Contradicting values of pressure for liquid cross-sections

[Logic hunter]

Consider a fixed horizontal tube of uniform cross section with pressure being 1atm at one of it's end and 5atm at the other (former due to 'open to atmosphere' and latter due to force on a piston), then liquid would flow towards low pressure end. By equation of continuity all cross sections will have same velocity. Since tube is horizontal all of them will have the same potential energy. So by bernoulli's theorem all of them should have same pressure, but the pressure at both the ends of the tube are different for liquid to flow. Also is this apparent 'contradiction' only true for the cross sections at the end and pressure is same for all other cross sections or is the pressure gradually decreasing on moving from one end of the tube to the other (towards the low pressure end).

 

[Geofleur]

I will take a stab at answering this, though this is not quite my area!

If there are no frictional forces within the tube to oppose the force due to the pressure gradient, then the fluid everywhere in the tube will accelerate toward the low pressure end. But then the density cannot be constant. Imagine a row of cars in which the car in front is always going faster than the car behind. The traffic must thin out in such a state of affairs.

On the other hand, if there are frictional forces within the tube, then Bernoulli's equation will not be quite right, because energy will be lost due to dissipation, and there is no term for that in Bernoulli's equation.

 

[Chestermiller]

I will take a stab at answering this, though this is not quite my area!

If there are no frictional forces within the tube to oppose the force due to the pressure gradient, then the fluid everywhere in the tube will accelerate toward the low pressure end. But then the density cannot be constant. Imagine a row of cars in which the car in front is always going faster than the car behind. The traffic must thin out in such a state of affairs.

On the other hand, if there are frictional forces within the tube, then Bernoulli's equation will not be quite right, because energy will be lost due to dissipation, and there is no term for that in Bernoulli's equation.

This is partly not correct. If the fluid is virtually incompressible (like water), the density will be constant. So, without frictional forces, the fluid will be accelerating, and the flow will be transient. This will mean that the pressure will vary linearly from the high end to the low end. The Bernoulli equation does not apply to transient flows.

If frictional forces are present, the Bernoulli equation will not just be "not quite right." It will be outright wrong. For steady flow, the Bernoulli equation would predict no pressure difference at all. With friction, the flow rate will adjust until the force difference from the pressures at the two ends is just balanced by the frictional shear force on the surface of the pipe.

 

[Geofleur]

Okay, I'm going to expose possibly a lot more confusion on my part but with the hope of learning something :)

If the pressure varies linearly, then fluid everywhere in the pipe has the same acceleration, yes? I was imagining another case where every parcel in a column of water has the same acceleration: Water falling from the tap. In that case, the stream of water narrows downstream. But what if you could force it not to narrow? Wouldn't the density of the water have to decrease?

 

[Dale]

Welcome back @Geofleur ! It is good to see you around again

 

[Geofleur]

Thanks - it's nice to be around again :)

 

[Chestermiller]

Okay, I'm going to expose possibly a lot more confusion on my part but with the hope of learning something :)

If the pressure varies linearly, then fluid everywhere in the pipe has the same acceleration, yes? I was imagining another case where every parcel in a column of water has the same acceleration: Water falling from the tap. In that case, the stream of water narrows downstream. But what if you could force it not to narrow? Wouldn't the density of the water have to decrease?

How would you force it not to narrow?

 

[Geofleur]

I have no idea :|

But I was thinking that the pipe scenario that the OP described would amount to just that if there were no friction, because in both situations the fluid at every point has the same acceleration, in one case proportional to the pressure gradient and in the other equal to gravitational acceleration. Since the fluid narrows in the latter case, I reasoned that it would have to also in the former (or the density would decrease, or...?). Would the fluid in such a pipe just peel away from the walls, or is my analogy not good?

 

[Chestermiller]

I guess I'm not too interested in the situations where the flow is accelerating; I'll leave that to you to take a shot at modeling. After you've analyzed it, I'll add any comments I might have.

The situation of more practical interest to me is where viscous friction enables you to establish a steady state flow. I'll gladly model that for the vertical case if you feel that that would be of interest.

Chet

 

[Geofleur]

Okay, it's a deal! I'll have a shot at modelling the accelerating case :)

I can think of scenarios where a model of viscous flow in a vertical pipe (or channel!) might be useful. For example, in models of geothermal energy extraction or flow in hydrothermal systems, fractures are sometimes modeled as vertical pipes or channels. At any rate, I'm interested!

