Review
How to iterate to find Vt (terminal settling velocity for gravimetric settling) or vr (terminal radial
velocity for inertial separation)
Example
Given:
• Particles of 2.5 µm diameter with particle density = 1600 kg/m3.
• Inertial separation with Uθ = 10.0 m/s and average radius of curvature rm = 0.25 m.
• Air at STP: ρ = 1.184 kg/m3 and µ = 1.849×10-5 kg/(m s).
To do: Calculate the terminal radial velocity for these particles:
Solution: The equations we need are:

r vr D p

π
 0.55  
Kn =
1 + Kn  2.514 + 0.80exp  −
, λ=
, C=
.
  , CD = CD(Re), where Re =
µ
Dp
0.499 8 ρ P
 Kn  

λ
µ
There are many equations for drag coefficient of a sphere, but the most useful ones are: Stoke’s flow approximation:
24
24
CD =
CD
for Re < 0.1 , and=
(1 + 0.0916 Re ) valid for any Re < 5 . I like to use this latter one always
Re
Re
2
24
CD
1 + 0.158Re 3 for 5 < Re < 1000 .
unless the Reynolds number exceeds 5, for which=
Re
Also, the equations for terminal settling velocity and terminal radial velocity are:
(
•
Gravimetric settling: Vt =
•
Inertial separation: vr =
In our =
example, λ
Kn
=
λ
=
Dp
m
π
=
0.499 8 ρ P
)
ρVt D p
4 ρp − ρ
C
gD p
, Re =
µ
3 ρ
CD
r vr D p
Uθ 2
C
4 rr
p −
Dp
, Re =
µ
rm
CD
3 r
 106 μm 
1.849 × 10-5 kg/(m s)
π
=

 0.06704 μm ,
0.499
8 (1.184 kg/m3 )(101325 N/m 2 )  m 

0.55  
0.06704 μm

= 0.026816 , and Cunningham = C =
1 + 0.026816  2.514 + 0.80exp  −
1.0674 .
 =
2.5 μm
 0.026816  

Now we need to set up our iteration. First we guess vr . Stokes approximation is typically a good first guess, but you
ρp − ρ 2 C
Dp g
can pick any guess you want – it just might take longer to converge. Stoke’s approximation is Vt =
µ
18
for gravimetric settling, and we substitute Uθ 2/rm in place of g for inertial separation, vr =
rr
p −
18
U 2  C
Dp 2  θ  .
 rm  m
So, for our problem at hand, a good first guess is
2  (10.0 m/s )
 U 2  C (1600 − 1.184 ) kg/m3
vr =
D p  θ  =( 2.5 × 10−6 m ) 
 0.25 m
18
18
 rm  m


1.0674
m

=
0.012819 .
−5
 1.849 × 10 kg/(m s)
s

But let’s show how we can converge on the correct answer even with a different first guess. Let’s pick 0.02 m/s. We
set up a table to iterate, using the following equations in sequence, and then using the new vr as our next guess:
rr
p −
Re =
2
r vr D p
Uθ 2
24
4 rr
C
p −
CD
Dp
,=
.
(1 + 0.0916 Re ) , vr =
µ
3 r
rm
CD
Re
This is most easily done in Excel, but I show it here “by hand” in a table:
2
vr (m/s)
0.02
0.01601
0.01432
0.013549
0.013178
0.012996
Re
0.0032017
0.002563
0.02293
0.002169
0.0021096
0.0020805
CD
7498.14
9366.47
10468.3
11066.8
11378.7
11567.9
Etc. After a few more iterations (not shown), we converge well enough to the final values of:
Re = 0.00205 , CD = 11699.2 , vr = 0.01282 m/s
New vr (m/s)
0.01601
0.01432
0.013549
0.013178
0.012996
0.012906