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AE/ME5102 Advanced Gas Dynamics: Notes1 Set 3

Instructor: J. BlandinoWorcester Polytechnic Institute, Fall 2015

Phase Space

At a time t, a particles position is specified by a position vector in Cartesian coor-dinates:

r= x1i+x2j+x3k

These coordinates (x1, x2, x3) define the particles location in physical space. Thecorresponding velocity vector

C= dr

dt =C1i+C2j+C3k

has coordinates (C1, C2, C3) which define the position in velocity space. The dy-namic state of the particle is specified by the position and velocity, which collectivelydefine the particles location inphase space, with six coordinates (x1, x2, x3, C1, C2, C3).

It will be useful to define a differential volume element in physical space (using thenotation from Ref [1]):

dVx = dx1dx2dx3

So a particle located at a position r, will reside in a volume element of physical

space, with an x coordinate that lies between x1 and x1+dx1, a y coordinate thatlies between x2 and x2+dx2, and a zcoordinate that lies between x3 and x3+dx3

1Notes are based on material from the course text, Ref. 1: Introduction to Physical Gas

Dynamics, Vincenti, W. and Kruger, C., Krieger Pub., Copyright 1965. Any figures used from

Ref.1 are so noted and copyrighted by Krieger Publishing Co.

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Figure 1: Volume element in physcial space [1]

Similarly, we can define a volume element in velocity space as:

dVC=dC1dC2dC3

Figure 2: Volume element in velocity space [1]

A volume element in phase spacecan therefore be written as:

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dVxdVC=dx1dx2dx3dC1dC2dC3

So particles in dVxdVC are a subset of those in dVx that have an x component ofvelocity betweenC1 and C1+dC1 etc.

Distribution Function

General Considerations

Thedistribution functionF(r, C, t) represents aphase space density. So if we define

dNm(r, C, t) as the number of particles with coordinates r, C in a volume elementdVxdVCof phase space, then

dNm(r, C, t) =F(r, C, t)dVxdVC

IfF is a function ofr inhomogeneous(i.e. spatially non-uniform with a particlenumber density n = n(r)). Note: shortly we will see how the number density anddistribution function are related.

If F is independent of r homogeneous (i.e. spatially uniform with n = n0 =constant)

If F is homogeneous, then F(r, C, t) F( C, t) and is referred to as a velocity

distribution function. Often, the t dependence is implied and we just write F( C).

For this case, dNm = F( C)dVCrepresents the number of particles with a velocity

vector Cin a volume element dVC.

Normalized Distribution Function

It is often more convenient to work with a normalizeddistribution function. We candefine this as

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f(Ci) =F(Ci)

Nm (1)

where we have written the velocity vector using indicial notation, i.e. Ci = C.Also, note that Nm is the total number of particles, with all possible positions andvelocities in the region or system of interest.

With this definition, the number of particles in a differential element of velocityspace will be given by

dNm= Nmf(Ci)dVC

Since all the particles will have some velocity, if we inte-grate over all possible velocities, we have to recover thenumber of particles present:

+

Nmf(Ci)dVC=Nm

Note that sincedVC=dC1dC2dC3, the integral above isactually three integrals, one over each velocity compo-nent.

Also, since Nm is just a constant, then the normalizeddistribution function has the important property that

+

f(Ci)dVC= 1

Mean Value of Particle Properties

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We will see later that many properties of the gas are functions of the particle velocity

(e.g. energy and momentum fluxes). If Q(Ci) represents some property which afunction of the particle velocity, then we can define the average or meanvalue ofthis property using the normalized distribution function:

Q=

Nm

QdNm

Nm=

+

Q(Ci)Nmf(Ci)dVC

Nm

Q=

+

Q(Ci)f(Ci)dVC

Relation to Number Density n

First, lets consider the case where the spatial distribution is nonuniform. Thenumber of particles in a volume element of phase space is then

dNm(ri, Ci) =F(ri, Ci)dVxdVC

where again, thet dependence ofF is not shown explic-itly.

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We can then write for the number density

dn(ri, Ci) = dNm(ri, Ci)

dVx=F(ri, Ci)dVC

Integrating over all possible velocities give the particlenumber density at the particular location

n(ri) =

F(ri, Ci)dVC

Ifn(ri) in turn, was integrated over all of the (physical)volume V , then we would recover the total number ofparticles present.

Nm=

V

n(ri)dVx =

V

F

(ri, Ci)dVxdVC

Now lets consider the case where the particle distribution is spatially uniformn(ri) =n0.

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Recall from Eq. 1

f(Ci) = F(Ci)

Nm=

F(Ci)

n0V

and the total number of particles in the system Nm canbe recovered from the normalized velocity distributionf(Ci)

Nm = n0V =V

n0f(Ci)dVC

In this course, we will generally assume that the particledistributions are spatially uniform.

Maxwellian Velocity Distribution

The Maxwellianvelocity distribution function is of particular importance because,as we will see, it describes the velocity distribution for a gas in equlibrium.

The derivation of the Maxwellian velocity distribution is presented in detail in Ref[1], Chapter 2, Sections 4 and 5. We will first consider an overview of the derivationand then consider some of the new concepts and terminology introduced. Thedetailed steps are covered as part of the reading assignment from Ref [1].

