AE5102_Notes Set 4
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Transcript of AE5102_Notes Set 4
AE/ME5102 Advanced Gas Dynamics: Notes1 Set 4Instructor: J. Blandino
Worcester Polytechnic Institute, Fall 2015
Maxwellian Velocity Distribution (continued)
f(Ci) =( m
2πkT
)3/2exp
[− m
2kT
(C2
1 + C22 + C2
3
)]f(Ci) is product of three probability functions Φ(Ci):
Φ(C1) =( m
2πkT
)1/2exp
(−mC
21
2kT
)
Figure 1: Maxwellian distribution function [1]
1Notes are based on material from the course text, Ref. 1: Introduction to Physical GasDynamics, Vincenti, W. and Kruger, C., Krieger Pub., Copyright 1965. Any figures used fromRef.1 are so noted and copyrighted by Krieger Publishing Co.
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Drift Velocity
If the gas has a bulk or drift velocity, ~vD = C1i + C2j + C3k, then the driftingMaxwellian is modified
f(Ci) =( m
2πkT
)3/2exp
[− m
2kT
[(C1 − C1
)2+(C2 − C2
)2+(C3 − C3
)2]]
Speed Distribution Function
Another useful form of distribution function is the speed distribution function (i.e.without angle dependence).
Consider velocity space in spherical coordinates (earlier we considered physical spacein spherical coordinates):
Figure 2: Element of a velocity space in spherical coordinate system [1]
An element of volume in velocity space dVC can then be written
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dVC = dC1dC2dC3
in Cartesian coordinates, or in spherical coordinates as
dVC = C2sinφdφdθdC
First, rewrite the Maxwellian distribution in terms of speed C, where C2 = C21 +
C22 + C2
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f(Ci) =( m
2πkT
)3/2e−
mC2
2kT
Now we integrate over all angles. :
χ(C)dC =
∫π
φ=0
∫2π
θ=0
( m
2πkT
)3/2e−
mC2
2kT C2 sinφdφdθdC
=
∫π
φ=0
sinφdφ
︸ ︷︷ ︸2
∫2π
θ=0
dθ
︸ ︷︷ ︸2π
( m
2πkT
)3/2e−
mC2
2kT C2dC
χ(C) = 4π( m
2πkT
)3/2C2e−
mC2
2kT
Now, the average value of a quantity that is a function of speed Q(C) will be givenby:
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Q(C) =
∫ +∞
0
Q(C)χ(C)dC
Note that the Appendix of Ref [1] has integrals of form
In(a) =
∫ +∞
0
xne−ax2
dx
Collision Rate
A collision rate is different from collision frequency. The rate gives the total numberof collisions, per unit time, per unit volume whereas the frequency gives the totalnumber of collisions per unit time for one molecule. From previous discussion, wefound for the collision rate:
n2d2f(Ci)f(Zi)g sinψ cosψdψdεdVcdVz
If now we have two different species na,nb with diameters da and db, the samederivation as for above results in a more general expression:
nanbd2abfa(Ci)fb(Zi)g sinψ cosψdψdεdVcdVz,
where dab = da+db2
is the average diameter and the relative speed is again given by
g =[(Z1 − C1)
2 + (Z2 − C2)2 + (Z3 − C3)
2]1/2Substituting the Maxwellian velocity distribution functions fa(Ci) and fb(Zi) intothe expression for collision rate, one gets the number of collisions per unit time involume dVcdVz within angles dψdε (per unit volume of physical space)
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nanb(mamb)
3/2
(2πkT )3d2abg exp
[− 1
2kT
(maC
2 +mbZ2)]
sinψ cosψdψdεdC1dC2dC3dZ1dZ2dZ3
Before we can integrate to get total collision rate ( integrated over all angles andvelocities) we need to express C2 and Z2 in terms of g. To do this, look at center ofmass frame (for x component)
Xcm1 =maXa1 +mbXb1
ma +mb
To get the velocity, differentiate(W1 = dX1
dt
):
W1 =maC1 +mbZ1
ma +mb
This step transforms the problem from one in terms of velocities Ci and Zi into onein terms of Wi and gi.
x-component of relative velocity: g1 = Z1 − C1.
