## Pipe Flow

This came up in the homework help section, and has me scratching my head.

Consider a vertical pipe open to atm at one end under steady flow.

The velocity in has to be equal to the velocity out because of the cont. equation.

The pressure at the bottom is zero, (atm), and the datum is at the bottom, (z=0).

That only leaves,

$$P + \gamma h = 0$$

h cant be negative, so that means the pressure must be equal and opposite to the hydrostatic head.

But then that mean's the fluid will flow in the direction of increasing pressure because the pressure will become less negative as you move down ?? <enter confusion>

Show me what I did wrong.

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 Admin I take it P is the pressure at the top of the pipe? There must be a force applied to have a fluid flow. Either the fluid is given momentum (from a fan or some device) or one puts energy (heat) into it and bouyancy causes a flow. The P at the top has to be less the P (1 atm) at bottom.

Yeah, everything at the bottom will become zero, and the velocity at the bottom will cancel the velocity at the top, so it drops out as well.

 The P at the top has to be less the P (1 atm) at bottom.
That's what the equation is saying. But that would imply reverse flow because the atm pressure would want to push the water up, and not down. (water is flowing down and out the tube)

## Pipe Flow

The OP didn't mention water. I thought it was an open pipe in air, and it was air flow.

The water flows down in gravity.

The head of water pushes down. Water displaces air.

The atmosphere provides a hydrostatic pressure, distinct from the head of water.

 Quote by Astronuc The OP didn't mention water. I thought it was an open pipe in air, and it was air flow. The water flows down in gravity. The head of water pushes down. Water displaces air. The atmosphere provides a hydrostatic pressure, distinct from the head of water.
Sorry, it can be any incompressible fluid.

Eh, I did a bad job in my OP, didnt I!

The problem is that if you go a step further and consider the top of the pipe to be pressurized, then you will get negative hydrostatic terms, which makes no sense.

Maybe the problem is that you cannot ignore the contraction coefficient?

 The head of water pushes down. Water displaces air.
This cant be true in steady flow.

 The water flows down in gravity.
Right, but it also flows in the direction of decreasing pressure, which is now a problem?

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 Quote by cyrusabdollahi Eh, I did a bad job in my OP, didnt I!
Yeah, you can say that again. I've got no idea what your system is!! Got a diagram? Which end's open, and what's at the other end?

It is a vertical pipe w/constant diameter. The bottom of the pipe is open and the water is flowing out of it. The top of the pipe keeps going on forever. Is that better?

I don't have a picture, I just made it up in my head. Just think of a pipe flowing out the bottom.

See post #3?
 Quote by me (water is flowing down and out the tube)

