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Hereinafter, the term “fluid” will be used to mean “ideal fluid” (i.e., an incompressible fluid without friction).

The fluid flows to the right through a horizontal pipe at a speed $v$. Then Bernoulli's equation will look like this: $$P_1+\frac{\rho v_1^2}{2}=P_2+\frac{\rho v_2^2}{2}$$

Let us select a portion of fluid A. Pressure forces ($F_1$ and $F_2$) act on part A from both sides of the fluid. According to Pascal's law, the action of external forces is transmitted equally in all directions.

  1. Forces F1 and F2 act on part A. How do you calculate the pressure inside part A? If two opposing forces F1 and F2 act on part A of the liquid, is P1 calculated as: $$\frac{F_1 + F_2}{S_1}$$ Is that true or not?

  2. Bernoulli's equation uses values P1 and P2, since these values are unique for a given section of fluid. It can be concluded that the pressure inside these sections (S1 and S2) is the same in all directions. That is, in section S1 there is pressure P1, which is the same in all directions. But how can there be a single pressure value at all points of a given section of fluid if the pressure of the fluid also depends on the depth of immersion?

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  • $\begingroup$ You say the fluid is moving at speed v. But the equation has two speeds, v1 and v2. So are you saying it’s moving at v, everywhere. Or are you say it’s moving at v1 at one location and at v2 (different value) at another location? Asking because sounds like you are saying P1 does not equal P2 so v1 does not equal v2, but that would contradict there being a single value of v everywhere. So I’m a little confused about the set up of the problem being asked about. $\endgroup$ Commented Nov 7 at 16:52
  • $\begingroup$ @kangermu, You're right, I did forget to explain why I wrote Bernoulli's equation. This equation uses the value of P1 in a given section (i.e., at a given speed). The question is, what is P1 in the equation if it turns out that the pressure in the same section is not the same at all points? I wanted to show that I don't understand how to find this P1 if the pressure is different at different points. I hope I was able to explain my idea. $\endgroup$ Commented Nov 7 at 17:30
  • $\begingroup$ "This is a contradiction..." No it is not. There is no contradiction, you are just confused about something. The pressure is the same everywhere since the velocity is the same everywhere. This is what the one equation you wrote already says. $\endgroup$ Commented Nov 7 at 17:33
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    $\begingroup$ You are already demonstrating that you do not understand Newton's 2nd Law when you assert that rightward moving implies that $F_1>F_2$ because if $v=$ const then there is no such inequality. $\endgroup$ Commented Nov 7 at 17:45
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    $\begingroup$ @Marmajuck, Pascal's law applies to static fluids. The Bernoulli equation applies to dynamic fluids. Is this the first time that you have studied the Bernoulli equation? $\endgroup$ Commented Nov 9 at 1:26

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In your "ideal" fluid with no friction, viscosity and compressibility, Bernoulli's equation describes the preservation of kinetic plus potential energy of the streaming fluid. In such a fluid, streaming in a horizontal tube of constant cross section, you don't need a higher pressure to the left of a volume to flow to the right, or vice versa. There is no force necessary for this flow to flow, it's just Newton's first law. (This in contrast to the viscous flow through a constant cross section pipe!) Also, the static pressure has no "direction".

Note added after last changes by OP(13 h ago):

You assume a constant flow to the right in a horizontal tube with changing cross section.

Point 1: You assume that two different opposing forces, $F_1$ and $F_2$, act on element A. This means that there is a resulting force $F_1-F_2$ on the element, so that the element would be accelerated in the constant cross section tube part. This contradicts the assumption of a constant velocity $v_1$ flow there, which requires that the forces $F_1$ and $F_2$ must cancel each other according to the constant pressure $p_1$ in this section. Your formula for $p_1$ is not correct!

Point 2: If with "depth of immersion" you mean the hydrostatic pressure change over the height of the horizontal tube, this can usually be neglected, but could in principle be accounted for in a more sophisticated calculation.

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  • $\begingroup$ Regarding point 1: how can we determine pressure P1 if we know forces F1 and F2 (F1 = F2)? I assumed that P1 = $$\frac{F_1 + F_2}{S_1}$$, since according to Pascal's law, pressure transmitted by external forces is transmitted equally in all directions, and forces F1 and F2 act on part A. $\endgroup$ Commented Nov 8 at 20:20
  • $\begingroup$ @Marmajuck - In the constant velocity flow in this unchanging cross section part of the tube you have everywhere the same pressure $p_1$. Thus $$F_1=S_1 p_1=F_2 =S_1 p_1$$ holds. This means that the sum of the two opposing forces on the element A is zero, as it should, because otherwise, according to Newton 2, the element would not have a constant velocity, but would be accelerated. $\endgroup$ Commented Nov 9 at 1:04
  • $\begingroup$ I don't understand why pressure (p1) is determined only by the values of forces F1 and F2 separately. If we press on the liquid from both sides with two pistons with forces F1 and F2 (F1 = F2, they are opposite in direction), the pressure will be the same as if we acted on the liquid only with piston F1 and replaced the second with a fixed wall. Why is this the case? $\endgroup$ Commented Nov 9 at 10:26
  • $\begingroup$ @Marmajuck - The situation with two pistons with different forces acting on fluid element A is incompatible with a $v_1=const$ flow to the right. If you have only element A with two pistons with different forces acting on it on each side, there would be a resultant force $F=F_1-F_2$ on the element which accelerates it in the direction of $F$ according to Newtons 2nd law. In the case of no acceleration ($v_1=const, including =0$), $F_1$ has to be equal to $F_2$ resulting in equal pressures $p_1=F_1/S_1=F_2/S_1$ at both pistons. In this case, there can only be one pressure $p_1$ in A. $\endgroup$ Commented Nov 9 at 15:24

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