Download Chapter 13: Fluids Mechanics

Chapter 13: Fluids Mechanics
Density
Definition of Density, ρ:
The density, ρ of a material is its mass, m per volume, V
ρ = m/V
SI unit: kg/m3
 “Fluids with different properties have different density”
Pressure
Definition of Pressure:
Pressure, P is force, F per area, A
:
P = F/A
Pressure is often given in the terms of Pascal (Pa).
1 Pa = 1 N/m2
1 atm = 1.013 x 105 Pa = 760 mm Hg
1 mm Hg = 1 torr = 133.32 Pa
SI unit: N/m2
Pressure & Depth
The top surface of the fluid is open to the atmosfera, with pressure Pat. If the crosssectional area of the container is A, the downward force exerted on the top surface by
the atmosfera is
Ftop = PatA
At the bottom of the container, the downward force is Ftop plus the weight of the fluid.
Recalling that
m =ρV
V = hA
Hence,
W = mg
= pVg
= p(hA)g
Fbottom = Ftop + W
= PatA + ρ(hA)g
Pbottom = Fbottom/A
= [PatA + ρ(hA)g] / A
= Pat + ρhg
Example:
The Titanic was found in 1985 lying on the bottom of the North Atlantic at a depth of
2.5 miles. What is the pressure at this depth? (ρ = 1025 kg/m3)
Solution:



1 mile =
2.5 miles
1609 meter
=
2.5 x 1609 m
=
4022.5 m
P
Pat + ρhg
1.01 x 105 Pa +
(1025 kg/m3)(9.81 m/s2)(4022.5 m)
4.1 x 107 Pa
=
=
=
Example:
A cubic box 20.00 cm on a side is completely immersed in a fluid. At the top of the
box the pressure is 105.0 kPa; at the bottom the pressure is 106.8 kPa. What is the
density of the fluid?
Solution:
The pressure at the top and bottom of the box are related by
P2  P1  hg

P2  P1 
hg
106.8 kPa  105.0 kPa

hg
106.8 kPa  105.0 kPa

9.81ms  2 0.2m 

 920 kg m

3
Barometer
Barometer is an equipment which is used to measure atmospheric pressure.
A fluid that is often used in barometer is mercury.
Mercury has a density,
ρ = 1.3595 x 104 kg/m3
Pat = ρhg
= (1.3595 x 104 kg/m3)(0.760 m)(9.81 m/s2)
= 1.013 x 105 Pa
Example:
A U-shaped tube is filled mostly with water, but a small amount of vegetable oil has
been added to one side. The density of the water is 1000 kg/m3 and the density of the
vegetable oil is 920 kg/m3. If the depth of the oil is 5.00 cm, what is the difference in
level, h between the top of the oil on one side of the U and the top of the water on the
other side?
Solution:
PA = Pat + ρwatergh1
PB = Pat + ρoilgh2
PA
Pat + ρwatergh1
ρwatergh1
 
h1  h2  oil
  water
h1
=
=
=
PB
Pat + ρoilgh2
ρoilgh2

 920 kg m 3 
  5.00 cm

3 
 1000 kg m 

=
=
=
h2 – h1
(5.00 – 4.60) cm
0.40 cm
Pascal’s Principle
An external pressure applied to an enclosed fluid is transmitted unchanged to every point
within the fluid.
A hydraulic lift works according to Pascal’s principle
A force F1 exerted on the small piston causes a much larger force, F2, to act on the large
piston.
When force F1 exerted on the small piston, this increases the pressure in that cylinder by
the amount of
F
P  1
A1
By Pascal’s Principle, the pressure in cylinder 2 increases by the same amount.
Therefore,
F2 = (∆P)A2
F1 F2

A1 A2
Example:
To inspect a 14,500 N car, it is raised with a hydraulic lift. If the radius of the small
piston is 4.0 cm, and the radius of the large piston is 17 cm, find the force that must be
exerted on the small piston to lift the car.
Solution:
According to Pascal’s Principle, an external pressure applied to an enclosed fluid is
transmitted unchanged to every point within the fluid.
A
F1  F2  1
 A2