 Does this help? The top is a section cut, that's all you need to know. What is above it is irrelevant.
 Recognitions: Gold Member Science Advisor The pressure at the exit down section is the atmospheric one. The pressure at the top is also the atmospheric one. There is no contradiction nowhere, BECAUSE you cannot apply hydrostatic pressure equilibrium in the vertical direction. Hydrostatic means static, and your fluid isn't static at all. Let's call the pipe section $$A$$, the atmospheric pressure $$P_a$$ and we have also the gravity $$g$$, the characteristic heigth of the pipe $$L$$, the fluid density $$\rho$$ and the dynamic viscosity $$\mu$$. And let's play dimensional analysis. Call the vertical upwards coordinate z, $$r$$ the radial component, and call the vertical velocity component $$u$$. The equations of motion are: $$\nabla\cdot \overline{v}=0$$ $$\frac{\partial u}{\partial t} +u\frac{\partial u}{\partial z}=-\frac{1}{\rho}\frac{\partial (P+\rho gz)}{\partial z}+\frac{\nu}{r}\frac{\partial}{\partial r}\left(r \frac{\partial u}{\partial r}\right)$$ The last equation is the z-Momentum equation. Note I could have not written $$P$$, because it's vertical gradient is identically zero in this problem. Note also that this problem is inherently unsteady. Let's talk about the order of magnitudes. Firstly, one expects to have as much velocity as the one caused by the gravitational force, so that $$U\sim \sqrt{gL}$$. Therefore, one defines the Reynolds Number $$Re=\sqrt{g/L}A/\nu$$. Non dimensionalizing $$u=u/U$$, $$z=z/L$$ and $$t=tU/L$$ the equations become: $$\nabla\cdot \overline{v}=0$$ $$\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial z}=-1+\frac{1}{rRe}\frac{\partial}{\partial r}\left(r \frac{\partial u}{\partial r}\right)$$ For $$Re>>1$$, that is, for very wide pipes, one can assume ideal flow, and the z-Momentum eqn at leading order (with errors $$O(1/Re)$$) is: $$\frac{\partial u}{\partial t}+ u\frac{\partial u}{\partial z}=\frac{\partial u}{\partial t}+1/2\frac{\partial u^2}{\partial z}= -1$$ which can be integrated from 0 to h (the level of water): $$\int_0^h \left\{\frac{\partial u}{\partial t}+1/2 \frac{\partial u^2}{\partial z}+1\right\}dz=h\frac{\partial u}{\partial t}+h=0$$ which is just a nondimensional Bernouilli equation. Observe that I have integrated the unsteady term directly, because the acceleration is homogeneous inside the fluid. Note that there is no variation of kinetic energy. Also note that from the equation of Continuity one obtains a relation between u and h: $$u=-\frac{dh}{dt}$$. And a second order linear differential equation arises for h: $$\frac{d^2h}{dt^2}+1=0$$. Then, $$h(t)=-t^2/2+h_o$$. Does not sound to you as the free-falling space law for a particle in a gravity field????. Yeah, the water is falling as a whole, as a rigid solid. To sum up, the hydrostatic balance $$\nabla P=\rho \overline{g}$$ does not hold in this system. If there is motion in the direction of the body force, there is no hydrostatic balance. I am still in good shape, am not?

 The pressure at the exit down section is the atmospheric one. The pressure at the top is also the atmospheric one. There is no contradiction nowhere, BECAUSE you cannot apply hydrostatic pressure equilibrium in the vertical direction. Hydrostatic means static, and your fluid isn't static at all.
Well, I said you could make the pressure at the top end whatever value you like, so it is not necessarily atm. Also, I did not apply hydrostatic equilibrium anywhere. I used the bernoulli equation which reduced itself to what I showed.

That is a great respons Clausius, but I asked a question about bernoulli and you replied with the navier stokes equations. Can you give a more explicit answer to the bernoulli? or if you have, restate it, because I can't see it.

 Recognitions: Science Advisor The fluid has to be accelerating. The velocities will not be equal at the two points, so they won't cancel out.

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 Quote by cyrusabdollahi Well, I said you could make the pressure at the top end whatever value you like, so it is not necessarily atm. Also, I did not apply hydrostatic equilibrium anywhere. I used the bernoulli equation which reduced itself to what I showed. That is a great respons Clausius, but I asked a question about bernoulli and you replied with the navier stokes equations. Can you give a more explicit answer to the bernoulli? or if you have, restate it, because I can't see it.
Hey cyrus. The Bernoulli equation arises automatically from my analysis. I mean, i am not answering you with another class of equations never seen before, the N-S equations suitably simplified give birth to the Bernoulli equation. Look again to my post and find the Bernoulli equation, which in this case is an UNSTEADY Bernoulli equation of the form:

$$\frac{\partial u}{\partial t}+\frac{\partial }{\partial z}\left(\frac{P}{\rho}+\frac{u^2}{2}+gz\left)=0$$

BTW: by non including the unsteady term you were actually assuming hydrostatic equilibrium even though you didn't want to.

 Quote by Fred The fluid has to be accelerating. The velocities will not be equal at the two points, so they won't cancel out.
That's false, and it repels the Continuity equation for incompressible flow. Your first sentence is true though, but your second one not quite.

I think you guys didn't understand my post.

