transport concentrates y tailings

TRANSPORT OF CONCENTRATES AND TAILINGS

2 Textbook of Solid-liquid Separation

Chapter 11 Transport of Tailings and Concentrates 3

CHAPTER 11 Equation Chapter 11 Section 1 TRANSPORT OF CONCENTRATES ANDTAILINGS

Within a mineral processing plant the ore and the product of the dressing operation and the concentrates and tailings, must be transported within a given unit operation, from one unit operation to another, from the plant to a port, where the concentrate is shipped to the world market and from the plant to a tailings pond to get rid of the waste materials. In the case of the crushing plant and the feed to the grinding plant, where the ore is essentially dry, the transport is made efficiently in belt conveyors. In the flotation plant and between flotation and solid-liquid separation in thickeners and filters the transport is made through pipelines and finally, flotation tailings is transported to tailing ponds through pipelines or channels.

The use of pipelines in the plant enables to transport a maximum load for a minimum of space using conventional centrifugal pumps and pipes that, in most cases, will not exceed 12 inches in diameter. For short distance tailing disposal or for long distance concentrate transportation, pipelines exhibit a remarkable degree of flexibility. No mater what the nature or complexity of the topography a pipeline can always be layout.

Slurries with particle size less than 270 mesh (50 microns) are called homogeneous, and they behave as fluids with increase density and particular rheology. A to diluted of such suspension requires a high transport velocity to prevent particles to settle. On the other hand a suspension with high concentration will behave as a fluid with yield stress and again high transport velocities will be required to maintain a turbulent regime that will prevent solid deposition. In either case, the system will operate beyond its economic point. The knowledge of the rheological properties of the slurries for the design of pipeline systems is therefore appreciated. The power consumption to pump 100 tph of homogeneous slurry horizontally is between 0.1 to 0.2 kW/ton-km (Condolios et al. 1967). Gravity transport of homogeneous slurries is possible if a gradient of at least 1.5 meters for each 100 meters is provided (Kleiman 1960).

Slurries with particle bigger than 270 mesh (50 microns) form heterogeneous or mixed slurries that form a vertical concentration profile and some bed formation while being transported. These materials are transported in suspension or by saltation or bed movement depending on the size of the particles. To prevent settling, higher velocities

4 Textbook of Solid-liquid Separation must be used and specific lower limits to this velocity must be established. Materials with particle size under 9 mesh (2 mm) and at least 20% of material under 270 mesh can be transported by centrifugal pumps with a power consumption, for 100 tph, of about 3 to 4 kW/ton-km (Condolios et al. 1967). Material with sizes over 9 mesh (2 mm) will require higher power consumption in the range of 6 to 12 kW/ton-km (Condolios et al. 1967) and will subject the pipe to severe wear.

11.1 TRANSPOR OF A FLUID THROUGH A PIPELINES

The incompressible stationary flow in a horizontal circular tube may be described by the following variables, the fluid density ( , )t r , velocity ( , )tv r , pressure ( , )p tr and extra stress ( , )E tT r . These field variables should obey the following field equations:

0 =v (11.1) ,E E ETp = + + =v v T g T T (11.2)

Since the tube is cylindrical there is axi-symmetry, therefore cylindrical coordinates will be used.

Continuity ( ), ) 0 , ( )z z zv r z v v rz = = (11.3)

Componente r: 0E

rzTp gr z

= + (11.4) Componente : 0 = p

(11.5)

( )1Componente z: 0= Erzp rTz r r + (11.6) If we call 0 0Lp p p = > the pressure drop between the extremes of the tube,

equation (11.6) may be written in the form:

( )1 Erzp rT Kz r r = = (11.7) Integrating by parts the left side of (11.7) yields:

0

00

, Lp L

Lp

p Kdz p p KL = = Defining 0 ,Lp p p =

Chapter 11 Transport of Tailings and Concentrates 5 Integrating by parts the right side of (11.7) yields:

( )Erzd rT Krdr= 1

2E

rzCT Krr

= +

Since at the tube axis the stress is finite, (0, )rzT z , for 0, 0r C= = , then:

1( )2

ErzT r Kr=

And substituting K from (11.8), yields the distribution of shear stress in a cylindrical tube:

1( )2

Erz

pT r rL

= , 1( )2

pr rL

= (11.9) where ( ) ( )Erzr T r = .

Fig. 11.1 Shear stress distribution for the flow in a cylindrical tube.

If we call ( )w R = the shear stress at the wall, from equation (11.9) we may write:

1 02w

p RL

= > (11.10) The ratio of shear stress at r and at the wall, is:

( )r


w

rR

= (11.11)

It is important to realize that equations (11.9) to (11.11) are valid for any type of fluids, since we have not invoked any type of constitutive equation for ErzT .

11.2 NEWTONIAN FLUIDS

For a Newtonian fluid, the constitutive equation for the extra stress ErzT is:

E zrzvTr

= (11.12) Substituting into (11.9) gives:

12

zvp rL r

=

Velocity distribution

Integrating with boundary condition ( ) 0zv R = at the wall gives:

2

2

2

21( )

2 21( ) 0

2 21

2 2

z

z

z

v p rr L

pv r r CLpv R R CL

pC RL

= = += + =

=

221( ) 1

4zpR rv r

L R =

(11.13)

The velocity distribution is parabolic as shown in figure 13.2.


Fig. 11.2 Velocity distribution for the flow of a Newtonian fluid in a circular tube.

