Download Lecture 28

Wednesday, November 18, 2009
46
The Final Exam Schedule is Final!
Friday, December 18, 1:30 - 4:30 pm
Frank Kennedy Brown & Gold Gyms
The whole course
30 multiple choice questions
Formula sheet provided
Seating (this info is on Aurora)
Brown Gym: A - S
Gold Gym: T - Z
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47
WileyPLUS Assignment 4
Chapters 8, 9, 10
Due: Friday, November 27 at 11 pm
Next Week: Tutorial & Test 4
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48
Fluids, continued...
Pascal’s Principle: a change in pressure is transmitted
equally throughout an enclosed fluid.
Archimedes’ Principle: the buoyant force acting on an
object is equal to the weight of fluid displaced by the
object.
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Clicker Question: Focus on Concepts, Question 8
Two objects, A and B, have the same volume and are
completely submerged in a liquid, although A is deeper
than B.
Which object, if either, experiences the greater buoyant
force?
A) Object B, because the closer an object is to the surface, the greater is the
buoyant force.
B) Both objects experience the same buoyant force.
C) Object A, because, being at a greater depth, it experiences a greater
pressure.
B) The buoyant force is equal to the weight of fluid displaced by the object
and the objects have the same volume.
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50
Clicker Question: Focus on Concepts, Question 11
Two beakers are filled to the same level with water. One of them has a
plastic ball floating in the water. If the beakers are placed on a scale, one
at a time, which weighs more?
A) The beaker without the plastic ball weighs more.
B) The beaker with the plastic ball weighs more.
C) The beakers have the same weight.
C) Archimedes: the weight of the ball is the same as the weight of the
“missing” water displaced by the ball. So the ball could be replaced by the
missing water without changing the total weight and then the picture would
be as on the left.
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Fluids in Motion
Streamline (orderly, or steady) flow – the velocity of a fluid at
some point does not change with time. Streamlines show the path
of fluid particles. Streamlines do not cross.
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52
Fluids in Motion
Unsteady flow – the velocity at a point in the fluid changes with time.
Turbulent flow is an extreme case of unsteady flow for a fastmoving fluid – the velocity changes erratically from moment to
moment, as at sharp obstacles or bends.
Viscous flow – a type of friction impeding the relative motion of
layers of a fluid, as in molasses.
Bernoulli’s equation, to follow, applies to streamline flow.
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Freely Flowing Fluids: The Equation of
Continuity
!m2
!m1
Volume = A1v1!t
Volume = A2v2!t
(flowing out)
(flowing in)
What goes in must come out
That is:
Mass flowing in = mass flowing out (Δm2 = Δm1)
So, "2 A2v2Δt = "1 A1v1Δt
and "2 A2v2 = "1 A1v1
Equation of continuity
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The Equation of Continuity
ρ1 A1 v1 = ρ2 A2 v2
!m
= " A v = mass flow rate
!t
If the fluid is incompressible, "1 = "2,
and the volume flow rate (volume per second) is
Q = v1A1 = v2A2
V = vAΔt
A
v
vΔt
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11.-/50: Oil is flowing at a speed of 1.22 m/s through a pipeline
with a radius of 0.305 m. How much oil flows in 1 day?
V = vAΔt
A
v
vΔt
The volume rate of flow (volume per second) is Q = vA
A = πr2 = π ! 0.3052 = 0.2923 m2.
So, Q = (1.22 m/s)(0.2923 m2) = 0.3565 m3/s.
In 24 hours, the flow is 3.08!104 m3.
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11.58/52: The volume rate of flow in an artery supplying the brain
is 3.6!10-6 m3/s. If the radius of the artery is 5.2 mm, determine
the average blood speed.
Q
3.6 × 10−6 m3 /s
v=
=
= 0.0424 m/s
A
π × 0.00522
Find the average blood speed if a constriction reduces the radius
of the artery by a factor of 3 (without reducing the flow rate).
v = Q/A, and r is reduced to r/3,
so the speed is increased by a factor of 32 = 9.
So, v = 9 ! 0.0424 = 0.381 m/s
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Bernoulli’s Equation
Q = vA = constant
F2 = P2A2
F1 = P1A1
Fluid of
constant
density
The fluid speeds up when it gets to the constriction. What is the force
that causes this acceleration?
There must be a drop in pressure force that accelerates the fluid to
the right.
F2 > F1, that is, P2A2 > P1A1
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Bernoulli’s Equation
Q = vA = constant
F2 = P2A
F1 = P1A
Pipe of constant
cross-sectional area
The fluid flows up hill at constant speed through a pipe of constant
area. Where does the force come from to push the fluid up the hill at
constant speed?
There must be a decrease in pressure that pushes the fluid up the
hill.
P2 > P1
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Bernoulli’s Equation
m
F1 = P1A1
F2 = P2A2
m
Follow a fluid element of mass m through the pipe from region 2 at
the left to region 1 at the right.
Work Wnc done by the pressure forces increases the mechanical
energy of the fluid element.
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Bernoulli’s Equation
m
m
Work-energy equation:
Wnc = !PE + !KE
1
1
= (mgy1 − mgy2) + ( mv21 − mv22)
2
2
What is the work done by non-conservative forces?
