M E 433
Professor John M. Cimbala
Lecture 24
Today, we will:
•
Discuss the Cunningham correction factor for drag coefficient
•
Discuss terminal settling velocity and equations for its prediction in quiescent air
•
Discuss aerodynamic equivalent diameter and how to apply it
Review of equations for particle motion so far:

  

Relative particle velocity = vr= v − U , where v = particle velocity and U = air velocity.
d
dd
1
Drag force on a spherical particle = Fdrag = − r CDp D p 2vr vr , where
8

r vr D p
24
24
CD
CD =
for Re < 0.1 ,=
.
(1 + 0.0916Re ) for 0.1 < Re < 5 , and Re =
µ
Re
Re
d
3
ddd
dv p D p
1
r p − r ) g − r CDp D p 2vr vr .
Equation of motion for a spherical particle: m p =
(
dt
6
8
Correction to account for free molecular effects for very small particles:
Use a correction factor or “fudge factor” called the Cunningham correction factor:
λ

µ
π
 0.55  
1 + Kn  2.514 + 0.80exp  −
C=
, and λ =
.
  , where Kn =
Dp
0.499 8 ρ P
 Kn  

d
1 C
dd
Corrected drag force on a spherical particle = Fdrag = − r D p D p 2vr vr .
8 C
Example: Variation of Cunningham correction factor with particle diameter
Given: Air at STP has mean free path λ = 0.06704 microns. Knudsen number is defined as
Kn = λ / D p . Cunningham correction factor is C =
1 + Kn  2.514 + 0.80exp ( −0.55 / Kn )  .
(a)To do:
Calculate C for particle diameter Dp = 0.1 microns
Solution:
(b)To do:
Calculate C for various values of particle diameter Dp from 0.001 to 100 microns
Solution: [I used Excel for the repetitive calculations]
Dp (µm)
0.001
0.0025
0.006
0.01
0.025
0.06
0.1
0.25
C
222.7
89.43
37.60
22.79
9.489
4.355
2.921
1.702
Vt (m/s)
6.56E-09
1.65E-08
3.98E-08
6.71E-08
1.75E-07
4.61E-07
8.60E-07
3.13E-06
Re
4.20E-13
2.63E-12
1.53E-11
4.30E-11
2.79E-10
1.77E-09
5.51E-09
5.01E-08
C
Dp (µm)
0.6
1.282
1
1.169
2.5
1.067
6
1.028
10
1.017
25
1.007
60
1.003
100
1.0017
Vt (m/s)
1.36E-05
3.44E-05
1.96E-04
1.09E-03
2.99E-03
1.85E-02
1.06E-01
0.295
Plot of Cunningham correction factor vs. particle diameter at STP:
Re
5.22E-07
2.20E-06
3.14E-05
4.19E-04
1.92E-03
2.96E-02
4.08E-01
1.89
Corrected equation of motion for a spherical particle:
d
3
ddd
dv p D p
1 C
r p − r ) g − r D p D p 2vr vr .
mp =
(
dt
6
8 C
=
m p ρρ
But for a spherical particle, we also know its mass,=
pV p
p
p Dp3
. Plugging this
6
mass into the equation of motion (after a little algebra), we get our final expression,
d
dv r p − r ddd
3 r CD 1
=
g−
vr vr .
dt
rp
4 r p C Dp
Simplest application – Terminal settling velocity in quiescent air
Terminal settling velocity: Vt =
ρVt D p
4 ρρ
C
p −
gD p
, but CD = CD(Re), where Re =
.
µ
CD
3 ρ
Example: Terminal settling velocity as a function of particle diameter
Given: Air at STP with Cunningham correction factors from previous calculations. For
ρρ
C
p −
Dp 2 ) g .
Stokes flow (Re less than about 0.1), CD = 24 / Re , and Vt =
(
18
µ
To do:
Calculate Vt for various values of particle diameter Dp. Also calculate Re to test our Stokes
flow approximation. At STP conditions, r = 1.184 kg/m3, and µ = 1.849 × 10-5 kg/(m s). Use
g = 9.807 m/s2 (gravitational constant), and rp = 1000 kg/m3 (unit density spheres). [Give
your answers to 3 significant digits, and be careful with units.]
Solution:
Table to be filled in during class:
Vt (m/s)
Re
Vt (m/s)
Re
C
C
Dp (µm)
Dp (µm)
0.001
222.7
0.6
1.282
0.0025
89.43
1
1.169
0.006
37.60
2.5
1.067
0.01
22.79
6
1.028
0.025
9.489
10
1.017
0.06
4.355
25
1.007
0.1
2.921
60
1.003
0.25
1.702
100
1.0017
Filled in Table:
C
Dp (µm)
0.001
222.7
0.0025
89.43
0.006
37.60
0.01
22.79
0.025
9.489
0.06
4.355
0.1
2.921
0.25
1.702
Vt (m/s)
6.56E-09
1.65E-08
3.98E-08
6.71E-08
1.75E-07
4.61E-07
8.60E-07
3.13E-06
Re
4.20E-13
2.63E-12
1.53E-11
4.30E-11
2.79E-10
1.77E-09
5.51E-09
5.01E-08
C
Dp (µm)
0.6
1.282
1
1.169
2.5
1.067
6
1.028
10
1.017
25
1.007
60
1.003
100
1.0017
Vt (m/s)
1.36E-05
3.44E-05
1.96E-04
1.09E-03
2.99E-03
1.85E-02
1.06E-01
0.295
Re
5.22E-07
2.20E-06
3.14E-05
4.19E-04
1.92E-03
2.96E-02
4.08E-01
1.89