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Fluids &
Bernoulli’s Equation
Chapter 10.8
Flow of Fluids
• There are two types of flow that fluids
can undergo;
• Laminar flow
• Turbulent flow.
Flow of Fluids
• There are two types of flow that fluids
can undergo;
We will be dealing
• Laminar flow
only with laminar flow
• Turbulent flow.
Flow of Fluids
• There are two types of flow that fluids
can undergo;
We will also be
• Laminar flow
assuming no friction
along the edges
• Turbulent flow.
Rate of Flow
• Defined as the volume of fluid that
passes a certain section in a given
time
R = Av
Where:
• R = the flow rate
• A = the cross-sectional area of the pipe
• v = the velocity of the fluid.
Rate of Flow
• Common units for rate of flow
are
3
• cubic feet per second (ft /s),
3
• cubic meters per second (m /s),
• gallons per second (gps),
• liters per second (L/s).
Rate of Flow
Fluid in = Fluid out:
Flow rate remains constant even if
the radius of the pipe changes
v1
A1
A2
v2
Rate of Flow
R = A1v1  A2v2
Flow rate remains constant even if
the radius of the pipe changes
v1
A1
A2
v2
Problem
• Water flows through a rubber
hose 2.0 cm in diameter at a
velocity of 4.0 m/s. If the hose
is coupled into a hose that has
a diameter of 3.5 cm, what is
the new speed of the fluid?
Answer
A1v1 = A2 v2 a
v2 
2
v1r1
2
r2
2
πr
1.0 cm 
m  2.0
 4.0
2
s 1.75
3.5 cm 
v =
1 1
2

2
πr
v
2 2
m
1.3
s
Area increases so velocity
decreases
Applications
Bernoulli’s Principle
• To accelerate a fluid as it goes into
the constriction, the pushing force in
the large diameter area must be
greater than the pushing force in the
constriction.
At point B, the
pushing force in
the x direction
has increased.
y
B
A
C
Bernoulli’s Principle
• To accelerate a fluid as it goes into
the constriction, the pushing force in
the large diameter area must be
greater than the pushing force in the
constriction.
y
PA – PB = ρgy
B
A
C
Bernoulli’s Principle
• When the speed of a fluid increases,
its internal pressure (pressure on the
sides of the container) decreases.
1 2
P   gy  v  const.
2
Where: const. is some constant, P is the pressure,
 is the fluid density, y is the height of the fluid,
and v is the fluid velocity
Bernoulli’s Principle
• We can analyze the flow of a fluid
through a system using this equation.
P1
v1
y
1
P2
v2
y
2
Bernoulli’s Principle
1 2
1
2
P1   gy1   v1  P2   gy2  v2
2
2
P1
v1
y
1
P2
v2
y
2
If no change in height
1 2
1
2
P1   v1  P2   v2
2
2
P1
v1
y
P2
v2
y
Problem
• An oddly shaped
tank is filled with
water to a depth of
1.20 m. Calculate
the pressure at point
B at the bottom of
the tank.
A
1.20 m
B
Problem
• The system is
static, so v is zero.
The equation
becomes-
A
1.20 m
P1   g y1  P2   g y2
B
Problem
• Assume the
pressure at point A
is zero, and the
equation becomes -
A
1.20 m
P1   g y1  P2   g y2
0  P2   g y2
P2    g y2
B
Problem
• Which
coincidentally is the
equation for
hydrostatic pressure
P   gh
0  P2   g y2
A
1.20 m
P2    g y2
B
Solution
P2   gy2
P2 
kg
 1.0 x 10 32
m
3
m

 9.8 2  1.20 m 
s 

4
1.18 x 10 Pa
Problem
• A container of water, diameter 12 cm, has a
small opening near the bottom that can be
unplugged so that the water can run out. If
the top of the tank is open to the
atmosphere, what is the exit speed of the
water leaving through the
hole. The water level is 15
cm above the bottom of the
container. The center of the
y2
3.0 diameter hole is 4.0 cm
from the bottom
Solution
• The water comes out the hole with a speed of
v1. The flow of water in the container, which
makes the surface level drop is very slow by
comparison. So slow that we can say that it is
 zero. So v2 = 0.
A2 P0 = P2
y
15 cm
4.0 cm
A1
P0
v1
Solution
• The pressure on the top of the surface is the
atmospheric pressure. The surface acting on the
water at the opening on the bottom is also the
atmospheric pressure (actually it is a tiny bit
bigger because it is slightly
A2 P0 = P2
lower, but the difference is
insignificant). So we can let
y
the two pressures equal each
A1
15 cm
other. So P1 = P2
4.0 cm
P0
v1
Solution
• Therefore Bernoulli’s equation becomes
1 2
 gy1   v1   gy2
2
A2 P0 = P2
And simplifies to:
1 2
gy1  v1  gy2
2
y
15 cm
4.0 cm
A1
P0
v1
Solution
m

v1  2  9.8 2   0.15 m 0.040 m 
s 


m
1.5
s