### solutions to chapter 3: multiple particle

```SOLUTIONS TO CHAPTER 3: MULTIPLE PARTICLE SYSTEMS
EXERCISE 3.1
A suspension in water of uniformly sized spheres of diameter 100 μm 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, 1000 kg/m3 and viscosity, 0.001 Pas)
The single particle terminal velocity of the spheres in water may be taken as 1.1mm/s.
Calculate:
a) The velocity at which the clear water/suspension interface settles;
b) The velocity at which the sediment/suspension interface rises.
SOLUTION TO EXERCISE 3.1
(a) Solids concentration of initial suspension, CB = 0.20
Text-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:
UpA CA − U pBC B
U int,AB =
C A − CB
Since CA = 0, the equation reduces to
U int,AB = U pB
UpB is the hindered settling velocity of particles relative to the vessel wall in batch
settling and is given by Text-Equation 3.24:
U p = UT ε n
To check whether Stokes Law is valid, we calculate the single particle Reynolds
number at the terminal velocity:
(100 × 10−6 ) × 1.1 × 10−3 × 1000
Re p =
= 0.11
0.001
This value is less than the limiting value for Stokes Law (0.3) and so Stokes Law
applies and therefore in Text-Equation 3.24 exponent n = 4.65.
The voidage of the initial suspension, εΒ = 1 - CB = 0.80
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS
Page 3.1
hence, U pB = 1.1 × 10 −3 × 0.804.65
= 3.897 x 10-4 m/s
Hence, the velocity of the interface between the initial suspension and the clear liquid
is 0.39 mm/s. The fact that the velocity is positive indicates that the interface is
moving downwards.
(b) Here again we apply Text-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:
UpBCB − U pSC S
U int,BS =
C B − CS
With CB = 0.20 and CS = 0.50 and since the velocity of the sediment, UpS is zero, we
have:
U int,BS =
UpB 0.20 − 0
0.20 − 0.50
= −0.6667UpB
And from part (a), we know that UpB = 0.39mm/s, and so U int,BS = −0.26 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
0.26mm/s.
EXERCISE 3.2
A height-time curve for the sedimentation of a suspension in a vertical cylindrical
vessel is shown in Text-Figure 3.E2.1. 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/m3 is in contact with the clear liquid.
c)
The velocity of the interface between the clear liquid and suspension of
concentration 240 kg/m3.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS
Page 3.2
d)
The velocity at which a layer of concentration 240 kg/m3 propagates upwards
from the base of the vessel.
e)
The concentration of the final sediment.
SOLUTION TO EXERCISE 3.2
(a) Since the initial suspension concentration is 150 kg/m3, 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.
0 − 80
Slope =
= −2.91 cm / s (the negative sign indicates downwards)
27.5 − 0
Therefore, velocity of the interface between clear liquid and suspension of
concentration 150 kg/m3 is 2.91 cm/s downwards.
(b) We must first find the point on the curve corresponding to the point at which a
suspension of concentration 240 kg/m3 interfaces with the clear suspension. From
Equation 3.38, with C = 240, CB = 150 and h0 = 80 cm, we find:
h1 =
150 × 80
= 50 cm
240
A line drawn through the point t = 0, h = h1 tangent to the curve locates the point on
the curve corresponding to the time at which a suspension of concentration 240 kg/m3
interfaces with the clear suspension (Solution Manual-Figure 3.2.1). The coordinates
of this point are
t = 22 sec, h = 33 cm.
(c) The velocity of this interface is the slope of the curve at this point:
slope of curve at 22 sec, 33 cm = −0.77 cm / s
downward velocity of interface = 0.77 cm/s
(d) From the consideration above, after 22 seconds the layer of concentration 240
kg/m3 has just reached the clear liquid interface and has travelled a distance of 33 cm
from the base of the vessel in this time.
h 33
Therefore, upward propagation velocity of this layer =
=
= 1.5 cm / s
t 23
(e) To find the concentration of the final sediment we again use Equation 3.38. The
value of h1 corresponding to the final sediment (h1S) is found by drawing a tangent to
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS
Page 3.3
the part of the curve corresponding to the final sediment and projecting it to the h
axis.
In this case h1S = 20 cm and so from Text-Equation 3.38,
final sediment concentration, C =
C0 h 0 150 × 80
3
=
= 600 kg / m
20
h1S
EXERCISE 3.3
A suspension in water of uniformly sized spheres of diameter 90 μm and density 1100
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, 1000 kg/m3 and viscosity, 0.001 Pas)
The single particle terminal velocity of the spheres in water may be taken as
0.44mm/s.
Calculate:
a)
The velocity at which the clear water/suspension interface settles;
b)
The velocity at which the sediment/suspension interface rises.
SOLUTION TO EXERCISE 3.3
(a) Solids concentration of initial suspension, CB = 0.20
Text-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:
UpA CA − U pBC B
U int,AB =
C A − CB
Since CA = 0, the equation reduces to
U int,AB = UpB
UpB is the hindered settling velocity of particles relative to the vessel wall in batch
settling and is given by Text-Equation 3.24:
U p = UT ε n
To check whether Stokes Law is valid, we calculate the single particle Reynolds
number at the terminal velocity:
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS
Page 3.4
(90 × 10− 6 ) × 0.44 × 10− 3 × 1000
Re p =
= 0.0396
0.001
This value is less than the limiting value for Stokes Law (0.3) and so Stokes Law
applies and therefore in Text-Equation 3.24 exponent n = 4.65.
The voidage of the initial suspension, εΒ = 1 - CB = 0.80
hence, U pB = 0.44 × 10 −3 × 0.804.65
= 1.56 x 10-4 m/s
Hence, the velocity of the interface between the initial suspension and the clear liquid
is 0.156 mm/s. The fact that the velocity is positive indicates that the interface is
moving downwards.
