lectures 3 and 5 on rtt

Louisiana Tech UniversityRuston, LA 71272

Reynolds Transport Theorem

Steven A. Jones

Biomedical Engineering

January 8, 2008


Things to File Away

• Divergence Theorem

• If the integral of some differential entity over an arbitrary sample volume is zero, then the differential entity itself is zero.


Conservation Laws

• Conservation of mass:Increase of mass = mass generated + mass flux

• Conservation of momentum

Increase of momentum = momentum generated + momentum flux

• Conservation of energy

Increase of energy = energy generated + energy fluxIf more mass goes in than comes out, mass accumulates (unless it is destroyed).

If we take in more calories than we use, we get fat.


Conservation Laws: Mathematically

CV CS

d ddV dA

dt dt

v n

All three conservation laws can be expressed mathematically as follows:

Increase of “entity per unit volume”

Production of the entity (e.g. mass, momentum, energy)

Flux of “entity per unit volume” out of the surface of the volume

(n is the outward normal)

is some property per unit volume. It could be density, or specific energy, or momentum per unit volume.

is some entity. It could be mass, energy or momentum.

Steven A. Jones

Sometimes we denote vectors as v with an arrow above it.


The Bowling Ball

If you are on a skateboard, traveling west and someone throws a bowling ball to you from the south, what happens to your westward velocity component?

you you ball you ball you and ball

you and ballyouyou and ball you

you ball you

v 0 vm m m m

vmv v

m m v

(You slow down).


Reynolds Transport Theorem: Mass

CV CS

dm ddV dA

dt dt v n

If we are concerned with the entity “mass,” then the “property” is mass per unit volume, i.e. density.

Effect of increased mass on density within the volume.

Production of mass within the volume

Flux of mass through the surface of the volume

Mass can be produced by:

1. Nuclear reactions.

2. Considering a certain species (e.g. production of ATP).

Steven A. Jones



Reynolds Transport Theorem: Momentum

xx xCV CS

d mv dv dV v dA

dt dt v n

If we are concerned with the entity “mass,” then the “property” is mass per unit volume, i.e. density.

Increase of momentum within the volume.

Production of momentum within the volume

Flux of momentum through the surface of the volume

Momentum can be produced by:

External Forces.

Steven A. Jones



Mass Conservation in an Alveolus

CV CS

dm ddV dA

dt dt v n

Density remains constant, but mass increases because the control volume (the alveolus) increases in size. Thus, the limits of the integration change with time.

Term 1: There is no production of mass.

Term 2: Density is constant, but the control volume is growing in time, so this term is positive.

Term 3: Flow of air is into the alveolus at the inlet, so this term is negative and cancels Term 2.

Control Volume (CV)

Control Surface CS


Mass Conservation in an Alveolus

O2

CO2

Can look separately at O2 and CO2.

N2, O2, CO2 and others.

Third term is different:

(Inflow of O2 from the bronchiole) – (Outflow of O2 into the capillary system)

CV CS

dm ddV dA

dt dt v n


Heating of a Closed Alveolus

Heat

Density can be “destroyed” through energy influx, but the transport theorem still holds.

Term 1 is zero. No mass is created inside the control volume.

Term 2 is zero. The decrease in density is cancelled by the increase in volume.

Term 3 is zero. There is no flux of mass through the walls.

CV CS

dm ddV dA

dt dt v n


Air Compressed into a Rigid Vessel

CV CS

dm ddV dA

dt dt v n

Density increases so mass increases while the control volume (vessel) remains constant.

Term 1: There is no production of mass in the container.

Term 2: There is an increase in the total mass of air in the container.

Term 3: There is flow of air into the alveolus at the inlet.

Region R(m)

Surface S(m)


Differential Form

xx x

CV CS

dm ddV dA

dt dt v n

xv

zv

yv

xv

zv

yv Along the 2 faces shown, vy and vz do not contribute to changes in the mass within the cube. Only vx contributes.

dz

dy

dx

The left hand term is production of mass. The first term on the right is an increase in density within the cube, and the second term on the right is the outward flux of fluid. If the control volume is stationary, then:

Because mass is not being created or destroyed, the left hand term is 0.

