12
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ Why do you say that pressure is inversely proportional to velocity ? $\endgroup$ – Tom-Tom Jan 28 '14 at 10:44
16
$\begingroup$

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:

  1. There's a change in cross-sectional area: $A_2 < A_1$
  2. Thanks to conservation of mass, (1) implies $v_2 > v_1$
  3. Thanks to Bernoulli, (2) implies $p_2 < p_1$

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.

$\endgroup$
  • 1
    $\begingroup$ Could you explain why $F_2 < F_1$? $\endgroup$ – Domi Apr 3 '15 at 9:36
  • $\begingroup$ Because $F = p \cdot A$ and both $p$ and $A$ are lower, as I showed in my answer. Try rereading the answer so you can come up with a more specific question. $\endgroup$ – André Chalella Apr 4 '15 at 0:53
  • $\begingroup$ Good explanation, but pay attention that you only consider the advection and not the diffusion effect. If you solve the Navier-Stokes equations for a fully-developed viscous flow in a straight channel, you will see that the pressure remains constant in y-direction (i.e. from bottom to top side of the channel) while the velocity changes along this direction, which can be understood by considering the diffusion terms. Therefore, we cannot say that the pressure is always inversely proportional to the velocity. $\endgroup$ – Mimi Jul 8 '15 at 9:40
  • $\begingroup$ Very nice explanation. Btw, I have one query : according to the definition of force, force is rate of change of momentum i.e F = mass flow rate × velocity. Here mass flow rate is constant (conservation of mass) and velocity is increasing, so force should increase! $\endgroup$ – Pandya Sep 23 '18 at 16:23
  • $\begingroup$ Ok. I found some explanation from this answer. $\endgroup$ – Pandya Sep 23 '18 at 17:31
5
$\begingroup$

The formula

$$Av=const$$

comes from

$$\rho_1A_1V_1=\rho_2A_2V_2 $$ (at reasonably low speeds, where fluid density can be assumed constant, $\rho_1=\rho_2$)

So you end up with:

$$A_1V_1=A_2V_2=constant $$ The units of this is the mass flow through area A per unit time (ie $kg/s$).

The formula
$$P=\frac{F}{A}$$

Is Pressure of a perpendicular Force applied on the Area of a surface. This Area is completely different to the one above, as there is no mass flowing through this one.

$\endgroup$
  • $\begingroup$ Can you clarify or explain the different areas? $\endgroup$ – Pandya Sep 23 '18 at 16:33
-2
$\begingroup$

AV = constant is could only in ideal condition for continuity flow. Real condition is to concern with losses due to internal surface of pipe roughness, pipe bend, sudden enlargement and contraction. A coefficient Cd values from 0.96 to 0.98 must be considered. Internal surface friction creates pressure drop. Pi = F/A is not constant by this reason.
Flow rate Q can be in the form of Q = cd.AV, pipe entrance ( developing region), pipe join, pipe bend and other physical conditions might cause significant losses which in turn affecting continuity in term of velocity. Flow rate indeed is constant, but somehow velocity may be reduced. Flow regime can also have influences on the case of flow velocity. One can rely on Moody Diagram to look into this. Where friction is defined perfectly.

$\endgroup$
  • $\begingroup$ -1. This has completely, utterly, nothing to do with the question. $\endgroup$ – André Chalella Sep 13 '14 at 10:13
  • $\begingroup$ And mass will always have to be conserved. What you are talking about is the Bernoulli equation with additional head losses, however this just increases the pressure losses, which in turn lowers the mean velocity/mass flow (of the entire system). This is similar to adding extra resistors into an electrical system, the amount of current coming is will still be equal to the amount of current coming out, but both will decrease if the overall resistance has increased. $\endgroup$ – fibonatic Sep 13 '14 at 15:26
-2
$\begingroup$

You have to think according to the situation. Consider a tank with high pressure, if we open the tap means water flows with high velocity and eventually after sometime the velocity will start to reduce because of pressure is decreasing accordingly. Velocity and pressure is directly proportional.

$\endgroup$
  • $\begingroup$ Proportional seems to suggest a linear relationship. Are you sure it's linear? I'd expect a non-linear relationship because kinetic energy is non-linear with velocity. I haven't worked out the math though. $\endgroup$ – Brandon Enright Aug 30 '14 at 5:20
  • $\begingroup$ -1. What you're saying is completely unrelated to the question. You're relating $v$ to a changing $p$, with no change in $A$ whatsoever. Changing $A$ is what the author is talking about. $\endgroup$ – André Chalella Sep 13 '14 at 10:12
-2
$\begingroup$

consider a cylider full of gas. Thus gas is exerting pressure on walls of cylinder while its velocity is zero. Now open the mouth of cylinder , u will c that gas is coming out with some velocity while the pressure on walls of cylinder is decreasing . same is the case with pipes. When fluid is passing through the pipe having constant area of cross section then that fluid exerts constant amount of pressure on walls of pipes and having constant velocity depending upon what Amount of pressure is applied on fluid from its starting point to make it move. Now consider a nozzle at the end of pipe. This sudden decrease in area will resist the fluid flow.now to make the constant massbflow rate , fluid will increase its speed . Now considering pressure on certain point on wall, it can only get decresed if mass of fluid on that point decreases. In cylinders , if we make the gas out then pressure decreases as I stated already ans in case of pipes , it decreases if every smallest element of fluid get less time time to stay on particular point in pipe . Thus pipe full of staying fluid will experience more pressure than pipe with moving fluid . Thus more is the velocity in pipe , lesser will be the pressure experienced by pipe. You can see this in bernoulli's apparatus. When water moves slowly in pipe , it goes higher in bernoulli's pipe than when it is moving fast . this is what actually happening when water flows through the nozzle. F/A has nothing to do with it . We are talking about pressure in walls of pipe by fluid , not the pressure by fluid on the target to which it strikes. If we talk about that then pressure will surely be high . Because of that high pressure , even iron caj be cut. But,because of high speed in nozzle , fluid cannot get sufficient time to exert force on walls of pipe. and if lesser is the force then less will be the pressure because force is directly proportional to pressure by relation P=F/A.

$\endgroup$
-3
$\begingroup$

Pressure and velocity have an inverse relation an example pumps are used to to increase the velocity of the fluids in case when water flows from a river to a canal the cross sectional area of the flowing water decreases and the velocity of the water increases the question is from where this energy comes the answer is it follows the law of conservation of energy and actually the pressure energy is converted into kinetic energy vice versa when this water again moves into the river its velocity decreases now the K.E is again converted into pressure energy.

$\endgroup$
  • 1
    $\begingroup$ Could you please put some periods, people will suffer reading this. $\endgroup$ – jinawee Dec 31 '14 at 18:31

Not the answer you're looking for? Browse other questions tagged or ask your own question.