Lecture-6

Motion of a Particle Through Fluid

(One dimensional Flow)

1

Equation of Motion of a spherical Particle(one dimensional Flow)

• On Board

2

Terminal Velocity

“Particle reaches a maximum velocity which is attainable under givencircumstances and is called terminal velocity”

𝑑𝑢

𝑑𝑡= g (

𝜌𝑝−𝜌𝑓

𝜌𝑝) −

𝐶𝐷𝑢𝑜2𝜌𝑓𝐴𝑝

2𝑚(1)

Under gravitational settling, ‘g’ is constant, drag increases with velocity, equation (1) shows that acceleration decreases with time and approaches zero.

3

Put du/dt = 0 in equation(1):

0 = g (𝜌𝑝−𝜌𝑓

𝜌𝑝) −

𝐶𝐷𝑢𝑜2𝜌𝑓𝐴𝑝

2𝑚

g (𝜌𝑝−𝜌𝑓

𝜌𝑝) =

𝐶𝐷𝑢𝑜2𝜌𝑓𝐴𝑝

2𝑚

2gm (𝜌𝑝−𝜌𝑓

𝜌𝑝)

1

𝐶𝐷𝜌𝑓𝐴𝑝= 𝑢𝑜

2

ut = 2gm (𝜌𝑝−𝜌𝑓

𝜌𝑝)

1

𝐶𝐷𝜌𝑓𝐴𝑝(2)

For case of centrifugal forces:

ut = ω 2rm𝜌𝑝−𝜌𝑓

𝜌𝑝

1

𝐶𝐷𝜌𝑓𝐴𝑝(3)

4

Evaluation of Terminal velocity(Influence of Drag Coefficient)

Determination of terminal velocity requires the numerical values of Drag Coefficient,“CD”.

Drag coefficient can be determined from the graph (Fig. 7.6) as a function of Reynold’snumber under some restricted conditions:

Particle must be a solid sphere.

Particle is far from other particles and from wall surface.

Moving with terminal velocity.

Problems Due to assumption Written above

Underestimate: The drag coefficients for accelerating particles are appreciably greater thanthose reported hence terminal velocity differs considerably than predicted.

Particle shape: Variation in particle shape and change in orientation during free motion ofparticles through fluid consumes energy and increases the effective drag on the particle. As aresult terminal velocity (especially disks and platelike particle) is less than would bepredicted from curves of fixed orientation.

5

The conditions during settling of particles also affect the value drag coefficient. Thesettling of spherical particle depends can be divided into two based on the environmentthe particle is facing.

Free settling:

When particle is at sufficient distance from the boundaries of container and fromother particles so that its fall is not affected by them, the process is called free settling(Drag is low)

Hindered settling:

If motion is impede by other particles and wall of the container then which willhappen, when the particles near each other even though, they may not actually be colliding,the process is called hindered settling (Drag is higher).

6

Being smaller in size poses another problem i.e. Brownian Movement (this is randommotion imparted to particle by collisions between the particle and the molecule ofsurrounding fluid)

It is appreciable at a particle size of about 2 -3 μm and pre-dominates over the gravitywith particle size 0.1 μm.

Brownian movement tends to suppress the effect of force of gravity, so settling does notoccur. But the application of centrifugal force reduces the relative effect of Brownianmovement.

7

Quantitative Analysis of motion of spherical particles:For sphere of diameter DP

m = π/6 Dp3 ρp

and

Ap = π/4 Dp2

So the settling velocity under the action of gravity: (putting into equation-2), It becomes:

ut = 4g (𝜌𝑝−𝜌𝑓

𝜌𝑓)

𝐷𝑝

3𝐶𝐷(4)

As general case the “terminal velocity” can be determined by trial and error after gettingRe,p (Particle Reynolds's Number) to get an initial estimate of CD (given as follows)

ut = 4g (𝜌𝑝−𝜌𝑓

𝜌𝑓)

𝐷𝑝

3𝐶𝐷Guess Re,p Read CD from Graph Calculate Ut

Recalculate Re,pCompare calculated Re,p

with Guess

No

YesTerminate 8

The previous algorithm can be bypassed for limiting cases of Reynolds number (either very low Reynolds number or very high Reynolds number.

