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Civil Engineering Hydraulics
Conservation of Momentum
Conservation of Momentum 2
Control Volume Revisited ¢ Previously, we considered developing a control
volume so that we could isolate mass flowing into and out of the control volume
¢ Our goal in developing this approach was to simplify complex problems into simpler problems for analysis
Wednesday, September 26, 2012
2
Conservation of Momentum 3
Control Volume Revisited ¢ We choose boundaries to isolate internal mass
flows from external mass flows ¢ We then extended this to consider energy
moving into and out of the control volume ¢ We developed a relationship conserving energy
using the Bernoulli equation when appropriate
Wednesday, September 26, 2012
Conservation of Momentum 4
Control Volume Revisited ¢ In a manner similar to that undertaken in Statics
and Dynamics, we looked at the control volume as being a boundary to isolate “internal” from “external” events
¢ Now we will add another element to our analysis and continue using the control volume approach to simplify analysis
Wednesday, September 26, 2012
3
Conservation of Momentum 5
Control Volume Revisited ¢ The development of a usable control volume is
critical in the solution of our problems ¢ Problems that are exceptionally computationally
complex can be reduced considerably by the consideration of a viable control volume
¢ Take the time to consider what the control volume provides before setting up your analysis
Wednesday, September 26, 2012
Conservation of Momentum 6
Linear Momentum ¢ In Statics, we considered systems that were not
under acceleration in any direction so we were able to make the statement
a = 0F = m a = 0
Wednesday, September 26, 2012
4
Conservation of Momentum 7
Linear Momentum ¢ In Dynamics, the systems considered no longer
were restricted to systems with an acceleration equal to 0 so you moved from the special case where the vector sum of the forces were equal to 0 to the more general case
F = m a
Wednesday, September 26, 2012
Conservation of Momentum 8
Linear Momentum ¢ The critical element of this expression is the
fact that both acceleration and force are vector expressions having both magnitude and direction
F = m a
Wednesday, September 26, 2012
5
Conservation of Momentum 9
Linear Momentum ¢ By definition, acceleration is the time rate of
change of velocity so if we consider constant mass we can rewrite the expression as
F = m a = m d
Vdt
Wednesday, September 26, 2012
Conservation of Momentum 10
Linear Momentum ¢ And since mass is a scalar quantity, we can
move it into the derivative to generalize the expression
F = m a = m d
Vdt
=d m
V( )
dt
Wednesday, September 26, 2012
6
Conservation of Momentum 11
Linear Momentum ¢ The last term represents the change of linear
momentum with respect to time ¢ Linear momentum being defined as the product
of a mass times its linear velocity
F = m a = m d
Vdt
=d m
V( )
dt
Wednesday, September 26, 2012
Conservation of Momentum 12
Linear Momentum ¢ This simple statement implies that a change in
the force acting on a body will change the momentum of the body and
¢ A change in momentum of a body will produce a force in response
F = m a = m d
Vdt
=d m
V( )
dtWednesday, September 26, 2012
7
Conservation of Momentum 13
Linear Momentum ¢ One is always connected with the other ¢ Notice that both force and velocity are vector
quantities and therefore they have both magnitude and direction
¢ That is a critical point
F = m a = m d
Vdt
=d m
V( )
dtWednesday, September 26, 2012
Conservation of Momentum 14
Angular Momentum ¢ If we look at the angular momentum of a body,
we consider the moment or torque that produces the rotation
¢ The expression is slightly different than for linear momentum
M = I
α
Wednesday, September 26, 2012
8
Conservation of Momentum 15
Angular Momentum ¢ M is the moment or torque causing the
rotational acceleration ¢ I is the moment of interia about the axis of
rotation ¢ And α is the angular acceleration about the axis