(b) Here again we apply Text-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:
UpBCB − U pSC S
U int,BS =
C B − CS
With CB = 0.20 and CS = 0.50 and since the velocity of the sediment, UpS is zero, we
have:
U int,BS =
UpB 0.20 − 0
0.20 − 0.50
= −0.6667UpB
And from part (a), we know that UpB = 0.156 mm/s, and so U int,BS = −0.104 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
0.104mm/s.
EXERCISE 3.4
A height-time curve for the sedimentation of a suspension in a vertical cylindrical
vessel is shown in Text-Figure 3.E2.1. The initial solids concentration of the
suspension is 200 kg/m3. Determine:
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS
Page 3.5
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/m3 is in contact with the clear liquid.
c)
The velocity of the interface between the clear liquid and suspension of
concentration 400 kg/m3.
d)
The velocity at which a layer of concentration 400 kg/m3 propagates upwards
from the base of the vessel.
e)
The concentration of the final sediment.
SOLUTION TO EXERCISE 3.4
(a) Since the initial suspension concentration is 200 kg/m3, 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 (Solution Manual-Figure 3.4.1).
0 − 80
Slope =
= −2.91 cm / s (value given in book is incorrect)
27.5 − 0
Therefore, velocity of the interface between clear liquid and suspension of
concentration 200 kg/m3 is 2.91 cm/s downwards.
(b) We must first find the point on the curve corresponding to the point at which a
suspension of concentration 400 kg/m3 interfaces with the clear suspension. From
Equation 3.38, with C = 400, CB = 200 and h0 = 80 cm, we find:
h1 =
200 × 80
= 40 cm
400
A line drawn through the point t = 0, h = h1 tangent to the curve locates the point on
the curve corresponding to the time at which a suspension of concentration 240 kg/m3
interfaces with the clear suspension (Solution Manual-Figure 3.4.1). The coordinates
of this point are
t = 32.5 sec, h = 27.5 cm.
(c) The velocity of this interface is the slope of the curve at this point:
slope of curve at 32.5 sec, 27.5 cm = −0.40 cm / s
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS
Page 3.6
downward velocity of interface = 0.4 cm/s
(d) From the consideration above, after 32.5 seconds the layer of concentration 400
kg/m3 has just reached the clear liquid interface and has travelled a distance of 27.5
cm from the base of the vessel in this time.
h 27.5
Therefore, upward propagation velocity of this layer = =
= 0.846 cm / s
t 32.5
(e) To find the concentration of the final sediment we again use Text-Equation 3.38.
The value of h1 corresponding to the final sediment (h1S) is found by drawing a
tangent to the part of the curve corresponding to the final sediment and projecting it to
the h axis.
In this case h1S = 20 cm and so from Text-Equation 3.38,
final sediment concentration, C =
C0 h 0 200 × 80
=
= 800 kg / m 3
20
h1S
EXERCISE 3.5
a) Spherical particles of uniform diameter, 40 μm and particle density 2000 kg/m3
form a suspension of solids volume fraction 0.32 in a liquid of density 880 kg/m3 and
viscosity 0.0008 Pas. 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 Text-Figure 3.E5.1. The initial concentration of the suspension for this test
is 0.12 m3/m3. 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/m3 propagates upwards
from the base of the vessel.
(iv)
the concentration of the final sediment.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS
Page 3.7
(v)
the velocity at which the sediment propagates upwards from the base.
SOLUTION TO EXERCISE 3.5
Part (a)
(i) Solids concentration of initial suspension, CB = 0.32
Text-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:
UpA CA − U pBC B
U int,AB =
C A − CB
Since CA = 0, the equation reduces to
U int,AB = UpB
UpB is the hindered settling velocity of particles relative to the vessel wall in batch
settling and is given by Text-Equation 3.24:
U p = UT ε n
Assuming Stokes Law applies, then n = 4.65 and the single particle terminal velocity
is given by Text-Equation 2.13 (see Chapter 2 - Single Particles in Fluids):
⎛ x2 (ρ p − ρ f )g ⎞
⎜
⎟
UT =
18μ
⎝
⎠
9.81× (40 × 10 −6 )2 × (2000 − 880)
18 × 0.0008
= 1.221 x 10-3 m/s
UT =
To check that assumption of Stokes Law is valid, we calculate the single particle
Reynolds number:
(40 × 10 −6 ) × 1.221 × 10 −3 × 880
Re p =
= 0.054
0.0008
This value is less than the limiting value for Stokes Law (0.3) and so the assumption
is valid. Hence in Text-Equation 3.24, exponent n = 4.65.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS
Page 3.8
The voidage of the initial suspension, εΒ = 1 - CB = 0.68
hence, U pB = 1.221 × 10 −3 × 0.684.65
= 2.03 x 10-4 m/s
Hence, the velocity of the interface between the initial suspension and the clear liquid
is 0.203 mm/s. The fact that the velocity is positive indicates that the interface is
moving downwards.
(ii) From Text-Equation 3.15, sedimentation volumetric flux, U ps = U p (1 − ε )
Hence, with Up = 0.203 mm/s and ε = 0.68,
volumetric flux, U ps = 0.203 × (1 − 0.68) = 0.065 mm3 / mm2
Part (b)
(i) Since the initial suspension concentration is 0.12 m3/m3, 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 (Solution Manual-Figure 3.5.1).
25 − 50
= −1.11 cm / s (the negative sign indicates downwards)
Slope =
22.5 − 0
Therefore, the velocity of the interface between clear liquid and a suspension of
concentration, 0.12 m3/m3 is 1.11 cm/s downwards.
(ii) We must first find the point on the curve corresponding to the point at which a
suspension of concentration 0.2 m3/m3 interfaces with the clear suspension. From
Text-Equation 3.38, with C = 0.2, CB = 0.12 and h0 = 50 cm, we find:
h1 =
0.12 × 50
= 30 cm
0.2
A line drawn through the point t = 0, h = h1 tangent to the curve locates the point on
the curve corresponding to the time at which a suspension of concentration 0.2 m3/m3
interfaces with the clear suspension (Solution Manual-Figure 3.5.1). The coordinates
of this point are
t = 35 sec, h = 18 cm.
