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If a fluid ( ideal) flows, say from left to right, then is the pressure at any point inside the fluid still independent of direction and has the same value no matter how I orient a surface element at that position?

If the above is true then, I am lead to an paradoxical condition. Say the fluid flows in cuboidal tube. If I consider a small cubical volume element then clearly the pressure on the left face of the cube is different from the right face of the cube but again since the volume element is in vertical equilibrium the variation of pressure with depth formula holds, so it means if the pressure on the open surface be P then the pressure at the two faces of the volume element are both (P + rdg) (where r is density, d is depth of that end and g is gravitational acceleration) implying they both have the same pressure.

Where is it that I am making mistake? Should I be considering the stress tensor instead of pressure?

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  • $\begingroup$ The equation you wrote applies to the case of a free surface (as you said), and your conclusion is correct if the free surface is horizontal. But, if the free surface is not horizontal, then the pressures on the right and left faces of the cuboid would not be equal, and there would be flow from the higher pressure side to the lower pressure side. $\endgroup$ Apr 9, 2019 at 19:42
  • $\begingroup$ But in the cuboidal tube of flow I consider, the free surface is still horizontal but yet there is pressure differences at the faces of some volume element, yet the vertical equilibrium condition leads to the contradiction that the pressure is same at the two faces of the volume element. What is going on? $\endgroup$
    – Metric
    Apr 9, 2019 at 20:45
  • $\begingroup$ If the free surface is horizontal, then there is no pressure difference between the faces of your volume element. So, you have no flow. $\endgroup$ Apr 9, 2019 at 20:47

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I consider a small cubical volume element then clearly the pressure on the left face of the cube is different from the right face of the cube...

I think that in the idealized case that you're presenting including a fluid which is an ideal liquid (i.e., an incompressible fluid with no viscosity) that the pressure on the left face of the cube should be equal to that on the right side of the cube. You can think of this in terms of Newton's F=ma. If a cubical volume element of fluid is just coasting along and flowing at constant speed with no acceleration, then the net force acting on it should be zero. So the pressures on both the left and right sides of the cube should be the same.

As for a stress tensor, I don't see why that would be needed here since the fluid is an ideal liquid with no viscosity. There will be no off-diagonal shear components to any stress tensor in the fluid.

(P.S.: Of course any real liquid will have some viscosity and so the pressures on both the left and right sides will not be exactly equal. Some pressure gradient will be required to keep a real fluid flowing at constant speed through a pipe. But, again, your question is for the limit of an ideal liquid.)

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  • $\begingroup$ But in streamlined flow or laminar flow of ideal fluid( which is perhaps the simplest possible flow) the velocity of the fluid changes from point to point,so clearly something must be causing that velocity change. In streamlined flow we clearly have a glow or velocity function U(X) so there must be some force causing it? Right? $\endgroup$
    – Metric
    Apr 9, 2019 at 20:43
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    $\begingroup$ Why do you think the velocity changes? With incompressible flow of constant area, continuity (I.e. conservation) requires constant flow. $\endgroup$ Apr 9, 2019 at 23:50
  • $\begingroup$ Oh damn you are right... The continuity equation means velocity doesn't change.. thanks a lot $\endgroup$
    – Metric
    Apr 10, 2019 at 3:24

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