High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize University Physics Textbook (Master) > Fluid Mechanics

Fluid Dynamics

Buoyancy Previous Section
ViscosityNext Section
Popular Courses
Find My Course
0:00 / 0:00

Bernoulli's Principle


If we have an ideal and incompressible fluid flowing at a steady rate, we have the Bernoulli's principle.


P+ρgh+12ρv2=constant\boxed{P+\rho gh+\frac{1}{2}\rho v^2=\text{constant}}
where
  • PP is the internal pressure
  • ρ\rho is the fluid density
  • gg is the gravitational acceleration (9.81 m/s2)
  • hh is the height of the point in the fluid, relative to some defined reference point
  • vv is the flow speed

So for any two points inside of our flow, we have:

P1+ρgh1+12ρv12=P2+ρgh2+12ρv22\boxed{P_1 + \rho g h_1 + \frac12 \rho v_1^2 = P_2 + \rho g h_2 + \frac12 \rho v_2^2 }


Wize Tip
You can think of the Bernoulli's principle as a form of energy conservation. The three terms corresponds to the internal energy (PP), the fluid potential energy (ρgh\rho g h), and the fluid kinetic energy (1/2(ρv2)1/2 (\rho v^2)).

Exam Tip
By picking two point inside of the flow and write down Bernoulli's equation for them, we can find missing information about one point by knowing those information about another point.



Now let's look at two points in the flow at the same height but with different cross sectional area (Points 1 an 2 in the upper pipe)


Wize Tip
Pressure is lower in the section of the tube in which the fluid moves faster!

Flow Rates and Continuity


To talk about moving fluids, we are going to define some flow rates, which will tell us how much fluid is moving per unit time.

Flow rate is usually defined as the rate of change in volume and it is shown by QQ:


Flow Rate=Q=ΔVΔt\boxed{\text{Flow Rate}=Q=\dfrac{\Delta V}{\Delta t}}

where VVis the volume of fluid.


If we assume that we have a steady flow and incompressible fluid, flow rate is constant and could be written in the following form:

Q = Av\boxed{Q\ =\ Av}


where AAis the cross-sectional area of the fluid (perpendicular to the flow direction), and vvis the flow speed.





Now by comparing any two points in the flow and putting their flow rates equal to each other we can find continuity equation:


A1v1=A2v2\boxed{A_1v_1=A_2v_2}





Wize Concept
All that the continuity equation is telling us is that the amount of fluid in must be the same as the fluid out. You can think of this like fluid conservation, just like how we have energy and momentum conservation.

Wize Tip
A wider pipe will have slower flow, and a thinner pipe will have faster flow.

Think about a garden hose with water coming out. If you put your thumb on half the opening of the pipe (thinning the cross-sectional area of the flow), the water will shoot out FASTER.

Exam Tip
If the cross sectional area of a pipe at one point is so much bigger than the cross sectional area of the rest of the pipe, we can assume that the velocity of the fluid in that point is approximately zero (for example the velocity of water in a huge tank!).
















Example: Venturi System


A Venturi flow-meter for a tube of water is shown below. There is stationary mercury in the U-tube, shown by the blue color, with density ρHg=13600\rho_{Hg}=13600 kg/m3 , and flowing water on the top pipe with density ρw=1000\rho_w=1000kg/m3 . We know that A1A2=5\dfrac{A_1}{A_2}=5 and v1=2v_1=2m/s. What is the difference in the height of mercury in the U-tube?



Note that P1>P2P_1>P_2 because P1P_1 has pushed down the mercury more compared to P2P_2.



→\to Here we have a static fluid (mercury) and a flowing fluid (water).
→\to So, we can use Pascal's principle for Mercury and Bernoulli's equation for water.


Write Bernoulli's Equation for the flow of water at points #1 and #2.

Here h1=h2h_1=h_2 no matter where the reference height is (it could be the bottom of the picture, or the line passing through these points, or any other horizontal line).

P1+12ρwv12+ρwgh1=P2+12ρwv22+ρwgh2P_1+\dfrac{1}{2}\rho_wv_1^2+\cancel{\rho_wgh_1}=P_2+\dfrac{1}{2}\rho_wv_2^2+\cancel{\rho_wgh_2}

Then the difference in pressures is given by:

P1−P2=12ρw(v22−v12)P_1-P_2=\dfrac{1}{2}\rho_w(v_2^2-v_1^2)

Let's use the continuity equation A1v1=A2v2A_1v_1=A_2v_2 to sub in v2=A1v1A2v_2=\dfrac{A_1v_1}{A_2} and get:

P1−P2=12ρw[(A1v1A2)2−v12]P_1-P_2=\dfrac{1}{2}\rho_w\bigg[\bigg(\dfrac{A_1v_1}{A_2}\bigg)^2-v_1^2\bigg]

=12ρw (A12A22−1) v12=\dfrac{1}{2}\rho_{w\ }\bigg(\dfrac{A_1^2}{A_2^2}-1\bigg)\ v_1^2


Write Pascal's principle for the stationary mercury: at point A at the surface of the lower branch on the left, and point B at the same level on the right branch.

PA=PBP_A=P_B

ρwgh+P1=ρHggh+P2\rho_{w} gh+P_1=\rho_{Hg}gh+P_2

where we have ignored the columns of water above the surface of the higher branch, since they're the same for both branches and so the pressure from them cancels out.

The difference in pressures is then:

P1−P2=ρHg gh−ρw ghP_1-P_2=\rho_{Hg}\ gh-\rho_{w} \ gh

=(ρHg−ρw) gh=(\rho_{Hg}-\rho_w)\ gh


Let's equate the two expressions we have for the pressure difference:

12ρw (A12A22−1) v12=(ρHg−ρw) gh\dfrac{1}{2}\rho_{w\ }\bigg(\dfrac{A_1^2}{A_2^2}-1\bigg)\ v_1^2=(\rho_{Hg}-\rho_w)\ gh


Isolate the height:

h=v12ρw2 (ρHg−ρw) g(A12A22−1)h=\dfrac{ v_1^2\rho_{w}}{2 \ (\rho_{Hg}-\rho_w)\ g}\bigg(\dfrac{A_1^2}{A_2^2}-1\bigg)

Put the numbers in:

=22×10002(13600−1000)(9.81)×(52−1)=\dfrac{2^2\times1000}{2(13600-1000)(9.81)} \times(5^2-1)

=0.388=0.388 (m)

Practice: Leaking Tank


A large cylindrical container is filled with water to a height hh and it is open to air on top. The density of the liquid is ρ\rho.
1) Find the velocity of the emerging fluid from the hole at the base of the container. (Assume that the cross sectional area of the container is much larger compared to the size of the hole)