  0.0040 m 2
 14,500 
2
  0.17 
 800 N




Buoyant Force
Fluids surrounding an object exert a force in the upward direction. The force is known as
buoyant force.
Consider a cubical block immersed in a fluid of density, ρ. The surrounding fluid exerts
normal forces on all of its faces. The horizontal forces pushing to the right and to the left
are equal, hence they cancel and have no effect on the block.
F2 > F1
The downward force exerted on the top face is less than the upward force exerted on the
lower face. This is because the pressure at the lower face is greater. Thus, the difference
in forces gives rise to a net upward force that is known as Buoyant Force.
To calculate the value of buoyant force:
 Firstly, assume that the cubical block is of length L on the side and the pressure on
the top surface is P1.
Force that acts on top surface, F1
F1
=
P1A
=
P1L2
Force that acts on bottom surface, F2
F2
=
P2A

P1+ρgL
P2
=
F2
= (P1+ρgL)L2
= P1L2 +ρgL3
= F1 + ρgL3
Buoyant force, Fb = F2 – F1
= ρgL3
Note that ρgL3 is the weight of fluid that would occupy the same volume as the cube.
Therefore, the buoyant force is equal to the weight of fluid that is displaced by the cube.
This phenomenon is known as Archimedes’ Principle.
Example:
A piece of wood with a density of 706 kg/m3 is tied with a string to the bottom of a
water filled container. The wood is completely immersed, and has a volume of 8.00 x
10-6 m3. What is the tension in the string?
Solution:
Apply Newton’s second law to the wood:
Fb = T + mg
Tension in the string:
T = Fb – mg
Weight of the wood:
mg = ρwoodVwoodg = (706)(8.00 x 10-6)(9.81)
= 0.0554 N
Buoyant force, Fb = weight of displaced fluid
= ρwaterVwaterg
= (1000) (8.00 x 10-6)(9.81)
= 0.0785 N
Therefore,
T
=
=
=
Fb + mg
0.0785 N – 0.0554 N
0.0231 N
Surface Tension
A fluid tends to pull inward on its surface, resulting in a surface of minimum area.
The surface of the fluid behaves much like an elastic membrane enclosing the fluid. This
kind of surface enable the insects to “stand” on the surface of the water.
13.4
Fluid Flow
Objectives :
a)
b)
identify the simplifications used in describing ideal fluid flow.
use the continuity equation and Bernoulli’s equation to explain common
effects of ideal fluid flow.
This photograph was taken in a water tunnel using hydrogen bubbles to visualize the flow
pattern around a cylinder.
The flow was started from rest, and at this instant the pattern shows the development of a
complex wake structure on the downstream side of the cylinder.
Four characteristics of an ideal fluid :
Condition 1 : Steady flow means that all the particles of a fluid have the same velocity
as they pass a given point.
Condition 2 : Irrotational flow means that a fluid element (a small volume of the fluid)
has no net angular velocity, which eliminates the possibility of whirlpools
and eddy currents. (The is non turbulent.)
Condition 3 : Nonviscous flow means that viscosity is negligible.
Condition 4 : Incompressible flow means that the fluid’s density is constant.
Equation of Continuity
If there no losses of fluid within a uniform tube, the mass of fluid flowing into the tube in
a given time must be equal to the mass flowing out of the tube in the same time (by the
conservation of mass).
Fig. a) mass enters tube,
Δm1
= ρ1ΔV1
= ρ1(A1Δx1)
= ρ1(A1v1Δt)
Fig. b) mass exits tube,
Δm2 = ρ2ΔV2
= ρ2(A2Δx2)
= ρ2(A2v2Δt)
since the mass is conserved,
Δm1 = Δm2
ρ1A1v1 = ρ2A2v2
or
ρAv = constant – equation of continuity
for an incompressible fluid, the density ρ is constant, so
A1v1 = A2v2
or Av = constant – flow rate equation
By the flow rate equation, the speed of a fluid is greater when the cross-sectional area of
the tube through which the fluid is flowing is smaller.
Think of a hose that is equipped with a nozzle such that the cross-sectional area of the
hose is made smaller.
Bernoulli’s Equation
a statement of the conservation of energy for a fluid.
Work-Energy Theorem ---» has great generality for fluid flow.
F1 does positive work :
motion
in the same direction as the fluid’s
F2 does negative work :
opposite to the fluid motion
The net work done by these forces is then,
Wnet
= F1Δx1 - F2Δx2
= (P1A1)(v1Δt) - (P2A2)(v2Δt)
The flow rate equation, requires that A1v1 = A2v2 ;
Wnet
= A1v1Δt (P1 – P2)
recall from the equation of continuity, Δm1 = Δm2
Therefore,
Wnet 
m

