Scientia Iranica B (2013) 20 (2), 351–358

Sharif University of Technology

Scientia Iranica
Transactions B: Mechanical Engineering
www.sciencedirect.com

Research note

Analytical study of single particle tracking in both free and

forced vortices
Alinaghi Salari ∗ , Mohsen Karmozdi, Reza Maddahian, Bahareh Firoozabadi
Mechanical Engineering Department, Sharif University of Technology, Azadi Ave. Tehran, 11155-9567, Iran

Received 8 May 2012; revised 11 November 2012; accepted 8 January 2013

KEYWORDS Abstract Today, the flow of gas–solid, solid–liquid and liquid–liquid mixtures is broadly used in many
Particle tracking; industries such as slurry transportation, propulsion, dredging and power generation equipment. In this
Free vortex; paper, single solid particle motion through free and forced vortices is analytically studied. The equations
Force vortex; are solved for cases in which the drag, pressure gradient, added mass, buoyancy and weight forces
Drag; are considered individually and simultaneously. Verification has been done for the value of ReP , which
Added mass; confirms the solution for the first t = 0.1 s. The results show that the most important force governing
Pressure gradient; particle motion is the pressure gradient force.
Buoyancy; © 2013 Sharif University of Technology. Production and hosting by Elsevier B.V.
Weight. Open access under CC BY license.

1. Introduction A review is placed on semi-empirical forms of the equation

of motion of rigid particles, bubbles, and droplets, which are
The goal of separating two immiscible materials has been widely used in engineering practices [7]. Rotation of a small
of great concern over the last few years. Also the flow of sphere suspended in a fluid vortex has also been widely
gas–solid and solid–liquid mixtures is so broadly used in many studied [8]. Some research has been done on mass transport
industries such as slurry transportation, propulsion, dredging, using a vortex ring [9], and some others have given a study of
power generation equipment and the pneumatic transport of vortex simulation for interaction and collision between a vortex
food grains [1]. ring and solid particles [10,11]. Some research has aslo been
In order to track a single particle in a fluid flow by the done to explain the pressure field in turbulent vortex rings [12].
Lagrangian method, it must be demonstrated that all forces The trajectory of an isolated solid particle dropped in the
acting on a particle are surely needed, and the relative amount core of a vertical vortex has been investigated theoretically and
of each force must be considered. experimentally, in order to analyze the effect of the history force
Many researchers examined the motion of a sphere settling
on the radial migration of the inclusion [13].
down under the gravity effect. An analytical solution can be
In this paper forces acting on a single solid spherical particle
derived only for a particle of Reynolds number less than 1
in both free and forced vortices are discussed.
(ReP < 1) [2–4]. Also, the equation of motion for a single
The particle shape is considered to be spherical for
particle in unsteady flow has been studied before [5].
Paths of small inertial particles in a steady Taylor vortex simplicity. Generally, a movable particle inside a viscous fluid
background flow were studied before [6]. flow is exposed to six forces, which are:

• Added mass & Basset

∗ Corresponding author. Tel.: +98 21 66165675. • Drag
E-mail address: a_salari@alum.sharif.edu (A. Salari). • Pressure gradient
Peer review under responsibility of Sharif University of Technology. • Buoyancy & Weight
• Lift
• Thermophoretic.

Now all forces are introduced.

1026-3098 © 2013 Sharif University of Technology. Production and hosting by Elsevier B.V. Open access under CC BY license.
doi:10.1016/j.scient.2013.02.011
352 A. Salari et al. / Scientia Iranica, Transactions B: Mechanical Engineering 20 (2013) 351–358

