SCREENING:
CONCEPTS
1. The material passing one screening surface and retained on a subsequent surface is called
a. intermediate material
c. plus material
b. minus
d. none of the above
2. In screen analysis, notation + 5 mm/ -10 mm means passing through
a. 10 mm screen and retained on 5 mm screen
b. 5 mm screen and retained on 10 mm screen
c. both 5 mm and 13 mm screens
d. neither 5 mm nor 10 mm screens
3. Optimum ratio of operating speed to critical speed of a trommel is
a. 0.33 – 0.45
c. 0.5 – 2
b. 1.33 – 1.45
d. 1.5 – 2.5
4. For sizing of fine materials, the most suitable equipment is a
a. trommel
c. shaking screen
b. grizzly
d. vibrating screen
5. Trommels separate a mixture of particles depending on their
a. size
c. screen size
b. wet ability
d. electrical and magnetic
6. A screen is said to be blinded when
a. oversizes are present in undersize fraction
b. undersizes are retained in oversize fraction
c. the screen is plugged with solid particles
d. its capacity is abruptly increased
7. Screen efficiency is
a. recovery rejection
b. recovery
c. rejection
d. none of these
8. The ratio of the area of opening in one screen (Taylor series) to that of openings in the next smaller
screen is
a. 1.5
b. 1
c. 2
d. none of these
9. Screen capacity is expressed in terms of
a. tons/h
b. tons/ft2
c. both a and b
10. 200 mesh screens means
a. 200 openings/cm2
b. 200 openings/inch
c. 200 openings/cm
d. 200 openings/inch2
d. tons/h-ft2
PROBLEMS
1.) A quartz mixture is screened through a 10-mesh screen. The cumulative screen analysis of feed,
overflow and underflow are given in the table. Calculate the mass ratios of the overflow and
underflow to feed and the overflow effectiveness of the screen.
Mesh
Dp (mm)
Feed
Overflow
Underflow
4
4.699
0
0
0
6
3.37
0.025
0.071
0
8
2.362
0.15
0.43
0
10
1.651
0.47
0.35
0.195
14
1.168
0.73
0.97
0.58
20
0.833
0.885
0.99
0.83
28
0.589
0.94
0.91
35
0.417
0.96
0.94
65
0.208
0.98
0.975
1
1
Pan
Answer: 66.91%
2.) The collection efficiency of a cyclone is 45% over the size range 0-5μm, 80% over the size range
5-10μm, and 96% for particles exceeding 10 μm. Calculate the efficiency of collection for a dust
with a mass distribution of 50% 0-5 μm, 30% 5-10μm and 20% above 10μm.
Given:
Size (μm)
0-5
5-10
>10
Efficiency (%)
45
80
96
For the dust :
Mass (%)
50
30
20
Answer: 65.7%
3.) The screen with a total feed to bottom of 75.1 tons/h are shaking screens with a capacity of 4
metric tons/m2-h-mm mesh size. How many square meters of screen are needed for each of the
screens 8 mesh to 14 mesh, if the feed to the first screen is 100tons/h?
Answer: 16.08m2
4.) The particle size distribution of the feed and collected solids for a gas cyclone are given below
Size range
1-5
5-10
10-15
15-20
20-25
25-30
Wt. of feed in the size 2.0
3.0
5.0
range
Wt. of collected solids 0.1
0.7
3.6
in the size range
What is the collection efficiency (%) of the gas cyclone?
6.0
3.0
1.0
5.5
2.9
1.0
Answer: 69%
5.) A sand mixture was screened through a standard 10-mesh screen. The mass fraction of the
oversize material in feed, overflow and underflow were found to be 0.38, 0.79 and 0.22
respectively. The screen effectiveness based on the oversize is?
Given:
10-mesh screen
Xf = 0.38
Xb = 0.22
Xd = 0.79
Solution:
𝐸=
=
(𝑋𝑓 − 𝑋𝑏 ) 𝑋𝑑
( )
(𝑋𝑑 − 𝑋𝑏 ) 𝑋𝑓
(0.38 − 0.22) 0.79
(
)
(0.79 − 0.22) 0.38
= 𝟎. 𝟓𝟖
SIZE REDUCTION:
CONCEPTS
1. Which of the following gives the crushing energy required to create new surface?
a. Taggart’s rule
c. Rittinger’s law
b. Fick’s Law
d. none of these
2. Size reduction mechanism used in Jaw crushers is
a. attrition
c. cutting
b. compression
d. impact
3. Feed size of ≥ 25 cm can be accepted by
a. ball mill
c. fluid energy mill
b. rod mill
d. jaw crusher
4. Maximum size reduction in a ball mill is done by
a. attrition
c. impact
b. compression
d. cutting
5. The main size reduction operation in ultra fine grinders is
a. cutting
c. compression
b. attrition
d. impact
6. The reduction ratio for grinders is defined as
a. Df/Dp
c. Df – Dp
b. Dp/Df
d. Dp – Df
7. The reduction ratio for fine grinders is
a. 5 – 10
b. 20 – 40
c. 10 – 20
d. as high as 100
8. A fluid energy mill is used for
a. cutting
b. grinding
c. ultragrinding
d. crushing
9. Wet grinding in a revolving mill
a. gives less wear on chamber walls than dry grinding
b. requires more energy than for dry grinding
c. increases capacity compared to dry grinding
d. complicates handling of the product compared to dry grinding
10. Cement clinker is reduced to fine size
a. Roll crusher
c. Tube mill
b. Ball mill
d. Hammer mill
PROBLEMS:
1. From measurements on a uniformly sized material from a dryer, it is inferred that the surface area of
the material is 1200 m2. If the density of the material is 1450 kg m-3 and the total weight is 360 kg
calculate the equivalent diameter of the particles if their value of l is 1.75.
Answer: 2200 microns
2. It is suspected that for a product of interest the oxidation reactions, which create off-flavours, are
surface reactions which proceed at a rate which is uniform with time, and if the shelf life of the product
is directly related to the percentage of the off-flavours that have been produced, estimate the
percentage reduction in shelf life consequent upon the size reductions of example 3, that is from 1 cm
to 0.3 cm and from 0.1 cm to 0.01 cm in diameter, assuming l = 1.5.
Answers: (a) 10:1 ; (b) 100:1
3.
If it is desired to reduce the separation time for milk to at least one week (before cream will rise to the
top), what maximum diameter of cream droplet would Stokes' Law predict to be necessary for the
homogenization to achieve? Assume the depth is 10 cm.
Answer: 0.0567 microns
4.
It is found that the energy required to reduce particles from a mean diameter of 1 cm to 0.3 cm is 11
kJ kg-1. Estimate the energy requirement to reduce the same particles from a diameter of 0.1 cm to
0.01 cm assuming:
(a) Kick's Law,
(b) Rittinger's Law,
(c) Bond's Equation.
Answer: (a) 21 kJkg-1 ; (b) 423 kJkg-1 ; (c) 91 kJkg-1
5. Find the power required for crushing 5 ton/hr of limestone (Rittinger’s number = 0.0765m 2/J) if the
specific surface areas of the feed and the product are 100 and 200 m 2/kg respectively. If the machine
consumes a power of 4hp, calculate its efficiency.
Given:
ton 1.39𝑘𝑔
=
hr
𝑠
𝐴𝑠𝑠𝑓 = 100m2/kg
𝐴𝑠𝑠𝑝 = 200m2/kg
Rittinger’s number = 0.0765m2/J
𝑚̇ = 5
Solution:
𝐴𝑠𝑠𝑝 − 𝐴𝑠𝑠𝑓
𝑃
= 𝐾𝑅 (𝐴𝑠𝑠𝑝 − 𝐴𝑠𝑠𝑓 ) =
𝑚̇
Rittinger’s number
𝑃
200 − 100
𝐽
=
= 1307.19
̇
0.0765
𝑘𝑔
1.39
𝐽
1816.99
𝑃 = 1816.99 =
= 2.43ℎ𝑝
𝑠
745.7
Efficiency =
2.43
4
𝑥 100 = 𝟔𝟎. 𝟕𝟓%
FLOTATION:
CONCEPTS
1. Froth Flotation is most suitable for treating
a. iron ores
c. quartz
b. sulfide ores
d. metal ores
2. In Froth Flotation, chemical agent added to cause air adherence is called
a. collector
c. modifier
b. frother
d. promoter
3. Pine oil used in forth flotation technique acts as a
a. collector
b. modifier
c. frother
d. activator
4. Which of the following is the most suitable for cleaning of fine coal dust (< 0.5 m)?
a. Through washer
c. Spiral separator
b. Baum Jig Washer
d. Froth Flotation
5. Any operation in which one solid is separated from another by floating one of them at or on the surface
of a fluid.
a. coagulation
c. centrifugation
b. flotation
d. sedimentation
6. The flotation agent that prevents coalescence of air bubbles as they travel to the surface of the water
is/are
a. collectors
c. frothing agent
b. promoters
d. modifying agent
7. A flotation modifier which assists in the selectivity or stop unwanted minerals from floating
a. depressants
c. alkalinity regulators
b. activators
d. promoters
8. An example of a collector for floatation of metallic sulfides and native metals is
a. xanthates
c. sodium sulfide
b. sodium silicate
d. sphalerite
9. Which of the following is an example of a deflocculant?
a. sulfuric acid
c. dithiophosphate
b. lignin sulforate
d. molybderite
10. Dispersants are important for the control of limes which sometimes interfere with the selectivity and
increase reagent consumption. Another term for dispersant is
a. deflocculant
c. frothers
b. depressants
d. regulators
PROBLEMS
1.) A copper ore initially contains 2.09% Cu. After carrying out a froth flotation
separation, the products are as shown in Table 1. Using this data, calculate:
(a) Ratio of concentration
(b) % Metal Recovery
(c) % Metal Loss
(d) % Weight Recovery, or % Yield
(e) Enrichment Ratio
Answers: (a) 10, (b) 5.7%, (c) 4.3%, (d) 10%, (e) 9.57
2.) A flotation plant processes 3000 tons/day of CuFeS 2. It produces 80 tons Cu concentrate assaying
25% Cu. If ore analyzes 0.7% Cu, the percent recovery is?
