# Bernoulli's Equation and decreasing pressure

1. Aug 3, 2009

### DmytriE

My mind is struggling trying to understand how a decrease in pressure can increase the velocity and vise versa for a fluid that is flowing through a tube of some kind. I've always thought that if you increase the amount of pressure then the velocity should also increase. It appears that they are inversely proportional.

P + rho*gh +1/2rho*v2 = K
Pressure = F/A = F/pi*r2
Density = m/V

I think to understand this problem density plays a part seeing as how it appears in two of the three parts of the equation.

Any help would be welcomed. Thanks!

2. Aug 3, 2009

### Cyrus

Consider water as the working fluid. I made this choice because it's incompressible, so density does not change. If you have a tank, then you are absolutely correct that the pressure due to waterline will increase the velocity at the bottom if there were an open valve. This is the hydrostatic term $$\rho gh$$ in the Bernoulli equation.

What the equation above is also telling you is that along a streamline, as the dynamic pressure $$\left(\frac{1}{2} \rho V^2\right)$$ goes up the static pressure goes down, $$P_{static}$$.

The apparent paradox is that the increased pressure (in the case of hydrostatics) occurs on the other side of the equality.

Last edited: Aug 4, 2009
3. Aug 3, 2009

### LeonhardEuler

Hi DmytriE. I can remember when that seemed counter-intuitive to me as well. But when you think about it harder, it makes perfect sense:

Remember, the pressure is a force per unit area. Think about an element of fluid (assume it is a cube for simplicity) that is in a larger volume of fluid with a constant pressure. On the left side it feels a pressure force to the right. On the right side it feels a pressure force to the left. If the pressure is constant, it feels no net force due to pressure. If the fluid is not moving, is will remain stationary. If it is moving, it will continue at constant velocity (All of this is assuming there are no other forces on the fluid, of course). It doesn't matter how high the pressure is.

Now consider the case where the pressure varies and the fluid flows. Keep in mind that the Bernoulli equation applies to different points along the flow path of a fluid and says the stuff on the left adds to a constant for all points on the same flow path. Imagine that the pressure increases as you go downstream ("with the fluid"). Now there is a pressure force on the front pushing back and a pressure force on the back pushing forward. If the pressure is greater in the front, then the net force pushes the fluid backward. Similarly, if the pressure is greater in the back, it will push the fluid forward.

This makes intuitive sense. Say you have a water balloon and you apply a lot of pressure to it. If you open the outlet, the pressure will cause the water to flow out. The velocity will be relatively small in the balloon where the pressure is high, but increases near the outlet where the pressure drops.

The final thing I will point out to make sure it makes sense is that the Bernoulli equation really does not say anything about the effect of the absolute pressure on the flow, but instead about the effect of pressure changes. The reason it seems intuitive that applying a high pressure to a fluid will make it go fast is that a high pressure typically allows for a high change in pressure when the fluid is exposed to the atmosphere. But if it is kept at a high pressure and no part of it is exposed to low pressure, the pressure will not cause it to flow.

4. Aug 4, 2009

### rcgldr

The equation is for an idealized situation. It would probably easier to explain Bernoulli principle by stating that pressure differentials coexist with accelerations within a fluid or a gas. A fluid or gas will accelerate from a higher pressure zone to a lower pressure zone. In an idealized case where no work is done within the transition, then Bernoulli's equation defines a relationship between the pressure and speed of the fluid or gas at any point within that flow between the pressure zones.

Bernoulli doesn't cover interactions between a solid and fluid or gas, or other situations when work is done. From this Nasa link:

at the exit, the velocity is greater than free stream because the propeller does work on the airflow. We can apply Bernoulli's equation to the air in front of the propeller and to the air behind the propeller. But we cannot apply Bernoulli's equation across the propeller disk because the work performed by the engine (by the propeller) violates an assumption used to derive the equation. :

propeller_analysis.htm

In the case of a propeller, there is a low pressure zone fore of the prop disk, and air accelerates towards this low pressure zone via a Bernoulli like reaction. Across the prop "disk", the speed remains about the same, but the pressure increases due to mechanical interaction between air and propeller, called a pressure jump. Just aft of the prop disk, the air has higher pressure, and continues to accelerate aft of the propeller. Part of the propeller's job is to reduce the amount of high pressure air just aft of the prop disk that would otherwise accelerate forwards to the low pressure zone just fore of the prop disk.

A wing does essentially the same thing, but the amount of induced wash is less (since the wing moves perpendicular to lift), and the pressure jump is less.

Any venturi based device is a good example of Bernoulli principle. Flow through a pipe is reduced in pressure as it flows due to friction between the pipe and fluid or gas, and viscosity within the fluid or gas. If the pipe has a narrowing section, then the flow has to speed up or otherwise mass would be accumulating. The faster moving flow in the narrower section has lower pressure and if this pressure is lower than some external pressure, it can be used to draw in an external source of fluid or gas.