1.1. Drag force 1.4. Buoyancy & weight force

The drag force (FD ) which act on a spherical particle inside The forces which are acted upon by gravity (g) and buoyancy
the flow far from the boundaries has been introduced by effect (FB ) are as follows:
Stokes [14].  
1
3 ρF mP FB = π D3 (ρF − ρP ) g . (6)
FD = CD (uF − uP ) |uF − uP | (1) 6
4 ρP D
where, for Reynolds numbers less than one (ReP < 1), the drag 1.5. Other forces
24
coefficient is CD = Re . The quantities m, D, ρ and u are mass,
P
diameter, density and velocity respectively. Subscripts, P and F, In this paper, for our analytical study the assumption of a
represent particle and fluid respectively. tiny particle is made, so the effects of slip-shear lift force and
slip-rotational lift force are neglected. Moreover it is assumed
1.2. Added mass/Basset force that the system is under an isothermal condition, and therefore,
the effect of thermophoretic force is not considered. As a
Unsteady forces due to acceleration of a moving body consequence, the forces that are more effective in the studied
relative to fluid will cause the added mass effect and the case, are as follows.
Basset force. In other words any increase or decrease in particle Drag, added mass, pressure gradient, buoyancy and weight.
velocity causes the fluid near it to accelerate either. This amount
of fluid is called added mass. The force, which appears due to 2. Main text
the lagging boundary layer development with changing relative
In this section, using Newton’s law of motion along three
velocity, is known as the Basset force [15]. The amount of these
directions of cylindrical coordinates, the particle path is
forces acting on an accelerating spherical particle in a viscous
obtained. For this goal, the forces which were introduced in the
fluid was obtained [16,17]. The added mass force (FA ) and Basset
previous section for ReP < 1 are used.
force (FB ) are as follows.
The problem is broken down into two distinct approaches:
− ∂∂utP′ ′
 t DuF
3 √ Dt ′
• Considering each force individually, in which the problem is
FB = D π ρF µF
2
√ dt , a 2D problem.
2 0 t − t′
  (2) • Considering all forces simultaneously, in which the problem
DuF duP is a 3D problem.
FA = 0.5CA mF −
Dt dt In all parts, the initial conditions are as follows:
2
0.132
where CA = 2.1 − , AC = |uF −uP |  and µ is viscosity. rt =0 = R

Ac 2 +0.12  d|uF −uP | 
D dt 
ṙt =0 = 0 (7)
The parameters were introduced previously. θt =0 = 0.
In order to determine Basset force, the integral needs to be −
→
calculated in the time during which the particle is moving along Steady state fluid flow velocity ( uF ) for a general vortex in
the path. As the calculation is time consuming, the Basset force cylindrical coordinates (êr , êθ , êz ) is considered as follows:
is usually neglected [18,19]. −
→
uF = cr α êθ + u0 êz (8)
Fluid will gain kinetic energy at the expense of the work
done by the accelerating particle. This phenomenon will cause where êθ and êz are tangential and axial unit vectors in
added mass force. For a single spherical particle inside an cylindrical coordinates, respectively. α is +1 for forced vortex
inviscid incompressible fluid, the added mass force is written and −1 for free vortex.
in Eq. (3). In order to study each force individually, tangential and
radial components of the equation of motion are coupled and
ρ F π D3 Dvrel
 
need to be solved simultaneously. But, buoyancy and weight are
FA = . (3) considered an axial component of the equation of motion, and
12 Dt
can be solved separately. Therefore, to study the effect of each
The added mass force can be simplified for a rigid particle as force on particle motion, the problem is broken into 4 parts. In
follows. each part, a single particle under the influence of only one force
ρ F π D3 ∂vrel ∂ uF in both free and forced vortices is studied. At last, the problem
 
FA = + uF . (4) is solved one more time and all these forces are considered
12 ∂t ∂x
simultaneously.
It was studied that the Basset force has only a weak effect Since drag, added mass and pressure gradient forces act on
in the region of a vortex center and gradually becomes more the plane of relative velocity, resultant particle motion would
important near a separatrix. Its principal effect is to change the be in the mentioned plane.
magnitude of the drift of particles towards either a vortex center During the solving process, equations of motion are simpli-
or a separatrix [20]. fied to a level at which the analytical solution can be achieved. In
some cases, where no analytical solution is obtained, a numer-
1.3. Pressure gradient force ical second order Runge–Kutta approach is used by MATLAB.