(iii) The velocity of this interface is the slope of the curve at this point:
slope of curve at 35 sec, 18 cm =−0.35 cm / s
downward velocity of interface = 0.35 cm/s
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS
Page 3.9
(iv) From the consideration above, after 35 seconds the layer of concentration 0.2
m3/m3 has just reached the clear liquid interface and has travelled a distance of 18 cm
from the base of the vessel in this time.
h 18
Therefore, upward propagation velocity of this layer = =
= 0.514 cm / s
t 35
(v) To find the concentration of the final sediment we again use Text-Equation 3.38.
The value of h1 corresponding to the final sediment (h1S) is found by drawing a
tangent to the part of the curve corresponding to the final sediment and projecting it to
the h axis (Solution Manual-Figure 3.5.1).
In this case h1S = 15 cm and so from Text-Equation 3.38,
final sediment concentration, C=
C0 h 0 0.12 × 50
=
= 0.4 m 3 / m3
15
h1S
(vi) At the end of the test (ts), we know that the surface of the sediment has risen from
h
15
h = 0 to h = h1S in a time ts. Hence sediment velocity = 1S =
= 0.3 cm / s
50
tS
EXERCISE 3.6
A height-time curve for the sedimentation of a suspension in a vertical cylindrical
vessel is shown in Text-Figure 3.E6.1. 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.
b)
the time from the start of the test at which the suspension of concentration
200 kg/m3 is in contact with the clear liquid.
c)
the velocity of the interface between the clear liquid and suspension of
concentration 200 kg/m3.
d)
the velocity at which a layer of concentration 200 kg/m3 propagates upwards
from the base of the vessel.
e)
the concentration of the final sediment.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.10
SOLUTION TO EXERCISE 3.6
(a) Since the initial suspension concentration is 100 kg/m3, 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.
Slope= −0.67 cm / s (negative sign indicates downwards)
Therefore, the velocity of the interface between clear liquid and suspension of
concentration 100 kg/m3 is 0.67 cm/s downwards.
(b) We must first find the point on the curve corresponding to the point at which a
suspension of concentration 200 kg/m3 interfaces with the clear suspension. From
Equation 3.38, with C = 200, CB = 100 and h0 = 80 cm, we find:
100 × 80
= 40 cm
200
A line drawn through the point t = 0, h = h1 tangent to the curve locates the point on
h1 =
the curve corresponding to the time at which a suspension of concentration 200 kg/m3
interfaces with the clear suspension (Solution Manual-Figure 3.6.1). The coordinates
of this point are
t = 140 sec, h = 26.5 cm.
(c) The velocity of this interface is the slope of the curve at this point:
slope of curve at 140 sec, 26.5 cm = −0.98 cm / s
downward velocity of interface = 0.98 cm/s
(d) From the consideration above, after 140 seconds the layer of concentration 200
kg/m3 has just reached the clear liquid interface and has travelled a distance of 26.5
cm from the base of the vessel in this time.
h 26.5
=
= 0.189 cm / s
t 140
(e) To find the concentration of the final sediment we again use Text-Equation 3.38.
The value of h1 corresponding to the final sediment (h1S) is found by drawing a
Therefore, upward propagation velocity of this layer =
tangent to the part of the curve corresponding to the final sediment and projecting it to
the h axis.
In this case h1S = 20 cm and so from Text-Equation 3.38,
final sediment concentration, C=
C0 h 0 100 × 80
3
=
= 400 kg / m
20
h1S
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.11
EXERCISE 3.7
A suspension in water of uniformly sized spheres of diameter 80 μm and density 1300
kg/m3 has a solids volume fraction of 0.10. The suspension settles to a bed of solids
volume fraction 0.4. (For water: density, 1000 kg/m3 and viscosity, 0.001 Pas)
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.
SOLUTION TO EXERCISE 3.7
(a) Solids concentration of initial suspension, CB = 0.10
Text-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:
UpA CA − U pBC B
U int,AB =
C A − CB
Since CA = 0, the equation reduces to
U int,AB = UpB
UpB is the hindered settling velocity of particles relative to the vessel wall in batch
settling and is given by Text-Equation 3.24:
U p = UT ε n
To check whether Stokes Law is valid, we calculate the single particle Reynolds
number at the terminal velocity:
xU Tρ f (80 × 10−6 ) × 1.0 × 10−3 × 1000
Re p =
=
= 0.08
μ
0.001
This value is less than the limiting value for Stokes Law (0.3) and so Stokes Law
applies and therefore, in Text-Equation 3.24, exponent n = 4.65.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.12
The voidage of the initial suspension, εΒ = 1 - CB = 0.90
hence, U pB = 1.0 × 10 −3 × 0.90 4.65
= 6.13 x 10-4 m/s
Hence, the velocity of the interface between the initial suspension and the clear liquid
is 0.613 mm/s. The fact that the velocity is positive indicates that the interface is
moving downwards.
(b) Here again we apply Text-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:
UpBCB − U pSC S
U int,BS =
C B − CS
With CB = 0.10 and CS = 0.40 and since the velocity of the sediment, UpS is zero, we
have:
U int,BS =
UpB 0.10 − 0
0.10 − 0.40
= −0.3333U pB
And from part (a), we know that UpB = 0.613 mm/s, and so U int,BS = −0.204 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 0.204
mm/s.
EXERCISE 3.8
A height-time curve for the sedimentation of a suspension in a vertical cylindrical
vessel is shown in Text-Figure 3.E6.1. 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/m3 is in contact with the clear liquid.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.13
c)
the velocity of the interface between the clear liquid and suspension of
concentration 200 kg/m3.
d)
the velocity at which a layer of concentration 200 kg/m3 propagates upwards
from the base of the vessel.
e)
the concentration of the final sediment.
