Transcript ME33: Fluid Flow Lecture 1: Information and

Chapter 6:
Momentum Analysis of Flow Systems
Siti Kamariah Md Sa’at
Pusat Pengajian Kejuruteraan Bioproses
September, 2009
NEWTON’S LAWS
Newton’s laws are relations between motions of bodies
and the forces acting on them.
First law: a body at rest remains at rest, and a body in motion
remains in motion at the same velocity in a straight path when
the net force acting on it is zero.
Second law: the acceleration of a body is proportional to the net
force acting on it and is inversely proportional to its mass.
Third law: when a body exerts a force on a second body, the
second body exerts an equal and opposite force on the first.
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Chapter 6: Momentum Analysis of Flow Systems
NEWTON’S LAWS AND CONSERVATION
OF MOMENTUM
For a rigid body of mass m, Newton’s second law is
expressed as
Therefore, Newton’s second law can also be stated as
the rate of change of the momentum of a body is equal
to the net force acting on the body.
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Chapter 6: Momentum Analysis of Flow Systems
NEWTON’S LAWS AND CONSERVATION
OF MOMENTUM
= Momentum
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The product of the mass and the
velocity of a body is called the
linear momentum.
Newton’s second law  the linear
momentum equation in fluid
mechanics
The momentum of a system is
conserved when it remains
constant  the conservation of
momentum principle.
Momentum is a vector. Its direction
is the direction of velocity.
Chapter 6: Momentum Analysis of Flow Systems
EXAMPLE 1-GRADUAL ACCELERATION
OF FLUID IN PIPELINE
Q: Water flows through a pipeline 60m long at velocity
1.8m/s when the pressure difference between the inlet
and outlet ends is 25 kN/m. What increase of pressure
difference is required to accelerate the water in pipe at
rate 0.02 m/s2?
Solution:
Lets
A= cross-sectional of the pipe
l = length of pipe
ρ = mass density of water
a = acceleration of water
δp = increase in pressure at inlet required to produce acceleration
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Chapter 6: Momentum Analysis of Flow Systems
EXAMPLE 1- GRADUAL ACCELERATION
OF FLUID IN PIPELINE
Not is steady flow state, consider a control mass comprising the
whole of the water in the pipe. By Newton’s Second Law:
1. Force due to δp,F = cross-sectional area x δp = A δp
2. Mass of water in pipe,m = Density x Volume
= ρ x Al
3. A δp = ρAl a
δp = ρl a
= 103 x 60 x 0.02 N/m2
= 1.2 kN/m2
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Chapter 6: Momentum Analysis of Flow Systems
LINEAR MOMENTUM EQUATION
Newton’s second law for a system of mass m subjected
to a force F is expressed as
During steady flow, the amount of momentum within the
control volume remains constant.
The net force acting on the control volume during steady
flow is equal to the difference between the rates of
outgoing and incoming momentum flows.
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Chapter 6: Momentum Analysis of Flow Systems
LINEAR MOMENTUM EQUATION
Consider a stream tube and assume steady non-uniform flow
A2
v2
A1
ρ2
v1
ρ1
v1
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Chapter 6: Momentum Analysis of Flow Systems
LINEAR MOMENTUM EQUATION
In time δt a volume of the fluid moves from the inlet
at a distance v1δt, so
volume entering the stream tube = area x distance
= A1 x v1δt
The mass entering,
mass entering stream tube = volume x density
= ρ1A1v1 δt
And momentum
momentum entering stream tube = mass velocity
= ρ1A1v1 δt v1
Similarly, at the exit, we get the expression:
momentum leaving stream tube = ρ2A2v2 δt v2
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Chapter 6: Momentum Analysis of Flow Systems
LINEAR MOMENTUM EQUATION
By Newton 2nd law
Force = rate of change of momentum
F
= (ρ2A2v2 δt v2 - ρ1A1v1 δt v1)
δt
We know from continuity that
Q= A1v1 = A2v2
And if we have fluid of constant density, ρ1 = ρ2 = ρ, then
F = Qρ (v2-v1)
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Chapter 6: Momentum Analysis of Flow Systems
LINEAR MOMENTUM EQUATION
An alternative derivation
From conservation of mass
mass into face 1 = mass out of face 2
we can write
rate of change of mass = m
˙ = dm/dt
= ρ1A1v1 = ρ2A2v2
The rate at which momentum enters face 1 is ρ1A1v1 v1 =m
˙ v1
The rate at which momentum leaves face 2 is ρ2A2v2 v2 =m
˙ v2
Thus the rate at which momentum changes across the stream
tube is
ρ2A2v2 v2 - ρ1A1v1 v1 = m
˙ v2 - m˙ v1
Force = rate of change of momentum
F
=m
˙ (v2-v1)
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Chapter 6: Momentum Analysis of Flow Systems
LINEAR MOMENTUM EQUATION
So, we know these two expression. Either
one is known as momentum equation:
F=m
˙ (v2-v1)
F = Qρ (v2-v1)
The momentum equation:
This force acts on the fluid
in the direction of the flow of the fluid
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Chapter 6: Momentum Analysis of Flow Systems
Momentum-Flux Correction Factor, b
Since the velocity across most inlets and outlets is not
uniform, the momentum-flux correction factor, b, is used
to patch-up the error in the algebraic form equation.