 

[Chestermiller]

Analysis of Steady Flow in Vertical Pipe

From a free body diagram on the plug of fluid situated between axial locations z and $z+\Delta z$ (where z is measured downward), the force balance on the fluid plug is:$$[P(z)-P(z+\Delta z)]\pi R^2+\pi R^2\rho g \Delta z-2\pi R\ \Delta z\ \tau=0$$where $\tau$ is the (upward) shear stress at the wall. Dividing by $\pi R^2\ \Delta z$ and taking the limit as $\Delta z$ approaches zero, we obtain:$$-\frac{dp}{dz}+\rho g=\frac{2}{R}\tau$$The shear stress at the wall is related to the cross-section average downward axial velocity v by: $$\tau=\frac{1}{2}\rho v^2 f$$where f is the Fanning friction factor. The Fanning friction factor is a function of the Reynolds number for the flow, $Re=\frac{\rho v D}{\mu}$, where $\mu$ is the fluid viscosity. For laminar flow (Re <2000) by $f=16/Re$. For larger Reynolds numbers, the flow is turbulent, and the Fanning friction factor is provided by the empirical Fanning correlation.

 

[cjl]

Okay, I'm going to expose possibly a lot more confusion on my part but with the hope of learning something :)

If the pressure varies linearly, then fluid everywhere in the pipe has the same acceleration, yes? I was imagining another case where every parcel in a column of water has the same acceleration: Water falling from the tap. In that case, the stream of water narrows downstream. But what if you could force it not to narrow? Wouldn't the density of the water have to decrease?

The water would only narrow if the acceleration caused the velocity to vary along the length of the pipe. In the case of a pipe of fixed diameter with such a gradient, what would happen is that the velocity would be the same everywhere along the pipe, but time-varying (so all of the fluid in the pipe would accelerate together).

 

[Geofleur]

Analysis of Unsteady Flow in a Horizontal Pipe

I took the 1D Euler equation down from the shelf and blew the dust off it: $$\rho \frac{\partial v}{\partial t}+\rho v\frac{\partial v}{\partial x} = -\frac{\partial p}{\partial x}.$$ We also need the 1D continuity equation: $$ \frac{\partial \rho}{\partial t}+ \frac{\partial}{\partial x}(\rho v) = 0.$$ I will assume that the density is constant and see what these equations predict. From the continuity equation, we get $$\frac{\partial v}{\partial x}=0,$$ implying that the velocity is at most a function of time. Incorporating this information into the Euler equation and assuming that the pressure is time independent gives $$\rho\frac{dv}{dt}=-\frac{dp}{dx}.$$ Let the high pressure end of the pipe be located at $x=0$ and the low pressure end at $x = L$, with the corresponding pressures being $p_0$ and $p_L$. Integrating the above equation from $x=0$ to $x=L$ and dividing by $\rho L$ leads to $$\frac{dv}{dt}=\frac{(p_0-p_L)}{\rho L}.$$ If the fluid is initially at rest, then integrating this equation gives the fluid velocity as $$v = \left(\frac{p_0-p_L}{\rho L}\right) t.$$

So all the fluid particles in the pipe would move together at the same speed at any given instant of time, and all would have the same acceleration (like cjl said!).

 

[Logic hunter]

Analysis of Unsteady Flow in a Horizontal Pipe

I took the 1D Euler equation down from the shelf and blew the dust off it: $$\rho \frac{\partial v}{\partial t}+\rho v\frac{\partial v}{\partial x} = -\frac{\partial p}{\partial x}.$$ We also need the 1D continuity equation: $$ \frac{\partial \rho}{\partial t}+ \frac{\partial}{\partial x}(\rho v) = 0.$$ I will assume that the density is constant and see what these equations predict. From the continuity equation, we get $$\frac{\partial v}{\partial x}=0,$$ implying that the velocity is at most a function of time. Incorporating this information into the Euler equation and assuming that the pressure is time independent gives $$\rho\frac{dv}{dt}=-\frac{dp}{dx}.$$ Let the high pressure end of the pipe be located at $x=0$ and the low pressure end at $x = L$, with the corresponding pressures being $p_0$ and $p_L$. Integrating the above equation from $x=0$ to $x=L$ and dividing by $\rho L$ leads to $$\frac{dv}{dt}=\frac{(p_0-p_L)}{\rho L}.$$ If the fluid is initially at rest, then integrating this equation gives the fluid velocity as $$v = \left(\frac{p_0-p_L}{\rho L}\right) t.$$

So all the fluid particles in the pipe would move together at the same speed at any given instant of time, and all would have the same acceleration (like cjl said!).

So what's the final conclusion about pressure in simpler terms.

 

[Chestermiller]

So what's the final conclusion about pressure in simpler terms.

Either way, it varies linearly with distance from the high pressure end to the low pressure end.

 

[cjl]

Bernoulli only applies to a steady flow, which this isn't.

 

[Logic hunter]

Bernoulli only applies to a steady flow, which this isn't.

How did you tell this isn't steady?

 

[cjl]

Because it is accelerating, as described before. More accurately, if you look at a single physical location within the flow, the velocity is changing over time (since as mentioned above, all the fluid in the tube will accelerate as a unit in this scenario).