phys222 – lssu – bazlurslide 1 chapter - 11 fluids
TRANSCRIPT
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PHYS222 – LSSU – Bazlur Slide 1
Chapter - 11
Fluids
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PHYS222 – LSSU – Bazlur Slide 2
Fluids 11.1 Mass Density11.2 Pressure11.3 Pressure and Depth in a Static Fluid11.4 Pressure Guages11.5 Pascal’e Principle11.6 Archimedes’ Principle11.7 Fluid in Motion11.8 The Equation of Continuity11.9 Bernoulli’s Equation11.10 Applications of Bernoulli’s Equation11.11 Viscous Flow
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PHYS222 – LSSU – Bazlur Slide 3
11.1 Mass Density
Definition of Mass Density
The mass density of a substance is the mass of a substance per unit volume:
V
m
SI Unit of Mass Density: kg / m3
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PHYS222 – LSSU – Bazlur Slide 4
11.1 Mass Density
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PHYS222 – LSSU – Bazlur Slide 5
11.1 Mass Density
Example-1: Blood as a Fraction of Body Weight
The body of a man whose weight is 690 N contains about5.2x10-3 m3 of blood.
(a) Find the blood’s weight and (b) express it as a percentage of the body weight.
kg 5.5mkg1060m102.5 333 Vm
V
m
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PHYS222 – LSSU – Bazlur Slide 6
11.1 Mass Density
N 54sm80.9kg 5.5 2 mgW(a)
(b) %8.7%100N 690
N 54Percentage
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PHYS222 – LSSU – Bazlur Slide 7
11.1 Mass Density
Definition of Weight Density
The Weight density of a substance is the Weight of a substance per unit volume:
V
mg
SI Unit of Weight Density: N / m3
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PHYS222 – LSSU – Bazlur Slide 8
Specific Gravity• Is relative density.• It is a ratio of the densities of two materials. • Dimensionless.
• Using water as a reference can be convenient, since density of water is (approximately) 1000 kg/m³ or 1 g/cm³ at 4C.
• Specific gravity of Mercury is 13.6, means it is 13.6 times massive or heavier than water.
Reference ofDensity
Substance ofDensity Gravity Specific
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PHYS222 – LSSU – Bazlur Slide 9
11.2 Pressure
A
FP
SI Unit of Pressure: 1 N/m2 = 1Pa
Pascal
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PHYS222 – LSSU – Bazlur Slide 10
Pressure• Pressure is force per unit area.
P = F/A• Pressure is not a vector quantity. • To calculate the pressure we only consider the
magnitude of the force.• The force generated by the pressure of the static fluid
is always perpendicular to the surface of the contact.
Pa = N/m2 = 1.450 x 10-4 lb/in2 (very small quantity)bar = 105 Pa = 100 kPaAtmospheric pressure is 101.3 kPa
= 1.013 bar = 1.450 lb/in2 = 760 torrlb/in2 or psi = 6.895 x 103 Pa
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PHYS222 – LSSU – Bazlur Slide 11
Pressure causes Perpendicular ForceThe forces of a liquid pressing against a surface add up
to a net force that is perpendicular to the surface.
Components parallel to the surface cancels out
Components vertical to the surface adds up
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PHYS222 – LSSU – Bazlur Slide 12
11.2 Pressure
Example-2: The Force on a Swimmer
Suppose the pressure acting on the backof a swimmer’s hand is 1.2x105 Pa. Thesurface area of the back of the hand is 8.4x10-3 m2.
(a) Determine the magnitude of the forcethat acts on it.
(b) Discuss the direction of the force.
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PHYS222 – LSSU – Bazlur Slide 13
11.2 Pressure
A
FP
N 100.1
m104.8 mN102.13
2325
PAF
Since the water pushes perpendicularly against the back of the hand, the forceis directed downward in the drawing.
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PHYS222 – LSSU – Bazlur Slide 14
Shape of the Dam• Would you prefer Dam-A or Dam-B?• Justify your answer.