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Bernoulli’s Equation
Volume V
m
The pressure difference
between the two ends of the
fluid element of mass m exerts
a net force on the mass –
(F + !F) − F = (P + !P)A − PA
So that, !F = !P × A
When the fluid element moves through its own
length, s, the net force does work on it:
!Wnc = !Fs = !P(A × s) = !P ×V
The work done by the whole pressure difference between the ends
of the pipe, P2 – P1 should then be:
Wnc = (P2 − P1) ×V
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Bernoulli’s Equation
Back to the work-energy equation: Wnc = !PE + !KE
1
1
Wnc = (P2 − P1) ×V = (mgy1 − mgy2) + ( mv21 − mv22)
2
2
Divide by V and use density " = m/V
1
1
P2 − P1 = (!gy1 − !gy2) + ( !v21 − !v22)
2
2
1
1
P1 + !gy1 + !v21 = P2 + !gy2 + !v22
2
2
Bernoulli’s Equation
• For streamline flow
• Streamlines form “virtual pipes”
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Bernoulli’s Equation
For streamline flow:
1
P + ρgy + ρv 2 = constant
2
If the speed of flow increases, then the pressure decreases
If the height increases, then the pressure decreases
When the speed of flow is constant, Bernoulli’s equation gives
same variation of pressure with depth as for a static fluid:
If v is constant, P2 = P1 + "g(y1 - y2)
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Bernoulli’s Equation
Q = vA = constant
F2 = P2A2
F1 = P1A1
Fluid of
constant
density
v increases, y is constant, so P must decrease
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Clicker Question: Focus on Concepts, Question 16
Blood flows through a section of a horizontal artery that is partially blocked by
a deposit along the artery wall. A hemoglobin molecule moves from the narrow
region into the wider region.
What happens to the pressure acting on the molecule as it moves from the
narrow to the wider region?
1
P + ρgy + ρv 2 = constant
2
A) The pressure decreases.
B) There is no change in the pressure.
C) The pressure increases.
C) The pressure increases because the speed decreases: P + "v2/2 = constant
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Clickers!
vA
vC
vB
Fluid is flowing from left to right through the pipe. Points A and B
are at the same height, but the cross-sectional areas of the pipe
differ. Points B and C are at different heights, but the crosssectional areas are the same.
Rank the pressures at the three points, from highest to lowest.
A) A and B (a tie), C
B) C, A and B (a tie)
C) B, C, A
D) C, B, A
E) A, B, C
Wednesday, November 18, 2009
vB > vA, what force speeds up the fluid?
C is higher than B, what force pushes
the fluid uphill?
E) PA > PB > PC
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Applications of Bernoulli’s Equation
Because of the enlargement of a blood vessel, the cross-sectional
area A1 of an aorta increases to A2 = 1.7A1. The speed of the
blood (" = 1060 kg/m3) through a normal portion of the aorta is v1
= 0.4 m/s. Assuming that the aorta is horizontal, find the amount
by which the pressure P2 in the enlarged region exceeds P1 in the
normal region.
First, what is the speed of the blood in the enlarged region?
Equation of continuity: v1 A1 = v2 A2
so, v2 =
v1 A1
0.4
=
= 0.235 m/s
A2
1.7
A1
v2
v1 = 0.4 m/s
P1
Bernoulli:
1
1
P1 + !v21 = P2 + !v22
2
2
A2 = 1.7A1
P2
(no change of height)
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1
1
P1 + !v21 = P2 + !v22
2
2
A1
A2 = 1.7A1
v1 = 0.4 m/s
1
So, P2 − P1 = ![v21 − v22]
2
v2 = 0.235 m/s
P1
P2
1
P2 − P1 = (1060 kg/m3)[0.42 − 0.2352] = 55 Pa
2
That is, the pressure is greater in the already weakened
enlarged section, putting greater stress on it.
The pressure must be larger because there has to be a
force that slows down the blood as it enters the enlarged
section.
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11.67/60: Water is circulating through a closed system of pipes in
a two-floor building.
On the first floor, the water has a gauge pressure of 3.4!105 Pa
and a speed of 2.1 m/s.
On the second floor, which is 4 m higher, the speed of the water
is 3.7 m/s. The speeds are different because the pipe diameters
are different.
What is the gauge pressure on the second floor?
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Bernoulli’s Equation
For streamline flow of an incompressible fluid (or close to
incompressible in some situations such as airflow over an airplane
wing)
1
P + !gy + !v2 = constant
2
Equation of continuity for fluid of constant density:
vA = constant
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Lift force on an airplane wing
Streamlines
Front of wing
Streamlines become
compressed over
curved top of wing.
Compare:
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11.63/59: An airplane wing is designed so that the speed of the air
across the top of the wing is 251 m/s when the speed of the air
below the wing is 225 m/s. The density of the air is 1.29 kg/m3.
Find the lift on a wing of area 24 m2.
Imagine streamlines with uniform air conditions in front of the
plane and that the streamlines divide and pass above and below the
wing.
1
1
1
P0 + !gy + !v20 = P1 + !gy + !v21 = P2 + !gy + !v22
2
2
2
In front of plane
Above wing
Below wing
1
1
So, P2 − P1 = !(v21 − v22) = × (1.29 kg/m3) × (2512 − 2252) = 7983 Pa
2
2
The net upward force is then: 7983xA = 7983!(24 m2) = 191,600 N
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View from above
Ball is moving to the right
Streamlines of air
moving relative to
the ball
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11.74/66: A liquid flows through a horizontal pipe of radius 0.02 m.
The pipe bends straight upward through a height of 10 m and joins
another horizontal pipe with radius 0.04 m.
What flow rate is needed to keep the pressures in the two
horizontal pipes the same?
h = 10 m
v1
v2
r2 = 0.04 m
r1 = 0.02 m
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