WORKED EXAMPLES WORKED EXAMPLE 3.1

Above feed

3.4 WORKED EXAMPLES WORKED EXAMPLE 3.1

A height–time curve for the sedimentation of a suspension, of initial suspension concentration 0.1, in vertical cylindrical vessel is shown in Figure 3W1.1. Determine:

(a) the velocity of the interface between clear liquid and suspension of concentration 0.1;

(b) the velocity of the interface between clear liquid and a suspension of concentration 0.175;

Below feed

Above feed Downward

flux below feed

Underflow flux line, slope L/A

Batch flux

Downward flux above feed

Overflow flux line, slope-V/A U_ps

U_ps= FC_F/A=

LC_L/A

=U_psmax

U_ps

CB CF CB=Cmax CL

Feed flux line, slope F/A

Figure 3.18 Alternative total flux plot shape; thickener at critical loading

68 MULTIPLE PARTICLE SYSTEMS

(d) the final sediment concentration.

Solution

(a) Since the initial suspension concentration is 0.1, the velocity required in this question is the velocity of the AB interface. This is given by the slope of the straight portion of the height–time curve.

Figure 3.19 Alternative total flux plot shape; overloaded thickener

(b) We must first find the point on the curve corresponding to the point at which a suspension of concentration 0.175 interfaces with the clear suspension. From Equation (3.38), withC ¼ 0:175, CB¼ 0:1 and h0¼ 40 cm, we find:

h1¼0:1 40

0:175 ¼ 22:85 cm

A line drawn through the pointt ¼ 0, h ¼ h1tangent to the curve locates the point on the curve corresponding to the time at which a suspension of concentration 0.175 interfaces with the clear suspension (Figure 3W1.2). The coordinates of this point are t ¼ 26 s, h ¼ 15 cm. The velocity of this interface is the slope of the curve at this point:

slope of curve at 26 s; 15 cm ¼15 22:85

26 0 ¼ 0:302 cm=s downward velocity of interface¼ 0:30 cm=s

(c) From the consideration above, after 26 s the layer of concentration 0.175 has just reached the clear liquid interface and has travelled a distance of 15 cm from the base of the vessel in this time.

Therefore, upward propagation velocity of this layer¼h t¼15

16¼ 0:577 cm=s

(d) To find the concentration of the final sediment we again use Equation (3.38). The value ofh1corresponding to the final sedimentðh1sÞ is found by drawing a tangent to the part of the curve corresponding to the final sediment and projecting it to theh axis.

In this caseh1S¼ 10 cm and so from Equation (3.38), final sediment concentration; C ¼C0h0

h1S ¼0:1 4:0 10 ¼ 0:4 Figure 3W1.1 Batch settling test; height–time curve

70 MULTIPLE PARTICLE SYSTEMS

WORKED EXAMPLE 3.2

A suspension in water of uniformly sized sphere (diameter 150mm, density 1140 kg/m³ has a solids concentration of 25% by volume. The suspension settles to a bed of solids concentration of 55% by volume. Calculate:

(a) the rate at which the water/suspension interface settles;

(b) the rate at which the sediment/suspension interface rises (assume water properties:

density, 1000 kg=m³; viscosity, 0.001 Pa s).

Solution

(a) Solids concentration of initial suspension,CB¼ 0:25

Equation (3.28) allows us to calculate the velocity of interfaces between suspensions of different concentrations.

The velocity of the interface between initial suspension (B) and clear liquid (A) is therefore:

Uint;AB¼UpACA UpBCB

CA CB

SinceCA¼ 0, the equation reduces to

Uint;AB¼ UpB

0 25

0 50

Time from start of test (s)

Height of interface of suspension with clear liquid (cm)

Tangent to curve at sediment concentration t = 26 s

h1S = 10 cm h1 = 22.85 cm

Slope = –0.302 cm/s Slope = –1.333 cm/s

h = 15 cm

75 100 125

Figure 3W1.2 Batch settling test

UpB is the hindered settling velocity of particles relative to the vessel wall in batch settling and is given by Equation (3.24):

Up¼ UTeⁿ

Assuming Stokes’ law applies, thenn ¼ 4:65 and the single particle terminal velocity is given by Equation (2.13) (see Chapter 2):

UT¼x²ðrp rfÞg 18m

UT¼9:81 ð150 10⁶Þ² ð1140 1000Þ 18 0:001

¼ 1:717 10³m=s

To check that the assumption of Stokes’ law is valid, we calculate the single particle Reynolds number:

Rep¼ð150 10⁶Þ 1:717 10³ 1000 0:001

¼ 0:258, which is less than the limiting value for Stokes’ law (0.3) and so the assumption is valid.