of rotation M = I
α
Wednesday, September 26, 2012
Conservation of Momentum 16
Angular Momentum ¢ If we use ω to represent angular velocity then
the expression can be rewritten as
M = I
α = I d
ϖdt
Wednesday, September 26, 2012
9
Conservation of Momentum 17
Angular Momentum ¢ If we use ω to represent angular velocity then
the expression can be rewritten as
M = I
α = I d
ϖdt
Wednesday, September 26, 2012
Conservation of Momentum 18
Angular Momentum ¢ Both expressions are similar in form but they do
differ in the units ¢ Both expressions do involve vector quantities
M = I
α = I d
ϖdt
F = m a = m d
Vdt
Wednesday, September 26, 2012
10
Conservation of Momentum 19
Force Balance ¢ In Statics, when we looked at the forces acting
on a free body diagram we had three types of forces l Externally applied loads l Reactive forces from supports l Weight
Wednesday, September 26, 2012
Conservation of Momentum 20
Force Balance ¢ On a control volume, the forces acting on the
system are expanded to include l Gravitational forces (weight) l Pressure forces l Viscous forces l And any other force acting on the system
Wednesday, September 26, 2012
11
Conservation of Momentum 21
Force Balance ¢ It will be critical to our analysis to be able to
evaluate both the magnitude and direction of forces acting on the system
¢ And to understand that we are dealing with vector quantities in this type of balance and so we have to use vector operations for our analysis
Wednesday, September 26, 2012
Conservation of Momentum 22
Application of Linear Momentum
¢ There are a number of theoretical developments of how the principle of the conservation of momentum is applied and an outline of all the special cases that can be utilized
Wednesday, September 26, 2012
12
Conservation of Momentum 23
EXAMPLE The Force to Hold a Deflector Elbow in Place
Wednesday, September 26, 2012
Conservation of Momentum 24
EXAMPLE The Force to Hold a Deflector Elbow in Place
¢ A reducing elbow is used to deflect water flow (mass flow) at a rate of 14 kg/s in a horizontal pipe upward 30° while accelerating it.
Wednesday, September 26, 2012
13
Conservation of Momentum 25
EXAMPLE The Force to Hold a Deflector Elbow in Place
¢ 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.
Wednesday, September 26, 2012
Conservation of Momentum 26
EXAMPLE The Force to Hold a Deflector Elbow in Place
¢ 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.
Wednesday, September 26, 2012
14
Conservation of Momentum 27
EXAMPLE The Force to Hold a Deflector Elbow in Place
¢ 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.
Wednesday, September 26, 2012
Conservation of Momentum 28
EXAMPLE The Force to Hold a Deflector Elbow in Place
They have already selected the control volume for you in this problem.
Wednesday, September 26, 2012
15
Conservation of Momentum 29
EXAMPLE The Force to Hold a Deflector Elbow in Place
Force directions are positive to the right and upward.
Wednesday, September 26, 2012
Conservation of Momentum 30
EXAMPLE The Force to Hold a Deflector Elbow in Place
What we know
kPa 1000Pa≡
ρwater 1000kg
m3:=
Area1 113cm2:= z1 0cm:= z2 30cm:=
Area2 7cm2:= mdot 14kgsec
:=
Wednesday, September 26, 2012
16
Conservation of Momentum 31
EXAMPLE The Force to Hold a Deflector Elbow in Place
We have three sets of expressions that we can use to analyze this system.
1. We know that the mass flow into the system must be equal to the mass flow out of the system.
2. We know that the energy in must equal to the energy out.
3. And we know that any change in momentum in the system must be reacted to by a force applied to the system.
Wednesday, September 26, 2012
Conservation of Momentum 32
EXAMPLE The Force to Hold a Deflector Elbow in Place
1. We know that the mass flow into the system must be equal to the mass flow out of the system.
m1 = m2
Wednesday, September 26, 2012
17
Conservation of Momentum 33
EXAMPLE The Force to Hold a Deflector Elbow in Place
We know that the mass flow into the system must be equal to the mass flow out of the system.
