aerodynamicsfluid dynamics
In a nozzle, the exit velocity increases as per continuity equation $Av=const$
as given by Bernoulli equation (incompressible fluid). Pressure is inversely proportional to velocity, so we have lower pressure at the exit of the nozzle. But as per definition of pressure, $P=F/A$, i.e., pressure is inversely proportional to the area which contradicts the above explanation on basis of continuity and Bernoulli equation.
Which is true?
What relation is true for compressible flow?
It is an empirically observable fact that subsonic jets (of which your water spout is an example) do in fact exit into a quiescent medium at the pressure of that very medium. The water will leave the nozzle at precisely the ambient pressure if the exit Mach number is less than 1.
What is the relation between pressure and velocity of a fluid in a closed pipe flow?
Bernoulli's equation:
By continuity equation velocity at all points is the same. Then shouldn't the pressure be same at both points?
No the pressure won't be the same at all points in the pipe. Considering Pascal's Law, a change in pressure at any point in an enclosed fluid at rest is transmitted undiminished to all points in the fluid. But it holds for static fluids essentially. And equation of continuity stems from the principle of conservation of mass but in this case the mass is not conserved per unit time. Consider 2 different cross-sections of the pipe and fluid flow through it per unit time is not the same since some component of $g$ along with some resultant force is acting on the water column at the 2 different cross-sections and the 2 different cross-sections have a distinct height difference.So equation of continuity is not VALID in this case and hence conclusions cannot be made from this.
Then how to find pressure difference between two points having height difference between them?
Use Bernoulli's principle and you get the relation $$P_2-P_1=(h_2-h_1)\rho g$$
The pipe is say tilted at an angle.
Use the components of pressure in that direction and use the above equation.
Best Answer
I disagree with the most voted answer, by CAGT. He says "This area is completely different to the one above", but this means nothing. The equation $p = {F \over A}$ mentioned by the author does hold, and there is no contradiction or paradox in it.
In fact, the equation $p = {F \over A}$ holds not only here but anywhere else in physics. You may write it in any situation, and it will always be true.
Let's begin with a small correction. Your $Av = \text{constant}$ equation is not Bernoulli, but mere conservation of mass. Here's Bernoulli. This is what gives, in your words, "pressure is inversely proportional to velocity." $${p \over \rho} + {v^2 \over 2} + gz = \text{constant}$$
So your problem is with $p = {F \over A}$. Well, there's no problem with it. What is really wrong with your thinking is that you're not paying attention to the equation: the force $F$ changes too.
Let's recap what happens in your situation:
Ok, now look at this.
The dark blue rectangle on the left is what we call an element. Like the rest of the flow in the bigger section, it flows with velocity $v_1$. It is delimited left and right by faces with area $A_1$. Note that, since the liquid left and right of it has pressure $p_1$, this element is compressed by forces $F_1 = p_1 A_1$ on each side.
Now to the element on the smaller section, which flows faster. Its cross-sectional area is smaller. The pressure left and right of it is also smaller. As a result, the forces compressing it, $F_2 = p_2 A_2$, are also smaller.
So, $p = {F \over A}$ still holds. Yes, when the situation changes, $A$ is smaller, which by itself would make $p$ bigger. However, as we saw above, then new $F$ is smaller than the old one too, which by itself would make $p$ smaller. The net effect of $p_2 < p_1$ (which we know beforehand from Bernoulli) means, therefore, simply that $F$ has diminished more than $A$ did.