The force that is a result of pressure gradient along fluid flow 2.1. Part 1: considering only drag force
around particle is called the pressure gradient force (FP), which
is as follows: The drag force which was introduced in Eq. (1) can be
  rewritten for ReP < 1 as follows:
1 DuF
FP = π D3 ρF . (5) −
→
FD = 3π Dµ ûF − ûP .
 
6 Dt (9)
 A. Salari et al. / Scientia Iranica, Transactions B: Mechanical Engineering 20 (2013) 351–358 353

A 2D particle velocity is considered in Eq. (10). Substituting Eq. (20) in Newton’s second law yields Eq. (21).
−
→
uP = ṙ êr + r θ̇ êθ (10) ρF π D3 
−c 2 r 2α−1 − r̈ + r θ̇ 2 êr + −2ṙ θ̇ − r θ̈ êθ
   
where êr is the radial unit vector in cylindrical coordinates. 12
−
→
And particle acceleration ( aP ) is defined in Eq. (11).

1

ρP π D3
r̈ − r θ̇ 2 êr + 2ṙ θ̇ + r θ̈ êθ .
    
= (21)
−
→ 6
aP = r̈ − r θ̇ 2 êr + 2ṙ θ̇ + r θ̈ êθ .
   
(11)
Eq. (21) can be broken up into tangential and radial components
Substituting Eqs. (9) and (11) in Newton’s second law yields Eq.
separately
(12).

3π Dµ cr α − r θ̇ êθ − ṙ êr r̈ − r θ̇ 2 = −ρF c 2 r 2α−1
  
  2ρ P + ρF (22)
1 2ṙ θ̇ + r θ̈ = 0.
ρP π D3 r̈ − r θ̇ 2 êr + 2ṙ θ̇ + r θ̈ êθ .
     
= (12)
6
The second equation in Eq. (22) can be easily solved by using
the initial conditions (Eq. (7)), which yields Eq. (23).
2.1.1. Forced vortex
In case of a forced vortex, α = +1, the tangential component r 2 θ̇ = const = R2 θ̇0 . (23)
of equation of motion (Eq. (12)) is neglected. Also, the angular Substituting Eq. (23) into the first equation in Eq. (22) yields
velocity of the particle is assumed to be constant at fluid angular Eq. (24).
velocity.
ρF
c 2 r 2α+2 − R2 θ̇0
2
r 3 r̈ + = 0.

(24)
θ̇ = c . (13) 2ρ P + ρF
The radial component is used to solve particle motion.

r̈ − rc 2 + Aṙ = 0 (14) 2.2.1. Forced vortex

18µ
Eq. (24) for α = +1 can be rearranged into Eq. (25).
where A = − ρ D2 .
P 2 ρF
R2 θ̇0

By using the initial conditions stated in Eq. (7), the solution − 2ρ P +ρF
c2r 4
is as follows. r̈ = . (25)
 r3
R Eq. (25) can be integrated by use of the initial conditions and
r = (β2 exp (β1 t ) − β1 exp (β2 t ))

β2 − β1 (15) r̈ = ṙ ddrṙ . Also it is assumed that initial velocity of the particle is
 θ = ct the same as the fluid velocity, which means that Rc = Rθ̇0 .
where: 
√ 
 ρP + ρF  ρP
 
−A ± A2 + 4c 2 4ρF
β1,2 = . (16) r =R  − cos ct . (26)
2 ρF ρF 2ρP + ρF

Then angular position is obtained by substituting Eq. (26) in

2.1.2. Free vortex Eq. (23) and integrating.
In case of a free vortex, α = −1, the tangential component of  
particle velocity is assumed to be constant at the fluid velocity. 2ρ P + ρF ρF
 
