Previous answers here to the question of pressure gradients created during liquid flow down a skinny (low Re) vertical tube are: a) Liquid flow in a vertical tube and b) Friction loss in a vertical pipe flow

These nice previous posts don't provide a "phenomenological" physical answer to the question of why pressure remains homogeneous (i.e. dP/dz = 0 everywhere) in the vertical tube. In the case of (a) the discussion incorrectly uses Bernoulli's Principle which is explicitly only for inviscid flows. In the case of (b) the leading answer from Chet Miller (using Darcy–Weisbach equation) is derived very nicely and rigorously up until the equation

$$\frac{d(P + \rho gz)}{dz}=\frac{f}{D}\frac{\rho v^2}{2}$$

and then after that equation the dP/dz is back-calculated to be zero for a vertical pipe but I don't see any "phenomenological" physical explanation of why dP/dz = 0. What "F = ma" cause-effect physical reason is there for dP/dz to be zero at all locations within the tube?

  • $\begingroup$ @ChetMiller is active daily on the site; if you have a question about one of the terms of his derivation, try posting a comment to his answer to get clarification. $\endgroup$ Aug 22, 2021 at 17:17
  • $\begingroup$ I tried to do exactly that. Stackexchange blocked me from posting a comment directly on Chet's derivation because I "don't have a stackexchange reputation of 50 or higher". $\endgroup$
    – dturn805
    Aug 22, 2021 at 21:08
  • $\begingroup$ Maybe you have overlooked the description "For a horizontal tube" between the second last and last equation, which you refer to. I think that explains it pretty well? And also it shows that in fact, gravity does not drop out.. $\endgroup$ Aug 22, 2021 at 23:29
  • $\begingroup$ Aha thank you @AtmosphericPrisonEscape. You're correct but my main point still stands. I reworded the question to resolve the error you pointed out. $\endgroup$
    – dturn805
    Aug 24, 2021 at 0:30
  • $\begingroup$ Is this a liquid without yeast in it so there will be no gas bubbles and so on ? $\endgroup$
    – Emil
    Aug 25, 2021 at 5:47

2 Answers 2


If we do a force balance on the fluid within the section of the tube between z and z + $\Delta z$, where z is measured upward, we get $$P(z+\Delta z)\pi \frac{D^2}{4}-P(z)\pi \frac{D^2}{4}+(\rho g\Delta z)\pi \frac{D^2}{4}=\tau_W\pi D\Delta z$$where $\tau_w$ is the upward viscous shear stress exerted by the wall on the fluid, and the 3rd term on the LHS is the weight of the fluid in the interval between z and z + $\Delta z$. Taking the limit as $\Delta z$ approaches zero, we obtain: $$\frac{dP}{dz}+\rho g =\frac{4}{D}\tau_w$$If we integrate this equation between z = 0 (the bottom of the tube where $P = p_{atm}$) and z = L (the top of the tube where, again, $P = p_{atm}$), we obtain $$\rho g =\frac{4}{D}\tau_w$$The LHS of this equation is the weight per unit volume of the liquid in the tube, and the RHS is the viscous frictional force of the wall per unit volume of the liquid in the tube.

Of course, you can apply a higher pressure at the bottom than at the top, causing a lower flow rate and wall shear stress, or apply a higher pressure at the top than at the bottom, causing a higher flow rate and wall shear stress. But, in all cases, the weight of the fluid helps to lower the pressure difference needed to maintain a desired flow rate.

Compressibilty has nothing to do with this.

For laminar flow of a viscous fluid, the shear stress at the wall is given by: $$\tau_w=\mu\frac{32Q}{\pi D^3}$$where Q is the downward volumetric flow rate and $\mu$ is the fluid viscosity. So for the general case where we have arbitrary pressures at the two ends of the tube, we have $$P(L)-P(0)=\mu\frac{128QL}{\pi D^4}-\rho gL$$