m1 = m2ρQ1 = ρQ2ρV1A1 = ρV2A2
water flow at a rate of 14 kg/s
This is dm/dt
cross-sectional area of the elbow is 113 cm2 at the inlet
This is A1
Wednesday, September 26, 2012
Conservation of Momentum 34
EXAMPLE The Force to Hold a Deflector Elbow in Place
We can use these along with our expression to solve for the velocity at the inlet of the system
m1 = ρV1A1
V1 =m1
ρA1=
14 kgs
1000 kgm 3 113cm
2 1m100cm
⎛⎝⎜
⎞⎠⎟
2
water flow at a rate of 14 kg/s
This is dm/dt
cross-sectional area of the elbow is 113 cm2 at the inlet
This is A1
Wednesday, September 26, 2012
18
Conservation of Momentum 35
EXAMPLE The Force to Hold a Deflector Elbow in Place
We can use these along with our expression to solve for the velocity at the inlet of the system
m1 = ρV1A1
V1 =m1
ρA1=
14 kgs
1000 kgm 3 113cm
2 1m100cm
⎛⎝⎜
⎞⎠⎟
2 water flow at a rate of 14 kg/s
This is dm/dt
cross-sectional area of the elbow is 113 cm2 at the inlet
This is A1
Velocity1mdot
ρwater Area1⋅:=
Velocity1 1.239ms
=
Wednesday, September 26, 2012
Conservation of Momentum 36
EXAMPLE The Force to Hold a Deflector Elbow in Place
And we can use the mass balance for the control volume to solve for the velocity at the outlet
ρV2A2 = ρV1A1water flow at a rate of 14 kg/s
This is dm/dt
cross-sectional area of the elbow is 113 cm2 at the inlet
This is A1
Velocity2Velocity1 Area1⋅
Area2:=
Velocity2 20ms
=
Wednesday, September 26, 2012
19
Conservation of Momentum 37
EXAMPLE The Force to Hold a Deflector Elbow in Place
There is nothing else that the conservation of mass expression can tell us for this problem
One of the things that the problem asks for is the gage pressure at the inlet, point 1
water flow at a rate of 14 kg/s
This is dm/dt
cross-sectional area of the elbow is 113 cm2 at the inlet
This is A1
Velocity2Velocity1 Area1⋅
Area2:=
Velocity2 20ms
=
Wednesday, September 26, 2012
Conservation of Momentum 38
EXAMPLE The Force to Hold a Deflector Elbow in Place
We can use the energy balance on the control volume to determine this pressure
gz1 +V1
2
2+P1ρ1
= gz2 +V2
2
2+P2ρ2
We know the elevations at point 1 and 2, the velocities at points 1 and 2, and the fact that the pressure at point 2 is atmospheric.
Wednesday, September 26, 2012
20
Conservation of Momentum 39
EXAMPLE The Force to Hold a Deflector Elbow in Place
We can use the energy balance on the control volume to determine this pressure
gz1 +v1
2
2+
p1
ρ= gz2 +
v22
2+
p2
ρz1 = 0, z2 = 30cmp2 = 0(gage)
v1 = 1.24 ms
,v2 = 20 ms
ρ = 1000 kgm3
v12
2+
p1
ρ= gz2 +
v22
2p1 = 100kPa
Wednesday, September 26, 2012
Conservation of Momentum 40
EXAMPLE The Force to Hold a Deflector Elbow in Place
Since we have a change in both magnitude and direction of velocity across the control volume, we have a change in momentum across the control volume
We can look at these in component form using the unit vectors to describe the inlet and outlet velocities.
V1 =V1
i
V2 =V2 cos300
i + sin300
k( )
Wednesday, September 26, 2012
21
Conservation of Momentum 41
EXAMPLE The Force to Hold a Deflector Elbow in Place
The change in momentum across the control volume can then be expressed as
V1 =V1
i
V2 =V2 cos30
0i + sin300
k( )
mV2 cos300i + sin300
k( )− mV1 i
Wednesday, September 26, 2012
Conservation of Momentum 42
EXAMPLE The Force to Hold a Deflector Elbow in Place
This change in momentum must be exactly offset by the forces acting on the system
F∑ = mV2 cos30
0i + sin300
k( )− mV1 i
Wednesday, September 26, 2012
22
Conservation of Momentum 43
EXAMPLE The Force to Hold a Deflector Elbow in Place
The forces acting on the system include the reaction forces needed to support the system as well as the pressure forces acting on the system.