The voidage of the initial suspension, eB¼ 1 CB¼ 0:75 hence; UpB¼ 1:717 10³ 0:75^4:65

¼ 0:45 10³m=s

Hence, the velocity of the interface between the initial suspension and the clear liquid is 0:45 mm=s. The fact that the velocity is positive indicates that the interface is moving downwards.

(b) Here again we apply Equation (3.28) to calculate the velocity of interfaces between suspensions of different concentrations.

The velocity of the interface between initial suspension (B) and sediment (S) is therefore

Uint; BS¼UpBCB UpSCS

CB CS

WithCB¼ 0:25 and CS¼ 0:55 and since the velocity of the sediment, UpS is zero, we have:

Uint;BS¼UpB0:25 0

0:25 0:55 ¼ 0:833UpB

And from part (a), we know thatUpB¼ 0:45mm/s, and so Uint;BS¼ 0:375 mm=s.

The negative sign signifies that the interface is moving upwards. So, the interface between initial suspension and sediment is moving upwards at a velocity of 0.375 mm/s.

72 MULTIPLE PARTICLE SYSTEMS

WORKED EXAMPLE 3.3

For the batch flux plot shown in Figure 3W3.1, the sediment has a solids concentration of 0.4 volume fraction of solids.

(a) Determine the range of initial suspension concentrations over which a zone of variable concentration is formed under batch settling conditions.

(b) Calculate and plot the concentration profile after 50 min in a batch settling test of a suspension with an initial concentration 0.1 volume fraction of solids, and initial suspension height of 100 cm.

Solution

(a) Determine the range of initial suspension concentrations by drawing a line through the pointC ¼ CS¼ 0:4, Ups¼ 0 tangent to the batch flux curve. This is shown as line XCS

in Figure 3W3.2. The range of initial suspension concentrations for which a zone of variable concentration is formed in batch settling (Type 2 settling) is defined by CBmin

andCBmax.CBminis the value ofC at which the line XCSintersects the settling curve and CBmaxis the value ofC at the tangent. From Figure 3W3.2, we see that CBmin¼ 0:036 and CBmax¼ 0:21.

(b) To calculate the concentration profile we must first determine the velocities of the interfaces between the zones A, B, E and S and hence find their positions after 50 min.

The line AB in Figure 3W3.2 joins the point representing A the clear liquid (0, 0) and the point B representing the initial suspension (0.1,Ups). The slope of line AB is equal to the

0.02

0.01

0.1 0.2 0.3 0.4

Volume fraction of solids, C Solids flux, Ups(mm/s)

Figure 3W3.1 Batch flux plot

velocity of the interface between zones A and B. From Figure 3W3.2, Uint; AB¼ þ0:166 mm=s or þ1:00 cm=min.

The slope of the line from point B tangent to the curve is equal to the velocity of the interface between the initial suspension B and the minimum value of the variable concentration zoneCEmin.

From Figure 3W3.2,

Uint; BEmin¼ 0:111 mm=s or 0:66 cm=min

The slope of the line tangent to the curve and passing through the point representing the sediment (pointC ¼ CS¼ 0:4, Ups¼ 0) is equal to the velocity of the interface between the maximum value of the variable concentration zoneCE_max and the sediment.

From Figure 3W3.2,

Uint; EmaxS¼ 0:0355 mm=s or 0:213 cm=min

Therefore, after 50 min the distances travelled by the interfaces will be:

AB interface 50:0 cmð1:00 50Þdownwards BEmininterface 33:2 cm upwards

EmaxS interface 10:6 cm upwards

Therefore, the positions of the interfaces (distance from the base of the test vessel) after 50 min will be

AB interface 50:0 cm BEmininterface 33:2 cm EmaxS interface 10:6 cm Volume fraction of solids, C Solids flux, Ups (mm/s)

Slope = velocity of interface of B with E_min

Slope = velocity of AB interface

Slope = velocity of interface of E_maxwith S

Figure 3W3.2 Graphical solution to batch settling problem in Worked Example 3.3

74 MULTIPLE PARTICLE SYSTEMS

From Figure 3W3.2 we determine the minimum and maximum values of suspension concentration in the variable zone

CE_min¼ 0:16 CEmax¼ 0:21

Using this information we can plot the concentration profile in the test vessel 50 min after the start of the test. A sketch of the profile is shown in Figure 3W3.3. The shape of the concentration profile within the variable concentration zone may be determined by the following method. Recalling that the slope of the batch flux plot (Figure 3W3.1) at a value of suspension concentration C is the velocity of a layer of suspension of that concentration, we find the slope at two or more values of concentration and then determine the positions of these layers after 50 min:

. Slope of batch flux plot at C ¼ 0:18 is 0:44 cm= min upwards.

