Sedimentation
Rep = d U ρf / µ
(2.1)
where:
d = particle diameter
µ = absolute viscosity of the fluid
At low fluid velocities (0<Rep<2) we have:
CD = 24/Rep and FD = 3πµUd
(2.4 and 2.2)
At intermediate fluid velocities (2<Rep<500) we have:
CD = 18.5/Rep0.6
(2.5)
Finally at high fluid velocities (500<Rep<2x105) we have:
CD = 0.44
FD = 3πµUd/Cc
(2.7)
Where Cc is the Cunningham slip function:
Many particles in the environment have irregular shapes (e.g. bacteria, clay minerals, precipitates, etc.).
When particles are nonspherical we can use a correction factor:
FD = χ 3πµUde
(2.9)
χ = shape factor or dynamic shape factor
de = equivalent-volume diameter
F g = M pg - M fg
(2.12)
where:
g = acceleration of gravity
Mf = mass of fluid displaced by the particle
Mf g = buoyancy force
F D = ½ C D ρ fU 2A
(2.3)
where:
CD = the steady drag coefficient
ρf = density of the fluid
A = projected area of the particle
U = free stream velocity (i.e. uniform velocity of the flow field at a distance far enough from the particle that
velocity is unaffected by the presence of the particle)
4/9
Stokes law
vt =
ρp − ρf
ρp
gτ
(2.17)
τ = particle characteristic time [T]
⎛ 18µ
τ=⎜
⎜ C ρ d2
⎝ c p
⎞
⎟
⎟
⎠
−1
(2.15)
For sphere in water:
24
for 0 ≤ Re ≤ 2
Re
18.5
CD = 0.6 for 2 ≤ Re ≤ 500
Re
CD = 0.44 for 500 ≤ Re ≤ 2 ×105
CD =
S = v 0τ
(2.41)
where S = stopping distance
For aerosol at steady state
M pg −
2 2
1 π CD ρ f d U
=0
8
Cc
Terminal velocity
⎛ 4gdCc ρ p ⎞
vt = ⎜
⎟
⎝ 3C D ρ f ⎠
substitute Re =
C D Re =
2
1/2
dvt ρ f
µ
C D d v ρ 2f
2 2
t
2
µ
=
4d 3ρ f ρ p gCc
3µ 2
5/9
Surface interactions
1. Debye Huckel Approximation ( potential away from a charged surface, valid for low potentials)
ψ = ψ 0 e−κ x , where,κ 2 =
e2
εε 0 kT
∑z n
2
i
i0
i
The ionic number concentration
⎛ mol ⎞ ⎛
L⎞ ⎛
1 ⎞
nio = ⎜ Ci
× ⎜ 1000 3 ⎟ × ⎜ N A
= 1000N ACi
⎟
L ⎠ ⎝
mol ⎟⎠
⎝
m ⎠ ⎝
substitute and find
−1/2
⎡
⎤
κ = 4.3× 10 ⎢ ∑ zi2Ci ⎥
⎣ i
⎦
−1/2
−1/2
1
Ionic strength I = ∑ zi2Cio ;κ −1 = 4.3× 10−10 ⎡⎣ 2I ⎤⎦ = 3.04 × 10−10 ⎡⎣ I ⎤⎦
2 i
−1
−10
(3.15,Clark)
2. Relationship between surface charge and surface potential
σ = εε 0κψ 0
(3.22, Clark)
3. Electrical Double layer interaction potential between two spheres with different ψ0
φEDL = φ =
R
⎤
⎛ 1 + exp ( −κ h ) ⎞
πεε 0 a1a2 (ψ 012 +ψ 022 ) ⎡ 2ψ 01ψ 02
ln
+
ln
1
−
exp(
−
2
κ
h
)
(
)
⎢ 2
⎥
⎜
⎟
2
⎜
⎟
(a1 + a2 )
⎢⎣ (ψ 01 +ψ 02 ) ⎝ 1 − exp ( −κ h ) ⎠
⎥⎦
(from Hogg et al, eq 21)
Where,
a1 and a2 are the radii of the two spherical particles, particle 1 and 2 respectively,
ψ01 and ψ02 are the surface potentials of the spherical particles 1 and 2.
h is the distance between the surfaces of the sphere.