Volume flow rate

The flow rate is given by 0

2 ( )R

zQ v r rdr= , then:

1 24

0

1 12

pR r r rQ dL R R R

=

418

p RQL

= (11.14)

Average velocity

The average velocity may be obtained from de volume flowrate by /zv Q A= , where 2A R= is the cross sectional area of the tube:

218z

pRvL

= (11.15)

The shear rate at the wall may be written in terms of the average velocity (11.15) and the shear stress at the wall (11.10) in the following way:

2

z

r R

v pRr L

=

=

Defining w z r Rv r == & and substituting (11.15) into this equation yields:


8 zwvD

=& (11.16) where 2D R= is the tube diameter Maximum velocity

The maximum velocity is obtains for 0r = , then, from (11.13):

21

4mpRv

L= (11.17)

Problem 11.1 Calculate the velocity distribution of water with viscosities, =1 cp in a tube of 1 inch

in diameter and 50 m in length, subjected to a pressure drop of 5 psi. Calculate also the flow rate, average and maximum velocity, the wall shear stress and shear rate.

Data are: m, 50 m, p=34935 Pa, 0.001 Pa-s0.0127 LR = ==

For 0.001 Pa-s = :

2 22 2

2

1 34935 0.0127( ) 1 1

4 4 0.001 50

( ) 28.2 1 m s

z

z

pR r rv r

L R R

rv r

R

= =

=

Maximum velocity 3.05 m smv =

Volume flowrate: 4 4

31 34935 0.0127 0.0071 m s8 8 0.001 50

p RQ

L

= = =

Average velocity: 2 2

0.007114.1 m s

3.14 0.0127zQ

vR= = =

Shear rate and shear stress at the wall

-18 8 14.1 4437 s

0.0127 20.001 4437 4.44 Pa

zw

w w

v

D

= = == = =

&

&


-0.015

-0.010

-0.005

0.000

0.005

0.010

0.015

0 10 20 30

Velocity v z, m/s

Rad

ius

r, m

Fig. 11.3 Velocity distribution for the flow of water in a pipe of 1 inch in diameter and 50 m long at a pressure of 5 psi.

Friction factor for Newtonian fluids

The friction factor, friction coefficient or resistance coefficient, is defined as the ratio of the friction at the wall to the dynamic pressure:

( )2(1 2)w zf v = (11.18) Since ( )1 2w pR L = , equation (11.10), substituting into this equation yields: 2 2

1 1 21 2 z z

pR p Rfv L L v

= =

Therefore, the pressure drop can be written in terms of the friction factor as:

2zLp f vR

= (11.19)

Substituting the value of zv from (11.15) into(11.19), we have:

2 28 16

z z

p R LfL v pR Dv

= =

10 Textbook of Solid-liquid Separation Using the definition of the Reynolds number Re zDv = , finally yields for laminar flow of a Newtonian fluid:

16 Ref = (11.20) Mechanical Energy Balance

The basis to calculate flow in conduits is the mechanical energy balance. This equation reads:

( ) ( )2 22 1 2 112 z z Lp v v z z hg g = + + (11.21) where 2 1 0p p p = > , p g is the pressure head, 2 2zv g is the velocity head and

Lh is the sum of the head loss in the pipe line and pipe line fittings. It is usual to give the head loss as numbers of velocity heads, 2 2zX v g . Table 11.1 gives the head loss for different fittings.

Table 11.1 Friction head losses Fitting X

45 elbow 0.3

90 elbow 0.7

90 square elbow 1.2

Exit from leg of T-piece 1.2

Entry into leg of T-piece 1.8

Unions and couplings small

Globe valve fully open 1.2-6.0

Gate valve fully open 0.15

Gate valve 3/4 open 1.0

Globe valve 1/2 open 4.0

Globe valve 1/4 open 16

Sudden expansion ( )( )221 21 D D Discharge into a large tank 1

Sudden contraction ( ) ( ) ( )6 4 22 1 2 1 2 10.7867 1.3322 0.1816 0.363X D D D D D D= + + Outlet of a large tank 0.5

Chapter 11 Transport of Tailings and Concentrates 11 Problem 11.2

Water flows under gravity from reservoir A to reservoir B, both of which are of large diameters. Estimate the flow rate through a 50 mm diameter and 75 m length pipe. Consider laminar flow. See figure.

Apply equation (11.21):

( ) ( )2 22 1 2 112 z z Lp v v z z hg g = + + 2 1 5 40 35 mz z = = 0p = 1 2z zv v= 1 2 35 mLh z z= = Head loss:

(pipe) (2gate valve) (2globe valve) (2elbows) (entrance)+ (outlet)L L L L L L Lh h h h h h h= + + + +

1

2

Otulet of large tank 0.5

Entrance to large tank 1.0

X

X

==

1 globe valve fully open 6.0

1 gate valve fully open 0.15glove

gate

X

X

==

2 elbows 0.7 2 1.4elbowsX = = 2

2

2

p 4L 1675 m 50 mm pipe = = ,

g D 2 Re

64 =

2

zpipe

z

z

vh f f

g

vLD v g

=


2

2

64 0.001 75 1.92 =

1000 0.05 2z

z z

vv v g

=

2

2

2

1.922 0.15 2 6.0 2 0.15 1 0.5

2

1.92 0.30 12.0 1.4 1 0.5

2

1.92 15.20

2

zL

z

z

z

z

z

vh

v g

vv g

vv g

= + + + + +

= + + + + +

= +

Then: 21.92

15.20 35 6.7 m s2

zz

z

vv

v g+ = =

1000 6.7 0.05Re 332928 4000 Turbulent regime0.001 = = >

The problem must be recalculated with parameters for the turbulent regime.