Hence, position of a layer of concentration 0.18 after 50 min is 22.0 cm from the base.

. Slope of batch flux plot at C ¼ 0:20 , is 0:27 cm= min upwards.

Hence, position of a layer of concentration 0.20 after 50 min is 13.3 cm from the base.

These two points are plotted on the concentration profile in order to determine the shape of the profile within the zone of variable concentration.

Figure 3W3.4 is a sketched plot of the height–time curve for this test constructed from the information above. The shape of the curved portion of the curve can again be Figure 3W3.3 Sketch of concentration profile in batch settling test vessel after 50 min

determined by plotting the positions of two or more layers of suspension of different concentration. The initial suspension concentration zone (B) ends when the AB line intersects the BEminline, both of which are plotted from a knowledge of their slopes.

The time for the end of the test is found in the following way. The end of the test is when the position of the EmaxS interface coincides with the height of the final sediment. The height of the final sediment may be found using Equation (3.38) [see part (d) of Worked Example 3.1]:

CShS¼ CBh0

wherehSis the height of the final sediment andh0is the initial height of the suspension (at the start of the test). With CS¼ 0:4, CB¼ 0:1 and h0¼ 100 cm, we find that hS¼ 25 cm. Plotting hSon Figure 3W3.4, we find that the EmaxS line intersects the final sediment line at about 120 min and so the test ends at this time.

WORKED EXAMPLE 3.4

Using the flux plot shown in Figure 3W4.1:

(a) Graphically determine the limiting feed concentration for a thickener of area 100 m² handling a feed rate of 0:019 m³=s and an underflow rate of 0:01 m³=s.

Under these conditions what will be the underflow concentration and the overflow concentration?

Figure 3W3.4 Sketch of height–time curve for the batch settling test in Worked Example 3.3

76 MULTIPLE PARTICLE SYSTEMS

(b) Under the same flow conditions as above, the feed concentration is increased to 0.2.

Estimate the solids concentration in the overflow, in the underflow, in the upflow section and in the downflow section of the thickener.

Solution

(a) Feed rate,F ¼ 0:019 m³=s Underflow rate,L ¼ 0:01 m³=s

Material balance gives, overflow rate,V ¼ F L ¼ 0:009 m³=s

Expressing these flows as fluxes based on the thickener areaðA ¼ 100 m²Þ:

A¼ 0:19 mm=s L

A¼ 0:10 mm=s V

A¼ 0:09 mm=s

The relationships between bulk flux and suspension concentration are then:

Feed flux¼ CF F A

Flux in underflow¼ CL L A

Flux in overflow¼ CV V A

Lines of slopeF/A, L/A and V/A drawn on the flux plot represent the fluxes in the feed, underflow and overflow, respectively (Figure 3W4.2). The total flux plot for the section below the feed point is found by adding the batch flux plot to the underflow flux line. The total flux plot for the section above the feed point is found by adding the batch

U_ps(mm/s)

Batch

flux C

0.03

0.02

0.01

0 0.1 0.2 0.3

Figure 3W4.1 Batch flux plot

flux plot to the overflow flux line (which is negative since it is an upward flux). These plots are shown in Figure 3W4.2.

The critical feed concentration is found where the feed flux line intersects the plot of total flux in the section below the feed (Figure 3.W4.2). This gives a critical feed flux of 0:0335 mm=s. The downflow section below the feed point is unable to take a flux greater than this. The corresponding feed concentration isCF_crit¼ 0:174.

The concentration in the downflow section,CBis also 0.174.

The corresponding concentration in the underflow is found where the critical flux line intersects the underflow flux line. This givesCL¼ 0:33.

(b) Referring now to Figure 3W4.3, if the feed flux is increased to 0.2, we see that the corresponding feed flux is 0:038 mm=s. At this feed concentration the downflow section

Figure 3W4.2 Total flux plot: solution to part (a) of Worked Example 3.4

78 MULTIPLE PARTICLE SYSTEMS

is only able to take a flux of 0:034 mm=s and gives an underflow concentration, CL¼ 0:34. The excess flux of 0:004 mm=s passes into the upflow section. This flux in the upflow section gives a concentration,CT¼ 0:2 and a corresponding concentration, CV¼ 0:044 in the overflow.