Therefore,
Momentum flux across an inlet or outlet:
Momentum-flux correction factor:
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Chapter 6: Momentum Analysis of Flow Systems
LINEAR MOMENTUM EQUATIONSTEADY FLOW
The net force acting on the control volume
during steady flow is equal to the difference
between the rates of outgoing and
incoming momentum flows. Therefore,
One inlet and one outlet
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Chapter 6: Momentum Analysis of Flow Systems
LINEAR MOMENTUM EQUATIONSTEADY FLOW ALONG COORDINATE
The previous analysis assumed the inlet and outlet velocities in the same
direction i.e. a one dimensional system.
What happens when this is not the case?
v2
v1
We consider the forces by resolving in the directions of the
co-ordinate axes.
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Chapter 6: Momentum Analysis of Flow Systems
LINEAR MOMENTUM EQUATIONSTEADY FLOW ALONG COORDINATE
The force in x-direction
Fx = m
˙ (v2 cos θ2 – v1 cos θ1)
=m
˙ (v2x– v1x )
The resultant foce can be
found by combining these
components
Or
Fx
= ρQ (v2 cos θ2 – v1 cos θ1)
= ρQ (v2x– v1x )
The force in y-direction
Fy = m
˙ (v2 sin θ2 – v1 sin θ1)
=m
˙ (v2y– v1y )
Or
Fy
= ρQ (v2 sin θ2 – v1 sin θ1)
= ρQ (v2y– v1y)
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Chapter 6: Momentum Analysis of Flow Systems
SUMMARY
Total Force on the fluid = rate of change of
momentum through the control volume
F =β m
˙ (vout - vin)
F = βQρ (vout - vin)
Remember!!!
We are working with vectors so F is in the direction of
the velocity
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Chapter 6: Momentum Analysis of Flow Systems
FORCE ACTING ON CONTROL VOLUME
Consist of two forces:
1. Body forces
- such as gravity, electric, magnetic force
2. Surface force
- such as pressure, viscous and reaction force
Total force acting on control volume is sum of body force
and surface force.
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Chapter 6: Momentum Analysis of Flow Systems
FORCE ACTING ON CONTROL VOLUME
Force is made up of these component:
1.
FR = force exerted on the fluid by any solid body touching the
control volume
2.
FB = Force exerted on the solid body (eg gravity)
3.
FP = Force exerted on the fluid by fluid pressure outside the
control volume
Total force is given by the sum of these three forces:
FT = FR + FB + FP
The force exerted by the fluid on the solid body touching the control
volume is opposite to FR
So, the reaction force, R is given by
R = - FR
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Chapter 6: Momentum Analysis of Flow Systems
APPLICATION OF THE MOMENTUM
EQUATION
FORCE DUE TO THE FLOW AROUND
THE PIPE BEND
A converging pipe bend lying in the horizontal plane turning
through an angle of θ.
A1
v1
ρ1
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Chapter 6: Momentum Analysis of Flow Systems
FORCE DUE TO THE FLOW AROUND
THE PIPE BEND
Why do we want to know the forces here?
As the fluid changes direction a force will act
on the bend.
This force can be very large in the case of
water supply pipes.
The bend must be held in place to prevent
breakage at the joints.
We need to know how much force a support
(thrust block) must withstand.
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Chapter 6: Momentum Analysis of Flow Systems
FORCE DUE TO THE FLOW AROUND
THE PIPE BEND
Step in analysis
1.
2.
3.
4.
5.
6.
Draw a control volume
Decide on coordinate axis system
Calculate the total force
Calculate the pressure force
Calculate the body force
Calculate the resultant force
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Chapter 6: Momentum Analysis of Flow Systems
EXAMPLE 2 –THE FORCE TO HOLD A
DEFLECTOR ELBOW
Q: A reducing elbow is used to deflect water flow at a rate
of 14 kg/s in a horizontal pipe upward 30° while
accelerating it.The elbow discharges water into the
atmosphere. The cross-sectional area of the elbow is
113 cm2 at the inlet and 7 cm2 at the outlet. The elevation
difference between the centers of the outlet and the inlet
is 30 cm. The weight of the elbow and the water in it is
considered to be negligible. Determine
(a) the gage pressure at the center of the inlet of the
elbow and
(b) the anchoring force needed to hold the elbow in
place.
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Chapter 6: Momentum Analysis of Flow Systems
EXAMPLE 2
Assumption
1. The flow is steady, and the frictional
effects are negligible.
2. The weight of the elbow and the
water in it is negligible.
3. The water is discharged to the
atmosphere, and thus the gage
pressure at the outlet is zero.