Outline of Derivation

1. Some new concepts and terms to be introduced:

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- detailed balancing

- velocity classfor a molecule

- depletionand replenishment collisions

2. Presentation of geometry used to describe collisions

- relative velocities

- sphere of influence

- line of centers

3. For a collision between two classes of molecules, write an expression for thefollowing:

- differential element of volume a molecule would haveto occupy to be involved in a collision with a moleculeof a different class

- frequency of collisions between molecules of two veloc-ity classes

- net collision rate of molecules into a class

4. In equilibrium, the net rate at which molecules enter or leave a class is zero.This requirement is imposed as a condition on f(Ci)

Principle of Detailed Balancing

The state of a gas is fully known when nand f(Ci) are known.

Repeated collisions produce continual changes in f(Ci).

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After a sufficient time, if undisturbed, the system of gas molecules reaches a steady-

state such that nand f remain statisticallythe same at every point for all time.

Analog birth and death rates do not alter the statis-tical age distribution of a large population.

This state of affairs is referred to as equilibrium

So we can state the problem we wish to solve as follows: For a population of bil-liard ball like, perfectly elastic molecules, undergoing continuous collisions, find arelation that must be satisfied by the distribution function when the system is inequilibrium.

To express this relation mathematically, we must first define a few more terms:

A molecule is said to be of a particular class, Ui if itsvelocity lies in the range

U1 to U1+dU1

U2 to U2+dU2U3 to U3+dU3

i.e. the velocity lies in a velocity space volume elementdVU.

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Similarly, a molecule is of class Yi if its velocity lies in

the range

Y1 to Y1+dY1

Y2 to Y2+dY2

Y3 to Y3+dY3

and the velocity lies in a velocity space volume elementdVY.

A depletioncollision for class Ui will remove a moleculefrom the volume of velocity space dVU that defines theclass.

Areplenishmentcollision for classUiwill add a moleculeto the volume of velocity space dVU that defines theclass.

In equilibrium, the frequency of depletion collisions isbalanced by the frequency of replenishment collisions.

Collision Geometry

We define a vector g to represent the relative velocity of molecules with velocitiesUi and Yi

The magnitude of this relative velocity can be expressed as

g= [(Y1 U1)2 + (Y2 U2)

2 + (Y3 U3)2]

1

2

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As before, we can define a sphere of in-fluence

centered on moleculeA

The line of centers L is a vector con-necting the centers of molecule A andB, originating in A

For now, we assume the molecules allhave the same diameter d. Molecule Awill undergo a collision whenever thecenter of molecule B comes within adistance d of A

The collision geometry is described in terms of the angles and velocities shown inFigure 3. Note that in this figure, the sphere of influence is centered about themolecule of class Ui.

Figure 3: Geometry used to describe collision between two molecules with arbitraryrelative velocity [1]

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The overall strategy is as follows:

Step 1: find an expression for the frequency of collisionsbetween molecules of two classes, lying in a specifiedrange of angles and velocity space

Step 2: use the principle of detailed balancing to imposea condition which the distribution function for the gasin equilibrium must satisfy

Step 3: use this imposed condition to find the correct

mathematical form for the distribution function of a gasin equilibrium.

We will discuss the Step 1 in lecture; you will review Steps 2 and 3 in the reading.

We refer to the angles shown in Figure3and consider a collision between a moleculeof class Ui and Yi, such that the line of centers L lies between and to +d and +d

This range of angles determines a differential area element d on the sphere ofinfluence

d= dsin d d d

d= d2sindd

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In a time interval dt, a collision will oc-

cur for any molecule of classYi

with itscenter located in a differential volume

diff vol =d2sincosddgdt

This is the volume of a slant cylinder(shown in Figure3).

Now we want to know, how many molecules of class Uiare present in a unit volume?These will be the potential targets of the molecules of class Yi.

Using our previously defined, normalized, velocity distribution functionf(we dontknow the mathematical form of it yet of course, that is what we are trying todetermine!), the number of molecules of class Ui that are present in a unit volumecan be expressed as

nf(Ui)dVU where dVU =dU1dU2dU3

And the total volume ofal lthe slant cylinders (one for each of the above molecules)per unit volume is

nd2f(Ui)gsincosdddVUdt

How many molecules of class Yi have their centers within this collection of slant

cylinders?

# of molec. of class Yi per unit volume total vol. of slant cylinders per unit vol.

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(nf(Yi)dVY)

(nd2

f(Ui)gsincosdddVUdt)

This expression gives the number of collisions occurring in a time dt per unit volume.To get the rate, we divide the above by dt(and re-arrange) to obtain the following:

n2d2f(Ui)f(Yi)gsincosdddVUdVY

which has units of collisions per unit volume per unit time.

This expression concludes Step 1 of the three steps we listed earlier.

It is worth pausing to think about what the above result represents. It gives usan expression, using the geometry we defined (Figure3), for the number of collisionsbetween molecules in class Ui and class Yi, per unit time, per unit volume. But itis restricted to onlythose molecules who lie in a small range of velocity specifiedbydVU and dVY, and onlythose molecules whose line of centers lie within a narrowrange of angles d and d.

In the derivation of the Maxwellian distribution for a gas in equilibrium, this resultis used to find an expression for the netrate of collisions into a particular class.

Remember that the Principle of Detailed Balancing tells us that for a gas in equi-librium, this net rate will be zero.

The details presented in the reading assignment involve applying the constraintimposed by the Principle of Detailed Balancing (Step 2), and then working out theexact mathematical form of the velocity distribution function for a gas in equilibrium(Step 3). This function, referred to as the Maxwelliandistribution, is found to havethe following form:

f(Ci) = m

2kT

32

exp

m

2kT(C21 +C

2

2 +C2

3)

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