Transition to turbulent regime

The flow of a Newtonian fluid may occur in laminar or turbulent regimes. The parameters defining the transition between laminar and turbulent flows are the friction factor (or coefficient of resistance) f and the Reynolds number Re. The Reynolds number has different form depending on the rheological model of the suspension. The friction factor is defined as the ratio of the overall pressure losses to the velocity pressure

Do to the overriding effect of the viscosity forces in laminar flow of Newtonian fluids, even flow past surface asperities appear to be smooth. Therefore, the roughness of the walls, unless it is very significant, does not affect the flow resistance. Under these conditions of the flow the friction coefficient is always a function of the Reynolds number alone.

As the Reynolds number increases, the inertia forces, which are proportional to the velocity squared, begin to dominate. Turbulent motion is initiated, which is characterized by the development of transverse velocity component giving rise to agitation of the fluid throughout the entire stream and to momentum exchange between randomly moving masses of fluid. All this causes a significant increase in the resistance to the motion in turbulent flow compared with the case of laminar flow.

When the surface of the walls is rough, separation occurs in the flow past roughness and resistance coefficient becomes a function of the Reynolds number and the relative roughness : D = (11.22)

Chapter 11 Transport of Tailings and Concentrates 13 where is the is the average height of the asperities and D is the tube diameter. Although for low velocity flows on rough tubes the friction factor decreases with an increase in Reynolds number, at higher velocities on rough tubes it shows an increase in the coefficient of friction with an increase in Reynolds number with constant relative roughness. This effect is explained by the fact that at low flows the viscous sublayer is larger than the roughness protuberances > and the fluid moves smoothly past the irregularities, while at higher velocities the sublayer becomes thinner than the roughness protuberances, < , which enhance the formation of vortices increasing the friction factor and pressure drop. The tubes may be considered smooth as long as the heights of the asperities are smaller than the thickness of the laminar sublayer. See the following figure 11.5.

Fig. 11.5 Flow past rough tube walls for different ratio of viscous sublayer to roughness asperities.

The friction factor was defined in the form: 24

1 2w

z

fv

= , where w is the friction

at the wall and zv is the average flow velocity. Experience show that the dependence of the friction factor f on the Reynolds number and the roughness of tubes, as determined experimentally by Nikuradse (1933), are given in three flow regimes, (1) laminar flow, (2) transition to turbulence and (3) rough walls regime.

First regime. In the first regime, with Reynolds numbers lower than 2100, f is independent of the roughness of the tube and is given by:

16 Ref = (11.23) Second regime. The second regime, called transition regime, consists of three segments of the resistance curve for uniform roughness and 2100 Re 4000< < : (1) In the first segment, the friction factor increases rapidly with Re but remains

independent of the roughness and is given by:

0.250.0791 Ref = (11.24) (2) The second segment includes friction factors for the lower values of

Reynolds numbers for different relative roughness. In this region the friction factor is given by Blasius equation or all roughness, equation (11.24).

(3) In the third segment, the friction factor starts form the Blasius curve and increases with the Reynolds number and diverging to different lines for different


constant relative roughness, and ends when the friction factor becomes a constant. See figure 11.6

Third regime. In the third regime, for Re 4000> , called quadratic regime, the friction factor become a different constant for each relative roughness, independent of the Reynolds number.

Nikuradse (1933) proposed the following equation for the resistance coefficient, defined as 4f = , in terms of the Reynolds number with the relative roughness as parameter: ( )1 1 11 log Re lga b c = + + (11.25) For 1 1 13.6 Re 10 0.8 2.0 0a b c = = + = (11.26) For 1 1 110 Re 20 0.068 1.13 0.87a b c = + = + = (11.27) For 1 1 120 Re 40 1.538 0 2.0a b c = + = = (11.28) For 1 1 140 Re 191.2 2.471 0.588 2.566a b c = + = = (11.29) For 1 1 1Re >191.2 1.138 0 2.0a b c = + = = (11.30)

These equations are represented graphically in the following figure 11.6 proposed by Nikuradse.

Fig. 11.6 Friction factor versus Reynolds number Re for tubes with uniform roughness, according to Nikuradse (1933).

Problem 11.3 Water flows under gravity from reservoir A to reservoir B, both of which are of large

diameters. Estimate the flow rate through a 50 mm diameter and 75 m length pipe considering the flow as turbulent. See figure 11.4.

From problem 11.2:

( ) ( )2 22 1 2 112 z z Lp v v z z hg g = + +

Chapter 11 Transport of Tailings and Concentrates 15 2 1 5 40 35 mz z = = 0p = 1 2z zv v= 1 2 35 mLh z z= = Head loss:

1

2

Otulet of large tank 0.5

Entrance to large tank 1.00

2 globe valve fully open 12.0

2 gate valve fully open 0.30glove

gate

X

X

X

X

====

2 elbows 1.40elbowsX =

( )22

2 2

2p 175 m 50 mm pipe = = , 0.4 log 2Re

g

4 =

2

4 75 = 6000

0.05 2 2

zpipe

z

z z

Lvh f f

Dg f

vLf

D g

v vf f

g g

= +

=

( )

( )

2

2

0.30 12.00 1.40 1.00 0.050 60002

15.20 60002

zL

zL

vh f

g

vh f

g

= + + + + +

= +

Entonces, ( ) 215.20 6000 352

zL

vh f

g= + =

( )1 0.4 log 2Re ff

= +

Assuming a value of Re=100.000 and using solver of Excel to obtain Lh and then, using the new values of Re and f to calculate a new value of f, the velocity and iterating, we get the following values:

Re 68.228 , 0.05893 , 1.36 m/szf v= = =

16 Textbook of Solid-liquid Separation 11.3 TRANSPORT OF SUSPENSIONS IN PIPELINES

The flow patterns of suspensions in tubes depend on the transport velocity. At low velocities, the particles form a bed at the bottom of the tube and are not transported by the fluid. As the velocity increases, particles at the surface of the bed start moving. At higher velocities, the sediment moves as a cloud in saltationary motion, and some particles will be suspended and carried away with the fluid. In these conditions the suspension receives de name of settling suspensions and the flow regime is called heterogeneous. Increasing the velocity further, all particles will be suspended and particles and fluid will behave as a homogeneous mixture. In this condition the suspension is called non-settling and the flow regime is called homogeneous. To each one of these behaviors correspond a pressure drop and the type of motion can be controlled by the pressure gradient.