SOLUTION TO EXERCISE 3.8
(a) Since the initial suspension concentration is 125 kg/m3, 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 (Solution Manual-Figure 3.8.1).
Slope = -0.67 cm / s (negative sign indicates downwards)
Therefore, the velocity of the interface between clear liquid and suspension of
concentration 125 kg/m3 is 0.67 cm/s downwards.
(b) We must first find the point on the curve corresponding to the point at which a
suspension of concentration 200 kg/m3 interfaces with the clear suspension. From
Text-Equation 3.38, with C = 200, CB = 125 and h0 = 80 cm, we find:
h1 =
125 × 80
= 50 cm
200
A line drawn through the point t = 0, h = h1 tangent to the curve locates the point on
the curve corresponding to the time at which a suspension of concentration 200 kg/m3
interfaces with the clear suspension (Solution Manual-Figure 3.8.1). The coordinates
of this point are
t = 80 sec, h = 35 cm.
(c) The velocity of this interface is the slope of the curve at this point:
slope of curve at 80 sec, 35 cm =−0.19 cm / s
downward velocity of interface = 0.19 cm/s
(d) From the consideration above, after 80 seconds the layer of concentration 200
kg/m3 has just reached the clear liquid interface and has travelled a distance of 35 cm
from the base of the vessel in this time.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.14
Therefore, upward propagation velocity of this layer =
h 35
=
= 0.4375 cm / s
t 80
(e) To find the concentration of the final sediment we again use Text-Equation 3.38.
The value of h1 corresponding to the final sediment (h1S) is found by drawing a
tangent to the part of the curve corresponding to the final sediment and projecting it to
the h axis.
In this case h1S = 20 cm and so from Text-Equation 3.38,
final sediment concentration, C=
C0 h 0 125 × 80
=
= 500 kg / m3
h1S
20
EXERCISE 3.9
Use the batch flux plot in Text-Figure 3.E9.1 to answer the following questions (Note
that the sediment concentration is 0.44 volume fraction).
(i) Determine the range of initial suspension concentration over which a variable
concentration zone is formed under batch settling conditions.
(ii) For a batch settling test using a suspension with an initial concentration 0.18
volume fraction and initial height 50cm, determine the settling velocity of the
interface between clear liquid and suspension of concentration 0.18 volume fraction.
(iii) Determine the position of this interface 20 minutes after the start of this test.
(iv) Produce a sketch showing the concentration zones in the settling test 20 minutes
after the start of this test.
SOLUTION TO EXERCISE 3.9
(a) Determine the range on initial suspension concentrations by drawing a line
through the point C = CS = 0.44, Ups = 0 tangent to the inflection point in the batch
flux curve. This is shown as line XCS in Solution Manual-Figure 3.9.1. 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 and CBmax. CBmin is the value
of C at which the line XCS intersects the settling curve and CBmax is the value of C at
the tangent to the inflection point. From Solution Manual-Figure 32.9.1, we see that
CBmin = 0.135 and CBmax = 0.318.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.15
(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 20
minutes.
The line AB in Solution Manual-Figure 3.9.1 joins the point representing A the clear
liquid (0, 0) and the point B representing the initial suspension (0.18, Ups). The slope
of line AB is equal to the velocity of the interface between zones A and B.
From Figure 2.9.1, U int,AB = +0.1333 mm / s or + 0.80 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 zone Emin.
From Solution Manual-Figure 3.9.1, U int,BE min = −0.104 mm / s or - 0.626 cm/min.
The slope of the line tangent to the curve and passing through the point representing
the sediment (point C = CS = 0.44, Ups = 0) is equal to the velocity of the interface
between the maximum value of the variable concentration zone CEmax and the
sediment.
From Solution Manual-Figure 3.9.1, U int,E max S = −0.082 mm / s or - 0.49 cm/min.
Therefore, after 20 minutes the distances travelled by the interfaces will be:
AB interface
16 cm (0.80 x 20) downwards
BEmin interface
12.5 cm (0.626 x 20) upwards
EmaxS interface
9.8 cm (0.49 x 20) upwards
Therefore the positions of the interfaces (distance from the base of the test vessel)
after 20 minutes will be:
AB interface
BEmin interface
EmaxS interface
34 cm
12.5 cm
9.8 cm
From Solution Manual-Figure 3.9.1 we determine the minimum and maximum values
of suspension concentration in the variable zone:
CEmin = 0.30
CEmax = 0.318
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.16
Using this information we can plot the concentration profile in the test vessel 20
minutes after the start of the test. A sketch of the profile is shown in Solution ManualFigure 3.9.2. 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 (Text-Figure 3.E3.1) 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 20
minutes. This allows us to determine the shape of the profile within the zone of
variable concentration.
EXERCISE 3.10
Consider the batch flux plot shown in Worked Example 3.3 (Figure 3.W3.1). 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 minutes of the batch
settling test with an initial suspension concentration of 0.08 and an initial
height of 100cm.
(c)
Estimate the height of the final sediment and the time at which the test is
complete.
SOLUTION TO EXERCISE 3.10
(a) Determine the range on initial suspension concentrations by drawing a line
through the point C = CS = 0.36, Ups = 0 tangent to the inflection point in the batch
flux curve. This is shown as line XCS in Solution Manual-Figure 3.10.1. 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 and CBmax. CBmin is the value
of C at which the line XCS intersects the settling curve and CBmax is the value of C at
the tangent to the inflection point. From Solution Manual-Figure 3.10.1, we see that
CBmin = 0.045 and CBmax = 0.20.
(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 40
minutes.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.17
The line AB in Solution Manual-Figure 3.10.1 joins the point representing A the clear
liquid (0, 0) and the point B representing the initial suspension (0.08, Ups). The slope
of line AB is equal to the velocity of the interface between zones A and B.
From Solution Manual-Figure 3.10.1, U int,AB = +0.2075 mm / s or + 1.245 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 zone Emin.
