# Bernoulli's Principle alternative view

1. Mar 28, 2013

Hi guys,

Could anyone explain Bernoulli's Principle to me so that it makes sense from an alternative point of view?

http://hyperphysics.phy-astr.gsu.edu/hbase/pber.html

I can make sense of the maths but I'm trying to understand what actually happens so I can visualise it.

At first it didn't make sense to me how the pressure could drop. To me it seemed like it would remain the same, halving the volume would mean the mass of fluid travels twice as fast and pressure remains the same.

I think I am starting to understand though, something to do with the force generated by a reduction of half volume produces a velocity of more than double?

As F= MA, the integration of V, V2/2 x ρ
let ρ = 1 and discard
If V = 5, then V2/2 = 12.5
If V = 10, then V2/2 = 50

So doubling the velocity required 4 times the force?
Am I close to getting it?

Cheers.

2. Mar 28, 2013

### Staff: Mentor

What you are missing is that there is more thane one type of pressure and Bernoulli's equation shows how they relate to each other. As your instinct tells you, total pressure must remain constant.

3. Mar 28, 2013

Sorry I was referring to the kinetic energy/pressure.

I'm just struggling to visualise it.

What would happen If the reduction in the pipe was a half, would the velocity be greater than double?
And this is where the pressure drop comes from?

4. Mar 28, 2013

Kinetic energy of the bulk fluid and static pressure are not the same thing. Basically, a given fluid has what is typically called "total pressure", and for an isentropic process, the total pressure is constant. If you want to, go through the dimensional analysis and you can see the units of pressure are equivalent to those of energy per unit volume. So, total pressure is essentially the total energy in a given parcel of fluid. For a steady flow, it is made up of three things: gravitational potential energy (the $\rho g h$ term in Bernoulli's equation), dynamic pressure (analogous to kinetic energy per unit volume; the $\rho V^2 /2$ term) and static pressure (perhaps most analogous to the potential energy in a spring if you consider the bulk fluid). Static pressure is the pressure you feel, for instance, from the atmosphere on you.

Basically then, if you do anything that causes the flow to speed up, you are increasing the kinetic energy of the fluid, meaning that it has to draw from one of the sources of potential energy. If the height isn't changing, this source is the static pressure.

5. Mar 28, 2013

### jim hardy

See if this one is any help.

How can water speed up without a push?
The pressure difference provides that push.
So between the wide and narrow sections of pipe must be a pressure difference in direction to accelerate the fluid. F=MA.

post #4 in above link was my attempt to make it more intuitive by simple mechanics.

See if you can improve it for the next guy ?

old jim

Last edited: Mar 28, 2013
6. Mar 28, 2013

Hmm I don't think I'm explaining myself very well, sorry about that.

As the liquid goes through the pipe, the area reduces and causes a squeeze on the water.. this squeeze tries to increase the pressure in the water, but as the water is free to move it gathers more speed instead. And in this way the pressure remains the same as it was originally. This is how I would think of it intuitively.

7. Mar 28, 2013

### jim hardy

It is very counterintuitive. I think we get imprinted as kids from playing with the garden hose, placing our finger over the end and feeling it push..

Maybe we can link it to something that is more intuitive, like F=MA.

Assume that pipe diameter is considerably larger than 1 cm.
Draw a free body diagram of a 1cm cube of water in the stream at horizontal centerline of pipe.
It has mass of ~1 gram.
As it flows from wide to narrow part of pipe it accelerates from V1 to V2.
It must speed up, else all the volume couldn't squeeze through the smaller area.

Sum forces on all six sides of the cube.
ƩF=MA
Forces up and down in plane of paper cancel out.
Forces into and out of paper cancel out.
So there's no acceleration in either of those two directions.

There IS acceleration from left to right, though.
So, opposing forces on left and right sides of our little cube must add to -MA.
So those two forces cannot be equal.
Since force on any side of the cube F= Pressure X Area,
and the pressures on left and right sides of the cube act on equal area, 1cm^2,
the pressures on left and right MUST be unequal.

So, in the converging part of the pipe is a pressure gradient that accelerates the water.
In diverging part of pipe is a pressure gradient that decelerates the water.

It cannot be otherwise.

can it?

8. Mar 28, 2013

I think what you are missing here is that you can't describe the motion of a fluid with a single equation. Given only the pipe and the fact that the flow is steady and inviscid, Bernoulli's equation can't solely determine the flow properties. You need to account for conservation of mass, so let's start there.

First we will call talk about a flow moving in one direction and the only thing that changes is the area of the duct but not, overall, the direction of the flow. This type of flow we call "quasi-one-dimensional". For an inviscid, quasi-one-dimensional flow, mass flow rate can be defined as
$$\dot{m} = \rho V A,$$
where, $\rho$ is the density of the fluid, $V$ is the velocity (or in a viscous flow, the average velocity), and $A$ is the cross-sectional area of the duct.

Now, in a steady-state flow you have to conserve mass, so for a given section of pipe, the mass flow rate going in has to equal the mass flow rate going out. Using the image from hyperphysics that jim hardy posted, that means that you can define the following relationship:
$$\dot{m}_1 = \dot{m}_2,$$
or
$$\rho_1 V_1 A_1 = \rho_2 V_2 A_2,$$
where the quantities are denoted with numbers corresponding to the inlet (1) and the throat (2) of the pipe in the picture.

Now, if the flow is incompressible, as with water or low-speed air, then the density drops out of the equation and we can state that
$$V_1 A_1 = V_2 A_2,$$
or alternatively
$$\dfrac{V_2}{V_1} = \dfrac{A_1}{A_2}.$$

This tells you that if $A_1 > A_2$, then the right side of that equation is greater than 1 and so $V_2 >V_1$. This simply means if the duct gets smaller, the flow gets faster and vice versa. That is why the flow speeds up. To this point, pressure wasn't involved.

Now that you have the velocity at the various locations in the duct using the continuity equation, you can calculate the pressure throughout the duct using Bernoulli. It makes sense then that, as the duct constricts and the water speeds up, something had to be exerting a force on that water to accelerate it. That force comes from what we call a pressure gradient, which is change in pressure along the length of the pipe. In this case, intuitively, the pressure would be decreasing as you move along the pipe and speed up, as jim hardy up there handily illustrated using forces.

9. Mar 31, 2013

So,
Increased velocity causes lower pressure, or rather pressure is used to provide the extra push, which lowers the pressure and speeds up the flow.

This is caused by the gradient, which is caused by the thinning of the pipe.

So.. and this is the bit I'm struggling with..

Is the pressure now lower in the thinnest part of the pipe than it was originally before the pipe stated thinning?

Does the thinning pipe cause higher pressure that gets turned into velocity so that we have..
Medium pressure, high pressure, medium pressure?
(wide pipe) (thinning pipe) (thin pipe)

10. Mar 31, 2013

The static pressure is now lower in that thinnest part of the pipe. Total pressure is the same under the assumptions of the Bernoulli equation. The highest pressure in this case is going to be where the pipe is the widest and will decrease as the pipe narrows and the flow speeds up.

11. Mar 31, 2013

### jim hardy

Now you're getting there.... introduce concept of total pressure....
Bernoulli says you can trade off between static pressure and velocity.

A pitot tube measures sum of them.

here's a link with an illustration:

http://www.grc.nasa.gov/WWW/k-12/airplane/pitot.html