ME2342_Sec4_CVAnalysis

7/29/2019 ME2342_Sec4_CVAnalysis

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P. S. Krueger ME/CEE 2342 4 - 1

ME/CEE 2342:

Paul S. Krueger

Associate Professor

Department of Mechanical Engineering

Southern Methodist UniversityDallas, TX 75275

[email protected]

(214) 768-1296Office: 301G Embrey

Fluid Mechanics

Section 4 Control Volume Analysis

[Chapters 5 & 6 in the text book]
mailto:[email protected]:[email protected]


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Sometimes we are only interested in the global behavior in a

specific region of the flow. We have two approaches to

analyze large chunks of flow:

1) Control Mass: Look at a fixed quantity of matter (a closed

system)2) Control Volume: Look at a specific region in space.

A control volume (CV) is a region we specify. It can move,change shape, or remain fixed depending on what we need.

Because CVs focus on what we want, they are usually more

convenient for analyzing fluid flows.

Mass and Energy Equations (Ch. 5)

System and Control Volume (CV) Concepts


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Conservation of Mass for a CVConsider flow through a diverging channel and select a fixed

region in the channel as a CV:


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Then, conservation of mass says that mass is not created or

destroyed. Physically

Rate of change

ofMCV=

Rate at which

mass enters Rate at which

mass leaves

Mathematically we write this as



To find the rate at which mass leaves the CV, consider flow

across a small region of the outlet:

Then the total mass crossing (2) during t is



Similarly,

Putting it all together:

This works for pipes with inlets and outlets. A more generalform is to combine the inlet and outlet integrals into an

integral over the entire control volume surface (CS):

( ) 0 =+

CSCV

dAdVt

nu



Notes:

The CS integral is an area integral (2D) over the surface

of the control volume is the outwardunit normal to the CSn

nu = speed normal to the CS. We use only thenormal component because we are only

interested in what actually crosses the CS.

0 >nu Outflow

0


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So far we have been considering only fixed CVs. But if the

surface of the CV is allowed to move, we have

( ) 0 =+

CS

r

CV

dAdVt

nV

where

CSr uuV

Is the velocity relative to the CS (uCS is the control surface

velocity specified by you!)


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( )dt

dBdAbdVb

t

sys

CS

r

CV

=+

nV

is called Reynolds Transport Theorem (RTT) and it relates

changes in system properties to changes in the CV.

For example,


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Conservation of Mass Examples:

1) Isolated Inlets and Outlets

Find U3


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Edges are placed where info is known ordesired.

Edges are perpendicular to u (if possible) at inlets andoutlets.

Notes on selecting a CV:

Now apply conservation of mass (COM):


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Thus,

Rearranging,


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Notice that for a general steady flow we have

=outin mm

&&

where

= dAVm n& and nu =nVIf the flow density is constant across each inlet and outlet

(though not necessarily the same at each inlet/outlet) we have

For steady, incompressible flow


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2) Flow through a garage

Find V(same for both windows). Assume incompressible,steady flow


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COM:

Outlets:

Inlet:

Combine:


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3) Non-Uniform Channel Flow

Find V. Assume incompressible, steady flow


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COM:

Inlet:

Outlet:


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In this frame of reference, the flow is steady and we write

COM as:

but now we compute asm&

Thus,


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Fi t L f Th d i f CV


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First Law of Thermodynamics for a CV

(Energy Equation)

For a system (fixed quantity of matter), the First Law of

Thermodynamics states

where

We can transform this into an expression for CVs using

RTT. In this class, however, we will consider only a specialtype of CV when dealing with the energy equation

S i l C St d i ibl fl th h


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Special Case: Steady, incompressible flow through astream tube (which is a CV with one inlet and one outlet)

Under these conditions the 1st Law says:

Types of energy:


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Types of energy:

So, we can write

Work:


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Work:

where

Recall,

On the walls:


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On the walls:

At the inlet and outlet:

Putting everything together:


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is called the kinetic energy correction factor and


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is called the kinetic energy correction factorandaccounts for non-uniformity of the flow:

= 1 for uniform flow

> 1 otherwise

Lets rewrite the current result into a form more convenient

for future calculations:

By experiment/experience (2nd Law of Thermodynamics) weknow that

To simplify the notation, lets define


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To simplify the notation, let s define

This gives the final form:

LTp

in

ave

sout

ave

s

hhhzg

Vpz

g

Vp+

+

+

=

+

+

22

22

which holds for steady, incompressible flow through a

streamtube.

Notes:


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The units of all terms in this expression are length (m, ft).

Notes:

Is called the total head and represents the height of astatic fluid particle with potential energy per unit mass

equal to gzVp ave ++ 22


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Energy equation:


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gy q

2) Fireboat Pump


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) p

What horsepower motor is required to drive the pump?

Begin by drawing and labeling the streamtube:


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Energy Equation:

Substitute known values:


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But:

We can find Q from

3) River Flow


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Determine the head loss due to the drag on the rocks in the

river

Energy equation:


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For the streamtube we have selected, the energy equation

b


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becomes


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Note:


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Bernoullis Equation: Special Case of


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Streamtube ResultsConsider a tube bounded by streamlines:

Note: Flow does not cross streamlines (it is tangent to the

streamlines) so this streamtube has only one inlet and one

outlet as before.