CV CS

d ddV dV

dt dt


Differential Form – Conservation of Mass

xx xv

zv

yv

xv

zv

yv

We can get a differential form if we convert the last integral to a volume integral. The divergence theorem says:dz

dy

dx

CS CV

dA dV v n v

soCV CV

dm ddV dV

dt dt v

CV CS

dm ddV dA

dt dt v n


Differential Form

xx x

dz

zvzdzzvdzz

dy

yvydyyvdyydx

xvxdxxvdxx

t

zz

yy

xx

0xv

zv

yv

xv

zv

yv

0

z

v

y

v

x

v

tzyx Continuity Equation,

Differential Form


Divergence

0

vt

If density is constant then .0 v

When is density constant?

Conservation of mass reduces to:


Constant Density

• Generally density is taken as constant when the Mach number Mv/c is much less than 1 (where c is the speed of sound).

• For biological and chemical applications, this condition is almost always true.

• For design of aircraft, changes in density cannot necessarily be ignored.

• In acoustics (but nobody pays any attention when I say this).


RTT Applied to Momentum

CV CS

d ddV dA

dt dt v n

Increase of “entity per unit volume”

Production of the entity (e.g. mass, momentum, energy)

Flux of “entity per unit volume” out of the surface of the volume


is momentum per unit volume.

is now momentum. m v

v



CV CS

d ddV dA

dt dt v n

Increase of “momentum per unit volume”

Production of the entity momentum

Flux of “momentum per unit volume” out of the surface of the volume


is momentum per unit volume.

is now momentum. m v

v



syst

CV CS

d m ddV dA

dt dt

vv v v n

This v is part of the property being transported.

This v transports the property.



syst

CV CS

d m ddV dA

dt dt

vv v v n

Momentum has three components. Therefore, this is really 3 equations.

Momentum is “produced” by external forces. Therefore, the first term represents the forces on the control volume.

CV CS

ddV dA

dt F v v v n


Example 3.7

v2

v1

CV CS

ddV dA

dt F v v v n

F What resultant force is required to hold the section of tubing in place?

Steady state

2 2 2 2 2 1 1 1 1 1CSdA A A F v v n v v n v v n

2 2 2 2 1 1 1 1 2 1A V A V m F v v v vWhite reduces to:


Example 3.7

v2

v1

F

2 2 2 2 1 1 1 1cosxF AV V AV V

2 2 2V v n

2 2 2 2 1 1 1 1

1 1 1 1

sin 0 sin

sin

yF AV V AV V

AV V

1 1 1V v n


Momentum and Pressure

PinPout

wall

syst

CV CS

d m ddV dA

dt dt

vv v v n

CS


Example 3.1 from White

CV

1

2

3

Section Type (kg/m2) V (m/s) A (m2) e (J/kg)

1 Inlet 800 5 2 300

2 Inlet 800 8 3 100

3 Outlet 800 17 2 150

Find the rate of change of energy in the control volume.


Example 3.1 Continued

3 3 1 1 2 2CVsyst

dE de dV e m e m e m

dt dt

If the system is in steady state (i.e. there is no change with time of the energy within the control volume), then the integral is zero. Thus, the loss of energy through the control surface must be balanced by a “production” of energy.

This example is a bit misleading because “production” may be considered to be a flux of energy through the control surface. However, production could also be caused by, for example, a chemical reaction.


Example 3.1 Continued

3 3 3 3 1 1 1 1 2 2 2 2syst

dEe AV e AV e AV

dt

-2.4 MW

-1.92 MW

4.08 MW

-2.4 + -1.92 = -4.32 (>4.08), so there is more energy coming in than going out. Therefore, the “box” must “destroy” the energy (e.g. by doing work).

lectures 3 and 5 on rtt

Documents

entity mass

production of mass

mass increases

volumeflux of mass

effect of increased

production of momentum

increase of momentum

control volume vessel