Conservation of Momentum - Memphis

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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

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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



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


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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



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


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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



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


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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



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


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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



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

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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


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


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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



Angular Momentum ¢  If we use ω to represent angular velocity then

the expression can be rewritten as

M = I

α = I d

ϖdt


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Angular Momentum ¢  If we use ω to represent angular velocity then

the expression can be rewritten as

M = I

α = I d

ϖdt



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


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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



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


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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



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


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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.


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¢ 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.


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¢ 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.




They have already selected the control volume for you in this problem.


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Force directions are positive to the right and upward.




What we know

kPa 1000Pa≡

ρwater 1000kg

m3:=

Area1 113cm2:= z1 0cm:= z2 30cm:=

Area2 7cm2:= mdot 14kgsec

:=


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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.




1.  We know that the mass flow into the system must be equal to the mass flow out of the system.

m1 = m2


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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




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


This is dm/dt


This is A1


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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


This is A1

Velocity1mdot

ρwater Area1⋅:=

Velocity1 1.239ms

=




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


This is A1

Velocity2Velocity1 Area1⋅

Area2:=

Velocity2 20ms

=


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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


This is dm/dt


This is A1

Velocity2Velocity1 Area1⋅

Area2:=

Velocity2 20ms

=




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.


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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




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( )


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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




This change in momentum must be exactly offset by the forces acting on the system

F∑ = mV2 cos30

0i + sin300

k( )− mV1 i


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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




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


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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




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


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Now we can substitute into the conservation of momentum expression to obtain.

Fpressure +

FR = mV2 cos30

0i + sin300

k( )− mV1 i


k( )− mV1 i = P1,gage Area1( ) i + FRx

i + FRz

k




Setting the coefficients in the i and k directions equal we have


k( )− mV1 i = P1,gage Area1( ) i + FRx

i + FRz

k

mV2 cos300 − mV1 = P1,gage Area1 + FRx

mV2 sin300 = FRz


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Isolating the unknowns and solving

mV2 cos300 − mV1 − P1,gage Area1 = FRx

mV2 sin300 = FRz




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=


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Homework 12-1


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.


Homework 12-2


2.  If the flowrate is doubled in the system shown in Problem 1, how are the forces to hold the system in place changed?


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Conservation of Momentum - Memphis