Dam-A Dam-B
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PHYS222 – LSSU – Bazlur Slide 15
11.2 Pressure
Atmospheric Pressure at Sea Level: 1.013x105 Pa = 1 atmosphere
Weight of the column of the air of 1m2 = 1.013x105 N
Mass of the column of the air of 1m2 = 1.013x105 N/g = 10326.2 kg
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PHYS222 – LSSU – Bazlur Slide 16
11.3 Pressure and Depth in a Static Fluid
012 mgAPAPFy
mgAPAP 12
Vm
Relation between Pressure and Depth
Since the column is in equilibrium,The sum of the vertical forces equal to zero
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PHYS222 – LSSU – Bazlur Slide 17
11.3 Pressure and Depth in a Static Fluid
VgAPAP 12
AhV
AhgAPAP 12
hgPP 12
The pressure increment = gh = weight density x heightPressure at a depth, h = P1 + gh = Atmospheric Pressure + Liquid Pressure
So, Liquid Pressure at a depth, h = gh = weight density x depth
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PHYS222 – LSSU – Bazlur Slide 18
PressureWater pressure acts perpendicular to the sides of a
container, and increases with increasing depth.
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PHYS222 – LSSU – Bazlur Slide 19
11.3 Pressure and Depth in a Static Fluid
Conceptual Example-3: The Hoover Dam
Lake Mead is the largest wholly artificial reservoir in the United States. The waterin the reservoir backs up behind the damfor a considerable distance (120 miles).
Suppose that all the water in Lake Meadwere removed except a relatively narrowvertical column.
Would the Hoover Dam still be neededto contain the water, or could a much lessmassive structure do the job?
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PHYS222 – LSSU – Bazlur Slide 20
11.3 Pressure and Depth in a Static Fluid
Example-4: The Swimming Hole
Points A and B are located a distance of 5.50 m beneath the surface of the water. Find the pressure at each of these two locations.
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PHYS222 – LSSU – Bazlur Slide 21
11.3 Pressure and Depth in a Static Fluid
Pa 1055.1
m 50.5sm80.9mkg1000.1Pa 1001.15
233
pressure catmospheri
52
P
ghPP 12
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PHYS222 – LSSU – Bazlur Slide 22
Where to put the pump?• Pump-X pushed the water up.• Pump-Y removes the air from the
pipe and air pressure outside pushes the water up.
• Pump-Y has a limit, what is that?
ghPatm
Pump-X Pump-Y
mm 10306m 306.10
sm80.9 mkg101.0
Pa 1001.1233
5
g
Ph atm
ghPatm
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PHYS222 – LSSU – Bazlur Slide 23
Pressure Gauges• One of the simplest pressure
gauges is the mercury barometer used for measuring atmospheric pressure.
• 760 mm of mercury
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PHYS222 – LSSU – Bazlur Slide 24
11.4 Pressure Gauges
ghPatm
mm 760m 760.0
sm80.9mkg1013.6
Pa 1001.1233
5
g
Ph atm
Point A and B are at the same depth, so,
PB = PA
Patm = P1 + gh
= 0 + gh
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PHYS222 – LSSU – Bazlur Slide 25
Open-tube manometer• The phrase “open-tube” refers to
the fact that one side of the U-tube is open to atmospheric pressure.
• The tube contains mercury• Its other side is connected to the
container whose pressure P2 is to be measured.
PB = PA
P2 = Patm + gh
P2 - Patm = gh
• P2 – Patm is called the guage pressure
• P2 is called the absolute pressure
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PHYS222 – LSSU – Bazlur Slide 26
11.4 Pressure Gauges
AB PPP 2
ghPPA 1
ghPP atm pressure gauge
2
absolute pressure
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PHYS222 – LSSU – Bazlur Slide 27
sphygmomanometer
blood-pressure instrument: an instrument used to measure blood pressure in an artery that consists of a pressure gauge, an inflatable cuff placed around the upper arm, and an inflator bulb or pressure pump.
Systolic pressure is 120 mm of mercury
Diastolic pressure is 80 mm of mercury
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PHYS222 – LSSU – Bazlur Slide 28
11.4 Pressure Gauges
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PHYS222 – LSSU – Bazlur Slide 29
11.5 Pascal’s Principle
Pascal’s Principle
Any change in the pressure applied to a completely enclosed fluid is transmitted undiminished to all parts of the fluid and enclosing walls.
