# Where does the pressure go when the fluid leaved the pipe?

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

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?

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.
• 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. Commented Jun 24 at 8:46
• 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 $v$) Commented Jun 24 at 8:50
• 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. Commented Jun 24 at 9:03