Substitute Z1 = g1 + C1 into expression for W1
W1 =maC1 +mbg1 +mbC1
ma +mb
⇒ C1 = W1 −mb
ma +mb
g1 (1)
Similarly
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Z1 = W1 +ma
ma +mb
g1 (2)
So the exponential term in the distribution functions becomes:
maC2 +mbZ
2 = (ma +mb)W2 +
mamb
ma +mb︸ ︷︷ ︸”reduced mass” m∗ab
g2
= (ma +mb)W2 +m∗abg
2
In order to transform variables of integration dC1dZ1 into dW1dg1 need to look atthe Jacobian:
dC1dZ1 =∂ (C1, Z1)
∂ (W1, g1)dW1dg1
where J = ∂(C1,Z1)∂(W1,g1)
represents the Jacobian determinant, formed from the partialderivative matrix.
Using the expressions given by equation (1) and equation (2):
∂ (C1, Z1)
∂ (W1, g1)=
∣∣∣∣∣ ∂C1
∂W1
∂C1
∂g1∂Z1
∂W1
∂Z1
∂g1
∣∣∣∣∣ =
∣∣∣∣1 −mbma+mb
1 mama+mb
∣∣∣∣ = 1
In the exponential term of the distribution function, W and g appear as speeds, notvelocities, i.e. (ma +mb)W
2 +m∗abg2
So, to facilitate the integrations we will need later, we convert velocity space volumedifferentials into differential volumes in terms of speed and angles
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dW1dW2dW3 =W 2 sinφWdφWdθWdW
and
dg1dg2dg3 =g2 sinφgdφgdθgdg
using same procedure we used for χ(C)
First, evaluate the integral over the center of mass velocity
nanb(mamb)
3/2
(2πkT )3d2abge
−m∗abg
2
2kT g2 sinφgdφgdθgdg
∫+∞
w=0
W 2e−(ma+mb)W
2
2kT dW
︸ ︷︷ ︸√π4
(2kT
ma+mb
)3/2
×
∫π
0
sinφWdφW
︸ ︷︷ ︸2
∫2π
0
dθW
︸ ︷︷ ︸2π
sinψ cosψdψdε
Also note that:
(mamb)3/2
(2πkT )6/2· 4π3/2
4·(
2kT
ma +mb
)3/2
=
(m∗ab
2πkT
)3/2
So after integration:
nanb
(m∗ab
2πkT
)3/2
d2abg3e−
m∗abg2
2kT sinφgdφgdθgdg sinψ cosψdψdε
Can now calculate Bimolecular Collision Rate Zab:
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Zab represents the number of collisions between particles of type a and b per unittime, per unit volume. Now we need to integrate over all possible relative velocities,i.e., we need to integrate the expression above with respect all possible values of g,θg, φg, ψ, ε.
nanb
(m∗ab
2πkT
)3/2
d2ab
∫+∞
g=0
g3e−m∗abg
2
2kT dg
︸ ︷︷ ︸12
(2kTm∗ab
)2
∫π
φg=0
sinφgdφg
︸ ︷︷ ︸2
×
∫2π
θg=0
dθg
︸ ︷︷ ︸
∫π/2
ψ=0
sinψ cosψdψ
︸ ︷︷ ︸12
∫2π
ε=0
dε
︸ ︷︷ ︸2π
Zab = nanb2
(2πkT
m∗ab
)1/2
d2ab
So the bimolecular collision rate is
Zab = nanb
√8πkT
m∗abd2ab
and has units of collisions/sec/m3
Note the difference between collision rate︸ ︷︷ ︸coll/sec/m3 for all
molecules of ”a” present
and collision frequency︸ ︷︷ ︸coll/sec for one
molecule of type ”a”
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Zab does not give collision rate of ”a” type molecules with ”a” type (or ”b” with”b”).
Cannot set na = nb and m∗ab = mb for example. This would count the collisionstwice, hence this introduces a factor of 1
2, so.
Zbb =n2b
2d2b
(8πkT
m∗bb
)1/2
where m∗bb =m2b
mb+mb= mb
2
This is the mutual collision rate, also with units of collisions/sec/m3
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