Upload
gunnar-sandifer
View
230
Download
3
Embed Size (px)
Citation preview
Chapter 9
Fluids
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 2
Chapter 9: Fluids• Introduction to Fluids
• Pressure
• Measurement of Pressure
• Pascal’s Principle
• Gravity and Fluid Pressure
• Archimedes’ Principle
• Continuity Equation
• Bernoulli’s Equation
• Viscosity and Viscous Drag
• Surface Tension
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 3
Pressure
Pressure arises from the collisions between the particles of a fluid with another object (container walls for example).
There is a momentum change (impulse) that is away from the container walls. There must be a force exerted on the
particle by the wall.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 4
Pressure is defined as .A
FP
The units of pressure are N/m2 and are called Pascals (Pa).
Note: 1 atmosphere (atm) = 101.3 kPa
By Newton’s 3rd Law, there is a force on the wall due to the particle.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 5
Example (text problem 9.1): Someone steps on your toe, exerting a force of 500 N on an area of 1.0 cm2. What is the average pressure on that area in atmospheres?
atm 49
Pa 10013.1
atm 1
N/m 1
Pa 1N/m 100.5
m 101.0
N 500
5226
24av
A
FP
242
2 m 100.1cm 100
m 1cm 0.1
A 500N person
weighs about 113 lbs.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 6
Gravity’s Effect on Fluid Pressure
An imaginary cylinder of fluid
FBD for the fluid cylinder
P1A
P2Aw
x
y
Imaginary cylinder can be any size
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 7
Apply Newton’s 2nd Law to the fluid cylinder. Since the fluids isn’t moving the net force is zero.
gdPP
gdPP
gdPP
gAdAPAP
wAPAPF
12
12
12
12
12
or
0
0
0
If P1 (the pressure at the top of the cylinder) is known, then the above expression can be used to find the variation of pressure with depth in a fluid.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 8
If the top of the fluid column is placed at the surface of the fluid, then P1 = Patm if the container is open.
gdPP atm
You noticed on the previous slide that the areas canceled out. Only the height matters since that is the direction of gravity.
Think of the pressure as a force density in N/m2
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 9
Example (text problem 9.19): At the surface of a freshwater lake, the pressure is 105 kPa. (a) What is the pressure increase in going 35.0 m below the surface?
atm 3.4kPa 343
m 35m/s 8.9kg/m 1000
23
atm
atm
gdPPP
gdPP
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 10
Example: The surface pressure on the planet Venus is 95 atm. How far below the surface of the ocean on Earth do you need to be to experience the same pressure? The density of seawater is
1025 kg/m3.
m 950
N/m 109.5m/s 8.9kg/m 1025
N/m 109.5atm 94
atm 1atm 95
2623
26
atm
d
d
gd
gd
gdPP
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 11
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 12
Measuring Pressure
A manometer is a U-shaped tube that is partially filled with liquid, usually Mercury (Hg).
Both ends of the tube are open to the atmosphere.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 13
A container of gas is connected to one end of the U-tube
If there is a pressure difference between the gas and the atmosphere, a force will be exerted on the fluid in the U-tube. This changes the equilibrium position of the fluid in the tube.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 14
Also
atmc PP At point C
B'B PP
The pressure at point B is the pressure of the gas.
gdP
gdPPPP
gdPPP
BCB
CBB
gauge
atm
'
From the figure:
gauge meas atmP = P - P
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 15
A Barometer
The atmosphere pushes on the container of mercury which forces mercury up the closed, inverted tube. The distance d is called the barometric pressure.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 16
Atmospheric pressure is equivalent to a column of mercury 76.0 cm tall.
gdPA
From the figure atmBA PPP
and
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 17
The Many Units of Pressure
1 ATM equals 1.013x105 N/m2
14.7 lbs/in2
1.013 bar
76 cm Hg
760 mm Hg
760 Torr
34 ft H2O
29.9 in Hg
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 18
Pascal’s Principle
A change in pressure at any point in a confined fluid is transmitted everywhere throughout the fluid. (This is useful in making a hydraulic lift.)