Although carburetors are sometimes used as examples of Bernoulli principle, you have the additional effect of a nozzle perpendicular to a flow, which creates a vortice, lowering the pressure further still. Some spray pumps also operate on this same nozzle effect, and home experiments using a straw perpendicular to a flow also rely on this nozzle effect and are not good examples of Bernoulli principle. In order to "sense" the pressure of a horizontal flow, you need something like a static port, which is a flush mounted pipe that "hides" inside a boundary layer between a surface and a horiztonal flow.

A good example of Bernoulli principle is an venturi based pump, connected to a water tap and used to drain water from aquarium. Here is an example:

http://andysworld.org.uk/aquablog/?postid=247

If you follow the USA patent, there's a link to images, but you'll need to install a TIFF viewer browser add-on to view it. If you follow the Candian patent:

http://brevets-patents.ic.gc.ca/opic-cipo/cpd/eng/patent/1245129/summary.html

Then figure 4 on drawing page 2 shows the flow when operating as a drain via venturi effect:

python_syphon_drawing_page_2.htm

Note that this device is essentially the same device as the one patented in 1933, I'm not sure why it was granted a patent.

Last edited: Aug 4, 2009
5. Aug 4, 2009

### DmytriE

I would like to quickly clarify to make sure I fully understand. Bernoulli's principle shows the relationship between velocity and pressure when a fluid (e.g. water) moves from an area of higher pressure to lower pressure.

When a fluid moves from an area of higher pressure to lower pressure the velocity increases because there is less force acting on the fluid in the opposite direction of the flow. The opposite, I hope, is true that when pressure increases there is a greater amount of force acting in the opposite direction of flow causing the velocity to decrease.

Could someone clarify what the P in bernoulli's equation is? Is it Pstatic and is that calculated through F/A or is there a different equation to solve for that? I know that rho*g*h is equivillent to mgh (fluid potential?) and 1/2rho*v2 to 1/2mv2 (fluid energy?).

Thanks for everyone's response!

6. Aug 4, 2009

### Staff: Mentor

Ehh, not exactly, and the rest of your post implies you're missing the critical point. The others gave good and complete responses, but in an effort to be more concise:

You haven't seemed to be understanding the critical point that there is more than one type of pressure (you used the word several times in your first post and didn't differentiate).
Bernoulli's principle states that the total pressure in a fluid stream is constant. So all that means is that when the fluid flows through a restriction, it trades one type of pressure (static pressure) for another (velocity pressure).
There are many forms of the equation, but one of the more common starting points is the one you showed first in your first post: Each term represents a different type of pressure. The first is static pressure ( P ) , the second is gravitational pressure ( rho*gh ), and the third is velocity pressure ( 1/2rho*v2 ).

7. Aug 5, 2009

### rcgldr

Yes, it's Pstatic, and the units are force / unit_area.

The SI unit for Pstatic is Pascal:

1 Pa = 1 N / m2 = 1 kg / (m s2)

For Pdynamic, to convert units to a pressure unit form:

1/2 ρ v2 = 1/2 (kg / m3) (m/s)2 = 1/2 kg / (m s2) = 1/2 Pa

Last edited: Aug 5, 2009
8. Aug 5, 2009

### DmytriE

So there are three different types of pressures that contribute to the overall pressure - static, gravitational, and velocity pressure. So I now understand this concept but to convince myself I tried to do calculations.

P$$_{1}$$ + $$\rho$$gh$$_{1}$$ + 1/2$$\rho$$v$$^{2}$$ = K.

Appologies for my terrible latex writing. I'm not familiar with this method of writing equations. :shy:

If we set this equal to the same equation but with different values I don't know where to input a different pressure. For example at sea level there is a pressure of 1atm and 3 meters down there is a pressure of 2 atm. Where could I account for this difference of pressure? Is this also static pressure? Once I know where to input those values into the equation I think I have a basic concept of Bernoulli's principle understood.

*The 1's are suppose to be subscripts so denote their values will be different from the opposite side.

9. Aug 5, 2009

### rcgldr

Note, you can use unicode (along with system font) to avoid using latex which has to generate an image. Just copy and past the unicode symbol.

pressure = energy / volume

energy = pressure x volume

For example the pressure energy of a cubic meter of some fluid or gas with 1 Pascal of pressure is 1 Joule.

1 Pa m3 = 1 (N / m2) m3 = 1 N m = 1 Joule

ρ g h is the gravitational potential energy / (unit volume) component of pressure.

1 atm = 101325 Pa
density of water at 20C = 998.2071 kg / m3[/SUP
g = 9.80665 m / s2
h for 2 atm ~= - 10.351 meters.

Assuming velocity = 0 in this case

1 atm + ρ g 0 = 2 atm + ρ g h

101325 Pa ~= 202650 Pa + (998.2071 kg / m3) (9.80665 m / s2) (-10.351 m)

Last edited: Aug 5, 2009