Answer: 95.24%
3.) Ground lead ore is to be concentrated by a single flotation process using 1.5 oz of reagent per ton
of ore. The feed concentrate and tailings have the following composition by weight on a dry basis
Feed %
Concentrate %
Tailings %
PbS
30
90
0.9
ZnS
25
3
35.6
SiO2
45
7
63.5
Water is fed to the cell at the rate of 1000 gallons per ton of wet concentrate with 99% of the water
leaving with the tailings and 1% with the concentrate. Find the mass of wet concentrate produced
per hour when ten tons of ore are fed to the cell / 24 hr. is?
Answer: 3.4
4.) From problem number 3 find the total water required in pounds per hour
Answer: 1185
5.) A typical flotation machine has the following specifications:
Number of cells = 4
Flotation time = 12min.
Cell Volume = 60 ft3
Hp per cell = 10hp
The material treated has the following specifications:
Pulp (mixture ore and water ) = 40% solids
Specific gravity of ore = 3
𝑇 𝑥 𝐶𝑎𝑝 𝑥 𝑑
𝑛=
𝑉 𝑥 1440
Where n= number of cells; V = volume in cu. Ft per cell; Cap = tons of dry ore / 24 hrs.; d= cu. Ft
of pulp (ore and water) containing one ton of solids.
Solution:
2000
2000 + 𝑥
𝑥 = 3000 𝐻2𝑂
𝐹 = 3000 + 2000 = 5000
2000
3000
𝑑=
+
= 58.76
3𝑥62.4 62.4
12(𝑥)(58.76)
4=
60(1440)
𝑐𝑎𝑝 = 𝟒𝟗𝟎. 𝟏𝟑
0.4 =
SEDIMENTATION
CONCEPTS:
1. Terminal velocity is
a. constant velocity with no acceleration
b. a fluctuating velocity
c. attained after moving one-half of total distance
d. none of these
2. In hindered settling, particles are
a. placed farther from the wall
b. not affected by other particles and the wall
c. near each other
d. none of these
3. Drag coefficient in hindered settling is
a. less than in free settling
b. equal to that in free settling
c. not necessarily quarter than in free settling
d. greater than in free settling
4. For the free settling of a spherical particle through a fluid, the slope of C D vs log NRe plot is
a. 1
c. – 1
b. 0.5
d. – 0.5
5. The terminal velocity of a small sphere settling in a viscous fluid varies as the
a. first power of its diameter
b. inverse of the fluid viscosity
c. inverse square of the diameter
d. square of the difference in specific weights of solid and fluid
6. Buoyant force
a. for non symmetrical bodies is not vertical
b. depends on the depth of the submergence of the floating body
c. depends on the weight of the floating body
d. none of these
7. Center of pressure in an immersed body is _______ the center of gravity
a. above
b. below
c. at
d. either above or below; depends on the liquid density
8. The line of action of the buoyant force passes through the center of gravity of the
a. submerged body
b. displaced volume of the fluid
c. volume of fluid vertically above the body
d. horizontal projection of the body
9. Drag is the force component exerted on an immersed object
a. passing the centroid of the body at 60° to the direction of motion
b. the component being parallel to the flow direction
c. the component being normal to the flow direction
d. none of these
10. Sphericity of raschig ring (whose length and diameter are equal) is
a. >1
c. <2
b. 1
d. 2
PROBLEMS
1.) Calculate the settling velocity of dust particles of (a) 60 mm and (b)10 mm diameter in air at 21°C
and 100 kPa pressure. Assume that the particles are spherical and of density 1280 kg m-3, and that
the viscosity of air = 1.8 x 10-5 N s m-2 and density of air = 1.2 kg m-3.
Answer: (a) 0.14m/s (b) 3.9 x 10-3m/s
2.) A mixture of Galena (ρ = 7500 kg/m3) and Silica (ρ = 2650 kg/m3) has size range between 0.08
mm to 0.7mm. a). What is the velocity of water needed to obtain a pure galena product? b) What is
the maximum size range of the galena product?
Answers: (a.) 0.44 m/s, (b.) 0.35 mm < Dp galena ≤ 0.70 mm
3.) Square mica plates, 1/32 in. thick and 0.01 sq. in. in area are falling randomly through oil with a
density of 55 lb/cu. Ft. and with viscosity of 15cp. The specific gravity of the mica is 3.0, the settling
velocity is?
Answer: 7.2cm/s
4.) A particle settles through a suspension containing 35% solids by weight. Given that the velocity
under free settling motion is ut = 0.00669 m/s. Find us.
Given:
[density of particle = 2800 kg/m3, Dp = 200 mesh, density of fluid = 996.5 kg/m3, viscosity of fluid =
0.8Cp]
Answer: 0.003036m/s
5.) A continuous separating tank is to be designed to follow after a water washing plant for liquid oil.
Estimate the necessary area for the tank if the oil, on leaving the washer, is in the form of globules
5.1 x 10-5 m diameter, the feed concentration is 4 kg water to 1 kg oil, and the leaving water is
effectively oil free. The feed rate is 1000 kg h-1, the density of the oil is 894 kg m-3 and the
temperature of the oil and of the water is 38°C. Assume Stokes' Law.
Given:
Viscosity of water = 0.7 x 10-3 N s m-2.
Density of water = 1000 kg m-3.
Diameter of globules = 5.1 x 10-5 m
Solution:
𝑣𝑚 =
𝑣𝑚 =
𝐷2𝑔(𝑟𝑝 − 𝑟𝑓)
18𝑚
(5.1 𝑥 10 − 5)2 𝑥 9.81 𝑥 (1000 − 894)
18 𝑥 0.7 𝑥 10 − 3
= 2.15 𝑥 10 − 4 𝑚 𝑠 − 1 = 0.77 𝑚 ℎ − 1.
and since F = 4 and L = 0, and dw/dt = flow of minor component = 1000/5 = 200 kg h-1,
𝐴 = 4 𝑥 200/(0.77 𝑥 1000)
= 𝟏. 𝟎 𝒎𝟐
CENTRIFUGATION
CONCEPTS
1. For separation of sugar solution from settled out mud we use
a. sparkler filter
c. centrifugal filter
b. plate & frame filter
d. rotary drum vacuum filter
2. Moisture can be removed from lubricating oil using
a. tubular centrifuge
c. sparkler filter
b. clarifier
d. vacuum leaf filter
3. Which of the following can be most effectively used for clarification of tube oil and printing ink?
a. sparkler filter
c. disc-bowl centrifuge
b. precoat filter
d. sharpless supercentrifuge
4.
a.
b.
c.
d.
If the radius of a basket centrifuge is halved and the rpm is doubled, then
linear speed of the basket is doubled
linear speed of the basket is halved
centrifugal force is doubled
capacity of centrifuge is increased
5. Where the difference in density of the two liquid phases to be separated is very small (as in milk
cream separator), the most suitable separator is
a disc bond centrifuge
c. batch basket centrifuge
b sharpless supercentrifuge
d. sparkler filter
6. Ultra centrifuges are used for the separation of __________ solid particles.
a
coarse
c. fine
b
colloidal
d. dissolved
7. Is created by moving a mass in a curved path and is exerted in the direction away from the center
of curvature of the path.
a
b
force
centrifugal force
c. settling
d. raising
8. Is the force applied to the moving mass in the direction toward the center of curvature which
causes the mass in the direction toward the center of curvature which causes the mass to travel in
a curved path.
a
centripetal force
c. rotational force
b
centrifugal force
d. filtration
9. Separation of isotopes is generally done using a/an __________ centrifuge.
a
ultra
c. both a & b
b
disk-bowl
d. neither a nor b
10. Uses the concept that an object whirled about an axis at a constant radial distance from the point is
acted on by a force.
a
filtration
c. centrifugal separation
b
sedimentation
d. none of these
PROBLEMS
1.) How many "g" can be obtained in a centrifuge which can spin a liquid at 2000 rev/min at a
maximum radius of 10 cm?
Answer: 450
2.) If a centrifuge is 3-ft diameter and rotates at 1,000 rpm, what must be the speed of a laboratory
centrifuge of 6-in diameter be ran if it is duplicate plant conditions?
Answer: 2449 rev/min
3.) A viscous solution containing particles with a density of 1461 kg/m3 is to be clarified by
centrifugation. The solution density is 801 kg/m3 and its viscosity is 100 cp. The centrifuge has a
bowl with r2 = 0.02225 m, r1 = 0.00715 m, and height b = 0.1970 m. Calculate the critical particle
diameter of the largest particles in the exit stream if N =23,000 rev/min and flow rate of 0.002832
m3/h.
Answer: 0.746 μm
4.) A centrifuge bowl is spinning at a constant speed of 2000 rpm. The radius of bowl in cm needed to
create a force of 455 g is.
Answer: 10.2
5.) If a cream separator has discharge radii of 5 cm and 7.5 cm and if the density of skim milk is 1032
kg m-3 and that of cream is 915 kg m-3, calculate the radius of the neutral zone so that the feed inlet
can be designed.
Given:
For skim milk, r1 = 0.075m, rA = 1032 kg m-3, cream r2 = 0.05 m, rB= 915 kg m-3
Solution:
(ρA𝑟12 − ρB𝑟22)
(ρA − ρB)
2
𝑟n = [1032 𝑥 (0.075)2 − 915 𝑥 (0.05)2] / (1032 − 915)
= 0.03 𝑚2
𝑟𝑛 = 0.17 𝑚
= 𝟏𝟕 𝒄𝒎
𝑟n2 =
TECHNOLOGICAL INSTITUTE OF THE PHILIPPINES
363 P. Casal St., Quiapo, Manila
Introduction to Particle Technology
Problem Set for Finals
Submitted by:
Ampon, Alexander S.