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PHYS222 – LSSU – Bazlur Slide 30
11.5 Pascal’s Principle
m 012 gPP
1
1
2
2
A
F
A
F
1
212 A
AFF
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PHYS222 – LSSU – Bazlur Slide 31
11.5 Pascal’s Principle
Example-7: A Car Lift
The input piston has a radius of 0.0120 mand the output plunger has a radius of 0.150 m.
The combined weight of the car and the plunger is 20500 N. Suppose that the inputpiston has a negligible weight and the bottomsurfaces of the piston and plunger are atthe same level. What is the required inputforce?
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PHYS222 – LSSU – Bazlur Slide 32
11.5 Pascal’s Principle
N 131m 150.0
m 0120.0N 20500 2
2
2
F
1
212 A
AFF
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PHYS222 – LSSU – Bazlur Slide 33
11.6 Archimedes’ Principle
APPAPAPFB 1212
ghPP 12
ghAFB
hAV
gVFB
fluiddisplaced
of mass
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PHYS222 – LSSU – Bazlur Slide 34
11.6 Archimedes’ Principle
Archimedes’ Principle
Any fluid applies a buoyant force to an object that is partiallyor completely immersed in it; the magnitude of the buoyantforce equals the weight of the fluid that the object displaces:
fluid displaced
ofWeight
fluid
forcebuoyant of Magnitude
WFB
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PHYS222 – LSSU – Bazlur Slide 35
11.6 Archimedes’ Principle
If the object is floating then the magnitude of the buoyant forceis equal to the magnitude of itsweight.
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PHYS222 – LSSU – Bazlur Slide 36
11.6 Archimedes’ Principle
Example-9: A Swimming Raft
The raft is made of solid squarepinewood. Determine whetherthe raft floats in water and ifso, how much of the raft is beneaththe surface.
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PHYS222 – LSSU – Bazlur Slide 37
11.6 Archimedes’ Principle
N 47000
sm80.9 m 8.4 mkg1000 233
max
gVVgF waterwaterB
m 8.4m 30.0m 0.4m 0.4 raftV
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PHYS222 – LSSU – Bazlur Slide 38
11.6 Archimedes’ Principle
N 47000
N 26000
sm80.9 m 8.4 mkg550 233
gVgmW raftpineraftraft
The raft floats!
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PHYS222 – LSSU – Bazlur Slide 39
11.6 Archimedes’ Principle
gVwaterwaterN 26000
Braft FW
If the raft is floating:
23 sm80.9 m 0.4 m 0.4 mkg1000N 26000 h
m 17.0sm80.9 m 0.4 m 0.4 mkg1000
N 2600023
h
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PHYS222 – LSSU – Bazlur Slide 40
11.6 Archimedes’ Principle
Conceptual Example-10: How Much Water is Neededto Float a Ship?
A ship floating in the ocean is a familiar sight. But is allthat water really necessary? Can an ocean vessel floatin the amount of water than a swimming pool contains?
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PHYS222 – LSSU – Bazlur Slide 41
11.6 Archimedes’ Principle
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PHYS222 – LSSU – Bazlur Slide 42
11.7 Fluids in Motion
In steady flow the velocity of the fluid particles at any point is constant as time passes.
Unsteady flow exists whenever the velocity of the fluid particles at a point changes as time passes.
Turbulent flow is an extreme kind of unsteady flow in which the velocity of the fluid particles at a point change erratically in both magnitude and direction.
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PHYS222 – LSSU – Bazlur Slide 43
11.7 Fluids in Motion
Fluid flow can be compressible or incompressible. Most liquids are nearly incompressible.
Fluid flow can be viscous or nonviscous.
An incompressible, nonviscous fluid is called an ideal fluid.
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PHYS222 – LSSU – Bazlur Slide 44
11.7 Fluids in Motion
When the flow is steady, streamlines are often used to representthe trajectories of the fluid particles.
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PHYS222 – LSSU – Bazlur Slide 45
11.7 Fluids in Motion
Making streamlines with dyeand smoke.
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PHYS222 – LSSU – Bazlur Slide 46
11.8 The Equation of Continuity
The mass of fluid per second that flows through a tube is calledthe mass flow rate.