TEST YOURSELF

3.1 The particle settling velocity in a fluid–particle suspension:

(a) increases with increasing ratio of particle diameter to characteristic system dimension;

(b) increases with increasing particle concentration;

Figure 3W4.3 Solution to part (b) of Worked Example 3.4

(d) none of the above.

3.2 The terminal velocity of a particle settling in a stagnant fluid:

(a) increases with increasing ratio of particle diameter to characteristic system dimension;

(b) increases with increasing solids concentration;

(d) none of the above.

3.3 In an overloaded thickener, the concentration in the bottom section of the thickener is equal to (when the total flux plot does not go through a minimum):

(a) the feed concentration;

(b) the overflow concentration;

(d) none of the above.

3.4 In an underloaded thickener, the concentration in the bottom section of the thickener is (when the total flux plot does not go through a minimum):

(a) greater than the feed concentration;

(b) less than the feed concentration;

3.5 In an underloaded thickener, the concentration in the overflow is (when the total flux plot does not go through a minimum):

(a) greater than the feed concentration;

(b) less than the feed concentration;

3.6 In an underloaded thickener, the concentration in the underflow is (when the total flux plot does not go through a minimum):

(a) greater than the concentration in the bottom section;

(b) less than the concentration in the bottom section;

80 MULTIPLE PARTICLE SYSTEMS

3.7 When a particle reaches terminal velocity:

(a) the particle acceleration is constant;

(b) the particle acceleration is zero;

(d) none of the above.

3.8 Which of the following do not influence hindered settling velocity?

(a) particle density;

(b) particle size;

(d) none of the above.

EXERCISES

3.1 A suspension in water of uniformly sized spheres of diameter 100mm and density 1200 kg=m³ has a solids volume fraction of 0.2. The suspension settles to a bed of solids volume fraction 0.5. (For water, density is 1000 kg=m³ and viscosity is 0.001 Pa s.)

The single particle terminal velocity of the spheres in water may be taken as 1:1 mm=s.

Calculate:

(a) the velocity at which the clear water/suspension interface settles;

(b) the velocity at which the sediment/suspension interface rises.

[Answer: (a) 0:39 mm=s; (b) 0:26 mm=s.]

3.2 A height–time curve for the sedimentation of a suspension in a vertical cylindrical vessel is shown in Figure 3E2.1. The initial solids concentration of the suspension is 150 kg=m³.

Determine:

(a) the velocity of the interface between clear liquid and suspension of concentration 150 kg=m³;

(b) the time from the start of the test at which the suspension of concentration 240 kg=m³is in contact with the clear liquid;

(d) the velocity at which a layer of concentration 240 kg=m³propagates upwards from the base of the vessel;

(e) the concentration of the final sediment.

[Answer: (a) 2:91 cm=s; (b) 22 s; (c) 0:77 cm=s downwards; (d) 1:50 cm=s upwards; (e) 600 kg=m³.]

3.3 A suspension in water of uniformly sized spheres of diameter 90mm and density 1100 kg=m³has a solids volume fraction of 0.2. The suspension settles to a bed of solids volume fraction 0.5. (For water, density is 1000 kg=m³and viscosity is 0:001 Pa s.)

The single particle terminal velocity of the spheres in water may be taken as 0:44 mm=s.

Calculate:

(a) the velocity at which the clear water=suspension interface settles;

(b) the velocity at which the sediment=suspension interface rises.

[Answer: (a) 0:156 mm=s; (b) 0:104 mm=s.]

3.4 A height–time curve for the sedimentation of a suspension in a vertical cylindrical vessel is shown in Figure 3E2.1. The initial solids concentration of the suspension is 200 kg=m³.

0 25 50 75 100 125 Time (s)

Height of suspension/clear liquid interface (cm)

Figure 3E2.1 Batch settling test results. Height–time curve for use in Exercises 3.2 and 3.4

82 MULTIPLE PARTICLE SYSTEMS

Determine:

(a) the velocity of the interface between clear liquid and suspension of concentration 200 kg=m³;

(b) the time from the start of the test at which the suspension of concentration 400 kg=m³is in contact with the clear liquid;

(d) the velocity at which a layer of concentration 400 kg=m³propagates upwards from the base of the vessel;

(e) the concentration of the final sediment.

[Answers: (a) 2.9 cm/s downwards; (b) 32.5 s; (c) 0.40 cm/s downwards; (d) 0.846 cm/s upwards; (e) 800 kg/m³.]

3.5

(a) Spherical particles of uniform diameter 40mm and particle density 2000 kg=m³form a suspension of solids volume fraction 0.32 in a liquid of density 880 kg=m³ and viscosity 0:0008 Pa s. Assuming Stokes’ law applies, calculate (i) the sedimentation velocity and (ii) the sedimentation volumetric flux for this suspension.

