I understand the idea of pressure in fluids at rest, but I find some difficulty in understanding the idea of "Static Pressure" in flowing fluids.

  1. What is the cause of "static pressure" in flowing fluids? How do we understand the idea of static pressure intuitively?

  2. I know that thermal molecular motion is NOT the cause of pressure in fluid statics, so it cannot be the case of pressure in fluid dynamics as well. If we disregard the idea of thermal motion then how do we understand the static pressure in flowing fluids?

  3. Suppose a fluid is flowing in a closed pipe of uniform cross-section, with uniform velocity, then what will be the static pressure in the fluid? In my understanding, it should be zero, as to move with uniform velocity there is no need for any external force and hence then pressure should also be zero. Is it right?

  4. Suppose fluid is flowing through variable cross-section areas then from Bernoulli's equation I know that there will be pressure gradient along with the flow. How do we understand the "Static Pressure" in this case? What causes it? How do we understand it intuitively?

  • $\begingroup$ Due to friction losses at a pipe wall and within the fluid itself, a pressure gradient must exist for flow to exist. $\endgroup$ – David White Dec 16 '19 at 18:16

Pressure is caused by constraining a fluid*

All** pressure is the result of constraining a fluid to a container. In other words, the walls of your container provide the force that "causes" there to be pressure in a fluid. An example of a fluid flowing in a constrained space is water in a pipe that changes diameters (your item #4). An example of an unconstrained fluid is a droplet of water in free-fall, like those created by astronauts on the ISS.

An intuitive explanation of pressure:

An excellent analogy to pressure is the normal force of a block on a frictionless incline. The incline constrains the block's motion by applying a normal force. When the incline is vertical, nothing is constrained and the normal force is zero. When a container constrains a fluid's motion, it applies a pressure that is felt everywhere in that fluid. Without a container, no pressure is applied**. See the below illustration:

constrained fluids

* Please only liquids and gases, no plasmas please :).

** we must exclude systems that violate a quasi-static equilibrium (e.g. shock waves, free expansion, etc.).

  • $\begingroup$ It's worth mentioning surface tension also constrains fluids, so droplets can have internal pressure as well, even in zero gravity. $\endgroup$ – KF Gauss Dec 17 '19 at 0:55
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    $\begingroup$ Can’t ever satisfy the literalists. Spherical cows only please. ;) $\endgroup$ – cms Dec 17 '19 at 0:56
  • $\begingroup$ I didn't say that to criticize your answer...I am saying that your description is correct even in zero gravity for droplets, just with a different type of "container". This is hardly being a "literalist" $\endgroup$ – KF Gauss Dec 17 '19 at 3:33
  • $\begingroup$ Winky face implies no rebuke. Your comment is appreciated as it helps not clutter up the answer with errata. $\endgroup$ – cms Dec 17 '19 at 3:59

1) Static pressure is the result of constraining a fluid, meaning a collection of particles, in a finite space. This will force the particles on a microscopic level to interact either in (elastic) collisions (such as in a dilute gas) or in more complex interactions of repulsive and attractive forces. The result of the presence of other particles is felt on a macroscopic level as a force per area (the particles push each other away), pressure.

2) Fluid statics is a model that artificially freezes the state of a liquid. In reality even macroscopic equilibrium never means uniformity but instead always allows for a certain amount of variance (Just think of the Brownian motion of particles in seemingly static fluids!), generally characterised by a certain Gaussian distributions. A fluid is simply a collection of particles that interact. Even in thermal equilibrium not all particles have the same energy content, the same kinetic energy, but they are stochastically distributed around a certain mean value. While at continuum level this fundamental property of restlessness is neglected and lumped into macroscopic properties such as pressure, viscosity or the equation of state, the fact the a fluid is never truly at rest, is the reason for any macroscopic property. As already pointed out the static pressure of flowing fluids is caused by molecular motion. In a moving fluid this motion is though - contrarily to a resting fluid - partially directed by the macroscopic flow velocity. As a result collisions will dominate in the flow direction while be significantly lower perpendicular to it. This results in a different perception of pressure (a certain anisotopy). The anisotropic part of pressure that is caused by the macroscopic flow with velocity $\vec u$ is termed dynamic pressure $p_d = \sum\limits_i \frac{\rho u_i u_i}{2}$ while the part that is isotropic (in all directions identical) is called the static pressure $p$ (It is also perceived by a element perpendicular to the flow direction). The combination of the two is felt at a stagnation point if the flow is slowed down isentropically: The stagnation pressure $p_s$ is the sum of the former two.

3) Such a completely uniform flow is impossible in a real world as any sort of friction introduces a flow profile that is zero towards the walls (no-slip boundary). This friction results in a loss of energy and thus a pressure gradient, a pressure drop. Theoretically one can look at flow as you describe it with an inviscid flow model (there is no friction, Euler equation, this model does not have to respect the no-slip boundary condition). The Bernoulli equation for such a flow would be

$$ p_1 + \rho \frac{u_1^2}{2} + \rho g h_1 = p_2 + \rho \frac{u_2^2}{2} + \rho g h_2 $$

For a velocity that stays the same between point 1 and 2 which also is at the same level $h_1 = h_2$ the static pressure has to take the same non-zero value. Its precise value can be determined by the equation of state, such as the ideal gas law

$$ p v = R_m T. $$

4) I have just given an answer to your fourth question on another post of yours. In short: Any pressure gradient can be interpreted by either reducing the mean free path (for a compressible ideal gas) or pressing the building bricks of an incompressible fluid (e.g. molecules) closer together exerting a increasing force and thus a larger pressure onto each other.


In real life, static pressure is often provided by a header tank mounted above your apparatus. The higher the tank the more pressure the fluid in your experiment is under.

This is static pressure.

Assuming the fluid has a constant density it makes no contribution to fluid motion. The static pressure is greater at greater depth by exactly the amount needed to support the fluid at rest.

tiny pin holes at different levels can be used to show the pressure difference. The lower the pinhole the stronger the leakage stream.

The pipes, pumps and motion of fluid in your experiment are unaffected by the height of the header tank, so the static pressure can usually be ignored, and only pressure differences are relevant.


So static pressure is not due to any external force when the fluid is at rest but if the fluid is in motion an additional static pressure is created by a part of external force to overcome the container's resultant friction(ie constraint) leading to motion of the fluid .This additional static force may be termed as fluid energy due to the additional static pressure, as such the Bernoulis equation remains as P(fluid energy or additional static pressure due to part of external force)+dynamic pressure+potential pressure =Constant( at any point moving in the fluid motion direction) . However these 3 pressures do not equalise the total external force which counters other forces or pressure in the moving fluid.


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