Pipe Flow

Cyrus

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,

[tex] P + \gamma h = 0 [/tex]

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.

Astronuc

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.

Cyrus

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.

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)

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.

Cyrus

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?

This cant be true in steady flow.

Right, but it also flows in the direction of decreasing pressure, which is now a problem?

Gokul43201

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?

Cyrus

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?

Cyrus

Does this help?



The top is a section cut, that's all you need to know. What is above it is irrelevant.

Clausius2

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 [tex]A[/tex], the atmospheric pressure [tex]P_a[/tex] and we have also the gravity [tex]g[/tex], the characteristic heigth of the pipe [tex]L[/tex], the fluid density [tex]\rho[/tex] and the dynamic viscosity [tex]\mu[/tex]. And let's play dimensional analysis. Call the vertical upwards coordinate z, [tex] r[/tex] the radial component, and call the vertical velocity component [tex]u[/tex].

The equations of motion are:
[tex]\nabla\cdot \overline{v}=0[/tex]
[tex]\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)[/tex]

The last equation is the z-Momentum equation. Note I could have not written [tex]P[/tex], 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 [tex]U\sim \sqrt{gL}[/tex]. Therefore, one defines the Reynolds Number [tex]Re=\sqrt{g/L}A/\nu[/tex]. Non dimensionalizing [tex]u=u/U[/tex], [tex]z=z/L[/tex] and [tex]t=tU/L[/tex] the equations become:

[tex]\nabla\cdot \overline{v}=0[/tex]
[tex]\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)[/tex]

For [tex]Re>>1[/tex], that is, for very wide pipes, one can assume ideal flow, and the z-Momentum eqn at leading order (with errors [tex]O(1/Re)[/tex]) is:

[tex]\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[/tex]

which can be integrated from 0 to h (the level of water):

[tex]\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[/tex] 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:

[tex]u=-\frac{dh}{dt}[/tex]. And a second order linear differential equation arises for h:

[tex]\frac{d^2h}{dt^2}+1=0[/tex].

Then, [tex]h(t)=-t^2/2+h_o[/tex]. 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 [tex]\nabla P=\rho \overline{g}[/tex] 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?

Cyrus

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.

FredGarvin

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

Clausius2

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:

[tex]\frac{\partial u}{\partial t}+\frac{\partial }{\partial z}\left(\frac{P}{\rho}+\frac{u^2}{2}+gz\left)=0[/tex]

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

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.

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.

Q_Goest

Hi cyrus,
Let me play back to you what I think you're saying and what your confusion is. First, you apply B's eq. to a verticle fluid column. In the case of zero velocity, you find the static pressure in any part of this column is equal to the pressure at the top of the column plus the head pressure. However, you also note that when the fluid is flowing (velocity > 0) and the bottom of the pipe is open to atmosphere, the same equation must hold, and thus if the pressure at the bottom of the column is atmospheric, you are suggesting the pressure above it is lower than atmospheric by the quantity rho*g*h (ie: head pressure). Thus, you note that the pressure above the opening in the bottom of the pipe is lower than Patm and you don't understand how that can be.

Note that in the static case of there being no flow in the pipe, the fluid is at a higher static pressure at the bottom than the top. Despite this pressure gradient, there is no flow going up the pipe. There is no flow at all. Flow can't be determined simply by saying the pressure is higher at one point so there is flow from a higher pressure location to a lower pressure location. The pressure due to head has no ability to force fluid to flow upwards because this upward pressure is balanced by the differential downward head pressure. Consider this however: if we remove static pressure head from the equation, we find the resulting pressure gradient is constant for a static condition. The static pressure in the verticle column of fluid when head pressure is neglected is constant.

B's equation is missleading, and unfortunately it doesn't include frictional pressure drop. Normally we add that part in, but we're generally not taught to do that in college. Why? Because B's eq. is actually a conservation of energy equation, and frictional losses don't conserve energy per se. B's eq. is simplifed to the point it doesn't represent reality. For it to represent reality, it has to include frictional pressure losses in the pipe which can be calculated by the Darcy Weisbach equation for example.

Now back to your example. For the case of steady flow (ie: not accelerating) the frictional pressure loss must be added into the equation. B's equation then becomes:

P1 = pgh + Pf + P2

Where P1 is the pressure at location 1 which I'll call the pipe outlet
P2 is the pressure some location above the outlet.
pgh is head pressure (density * g * h)
Pf is the frictional pressure loss which should be a negative term since we loose static pressure due to frictional losses along a pipe.

We know pgh from the fluid's density and height. We find Pf from the Darcey Weisbach equation or equivalent equation, and if we know P1 or P2, we can calulate the other. This doesn't mean that P2 is necessarily lower or higher than P1. For your example, you probably have some ideas in mind as to what's happening, but if you work them out I suspect you'll resolve the problem if you simply realize that P1 and P2 are only related to each other through the fluid head and frictional pressure losses (see my examples below). For a real situation, you need to include frictional pressure losses and you need to identify what pressure you may have going into the pipe which may be higher or lower than atmospheric pressure.

Examples:
1) If you say the pipe is open to atmosphere at the top and bottom, then you're saying that P1 = P2 and then the head pressure is equal to the frictional pressure loss.
2) If you say the pipe is open to atmosphere at the bottom and head pressure is larger than the absolute value of frictional pressure loss, then the pressure at the top of the pipe (P2) is below atmospheric pressure.
3) If you say the pipe is open to atmosphere at the bottom and head pressure is smaller than the absolute value of frictional pressure loss, then the top of the pipe is above atmospheric pressure.

Clausius2

The flow is accelerating, but the acceleration [tex]\frac{\partial u}{\partial t}[/tex] 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).

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.

Q_Goest

Hi Clausius.
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.

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.

Clausius2

The flow accelerates until there is no fluid inside the pipe, which I am assuming as the domain of integration. I told you, there is no way of looking for an steady state in this problem, because there are no disparity of time scales. It is not the same problem than the gasoline one. Well it could be. Imagine you have a cylindrical can, you open the upper side making a hole of the same diameter than the can one, and you put instantaneously the can upside down. There is no steady regime nowhere. It is the same case than if I throw a brick from a plane, the brick accelerates with the gravitational acceleration. In the pipe problem, the acceleration is constant in time and equal to the gravitational acceleration.

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 [tex]L\sim c^2/g[/tex], 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.