Transcript Fluids in Motion

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
vd v d
2
1 1
2
2 2
R = A1v1 = A2v2
A2
v1
v2
v2
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:
vd v d
2
1 1
2
2 2
2
2
v
d
(4
m/s)(2
cm)
d22  1 1 
2
v2
(20 cm)
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 
d
2
1
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:
vd v d
2
1 1
2
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  ½rv12  P2  r gh2  ½rv22
v2
v1
Bernoulli’s Theorem:
P1  r gh1  ½ rv  Const
2
1
h2
h1
Bernoulli’s Theorem (Horizontal Pipe):
P1  r gh1  ½rv12  P2  r gh2  ½rv22
Horizontal Pipe (h1 = h2)
P1  P2  ½ rv22  ½ rv12
v1 h
v2
r
h1 = h 2
Now, since the difference in pressure DP = rgh,
Horizontal
Pipe
DP  r gh  ½ rv22  ½ rv12
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  ½rv  ½rv
2
2
2
1
Cancel r, then clear fractions:
h
v1 = 4 m/s
v2
r
h = 12 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  ½rv12  P2  r gh2  ½rv22
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  ½rv  P2  r gh2  ½rv
2
1
Consider right figure. If
surface v2  0 and P1=
P2 and v1 = v we have:
Torricelli’s theorem:
v  2gh
2
2
v2  0
h2 h
h1
v  2gh
Interesting Example of
Torricelli’s Theorem:
Torricelli’s theorem:
v  2gh
• 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  2gh
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  ½rv12  P2  r gh2  ½rv22
Strategies (Continued)
P1  r gh1  ½rv12  P2  r gh2  ½rv22
• 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 rv22  ½ rv12
Strategies (Continued)
P1  r gh1  ½rv12  P2  r gh2  ½rv22
• For no change in pressure, P1 = P2 and we have:
Torricelli’s Theorem
v  2gh
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.
R = 30 L/s = 0.030 m3/s
B
D
R=30 L/s
A   R2 ; R 
2
8m
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
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  v2d22
R  v1 A1  v2 A2
Fluid at Rest:
PA - PB = rgh
Horizontal Pipe (h1 = h2)
P1  P2  ½ rv22  ½ rv12
Bernoulli’s Theorem:
P1  r gh1  ½ rv12  Constant
Torricelli’s theorem:
v  2gh
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  ½rv12  P2  r gh2  ½rv22