4. The flow is turbulent and fully
developed at both the inlet and outlet
of the control volume, and we take
the momentum-flux correction
factor,β to be 1.03.
Figure 2
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Chapter 6: Momentum Analysis of Flow Systems
EXAMPLE 2
Density,ρ = 1000 kg/m3
The continuity equation for this one-inlet, one-outlet, steady-flow
system is
=14 kg/s.
Noting that
, the inlet and outlet velocities of water
are
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Chapter 6: Momentum Analysis of Flow Systems
EXAMPLE 2
Step 1: Draw Control volume as Figure 2.
We take the elbow as the control volume and designate the inlet
by 1 and the outlet by 2.
Step 2: Decide on coordinate axis system
Take the x- and z-coordinates as shown.
Step 3: Calculate the total force
FTx = m
˙ (v2 cos 30o - v1) β
= 14 kg/s x (20cos 30o – 1.24 m/s) (1N/1kg.m/s2)(1.03)
= 231.88 N
FTz = m
˙ v2 sin 30o β
= 14 kg/s x 1.24 x 20sin 30 x (1N/1kg.m/s2)(1.03)
= 144.2 N
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Chapter 6: Momentum Analysis of Flow Systems
EXAMPLE 2 (a)
Step 4: Calculate the pressure force
Using Bernoulli equation, pressure can be calculated if we know
pressure in inlet/outlet.
Taking the center of the inlet cross section as the reference level
(z1= 0) and noting that P2 = Patm, the Bernoulli equation for a
streamline going through the center of the elbow is expressed as
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Chapter 6: Momentum Analysis of Flow Systems
EXAMPLE 2 (b)
Calculate the pressure force:
Fp = Pressure force at 1 – Pressure force at 2
Fpx = P1A1 cos 0 – P2A2 cosθ = P1A1 – P2A2 cos 30
= P1A1 = 202.2 kN/m2 x 0.0113 m2 = 2.285 kN
Fpz = P1A1 sin 0 – P2A2 sin θ = – P2A2 sin 30
=0
since P2 = Patm
Step 5: Calculate the body force
There is no body force in x or z direction.
The body force is acting on y-direction. So not to be considered.
FBx = FBz = 0
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Chapter 6: Momentum Analysis of Flow Systems
EXAMPLE 2 (b)
Step 6: Calculate the resultant force
We let the x- and z-components of the anchoring force of the elbow
be FRx and FRz, and assume them to be in the positive direction.
Solving for FRx and FRz, and substituting the calculated values,
FRx = FTx – FPx – FBx
= 231.88 – 2285 - 0
= -2053.12 N
FRZ = FTz –FPz – FBz
= 144.2 -0 – 0
= 144.2 N
And the resultant force on the fluid is given by
FR =√( FRx2 – FRx2) = √((2053.12)2 – (144.2)2) = 2048 N
The force of the bend is the same magnitude but in opposite direction
R = -FR = - 2048 N
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Chapter 6: Momentum Analysis of Flow Systems
EXAMPLE 3- WATER JET STRIKING A
STATIONARY PLATE
Q: Water accelerated by a nozzle to an average speed of
20 m/s and strikes a stationary vertical plate at rate of 10
kg/s with a normal velocity 20 m/s. After the strike, the
water stream splatters off in all directions in the plane of
the plate. Determine the force needed to prevent the
plate from moving horizontally due to the water stream.
Figure 3
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Chapter 6: Momentum Analysis of Flow Systems
EXAMPLE 3
Solution:
Assumption
1.
The flow of water at nozzle outlet is steady.
2.
The water splatters in directions normal to the approach
direction of the water jet.
3.
β=1
•
We want to find the reaction force of the plate, so that the plate
stay in the position.
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Chapter 6: Momentum Analysis of Flow Systems
EXAMPLE 3
• Step 1 & 2 : Control volume and Coordinate axis
shown in Figure 3
• Step 3: Calculate the total force
In the x-direction :
FTx = β m
˙ (v2x – v1x),
=βm
˙ v1x
= (1) (10 kg/s) (20 m/s) (1N/1kg.m/s2) = 200 N
The system is symetrical, the force in y-direction is not considered.
• Step 4: Calculate the pressure force
The pressures at both the inlet and the outlets to the control volume
are atmospheric. The pressure force is zero.
FPx =FPy =0
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Chapter 6: Momentum Analysis of Flow Systems
EXAMPLE 3
•
•
Step 5: Calculate the body force
As the control volume is small we can ignore the body force due to
gravity.
FBx = FBy = 0
Step 6: Calculate the resultant force
FRx = FTx – FPx – FBx
= 200 – 0 - 0
= 200 N
The force on the plane is the same magnitude but in the opposite
direction
R = - FRx = -200 N
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Chapter 6: Momentum Analysis of Flow Systems
Force on angle plane
If the plane were at an angle the analysis is the same. It is usually most
convenient to choose the axis system normal to the plate.
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Chapter 6: Momentum Analysis of Flow Systems