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Chapter 15B - Fluids in Motion
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
©
2007
Fluid Motion
Paul E. Tippens
The lower falls at
Yellowstone
National Park: the
water at the top of
the falls passes
through a narrow
slot, causing the
velocity to increase
at that point. In this
chapter, we will
study the physics of
fluids in motion.
Objectives: After completing this
module, you should be able to:
• Define the rate of flow for a fluid and
solve problems using velocity and crosssection.
• Write and apply Bernoulli’s equation for
the general case and apply for (a) a fluid
at rest, (b) a fluid at constant pressure,
and (c) flow through a horizontal pipe.
Fluids in Motion
All fluids are assumed
in this treatment to
exhibit streamline flow.
• Streamline flow is the motion of a fluid in
which every particle in the fluid follows the
same path past a particular point as that
followed by previous particles.
Assumptions for Fluid Flow:
• All fluids move with streamline flow.
• The fluids are incompressible.
• There is no internal friction.
Streamline flow
Turbulent flow
Rate of Flow
The rate of flow R is defined as the volume V of a fluid
that passes a certain cross-section A per unit of time t.
The volume V of fluid is given by
the product of area A and vt:
A
V  Avt
vt
Volume = A(vt)
Avt
R
 vA
t
Rate of flow = velocity x area
Constant Rate of Flow
For an incompressible, frictionless fluid, the velocity
increases when the cross-section decreases:
R  v1 A1  v2 A2
A1
v d  v2 d
2
1 1
R = A1v1 = A2v2
A2
v1
v2
v2
2
2
Example 1: Water flows through a rubber
hose 2 cm in diameter at a velocity of 4 m/s.
What must be the diameter of the nozzle in
order that the water emerge at 16 m/s?
The area is proportional to
the square of diameter, so:
v1d12  v2 d 22
2
1 1
vd
(4 m/s)(2 cm)
d 

v2
(20 cm) 2
2
2
2
d2 = 0.894 cm
Example 1 (Cont.): Water flows through a
rubber hose 2 cm in diameter at a velocity of
4 m/s. What is the rate of flow in m3/min?
R  v1 A1  v2 A2
R  v1 A1 ;
A1 
 d12
4
2
2
 d1 (4 m/s) (0.02 m)
R1  v1