Low Reynolds number(less than 1):

Drag Coefficient varies inversely with Re,p and the pertinent equations are given below

CD = 24 / Re,p (5)

FD = (3πμutDP)/gc (6)

and

𝑢𝑡 =𝑔(𝜌𝑝−𝜌𝑓)𝐷𝑝

2

18𝜇(7) (Stokes law)

Above equation (stoke’s Law) can be applied for the Re,p < 1.

At Re,p = 1 , CD = 26.5 instead of 24 in the above equation.

Stokes law gives about 5% error.

The stokes law can be modified to predict the velocity of a small sphere in a centrifugal field bysubstituting 𝑟𝜔2 for g

9

• Higher Reynolds Number (1000 < Re,p < 200,000 )

For 1000 < Re,p < 200,000 the drag coefficient is constant then:

CD = 0.44 (8)

FD = (0.055 ρπDp2 u2

t )/ gc (9)

Ut = 1.75 g (𝜌𝑝−𝜌𝑓

𝜌𝑓) 𝐷𝑝 (10)

Newtonian Law applicable for large particle falling in gas or low viscosity fluids.

The terminal velocity, ut, varies with 𝑫𝒑𝟐 in stokes’ region whereas in the Newton’s

region it varies with 𝑫𝒑𝟎.𝟓

10

Criterion for settling regimes

To identify the range in which motion of the particle lies, the velocity term is eliminated from Reynoldsnumber by substituting ut from equation 7, For Stock’s Region

Re,p =𝐷𝑝𝑢𝑡𝜌

µ=𝐷𝑝

µ(𝜌𝑓𝑔𝐷𝑝

2(𝜌𝑝−𝜌𝑓)

18µ) (11)

=𝜌𝑓𝑔𝐷𝑝

3(𝜌𝑝−𝜌𝑓)

18 µ2

K = 𝐷𝑝 (𝜌𝑓𝑔(𝜌𝑝−𝜌𝑓)

µ2)1/3 (12)

Then the equation (12) will become

𝑅𝑒𝑝 =1

18𝐾3

Setting Re,p = 1 gives K = 2.6, for know size of particle ‘K’ can be calculated from equation (12) If‘K’ is less than ‘2.6’ then Stoke’s law applicable.

11

• Substituting for Ut from equation (10) into equation (11) gives the following result

Re,p = 1.75 K1.5

setting Re,p = 1000

K = 68.9 and

Re,p = 200,000 gives K = 2360

So;

68.9 < K < 2360 Newton’s law would be applicable

When the “K” is greater than 2360, the drag coefficient may change abruptly with small changes in the fluid velocity.

Above this range and between 2.6<K<68.9 the terminal velocity can be calculated by trial and error procedure indicated on the slide # 8

12

Algorithm (Criteria of Settling)

After having all properties

Find values of ‘K’

Stoke’s Region K < 2.6 Newton’s Region 2.6 < K < 68.9

In both cases assume ‘Re,p’

Calculate ‘ut‘ (from equation 4)

Check ‘Re,p’

YES NO

If the K > 2360 or 2.6<K<68.9

13

Hindered Settling

15

In Hindered Settling, the velocity of the particle is affected by the presence of nearby particles, sonormal drag correlations do not apply.

Particles in settling displaced liquid, which flows upward and makes the particle velocity relative tofluid greater than the absolute settling velocity.

For Uniform suspensions the settling velocity can be determined by following empirical correlationdeveloped by Maude & Whitmore

𝒖𝒔𝒖𝒕

= Ɛ 𝒏

For Small particles For Large particles

n = 4.6 Stoke’s Regionn = 2.5 Newton’s Region

= 0.62 (Ɛ = 0.9)𝑢𝑠𝑢𝑡

= 0.095 (Ɛ = 0.6)𝑢𝑠𝑢𝑡

= 0.77 (Ɛ = 0.9)𝑢𝑠𝑢𝑡

= 0.28 (Ɛ = 0.6)𝑢𝑠𝑢𝑡

Here Ɛ = void fraction

For Fine Suspension‘ut’ should be calculated by using density and viscosity of fine suspension.

‘Ɛ’ can be taken as volume fraction of suspension instead of void fraction.

e.g; suspension of very fine sand in water are used in seperating coal from heavy mineral,and the density of suspension is adjusted to a value slightly greater than that of coal to makethe coal particles rise to the surface, while the mineral articles sink to the bottom.

In that case suspension effective viscosity can be calculated as:

µ𝑠µ

=1 + 0.5(1 − Ɛ)

Ɛ4

The above equation applies only when ε < 0.6 and is most accurate when ε > 0.9