4. Van der Waals forces between two spherical particles
Aa
12h(1 + 14h / λ )
A: Hamaker constant; λ : dielectric constant
Vvdw( J ) = −
Fluid mechanics
Equation of motion
∂
ρ v = − [∇ ⋅ ρ vv ] − [∇ ⋅τ ] − ∇P + ρ g
∂t
6/9
For Newtonian and incompressible fluid, we have Navier-Stokes equation
Dv
= µ∇ 2 v − ∇P + ρ g
Dt
2
∇ laplacian; ∇ gradient; D material or substantive derivative
ρ
Equation of continuity in cylindrical coordinates
∂ρ 1 ∂
1 ∂
∂
+
( ρ rvr ) +
( ρ vθ ) + ( ρ vz ) = 0
∂t r ∂r
r ∂θ
∂z
Equation of continuity in rectangular coordinates
∂ρ ∂
∂
∂
+ ( ρ vx ) + ( ρ vy ) + ( ρ vz ) = 0
∂t ∂x
∂y
∂z
Equation of motion in cylindrical coordinates
z − component
⎡ ∂vz
⎡ 1 ∂ ⎛ ∂vz
∂v v ∂v
∂v ⎤
+ vr z + θ z + vz z ⎥ = µ ⎢
⎜r
∂r r ∂θ
∂z ⎦
⎣ ∂t
⎣ r ∂r ⎝ ∂r
ρ⎢
2
2
⎞ 1 ∂ vz ∂ vz ⎤ ∂P
+
+
+ ρg
⎥−
⎟
2
∂z 2 ⎦ ∂z
⎠ r ∂θ
Equation of motion in rectangular coordinates
x − component
⎛ ∂vx
∂v
∂v
∂v
+ vx x + v y x + vz x
∂x
∂y
∂z
⎝ ∂t
ρ⎜
⎛ ∂ 2v ∂ 2v ∂ 2v
= µ ⎜ 2x + 2x + 2x
∂y
∂z
⎝ ∂x
⎛ ∂τ xx ∂τ yx ∂τ zx
⎞
+
+
⎟ = −⎜
∂y
∂z
⎠
⎝ ∂x
⎞ ∂P
+ ρ gx
⎟−
⎠ ∂x
⎞ ∂P
∂P
+ ρ g x = µ∇ 2 vx −
+ ρ gx
⎟−
∂x
⎠ ∂x
Newton’s law of viscosity for laminar flow
⎛ ∂v ∂v ⎞
shear stress τ i j = − µ ⎜ i + j ⎟
⎜ ∂x
⎟
⎝ j ∂xi ⎠
⎛ ∂v ⎞ 2
normal stress τ ii = −2µ ⎜ i ⎟ + µ ( ∇ ⋅ v )
⎝ ∂xi ⎠ 3
7/9
For flow through a pipe
2π R
total volumetric flow rate
vz,ave =
=
cross section area
∫ ∫ v r dr dθ
z
0 0
2π R
=
∫ ∫ r dr dθ
0 0
mass flow rate=cross section area × density × vz,ave =
the z-component of the fluid force on the wetted surface of the pipe is the shear stress
intergrated over the wetted area
⎛
dv ⎞
Fz = (2π RL) ⎜ − µ z ⎟
dr ⎠ r= R
⎝
Stokes flow
We express the velocity components in spherical coordinate using a stream function ψ as follows:
1 ∂ψ
r sin θ ∂θ
1 ∂ψ
vθ = +
r sin θ ∂r
vφ = 0
vr = −
2
We can obtain the following solution
⎡⎛ a ⎞ 3 ⎛ a ⎞ ⎤
U
vr = − cos θ ⎢⎜ ⎟ − 3 ⎜ ⎟ + 2⎥
2
⎝ r ⎠ ⎥⎦
⎢⎣⎝ r ⎠
(5.52)
⎡⎛ a ⎞3 ⎛ a ⎞ ⎤
U
vθ = − sin θ ⎢⎜ ⎟ + 3 ⎜ ⎟ − 4⎥
4
⎝ r ⎠ ⎦⎥
⎣⎢⎝ r ⎠
(5.53)
The pressure distribution at the surface of the sphere is
3µU
P − P∞ =
2a
2
⎛a⎞
⎜ ⎟ cos θ
⎝r⎠
(5.54)
For a particle, shear stress
⎡ ∂ ⎛ v ⎞ 1 ⎛ ∂v ⎞ ⎤
τ rθ = µ ⎢ r ⎜ θ ⎟ + ⎜ r ⎟ ⎥
⎣ ∂r ⎝ r ⎠ r ⎝ ∂θ ⎠ ⎦
8/9
Solution for Bernoulli equation:
P v2
+
+ gz = cons tan t
ρ 2
where:
z = elevation above some fixed datum
constant = sometimes known as Bernoulli’s constant, is constant along streamlines of constant velocity.
9/9