College Physics 140 Chapter 9: Fluids
We will be investigating the movement and dynamics of fluids!
Chapter 9: Fluids
• Introduction to Fluids
• Pressure
• Pascal’s Principle
• Gravity and Fluid Pressure
• Measurement of Pressure
• Archimedes’ Principle
• Continuity Equation
• Bernoulli’s Equation
• Viscosity and Viscous Drag
• Surface Tension Fluids
A liquid will flow to take the shape of the container that holds it. A gas will completely fill its container.
Fluids are easily deformable by external forces.
A liquid is incompressible. Its volume is fixed and is impossible to change.
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. By Newton’s 3rd Law, there is a force on the wall due to the particle.
Pressure is defined as P
The units of pressure are N/m2 and are called Pascals (Pa).
Note: 1 atmosphere (atm) = 101.3 kPa F
= .
A
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?
2
$
2! 1 m
−4
2
1.0 cm #
& = 1.0 ×10 m
" 100 cm %
F
500 N
=
A 1.0 ×10 −4 m 2
&
1 atm
6
2 # 1 Pa & #
= 5.0 ×10 N/m %
(
2 (%
5
$ 1 N/m '$ 1.013×10 Pa '
= 49 atm
Pav =
Pascal’s Principle
A change in pressure at any point in a confined fluid is transmi_ed everywhere throughout the fluid. (This is useful in making a hydraulic lift.)
The applied force is transmi_ed to the piston of cross-­‐‑sectional area A2 here.
Mathematically, ΔP at point 1 = ΔP at point 2
F1 F2
=
A1 A 2
" A2 %
F2 = $ ' F1
# A1 &
Apply a force F1 here to a piston of cross-­‐‑
sectional area A1.
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. Using Pascal’s Principle:
A2 A1
1
10
100
F2
500 N
5000 N
50,000 N
The work done pressing the smaller piston (1) equals the work done by the larger piston (2).
F1d1 = F2 d2
Using an hydraulic lift reduces the amount of force needed to lift a load, but the work done is the same.
Example: In the previous example, for the case A2/A1 = 10, it was found that F2/
F1 = 10. If the larger piston needs to rise by 1 m, how far must the smaller piston be depressed?
Using the result on the previous slide,
F2
d1 = d2 = 10 m
F1
Gravity’s Effect on Fluid Pressure
FBD for the fluid cylinder
y
A cylinder of fluid
P1A
x
w
Apply Newton’s 2nd Law to the fluid cylinder:
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.
P2A
∑F = P A− PA− w = 0
2
1
P2 A − P1 A − (ρAd )g = 0
P2 − P1 − ρgd = 0
∴ P2 − P1 = ρgd
or P2 = P1 + ρgd
Gravity’s Effect on Fluid Pressure
FBD for the fluid cylinder
y
A cylinder of fluid
P1A
x
w
P2A
If the top of the fluid column is placed at the surface of the fluid, then P1 = Patm if the container is open.
P = Patm + ρ gd
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?
P = Patm + ρ gd
ΔP = P − Patm = ρ gd
= (1000 kg/m 3 ) ( 9.8 m/s2 ) (35 m )
= 343 kPa = 3.4 atm
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. P = Patm + ρ gd
95 atm = 1 atm + ρ gd
ρ gd = 94 atm = 9.5 ×10 6 N/m 2
(1025 kg/m ) (9.8 m/s ) d = 9.5 ×10
3
2
d = 950 m
6
N/m
2
Measuring Pressure
A manometer is a U-­‐‑
shaped tube that is partially filled with liquid.
Both ends of the tube are open to the atmosphere.
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.
From the figure: At point C
Also
Pc = Patm
PB = PB'
The pressure at point B is the pressure of the gas.
PB = PB' = PC + ρ gd
PB − PC = PB − Patm = ρ gd
Pgauge = ρ gd
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.
From the figure
and
PA = PB = Patm
PA = ρ gd
Atmospheric pressure is equivalent to a column of mercury 76.0 cm tall.
Example (text problem 9.22): An IV is connected to a patient’s vein. The blood pressure in the vein has a gauge pressure of 12 mm of mercury. At least how far above the vein must the IV bag be placed in order for fluid to flow into the vein? Assume that the density of the IV fluid is the same as blood. Blood:
Pgauge = ρ Hg gh1
h1 = 12 mm
IV:
The pressure is equivalent to raising a column of mercury 12 mm tall.
Pgauge = ρ blood gh2
At a minimum, the gauge pressures must be equal. When h2 is large enough, fluid will flow from high pressure to low pressure.
Pgauge = ρ Hg gh1 = ρ blood gh2
ρ Hg gh1
h2 =
ρ blood g
! ρ Hg $
! 13, 600 kg/m 3 $
=#
12 mm )
& h1 = #
3 &(
" 1060 kg/m %
" ρ blood %
= 154 mm
Archimedes’ Principle
An FBD for an object floating submerged in a fluid.
y F 1
The total force on the block due to the fluid is called the buoyant force. FB = F2 − F1
x
where F2 > F1
w
The magnitude of the buoyant force is:F
B
= F2 − F1
= P2 A − P1 A
From before:
P2 − P1 = ρ gd
The result is
FB = ρ gdA = ρ gV
= ( P2 − P1 ) A
F2
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. FB = ρ gV
Example (text problem 9.28): A flat-­‐‑bo_omed barge loaded with coal has a mass of 3.0×105 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?
y
FBD for the barge
Apply Newton’s 2nd Law to the barge:
∑F = F
FB
B
−w=0
FB = w
x
w
mw g = ( ρ wVw ) g = mb g
ρ wVw = mb
ρ w ( Ad ) = mb
mb
3.0 ×10 5 kg
d=
=
= 1.5 m
3
ρ w A (1000 kg/m ) ( 20.0 m*10.0 m )
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.)
y
FBD for the metal
Apply Newton’s 2nd Law to the piece of metal:
FB
∑F = F
B
x
w
Solve for a:
− w = ma
The magnitude of the buoyant force equals the weight of the fluid displaced by the metal.
