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I have an elementary question on Bernoulli's law. Consider the pipe as depicted below.

Flow through a widening pipe

At point $1$, the pipe has the cross-section $A_1$ and the incompressible, non-viscous fluid is flowing with the speed $v_1$. The pipe lies horizontally and we assume no external pressure, so we have only dynamic pressure:

$$ p = \frac{1}{2} \rho v_1^2 $$

At point $2$, the pipe is wider, having the cross-section $A_2$, so the fluid needs to slow down in order to keep the flow constant:

$$ v_2 = \frac{A_1}{A_2} v_1 $$

According to the Bernoulli's law, this must lead to the increase of the static pressure inside the pipe, in order to keep the total pressure, static+dynamic, constant:

$$ p = \frac{1}{2} \rho v_1^2 = p_2 + \frac{1}{2} \rho v_2^2 $$

Now, to my understanding of the Pascal's law, $p_2$ must act in all directions, right? If yes, what happens to the pressure $p_2$ at point $3$, once the fluid leaves the pipe? I understand that it carries potential energy, and energy must be conserved. So does the pressure force the fluid to spread sidewards and up/downwards, adding a transversal velocity component to the fluid's flow? Does the fluid stream disintegrate into drops? Anything else?

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If the pressure at $3$ and $1$ are both atmospheric, the fluid is not going to move. If you want the fluid to move the pressure at $1$ will have to be larger than atmospheric.

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  • $\begingroup$ Would it be unrealistic to just assume that the fluid is flowing at point $1$, no matter why? Perhaps it came initially from a reservoir positioned above, so that hydrostatic pressure due to gravity caused it to move, but all that pressure has been converted into the dynamic pressure. $\endgroup$
    – Igor F.
    Commented Jun 24 at 8:46
  • $\begingroup$ As your Bernoulli calculation shows ,the fluid at 1 and 3 must have different pressures. If 3 is open to the air, the pressure at 3 is 1atm. (Bernoulli assumes no friction, incompressible, and time-independent $v$) $\endgroup$
    – mike stone
    Commented Jun 24 at 8:50
  • $\begingroup$ The assumption of time independent $v$ shopws what the problem is: If you assume that the fluid is intially moving, than, with the equal pressures it will be slowing down in time. Then you will get a consistent Bernoulli after you add the $\partial \phi/\partial t$ term that the time-dependent Bernoulli requires. $\endgroup$
    – mike stone
    Commented Jun 24 at 9:03
  • $\begingroup$ @IgorF. the driving force for fluid flow is a pressure difference. Therefore, it is indeed unrealistic to assume that there is no pressure difference between points 1 and 3. $\endgroup$
    – David White
    Commented Jun 24 at 16:43

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