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 19
Apply a force F1 here to a piston of cross-sectional area A1.
The applied force is transmitted to the piston of cross-sectional area A2 here.
In these problems neglect pressure due to columns of fluid.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 20
Mathematically,
11
22
2
2
1
1
A
A
A
F
A
F
2point at 1point at
FF
PP
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 21
Example: Assume that a force of 500 N (about 110 lbs) is applied to the smaller piston in the previous figure. For each case, compute the force on the larger piston if the ratio of the piston areas (A2/A1)
are 1, 10, and 100.
50,000 N100
5000 N10
500 N1
F212 AA
Using Pascal’s Principle:
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 22
Archimedes’ Principle
An FBD for an object floating submerged in a fluid.
The total force on the block due to
the fluid is called the buoyant force.
12
12
where FF
FFF
B
wF2
F1
x
y
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 23
The magnitude of the buoyant force is:
APP
APAP
FFFB
12
12
12
gdPP 12From before:
gVgdAFB The result is
Buoyant force = the weight of the fluid displaced
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 24
gVFB
Archimedes’ Principle: A fluid exerts an upward buoyant force on a submerged object equal in magnitude to the weight of the
volume of fluid displaced by the object.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 25
Example (text problem 9.28): A flat-bottomed barge loaded with coal has a mass of 3.0105 kg. The barge is 20.0 m long and 10.0 m wide. It floats in fresh water. What is the depth of the barge below the waterline?
x
y
w
FB
FBD for the barge
Apply Newton’s 2nd Law to the barge:
bw
bww
bwww
B
B
mAd
mV
gmgVgm
wF
wFF
0
m 5.1m 10.0*m 0.20kg/m 1000
kg 100.33
5
A
md
w
b
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 26
Example (text problem 9.40): A piece of metal is released under water. The volume of the metal is 50.0 cm3 and its specific gravity is 5.0. What is its initial acceleration? (Note: when v = 0, there is no drag force.)
FBD for the metal mawFF B
VgFB water
The magnitude of the buoyant force equals the weight of the fluid displaced by the metal.
Solve for a:
1
ρ
ρ
ρ
ρ
objectobject
water
objectobject
water
V
Vgg
V
Vgg
m
Fa B
Apply Newton’s 2nd Law to the piece of metal:
x
y
w
FB
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 27
Since the object is completely submerged V=Vobject.
water
gravity specific
where water = 1000 kg/m3 is the density of water at 4 °C.
Given 0.5gravity specificwater
object
2
objectobject
water m/s 8.710.5
11
..
11
ρ
ρ
g
GSg
V
Vga
Example continued:
The sign is minus because gravity acts down. BF causes a < g.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 28
Fluid Flow
A moving fluid will exert forces parallel to the surface over which it moves, unlike a static fluid. This gives rise to a viscous force that impedes the forward motion of the fluid.
A steady flow is one where the velocity at a given point in a fluid is constant.
V1 = constant
V2 = constant
v1v2
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 29
Steady flow is laminar; the fluid flows in layers. The path that the fluid in these layers takes is called a streamline.
An ideal fluid is incompressible, undergoes laminar flow, and has no viscosity.
Streamlines do not cross.
Crossing streamlines would indicate a volume of fluid with two different velocities at the same time.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 30
The Continuity Equation—Conservation of Mass
The amount of mass that flows though the cross-sectional area A1
is the same as the mass that flows through cross-sectional area A2.