BSChE – 5th year
Submitted to:
Engr. Robert Delfin
Date Submitted:
March 24, 2015
Concept
I. Screening
1. With increase in the capacity of screens, the screen effectiveness
A.
remains unchanged
B.
increases
C.
decreases
D.
decreases exponentially
2. As particle size is reduced
A.
screening becomes progressively more difficult.
B.
screening becomes progressively easier.
C.
capacity and effectiveness of the screen is increased.
D.
none of these.
3. In screen analysis, the notation +5 mm/-10 mm means particles passing through
A.
10 mm screen and retained on 5 mm screen.
B.
5 mm screen and retained on 10 mm screen.
C.
both 5 mm and 10 mm screens.
D.
neither 5 mm nor 10 mm screen.
4. Screen capacity is expressed in terms of
A.
tons/hr
B.
tons/ft2
C.
both (a) & (b)
D.
tons/hr-ft2
5. A screen is said to be blinded, when the
A.
oversizes are present in undersize fraction.
B.
undersizes are retained in oversize fraction.
C.
screen is plugged with solid particles.
D.
screen capacity is abruptly increased.
6. For sizing of fine materials, the most suitable equipment is a
7
A.
trommel
B.
grizzly
C.
shaking screen
D.
vibrating screen
Which of the following is not categorised as a "mechanical operation" ?
A.
Agitation
B.
Filtration
C.
Size enlargement
D.
Humidification
8. 200 mesh screen means 200 openings per
A.
cm2
B.
cm
C.
inch
D.
inch2
9. Screen capacity is not a function of
A.
its openings size.
B.
screening mechanism.
C.
screening surface.
D.
atmospheric humidity.
10. increasing the capacity of a screen __________ the screen effectiveness.
A.
decreases.
B.
increases
C.
does not effect
D.
none of these
II. Size Reduction
1. In a size reduction crushing operation, the feed size is 10 to 50 mm, while the product size is 2 to 10
mm. This is a case of __________ crushing.
A.
primary
B.
secondary
C.
fine
D.
ultrafine
2. Size reduction of __________ is accomplished in steam heated rollers and roll crushers.
A.
resins
B.
gums
C.
hard rubber
D.
waxes
3. Size reduction of the __________ can be suitably done by ball mills, crushing rolls and rod mills.
A.
metalliferrous ores
B.
non-metallic ores
C.
basic slags
D.
asbestos & mica
4. The main size reduction operation in ultrafine grinders is
A.
cutting
B.
attrition
C.
compression
D.
impact
5. In case of a hammer crusher,
A.
crushing takes place by impact breaking.
B.
maximum acceptable feed size is 30 cms.
C.
reduction ratio can be varied by adjusting the distance from cage to hammers.
D.
all (a), (b) and (c).
6. Out of the following size reduction equipments, the maximum feed size can be accepted by the
A.
tube mill
B.
ball mill
C.
jaw crusher
D.
jet pulveriser
7. Which of the following gives the work required for size reduction of coal to -200 mesh in a ball mill most
accurately?
A.
Rittinger's law
B.
Kick's law
C.
Bond's law
D.
none of these
8. Electrical energy consumed by a jaw crusher is not a function of the
A.
average feed size
B.
average product size
C.
machine capacity
D.
none of these
9. Size reduction mechanism used in Jaw crushers is
A.
attrition
B.
compression
C.
cutting
D.
impact
10. Production rate __________ with increased fineness, with a given energy input to the size reduction
machine.
A.
decreases
B.
increases
C.
remains unchanged
D.
may increase or decrease; depends on the machine
III. Sedimentation
1. Two particles are called to be equal settling, if they are having the same.
A. size.
B. specific gravity.
C.terminal velocities in the same fluid & in the same field of force.
D.none of these.
2. Solid particles separation based on the difference in their flow velocities through fluids is termed as the
A. clarification
B. classification
C.elutriation
D.sedimentation
3. Gravity settling process is not involved in the working of a
A. hydrocyclone
B. classifier
C.dorr-thickener
D.sedimentation tank
4. Traces of solids are removed from, liquid in a
A. classifier
B. clarifier
C.sparkler filter
D.rotary vacuum filter
5. For separating particles of different densities, the differential settling method uses a liquid sorting
medium of density.
A. intermediate between those of the light and the heavy once.
B. less than that of either one
C. greater than that of either one.
D. of any arbitrary value.
6. A particle attaines its terminal settling velocity when
A. gravity force + drag force – buoyancy force
B. gravity force - drag force – buoyancy force
C. buoyancy force + drag force = gravity force
D. drag force = buoyancy force
7. Stokes equation is valid in the Reynolds number range
A. 0.01 to 0.1
B. 0.1 to 2
C. 2 to 10
D. 10 to 100
8. A suspension of glass beads in ethylene glycol has a hindered settling velocity of 1.7 mm/s while the
terminal settling velocity of a glass bead in ethylene glycol is 17 mm/s. If the Richardson-Zaki hindered
settling index is 4.5, the volume fraction of solids in the suspension is
A. 0.1
B. 0.4
C. 0.6
D. None of these
9. A 30% (by volume) suspension of spherical sand particles in a viscous oil has hindered settling velocity
of 4.44 x 10^-6 m/s. If the Richardson-Zaki hindered settling index iss 4.5, then the terminal velocity of sand
is
A. 0.9 𝜇m/s
B. 1 mm/s
C. 22.1 𝜇m/s
D. 0.02 𝜇m/s
10. For a particle settling in water at its terminal settling velocity, which of the following is true?
A. Buoyancy = weight + drag
B. weight = buoyancy + drag
C. drag = buoyancy + weight
D. drag = weight
IV. Flotation
1. Froth floatation is the most suitable for treating
A. iron ores
B. sulphide ores
C.quartzite
D.none of these
2. Pine oil used in froth floatation technique acts as a/an
A. collector
B. modifier
C. frother
D.activator
3. Which of the following is the most suitable for cleaning of fine coal dust (<0.5 mm) ?
A. Trough washer
B. Baum jig washer
C.Spiral separator
D.Froth floatation
4. dispersants are important for the control of slimes which sometimes interfere with the selectivity and
increase reagent consumption. Another term for dispersants is
A. deflocculant
B. depressants
C. frothers
D. regulators
5. which of the following is an example of a deflocculant?
A. sulfuric acid
B. lignin sulforate
C. dithiophosphate
D. molybderite
6. an example of a collector for flotation of metallic sulfides and native metals is
A. xanthates
B. sodium silicates
C. sodium sulfides
D. sphalerite
7. What is the selectivity index, if the grade of tailings & concentrate is the same?
A. 0
B. ∞
C. 1
D. 0.5
8. Any operation in which one solid is separated from another by floating one of them at or on the surface
of a fluid
A. coagulation
B. flotation
C. centrifugation
D. sedimentation
9. a flotation modifier which assist in the selectivity (sharpness of separation ) or stop unwanted minerals
from floating
A. collector
B. modifier
C. frother
D. activator
10 the flotation agent that prevents coalescence of air bubbles as they travel to the surface of the water
is/are
A. collector
B. frothing agent
C. promoters
D. modifying agent
V. Centrifugation
1. Tabular bowl centrifuges as compared to disk bowl centrifuges
A. operate at higher speed.
B. employ bowl of larger diameter.
C.can not be operated under pressure/vacuum.
D.can't be used for separation of fine suspended solids from a liquid.
2. __________ centrifuge is normally used in sugar mills.
A. Tubular bowl
B. Disc-bowl
C.Suspended batch basket
D.Perforated horizontal basket continuous
3. If a force greater than that of gravity is used to separate solids & fluids of different densities, the process
is termed as the
A. sedimentation
B. flocculation
C.dispersion
D.centrifugation
4. A centrifuge of diameter 0.2 m in a pilot plant rotates at a speed of 50 Hz in order to achieve effective
separation. If this centrifuge is scaled up to a diameter of 1 m in the chemical plant, and the same
separation factor is to be achieved, what is the rotational speed of the scaled up centrifuge?
A. 15 Hz
B. 22.36 Hz
C. 30 Hz
D. 44.72 Hz
5. if the radius of a basket centrifuge is halved and the rpm is doubled, then the
A. linear speed of the basket is doubled
B. linear speed of the basket is halved
C. batch basket centrifuge
D. sparkler filter
6. Which of the following can be most effectively used for clarification of lube oil and printing ink?
A. Sparkler filter
B. Precoat filter
C. Disc-bowl centrifuge
D. Sharpies super-centrifuge
7. Ultracentrifuges running at speeds upto 100000 rpm is normally used for the
A. separation of isotopes based on their density or molecular weights difference.
B. concentration of rubber latex
C. separation of cream from milk.
D. dewaxing of lubricating oil.
8. Separation of isotopes is generally done using a/an __________ centrifuge
A. ultra
B. disk-bowl
C. both (a) & (b)
D. neither (a) nor (b)
9. Where the density difference of the two liquid phase to be separated is very small (as in milk cream
separator), the most suitable separator is a
A. disc bowl centrifuge.
B. sharpies supercentrifuge.
C. batch basket centrifuge
D. sparkler filter
3) Ultra centrifuges are used for the separation of __________ solid particles.