When a pressure is applied to the fluid in a pipe filled with sediment, the pressure drop diminishes slightly as the particles begin to be suspended and move. Between the initiation of the particle motion and the complete suspension of the particles, a range of velocities exist, for each feed concentration F , for which the pressure drop is a minimum and after which it increases. This range of velocities receives the name of critical transport velocities.

As discussed above, the flow pattern in the transport of suspensions in a tube is closely related to the feed suspension concentration. When particles begin to move above a stationary bed, the fraction of the feed concentration F , is F with values 0 0.2F < < . Motion of the bed yield fraction of concentrations in the range 0.2 0.7F < < . Partial suspension gives fraction of concentration 0.7 1F < < and complete suspensions of particles give 1F = .

Table 11.1 Concentration ratio for different flow patterns

Mixture velocity

Miv

Flow pattern Fraction of concentration

F 1Mv Homogeneous suspension 1.0

2Mv Asymmetric suspension 0.7 1.0

3Mv living bed with asymmetric suspension 0.2 0.7

4Mv Stationary bed with some particles in suspension 0 0.2


Fig. 11.7 Particle behavior in the flow in a tube.

A heterogeneous regime occurs when some of the particles in the suspension are bigger than those that would give a homogeneous regime and will settle. Nevertheless they will not form a bed at the bottom of the conduct but rather a concentration gradient will be established. Since the segregating particles are the coarser ones, the fluid will not change its rheological characteristics. This is the usual regime in industrial applications. Here only experience counts.

Figure 11.7 shows that a minimum pressure drop exists at a certain flow velocities for each feed suspension concentration. These are the velocities that just prevent the formation of a bed. If the minimum for each concentration are connected, a curve of the limiting deposit velocity ( )Lv is obtained (see figure 11.8). The optimum velocity of a heterogeneous flow is that producing the minimum pressure drop without bed formation and that is just to the right of the limiting velocity.

Since the limiting deposit velocity is a function of the suspension concentration, the first step in their calculation is to obtain the settling velocity as a function of concentration.

Limiting deposit velocity

The limiting deposit velocity is related to the settling velocity at the transport concentration. Therefore it is important to know the settling velocity of the particles at several concentrations. This may be obtained from laboratory experiments or by


Fig. 11.8 Head loss versus average transport velocity for 0.44 mm fine sand (Condolios and Chapus 1963).

calculations from sedimentation models. A convenient such model was proposed by Concha and Almendra (1979a) and was discussed in chapter 4 of this book.

The settling velocity of suspensions can be expressed in the form:

( ) ( ) ( )( )21 23 2 3 220 52 1 0 0921 1p q p.u* f f . f d *d * = + (11.31) * * and d ud u

P Q= = (11.32)

1 3 1 32

2

3 44 3

f f

f f

gP y Q

g

= = (11.33)

( ) ( ) ( ) ( )2 033 0 1671 1. .p qf , f = = (11.34) where and d u are the diameter and settling velocity of the particle in the suspension,

and P Q are parameters of the solid-liquid system and ( ) and ( )p qf f are the correction factor for concentration.

Correlations for the limiting deposit velocity

The simplest criterion to avoid a heterogeneous flow regime (Faddick 1986) is to require the flow to be turbulent and that the coarser particles of the system are in

Chapter 11 Transport of Tailings and Concentrates 19 Newtons drag regime. These requirements are obeyed if the flow Reynolds number Re and the particle Reynolds number Re p are:

pRe 2100 and Re 1000fz duDv

= > = > (11.35)

where and f are the water density and viscosity, and are the fluid density and viscosity (if slurry behaving as Newtonian fluid), and D d are diameters of the pipe and the particle respectively and and zv u are the average flow velocity and the particle settling velocity respectively.

Problem 11.4 Design a pipe for the flow of 600 tph of magnetite mineral slurry which behaves as a Newtonian fluid with density 1667 kg/m3and viscosity 5 mPa-s. The magnetite density is 5000 kg/m3 ant its maximum particle size is 5mm. Make sure that the flow regime is heterogeneous.

Volume flow is 3600 1000

0.100 m s1667

FQ

= = =

Particle size: 0.005 md = Magnetite density 35000 kg m

s =

Water density 31000 kg mf

=

% solids: 50% solid by weight100 ( ) 100 5000 (1667 1000)

( ) 1667 (5000 1000)s f

s f

w

= = =

Pulp concentration:

0.1671000 50

(100 ) 5000 (100 50) 1000 50f

s f

w

w w = = =

+ +

(1 / 3 )2

5 1 3

1 323 0.001

2.674 10 m4 (5000 1000) 1000 9.81

34

f

f

Pg

= =

( )(1 / 3 ) 1 321 32 4 (5000 1000) 0.001 9.81 0.0374 ( m s )3 100043 ffQ g = = = *

5

0.005187.01

2.674 10d

dP

= = =


( )( ) ( )( )2 21 2 1 2* *3 2 *3 2*20.52 20.521 0.0921 1 1 0.0921 187 118722.698

u dd

= + = + =

* 0.84922.689 0.03740 m su u Q= = =

Re 4243.0 > 1000fpud

= =

Select an average transport velocity: 2.0 m szv =

1 24 4 0.100

0.2524 m s3.14 0.304

10.0 inchesz

Q

vD

D

= = =

=

5Re1667 2.00 0.2524

1.6827 100.005

> 2100zfv D

= = =

Both Reynolds numbers fulfill the conditions for a heterogeneous flow.