From Solution Manual-Figure 3.10.1, U int,BE min = −0.085 mm / s or - 0.51 cm/min.
The slope of the line tangent to the curve and passing through the point representing
the sediment (point C = CS = 0.36, Ups = 0) is equal to the velocity of the interface
between the maximum value of the variable concentration zone CEmax and the
sediment.
From Solution Manual-Figure 3.10.1, U int,E max S = −0.043 mm / s or - 0.26 cm/min.
Therefore, after 40 minutes the distances travelled by the interfaces will be:
AB interface
49.8 cm (1.245 x 40) downwards
BEmin interface
20.4 cm (0.51 x 40) upwards
EmaxS interface
10.4 cm (0.26 x 40) upwards
Therefore the positions of the interfaces (distance from the base of the test vessel)
after 40 minutes will be:
AB interface
BEmin interface
EmaxS interface
50.2 (100 - 49.8) cm
20.4 cm
10.4 cm
From Solution Manual-Figure 3.10.1 we determine the minimum and maximum
values of suspension concentration in the variable zone:
CEmin = 0.17
CEmax = 0.20
Using this information we can plot the concentration profile in the test vessel 40
minutes after the start of the test. A sketch of the profile is shown in Solution ManualFigure 3.10.2. The shape of the concentration profile within the variable
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.18
concentration zone may be determined by the following method. Recalling that the
slope of the batch flux plot 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 value of
concentration and then determine the positions of these layers after 40 minutes. These
points are plotted on the concentration profile in order to determine the shape of the
profile within the zone of variable concentration.
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 Text-Equation 3.38
[see part (iv) of Worked Example 3.1]:
CS h S = C Bh0
where hS is the height of the final sediment and h0 is the initial height of the
suspension (at the start of the test). With CS = 0.36, CB = 0.08 and h0 = 100cm, we
find that hS = 22.2 cm. The velocity of the sediment interface with the variable
concentration zone was found above to be 0.26 cm/min upwards (U int, E max S ).
Hence the test ends when the sediment interface has travelled a distance hS at this
velocity:
i.e. time to end of test =
hS
U int,E max S
=
22.2
= 85.5 minutes
0.26
EXERCISE 3.11
The batch and continuous flux plots supplied in Text-Figure 3.E11.1 are for a
thickener of area 200m2 handling a feed rate of 0.04m3/s and an underflow rate of
0.025m3/s.
Using these plots, graphically determine the critical or limiting feed concentration for
this thickener.
If the feed concentration is 0.18 m3/m3, determine the solids concentrations in the
overflow, underflow, in the regions above and below the feed well.
Under the same flow rate conditions in the same thickener, the feed concentration
increases to 0.24. Estimate the new solids concentration in the overflow and the
underflow once steady state has been reached.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.19
SOLUTION TO EXERCISE 3.11:
Part (a)
Feed rate, F = 0.04 m3/s
Underflow rate, L = 0.025 m3/s
Material balance gives, underflow rate, V = F - L = 0.015 m3/s
Expressing these flows as fluxes based on the thickener area (A = 200 m2):
F
= 0.2 mm / s
A
L
= 0.125 mm / s
A
V
= 0.075 mm / s
A
The variation in fluxes with concentration of the suspension are then:
⎛F
Feed flux = C F ⎝ ⎞⎠
A
⎛L
Flux in underflow = C L ⎝ ⎞⎠
A
⎛ V⎞
Flux in overflow = C V ⎝ ⎠
A
Lines of slope F/A, L/A and -V/A drawn on the flux plot represent the fluxes in the
feed, underflow and overflow respectively (Solution Manual-Figure 3.11.1). The total
flux plot for the section below the feed point is found by adding batch flux plot to the
underflow flux line (this has already been done in Solution Manual-Figure 3.11.1).
The total flux plot for the section above the feed point is found by adding the batch
flux plot to the underflow flux line (which is negative since it is an upward flux) (this
has also already been done in Solution Manual-Figure 3.11.1). These plots are shown
in Solution Manual-Figure 3.11.1.
In this case the total flux curve has a minimum in it and so this corresponds to the
maximum flux that the downflow section can take. The critical feed flux is therefore
0.041 mm/s (Solution Manual-Figure 3.11.1). The corresponding feed concentration
is CFcrit = 0.21.
The corresponding concentration in the downflow section, CB is also 0.21.
The corresponding concentration in the underflow is found where the critical flux line
intersects the underflow flux line. This gives CL = 0.34.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.20
Part (b):
Referring again to Solution Manual-Figure 3.11.1, if the feed flux is increased to
0.18, we see that the corresponding feed flux is about 0.036 mm/s. The thickener is
underloaded since the feed concentration is less than the critical value. All the feed
passes through the downflow section giving a concentration CB = 0.087 (where
constant flux 0.036 intersects the total down flux curve) and an underflow
concentration CL = 0.29 (where the constant flux 0.036 intersect the underflow line).
The concentration in the upward flow section CT and the overflow concentration CV
are both zero.
(c) Referring to Solution Manual-Figure 3.11.1, if the feed flux is increased to 0.24,
the thickener is now overloaded and we see that the corresponding feed flux is about
0.048 mm/s. At this feed concentration the downflow section is only able to take a
flux of 0.042 mm/s and gives an underflow concentration, CL = 0.34. The excess flux
of 0.006 mm/s passes into the upflow section. This flux in the upflow section gives a
concentration, CT = 0.24 and a corresponding concentration, CV = 0.08 in the
overflow.
EXERCISE 3.12
(a) Using the batch flux plot data given in Text-Table 3.E12.1, graphically determine
the limiting feed concentration for a thickener of area 300 m2 handling a feed rate of
0.03 m3/s and with an underflow rate of 0.015 m3/s. Determine the underflow
concentration and overflow concentration under these conditions. Sketch a possible
concentration profile in the thickener clearly indicating the positions of the overflow
launder, the feed well and the point of underflow withdrawal (neglect the conical base
of the thickener).