1
θ = 2arctan tan ct . (27)
c ρF 2 2ρ P + ρF
r˙θ = . (17)
r
After differentiation, it yields Eq. (18).
2.2.2. Free vortex
−2c ṙ Eq. (24) for α = −1 can be rearranged.
θ̈ = . (18)
r3 2 ρF
R2 θ̇0

After substituting Eq. (18) in Eq. (12), the equation of tangential − 2ρ P +ρF
c2
r̈ = . (28)
component is an obvious solution. So, the radial component r3
yields Eq. (19). Eq. (28) can be integrated by use of initial conditions and r̈ =
dṙ
r r̈ + Ar ṙ − c = 0. 2
(19) ṙ dr . Similarly, it is assumed that the initial angular velocity of

There is no analytic solution for this equation, therefore, a

the particle is the same as the fluid velocity, so c
R
= Rθ̇0 .
numerical method is used. 
2ρP c2
 
r =R 1+ t 2. (29)
2.2. Part 2: considering only added mass force 2ρ P + ρF R4

The added mass force introduced in Eq. (3) can be rewritten Then, the angular position is obtained by substituting Eq. (26) in
for cylindrical coordinates as follows. Eq. (23) and integrating.

ρF π D3 
   
−
→ 2ρP c
FA = −c 2 r 2α−1 − r̈ arctan 2ρ P +ρF R2
t
12 θ=  . (30)
2ρP
+ r θ̇ 2 êr + −2ṙ θ̇ − r θ̈ êθ .
   
(20) 2ρ P +ρF
354 A. Salari et al. / Scientia Iranica, Transactions B: Mechanical Engineering 20 (2013) 351–358

2.3. Part 3: considering only pressure gradient force of forces, so we have Eq. (40).

cr α − r θ̇ êθ + ṙ êr

Fr ,θ = 3π Dµ
  
The pressure gradient force introduced in Eq. (5) can be
rewritten for cylindrical coordinates as follows.
ρF π D3 
−c 2 r 2α−1 − r̈ + r θ̇ 2 êr + −2ṙ θ̇ − r θ̈ êθ
   
1 2 2α−1
+
FP = − π D ρF c r 3
êr . (31) 12
6 1
Substituting Eq. (31) in Newton’s second law yields Eq. (32). − π D3 ρF c 2 r 2α−1 êr . (40)
6
1
− π D3 ρF c 2 r 2α−1 êr Using Newton’s second law and Eq. (40), the equation of motion
6
  is achieved.
1
ρP π D3 r̈ − r θ̇ 2 êr + 2ṙ θ̇ + r θ̈ êθ .
    
= (32)
36µ

ρF

6
r̈ − r θ̇ 2 = − ṙ − 3c 2 r 2α−1 (41)
Eq. (32) can be broken up into tangential and radial components (ρF + 2ρP ) D2 ρF + 2ρP
separately 36µ  α
2ṙ θ̇ + r θ̈ = cr − r θ̇ .

 (42)
r̈ − r θ̇ 2 = −ρF c 2 r 2α−1 (ρF + 2ρP ) D2
ρP (33)
2ṙ θ̇ + r θ̈ = 0.

2.4.1.1. Forced vortex. In order to determine the analytical
The second equation in Eq. (33) can be easily solved by using solution of the equations, the particle velocity is assumed to be
the initial conditions (Eq. (7)), which yields Eq. (34). the same as the fluid flow velocity around it.
r 2 θ̇ = const = R2 θ̇0 . (34)
r θ̇ = cr α . (43)
Substituting Eq. (34) into the first equation in Eq. (33) yields
Eq. (35). By substituting α = +1 in Eqs. (41), (42) and solving the
ρF 2 2α+2  2 2 equations with initial conditions stated in Eq. (7), the solution
3
r r̈ + c r − R θ̇0 = 0. (35) is obtained.
ρP
The general form of Eq. (35) is the same as Eq. (24), so only the