It has not been possible to establish the limiting deposition velocity from fundamentals, but many empirical correlations have been proposed for particles in the range of 50 m to 5 mm in pipes from 50 mm (2 in) to 300 mm (12 in) and concentrations from 6 to 44% solid by volume, by Durand (1953), Spell (1955), Newitt et al. (1955), Cairns et al. (1960), Govier (1961), Schulz (1962), Sinclair (1962), Condolios and Chapus (1963), Yufin and Lopasin (1966), Zandi and Govatos (1967), Babcock (1968), Shook (1969), Bain and Bonnington (1970),Charles (1970), Wasp et al. (1977), Wilson (1979), Thomas (1979), Oroskar and Turian (1980), Gillies and Shook (1991), Chien (1993).

The values of the limiting deposition velocity predicted by these many correlations are so widely scattered that it is difficult to choose one as the best. We suggest using the correlation of Condolios and Chapus (1963) due to it simplicity:

0.1493.0Lv gD= (11.36) Figure 11.8 shows the limiting deposition velocity versus the pulp concentration

with pipe diameter as parameter and figure 11.9 shows the correlation versus pipe diameter with concentration as parameter.


0

1

2

3

4

5

6

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

Volume fraction of solids

Lim

iting

dep

osit

velo

city

vL in

m/s

D= 2 inches 4 inches 6 inches 8 inches 10 inches 12 inches

Fig. 11.9 Correlations for the limiting deposition velocity versus concentration for a copper tailing of an average particle size of 75 mm with the pipe diameter as a parameter, according to Condolios and Chapus (1963).

0

1

2

3

4

5

6

0.0 2.0 4.0 6.0 8.0 10.0 12.0

Pipeline diameter D in inches

Lim

iting

dep

osit

velo

city

vL m

/s

"FI=0.1"" =0.2"" =0.3"" =0.4"" =0.6"" =0.7"

Fig 11.10. Critical velocity for the transport of a copper tailings with an average particle size of 75m and suspension concentration as a parameter, according to Condolios and Chapus (1963).

22 Textbook of Solid-liquid Separation Scale-up of limiting deposition velocity

If the limiting deposit velocity Lov is known for a pipe of diameter oD , the limiting deposit velocity Lv for a diameter D may be calculated from Roco (1977) for pipes from 3 to 6 inches and % solids by volume of 0.1 to 60%:

0.4

L

Lo o

v Dv D

=

Problem 11.5 Design a pipeline for transporting 50 tph of a copper concentrates, with density

33800 kg ms = at 55% solid. Calculate the liming deposit velocity if the maximum particle size of the concentrate is 100 m. Calculate what would be the limiting velocity in a similar pipeline but for a 3 inch pipe.

Particle size: 41.00 10 md = Magnetite density 33800 kg m

s =

Water density 31000 kg mf

=

% solids: 55.0% solid by weightw = Pulp density:

3100 100 3800 1000 1681.4 kg m(100 ) 3800 (100 55) 1000 55

s f

s fw w

= = = + +

Pulp concentration:

0.2431000 55

(100 ) 3800 (100 55) 1000 55f

s f

w

w w = = =

+ +

Volume flow is 3220 1000

0.03635 m s1681.4

FQ

= = =

Select an average transport velocity: 2.0 m szv =

1 24 4 0.03635

0.1522 3.14 2.0

m

6.0 inchesz

Q

vD

D

= = =

=

( ) ( )0.4 0.43.0 6 3 2.5 m sL Lo ov v D D= = =

Chapter 11 Transport of Tailings and Concentrates 23 11.3.1 Flow of homogeneous suspensions

Homogeneous suspensions may be treated as non-Newtonian fluids, therefore we will consider the flow of Bingham, Potential-law and Herschel-Bulckley fluids in a tube.

Bingham Fluid

A Bingham fluid has the following constitutive equation for the shear stress in cylindrical coordinates:

0

0

for 0( )

for 0

zE

rz zz

v rT r vK v r

r

== + is a constant called plastic viscosity and ( ) 0ErzT r < . Defining the shear stress ErzT = , the yield stress 0y = and the shear rate zv r = & , we can write (11.37) in the usual form:

for 0for 0

y

y K

== + >&

& & (11.38)

For any fluid:

12

zy

v pK rr L

+ = (11.39) Calling yR the radius for which the stress is y , we have:

12y y

p RL

= (11.40)

From (11.10), the stress at the wall is given by 12w

p RL

= , therefore the relationship between the yield stress y and the wall shear stress w is:

y yw

RR

= (11.41)


For ( ) yr > , from (11.39) we have:


12

yzv p rr KL K

=

and using (11.40) for y , results in:

( )12

zy

v p R rr KL

= (11.42) Integrating this expression yields:

21 12 2z y

pv R r r CKL = +

For , ( ) 0zr R v R= = , therefore:

21 12 2y

pC R R RKL =

and 2 21 1 1 1( )2 2 2 2z y y

p pv r R r r R R RKL KL =

Then, 221 1( ) 1 1 , for

2 2y

z y

RpR r rv r R r RKL R R R

= (11.43)

