Post on 12-May-2018
Chapter 11 - Fluids in Motion
Sections 7 - 9
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.
Fluid Motion
Paul E. Tippens
Objectives: After completing this
module, you should be able to:
• Define the rate of flow for a fluid and solve problems using velocity and cross-section.
• 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:
Streamline flow Turbulent flow
• All fluids move with streamline flow.
• The fluids are incompressible.
• There is no internal friction.
Rate of Flow The rate of flow 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 Δl, but Δl = vΔt, so V = A Δl, or
Δl = the distance traveled
Volume = A(vt)
A
tAvV
mV tAv
m
Av
t
m
Which is the equation for the MASS FLOW RATE.
Constant Rate of Flow The MASS that flows into a region = The MASS that flows out of a region. For an incompressible, frictionless fluid, the velocity increases when the cross-section decreases:
A1
A2
R = A1v1 = A2v2
v1
v2
v2
tvAtvA 222111
Continuity of Equation The,vAvA 2211
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?
2
2
21
2
1
221
vrvr
vAvA v
cm.cm.r;cm.r
)s/m(
)s/m()cm(
v
vrr
50250250
16
41
22
2
22
2
2
2
1
2
12
2
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 = 0.00126 m3/s
R1 = 0.0754 m3/min
1
2
11 vrvAt
mR
)s/m()m.(vrR 4010 2
1
2
11vAR
min 1
s60x
s
m 0.00126 R
3
1
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.
• Be sure to use consistent units for area and velocity.
2211 vAvA
Problem Strategy (Continued):
• Since the area A of a pipe is proportional to its diameter d, a more useful equation is:
• 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.
2 2
1 1 2 2v d v d
The Venturi Meter
The higher velocity in the constriction B causes a difference of
pressure between points A and B.
PA - PB = gh
h
A B
C
11.9 Bernoulli's Principle
• The Swiss Physicist Daniel Bernoulli, was interested in how the velocity changes as the fluid moves through a pipe of different area. He especially wanted to incorporate pressure into his idea as well. Conceptually, his principle is stated as:
• "If the velocity of a fluid increases, the pressure decreases and vice versa
Work in Moving a Volume of Fluid We will break it up into small sections: WORK is equal to FORCE
times DISPLACEMENT , (in this case, the length of the section the fluid travels. The grey area.)
FxW
WORK is done by the Section
One flowing FORWARD
111 lFW As well as Section Two working
AGAINST it, lFW 22
The work is negative due to Newton's Third Law and working
against gravity. Work must go into or onto moving it.
The Pressure formula can be inserted for the force and solving for F.
PAFA
FP
Substituting PA in for F, you will get the following two
equations:
lAPW 111 2222 lAPW and
Since WORK is energy we have to ask ourselves if there is any
other energy. Since the water rises it has HEIGHT and thus
POTENTIAL ENERGY.
)yy(mgW 123
Setting all three WORKS equal to the NET WORK.
.
,,
,,
12222111net
321net
321net
mgymgylAPlAPW
soWWWW
soWWWW
WHAT DOES THE NET WORK EQUAL TO?
A CHANGE IN KINETIC ENERGY!
12222111
2
2
2
1
12222111
2
2
2
1net
mgymgylAPlAPmv2
1mv
2
1
mgymgylAPlAPKE
mv2
1mv
2
1KEKEW
,
Consider that Density = Mass per unit Volume, AND
that VOLUME is equal to AREA time LENGTH , then substituting
and solving for m, we get:
lAm
We will now substitute for the MASS in the equation with energy.
12222111
2
2
2
1mgymgylAPlAPmv
2
1mv
2
1 To get:
12222111
2
o
2ρAlgyρAlgyΔlAPΔlAPρAΔlv
2
1ρAΔlv
2
1
We can now cancel out the AREA and LENGTH in all cases
because AREA times LENGTH equals VOLUME, and volume
remains constant in the pipe.
12222111
2
2
2
1gylAρgylAρlΔAPlΔAPvlΔAρ
2
1vlAρ
2
1
Leaving:
1221
2
2
2
1ρgyρgyPPρv
2
1ρv
2
1
Moving everything related, initials and finals together, to one
side results in:
2
2
2211ρgyρv
2
1Pρgyρv
2
1P 2
1
What this basically shows is that Conservation of Energy holds
true within a fluid and that if you add the PRESSURE, the
KINETIC ENERGY(in terms of density) and POTENTIAL
ENERGY(in terms of density) you get the SAME VALUE
anywhere along a streamline.
Constantρgyρv2
1P
1
2
11
Which is known a Bernoulli’s Equation!
Bernoulli’s Theorem (Horizontal Pipe):
2 2
1 1 1 2 2 2½ ½P gh v P gh v
h1 = h2
v1 v2 Horizontal Pipe (h1 = h2)
2 2
1 2 2 1½ ½P P v v
h
Now, since the difference in pressure P = gh,
2 2
2 1½ ½P gh v v Horizontal Pipe
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?
v1 = 4 m/s
v2 h
h = 6 cm
2 2
2 1½ ½P gh v v
Bernoulli’s Equation (h1 = h2)
2gh = v22 - v1
2 Cancel ρ, then clear fractions:
2 2 2
2 12 2(9.8 m/s )(0.12 m) (4 m/s)v gh v
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:
2 2
1 1 1 2 2 2½ ½P gh v P gh v
P1 - P2 = gh2 - gh1 P = g(h2 - h1)
h
= 1000 kg/m3
This is the same relation seen earlier for finding the pressure P at a given depth h = (h2 - h1) in a fluid.
Torricelli’s Theorem
2v gh
h1
h2 h
When there is no change of pressure, P1 = P2.
2 2
1 1 1 2 2 2½ ½P gh v P gh v
Consider right figure. If surface v2 0 and P1= P2 and v1 = v we have:
Torricelli’s theorem:
2v gh
v2 0
Interesting Example of Torricelli’s
Theorem:
v
v
v
Torricelli’s theorem:
2v gh
• Discharge velocity
increases with depth.
• Holes equidistant above and below midpoint
will have same horizontal range.
• Maximum range is in the middle.
Example 4: A dam springs a leak at a point
20 m below the surface. What is the
emergent velocity?
2v ghh
Torricelli’s theorem:
2v gh
Given: h = 20 m
g = 9.8 m/s2
22(9.8 m/s )(20 m)v
v = 19.8 m/s2
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 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.
Conceptual Applications
In general, things tend to move TOWARD areas of low pressure. When the velocity of a fluid increases, this creates a NET FORCE toward the low pressure area. Most of the time it happens when the AREA decreases. Example: Putting your thumb over a water hose.
Another example is an airfoil.
The shape of the top of the airfoil causes the streamlines to scrunch together and thus increasing their velocity. The airfoil then moves upward to the low pressure area created.
Example Problem
A large storage tank filled with water develops a small hole in its side at a point 16 m below the water level. If the rate of flow from the leak is 2.5 x 10-3 m3/min, determine (a) the speed at which the water leaves the hole and (b) the diameter of the hole.
Answer: a. 17.71 m/s; b. 0.00173 m or 1.73 mm
Example Problem
• Water flows through a fire hose of diameter 6.35 cm at a rate of 0.012 m3/s. The fire hose ends in a nozzle of inner diameter 2.2 cm. What is the velocity with which the water exits the nozzle?
• Answer: 31.6 m/s
Assignment
• Pages 353 – 354,
• #54, 55, 56, 62, 67, and 72