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 Quote by Clausius2 That's false, and it repels the Continuity equation for incompressible flow. Your first sentence is true though, but your second one not quite.
I guess I'll have to digest it a bit more. What is escaping me right now is how it is violating continuity. We know that the flow area of a stream will decrease as the fluid accelerates. If the fluid is accelerating in a linear fashion, then how can the velocities not be different? I think...I need to look at your post again.

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 Quote by FredGarvin I guess I'll have to digest it a bit more. What is escaping me right now is how it is violating continuity. We know that the flow area of a stream will decrease as the fluid accelerates. If the fluid is accelerating in a linear fashion, then how can the velocities not be different? I think...I need to look at your post again.

The flow is accelerating, but the acceleration $$\frac{\partial u}{\partial t}$$ as I said is homogeneous through the entire fluid, so it does not depend on z. The equation of Continuity says that UA=constant in the whole pipe, IFFFF the density remains constant. IFFF the acceleration would depend on the height, the velocity would too, and the flow wouldn't be incompressible at all (the column of water would behave as a spring). But experimentally that's not the case, it behaves like approximately a brick, so U=constant throughout the fluid but NOT in time. Moreover, the acceleration in incompressible flow is instantaneously felt by any part of the fluid, because acceleration is propagated via pressure waves, and pressure waves travel at the speed of sound (infinite in incompressible flow).

 Quote by QGoest For the case of steady flow
I appreciate your comments, but this is NOT a steady flow, as well as the free falling of a brick is not steady. A problem that could be treated as a quasisteady one could be the discharge of water FROM a vessel through a pipeline of diameter much smaller than the vessel diameter, because the time scale variation of the vessel water height can be approximated as very long compared with the local flow times at the exit of the pipe.

Still you can apply Bernoulli, but not the usual Bernoulli. You have to employ the unsteady Bernoulli equation. I did it, solving the problem, and it showed us, as it was sensible, that it is consistent with the classical mechanics of a falling body. The static pressure is the atmospheric pressure throughout the entire pipeline, as it couldn't be otherwise.

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Hi Clausius.
 but this is NOT a steady flow,
Yes, we can say it's an accelerating flow. As soon as one allows flow to begin, the fluid must accelerate since it begins by standing still.

However, the flow doesn't accelerate forever. The flow is likely to come to some equilibrium unless it undergoes deceleration and oscilates around some nominal value which is especially likely for the case of gasoline being poured from a small can for example. In this case, the pressure in the can decays as gasoline is poured into our lawn mower and flow slows down. Air then enters the can by traveling up the nozzle (pipe) which repressurizes the tank, and the gasoline accelerates again. But I don't think this is what the OP is concerned with.

The OP doesn't state whether they are concerned with the equilibrium state or the non-equilibrium, transient condition, during which flow must accelerate and potentially decelerate. I've assumed the OP was interested in the steady state condition, not the transient. There is a perfectly valid steady state condition of interest, and that's what I've provided an explanation for.

 Moreover, the acceleration in incompressible flow is instantaneously felt by any part of the fluid, because acceleration is propagated via pressure waves, and pressure waves travel at the speed of sound (infinite in incompressible flow).
Note that although we might model something as having infinite pressure wave velocity, that isn't the reality of it. Modeling it this way can work if the distances the wave must travel are small, or pressure waves are small compared to gross pressure changes needed to accelerate the fluid, but we must remember these are only simplifying assumptions, just as using B's eq. without frictional pressure losses, or without acceleration pressure losses are simplifying assumptions.

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Hey Q, we are engineers, and as engineer one needs to discard, via dimensional analysis, what is important for my problem and what is a second order effect for my problem. The compressibility of the fluid here plays no role UNLESS the pipe has a length of order $$L\sim c^2/g$$, where c=1500 m/s in water (speed of sound). Do the numbers and tell me if you know a vertical pipeline of that dimensions. If that is not the case, the time spent by a pressure wave to travel from the bottom to the top of the pipeline is doubtless so small compared with the rest of the times scales that is not worthy at all to include this time range in the formulation. The unsteadiness here is due to the acceleration of the bulk motion, not due to the propagation of pressure waves.