Introducing (11.41), we have the alternative expression:

221 1( ) 1 1 , for

2 2y

z yw

pR r rv r R r RKL R R

= (11.44)

For ( ) yr 221 1( ) 1 1

2 2y y y

z

R R RpRv rKL R R R

=

221( ) 1 , for 0

4y

z y

RpRv r r RKL R

= (11.45)

Using (11.41) we obtain the alternative expression:

221( ) 1 , for 0

4y

z yw

pRv r r RKL

= (11.46)

Chapter 11 Transport of Tailings and Concentrates 25 Volume flow rate

The volume flow rate is given by 0

2 ( )R

zQ v r rdr= , then: 2 22

0

1 11 1 12 2

y

y

R R

y y

R

R RpR r rQ rdr rdrKL R R R R

= + 1 224

0

1 11 1 12 2

y

y

R R

y y

R R

R RpR r r r r r rd dKL R R R R R R R R

= + 44 4 11

8 3 3y yR RpRQ

KL R R = +

or 44 4 11

8 3 3y y

w w

pRQKL

= + (11.47)

Average velocity

The average velocity is given by 2/zv Q R= , then:

421 4 11

8 3 3y y

z

R RpRvKL R R

= + (11.48)

421 4 11

8 3 3y y

zW W

pRvKL

= + (11.49)

Using a similar procedure than in the case of Newtonian fluids, we have:

4

8 1 4 114 3 3

y yz R Rv pDD KL R R

= + (11.50)

4

8 4 113 3

y ywz

w w

vD K

= + (11.51)

Maximum velocity

From (11.45) the maximum velocity is that for 0 yr R

21 1 2

4y

m

RpRvKL R

= or

21 1 24

ym

w

pRvKL

= (11.52)

26 Textbook of Solid-liquid Separation Pressure drop

14

4

8 4 113 3

y y

w w

K L QpR

= +

Problem 11.7 For three different homogeneous clay suspensions, which may be represented by the

Bingham model in the range -110 100s&< < with 5, 10 y 15 Pay = respectively and a plastic viscosity of 150 mPa-s = , flowing in a cylindrical tube of 4 cm in diameter and 200 m in length. Calculate the pressure drop necessary to transport 100 liter per minute of a suspension with a plastic viscosity of 150 mPa-s.

The shear stress versus shear rate for this material is given in figure 11.11.

0

5

10

15

20

25

30

35

0 20 40 60 80 100 120

Shear rate , s-1

She

ar s

tress

, P

a

5 Pa10 Pa15 Pa

Fig. 11.11 Shear stress versus shear rate for a Bingham model of clays with plastic viscosity of 150 mPa-s and Yield stresses of 5, 10 y 15 Pa.

3100 (60 *1000) 0.001666 m sQ = = 14

4

8 4 113 3

y y

w w

K L QpR

= +

The velocity distribution is given by: 421 4 11

8 3 3y y

z

R RpRvKL R R

= + , then:


K(Pa-s)= 0.15 0.15 0.15L(m)= 200 200 200

y(Pa)= 5 10 15R (m)= 0.02 0.02 0.02

Q(m3/s)= 0.001666667 0.00166667 0.00166667vzav (m/s)= 1.327 1.327 1.327

w(Pa)= 22.85 36.05 47.62p(Pa)= 1.580E+06 1.981E+06 2.317E+06Ry (m)= 0.0044 0.0055 0.0063

Ry (inch)= 0.1723 0.2184 0.2480

-0.025

-0.020

-0.015

-0.010

-0.005

0.000

0.005

0.010

0.015

0.020

0.025

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5

Velocity v z(r ) m/s

Pip

e ra

dius

r, m

5 Pa10 Pa15 Pa

Fig. 11.12 Velocity distributions for a Bingham model of clays with plastic viscosity of 150 mPa-s and Yield stresses of 5, 10 y 15 Pa.

Power Law Fluids

The constitutive equation for the shear stress of power low fluids flowing in a circular tube is:

1

( )n

E z zrz

v vT r mr r

= (11.53)

where m is the consistency index and n is the power index. Note that ErzT has the sign of zv z .

28 Textbook of Solid-liquid Separation Velocity distribution

Replacing (11.53) in (11.39) we have:

1 1

2

nz zv v pm r

r r L

=

Since 0m > , 0zv r < and 0p > , we can write zv r in the form:

1

2

nzv p r

r mL = (11.54)

Integrating and using the condition ( ) 0zv R = , yields

1 ( 1)

( ) 11 2

n n n

znR pR rv r

n mL R

+ = + (11.55)

Problem 11.8 Calculate the velocity distribution for a power law fluid with consistency index

n3.0 Pa-sm = and power law index of 0.20; 0.33; 0.50; 1.0 and 3.0n = . Figure 11.13 shows the result.

-0.015

-0.010

-0.005

0.000

0.005

0.010

0.015

0.0 0.5 1.0 1.5 2.0 2.5 3.0

Velocity distribution v z(r) m/s

Pip

e ra

dius

r, m

n=0.20 n=0.33 n=0.50

n=1.00

n=2.00

Fig. 11.13 Velocity distribution of power low fluids with consistency index

n3 Pa-sm = and power low indices 0.20; 0.33; 0.50; 1 and 3.

