Main content

### Course: Physics archive > Unit 9

Lesson 3: Fluid Dynamics# Venturi effect and Pitot tubes

The Venturi effect demonstrates how fluid speed increases and pressure decreases in constricted areas. This video breaks down Bernoulli's equation, revealing the surprising principle that faster fluid equals lower pressure. It also introduces the Pitot tube, a practical tool for measuring fluid velocity.

## Want to join the conversation?

- The Venturi effect and our circulatory/venous system seem to directly defy each other.

When we need to increase our blood pressure our veins constrict, and when we want to decrease blood pressure our veins dilate. This is completely opposite of what the Venturi effect is stating. Can someone shed some light on this please?(39 votes)- Blood flow in our bodies are turbulent, unsteady (pumping nature of heart) and does not agree with assumptions we have adopted to derive Bernoulli's laws. Hence this behavior. Please correct me if I am wrong.(26 votes)

- At8:30on the top of tube where hole is at right angle, doen't some air flow in by turbulence or something?(13 votes)
- I think the previous answer may be misleading: volume is not a conserved quantity because air is compressible. It is unnecessary for the air to leave at the same rate. The pressure in the tube will rise when turbulent flows enter from the side. These flows, however, are typically very minor because pitot tube is used in laminar flow regimes. In turbulent flows, a correction factor is typically used, as well as signal processing and time averaging of measurements. This works because turbulent flow regimes are oscillatory in nature, and averaging over time will yield net pressure. The correction factors account for the fact that the particle velocity in fluid flow regimes may be faster than the free stream velocity, as they move along curved paths.(9 votes)

- how can you say that air inside the pitot tube(Stagnation chamber ) is now moving.the whole PITOT table has some velocity with respect to ground...........7:58(2 votes)
- and the earth is spinning on an axis, which is orbiting around the sun, which is moving around in the galaxy.... etc.

You pick a frame of reference and set it to zero. So your point of observation is from the airplane, and you are just observing the speed of the air with respect to the airplane. (since the flow of air over the wing is really what is important anyways). so the air in the pitot tube is not moving with respect to the plane(20 votes)

- Why in capillaries do you have much lower pressure and lower velocity as opposed to arteries where you have much higher pressure and higher velocity.(6 votes)
- Just a clarification then. For Venturi's Effect, there will be a decreased pressure at a location of smaller radius, but the volume won't change, because fluids are incompressible.

I'm guessing this would be different for gases though, because then Boyle's Law would come into effect P1V1 = P2V2?(5 votes)- they are two separate sets of pressures

1. in Venturi's effect (and formulas for fluid in general)

pressures are acting upon the fluid from outside (e.g. from the atmosphere left and right side of the fluid)

2. in Boyle's law (and formulas for gas in general)

pressures are acting by the gas to the surface around it (e.g. the walls of a container or cylinder)

3. actual cases

1) for the first case, volume won't change as you said even though the pressure from outside to the fluid is decreasing (cause they are not "directly" related)

2) for the second case, when volume is constant, pressure is too. no change. and when volume of gas increases, pressure by the molecules inside gas toward the walls decreases. (cause the same number of them should push away a larger area with same force)

in short, two pressures are different. not the fundamental natures of two concepts (fluid and gas) are(0 votes)

- doesn't air from higher pressure flow toward lower pressure regions? Then, how can there be stagnant air at all?(4 votes)
- Can someone solve this question? :

A pitot tube is mounted on an airplane to measure the speed of the plane. The tube contains alcohol and shows a level difference of 40 cm. What is the speed of the plane relative to the air? Given that specific gravity of alcohol=0.8, density of air=1kg/m^3, g=10m/s^2.(3 votes)- So if we start with the equation P,s = P,1 + 1/2qv^2 where q = roe we can rearange it to sqrt((2*P,s-P,1)/q) then the pressure difference is equal to the change in height of the alochol times its density times gravity. Basically deltaP = 800 kg/m^3 * 10 m/s^2 * 0.4m this is equal to 3200 pascals. We can subsitute this for P,s-P,1 in the firsty equation getting sqrt((2*3200)/airdesnity) = v this resolves as sqrt(2*3200/1000) = 80 m/s(0 votes)