4
4
m3  1 min 
R1  0.00126


min  60 s 
R1 = 0.00126 m3/s
R1 = 0.0754 m3/min
Problem Strategy for Rate of Flow:
• Read, draw, and label given information.
• The rate of flow R is volume per unit time.
• When cross-section changes, R is constant.
R  v1 A1  v2 A2
• Be sure to use consistent units for area
and velocity.
Problem Strategy (Continued):
• Since the area A of a pipe is proportional to its
diameter d, a more useful equation is:
v d  v2 d
2
1 1
2
2
• The units of area, velocity, or diameter chosen
for one section of pipe must be consistent with
those used for any other section of pipe.
The Venturi Meter
h
A
B
C
The higher velocity in the constriction B causes a
difference of pressure between points A and B.
PA - PB = rgh
Demonstrations of the Venturi Principle
Examples of the Venturi Effect
The increase in air velocity produces a difference
of pressure that exerts the forces shown.
Work in Moving a
Volume of Fluid
A1
P1
Volume
F1
V
P1  ; F1  P1 A1
A1
A1
P1
F1
A2
P2
Note
differences in
pressure DP
and area DA
F2
P2  ; F2  P2 A2
A2
A2
P2 , F2
h
Fluid is raised
to a height h.
Work on a Fluid (Cont.)
v2
F1 = P1A1
A2
v1
A1
h1
s1
F2 = P2A2
s2
h2
Net work done on
fluid is sum of work
done by input force
Fi less the work done
by resisting force F2,
as shown in figure.
Net Work = P1V - P2V = (P1 - P2) V
Conservation of Energy
v2 F2 = P2A2
Kinetic Energy K:
DK  ½mv22  ½mv12
Potential Energy U:
F1 = P1A1
A2
v1
A1
DU  mgh2  mgh1
h1
Net Work = DK + DU
also
s2
h2
s1
Net Work = (P1 - P2)V
( P1  P2 )V  (½mv  ½mv )  (mgh2  mgh2 )
2
2
2
1
Conservation of Energy
( P1  P2 )V  (½mv  ½mv )  (mgh2  mgh2 )
2
2
2
1
Divide by V, recall that density r  m/V, then simplify:
P1  r gh1  ½ r v12  P2  r gh2  ½ r v22
v2
v1
Bernoulli’s Theorem:
P1  r gh1  ½ r v  Const
2
1
h2
h1
Bernoulli’s Theorem (Horizontal Pipe):
P1  r gh1  ½ r v12  P2  r gh2  ½ r v22
Horizontal Pipe (h1 = h2)
P1  P2  ½ r v22  ½ r v12
v1 h
v2
r
h1 = h 2
Now, since the difference in pressure DP = rgh,
Horizontal
Pipe
DP  r gh  ½ r v22  ½ r v12
Example 3: Water flowing at 4 m/s passes through
a Venturi tube as shown. If h = 12 cm, what is the
velocity of the water in the constriction?
Bernoulli’s Equation (h1 = h2)
DP  r gh  ½ r v  ½ r v
2
2
2
1
Cancel r, then clear fractions:
h
v1 = 4 m/s
v2
r
h = 6 cm
2gh = v22 - v12
v2  2 gh  v12  2(9.8 m/s 2 )(0.12 m)  (4 m/s) 2
v2 = 4.28 m/s
Note that density is not a factor.
Bernoulli’s Theorem for Fluids at Rest.
For many situations, the fluid remains at rest so that
v1 and v2 are zero. In such cases we have:
P1  r gh1  ½ r v12  P2  r gh2  ½ r v22
P1 - P2 = rgh2 - rgh1
This is the same relation
seen earlier for finding the
pressure P at a given depth
h = (h2 - h1) in a fluid.
DP = rg(h2 - h1)
h
r = 1000
kg/m3
Torricelli’s Theorem
When there is no change of pressure, P1 = P2.
P1  r gh1  ½ r v  P2  r gh2  ½ r v
2
1
Consider right figure. If
surface v2  0 and P1=
P2 and v1 = v we have:
Torricelli’s theorem:
v  2 gh
2
2
v2  0
h2 h
h1
v  2 gh
Interesting Example of
Torricelli’s Theorem:
Torricelli’s theorem:
v  2 gh
• Discharge velocity
increases with depth.
v
v
v
• Maximum range is in the middle.
• Holes equidistant above and below midpoint
will have same horizontal range.
Example 4: A dam springs a leak
at a point 20 m below the surface.
What is the emergent velocity?
Torricelli’s theorem:
v  2 gh
h
Given: h = 20 m
g = 9.8 m/s2
v  2(9.8 m/s 2 )(20 m)
v = 19.8 m/s2
v  2gh
Strategies for Bernoulli’s Equation:
• Read, draw, and label a rough sketch with givens.
• The height h of a fluid is from a common reference
point to the center of mass of the fluid.
• In Bernoulli’s equation, the density r is mass
density and the appropriate units are kg/m3.
• Write Bernoulli’s equation for the problem and
simplify by eliminating those factors that do not
change.
P1  r gh1  ½ r v12  P2  r gh2  ½ r v22
Strategies (Continued)
P1  r gh1  ½ r v12  P2  r gh2  ½ r v22
• For a stationary fluid, v1 = v2 and we have:
DP = rg(h2 - h1)
r = 1000
h kg/m3
• For a horizontal pipe, h1 = h2 and we obtain:
P1  P2  ½ r v22  ½ r v12
Strategies (Continued)
P1  r gh1  ½ r v12  P2  r gh2  ½ r v22
• For no change in pressure, P1 = P2 and we have:
Torricelli’s Theorem
v  2 gh
General Example: Water flows through the pipe at
the rate of 30 L/s. The absolute pressure at point A is
200 kPa, and the point B is 8 m higher than point A.
The lower section of pipe has a diameter of 16 cm and
the upper section narrows to a diameter of 10 cm.
Find the velocities of the stream at points A and B.
B
R = 30 L/s = 0.030 m3/s
A   R2 ;
D
R
2
AA = (0.08 m)2 = 0.0201 m3
AB = (0.05 m)2 = 0.00785 m3
R 0.030 m 3 /s
vA 

 1.49 m/s;
2
AA 0.0201 m
vA = 1.49 m/s
R=30 L/s
8m
A
R 0.030 m 3 /s
v2 

 3.82 m/s
2
A2 0.00785 m
vB = 3.82 m/s
General Example (Cont.): Next find the absolute
pressure at Point B.
Given: vA = 1.49 m/s
vB = 3.82 m/s
PA = 200 kPa
hB - hA = 8 m
B
R=30 L/s
8m
A
Consider the height hA = 0 for reference purposes.
0
PA + rghA +½rvA2 = PB + rghB + ½rvB2
PB = PA + ½rvA2 - rghB - ½rvB2
PB = 200,000 Pa + 1113 Pa
–78,400 Pa – 7296 Pa
PB = 200,000
Pa + ½(1000 kg/m3)(1.49 m/s)2
– (1000 kg/m3)(9.8 m/s2)(8 m) - ½(1000 kg/m3)(3.82 m/s)2
PB = 115 kPa
Summary
Streamline Fluid Flow in Pipe:
v1d12  v2 d 22
R  v1 A1  v2 A2
Fluid at Rest:
PA - PB = rgh
Horizontal Pipe (h1 = h2)
P1  P2  ½ r v22  ½ r v12
Bernoulli’s Theorem:
P1  r gh1  ½ r v12  Constant
Torricelli’s theorem:
v  2 gh
Summary: Bernoulli’s Theorem
• Read, draw, and label a rough sketch with givens.
• The height h of a fluid is from a common reference
point to the center of mass of the fluid.
• In Bernoulli’s equation, the density r is mass density
and the appropriate units are kg/m3.
• Write Bernoulli’s equation for the problem and
simplify by eliminating those factors that do not
change.
P1  r gh1  ½ r v12  P2  r gh2  ½ r v22
CONCLUSION: Chapter 15B
Fluids in Motion