For steady incompressible flow with no shafts crossing the

streamtube the energy equation gives:


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streamtube, the energy equation gives:

Now assume that viscous effects are negligible along the

streamtube. This means that

This is often satisfied if the streamtube is

not near any walls

not within a pipe

Finally, let A 0 so that the streamtube reduces to astreamline In this limit


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streamline. In this limit

Thus, we arrive at Bernoullis Equation:

1

211

2

222

22z

g

Vpz

g

Vp

ss

++

=++

1

2

112

2

222

1

2

1zVpzVp ss ++=++

constant2

1 2 =++ zVp s

or,

or,

The assumptions used to arrive at Bernoullis equation are


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Note that will, in general, be a different

constant on different streamlines (SLs). That is, we cannot

apply Bernoullis equation across streamlines in general (see5-4 in the textbook). In a special case, however, the

difference in the constant across streamlines is easy to

determine

zVp s++ 22


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Alternatively, applying Bernoullis equation on two streamlines

we have


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we have

Using the force balance result we have

That is, is the same constant on all streamlines

for straight, horizontal streamlines

zVp s++ 22

Bernoullis Equation Examples:


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1) Pitot Tube (Velocity Measurement)

We want to use Bernoullis equation to relate V to p1 and p2.

To use Bernoullis equation, first draw and labelthe

streamline you will be using with Bernoullis equation.

Bernoullis equation applied to our chosen SL gives


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From our previous results we know

Combine:

Definitions:


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Static Pressure:

Dynamic Pressure:

Total Pressure:

Stagnation Point:

Example: Pitot-Static Probe


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Pitot Tube Example: Boeing B-52


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Static Port

Pitot Tube

Pitot Tube Example: Coast Guard HU-25 FalconJet


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Pitot Tube

Static Port

2) Free Jet


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Find the jet velocity at (2). Lets apply Bernoullis equation:

What is p2? Consider a close-up view of the nozzle:


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The streamlines are straight and horizontal, so the pressure

variation across the streamlines is hydrostatic. But, the fluid is

in free fall here with nothing to support its weight, so thepressure change across streamlines should be zero! Thus,

the pressure inside the jet equals the pressure outside the jet:

atmjet pp = Free-Jet Condition


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3) Venturi Tube (Flow Rate Meter)


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Begin with Bernoullis equation between (1) and (2):

But V1 and V2 are related through COM:


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Substituting into Bernoullis equation:

Or, in terms of volume flow rate Q:

Note:


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Words of caution: If you cant draw a streamline between two

points, you (probably) cant use Bernoullis equation.


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Newtons 2nd Law for a CV (N2 or

Momentum Equation)


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Momentum Equation)

For a particle, Newtons 2nd Law is

= aF mparticleFor a system of particles, we write

=dt

d syssys

PF

where Psys is the total momentum of the system. We can use

RTT to transform this into an equation for a CV, or just write

down by inspection:

Net force on

the CV=

Rate of change

of momentum

in the CV

Net rate at which

momentum

leaves the CV

Mathematically we have

( )+

dAdV nVuuF


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( )

+

=CS

r

CV

CV dAdV

t

nVuuF

where Vr is the velocity relative to the CS as defined earlier.

Note that u is momentum per unit volume.

Forces on a CV:

Surface Forces:

Normal (Fn):


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Normal (Fn):

Shear (Fs):

Body Forces: Weight


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External: Supports crossing CV surface, etc.

Putting it all together:

++= extsCVCS

CV dVgdAp FFznF

Just remember, there are fourforces you need to account for: Pressure

Weight

External

Shear (mostly seen in pipe flow and

boundary layers)

N2 Examples:

1) Force on a Tank


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1) Force on a Tank

Assume incompressible and steady flow. What reaction force

from the ground is necessary to hold the tank in place?

Step 1: Draw CV & FBD


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Forces:

Step 2: Use available tools to find unknowns

Bernoullis Equation: Find unknown velocities in


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frictionless flow (assuming steady and incompressible) Conservation of Mass: Find unknown velocity or area

(recall that COM relates velocities and areas for steady,

incompressible flow)In this case we use Bernoullis equation to obtain:

Step 3: Determine CVF

Step 4: N2 (Momentum Equation)


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Now substitute the results from step 3:


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Equate components:


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For non-uniform flow we have to be more careful.

Returning to N2 and imposing the steady flow assumption

we have


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we have

If the flow is parallel (i.e., all in the same direction at a

given inlet/outlet) then

( ) ( ) =inaveoutaveCV mm

VVF &&

Here we have used the following definitions:


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ForV= Vn,

= dAV

V

A ave

2

1

Notes:

is called the momentum-flux correction factor.


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(Vave)in and (Vave)out are not necessarily equal even if

|(Vave)in| = |(Vave)out|.

= 1 for uniform flow

> 1 otherwise


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Step 2: Unknown Velocity Use COM


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Step 3: Forces (look at x-y plane only)

Step 4: N2 (Momentum Equation)


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Equating Components:


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3) Jet Pump ( 1)


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Assume incompressible and steady flow. Find the pressure

difference between (1) and (3) ignoring shear on the walls.

Does shear increase or decrease the pressure difference?

Step 1: Draw CV & FBD


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Forces:

Step 2: [Unnecessary: no unknowns]


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Step 3: Forces (x-component only)

Step 4: N2 (x-component only)

By COM:

Because the inlet and outlet areas are the same we also have


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Newtons 2nd Law for a Moving, Non-deforming CV

Using a moving, non-deforming CV is most appropriate when

th i f i t t i i t t t l it I thi


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the region of interest in moving at a constant velocity. In thiscase, uCS is the same over the entire CV:

If the flow appears steady in the moving CV, then N2 reduces

to

Also, COM gives


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Combining COM and N2 we have


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Example: Jet Thrust (Moving CV)


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P. S. Krueger ME/CEE 2342 4 - 91Find the thrust force of the engine on the airplane (FT).


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Step 2: [Unnecessary: no unknowns]

Step 3: Forces (x component only)


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Step 3: Forces (x-component only)Step 4: N2 (x-component only)

ME2342_Sec4_CVAnalysis

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