(b) A height–time curve for the sedimentation of a suspension in a cylindrical vessel is shown in Figure 3E5.1. The initial concentration of the suspension for this test is 0:12 m³=m³.

0 25 50 75 Time (s)

Height of suspension/clear liquid interface (cm)

Figure 3E5.1 Batch settling test results. Height–time curve for use in Exercise 3.5

Calculate:

(i) the velocity of the interface between clear liquid and a suspension of concentration, 0:12 m³=m³;

(ii) the velocity of the interface between clear liquid and a suspension of concentration 0:2 m³=m³;

(iii) the velocity at which a layer of concentration, 0:2 m³=m³propagates upwards from the base of the vessel;

(iv) the concentration of the final sediment;

(v) the velocity at which the sediment propagates upwards from the base.

[Answer: (a)(i) 0:203 mm=s, (ii) 0:065 mm=s, (b)(i) 1:11 cm=s downwards, (ii) 0:345 cm=s downwards, (iii) 0:514 cm=s upwards, (iv) 0.4, (v) 0:30 cm=s upwards.]

3.6 A height–time curve for the sedimentation of a suspension in a vertical cylindrical vessel is shown in Figure 2E6.1. The initial solids concentration of the suspension is 100 kg=m³.

Determine:

(a) the velocity of the interface between clear liquid and suspension of concentration 100 kg=m³;

Time (s)

0 100 200 300 400 500 75

Height of interface between suspension and clear liquid (cm) 0

Figure 3E6.1 Batch settling test results. Height–time curve for use in Exercises 3.6 and 3.8

84 MULTIPLE PARTICLE SYSTEMS

(b) the time from the start of the test at which the suspension of concentration 200 kg=m³is in contact with the clear liquid;

(d) the velocity at which a layer of concentration 200 kg=m³propagates upwards from the base of the vessel;

(e) the concentration of the final sediment.

[Answer: (a) 0:667 cm=s downwards; (b) 140 s; (c) 0:0976 cm=s downwards; (d) 0:189 cm=s upwards; (e) 400 kg=m³.]

3.7 A suspension in water of uniformly sized spheres of diameter 80mm and density 1300 kg=m³has a solids volume fraction of 0.10. The suspension settles to a bed of solids volume fraction 0.4. (For water, density is 1000 kg=m³and viscosity is 0:001 Pa s.) The single particle terminal velocity of the spheres under these conditions is 1:0 mm=s.

Calculate:

(a) the velocity at which the clear water=suspension interface settles;

(b) the velocity at which the sediment/suspension interface rises.

[Answer: (a) 0:613 mm=s; (b) 0:204 mm=s.]

3.8 A height–time curve for the sedimentation of a suspension in a vertical cylindrical vessel is shown in Figure 3E6.1. The initial solids concentration of the suspension is 125 kg=m³.

Determine:

(a) the velocity of the interface between clear liquid and suspension of concentration 125 kg=m³;

(b) the time from the start of the test at which the suspension of concentration 200 kg=m³is in contact with the clear liquid;

(d) the velocity at which a layer of concentration 200 kg=m³propagates upwards from the base of the vessel;

(e) the concentration of the final sediment.

[Answer: (a) 0:667 cm=s downwards; (b) 80 s; (c) 0:192 cm=s downwards; (d) 0:438 cm=s upwards; (e) 500 kg=m³.]

3.9 Use the batch flux plot in Figure 3E9.1 to answer the following questions. (Note that the sediment concentration is 0.44 volume fraction.)

(a) Determine the range of initial suspension concentration over which a variable concentration zone is formed under batch settling conditions.

(b) For a batch settling test using a suspension with an initial concentration 0.18 volume fraction and initial height 50 cm, determine the settling velocity of the interface between clear liquid and suspension of concentration 0.18 volume fraction.

(d) Produce a sketch showing the concentration zones in the settling test 20 min after the start of this test.

[Answer: (a) 0.135 to 0.318; (b) 0:80 cm= min; (c) 34 cm from base; (d) BE interface is 12.5 cm from base.]

3.10 Consider the batch flux plot shown in (Figure 3W3.1). Given that the final sediment concentration is 0.36 volume fraction:

(a) determine the range of initial suspension concentration over which a variable concentration zone is formed under batch settling conditions;

(b) calculate and sketch the concentration profile after 40 min of the batch settling test with an initial suspension concentration of 0.08 and an initial height of 100 cm;

In document Introduction to Particle Technology, 2nd Ed [Martin Rhodes] (Page 86-109)