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PHYS222 – LSSU – Bazlur Slide 47
11.8 The Equation of Continuity
2222 vAt
m
111
1 vAt
m
distance
tvAVm
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PHYS222 – LSSU – Bazlur Slide 48
11.8 The Equation of Continuity
222111 vAvA
Equation Of Continuity
The mass flow rate has the same value at every position along a tube that has a single entry and a single exit for fluid flow.
SI Unit of Mass Flow Rate: kg/s
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PHYS222 – LSSU – Bazlur Slide 49
11.8 The Equation of Continuity
Incompressible fluid: 2211 vAvA
Volume flow rate Q: AvQ
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PHYS222 – LSSU – Bazlur Slide 50
11.8 The Equation of Continuity
Example-12: A Garden Hose
A garden hose has an unobstructed openingwith a cross sectional area of 2.85x10-4m2. It fills a bucket with a volume of 8.00x10-3m3
in 30 seconds.
Find the speed of the water that leaves the hosethrough (a) the unobstructed opening and (b) an obstructedopening with half as much area.
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PHYS222 – LSSU – Bazlur Slide 51
11.8 The Equation of Continuity
AvQ
sm936.0
m102.85
s 30.0m1000.824-
33
A
Qv
(a)
(b) 2211 vAvA
sm87.1sm936.0212
12 v
A
Av
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PHYS222 – LSSU – Bazlur Slide 52
Bernoulli’s EquationFor steady flow,
The speed,
Pressure, and
Elevation
of an incompressible and nonviscous fluid are related by Bernoulli’s equation.
2222
121
212
11 gyvPgyvP
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PHYS222 – LSSU – Bazlur Slide 53
11.9 Bernoulli’s Equation
The fluid accelerates toward the lower pressure regions.
According to the pressure-depthrelationship, the pressure is lowerat higher levels, provided the areaof the pipe does not change.
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PHYS222 – LSSU – Bazlur Slide 54
Bernoulli’s Equation• Bernoulli’s Equation from Work-energy theorem:• Work done on an object by external non-conservative forces is
equal to the change in total mechanical energy.• The pressure within a fluid is caused by collisional forces, which
are non-conservative.• Therefore, when a fluid is accelerated because of a difference
in pressures, work is being done by non-conservative forces.
2222
11
212
1
21
0fnc
mgymvmgymv
EE
EEW
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PHYS222 – LSSU – Bazlur Slide 55
11.9 Bernoulli’s Equation
2222
11
212
1nc mgymvmgymvW
VPPAsPsFsFW 12
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PHYS222 – LSSU – Bazlur Slide 56
Bernoulli’s Equation• Whenever a fluid is flowing in a horizontal pipe and encounters
a region of reduced cross-sectional area, the pressure of the fluid drops.
• Can be explained by the Newton’s 2nd law.• When moving from the wider region 2 to the narrower region 1,
the fluid speeds up according to the equation of continuity.• According to the 2nd law, th eaccelerating fluid must be
subjected to an unbalanced force.• There can be an unbalanced force only if the pressure in region
2 is higher than the pressure in region 1.
2222
11
212
1
21
0fnc
mgymvmgymv
EE
EEW
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PHYS222 – LSSU – Bazlur Slide 57
Bernoulli’s Equation• Explanation:• In figure-b on the top surface of the blue fluid element, the
surrounding fluid exerts a pressure P.• This pressure give rise to a force, F = P/A• On the botom of the blue fluid element, the pressure is slightly
higher, P+P.• As a result, the force on the bottom surface has a magnitude of
F+ F = (P+P)/A.• The net force pushing the fluid element up the pipe is F=P/A
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PHYS222 – LSSU – Bazlur Slide 58
Bernoulli’s Equation• When the fluid element moves through its own length s, the
work done is the
W = F s = P A s = P V
The total work done on the fluid element in moving it from region 2 to region 1is the sum of the small increments of work P V done as the element moves along the pipe.