Table 3.E12.1: Batch flux test data for Exercise 3.12
C
0.01 0.02 0.04
5.0
Flux
mm/s
(x 103)
C
0.22
9.1
13.6
0.24
0.06
0.08
0.10
0.12
0.14
0.16
0.18 0.20
15.7
16.4
16.4
15.7
13.3
10.0
8.3
0.26
0.28
0.30
0.32
0.34
0.36
0.38
7.3
0.40
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.21
Flux
mm/s
(x 103)
6.7
5.6
5.1
4.5
4.2
3.8
3.5
3.3
3.0
2.9
(b) Under the same flow conditions as above, the concentration in the feed increases
to 110% of the limiting value. Estimate the solids concentration in the overflow, in
the underflow, in the section of the thickener above the feed well and in the section
below the feed well.
SOLUTION TO EXERCISE 3.12:
Part (a)
The batch flux data is first plotted with concentration as abscissa (see Solution
Manual-Figure 3.12.1).
Feed rate, F = 0.03 m3/s
Underflow rate, L = 0.015 m3/s
Material balance gives, underflow rate, V = F - L = 0.015 m3/s
Expressing these flows as fluxes based on the thickener area (A = 300 m2):
F
= 0.1 mm / s
A
L
= 0.05 mm / s
A
V
= 0.05 mm / s
A
The variations in fluxes with concentration of the suspension are then:
⎛F
Feed flux = C F ⎝ ⎞⎠
A
⎛L
Flux in underflow = C L ⎝ ⎞⎠
A
⎛ V⎞
Flux in overflow = C V ⎝ ⎠
A
Lines of slope F/A, L/A and -V/A drawn on the flux plot represent the fluxes in the
feed, underflow and overflow respectively (Solution Manual-Figure 3.12.1). The total
flux plot for the section below the feed well is found by adding batch flux plot to the
underflow flux line. The total flux plot for the section above the feed well is found by
adding the batch flux plot to the overflow flux line (which is negative since it is an
upward flux). These plots are shown in Solution Manual-Figure 3.12.1.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.22
In this case the total flux curve has a minimum in it and so this corresponds to the
maximum flux that the downflow section can take. The critical feed flux is therefore
0.017 mm/s (Solution Manual-Figure 3.12.1). The corresponding feed concentration
is CFcrit = 0.17.
The corresponding concentration in the underflow is found where the critical flux line
intersects the underflow flux line (slope L/A). This gives CL = 0.34.
There are two possible concentrations in the downflow section which could give rise
to the critical flux (0.017 mm/s); these are CBmin = 0.05 and CBmax = 0.19.
Under critical conditions the flux in the upflow section (CT) is zero and so the
overflow concentration CV is zero.
Part (b):
Referring now to Solution Manual-Figure 3.12.2, if the feed flux concentration is
increased to 0.187 (110% of CFcrit), we see that the corresponding feed flux is about
0.0187 mm/s. The thickener is overloaded since the feed concentration is greater than
the critical value. The excess flux (0.0187 - 0.017 = 0.0017 mm/s) passes to the
upflow section above the feed well. Solution Manual-Figure 3.12.2 shows that this
excess flux gives rise to a concentration above the feed well of CT = 0.19 and an
overflow concentration of CV = 0.034. Passing through the downflow section below
the feed well is the critical flux, which gives rise to a concentration in this section
CBmin = 0.19 or CBmax = 0.05 and an underflow concentration of CL = 0.34.
EXERCISE 3.13:
Uniformly sized spheres of diameter 50 μm and density 1500 kg/m3 are uniformly
suspended in a liquid of density 1000 kg/m3 and viscosity 0.002 Pas. The resulting
suspension has a solids volume fraction of 0.30.
The single particle terminal velocity of the spheres in this liquid may be taken as
0.00034 m/s (Rep < 0.3). Calculate the velocity at which the clear water/suspension
interface settles.
SOLUTION TO EXERCISE 3.13:
(a) Solids concentration of initial suspension, CB = 0.30
Text-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:
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.23
U int,AB =
UpA CA − U pBC B
C A − CB
Since CA = 0, the equation reduces to
U int,AB = UpB
UpB is the hindered settling velocity of particles relative to the vessel wall in batch
settling and is given by Text-Equation 3.24:
U p = UT ε n
To check whether Stokes Law is valid, we calculate the single particle Reynolds
number at the terminal velocity:
Re p =
(50 × 10 −6 ) × 3.4 × 10 −4 × 1000
= 0.0085
0.002
This value is less than the limiting value for Stokes Law (0.3) and so Stokes Law
applies and therefore in Text-Equation 3.24 exponent n = 4.65.
The voidage of the initial suspension, εΒ = 1 - CB = 0.70
hence, UpB = 3.4 × 10 −4 × 0.70 4.65
= 6.474 x 10-5 m/s
Hence, the velocity of the interface between the initial suspension and the clear liquid
is 0.0647 mm/s. The fact that the velocity is positive indicates that the interface is
moving downwards.
EXERCISE 3.14:
Calculate the settling velocity of glass spheres having a diameter of 155 microns in
water at 293K. The slurry contains 60 wt % solids. The density of the glass spheres
is 2467 kg/m3.
How does the settling velocity change if the particles have a sphericity of 0.3 and an
equivalent diameter of 155 microns?