A. coarse
B. fine
C. colloidal
D. dissolved
Answers:
I. Screening
1. c
2. a
3. a
4. d
5. c
6. a
7. d
8. c
9. d
10. a
II. Size Reduction
1. c
2. c
3. a
4. b
5. d
6. c
7. a
8. d
9. b
10. a
III. Sedimentation
1. c
2. b
3. a
4. b
5. a
6. b or c
7. a
8. c
9. b
10. b
IV. Flotation
1. b
2. c
3. d
4. a
5. b
6. a
7. c
8. c
9. c
10. b
V. Centrifugation
1. a
2. c
3. d
4. b
5. a or c
6. d
7. a
8. a
9. a
10. c
Problems without solution but with answer:
I. Screening
1. A sand mixture was screened through a standard 10-mesh screen. The mass fraction of the oversize
material in feed, overflow and underflow were found to be 0.38, 0.79 and 0.22 respectively. The screen
effectiveness based on the oversize is. ANSWER: Screen effectiveness = 0.58
II. Size Reduction
1. Particles of average feed size 25 x 10^-4 m are crushed to an average product size of 5 x10^-4 m at the
rate of 15 tons per hour. At this rate the crusher consumes 32 kW of power of which 2 kW are required for
running the mill empty. What would be the power consumption if 10 tons per hour of this product is further
crushed 1 x10^-4 m size in the same mill? Assume the Rittinger’s law is applicable. ANSWER: P = 100 kW
2. The energy required per unit mass to grind limestone particles of very large size to 100 Mm is 12.7
kWh/ton. An estimate (using Bond’s Law) of the energy to grind the particles from a very large size to 50
mm is? ANSWER: E2 = 18 kWh/ton
3. It is desired to crush 10 ton/h of iron ore hematite. The size of the feed is such that 80% passes a 3-in.
(76.2-m screen ad 80% of the product is to pass a 1/8 – in. (3.175-mm) screen. Calculate the gross power
required. Use a work index Ei for iron ore hematite of 12.68 (P1).
ANSWER: P = 24.1 hp(17.96 kW)
4. In crushing a certain ore, the feed is such that 80 is less than 50.8 mm in size and the product size is
such that 80% is less than 6.35 mm. he power required is 89.5 kW. What will be the power required using
the same feed so that 800% is less than 3.18 mm? Use the bond equation.(Hint: The work index Ei is
unknown, but it can be determined using the original experimental data in terms of T. In the equation for the
new size, the same unknowns appear. Dividing one equation by the other will eliminate these unknowns.)
ANSWER: 146.7 kW
5. Sugar is ground from crystals of which it is acceptable that 80% pass a 500 mm sieve(US Standard
Sieve No.35), down to a size in which it is acceptable that 80% passes a 88 mm (No.170) sieve, and a 5horsepower motor is found just sufficient for the required throughput. If the requirements are changed such
that the grinding is only down to 80% through a 125 mm (No.120) sieve but the throughput is to be
increased by 50% would the existing motor have sufficient power to operate the grinder? Assume Bond's
equation.
ANSWER:
x = 5.4 horsepower
III. Sedimentation
1. Many animal cells can be cultivated on the external surface of dextran beads. These cell-laden beads or
“microcarriers” have a density of 1.02 g/cm3 and a diameter of 150 μm. A 50-liter stirred tank is used to
cultivate cells grown on microcarriers to produce a viral vaccine. After growth, the stirring is stopped and
the microcarriers are allowed to settle. The microcarrierfree fluid is then withdrawn to isolate the vaccine.
The tank has a liquid height to diameter ratio of 1.5; the carrier-free fluid has a density of 1.00 g/cm3 and a
viscosity of 1.1 cP. Estimate the settling time by assuming that these beads quickly reach their maximum
terminal velocity. ANSWER: 2379 seconds
2. Oil droplets having a diameter of 20 μm (0.020 mm) are to be settled from air at temperature of 37.8°C
and 101.3 kPa pressure. The density of the oil is 900 kg/m3. Calculate the terminal settling velocity of the
droplets. Air at 37.8°C: ρ = 1.137 kg/m3, μ = 1.90 X 10-5 Pa•s ANSWER: 0.0103 m/s
3. A particle settles through a suspension containing 35% solids by weight. Given that the velocity under
free settling motion is ut = 0.00669 m/s. Find Us.
[density of particle = 2800 kg/m3, Dp = 200 mesh, density of fluid = 996.5 kg/m3, viscosity of fluid = 0.8Cp]
ANSWER: 0.003036 m/s
4. A mixture of Galena (ρ = 7500 kg/m3) and Silica (ρ = 2650 kg/m3) has size range between 0.08 mm to
0.7mm.
a). What is the velocity of water needed to obtain a pure galena product? [ANSWER: 0.44 m/s]
b) What is the maximum size range of the galena product? [ANSWER: 0.35 mm < Dp galena ≤ 0.70 mm ]
5. A suspension of uniform particles in water at a concentration of 500 kg of solids per cubic meter of slurry
is settling in a tank. Density of the particles is 2500 kg/m3 and the terminal velocity of a single particle is 20
cm/s. What will be thee settling velocity of suspension? Richardson and Zaki index is 4.6
ANSWER: Us = 7.16 cm/s
IV. Flotation
V. Centrifugation
1. If a centrifuge is 3-ft diameter and rotates at 1,000 rpm, what must be the speed of a laboratory
centrifuge of 6-in diameter be ran if it is duplicate plant conditions? ANSWER: 2449 rev/min
2. A viscous solution containing particles with a density of 1461 kg/m3 is to be clarified by centrifugation.
The solution density is 801 kg/m3 and its viscosity is 100 cp. The centrifuge has a bowl with r2 = 0.02225
m, r1 = 0.00715 m, and height b = 0.1970 m. Calculate the critical particle diameter of the largest particles
in the exit stream if N = 23,000 rev/min and flow rate of 0.002832 m3/h. ANSWER: 0.746 μm
3. What is the capacity in cubic meters per hour of a clarifying centrifuge operating under these conditions?
Diameter of the bowl = 600 mm
Thickness of liquid layer = 75 mm
Depth of bowl = 400 mm
Speed = 1200 rpm
SG of liquid = 1.2
SG of solid = 1.6
Viscosity of liquid = 2 cp
Cut-size of particles = 30 μm
ANSWER: 210 m3/h
4. A single batch settling test was made on limestone slurry. The test was made on 236 g limestone per L
slurry. Find the area required for a feed rate of 3.785M liters/day. Sludge concentration is 700 g/L.
Underflow velocity is 500 cm/day. Result of the test is given below:
ANSWER: 430.8 m2
5. If a particle of mass m is rotating at radius x with an angular velocity ω, it is subjected to a centrifugal
force mxω2 in a radial direction and a gravitational force mg in a vertical direction. The ratio of the
centrifugal to gravitational forces, xω2/g, is a measure of the separating power of the machine, and in order
to duplicate conditions this must be the same in both machines.
ANSWER: speed of rotation = 49 Hz
Problems with solution and answer:
I. Screening
II. Size Reduction
In an analysis of ground salt using Tyler sieves, it was found that 38% of the total salt passed through a 7
mesh sieve and was caught on a 9 mesh sieve. For one of the finer fractions, 5% passed an 80 mesh
sieve but was retained on a 115 mesh sieve. Estimate the surface areas of these two fractions in a 5 kg
sample of the salt, if the density of salt is 1050 kg m -3 and the shape factor () is 1.75.
Aperture of Tyler sieves, 7 mesh = 2.83 mm, 9 mesh = 2.00 mm, 80 mesh = 0.177 mm, 115 mesh =
0.125 mm.
Mean aperture 7 and 9 mesh
= 2.41 mm = 2.4 x 10-3m
Mean aperture 80 and 115 mesh = 0.151 mm = 0.151 x 10-3m
Now from Eqn. (11.6)
A1 = (6 x 1.75 x 0.38 x 5)/(1050 x 2.41 x 10-3)
= 7.88 m2
A2 = (6 x 1.75 x 0.05 x 5)/(1050 x 0.151 x 10 -3)
= 16.6 m2.
III. Sedimentation
1. Many animal cells can be cultivated on the external surface of dextran beads. These cell-laden beads or
“microcarriers” have a density of 1.02 g/cm3 and a diameter of 150 mm. A 50-liter stirred tank is used to
cultivate cells grown on microcarriers to produce a viral vaccine. After growth, the stirring is stopped and
the microcarriers are allowed to settle. The microcarrier-free fluid is then withdrawn to isolate the vaccine.
The tank has a liquid height to diameter ratio of 1.5; the carrier-free fluid has a density of 1.00 g/cm3 and a
viscosity of 1.1 cP. (a) Estimate the settling time by assuming that these beads quickly reach their
maximum terminal velocity. (b) Estimate the time to reach this velocity.
Solution:
Data: d = 150 mm = 0.015 cm; m = 1.1 cP = 0.011 g/cm-s; rs = 1.02 g/cm3; r = 1.00 g/cm3; g = 980 cm/s2
(a) Estimate the settling time by assuming that these beads quickly reach their maximum terminal velocity.
d2
vg =
(  s −  ) g → vg = 0.022 cm/s
18 
Check: N Re =
vd 1 0.022  0.015
=
= 0.03  1

0.011
 d t2 
 h 
  h =   h = 50 L
=
Liquid volume, V 
4  1.5 
 4 
2
 h = 52.3 cm  Settling time =
h
52 .3 cm
=
= 2379 s
vg 0.22 cm/s
(b) Estimate the time to reach the terminal velocity.
Force balance: m
m=
d 3  s
6
dv
mg
v 2
= mg −
− CD A
dt
s
2
; CD A
v 2
 24   d 2  v 2 

 = 3dva

= 
2
 vd  4  2 
dv
g 18v
=g−
−
dt
s d 2 s

dv 18 

+ 2 v = 1 −  g (I.C.: t = 0, v = 0)
dt d  s
 s 
v=

 − 18  
d2
(  s −  ) g 1 − exp  2 t 
18 
 d  s 

At steady state (t → ), v = vg =
 When
d2
(s −  ) g
18 
18t
 1 , v = vg
d 2 s
 When t >> 1.16  10-3 s, v = vg
 For v = 0.99vg, t = 5.34  10-3 s
IV. Flotation
V. Centrifugation
1. A centrifuge of diameter 0.2 m in a pilot plant rotates at a speed of 50 Hz in order to achieve effective
separation. If this centrifuge is scaled up to a diameter of 1 m in the chemical plant, and the same
separation factor is to be achieved, what is the rotational speed of the scaled up centrifuge?
