1. Mar 4, 2015

### FeynmanFtw

Now I've read a few explanations online already in the hopes of getting to grips with this phenomenon, but it seems they all bang on about the same thing (higher velocity in the narrow portion due to the principle of continuity), though none really explain WHY pressure drops in the narrow portion (i..e where the velocity of the fluid flow increases).

Could anybody shed further light on this, preferably in terms of molecules? Why would a greater number of velocity components in a direction parallel to the tube walls mean less/weaker collisions with the tube walls?

edit: perp corrected to parallel.

Last edited: Mar 4, 2015
2. Mar 4, 2015

### Staff: Mentor

For an intuitive model, consider that if the temperature of the gas doesn't change as it passes through the venturi, then the average speed of the molecules will remain the same. The speed is the magnitude of the velocity vector, so if the speed of the average molecule is the same while the component of the velocity in the parallel direction increases on average, then the perpendicular components of the velocity must go down on average. That means fewer and less energetic collisions of the molecules against the walls, and hence less pressure on the walls.

3. Mar 4, 2015

### FeynmanFtw

This has already helped quite a bit, and is something I was beginning to assume before I posted this thread, so thank you! However it begs the question: why do the perpendicular components go down? What causes them to be reduced?

4. Mar 4, 2015

### cjl

While intuitive, this explanation has the unfortunate downside of being wrong.

Temperature is based on the random component of the velocities of the molecules - bulk motion is not included by definition.

The real answer is that the gas must be flowing faster because of the lower cross sectional area. Assuming no density change, the only way to maintain the same volumetric flow rate is to increase the flow speed. Now consider the perspective of a gas element flowing through this venturi. As it approaches the restriction, it must accelerate, which requires a force. The only way that a force could be accelerating it though is if the pressure is higher behind it and lower ahead. Thus, if a flow accelerates (without a pump or fan or something adding energy to the flow), the pressure must be decreasing, since the flow must be going from a region of higher pressure to a region of lower pressure. Similarly, once it gets past the restriction, it is decelerating, and thus the pressure ahead must be higher and the pressure behind lower.

5. Mar 4, 2015

### FeynmanFtw

Yes, but this is just a causal interpretation. i.e. Fluid flows faster in a narrower section therefore the pressure in this region must be lower.

I've seen the explanations regarding pressure differences, forces and whatnot, but never an explanation when looking at the particles themselves.

6. Mar 4, 2015

### 256bits

Well, I guess you should be familiar with the kinetic theory on random molecular movement and how static pressure arises from the exchange of momentum of the molecules with the sides of a container from these random movements.. And also the Bernouilli equation for fluid flow, which is just an energy balance of the fluid from one location to the next. Bernouilli divides the energy of the fluid into components: a kinematic pressure component ( from the bulk velocity of the fluid ), a static pressure component ( from the random molecular motions ), and a gravity component ( from the in elevation ie mgh ).

Both explanations given have some element that satisfies your question, but not totally.

Cjt, though is explaining how an acceleration of the fluid will change the kinematic portion of Bernouiili, ( but not the static pressure component ). So if the kinetic pressure goes up, the static pressure has to go down, to keep energy balances in order.

Nugatory seems to refer to the bulk velocity, broken into its with parallel and perpendicular parts, but here, the bulk perpendicular component is normally considered to be zero.

Take a pipe with a venturi and label the z-axis as allong its length and the y-axis along it breadth is a 2-d representation.

With NO flow, you will have to agree that the static pressure in the larger section and in the constricted section are the same. In the larger section of pipe, take a small length L along the z-axis comprising N number of molecules in this enclosed volume V. With random MOLECULAR velocity Vmy striking the walls in a time t ( simplified kinetic theory ), a static pressure P will arise from these collisions. In the constricted section, the same length L has fewer molecules in this volume ( the volume is smaller), but the molecules can strike the wall more times ( in the y-direction, the distance from wall to wall has decreased ) , so the same pressure P.