SOLUTION TO EXERCISE 3.14:
In order to calculate the settling velocity o0f the slurry, the single particle settling
velocity must first be calculated.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.24
To calculate UT for a given particle size x, calculate the group:
C D Re2p
3
4 x ρ f (ρ p − ρf )g
=
μ2
3
=
4 ⎡ (155 × 10 −6 )3 × 1000 × (2467 − 1000 ) × 9.81⎤
⎥
⎢
3 ⎢⎣
(0.001)2
⎥⎦
= 71.4
Using the standard drag curve (Figure 2.3) for spherical particles (ψ = 1.0), Rep = 2
The terminal velocity UT may be calculated from:
Re p = 2 =
ρ f x v UT
μ
Hence, terminal velocity, UT = 0.0129 m/s
Settling velocity of the suspension = Up = UT ε n
Where n is given by:
And Ar =
4.8 − n
= 0.043 Ar 0.57
n − 2.4
x 3 ρ f (ρ p − ρ f )g
μ
2
(155 × 10 )
=
−6 3
× 1000 × (2467 − 1000 ) × 9.81
(0.001)2
= 53.5
Which gives n = 4.1
NOTE: solids volume fraction of the slurry is not 0.6. 100 kg of slurry contain 60 kg
of glass spheres and 40 kg water. So the slurry contains 0.0243 m3 glass spheres and
0.04 m3 water. The solids volume fraction of the slurry is therefore 0.38 and ε = 0.62.
Settling velocity of the suspension = Up = UT ε n = 0.0129 × (0.62)4.1 = 1.8 × 10 −3 m/s
If the particles have a sphericity of 0.3, we need to go back to the standard drag curve
for ψ = 0.3. In which case, Rep = 0.3 and so:
The terminal velocity UT may be calculated from:
Re p = 0.3 =
ρ f x v UT
μ
Hence, terminal velocity, UT = 0.00194 m/s
Since Rep = 0.3 , n = 4.65 and so:
Settling velocity of the suspension = UT ε n = 0.00194 × (0.62)4.65 = 2.1 × 10 −4 m/s
EXERCISE 3.15:
Develop an expression to determine the time it takes for a particle settling in a liquid
to reach 99% of its terminal velocity.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.25
SOLUTION TO EXERCISE 3.15:
Start with the force balance on a single particle:
Acceleration force = buoyancy force - gravity force - drag force
Thus for a particle in the Stokes region:
πx 3 du
πx 3
= (ρp − ρ f )
g − 3πxμu
6
6 dt
du B
A
+ u= ,
dt C
C
πx 3
πx 3
g , B = 3πxμu and C = ρp
where A = (ρ p − ρ f )
6
6
ma = ρp
Solve for u:
Bt Bu
du
A Bt
+e C
= e C
dt
C
C
Bt
d⎛⎜ ue C ⎞⎟
⎝
⎠ = A e Bt C
dt
C
Bt
Bt
A
ue C =
e C
C
Bt
A C Bt C
ue C =
e
+ C1
CB
−Bt
A
u = + C1e C
C
e
Bt
C
∫
Apply the initial condition to find C1:
At t = 0, u = 0.
A
B
−Bt
A
Hence, u = ⎛⎜1 − e C ⎞⎟
B⎝
⎠
So C1 = −
Examining this expressions, we see that as t → ∞ , u →
So we need to know the value of t at which, u = 0.99
0.99
A
(terminal velocity).
B
A
B
−Bt
A A⎛
= ⎜1 − e C ⎞⎟
B B⎝
⎠
Solving for t:
ln(0.01) = −
Gives t =
Bt
C
4.6ρ p x 2
18μ
Therefore, time for a particle to reach 99% of its terminal velocity in the Stokes
region is: t =
4.6ρ p x 2
18μ
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.26
EXERCISE 3.16:
A suspension in water of uniformly-sized spheres (diameter = 150 μm and density =
1140 kg/m3) has a solids concentration of 25% by volume. The suspension settles to
a bed of solids concentration 62% by volume. Calculate the rate at which the spheres
settle in the suspension. Calculate the rate at which the settled bed height rises.
SOLUTION TO EXERCISE 3.16:
(a) Solids concentration of initial suspension, CB = 0.25
Text-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:
UpA CA − U pBC B
U int,AB =
C A − CB
Since CA = 0, the equation reduces to
U int,AB = UpB
UpB is the hindered settling velocity of particles relative to the vessel wall in batch
settling and is given by Text-Equation 3.24:
U p = UT ε n
Assuming first that Stokes Law applies, then:
UT =
x 2 (ρ p − ρ f )g
18μ
(150 × 10 )
=
× (1140 − 1000 ) × 9.81
= 0.001717
18 × 0.001
−6 2
Checking the Reynolds number,
Re p =
xUT ρ f (150 × 10 −6 ) × 0.001717 × 1000
=
= 0.2576
μ
0.001
Reynolds number is less than 0.3, so Stokes Law assumption is valid.
Therefore, in Text-Equation 3.24, exponent n = 4.65.
The voidage of the initial suspension, εΒ = 1 - CB = 0.75
hence, UpB = 0.001717 × 0.75 4.65
= 0.0004506 m/s
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.27
Hence, the velocity of the interface between the initial suspension and the clear liquid
is 0.451 mm/s. The fact that the velocity is positive indicates that the interface is
moving downwards.
(b) Here again we apply Text-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:
UpBCB − U pSC S
U int,BS =
C B − CS
With CB = 0.25 and CS = 0.62 and since the velocity of the sediment, UpS is zero, we
have:
Uint,BS =
UpB 0.25 − 0
0.25 − 0.62
= −0.6757UpB
And from part (a), we know that UpB = 0.451 mm/s, and so Uint,BS = −0.305 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 0.305
mm/s.
EXERCISE 3.17:
If 20 micron particles with a density of 2000 kg/m3 are suspended in a liquid with a
density of 900 kg/m3 at a concentration of 50 kg/m3, what is the solids volume
fraction of the suspension? What is the bulk density of the suspension?