R
r = (β2 exp (β1 t ) − β1 exp (β2 t ))

final solution is presented.
β2 − β1 (44)
 θ = ct
2.3.1. Forced vortex
18ρF uF
β1,2 = −
 F + 2ρP ) D
(ρ
   2

 ρP + ρF  ρP − ρF 4ρF
18ρF uF ρP − ρF
2
r =R − cos ct (36)
   
2ρF 2ρF ρP ± + 2 c2. (45)
(ρF + 2ρP ) D2 ρF + 2ρP
ρP 1 ρF
   
θ = 2arctan tan ct . (37)
ρF 2 ρP 2.4.1.2. Free vortex. In this case, particle velocity is assumed
to be the same as the fluid flow velocity around it, too. By
2.3.2. Free vortex substituting α = −1, the equation of motion is simplified to
Eq. (46).

ρP − ρF c 2 2 36µ ρF − ρP
   
r =R 1+ t (38) r̈ + ṙ + 2c 2 r −3 = 0. (46)
ρP R4 (ρF + 2ρP ) D2 ρF + 2ρP
   
ρP −ρF c Unfortunately, this equation cannot be analytically solved at all.
arctan ρP R2
t
θ=  . (39) A numerical solution must be used instead.
ρP −ρF
ρP
2.4.2. Axial
For the axial direction of motion, buoyancy and weight
2.4. Part 4: considering all forces simultaneously
forces are included. Also, the solution is independent of the fluid
By assuming the axial component of cylindrical coordinates flow regime (forced or free vortex).
along with gravitational force, the effects of buoyancy and The axial component of the equation of motion is as follows:
weight forces on tangential and radial components are elimi-  
 1
nated. Therefore, in cases where all forces act on a particle, the Fa = mP + mF z̈ + 3π Dµż
problem should be solved in a 2 dimensional plane. Then, tan- 2
gential and radial components will be solved independent of the 
1

axial component. + π D3 (ρP − ρF ) g . (47)
6
2.4.1. Radial & tangential Initial conditions are as follows:
The resultant forces in a radial direction are achieved from 
Eq. (40). It should be noted that the buoyancy and weight z = z0
(48)
forces are not considered in radial and tangential components ż = u0 .
 A. Salari et al. / Scientia Iranica, Transactions B: Mechanical Engineering 20 (2013) 351–358 355

The analytical solution of Eq. (47) is as follows.

(ρF − ρP ) D2 g
 
z = z0 − u0 + t
18ρF uF
(ρF − ρP ) (ρF + 2ρ P ) D4 g −36ρF uF
   
− exp t .
648 (ρF + 2ρP ) D2
(49)

2.5. The ratio of radial forces

The portion of each force in the total radial force acting on

the particle is calculated as follows:
FD
Drag ratio =
Ftotal
−3π Dµṙ
= (50)
−3π Dµṙ − 1
r̈ − r θ̇ 2 − 32 mF c 2 r 2α−1
 
m
2 F
FA Figure 1: Pressure gradient ratio as a function of time for free & forced vortices
Added mass ratio = (Case 1).
Ftotal
− 21 mF c 2 r 2α−1 + r̈ − r θ̇ 2
 
= (51)
−3π Dµṙ − 21 mF r̈ − r θ̇ 2 − 32 mF c 2 r 2α−1
 

FP
Pressure gradient ratio =
Ftotal
−mF c 2 r 2α−1
= . (52)
−3π Dµṙ − 21 mF r̈ − r θ̇ 2 − 32 mF c 2 r 2α−1
 

2.5.1. Forced vortex

In this case Eqs. (50)–(52) can be simplified by using Eqs. (44)
and (45).

exp (β1 t ) − exp (β2 t )