Chapter 11 Transport of Tailings and Concentrates 29 Volume flow rate


2R

zQ v rdr= , then substituting (11.55) and integrating:

1 ( 1)

0

2 11 2

R n n nnR pR rQ rdrn mL R

+ = +

1 11 (2 1)3

0 0

21 2

n n nnR pR r r r rQ d dn mL R R R R

+ = +

( )( )( )( )11 2 2 1 1

3

0

1 121 2 2 2 1 1

n n nn pR r rQ Rn mL R Rn n

+ + = + + +

( )11

2

3 1 2

nnn pRQ Rn mL

+ = +

(11.56)

Average velocity

The average velocity is given by 2zv Q R= , then:

( )11

3 1 2

nn

zn pRv

n mL

+ = + (11.57)

( )18 4

3 1 2

nzv n pR

D n mL = + (11.58)

Using (11.54) for r R= :

1

2

nz

wr R

v pRr mL

=

= = &

( )3 1 84

zw

n vn D

+=& (11.59) Maximum velocity

The maximum velocity is obtained from (11.55) for 0r = , then:

11

1 2

nn n

mn pRv

n mL

+ = + (11.60)

30 Textbook of Solid-liquid Separation Pressure drop

From (11.56)

( )1 23 12n

n

nmL QpR n R+

+ = (11.61)

Wall shear stress and Reynolds number

Defining the friction coefficient in the same way as for Newtonian fluids in laminar flow, 16 ReDf = , we can define a Plastic Reynolds number RePL for a Power Law fluid, where w is the wall shear stress. Then we have: w 2

16 1,1 2 Re 2D wz PL

pf Rv L

= = = (11.62)

216Re zPL

LvpR

=

( )12 1

1

2 3 1 3 1 2

n nn

z zn

n pR mL nv p vn mL R n

++

+ = = +

Substituting in the previous equation gives:

( )216Re

3 12

n nz

PL n

R vn

mn

= +

2

1

Re3 18

4

n nz

PL nn

v Dnm

n

= + (11.63)

This definition of the Reynolds number was given by Metzner and Reed (1959).

Problem 11.9

A polyacrilamide solution of 31.074 kg m = in density is to be pumped through a 2.54 cm diameter and 10 m length tube at a rate of 2.500 kg h . Measurement in the laboratory showed that the fluid may be represented with the power law model with

0.53 Pa-s y 0.5m n= = . Calculate the necessary pressure to maintain the flow and calculate the velocity distribution, average and maximum velocity.

Chapter 11 Transport of Tailings and Concentrates 31 Volume flow

4 32500 6.466 10 m s3600 1074 3600F

Q = = =

Pressure drop:

1 2 0.5 1) 2(

2 3 1 2 3 10 3 0.5 1 0.0006466105.920 Pa

0.0127 0.5 3.14 0.0127

n

n

mL n Qp

R n R+ ++ + = = =

Velocity distribution: 2 ( 1) ( 1)0.5 0.0127 105920 0.0127

( ) 1 11.5 2 3 10

2.12787n n n n

z

r rv r

R R

+ + = =

Average velocity

2 2

0.00064661.277 m s

3.14 0.0127zQ

vR= = =

Maximum velocity:

(0) 2.13 m szmv v ==

-0.015

-0.010

-0.005

0.000

0.005

0.010

0.015

0.0 0.5 1.0 1.5 2.0 2.5

Velocity distribution v z(r ), m/s

Pip

e di

amet

er r

, m

Fig. 11.14 Velocity distribution for a polyacrylamides solution with a power low model: 0.53 Pa-s y 0.5m n= = .

32 Textbook of Solid-liquid Separation Transition to turbulent regime

As in the case of Newtonian fluids, the friction factor gives the transition from laminar to turbulent flow. The following figure shows the friction factor as a function of Metzners Reynolds number for different values of the power function n.

Fig. 11.15 Friction factor as a function of Metzners Reynolds number for different values of the power function n (Chabra and Richardson 200x).

Problem 11.10

A non-Newtonian fluid with density equal to that of water, flows in a 300 mm diameter and 50 m long tube at a rate of 300 kg/s. Rheological measurements yield the following power law parameters: 0.32.74 Pa-s and =0.30m n= . Determine the necessary power of a pump and the wall shear stress.

Average velocity: 2 2

300 10004.24 m s

3.14 (0.15)zQ

vR= = =

The transitional, or critical, Reynolds number is Re 2100 .

2

1

Re 21003 1

84

n n

cMRc n

n

v D

nm

n

= =+

then, the critical velocity, that is, the velocity at which the flow changers from laminar to turbulent is:

1 /( 2 )

1 0.627 m s3 1

8 21004

n

c

nn nv

nm D

n

= =+


Since the average velocity 4.1 m/s is greater than the critical velocity 0.627 m/s, the regime is turbulent. The actual Reynolds number is:

( ) ( )

( ) ( )2 0.3 0.32

0.3 1

1

1000 4.24 0.3Re 89.324

3 1 8 2.748

4

(3 0.3 1) / 4 0.3

n n

zMR n

n

v D

nm

n

= = =+ +

With MRRe 7740.5 and 0.3n= = , from figure 11.15, we get a friction factor 0.0029f = . The value of the pressure drop necessary to to produce the flow is obtained from the friction factor

by definition: 1w ;

2 221 2

p p Rf R fw L Lv vz z

= = = ;

therefore: ( ) 502 22 2 1000 4.27 0.0029 17.412 Pa0.3

Lp v fz D

= = = .

and, the pump power Po is: 0.30 17.412 5.224 WPo Q P= = = .

The wall shear stress is 1 17.412 0.15 26.1 Pa2 2 50

pRw L

= = = .