P1  P2 
The net work done on the system by the external forces (nonconservative work) must be
equal to the change in the total mechanical energy.
Kinetic Energy :
ΔK = ½Δm(v22 – v12)
Gravitational Potential Energy :
ΔU = Δmg(y2 – y1)
thus,
Wnet
m

= ΔK + ΔU
( P1  P2 ) = ½Δm(v22 – v12) + Δmg(y2 – y1)
cancelling each Δm
P1 + ½ρv12 + ρgy1 = P2 + ½ρv22 + ρgy2
P + ½ρv2 + ρgy = constant - Bernoulli’s Equation
Bernoulli’s equation can be applied to many situations.
1.
Venturi meter
In a region of smaller cross-sectional area, the flow speed is greater; the pressure
in that region is lower than in other regions.
The pressure P1 is greater than the pressure P2, because v1 < v2.
This device can be used to measure the speed of fluid flow.
2.
Airplane lift
Because of the shape and orientation of an airfoil or airplane wing, the air
streamlines are closer together, and the air speed is greater above the wing than
below it. The resulting pressure difference supplies an upward force, or lift.
Streamline flow around an airplane wing.
The pressure above is less than the pressure below, and there is a dynamic lift
force upward.
3.
Fluid flow from a tank
If we assume that the cross-sectional area of the tank is large relative to that of the
hole (A2 » A1), then the water level drops very slowly and we can assume v2 ≈ 0.
Let us apply Bernoulli’s equation to points 1 and 2.
If we note that P1 = P2 at the hole, we get v1 =
2 gh
Fig. (a) :The airstream around a nonrotating tennis ball moving from right to left. The
streamlines represent the flow relative to the tennis ball. Note the symmetric region of
turbulence behind the ball.
Fig. (b) :Curve Balls
The airstream around a spinning tennis ball. The ball experiences a deflecting
force because of the Bernoulli effect.
13.5
Viscosity
objectives : to discuss fluid viscosity.
Viscosity is a fluid’s internal resistance to flow.
(All real fluids have a nonzero viscosity.)
Fig. (a) :
Streamlines never cross and are closer together in regions of greater fluid
velocity. The stationary paddle wheel indicates that the flow is irrotational, or
without whirlpools and eddy currents
Fig. (b) :
The smoke from an extinguished candle begins to rise in nearly streamline flow,
but quickly becomes rotational and turbulent.
Fig a) internal friction causes the layers of the fluid to move relative to each
other in response to a shear stress; called laminar flow.
Fig b) at higher velocities, the flow becomes rotational (the speed of the fluid is
less near the walls of the pipe than near the centre because of frictional
drag between the walls and the fluid); called turbulent.
The viscosities of some fluids are listed in Table below.
The greater the viscosity of the liquid, which is easier to visualize than that of the gas,
the greater is the shear stress required to get the layers of the liquid to slide along each
other.
Example : the large viscosity of glycerin compared to that of water.
Poiseuille’s Law
When a fluid flows through a pipe, there is frictional drag between the liquid and the
walls, and the fluid velocity is greater toward the center of the pipe.
refer Fig. b) : this effect makes a difference in a fluid’s
average flow rate
The flow rate depends on the properties of the fluid and the dimensions of the pipe, as
well as on the pressure difference (ΔP) between the ends of the pipe.
where,
ΔP is pressure difference,
r is radius,
L is length and
η is viscosity.
Terminal velocity
When things move through a fluid, drag increases with speed.
This is because they encounter more fluid per second and hit it harder.
For low velocities, drag is proportional to speed, and for higher velocities it increases
with velocity squared.
In either case the motion will differ from free fall.
The resultant downward force will now be the difference between weight W (= mg)
and drag D (which increases with velocity) so the equation of motion is:
resultant F
=W–D
= mg –D
= ma
so the acceleration is
a
mg  D
m
or,
ag
D
m
When v = 0, D = 0 so a = g. (This is the initial acceleration.)
As v increases, D increases, so a falls.
Eventually a velocity is reached when D = mg, and then a = 0.
The object continues to fall at a constant terminal velocity.
Stoke’s Law
In1845 a scientist named George Stokes found that the magnitude of the resistive force,
Fr on a very small spherical object of radius r falling slowly through a fluid of viscosity η
with terminal velocity, vT is given by
Fr = 6πηrvT
This equation, called Stoke’s Law, has many important applications.
For example, it describes the sedimentation of particulate matter in blood samples.