Solution:
𝜔2 𝑟
Separation factor S =
𝑔
𝜔1 2 𝑟1
𝑔
=
𝜔2 2 𝑟2
𝑔
𝑟
0.2
𝜔2 = 𝜔1 √𝑟1 = 50 x √ 1 = 22.36 Hz
2
TECHNOLOGICAL INSTITUTE OF THE PHILIPPINES
Quiapo, Manila
Introduction to Particle Technology
Problem Set
Submitted by:
Bakal, Hubert A.
Submitted to:
Engr. Robert Delfin
SCREENING
1.) The results of a screen analysis are tabulated to show the _______________ of each screen
increment as a function of the mesh size range of the increment.
a. Mass fraction
b. Mole fraction
c. Volumetric Fraction
d. Efficiency
2.) The ratio of the actual mesh dimension of any screen to that of the next-smaller screen.
a. 1.189
b. 1.41
c. 1.567
d. None of the above
3.) It is useful for sizing particles with diameters greater than about 44 microns (325-mesh)
a. Wet screening
b. Dry screening
c. Rotary screening
d. None of the above
4.) A device that measures changes in the resistivity of an electrolyte as it carries the particles one by
one through a small orifice.
a. Screen mesh
b. UV-Vis
c. Coulter counter
d. Gas Chromatography
5.) Screens that have two screen, one above the other, held in a casing inclined at an angle between
16o and 30o with the horizontal.
a. Stationary Screen
b. Grizzlies
c. Gyrating Screen
d. Vibrating Screen
6.) It is measure by the mass of material that can be fed per unit time to a unit area of screen.
a. Screen capacity
b. Screen efficiency
c. Screen analysis
d. Screening
7.) It has a grid of parallel metal bars set in an inclined stationary frame and its slope and the path of
the material are usually parallel to the length of the bars.
a. Grizzlies
b. Gyrating Screen
c. Vibrating Screen
d. Rotary Screen
8.) A single screen can make but a single separation into _________ fractions.
a. One
b. Two
c. Three
d. Fine, Intermediate and Coarse
9.) These screens would sharply separate the feed mixture in such a way that the smallest particle in
the overflow would be just larger than the largest particle in the underflow.
a. Grizzlies
b. Actual Screen
c. Gyrating Screen
d. Ideal Screen
10.) These screens do not give a perfect separation about the cut diameter and the closest separations
are obtained with spherical particles on standard testing screens but even here is an overlap
between the smallest particles in the overflow and the largest ones in the underflow.
a. Grizzlies
b. Actual Screen
c. Gyrating Screen
d. Ideal Screen
11.) The screen analysis shown in the table applies to a sample of crushed quartz. The density of
particles is 2,650 kg/m3 and the shape factors are a =0.8 and sphericity=0.571. For the material
between 4 mesh and 200 mesh in particles size, calculate (a) A w in mm2 per gram and Nw in
particles per gram, (b) Dv, (c) Ds, (d) Dw (e) Ni for the 150/200 mesh increment.
Mesh
4
6
7
10
14
20
28
35
48
65
100
150
200
Pan
Solutions:
Screen
Opening
4.699
3.327
2.362
1.651
1.168
0.833
0.589
0.417
0.295
0.208
0.147
0.104
0.074
0
Mass
fraction
Retained
0.0000
0.0251
0.1250
0.3207
0.2570
0.1590
0.0538
0.0210
0.0102
0.0077
0.0058
0.0041
0.0031
0.0075
Average Particle
Diameter in Increment
4.013
2.8445
2.0065
1.4095
1.0005
0.711
0.503
0.356
0.2515
0.1775
0.1255
0.089
0.037
Cumulative
Fraction
Smaller than Dpi
1.0000
0.9749
0.8499
0.5292
0.2722
0.1132
0.0594
0.0384
0.0282
0.0205
0.0147
0.0106
0.0075
0.0000
a.) 𝐴𝑤 =
𝑁𝑤 =
b.) 𝐷𝑣 =
3965 𝑥 0.8284
0.9925
= 𝟑𝟑𝟎𝟗
𝒎𝒎𝟐
𝒈
471.7 𝑥 8.8296
𝒑𝒂𝒓𝒕𝒊𝒄𝒍𝒆𝒔
= 𝟒𝟏𝟗𝟔
0.9925
𝒈
1
1
= 𝟎. 𝟒𝟖𝟑𝟖 𝒎𝒎
(8.8296)3
1
c.) 𝐷𝑠 = 0.8284 = 𝟏. 𝟐𝟎𝟕 𝒎𝒎
d.) ∑ 𝑋𝑖 𝐷𝑝𝑖 = 𝐷𝑤 = 𝟏. 𝟔𝟕𝟕 𝒎𝒎
𝑋
e.) 𝑁2 = 𝑎𝜌 2𝐷3 =
𝑝 𝑝
0.0031
0.8 𝑥 0.00265 𝑥 0.0893
= 𝟐𝟎𝟕𝟒
𝒑𝒂𝒓𝒕𝒊𝒄𝒍𝒆𝒔
𝒈
12.) It is desired to separate a mixture of sugar crystals into two fractions, a coarse fraction retained on
an 8-mesh screen and a fine fraction passing through it. Screen analysis of feed, coarse and fine
fraction shows
Mass fraction of +8 particles in feed = 0.46
In coarse material = 0.88
In fine fraction = 0.32
The overall efficiency of the screen per 100 kg of feed is?
Answer:
45.17%
13.) Limestone is crushed by six units operating in parallel and the products separated by six 35 mesh
screen also in parallel into two fractions. The effective dimensions of each screen is 6 ft by 20 ft.
The common undersize from the screen comes out at the rate of 50 tons/hr. assume no losses.
Mesh
Size
6/8
8/10
10/25
25/35
35/48
48/65
65/100
100/150
Feed
Size
0.075
0.125
0.100
0.125
0.125
0.175
0.225
0.050
Oversize Undersize
0.080
0.020
0.145
0.055
0.170
0.090
0.150
0.085
0.280
0.500
0.175
0.150
0.250
0.100
Determine the efficiency of the screen operation and the capacity of each screen in lb/24-hr-ft2.
Answer:
E = 52.82 %
C = 5757.33 lb/24-hr-ft2
14.) 1800 pounds of dolomite per hour is produced by crushing and then screening through a 14-mesh
screen. The screen analysis is as follows:
Mesh
Size
4 on
8 on
14 on
Feed
Size
14.300
20.000
20.000
28 on
48 on
100 on
100 on
through
28.500
8.600
5.700
2.860
Oversize
40.000
30.000
20.000
10.000
Undersize
20.000
28.000
24.000
0 through
24
Calculate the efficiency and the total load to the crusher.
Answer:
F = 6315.79 lb/hr
E=62.42%
15.) Granular feldspar is produced by beneficiation of high alumina river sand. After screening, drying
and magnetic separation, the recovery is only 21 %. The screen analysis of this product on a dry
basis is as shown:
US Standard
Sieve
Percent
-10 +20
-20 +40
-40 +60
-60 +80
-80 +100
-100 +120
-120
1.5
24
30
18.5
12.5
8
5.5
In a specific application, this product must be reprocessed to remove grains finer than 100 mesh
where 2.0 MT reprocessed product is required. The MT of river sand with 10 % moisture that would have to
be beneficiated is?
Answer:
F = 12.23 MT
16.) Fine silica is fed at 1500 lbs/hr to a double deck vibrating screen combination to obtain a 48/65
mesh product. The silica feed is introduced into the upper screen of the 48 mesh and the product is
discharged off the surface of the lower screen of 65 mesh. During the screening operation the ratio
of laboratory analysis of different fractions O = 2, P =1, U = 0.5.
Screen Mesh
10/14 to 28/35
35/48
48/65
65/100
100/150 to
150/200
Feed
Mass
Fraction
0.2821
0.2580
0.2810
0.0910
Oversize
Mass
Fraction
0.5850
0.3370
0.0660
0.0050
Product
Mass Fraction
0.3385
0.3220
0.5260
0.0670
Undersize
Mass
Fraction
0.0045
0.0036
0.3440
0.2990
0.0870
0.0060
0.0260
0.3530
The effectiveness of screening equipment is?
If the screen measures 5 ft x 8 ft each, the capacity in MT/day-ft2-mm of the mesh screen on the
basis of a perfectly functioning 48 mesh screen is?
Answer:
E=43.41%
C=0.901 MT/day-ft2-mm
SIZE REDUCTION
1.) An equipment that does heavy work of breaking large pieces of solids into small lumps.
a. Mills
b. Grinders
c. Crushers
d. Cutting Machines
2.) Refer to variety of size reduction machine for intermediate duty, used secondly after the crusher for
further reduction.
a. Ultrafine Grinders
b. Grinders
c. Mill
d. Secondary Crusher
3.) Mills that reduces solids to such fine particles averaging from 1 to 20 microns in size.
a. Ultrafine Grinders
b. Fine Mills
c. Fine Cutting Machine
d. None of the above
4.) Produces cubes, thin squares or diamonds
a. Mills
b. Grinders
c. Crusher
d. Cutting machine
5.) Used for the segregation of grinding units in single chamber.
a. Conical ball mill
b. Compartment mill
c. Rod mill
d. Attrition mill
6.) Intense fluid shear in a high velocity stream is used to disperse particles or liquid droplets to form a
stable suspension of emulsion.
a. Agitated mill
b. Fluidized bed
c. Fluid energy mill
d. Colloid mill
7.) Looked like jaw crusher with circular jaws, between which materials are being crushed at some
point at all times.
a. Roll Crusher
b. Gyratory Crusher
c. Impactor
d. Tumbling Mill
8.) A cylindrical shell slowly turning about horizontal axes and filled to about half of its volume with a
solid grinding medium.
a. Rod mill
b. Attrition mill
c. Tumbling mill
d. Ball mill
9.) Particles are suspended in a high velocity gas stream.
a. Fluidized bed
b. Fluid energy mill
c. Colloid mill
d. Agitated mill
10.) Crusher admitted with or between two jaws, set to form a V open at the top.
a. Gyratory crusher
b. Jaw crusher
c. Roll crusher
d. Impactor
11.) A material is crushed in a blaked jaw crusher and the average size of particles reduced from
50mm to 10mm with the consumption of energy at the rate of 13 KW/kg/s. The consumption of
energy needed to crush the same material of an average size of 75mm to an average size of
25mm is? Assuming Kick’s Law applies.