Drag ratio = 3π Dµβ1 β2 (53)
α1 exp (β1 t ) − α2 exp (β2 t )
1 β1 exp (β1 t ) − β2 exp (β2 t )
A.M. ratio = mF β1 β2 (54)
2 α1 exp (β1 t ) − α2 exp (β2 t )
β2 exp (β1 t ) − β1 exp (β2 t )
P.G. ratio = mF ω2 (55) Figure 2: Added mass ratio as a function of time for free & forced vortices
α1 exp (β1 t ) − α2 exp (β2 t ) (Case 1).
where
The main assumption, which was made at the beginning
1
α1 = 3π Dµβ1 β2 + mF β1 2 β2 + mF ω2 β2 (56) of the paper, is that ReP < 1. Now in order to verify this
2 assumption, ReP changes are plotted for both free and forced
1 vortices, in two cases.
α2 = 3π Dµβ1 β2 + mF β2 2 β1 + mF ω2 β1 . (57) Figures 11–14 depict ReP variations by passing time. The ReP
2
value is smaller than the one for time up to 0.1 s. It means that
The amount of β1,2 was determined in Eq. (45).
the assumed condition, ReP < 1, is true for the first 0.1 s after
The portion of each force in the total radial force acting
the solution is started.
on the particle in free vortex is calculated using numerical
Figures 4 and 9 depict the forces as a function of the time for
methods.
forced vortex.
Figures 5 and 10 show the forces as a function of the time for
3. Results
free vortex.
Figures 1 and 6 show the pressure gradient force as a
The results are shown for two cases, which are as follows.
function of time.
ReF is Reynolds number for fluid flow.
It is obvious in Figure 1 that the pressure gradient ratio for
Case 1 Case 2 a forced vortex is independent from time 0.02 s after particle
release. In a free vortex, the pressure gradient has a decreasing
ρF > ρP ρF < ρP manner by time.
ReF = 66666.67 ReF = 53333.33 Figures 2 and 7 show the added mass force as a function of
time.
356 A. Salari et al. / Scientia Iranica, Transactions B: Mechanical Engineering 20 (2013) 351–358

Figure 3: Drag ratio as a function of time for free & forced vortices (Case 1).

Figure 6: Pressure gradient ratio as a function of time for free & forced vortices
(Case 2).

Figure 4: All forces ratio as a function of time for forced vortex (Case 1).

Figure 7: Added mass ratio as a function of time for free & forced vortices
(Case 2).

The added mass ratio shown in Figure 2 has a simultaneous

decreasing manner for both free and forced vortices.
Figures 3 and 8 show the drag force as a function of time.
Figure 3 shows how drag force rapidly gets a constant value
for a forced vortex, but has an increasing manner for a free
vortex.
As obvious in Figures 4 and 9, for a forced vortex, after the
initial period, two main forces, pressure gradient and drag are
governing particle motion. The pressure gradient is the main
remaining force, which contains 80% of total forces acting on a
particle. The added mass force vanishes after the initial period
and does not have any role to play in particle motion.
In a free vortex, the most influential force is the pressure
gradient which governs the system.
In a forced vortex, the time that is needed in order for the
forces to achieve constant values for a denser particle (Case 2)
is less than the time needed for a particle in Case 1. It is obvious
Figure 5: All forces ratio as a function of time for free vortex (Case 1).
by comparing Figures 4 and 9.
 A. Salari et al. / Scientia Iranica, Transactions B: Mechanical Engineering 20 (2013) 351–358 357

Figure 8: Drag ratio as a function of time for free & forced vortices (Case 2).
Figure 11: ReP changes as a function of time for forced vortex (Case 1).

Figure 12: ReP changes as a function of time for free vortex (Case 1).
Figure 9: All forces ratio as a function of time for forced vortex (Case 2).