Of course there is molecular motion in the z- and the x- directions, but these motions do not contribute to the pressure in the y-direction, which is what we are interested in. Did I say it was a simplified version of kinetic molecular theory of pressure?

Once the fluid is flowing, the random molecular motion should be the same - right assumption? But on top of this we have the bulk velocity Vz of the fluid in the z-direction.
Here we can do a similar analysis as above, except we have to take into acount the distance d the fluid has moved in time t, where d is just Vz/t.

In the pipe: d1 = Vz(pipe)/t
In the venturi section: d2 = Vz(venturi)/t

Of course Vz(venturi) > Vz(pipe), which means d2>d1.
So while the section of length L of fluid in the pipe has moved forward a distance d1, in the venturi section the same length has moved forward a greater distance d2. Now, static pressure should decrease in the venturi since the swept area for the molecules to strike the walls has increased in time t, in some inverse relation to the ration d1/d2, going from the pipe section to the venturi.

A somewhat convoluted explanation, I do agree, but I hope this picture helps you out.

This should make sense.

7. Mar 5, 2015

### A.T.

On particle level it's always about probabilities . The particles, that have more momentum along the flow direction than perpendicular to it, are more likely to go through the narrow part.

Last edited: Mar 5, 2015
8. Mar 5, 2015

### Delta²

Suppose we go back in time in 1750s and we want to explain venturi effect. We know nothing about the microscopic model of matter about molecules, atoms and such. Still we can explain it using conservation of mass and energy. Conservation of mass explains the continuity equation and conservation of energy explains the bernoulli principle.

I think at the core of our intuitive understanding today (and where i believe the problem is lying) is the modeling of pressure in the walls of pipe as a result of a large number of collisions of the air molecules with the pipe's walls molecules. Back at 1750s pressure was modeled as a force that the air was exerting on the walls of the pipe, nothing about velocities of molecules and collisions. Sometimes times knowing less makes things more simple to explain :)

9. Mar 8, 2015

### FeynmanFtw

I believe you've nailed it tbh. So am I to understand (assuming I've got this), that the particle velocities causing static pressure haven't really changed, it's that if we look at the simple idea of pressure = force/area, in some time t, as you've stated, within the venturi pipe the effective area for the particles to collide with has increased.

To illustrate what I'm saying, imagine we have a ping pong ball bouncing of the bottom side (let's just hypothetically examine this portion), and moving slightly from left to right. In some time t within the larger section of the pipe, it will collide n times with some small length Z of the pipe. Now we go to the venturi section. The up/down velocity hasn't changed (hence the number of times it strikes the bottom wall hasn't changed, or it might increase a touch...not 100% sure about this bit but regardless), but now the length Z has increased several fold, which reducing the effective number of collisions per unit area and hence static pressure.

Am I close?

10. Mar 12, 2015

### 256bits

Sorry for the late reply., Feynman.
I still am debating with myself as to whether that is a fully qualified explanation for the decrease in pressure, as it really does not have is a complete enough rigor . It is though similar to what Bernouilli had I mind himself from my understanding.

What we do know is that, if one believes in atoms and molecules and the kinetic theory for pressure, the momentum exchange perpendicular to the direction of flow due to their random motions has to change when the fluid is moving.

Your questions is a good one.

11. Mar 15, 2015

### FeynmanFtw

Thank you.

I've always tried to attack problems from several different angles as in my opinion, this helps bolster your true understanding of a phenomenon at work. If ever one has read any books by Feynman for example, they'll get what I mean.

As for your approach, it makes the most sense so far. I've tried to fault it (as any sceptic would), but haven't found a solid flaw yet. So far, so good!

12. Mar 16, 2015

### cjl

That's not necessarily true though. Thermal motion is considered distinct from bulk motion, and temperature is defined by the average kinetic energy of the molecules after the bulk motion is subtracted from each molecule's individual velocity. As a result, this explanation is really insufficient, even though at the surface it seems to make sense.