Solution:
𝐸𝐴 = 𝐾𝐾 ln
13
𝑋1
𝑋2
𝐾𝑊
50𝑚𝑚
)
= 𝐾𝐾 ln (
𝑘𝑔
10𝑚𝑚
𝑠
𝐾𝐾 = 8.08
𝐸𝐵 = 8.08 ln (
𝑬𝑩 = 𝟖. 𝟖𝟖
75𝑚𝑚
)
25𝑚𝑚
𝑲𝑾
𝒌𝒈
𝒔
12.) What is the power required to crush 100 ton/hr of limestone if 80 percent of the feed passes a 2-in
screen and 80 percent of the product a 1/8 – in. screen?
Work index = 12.74
Specific Gravity = 2.66
Answer:
P= 168.6 kW
13.) In crushing a certain ore, the feed is such 80% is less than 50.8 mm in size and the product size is
such that 80 % is less than 6.35 mm. the power required is 89.5 kW. Use Bond equation. The
power required using the same feed so that 80% is less than 3.18 mm is?
Answer:
146.7 kW
14.) Work index is defined as the gross energy requirement in kWh/ton of feed needed to reduce very
large feed to such a size that 80% of the product passes a 100 microns screen. What is the work
index of gypsum rock?
Answer:
6.73
15.) Shape factors are important in characterizing particles such as those encountered in various unit
operations. The specific surface, in cm2/g for quartz with a diameter of 2in, density of 165 lb/ft 3 and
a shape factor of 10 is?
Answer:
4.47
16.) A 6000 lb of a material goes through a crusher and grinder per hour in succession. Screen analysis
from the crusher shows a surface are of product of 500 ft2 per lb. screen analysis of the grinder
product indicates a surface are of 44200 ft 2 per lb. the Rittinger number of the material processed
is 163 in2 per ft-lbf. The total power to be delivered to the equipment is?
Answer:
38.4 hp
FLOTATION
1.) An example of collector for flotation of metallic sulfides and native metals
a. Xanthates
b. Potassium Sulfide
c. Carbon Sulfide
d. Sodium Sulfide
2.) The flotation agent that prevents coalescence of air bubbles as they travel to the surface of the
water.
a. Modifying agent
b. Frothing agent
c. Defoaming agent
d. Promoters
3.) Includes any operation in which one solid is separated from another by floating one of them on the
surface of the fluid.
a. Sedimentation
b. Flocculation
c. Flotation
d. Centrifugation
4.) Added to strengthen temporarily covering film of the air bubbles.
a. Frothers
b. Promoters
c. Collectors
d. Activators
5.) It serves to stabilize the froth and holds the minerals until the froth can be scraped off into the
concentrate launder
a. Frothers
b. Promoters
c. Collectors
d. Activators
6.) Prevents the absorption of a collector by a mineral particle and thereof inhibit the flotation of
minerals.
a. Inhibitor
b. Depressants
c. Conditioners
d. Frothers
7.) Involves chemical treatment of ore pulp to create conditions favorable for the attachment of certain
mineral particles to the air bubbles then the air bubbles carry the selected minerals on the surface
of the pulp which is skimmed off while the other minerals submerged at the bottom.
a. Conditioning
b. Flotation Process
c. Cleaning
d. Roughing Process
8.) Solid to water ratio is expressed as
a. L/D
b. L/W
c. S/W
d. L/S
9.) Enhances the absorption of a collector by mineral particles
a. Promoters
b. Collectors
c. Activators
d. Regulators
10.) Substances that are added to maintain the proper pH.
a. Promoters
b. Frothers
c. Defoaming agent
d. Conditioners
11.) Ground lead ore is to be concentrated by a single flotation process using 1.5 oz of reagent per ton
of ore. The feed concentrate and tailings have the following composition by weight on a dry basis:
PbS
ZnS
SiO2
Feed %
30.000
25.000
45.000
Concentrate
%
90.000
3.000
7.000
Tailings %
0.900
35.600
63.500
Water is fed to the cell at the rate of 1000 gallons per ton of wet concentrate with 99% of the water
leaving with the tailings and 1% of the concentrate.
What is the mass of wet concentrate produced per hour when ten tons of ore are fed to the cell per
24 hours?
Solutions:
𝑂𝑀𝐵:
10 = 𝑇 + 𝐶
𝑆𝐵: 0.3(10) = 0.009𝑇 + 0.9𝐶
𝑇 = 6.73 ; 𝐶 = 3.27
𝑔𝑎𝑙
1𝑓𝑡 3
𝑙𝑏
1 𝑡𝑜𝑛
(
)
) (0.01)
𝑊𝐶 = 3.27 + 1000
𝑊𝐶 (
) (62.4 3 ) (
𝑡𝑜𝑛 − 𝑊𝐶
7.48 𝑔𝑎𝑙
𝑓𝑡
2000𝑙𝑏
𝑾𝑪 = 𝟑. 𝟒𝟏 𝒕𝒐𝒏𝒔/𝒅𝒂𝒚
𝑔𝑎𝑙
𝑡𝑜𝑛𝑠
1𝑓𝑡 3
𝑙𝑏𝑠
𝑑𝑎𝑦
(4.41
)(
)
𝐻2 𝑂 𝑛𝑒𝑒𝑑𝑒𝑑 = 1000
) (62.5 3 ) (1
𝑡𝑜𝑛 − 𝑊𝐶
𝑑𝑎𝑦 7.48 𝑔𝑎𝑙
𝑓𝑡
24 ℎ𝑟𝑠
𝒍𝒃𝒔
𝒉𝒓
12.) What is the total water required in pounds per hour in the previous problem?
Answer:
1185.29 lbs/hr
𝑯𝟐 𝑶 𝒏𝒆𝒆𝒅𝒆𝒅 = 𝟏𝟏𝟖𝟓. 𝟐𝟗
13.) A flotation plant processes 3000 tons/day of CuFeS 2. It produces 80 tons of Cu concentrate
assaying 25 % Cu. If ore analyzer 0.7 % Cu, the percent recovery is?
Answer:
95.24%
14.) A typical flotation machine has the following specifications:
Number of cells = 4
Cell volume = 60 cu. Ft
Flotation time = 12 min
Hp per cell = 10 Hp
The material treated has the following specifications:
Pulp (mixture of ore and water) = 40 % solids
Specific Gravity of ore = 3
𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛: 𝑛 =
𝑇𝑥𝐶𝑎𝑝𝑥𝑑
𝑉𝑥1440
Where n= number of cells
V=volume in cu. Ft per cell
Cap = tons of dry ore per 24 hrs
d = cu ft of pulp (ore and water) containing one ton (2000 lbs) of solids
What is the capacity of the machine in tons of dry ore per 24 hours?
Answer:
490
15.) A flotation section of a mining company is extracting CuS from covellite ores. The ore consists of
5% CuS and 95% gangue, which may be assumed to be SiO 2. The following data are given:
Feed
% CuS
5
% SiO2
95
Concentrate
Rougher Tailings
Scavenger
Concentrate
Final Tailings
85
1
15
99
10
0.3
90
99.7
Laboratory experiments indicated that the water to solid ratio, L/S=2 and the contact tie is 10 min.
in the rougher; L/S=4, contact time = 18 min in the scavenger. On the basis of 300 tons per day of ore
treated.
Data:
Density of SiO2 = 2.65 g/cc
Density of CuS = 4.6 g/cc
Calculate the volume needed for the rougher.
Answer:
169.8 ft3
16.) Calculate the volume needed for the scavenger in the previous problem.
Answer:
535 ft3
SEDIMENTATION
1.) The separation of a suspension into a supernatant clear fluid and a rather dense slurry containing a
higher concentration of solid.
a. Centrifugation
b. Flocculation
c. Sedimentation
d. Flotation
2.) Large tanks which receives the suspension or dilute slurry at the center or side, permit the overflow
of supernatant liquid and produce sludge from the bottom of the tank.
a. Clarifier
b. Thickener
c. Thinner
d. Classifier
3.) An installation of one thickener directly above another, which may be operated independently on
the same or different feeds or in series.
a. Tray thickener
b. Filter thickener
c. Clarifier
d. Continuous thickener
4.) The combination of filter and thickener in which feed is introduced to the tank and the solution is
withdrawn through a submerged filtering medium or cloth.
a. Filter thickener
b. Clarifier
c. Continuous thickener
d. Tray thickener
5.) Is conducted in inverted cones, or in cylindrical or rectangular tanks or vessels equipped with
slowly revolving rakes for moving the thickened sludge to the central discharge.
a. Continuous thickener
b. Filter thickener
c. Continuous sedimentation
d. Tray thickener
6.) At the start of a batch sedimentation, the concentration of solids is _____________ throughout the
cylinder.
a. Varying
b. Constant
c. Zero
d. Undetermined
7.) After the process begins, all particles of suspended solid fall through the fluid at their
____________ velocities under existing conditions of hindered settling.
a. Minimum
b. Maximum
c. Optimum
d. Settling
8.) Tanks have been and still are widely used and operate in a manner described for small-scale
laboratory graduate. The tank is filled and the slurry is allowed to settle for desired time. The
thickened material may be removed through a valve in the bottom of the tank or the clarified
solution may be withdrawn either by lowering a swing siphon or by successive opening of draw-off
connections.
a. Continuous sedimentation
b. Simple batch settling
c. Mutliple tank settling
d. Tray Thickener
9.) The more dense slurries settle at ______________ rate indicating mutual interference of particles
in hindered settling and the deviations to be expected from settling rates estimated from the
behavior of individual particles in free settling.
a. Faster
b. Slower
c. Intermediate
d. Very Slow
10.) Is determined by the settling rate of suspended solids
a. Thickener capacity
b. Clarification capacity
c. Classifying capacity
d. Settling capacity
11.) A sphere of diameter 10 mm and density 7700 kg/m3 falls under gravity at terminal conditions
through a liquid of density 900kg/m3 in a tube of diameter 12 mm. the measured terminal velocity of
the particle is 1.6 mm/s. calculate the viscosity of the fluid. Verify that Stoke’s Law applies at N Re =
0.3.