Herschel-Bulckley Fluid

For the flow in circular pipes, Herschel-Bulckley fluids have a constitutive equation of the form:

0

1

0

for 0( )

for 0

z

Enrz

z z z

vr

T rv v vmr r r

= + &

& & (11.65)


For ( ) ; :y yr R r R

1 1

2

nz z

yv v pm rr r L

=

Since 0, p>0, 0 and 0y zm v r > > < , The parenthesis is positive and this equation may be written in the form:


( )1 12 2

n nyz

yv p pr r Rr mL m mL

= = (11.66)

Integrating with boundary condition ( ) 0zv R =

( )( )

( )

11

1( 1)

1

1( 1)

1

( )2

( ) 02 1

C2 1

nn

z y

nn n

z y

nn n

y

pv r r R drmL

p nv R R R CmL n

p n R RmL n

+

+

= = + = +

= +

( )1 1( )2

nn

z ypv r r R dr

mL =

( ) ( )1 1( 1) ( 1)( )2 1 2 1

n nn n n n

z y yp n p nv r r R R R

mL n mL n+ + = + +

( 1) ( 1)1

( ) 12 1

n n n nny y

z

R RpR nR rv rmL n R R R

+ + = + (11.67)

For ( )( ) ; 0 , where :y y y yr r R R < < < = ( )0 ( )z z z yv v r v Rr = = (11.68) From (11.67),

( 1)1

( ) 12 1

n nny

z y

RpR nRv RmL n R

+ = + for 0 yr R (11.69)

Problem 11.11

Determine the velocity distribution of Herschel-Bulkley fluids with yield stresses of 5, 10 and 15 Pay = , consistency index m=3 Pa-sn and power low indices 0.50. The result is given in figure 11.16.

Problem 11.12

Determine the velocity distribution of Herschel-Bulkley fluids with yield stress of 15 Pay = , consistency index m=3 Pa-sn and power low indices 0.20, 0.33 add 0.50.

The result is given in figure 11.17.


-0.015

-0.010

-0.005

0.000

0.005

0.010

0.015

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

Velocity distribution v z(r), m/s

Pip

e ra

dius

r, m

ty=15, m=3, n=0.5ty=10, m=3, n=0.5ty=15, m=3, n=0.5

Fig. 11.16 Velocity distribution of Herschel-Bulkley fluids with yield stresses y=5; 10 and 15; Pa, consistency index m=3 Pa-sn and power low index 0.50.

-0.015

-0.010

-0.005

0.000

0.005

0.010

0.015

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6

Velocity distribution v z(r ), m/s

Pip

e ra

dius

r, m

n=0.5n=0.33n=0.20

Fig. 11.17 Velocity distribution of Herschel-Bulkley fluids with yield stress

15y = , consistency index m=3 Pa-sn and power index 0.20,0.33 and 0.5n = .

36 Textbook of Solid-liquid Separation Volume Flow rate


2R

zQ v rdr= , then substituting (11.67) and (11.69) into this equation and integrating yields:

( 1)

1( 1) 02

( 1) ( 1)1

1

22 1

1

y

y

n nR Ry

nn

n n n ny y

R R

R rdR RpR nQ R

mL n R Rr rdR R R R

+

++ +

= + +

( )( 1) 2 1

1( 1)2

( 1) 2

112

22 1 11

2

y

n nmy y

nn R R

n ny y

R Ra bx xdx

R RpR nQ RmL n R R

R R

+

+

+

+ + = + +

If we make , , 1, ( 1)yx r R a R R b m n n= = = = +

( )1( 1) 1

222 1 y

nnm

R R

pR nQ R a bx xdxmL n

+ = + + +

Using the Dwight Table of Integrals and other mathematical functions (MacMillan Co, New York, 1966), results in: Dwight,

( ) 11 1( 1)222 1 ( 1)

y

n mn

R R

a bxpR nQ RmL n b m

++ + = + +

1(2 1)

1( 1)2 2 2 12 1

y

n ny

nn

R R

RrR RpR nR nmL n

n

+

+

= ++

( )( )(2 1)1( 1) 2

22 12 1 2 1

n nnnyRpR nQ R

mL n n R

++ = + + (11.70)

Chapter 11 Transport of Tailings and Concentrates 37 Average velocity

The average velocity is given by 2zv Q R= , then:

( )( )(2 1)1( 1) 2

2 12 1 2 1

n nnny

z

RpR nvmL n n R

++ = + + (11.71)

Maximum velocity The maximum velocity is obtained for 0r = , then:

( 1)1

12 1

n nny

m

RpR nRvmL n R

+ = + (11.72)

11.3.2 Flow in a heterogeneous regime

In a heterogeneous regime the head loss mJ may be composed of two contributions, one to maintain the turbulent flow of the fluid LJ and another to maintain the particles in suspension sJ .

m L sJ J J= + (1.73) where LJ and sJ are measured in meters of liquid column per meter of pipe length ( )J h L p gL= = evaluated at the mixture average velocity.

Several empirical correlations have been proposed: Durand (1953), Condolios and Chapus (1963), Newitt (1955), Zandi y Govatos (1967) and Wasp (1977), but unfortunately they again give different results. We will suggest using Newitt relationship (1955) because it was extended for all regime, as we will see later:

Newitt (1955) ( )1 331 1100 , for 17 1800m L zz f

gDuJ J u v gDuv

= + <


( )1000 60

0.283100 3800(100 60) 1000 60

f

s f

w

w w

= = = + +

6

* *5 2

75 102.49,

3.0123 10 3.3198 10d u u

d uP Q

= = = = =

( )( ) ( )( )3 2 3 22 21 2 1 2* ***

20.52 20.521 0.0921 1 0.0921 2.49 0.2297

2.490.2297 0.0332 0.00763 m s

u dd

u u Q

= + = + == = =

( )1 3 17 18000.130 2.5 3.014

zu v gDu

transport concentrates y tailings

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