This sum amounts to Wnc = (P2 – P1)V
Wnc = (P2 – P1)V = E1 – E2
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PHYS222 – LSSU – Bazlur Slide 59
11.9 Bernoulli’s Equation
2222
11
212
112 mgymvmgymvVPP
2222
11
212
112 gyvgyvPP
Bernoulli’s Equation
In steady flow of a nonviscous, incompressible fluid, the pressure, the fluid speed, and the elevation at two points are related by:
2222
121
212
11 gyvPgyvP
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PHYS222 – LSSU – Bazlur Slide 60
Bernoulli’s EquationThe term
has a constant value at all positions in the flow.
constant221 gyvP
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PHYS222 – LSSU – Bazlur Slide 61
Bernoulli’s EquationBernoulli’s Equation
reduces to the result for static fluids (v1=v2), as it is when the cross-sectional area remains constant.
2222
121
212
11 gyvPgyvP
ghPP
yygPP
gygyPP
gyPgyP
12
2112
2112
2211
)(
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PHYS222 – LSSU – Bazlur Slide 62
11.10 Applications of Bernoulli’s Equation
Conceptual Example-14: Tarpaulins and Bernoulli’s Equation
When the truck is stationary, the tarpaulin lies flat, but it bulges outwardwhen the truck is speeding downthe highway.
Account for this behavior.
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PHYS222 – LSSU – Bazlur Slide 63
Bernoulli’s EquationBernoulli’s Equation
reduces to
When all parts have the same elevation (y1 = y2).
The term P+ ½ remains constant throughout a horizontal pipe.
If increases, P decreases and vice versa.
2222
121
212
11 gyvPgyvP
222
12
212
11 vPvP
constant221 vP
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PHYS222 – LSSU – Bazlur Slide 64
11.10 Applications of Bernoulli’s Equation
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PHYS222 – LSSU – Bazlur Slide 65
11.10 Applications of Bernoulli’s Equation
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PHYS222 – LSSU – Bazlur Slide 66
11.10 Applications of Bernoulli’s Equation
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PHYS222 – LSSU – Bazlur Slide 67
11.10 Applications of Bernoulli’s Equation
Example-16: Efflux Speed
The tank is open to the atmosphere atthe top. Find and expression for the speed of the liquid leaving the pipe atthe bottom.
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PHYS222 – LSSU – Bazlur Slide 68
11.10 Applications of Bernoulli’s Equation
2222
121
212
11 gyvPgyvP
atmPPP 2102 v
hyy 12
ghv 212
1
ghv 21
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PHYS222 – LSSU – Bazlur Slide 69
11.11 Viscous Flow
Flow of an ideal fluid.
Flow of a viscous fluid.
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PHYS222 – LSSU – Bazlur Slide 70
11.11 Viscous Flow
Force Needed to Move a Layer of Viscous Fluid withConstant Velocity
The magnitude of the tangential force required to move a fluid layer at a constant speed is given by:
y
AvF
coefficient of viscosity
SI Unit of Viscosity: Pa·s
Common Unit of Viscosity: poise (P)
1 poise (P) = 0.1 Pa·s
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PHYS222 – LSSU – Bazlur Slide 71
11.11 Viscous Flow
Poiseuille’s Law
The volume flow rate is given by: L
PPRQ
812
4
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PHYS222 – LSSU – Bazlur Slide 72
11.11 Viscous Flow
Example-17: Giving and Injection
A syringe is filled with a solution whose viscosity is 1.5x10-3 Pa·s. The internalradius of the needle is 4.0x10-4m.
The gauge pressure in the vein is 1900 Pa.What force must be applied to the plunger,so that 1.0x10-6m3 of fluid can be injected in 3.0 s?
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PHYS222 – LSSU – Bazlur Slide 73
11.11 Viscous Flow
Pa 1200
m104.0
s 0.3m100.1m 025.0sPa105.18
8
44-
363
42
R
LQPP
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PHYS222 – LSSU – Bazlur Slide 74
11.11 Viscous Flow
Pa 120012 PP
Pa 19001 P
Pa 31002 P
N25.0m100.8Pa 3100 252 APF
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PHYS222 – LSSU – Bazlur Slide 75
ForceA
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PHYS222 – LSSU – Bazlur Slide 76
ForceA
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PHYS222 – LSSU – Bazlur Slide 77
ForceA
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PHYS222 – LSSU – Bazlur Slide 78
ForceA