice, all points on that plane are at the same elevation.

https://imgur.com/a/6MCh7

This one, In the center, rho g (D/2) but left or right from the center h changes. What is wrong with this argument?

 

[russ_watters]

https://imgur.com/a/6MCh7

This one, In the center, rho g (D/2) but left or right from the center h changes. What is wrong with this argument?

All of the points on the plane are at the same elevation. In hydrostatic pressure, the shape of the container is not relevant. h is measured versus a single chosen reference elevation.

 

[Conductivity]

Two last question and I would be really grateful.

1) So since now we agreed that the pressure at a point is equal to P + rho g h where h is the height. I assumed a rectangular pipe (For simplicity) whose elevation changed and area. In the normal derivation of brenoulli's principle, We calculate the external work that is applied on both sides of the section of fluid we take (ref http://www.4physics.com/phy_demo/bernoulli-effect-equation.html) and the work of gravity can be easily found with potential energy. However I tried calculating the work of gravity using the pressure difference we found. I integrated over the surface of each side and I got the correct value but instead of negative work I got a positive one. What am I doing wrong here?

https://i.imgur.com/TwInjDb.png
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)2) Is there is a proof that hydrostatic pressure doesn't depend on the shape but only the depth? I know the normal proof for a column of fluid but is there a general one? I assume we can prove it by saying if it is rho g h at one point then it must be the same for all points with the same depth or you will get horizontal movement?

 

[Chestermiller]

Two last question and I would be really grateful.

1) So since now we agreed that the pressure at a point is equal to P + rho g h where h is the height. I assumed a rectangular pipe (For simplicity) whose elevation changed and area. In the normal derivation of brenoulli's principle, We calculate the external work that is applied on both sides of the section of fluid we take (ref http://www.4physics.com/phy_demo/bernoulli-effect-equation.html) and the work of gravity can be easily found with potential energy. However I tried calculating the work of gravity using the pressure difference we found. I integrated over the surface of each side and I got the correct value but instead of negative work I got a positive one. What am I doing wrong here?

https://i.imgur.com/TwInjDb.png
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)2) Is there is a proof that hydrostatic pressure doesn't depend on the shape but only the depth? I know the normal proof for a column of fluid but is there a general one? I assume we can prove it by saying if it is rho g h at one point then it must be the same for all points with the same depth or you will get horizontal movement?

Even if it is moving horizontally, the pressures being equal at the same depth means that the velocity is not changing horizontally.

 

[Conductivity]

Even if it is moving horizontally, the pressures being equal at the same depth means that the velocity is not changing horizontally.

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

 

[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.