Faster Slower
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 31
is the mass flow rate (units kg/s)Avt
m
Avt
V
is the volume flow rate (units m3/s)
222111 vAvA
The continuity equation is
If the fluid is incompressible, then 1= 2.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 32
Example (text problem 9.41): A garden hose of inner radius 1.0 cm carries water at 2.0 m/s. The nozzle at the end has radius 0.20 cm. How fast does the water move through the constriction?
m/s 50m/s 0.2cm 0.20
cm 0.12
122
21
12
12
2211
vr
rv
A
Av
vAvA
Simple ratios
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 33
Bernoulli’s Equation
Bernoulli’s equation is a statement of energy conservation.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 34
2222
2111 2
1
2
1vgyPvgyP
Potential energy per unit volume
Kinetic energy per unit volume
Work per unit volume done by the fluid
Points 1 and 2 must be on the same streamline
This is the most general equation
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 35
Example (text problem 9.49): A nozzle is connected to a horizontal hose. The nozzle shoots out water moving at 25 m/s. What is the gauge pressure of the water in the hose? Neglect viscosity and assume that the diameter of the nozzle is much smaller than the inner diameter of the hose.
2222
2111 2
1
2
1vgyPvgyP
Let point 1 be inside the hose and point 2 be outside the nozzle.
The hose is horizontal so y1 = y2. Also P2 = Patm.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 36
21
22atm1
22atm
211
2
1
2
12
1
2
1
vvPP
vPvP
Substituting:
v2 = 25 m/s and v1 is unknown. Use the continuity equation.
2
2
1
222
1
2
2
21
21
2
2v
d
dv
d
d
vA
Av
Since d2<<d1 it is true that v1<<v2.
Example continued:
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 37
Pa 101.3
m/s 25kg/m 10002
12
1
2
12
1
2
1
5
23
22
21
22
21
22atm1
vvv
vvPP
Example continued:
1Since v 0
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 38
Viscosity
A real fluid has viscosity (fluid friction). This implies a pressure difference needs to be maintained across the ends of a pipe for fluid to flow.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 39
Viscosity also causes the existence of a velocity gradient across a pipe. A fluid flows more rapidly in the center of the pipe and more slowly closer to the walls of the pipe.
The volume flow rate for laminar flow of a viscous fluid is given by Poiseuille’s Law.
4
8r
LP
t
V
where is the viscosity
4th power
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 40
Example (text problem 9.55): A hypodermic syringe attached to a needle has an internal radius of 0.300 mm and a length of 3.00 cm. The needle is filled with a solution of viscosity 2.0010-3 Pa sec; it is injected into a vein at a gauge pressure of 16.0 mm Hg. Neglect the extra pressure required to accelerate the fluid from the syringe into the entrance needle.
(a) What must the pressure of the fluid in the syringe be in order to inject the solution at a rate of 0.250 mL/sec?
Solve Poiseuille’s Law for the pressure difference:
Pa 4716
seccm 250.0cm 103.0
cm 00.3sec Pa 1000.288 341
3
4
t
V
r
LP
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 41
Example continued:
This pressure difference is between the fluid in the syringe
and the fluid in the vein. The pressure in the syringe is
Pa 6860Pa 4720Pa 2140
PPP
PPP
vs
vs
51.013x10 Pa16 (mm Hg)× = 2132 Pa
760 mm Hg
Conversion:
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 42
(b) What force must be applied to the plunger, which has an area of 1.00 cm2?
Example continued:
The result of (a) gives the force per unit area on the plunger so the force is just F = PA = 0.686 N.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 43
Viscous Drag
The viscous drag force on a sphere is given by Stokes’ law.
rvFD 6
Where is the viscosity of the fluid that the sphere is falling through, r is the radius of the sphere, and v is the velocity of the sphere.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 44
Example (text problem 9.62): A sphere of radius 1.0 cm is dropped into a glass cylinder filled with a viscous liquid. The mass of the sphere is 12.0 g and the density of the liquid is 1200 kg/m3. The sphere reaches a terminal speed of 0.15 m/s. What is the viscosity
of the liquid?
FBD for sphere mawFFF BD FD
wx
y
FB
Apply Newton’s Second Law to the sphere
Drag Buoyant
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 45
Example continued:
When v = vterminal , a = 0 and
sec Pa 4.26
06
06
06
0
t
sls
sslt
sllt
slt
BD
rv
gVgm
gmgVrv
gmgVrv
gmgmrv
wFF
Solving for
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 46
Surface Tension
The surface of a fluid acts like a stretched membrane (imagine standing on a trampoline). There is a force along the surface of the fluid.