- Is there a reason that the Pitot tube would not be mounted to the object (for which it was to determine velocity) with the cross section in the video's diagram in the horizontal dimension? David said that the height difference from end to end is negligible, but why not minimize it?(1 vote)
- Good question. You're on the right track - if you want to, you can rotate the tube about its axis all you want. But remember, we want to minimize our impact on the free stream flow, right? So however you're mounting this thing, you should have the second chamber vent to the opposite side. That's more important than the microscopic difference in ambient pressure over an inch or so.(3 votes)

- Where is the Venturi effect mimicked in the body? It makes sense to me that velocity and pressure are inversely related, but what I don't understand is how when the diameter is smaller, can there be less pressure? What is the relationship between the diameter of a blood vessel and its pressure?(2 votes)
- @calebaamy I understand what you are saying up until the entire system being under higher pressure, why is that?(1 vote)

- How's the Bernouli equation used in case of air? I mean air is a just a mixture of gases and gases are highly compressible, and Bernouli's equation is applied for incompressible fluids.(1 vote)
- Well, I would say, use the equation to work out the pressures, velocities etc AS IF it were in-compressible, and then use your answer as an approximation. Chances are, if you were working out these values n making a new machine or process you would need to do experiments to find the real-life values anyway...(2 votes)

## Video transcript

- Let's talk about the Venturi effect. This has to do with water or any fluid flowing through a pipe. And it turns out, let's say
this water's flowing right here. Minding its own business, having
a good day for that matter, when it meets a constriction. What's gonna happen here? Well, the water's gotta keep flowing, but it's gonna start flowing faster through the constricted region. And the reason is, well,
there's a certain amount of fluid that's flowing through this pipe. Let's say all the fluid
in this region right here. Let's say this front part of the water. I mean, this whole thing's filled up, but just say this
cross-section of the water happened to make it from this back portion to this front portion in, I don't know, let's just say one second. So this entire volume
moved through this section of the pipe in one second. Well, there's a law in physics
that says that same volume's gotta make it through
each portion of this pipe. Because if it didn't, where's it gonna go? This pipe would have
to break or something. This water's gotta go somewhere. If that much flowed
through here in one second, then this much has to flow through this little tiny region in one second, but the only way that
that's possible is for this front surface, instead
of just traveling from there to there in one
second, the front surface is gonna have to change it's shape. But the front part of
the water's gonna have to travel from here to here
maybe in 1/4 of a second because all of this has
gotta cram through here in the same amount of time. Because that water's
still coming behind it. There's more water coming. And the volume flow rate
has got to stay the same. The volume per time
flowing through one region of the pipe has got to be the same as the volume flow rate through
some other region of the pipe because this water's got to go somewhere. It doesn't just disappear in here. It's gotta keep flowing. That means... The important part is
the water flows faster through the constricted region. Sometimes much faster through
the constricted region. The smaller this is compared
to this original radius, the faster the fluid
will flow through here. Why do we care? Well, because faster moving
fluid also means lower pressure. Why does faster moving
fluid mean lower pressure? Well, if we look at
the Bernoulli equation, Bernoulli's equation says
P one plus row g h one plus 1/2 row v one squared equals P two plus row g h two plus 1/2 row v two squared. Oh my goodness this looks frightening, but look at P one, we just
pick some point in the pipe. Let's just pick this point right here. We'll call that point one. So all these ones, this whole
side refers to that point. Let's just pick point two right here. All this whole side refers to that point. Now, notice something. These are basically the same height, and assume height's not
really a big difference here. So let's cross out the heights, because they're the same heights. We don't have to worry about that. This says that, alright, if
there's some pressure at one and some velocity of the water at one, you can plug those in
here and get this side. And now look at over here. We know that the velocity
at two is bigger. We just said that, it
has to be because the volume flow rate's got to stay the same. So this speeds up in here. So this is bigger, this quantity here. But we know the whole
thing equals this side. So if this term increased,
that means that the pressure's got to decrease so that when they add up they get the same as this side over here. This is actually called
Bernoulli's Principle. Bernoulli's Principle
says that when a fluid speeds up, it's pressure goes down. It's totally counter-intuitive. We always expect the opposite. We think fast moving fluid, that's gotta have a lot of pressure, but
it's the exact opposite. Fast moving fluid actually
has a smaller pressure and it's due to Bernoulli's equation. And this is what causes
the Venturi effect. The Venturi effect refers
