This document discusses the principles of continuity, Bernoulli's principle, and their applications to fluid flow problems. It begins by defining assumptions about incompressible, steady, and non-viscous fluid flow. It then explains continuity of fluid flow, relating mass and volume flow rates. Bernoulli's principle is introduced, relating pressure, height, and flow speed. Examples show applying these principles to problems involving changing pipe diameters, fluid velocities and pressures.

1. Continuity of fluid flow
&
Bernoulli’s PrinCiPle

2. The Flow of a Fluid
The flow of a fluid is very complicated. Many aspects of fluid flow are not
well understood yet. We will limit our study to these three assumptions:
1) Incompressible fluids – this means the fluid does not change its density.
2) Steady (laminar) flow – this means the fluid flows in layers (streamlines) rather
than in a chaotic fashion as in turbulent flow.
3) Non-viscous fluids – this means there is no friction in the fluid. A non-
viscous fluid would be like water as opposed to pancake
batter.

3. Mass Flow RateMass Flow Rate
(kg/s)(kg/s)
Volume Flow RateVolume Flow Rate
(m(m33
/s)/s)
A ≡ Cross-sectional Area
(m2
)
m ≡ mass (kg)
t ≡ time (s)
v ≡ velocity (m/s)
V ≡ volume (m3
)
ρ ≡ density (kg/m3
)
Continuity of Fluid FlowContinuity of Fluid Flow
As a fluid flows through a pipe which changes its
cross-sectional area, its mass flow rate must remain
constant.
If the fluid is incompressible, then the volume flow rate must remain constant.

4. Relates the pressure (Pa), the height of the fluid
(m) and flow speed of the fluid (m/s) at one
point in a laminar flow to another point in a
laminar flow.
A special case of Bernoulli’s Principle occurs
when the flow speed of a fluid is zero at one
point in the flow and the pressure at two points
is the same. In this case Toricelli’s Equation
gives the speed of the fluid at the other point.

5. Momentum & Energy Example 29: Fluid Velocity & Pressure in a Pipe
Water is flowing through a 19-cm diameter pipe at 75-m/s. The pipe’s diameter decreases
to 10-cm. The pipe is level.
a)What is the speed of the air in the 10-cm diameter pipe?
I’m baaaaaaaaaaack!!!!!!!
First, draw a picture of the pipe and
label it. Don’t forget to convert the
diameters to meters.
D1 = 0.19-m D2 = 0.10-m
v1 = 75-m/s v2
Now apply continuity of fluid flow
and solve for v2, just like all good
AP Physics students do!!!
Calculate the areas, A1 & A2. Remember to use
the radius of each rather than the diameter.
Finally, substitute into the equation for
v2 and simplify.
Change this

6. Momentum & Energy Example 29: Fluid Velocity & Pressure in a Pipe
Water is flowing through a 19-cm diameter pipe at 75-m/s. The pipe’s diameter decreases
to 10-cm. The pipe is level.
b)If you wanted the airflow to reach the speed of sound (345-m/s), what would the
diameter of the smaller pipe need to be?
Draw and label the picture!!
D1 = 0.19-m D2
v1 = 75-m/s v2=345-m/s
Apply continuity of fluid flow and solve for
the area at point 2, A2.
Calculate A1, and substitute into the
equation to find A2.
Now find the radius and then the diameter.

7. Momentum & Energy Example 29: Fluid Velocity & Pressure in a Pipe
Water is flowing through a 19-cm diameter pipe at 75-m/s. The pipe’s diameter decreases
to 10-cm. The pipe is level.
c)The pressure in the large diameter pipe is 1.2-atm. If the 10-cm pipe is placed 1.5-m
above the 19-cm diameter pipe, what will the pressure in the 10-cm pipe be?
Draw and label the diagram. Continuity of
fluid flow still gives you a speed of 270-m/s
at the 10-cm end.
h =1.5-mv1 = 75-m/s
v2 = 270-m/s
Convert the pressure to Pa.
P1 = 1.2 X 105
-Pa
Write out Bernoulli’s Equation. We will
assume the density of the water remains
constant. (1000-kg/m3
)
Since h1 = 0, we can drop this term
as we solve for P2.
Substitute in and simplify.

8. Momentum & Energy Example 30: Water Flow from a Tank
A water tank with a valve at the bottom is shown below. Assume the cross-sectional area
at A is very large compared with that at B, and the acceleration of gravity is 9.81-m/s2
.
a) What is the speed the water leaves the spigot at B after the valve is opened?
Apply Bernoulli’s principle.
At A & B, the pressure is the same so we
can eliminate these terms.
We need to find hB using a bit of
trigonometry.
We can also assume the speed at
A is zero.
vA = 0-m/s
Solve the equation for vB.This is Torricelli’s Equation.Substitute and find vB.

9. Momentum & Energy Example 30: Water Flow from a Tank
A water tank with a valve at the bottom is shown below. Assume the cross-sectional area
at A is very large compared with that at B, and the acceleration of gravity is 9.81-m/s2
.
b) What is the maximum height above the opening of the spigot (∆ymax
) attained
by the water stream coming out of the spigot at B after the valve is opened?
This is just a projectile motion problem.
So we can use our equations of motion.
At maximum height vy = 0, so solving
for ∆ymax we get this equation.
We need to use trigonometry, to find
voy. then substitute to find the
solution.
10-m/s
voy = 10sin(49) = 7.5-m/s)

10. Momentum & Energy Example 30: Water Flow from a Tank
A water tank with a valve at the bottom is shown below. Assume the cross-sectional area
at A is very large compared with that at B, and the acceleration of gravity is 9.81-m/s2
.
b) What is the maximum height above the opening of the spigot (∆ymax
) attained
by the water stream coming out of the spigot at B after the valve is opened?
This is just a projectile motion problem.
So we can use our equations of motion.
At maximum height vy = 0, so solving
for ∆ymax we get this equation.
We need to use trigonometry, to find
voy. then substitute to find the
solution.
10-m/s
voy = 10sin(49) = 7.5-m/s)