Figure 10: All forces ratio as a function of time for free vortex (Case 2). Figure 13: ReP changes as a function of time for forced vortex (Case 2).
358 A. Salari et al. / Scientia Iranica, Transactions B: Mechanical Engineering 20 (2013) 351–358

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Figure 14: ReP changes as a function of time for free vortex (Case 2). turbulent flows using the Euler/Lagrange-method (Modellierung und
numerische berechnung von partikel beladenen turbulenten strömungen
mithilfe des Euler/Lagrange-verfahrens), Berichteaus der Strömungstechnik
A brief look at Figures 5 and 10 shows that in Case 2, more (1996).
time is needed in order for the forces to acheive the same values [19] Sommerfeld, M., Kohnen, G. and Ruger, M. Some open questions and
as Case 1. inconsistencies of lagrangian particle dispersion models, Proc. Ninth Symp.
On Turbulent Shear Flows, Paper 15.1, Kyoto (1993).
[20] Yannacopoulos, A.N., Rowlands, G. and King, G.P. ‘‘Influence of particle
4. Conclusions inertia and Basset force on tracer dynamics: analytic results in the small-
inertia limit’’, Phys. Rev. E, 55(4), pp. 4148–4157 (1997).
In this paper, single solid particle motion through free and
forced vortices is studied. Though the results are not limited
Alinaghi Salari received his first MS degree in Mechanical Engineering from
to particle material but only depend on particle density, the Sharif University of Technology, Tehran, Iran, and is currently an MS degree
equations are solved for two cases, in which fluid density is student of Electrical Engineering at the University of Calgary, Canada. His
greater and smaller than particle density. Verification has been research interests include: MEMS, microfluidics, and bio-engineering. He has
worked on Micro-bubble generation for Ultrasonography in his MSc period. He
done for the ReP value, which confirms the solution for the first has also worked on Micropumps and Droplet Breakup in Microfluidics.
t = 0.1 s.
The results show that the most important force governing
particle motion is the pressure gradient force. Mohsen Karmozdi received his MS degree in Mechanical Engineering from
The present work can be used in a wide area of engineering Sharif University of Technology, Tehran, Iran, where he is currently a PhD degree
student in the same subject. His research interests include: fluid mechanics in
fields like cyclones. microfluids, energy auditing, MEMS and micropumps.

References
Reza Maddahian received his PhD degree in Mechanical Engineering from
Sharif University of Technology, Tehran, Iran, in 2012. He has published several
[1] Pagalthivarthi, K.V. and Gupta, P.K. ‘‘Particle tracking in rotating channel
papers in numerical fluid mechanics, and is currently working on the design
flow’’, Indian J. Eng. Mater. Sci., 15(5), pp. 365–376 (2008).
and optimization of static and rotary de-oiling hydro-cyclones for separation of
[2] Basset, A.B., Treatise on Hydrodynamics, 2, Deighton Bell, London
oil from water. His research interests include: heat transfer, swirling flows, and
pp. 285–297 (1888).
multiphase flows.
[3] Boussinesq, J. ‘‘Analytical theory of heat (Theorie analytique de la
chaleur)’’, In L Ecole Polytechnique, 2, p. 224. Gauthiers Villars, Paris (1903).
[4] Oseen, C.W., Hydrodynamics (Hydrodynamik), Akademische Verlagsge-
sellschaft, Leipzig p. 132 (1927). Bahareh Firoozabadi received her PhD degree in Mechanical Engineering
[5] Maxey, M.R. and Riley, J.J. ‘‘Equation of motion for a small rigid sphere in from Sharif University of Technology, Tehran, Iran, where she is currently
a non-uniform flow’’, Phys. Fluids, 26(4), pp. 883–889 (1983). Associate Professor in the School of Mechanical Engineering. Her research
[6] Henderson, K., Gwynllywa, D. and Barenghi, C. ‘‘Particle tracking in interests include: fluid mechanics in density currents, focusing on bio fluid
Taylor–Couette flow’’, Eur. J. Mech. B, 26, pp. 738–748 (2007). mechanics, and porous media. She teaches fluid mechanics and gas dynamics
[7] Michaelides, E.E. ‘‘Review – The transient equation of motion for particles, for undergraduates, and viscous flow, advanced fluid mechanics, continuum
bubbles and droplets’’, J. Fluids Eng., 119(2), pp. 233–247 (1997). mechanics and bio fluid mechanics for graduate students.