FB = ρ waterVg
⎛ ρ waterV
⎞
ρ waterVg
FB
a=
−g =
− g = g ⎜
− 1⎟
⎜ ρ V
⎟
m
ρobjectVobject
object
object
⎝
⎠
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.)
y
FBD for the metal
Since the object is completely submerged V=Vobject.
FB
specific gravity =
x
w
ρ
ρ water
where ρwater = 1000 kg/m3 is the density of water at 4 °C.
ρ object
Given specific gravity =
= 5.0
ρ water
⎛ ρ waterV
⎞
⎛ 1
⎞
⎛ 1
⎞
⎜
⎟
a=g
− 1 = g ⎜
− 1⎟ = g ⎜
− 1⎟ = −7.8 m/s2
⎜ ρ V
⎟
⎝ S .G. ⎠
⎝ 5.0 ⎠
object
object
⎝
⎠
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
v1≠v2
Steady flow is laminar; the fluid flows in layers. The path that the fluid in these layers takes is called a streamline. Streamlines do not cross.
An ideal fluid is incompressible, undergoes laminar flow, and has no viscosity.
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. Δm
= ρAv
Δt
ΔV
= Av
Δt
is the mass flow rate (units kg/s)
is the volume flow rate (units m3/s)
ρ1 A1v1 = ρ2 A2v2
The continuity equation is
If the fluid is incompressible, then ρ1= ρ2.
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?
A1v1 = A2 v2
! A1 $
! π r12 $
v2 = # & v1 = # 2 & v1
" A2 %
" π r2 %
2
! 1.0 cm $
=#
& ( 2.0 m/s) = 50 m/s
" 0.20 cm %
Bernoulli’s Equation
Bernoulli’s equation is a statement of energy conservation.
1 2
1 2
P1 + ρ gy1 + ρ v1 = P2 + ρ gy2 + ρ v2
2
2
Work per unit volume done by the fluid
Potential energy per unit volume
Kinetic energy per unit volume
Points 1 and 2 must be on the same streamline
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.
Let point 1 be inside the hose and point 2 be outside the nozzle.
1 2
1 2
P1 + ρgy1 + ρv1 = P2 + ρgy 2 + ρv2
2
2
The hose is horizontal so y1 = y2. Also P2 = Patm.
Substituting:
1 2
1 2
P1 + ρ v1 = Patm + ρ v2
2
2
1 2 1 2
P1 − Patm = ρ v2 − ρ v1
2
2
v2 = 25 m/s and v1 is unknown. Use the continuity equation. ! ! d $2 $
2
2
#
π
#
& &
! A2 $
!
$
d
"2% &
v1 = # & v2 = #
v2 = # 2 & v2
2
# !d $ &
" A1 %
" d1 %
1
#π# & &
" "2% %
Since d2<<d1 it is true that v1<<v2.
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.
Let point 1 be inside the hose and point 2 be outside the nozzle.
1 2
1 2
P1 + ρgy1 + ρv1 = P2 + ρgy 2 + ρv2
2
2
The hose is horizontal so y1 = y2. Also P2 = Patm.
P1 − Patm
1 2 1 2
= ρ v2 − ρ v1
2
2
1
1
= ρ ( v22 − v12 ) ≈ ρ v22
2
2
1
2
3
= (1000 kg/m ) ( 25 m/s)
2
= 3.1×10 5 Pa
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. 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.
ΔV π ΔP L 4
=
r
Δt 8 η
where η is the viscosity
Example (text problem 9.55): A hypodermic syringe a_ached 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 2.00×10-­‐‑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:
ΔP =
8η L ΔV
=
4
π r Δt
8 ( 2.00 ×10 −3 Pa sec) (3.00 cm )
π ( 0.3×10 cm )
−1
4
⋅ 0.250 cm 3 sec
= 4716 Pa
This pressure difference is between the fluid in the syringe and the fluid in the vein. The pressure in the syringe is ΔP = Ps − Pv
Ps = Pv + ΔP
= 2140 Pa + 4720 Pa = 6860 Pa
Example (text problem 9.55): A hypodermic syringe a_ached 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 2.00×10-­‐‑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.
(b) What force must be applied to the plunger, which has an area of 1.00 cm2?
The result of (a) gives the force per unit area on the plunger so the force is just F = PA = 0.686 N.
Viscous Drag
The viscous drag force on a sphere is given by Stokes’ law.
FD = 6πηrv
Where η is the viscosity of the fluid, r is the radius of the sphere, and v is the velocity of the sphere.
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? y
FBD for sphere
Apply Newton’s Second Law to the sphere
FB FD
∑F = F
D
x
+ FB − w = ma
w
FD + FB − w = 0
When v = vterminal, a = 0 and 6 πηrv + m g − m g = 0
t
l
s
6πηrvt + ρlVl g − ms g = 0
6πηrvt + ρlVs g − ms g = 0
Solving for η η
ms g − ρlVs g
=
= 2.4 Pa sec
6π rvt
Surface Tension
The surface of a fluid acts like a 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. 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. ! 2γ $ 2
F = PA = # & π r = 9 ×10 −6 N
" r %
(b) What is the maximum mass of this water strider so that it can keep from breaking through the water surface?
To be in equilibrium, each leg must support one-­‐‑sixth the weight of the insect.
1
6F
F = w or m =
= 5 ×10 −6 kg
6
g