to the fact that if you have a tube and you want
a smaller pressure region, you want the pressure
to drop for some reason, which actually comes up in a lot of cases, just cause a narrow
constriction in that tube. In this narrow constriction,
faster moving fluid, and it'll cause a lower pressure. This is the idea behind
the Venturi effect. So the Venturi effect basically says for a constriction in a pipe, you're
gonna get a lower pressure. While we're talking about fluid flow, we should talk about one more thing. Let me get rid of this here. Imagine you just had a brick wall with fluid flowing towards it. Maybe it's air here. So you've got some fluid
flowing towards this brick wall. This seems like a really dumb example of Bernoulli's principle
but I'm going somewhere with this so stay with me. This is flowing towards here. What's going to happen? Well, it can't go through the wall. It's gotta go somewhere. Maybe this just goes up like that and this, you know, I'm gonna go this way. It's closer to go that way. This side maybe just goes down. This is actually kind of what happens. But there'll be a portion in the middle that basically just terminates. It hits here and kind of just gets stuck. So there'll be some air
right near here in the middle where it's just not moving. What if we wanted to
know what the pressure was there, based on the variables
involved in this problem? We could use Bernoulli's equation again. Pick two points, let's
pick this one, point one. Let's pick this one, point two. Use Bernoulli's equation, it says this, and again let's say these
are basically the same height so that height is not a big factor. And if these terms are the same, then we can just cross
them out because we can subtract them from both
sides, they're identical. Now, what can we say? We know the velocity of the air at two. It's not moving, got stuck here. It got stagnant. And so v two is just zero. And we get this statement
that the pressure at two, which is sometimes
called the stagnation pressure, so I'm gonna call it
the stagnation pressure, because the air right here
gets stuck and it's not moving. You might object, you
might say, "Wait, hold on. "I thought the air had to go somewhere?" Well, it's all going somewhere. The point is, there's some air
right here that gets stuck. It get stuck and air starts passing it by. And so, what's this pressure here? Well, up here we just read it off. All these went away. P two, which is what I'm
calling the stagnation pressure, has gotta equal P one,
the pressure over here, plus 1/2 row v one squared
and we get this formula. You might think, "Why
would we care about this? "Who is regularly shooting
air at a brick wall?" People do it all the time,
because you can build a pretty important instrument
with this called a Pitot tube. And the Pitot tube looks
something like this. Let's get rid of that. So why would someone use this system? It's called a Pitot tube. People use it to measure fluid velocity or, if you're moving through the fluid, it's a way to measure your
velocity or your speed. So what happens is you've got this set up. Let's say you're in an airplane. You mount this on the airplane. You're flying through the
fluid, which is the air. So that mean air is rushing
towards this section here. Rushing past you, let's say
you're flying to the left. So you'll notice air flying past you. A Pitot tube always has
this section that's facing into the wind or into the air. This air would be directed
straight toward this region, and the key is this is
blocked off at the end. So there's air in here,
but it can't be moving. The air in this section
can't be moving all the way to the front because, I
mean, where's it gonna go? We just said if fluid flows
it, fluid's gotta flow out. There's no out here. And then there's another region. Up here you've got a second chamber where the air flows over the top. And this is directed at a
right angle to that air flow. You've got another chamber. And again, in here, fluid's not flowing. The key is this gives you a
way to determine the difference between the pressure here
and the pressure there. If you had some sort of membrane in here, something dividing these
two sections that could tell you the pressure differential, right? If the pressure on this
side is a little bigger than the pressure on this side,
and this would bow outward, one of these is measuring
the pressure here and one of them is measuring
the pressure there. What is the... Mathematically, what's the relationship? It's the one we just found. Right here, this is the
stagnation pressure, right? The air's not moving in
here, it flowed straight in. We know the v is zero right here. And so the stagnation pressure
equals the pressure up here. Again, I'm assuming there's
very little height difference. Let's say this is a very small device and it's not like 10 meters tall. So any height differences are minuscule, and we would just have
our same equation before. This would just equal the
pressure plus 1/2 row v squared. And this is how you determine the velocity or the speed, because now we
can just solve this for v. I'd get that v one equals P
s, the stagnation pressure, minus the pressure at
one, that whole thing, times two, divided by
the density of the air and then a square root because you have to solve for the v one. So this device lets you determine this pressure differential right here, check. You need to know what
the density of air is and this gives you a way to determine the velocity of the fluid, or in other words, the velocity of your aircraft
flying through the air.