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Let there be a cylindrical pipe through which a fluid is flowing.

Between 2 points at the same height: $$p_1+\frac12\rho(v_1)^2 = p_2+\frac12\rho(v_2)^2$$ Thus, $$p_1 - p_2 = \frac12\rho[(v_2)^2 - (v_1)^2] = \frac12\rho[(v_2+v_1)(v_2-v_1)]$$

Now, if I am using the same equation from the frame of an observer moving opposite to the fluid with speed $v_3$ m/s, then $$p_1 - p_2 = \frac12\rho[\{(v_2+v_3)+(v_1+v_3)\}\{(v_2+v_3)-(v_1+v_3)\}]$$ $$p_1 - p_2 = \frac12\rho[(v_2+v_1+2v_3)(v_2-v_1)]$$

This means that the pressures at those points are changing with respect to different observers, but fundamentally, they should remain same, as $P=\frac{dF}{dA}$ always, and neither force nor cross-sectional area changes.

Kindly help solving this contradiction.

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  • $\begingroup$ The Bernoulli equation is derived for and applies to stable flows, where velocity field in space is constant in time. If this is so in the first frame, and the flow is not a uniform field, in the other frame moving towards the fluid, the velocity field is not constant in time, so the simple Bernoulli equation does not apply there. $\endgroup$ Feb 12 at 22:18

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The velocities of fluid in question are relative to the pipe, so they do not depend on how the observer moves (both pipes and fluid velocities undergo the Galilean transformation - not that we are dealing with non-relativistic fluid mechanics here.)

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  • $\begingroup$ Does that mean that if the pipe was moving, it would change the pressure difference between two points ? $\endgroup$ Feb 12 at 15:26
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    $\begingroup$ @wonderingwhy no, it means the opposite. Again, we are dealing with relative velocities - whether the fluid is moving in respect to the pipe, or the pipe is moving in respect to the fluid doesn't matter. $\endgroup$
    – Roger V.
    Feb 12 at 15:51
  • $\begingroup$ @RogerV. What do you mean by, "it means the opposite"? Assuming that the moving of pipe wouldn't affect the liquid flow due to any kind of drag, etc, what happens to the pressure difference, if the container moves? or, back to the answer, how would the equation get affected by the movement of the container (that would change the fluid's speed relative to the container)? $\endgroup$ Feb 13 at 12:22
  • $\begingroup$ @ShubhamGoel suppose in your reference frame the velocity of the pipe is $u$, whereas the velocity of liquid is $w$. The relative velocity in the Bernoulli equation is $v = w-u$. So switching to another reference frame, $u'=u+U, w'=w+U$ does not change the relative velocity; $w'-u'=w-u=v$. $\endgroup$
    – Roger V.
    Feb 13 at 12:28
  • $\begingroup$ @RogerV. But why are the velocities in the equation relative to the container? $\endgroup$ Feb 14 at 8:49

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