Solutions:
𝑔 𝐷𝑝2 (𝜌𝑝 − 𝜌)
𝑣𝑡 =
18𝜇
9.81𝑚
(10𝑥10−3 𝑚)2 (7700 − 900)𝑘𝑔
𝑠2
1𝑚
𝑚3
(
)
𝑣𝑡 =
𝑘𝑔
1000𝑚𝑚
18 ( 𝜇 𝑚 − 𝑠)
𝝁 = 𝟐𝟑𝟏. 𝟔𝟐𝟓
𝒌𝒈
𝒎−𝒔
12.) Oil droplets having a diameter of 20 microns are to be settled from air at temperature of 37.8
centigrade and 101.3 kPa pressure. The density of the oil is 900 kg/m 3. Calculate the terminal
settling velocity of the droplets.
Answer:
0.0103 m/s
13.) Calculate the settling velocity of glass spheres having a diameter of 1.554 x 10 -4 m in water at
293.2K. the slurry contains 60 wt % solids. The density of the glass sphere is 2467 kg/m 3.
Answer:
5.03 x 10-3 ft/s
14.) A mixture of silica (B) and galena (A) solid particles having a size range of 5.21 x 10 -6 to 2.50 x 10-5
m is to be separated by the hydraulic classification sing free settling condition in water at 293.2
K(B1). The specific gravity of silica is 2.65 and that of galena is 7.5. Calculate the size range of the
various fractions obtained in the settling. If the settling is in the laminar region, the drag coefficients
will be reasonably close to that for spheres.
Answer:
1.03 x 10-5 m for B ; and 1.26 x 10-5 for A
15.) Determine the terminal settling velocity of dust particles having a diameter of 60 microns at 294.3 K
and 101.32 kPa. The dust particles can be considered spherical with a density of 1280 kg/m 3. For
air: M=0.01828 x 10-5 Pa-s and P = 1.202 kg/m3.
Answer:
0.14 m/s
16.) A random handful of silica particles ranging in size from 28 mesh to 200 mesh is thrown to a very
deep body of water (without tides or turbulence).
Data:
Viscosity of water = 0.01 P
28 mesh = 0.0589 cm
Density of silica = 2.65 g/cc
200 mesh = 0.0074 cm
What is the distance between the largest and smallest particle after 10 mins.
Answer:
4700 cm
CENTRIFUGATION
1.) If centrifuge is used for sedimentation, a particle of a given size can be removed from the liquid in
the bowl if there is sufficient __________________ of the particle in the bowl for the particle to
reach the wall.
a. Residence time
b. Terminal velocity
c. Settling velocity
d. Contact time
2.) An equal and opposite force that is exerted when an object is being rotated in a cylindrical
container.
a. Centripetal force
b. Centrifugal force
c. Rotational force
d. Compressive force
3.) A bowl that is tall and has a narrow diameter ranging from 100 to 150 mm.
a. Tubular centrifuge
b. Disk bowl centrifuge
c. Ultra centrifuge
d. None of the above
4.) Often used in liquid-liquid separations. The feed enters the actual compartment at the bottom and
travels upward through vertically spaced feed holes, filling the spaces between the discs.
a. Tubular centrifuge
b. Disk bowl centrifuge
c. Ultra centrifuge
d. None of the above
5.) It is a characteristic of the centrifuge for scale up.
a. Sigma value
b. Radius
c. Thickness
d. Depth
6.) The diameter of that particle which just reaches one-half the distance between r1 and r2.
a. One-half point
b. Thickness
c. Cut point
d. Cut diameter
7.) A particle of this size moves a distance of y= (r2 – r1)/2 during the settling time allowed.
a. One-half point
b. Thickness
c. Cut point
d. Cut diameter
8.) The volume of the liquid in the bowl V divided by the volumetric flow rate q.
a. Residence time
b. Contact time
c. Time of centrifugation
d. Settling time
9.) The scroll is cut away in the feed and sedimentation sections to form a ribbon conveyor operating
near the bowl wall, leaving the liquid surface undisturbed.
a. Axial flow conveyor centrifuge
b. Parallel flow conveyor centrifuge
c. Conveyor centrifuge
d. Full-scroll centrifuge
10.) The liquid flows in a spiral path against the motion of the conveyor. Under these conditions the
boundary layer is as much as 10 times as thick as it is when the layer flows only in an axial
direction.
a. Parallel flow conveyor centrifuge
b. Conveyor centrifuge
c. Full-scroll centrifuge
d. Axial flow conveyor centrifuge
11.) A centrifuge having a radius of the bowl of 0.1016 m is rotating at 100 rev/min. Calculate (a) the
centrifugal force developed in terms of gravity forces (b) compare this force to that for a bowl with a
radius of 0.2302 m rotating at the same revolutions per minute.
Solution:
𝐹𝐶
= 0.001118𝑟𝑁 2
𝐹𝑔
𝐹𝐶
= 0.001118 (0.1016)(1000)2
𝐹𝑔
𝑭𝑪
= 𝟏𝟏𝟑. 𝟔 𝒈𝒓𝒂𝒗𝒊𝒕𝒊𝒆𝒔
𝑭𝒈
𝐹𝐶
= 0.001118 (0.2032)(1000)2
𝐹𝑔
𝑭𝑪
= 𝟐𝟐𝟕. 𝟐 𝒈𝒓𝒂𝒗𝒊𝒕𝒊𝒆𝒔
𝑭𝒈
12.) A viscous solution containing particles with a density of 1461 kg/m 3 is to be clarified by
centrifugation. The solution density is 801 kg/m3 and its viscosity is 100 cP. The centrifuge has a
bowl with r2 = 0.00716 m and height b=0.1970 m. Calculate the critical particle diameter of the
largest particles in the exit stream if N=23 000 rev/min and the flow rate q = 0.002832 m 3/h.
Answer:
0.746 microns
13.) In a vegetable oil refining process, an aqueous phase is being separated from the oil phase in a
centrifuge. The density of the oil is 919.5 kg/m3 and that of the aqueous phase is 980.3 kg/m3. The
radius r1 for overflow of the light liquid has been set at 10.16 mm and the outlet for the heavy liquid
at 10.414 mm. calculate the location of the interface in the centrifuge.
Answer:
13.75 mm
14.) The terminal velocity of the 10 micron particles with SG = 1.2 clarified out of the water in the
laboratory centrifuge (D= 6 in) that has 3 ft in diameter and rotates at 1000 rpm is?
Answer:
556 ft/s
15.) If a centrifuge is 3 ft diameter and rotates at 1000 rpm the speed of a laboratory centrifuge of 6 inc
diameter be ran if it is to duplicate plant condition is?
Answer:
2449 rpm
16.) A centrifuge with a bowl which is 500 mm long and has an inside radius of 50.5 mm is to be used
to separate crystals from a dilute aqueous mother liquor. The optimum speed of rotation for the
centrifuge is 60000 rpm and the discharge weir is adjusted so that the depth of liquid at the bowl
wall is 38.5 mm. the crystals are approximately spherical and none are smaller than 2 microns in
diameter. The maximum volumetric flow rate in m3/s of the mother liquor that can be processed by
this centrifuge if all the crystals have to be removed is?
Answer:
0.054
References
Brown G. (1950). Unit Operations
Geankoplis, C. (2003). Principles of Transport Processes and Separation Processes (4th ed)
McCabe, W., & Smith, J. (2006). Unit operations of chemical engineering (7th ed)
Olano, S., Bungay, V., Centeno, C., Medina, L., & Salazar, C. (2008). Reviewer for Chemical Engineering
Licensure Examination (2nd ed)
Technological institute of the Philippines
Quiapo, Manila
Chemical engineering Department
Introduction to Particle Technology
Problem Set
Submitted by:
Bautista, Keziah Lynn S.
0912117
Submitted to:
Engr. Robert Delfin
March 24, 2015
Sedimentation
Concept
1. A process where particles are separated from the fluid by gravitational forces acting on the
particles.
a) Sedimentation
b) Size Reduction
c) Screening
d) Flotation
2. The purpose of sedimentation is/are:
a) Remove the particles from the fluid stream
b) Particles are recovered as the product
c) Separate particles into fractions differing in size or in density.
d) All of the given
3. When a particle is at a sufficient distance from the walls of the container and from other
particles so that its fall is not affected by them, the process is called?
a) Hindered Settling
b) Free Settling
c) Liquid – liquid extraction
d) Terminal Settling
4. When particles are crowded, they settle at a lower rate and the process is called?
a) Hindered Settling
b) Free Settling
c) Liquid – liquid extraction
d) Terminal Settling
5. The separation of dilute slurry or suspension by gravity settling into a clear fluid and a slurry of
higher solids content is called?
a) Sedimentation
b) Size Reduction
c) Screening
d) Flotation
6. Devices for the separation of solid particles into several fractions based upon their rates of flow
or settling through fluids are known as?
a) Sink-and-float methods
b) Classifiers
c) Differential settling methods
d) None of the given
7. A liquid is used whose density is intermediate between that of the heavy or high-density
material and the light-density material.
a) Sink-and-float methods
b) Classifiers
c) Differential settling methods
d) None of the given
8. The separation of solid particles into several size fractions based upon the settling velocities in
medium is called?
a) Sink-and-float methods
b) Classifiers
c) Differential settling
d) None of the given
9. Consists of a series of conical vessels of increasing diameter in the direction of flow. The slurry
enters the first vessels, where the largest and faster-settling particles are separated.
a) Spitzkasten classifier
b) Simple gravity settling
c) Sedimentation thickener
d) None of the given
10. Industrially, sedimentation operations are often carried out continuously in equipment called?
a) Thickener
b) Simple gravity settling
c) Sedimentation thickener
d) None of the given
Problem Solving
1.