The surface tension is a force per unit length.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 47
Example (text problem 9.70): Assume a water strider has a roughly circular foot of radius 0.02 mm. The water strider has 6 legs.
(a) What is the maximum possible upward force on the foot due to the surface tension of the water?
The water strider will be able to walk on water if the net upward force exerted by the water equals the weight of the insect. The upward force is supplied by
the water’s surface tension.
N 1092 62
r
rPAF
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 48
(b) What is the maximum mass of this water strider so that it can keep from breaking through the water surface?
Example continued:
To be in equilibrium, each leg must support one-sixth the weight of the insect.
kg 1056
or 6
1 6g
FmwF
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 49
Summary
•Pressure and its Variation with Depth
•Pascal’s Principle
•Archimedes Principle
•Continuity Equation (conservation of mass)
•Bernoulli’s Equation (conservation of energy)
•Viscosity and Viscous Drag
•Surface Tension
Quick Questions
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 51
Consider a boat loaded with scrap iron in a swimming pool. If the iron is thrown overboard into the pool, will the water level at the edge of the pool rise, fall, or remain unchanged?
1. Rise2. Fall3. Remain unchanged
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 52
1. Rise2. Fall3. Remain unchanged
Consider a boat loaded with scrap iron in a swimming pool. If the iron is thrown overboard into the pool, will the water level at the edge of the pool rise, fall, or remain unchanged?
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 53
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 54
In the presence of air, the small iron ball and large plastic ball balance each other. When air is evacuated from the container, the larger ball
1. rises.2. falls.3. remains in place.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 55
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 56
The weight of the stand and suspended solid iron ball is equal to the weight of the container of water as shown above. When the ball is lowered into the water the balance is upset. The amount of weight that must be added to the left side to restore balance, compared to the weight of water displaced by the ball, would be
1. half. 2. the same.3. twice. 4. more than twice.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 57
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 58
A pair of identical balloons are inflated with air and suspended on theends of a stick that is horizontally balanced. When the balloon on the left is punctured, the balance of the stick is
1. upset and the stick rotates clockwise.
2. upset and the stick rotates counter-clockwise.
3. unchanged.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 59
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 60
Consider an air-filled balloon weighted so that it is on the verge of sinking—that is, its overall density just equals that of water.Now if you push it beneath the surface, it will
1. sink.2. return to the surface.3. stay at the depth to
which it is pushed.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 61
Extras
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 63
The density of the block of wood floating in water is
1. greater than the density of water.2. equal to the density of water.3. less than half that of water.4. more than half the density of water. 5. … not enough information is given.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 64
1. greater than the density of water.2. equal to the density of water.3. less than half that of water.4. more than half the density of water. 5. … not enough information is given.
The density of the block of wood floating in water is
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 65
1. rises.2. falls.3. remains in place.
In the presence of air, the small iron ball and large plastic ball balance each other. When air is evacuated from the container, the larger ball
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 66
1. half. 2. the same.3. twice. 4. more than twice.
The weight of the stand and suspended solid iron ball is equal to the weight of the container of water as shown above. When the ball is lowered into the water the balance is upset. The amount of weight that must be added to the left side to restore balance, compared to the weight of water displaced by the ball, would be
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 67
1. upset and the stick rotates clockwise.
2. upset and the stick rotates counter-clockwise.
3. unchanged.
A pair of identical balloons are inflated with air and suspended on theends of a stick that is horizontally balanced. When the balloon on the left is punctured, the balance of the stick is
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 68
1. sink.2. return to the surface.3. stay at the depth to
which it is pushed.
Consider an air-filled balloon weighted so that it is on the verge of sinking—that is, its overall density just equals that of water.Now if you push it beneath the surface, it will