SOLUTION TO EXERCISE 3.17:
Assuming 1 m3 of suspension:
⎞
⎟
⎟
⎠
volume
particles
⎞
⎛
Therefore solids volume fraction = ⎜
⎟ = 0.025
total
volume
⎠
⎝
⎛
50kg
3
⎝ 2000kg / m
50 kg of particles occupies 0.025 m3 ⎜⎜
Bulk density of suspension =
ερ f + (1 − ε )ρp = 0.975 × 900 + 0.025 × 2000 = 927.5 kg/m
3
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.28
EXERCISE 3.18:
Given Figure 3E.18.1 for the height-time curve for the sedimentation of a suspension
in a vertical cylindrical vessel with an initial uniform solids concentration of 100
kg/m3.
a) What is the velocity at which the sediment/suspension interface rises?
b) What is the velocity of the interface between the clear liquid and suspension of
concentration 133 kg/m3?
c) What is the velocity at which a layer of concentration 133 kg/m3 propagates
upwards from the base of the vessel?
d) At what time does the sediment/suspension interface start rising?
e) At what time is the concentration of the suspension in contact with the clear liquid
no longer 100 kg/m3?
SOLUTION TO EXERCISE 3.18:
(a) At the end of the test we know that the surface of the sediment has risen from
h = 0 to hs = 30 cm in 20 seconds.
Hence the sediment velocity is:
30
= +1.66cm / s (upward)
20
(b) Referring to Solution-Manual-Figure 3.18.1, we must first find the point on the
curve corresponding to the point at which a suspension of concentration 133 kg/m3
interfaces with the clear suspension. From Text-Equation 3.38, with C = 133, CB =
100 and h0 = 100 cm, we find:
h1 =
100 × 100
= 75 cm
133
A line drawn through the point t = 0, h = h1 tangent to the curve locates the point on
the curve corresponding to the time at which a suspension of concentration 133 kg/m3
interfaces with the clear suspension (Solution Manual-Figure 3.18.1). The coordinates
of this point are
t = 13.5 sec, h = 37.5 cm.
The velocity of this interface is the slope of the curve at this point:
slope of curve at 13.5 sec, 37.5 cm = − 2.78 cm / s
downward velocity of interface = 2.78 cm/s
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.29
(c) From the consideration above, after 13.5 seconds the layer of concentration 133
kg/m3 has just reached the clear liquid interface and has travelled a distance of 37.5
cm from the base of the vessel in this time.
Therefore, upward propagation velocity of this layer =
h 37.5
=
= +2.78 cm / s
t 13.5
(d) At time = 0
(e) This is the time at which the AB interface disappears. In the figure, this is the time
at which the slope of the height versus time curve stops being a straight line. This
time is 10 seconds.
EXERCISE 3.19:
Given Figure 3E19.1 for the fluxes below the feed in a thickener of area 300 m2 and a
feed solids volume concentration of 0.1,
a) What is the concentration of solids in the top section of the thickener?
b) What is the concentration of solids in the bottom section of the thickener?
c) What is the concentration of solids exiting the thickener?
d) What is the flux due to bulk flow below the feed?
e) What is the flux due to settling below the feed?
SOLUTION TO EXERCISE 3.19:
From Solution Manual-Figure 3.19.1:
(a) CT = 0 for an underloaded thickener
(b) CB = 0.04 (from Figure)
(c) CL = 0.25 (from Figure)
(d) Flux due to bulk flow = 0.0015 mm/s (from Figure – underflow flux)
(d) Continuous total downward flux = 0.01 mm/s
Total downward flux = flux due to bulk flow + flux due to settling.
Therefore flux due to settling = 0.01 – 0.0015 = 0.0085 mm/s
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.30
Figure 3.E2.1: Batch settling test results. Height versus time curve for use
in Exercises 3.2 and 3.4
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.31
Figure 3.E5.1: Batch settling test results. Height versus time curve for use
in Exercise 3.5.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.32
Figure 3.E6.1: Batch settling test results. Height versus time curve for use
in Exercises 3.6 and 3.8.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.33
Figure 3.E9.1: Batch flux plot for use in Exercise 3.9.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.34
Figure 3.W3.1: Batch flux plot for use in Worked Example 3.3 and Exercise 3.10.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.35
Figure 3.E11.1: Flux plots for use in Exercise 3.11.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.36
Figure 3.2.1: Batch settling test result; solution to Exercise 3.2.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.37
Figure 3.4.1: Batch settling test results; solution to Exercise 3.4.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.38
Figure 3.5.1: Batch settling test results; solution to Exercise 3.5.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.39
Figure 3.6.1: Batch settling test results; solution to Exercise 3.6.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.40
Figure 3.8.1: Batch settling test results; solution to Exercise 3.8.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.41
Figure 3.9.1: Batch flux plot; solution to Exercise 3.9.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.42
Figure 3.9.2: Sketch of concentration profile in batch settling test vessel
after 20 minutes; solution to Exercise 3.9.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.43
Figure 3.10.1: Batch flux plot; solution to Exercise 3.10.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.44
Figure 3.10.2: Sketch of concentration profile in batch settling test vessel
after 40 minutes. Solution to Exercise 3.10.
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Figure 3.11.1: Batch and continuous flux plots; solution to Exercise 3.11.
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Figure 3.12.1: Flux plots and solution for exercise 3.12.
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Figure 3.12.2: Flux plots and solution for exercise 3.12 – continued.
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.48
Height
100
(cm)
slope = +2.78 cm/s
h1 = 75 cm
50
slope = +1.66 cm/s
h = 37.5 cm
slope = -2.78 cm/s
10
20
30
40
Time
(s)
t = 13.5 s
Figure A
Figure: Solution Manual-Figure 3.18.1
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.49
Downward Flux
Below Feed
U PS
Feed Flux
⎛ mm ⎞
⎜
⎟
⎝ s ⎠
0.03
0.02
Underflow Flux
0.01
0.0015 mm/s
CB
0.1
CF
0.2
CL
0.3
0.4
C
Solution Manual-Figure
3.19.1
– Solution to Exercise 3.19
Figure
B
SOLUTIONS TO CHAPTER 3 EXERCISES: MULTIPLE PARTICLE SYSTEMS Page 3.50
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