The problem statement provided the reasoning behind neglecting the weight of the water and the weight of the value.
F∑ = mV2 cos30
0i + sin300
k( )− mV1 i
Wednesday, September 26, 2012
Conservation of Momentum 44
EXAMPLE The Force to Hold a Deflector Elbow in Place
Since atmospheric pressure is constant at all points in the system, we can only consider the forces developed by gage pressures.
Again, we only consider these at the boundaries of the control volume so the only force developed by pressure is that developed normal to the flow into the system at 1.
F∑ = mV2 cos30
0i + sin300
k( )− mV1 i
Wednesday, September 26, 2012
23
Conservation of Momentum 45
EXAMPLE The Force to Hold a Deflector Elbow in Place
From our diagram you can see that the direction of this force is in the positive i direction.
F∑ = mV2 cos300
i + sin300
k( )− mV1
i
Fpressure = p1,gage Area1( )i
Wednesday, September 26, 2012
Conservation of Momentum 46
EXAMPLE The Force to Hold a Deflector Elbow in Place
The only other force in the system will be the support force, our unknown.
F∑ = mV2 cos300
i + sin300
k( )− mV1
i
Fpressure = p1,gage Area1( )iFR = FRx
i + FRz
k
Wednesday, September 26, 2012
24
Conservation of Momentum 47
EXAMPLE The Force to Hold a Deflector Elbow in Place
Now we can substitute into the conservation of momentum expression to obtain.
Fpressure +
FR = mV2 cos30
0i + sin300
k( )− mV1 i
mV2 cos300i + sin300
k( )− mV1 i = P1,gage Area1( ) i + FRx
i + FRz
k
Wednesday, September 26, 2012
Conservation of Momentum 48
EXAMPLE The Force to Hold a Deflector Elbow in Place
Setting the coefficients in the i and k directions equal we have
mV2 cos300i + sin300
k( )− mV1 i = P1,gage Area1( ) i + FRx
i + FRz
k
mV2 cos300 − mV1 = P1,gage Area1 + FRx
mV2 sin300 = FRz
Wednesday, September 26, 2012
25
Conservation of Momentum 49
EXAMPLE The Force to Hold a Deflector Elbow in Place
Isolating the unknowns and solving
mV2 cos300 − mV1 − P1,gage Area1 = FRx
mV2 sin300 = FRz
Wednesday, September 26, 2012
Conservation of Momentum 50
EXAMPLE The Force to Hold a Deflector Elbow in Place
Isolating the unknowns and solving
02 1 1, 1
02
cos30
sin30x
z
gage R
R
mV mV P Area F
mV F
− − =
=
& &
&
FRx mdot− Velocity1⋅ mdot Velocity2⋅ cos 30deg( )⋅+ P1gage Area1⋅−:=
FRx 2059.43− N=
FRy mdot− Velocity2⋅ sin 30deg( )⋅:=
FRy 140− N=
Wednesday, September 26, 2012
26
Homework 12-1
Conservation of Momentum 51
1. A reversing elbow is placed such that water makes a 180° U-turn before it is discharged. The elevation difference between the centers of the inlet and the exit sections is 0.3 m. Determine the anchoring force needed to hold the elbow in place. The water flows at a rate of 14 kg/s. 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 weight of the elbow is negligible.
Wednesday, September 26, 2012
Homework 12-2
Conservation of Momentum 52
2. If the flowrate is doubled in the system shown in Problem 1, how are the forces to hold the system in place changed?
Wednesday, September 26, 2012