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    $\begingroup$ You cannot "integrate this equation" between z = 0 and z = L if you don't already know the pressure profile. The whole point of this post is that we don't know the profile of pressure (or the profile of the pressure gradient). Therefore you cannot rigorously move from the $\frac{dP}{dz}+\rho g =\frac{4}{D}\tau_w$ equation to the $\rho g =\frac{4}{D}\tau_w$ equation via integration. That integral is unknown, given the a priori information at hand. $\endgroup$
    – dturn805
    Aug 22, 2021 at 21:11
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    $\begingroup$ Of course I can integrate. This equation follows directly from the solution to the Navier Stokes equations, which reveal all the characteristics of the pressure variation we need. I guess you are not familiar with the Navier Stokes equations or how to solve them? $\endgroup$ Aug 22, 2021 at 21:49
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    $\begingroup$ Oh yeah? Well let’s see your approach to solving the NS equations for steady flow through a vertical tube. $\endgroup$ Aug 24, 2021 at 0:31
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    $\begingroup$ Lemme guess. You're not an engineer? You're a mathematician? $\endgroup$ Aug 24, 2021 at 0:41
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    $\begingroup$ How come the only approximation you are attacking is that the fluid is incompressible. Why not the effect of elevation on g? Or the effect of the phases of Jupiter's moons on g? Or the effect of viscous heating on temperature? Or the effects of tube roughness on the flow? $\endgroup$ Aug 24, 2021 at 0:49

Summary: I show why the only possible pressure gradient can be a constant gradient. And that neither end can be above atmospheric. Therefore by contradiction there cannot be a pressure gradient. That’s toward the bottom. Most of the rest is to gain intuitive understanding.

Overall Intuition:

A volume of gas that does not have any external pressure on it will expand to fill all available space. A volume of liquid that does not have any pressure on it will expand to its zero-pressure specific volume, $v_{0*}$. (We don’t generally think of liquids as being compressible because they are not very compressible, but they are.)

If a liquid is confined at the bottom, then the fluid layer on the bottom is confined and faces the pressure of the weight of the liquid above it. It cannot expand downward even the tiny amount necessary to achieve the $v_{0*}$ volume. This layer of liquid is confined under pressure.

However if the bottom is removed, the entire volume of liquid is free to expand the tiny amount necessary for the full height of it to be at $v_{0*}$.

The only possible resistance to the expansion of the whole column is the viscous sheer forces of flow in the tube. But because the compression is only ~$0.01$%, the expansion of a $10m$ tube of liquid will only be one millimeter. This will happen quickly without generating enough velocity for viscous forces to hamper it (Viscous force is an increasing function of velocity, and $F_{\mu}(0)=0$).

All of this is true regardless of whether and how much the fluid is flowing. It is still a column of fluid and will require external net pressure to be compressed. If a bottom is provided, then the weight provides the pressure from above to compress it and the container bottom provides the pressure from below and the tube sides provide pressure from the sides. Barring force from all directions, the liquid can expand and will not have any pressure.


You may still be thinking that there could be a gradient and dynamics somehow. But consider each layer as below.

If we consider a small control mass of fluid of height $L_z$ inside the tube (a control volume is a defined section of space where mass can move into and/or out of the control volume; a control mass is a quantity of actual, specific fluid particles), then it will be flowing down the tube due to gravity at velocity $v$, with a viscous resistance force per unit length $f_{\mu}=\frac{F_{\mu}}{h}$, and will be governed by:

$$F= f_{\mu}(v) L_z + L_z \frac{dp}{dz} - gA \rho L_z = ma=0$$

Note this is true for all $z$. That implies $v$ is constant. And therefore $ \frac{dp}{dz}$ is constant (because both $gA \rho $ and $f_{\mu}(v) $ are constant; true even for $a \neq 0$).

All this means that the only possible pressure gradient is a constant pressure gradient. If we have a constant pressure gradient then either the top layer or bottom layer or both is above atmospheric. If, $wlog$, the bottom layer is above atmospheric, then it will expand against the air until it is at $v_{0*}$ and no longer be so, which shows by contradiction there could not have been a gradient. If neither end can be above atmospheric and the only possible pressure gradient is constant, then there cannot be a pressure gradient.

More intuitively, the fluid layer at each end can expand instantly, and the layer next to it can also expand pushing the layer below it the tiny amount necessary, etc. Even the middle layer in the center of the tube can expand and push half the liquid far enough to be able to expand, because the needed displacement is so low and flow only resists velocity.


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