A particle settles through a suspension containing 35% solids by weight. Given that the velocity
under these settling motion is vt = 0.00669 m/s. Find vs.
Given:
𝜌𝑝 = 2800 𝑘𝑔/𝑚3 𝜌 = 998 𝑘𝑔/𝑚3 𝐷𝑝 = 200 𝑚𝑒𝑠ℎ𝜇 = 0.8 𝑐𝑃
Answer: 0.003036
𝑚
𝑠
2. A mixture of Galena 𝜌 = 7500 𝑘𝑔/𝑚3 and Silica 𝜌 = 2650 𝑘𝑔/𝑚3 has size range between
0.08 mm to 0.7mm. What is the velocity of water needed to obtain a pure galena product?
Answer: 0.44
3.
𝑚
𝑠
In problem 2, what is the maximum size range of the galena product?
Answer: 0.35mm <Dp galena ≤ 0.70 mm
4.
In a laboratory test, the following data were obtained: Initial Height = 30 cm, Critical Height = 16
cm, Height = 10 cm, after 145 min, height at infinite time is 4 cm, velocity of settling is 0.2 cm/min.
For a batch cylindrical tank with a volume of 1 m3, what is the height if it is twice the diameter?
Answer: 172 𝑐𝑚
5.
In problem 4, find the settling time for the particles to settle to a height which is 20% of the original.
Answer: 1513.08 𝑚𝑖𝑛
6. Many animal cells can be cultivated on the external surface of dextran beads. These cell-laden
beads or “microcarriers” have a density of 1.02 g/cm3 and a diameter of 150 𝜇m. A 50=liter stirred
tank is used to cultivate cells grown on microcarriers to produce a viral vaccine. After growth, the
stirring is stopped and the microcarriers are allowed to settle. The microcarrier-free fluid is then
withdrawn to isolate the vaccine. The tank has a liquid height to diameter ratio of 1.5; the carrier
free fluid has a density of 1.00 g/cm3 and a viscosity of 1.1 cP. Estimate the settling time by
assuming that these beads quickly reach their maximum terminal velocity.
Given:
Dextran Beads
Fluid
Tank
𝜌𝑝 = 1.02 𝑔/𝑐𝑚3 𝜌 = 1 𝑔/𝑐𝑚3 𝑣 = 50 𝐿
𝐷𝑝 = 150 𝜇𝑚 𝜇 = 1.1 𝑐𝑃
𝐻
𝐷
= 1.5
Required: T =?
Solution:
*Assume Stokes Law
𝑣𝑡 =
𝑣𝑡 =
𝑔𝐷𝑝2 (𝜌𝑝 − 𝜌)
18𝜇
9.81(150𝑥10−6 )2 (1.02 − 1) (
1.1
18 (1000)
𝑣𝑡 = 2.23𝑥10−4
𝑁𝑅𝑒 =
100 3
1
) (
)
1
1000
𝑚
𝑠
𝐷𝑣𝜌
𝜇
1
(150𝑥10−6 )(2.23𝑥10−4 )(1)(100)3 (
1000)
𝑁𝑅𝑒 =
1.1
1000
𝑁𝑅𝑒 = 0.03
𝑣=
𝜋 2
(𝐷 ). 𝐻
4
𝜋 𝐻 2
𝑣 = ( ) .𝐻
4 1.5
𝐻 = 0.523 𝑚
𝑡 = 0.523 𝑚 (
𝑆
) = 2346 𝑆
2.23𝑥10−4 𝑚
Size Reduction
Concept
1. The term applied to all ways in which particles of solids are cut or broken into smaller pieces
a) Size reduction
b) Screening
c) Comminution
d) Crushing
2. States that the energy required for crushing is proportional to the new surface created.
a) Rittinger’s law
b) Kick’s law
c) Bonds law
d) Energy law
3. It is defined as the efficiency of technical grinding compared with that of laboratory crushing
experiments.
a) Grinding efficiency
b) Bond work index
c) Practical energy efficiency
d) None of the given
4. Equivalent diameter of a particle is the diameter of the sphere having the same
a) Ratio of surface to volume as the actual volume
b) Ratio of volume to surface as the particle
c) Volume as the particle
d) None of the given
5. For coarse reduction of hard solids, use
a) Impact
b) Attrition
c) Compression
d) Cutting
6. Cement clinker is commonly reduced t fine size using a
7.
8.
9.
10.
a) Roll crusher
b) Ball mill
c) Tube mill
d) Hammer mill
Soft and non—abrasive materials can be made into fines by
a) Attrition
b) Compression
c) Cutting
d) None of the given
A fluid energy mill is used for
a) Cutting
b) Grinding
c) Ultragrinding
d) Crushing
Size reduction is important in chemical engineering since
a) It prevents chemical engineers from becoming overweight
b) It makes products to become uniform in size
c) It prepares raw materials of the desired sizes prior to processing
d) None of the given
The hardness of a mineral is a criterion of its resistance to crushing. Which of the following is an
example of a hard material?
a) Talc
b) Calcite
c) Sapphire
d) Feldspar
Problem Solving
1. From measurements on a uniformly sized material from a dryer, it is inferred that the surface area
of the material is 1200 m2. If the density of the material is 1450 kg m-3 and the total weight is 360
kg calculate the equivalent diameter of the particles if their value of  is 1.75.
Answer: 2200 microns
2. Calculate the shape factors ~ for model systems in which the particles are:
a) Cylinders with L = 2D,
b) Tetrahedral with their sides being equilateral triangles (the volume of a tetrahedron being
the area of the base multiplied by 1/3 the vertical height)
c) Estimate the specific surface area of a powder consisting of equal numbers of the above
two shapes in which there are 4 x 103 particles kg-1. The cylinders have a density of 1330
kg m-3 and the tetrahedral density of 1500 kg m-3.
Answer: (a) 0.83;(b) 2.4;(c) 0.81 m2kg-1
3. It is found that the energy required to reduce particles from a mean diameter of 1 cm to 0.3 cm is
11 kJ kg-1. Estimate the energy requirement to reduce the same particles from a diameter of 0.1
cm to 0.01 cm assuming:
a) Kick's Law,
b) Rittinger's Law,
c) Bond's Equation.
Answer: (a) 21 kJkg-1; (b) 423 kJkgkg-1; (c) 91 kJkg-1
4. It is suspected that for a product of interest the oxidation reactions, which create off-flavours, are
surface reactions which proceed at a rate which is uniform with time, and if the shelf life of the
product is directly related to the percentage of the off-flavours that have been produced, estimate
the percentage reduction in shelf life consequent upon the size reductions of example 3, that is
from 1 cm to 0.3 cm and from 0.1 cm to 0.01 cm in diameter, assuming = 1.5.
Answer: (a) 10:1; (b) 100:1
5. If it is desired to reduce the separation time for milk to at least one week (before cream will rise to
the top), what maximum diameter of cream droplet would Stokes' Law predict to be necessary for
the homogenization to achieve? Assume the depth is 10 cm.
Answer: 0.0567 microns
6. In crushing a certain ore, the feed is such that 80% is less than 50.8 mm in size and the product
size is such that 80% is less than 6.35 mm. the power required is 89.5 kW. What will be the power
required using the same feed so that 80% is less than 3.18 mm? Use the Bond equation. (Hint:
The work index Ei is unknown, but it can be determined using the original experimental data in
terms of T. In the equation for the new size, the same unknowns appear. Dividing one equation by
the other will eliminate these unknowns.)
Given:
𝑋𝑓1 = 50.8 𝑚𝑚
𝑋𝑝1 = 6.35 𝑚𝑚
𝑃1 = 89.5 𝑘𝑊
𝑋𝑝2 = 3.18 𝑚𝑚
Required: P2=?
Solution:
𝑃1
1
1
= 0.3162 𝐸𝑖 [
−
]
𝑇
√𝑋2 √𝑋1
89.5
1
1
= 0.3162 𝐸𝑖 [
−
]
𝑇
√6.35 √50.8
(𝐸𝑖 )(𝑇) = 1103.30
𝑃2 = 0.3162(1103.30)[
1
√3.18
−
1
]
√50.8
𝑃2 = 146.69 𝑘𝑊
Centrifugation
Concept
1. A mechanical process of separating multi-phase mixture via the use of centrifugal force.
a) Sedimentation
b) Filtration
c) Size reduction
d) Centrifugation
2. Settling process due to the difference in densities of the solid and fluid media.
a) Stokes law
b) Sedimentation
c) Clarifying centrifuge
d) Classifier
3. The most common type of equipment used in clarifying centrifuge is?
a) Sharples centrifuge
b) Disk centrifuge
c) Suspended basket centrifuge
d) None of the given
4. It is used to enhance settling time of particles.
a) Centrifugal separators
b) Disk centrifuge
c) Suspended basket centrifuge
d) None of the given
5. The particles leaves the bowl with the liquid if
a) rb = r2
b) r2 >rb
c) rb<r2
d) None of the given
6. For separation of sugar solution from settled out mud we use
a) sparkler filter
b) plate & frame filter
c) centrifugal filter
d) rotary drum vacuum filter
7. Moisture can be removed from lubricating oil using
a) tubular centrifuge
b) clarifier
c) sparkler filter
d) vacuum leaf filter
8. Which of the following can be most effectively used for clarification of tube oil and printing ink?
a) sparkler filter
b) precoat filter
c) disc-bowl centrifuge
d) sharpless supercentrifuge
9. If the radius of a basket centrifuge is halved and the rpm is doubled, then
a) linear speed of the basket is doubled
b) speed of the basket is halved
c) centrifugal force is doubled
d) capacity of centrifuge is increased
10. Where the difference in density of the two liquid phases to be separated is very small (as in milk
cream separator), the most suitable separator is
a) disc bond centrifuge
b) sharpless supercentrifuge
c) batch basket centrifuge
d) sparkler filter
Problem Solving
1. A viscous solution containing particles with a density of 1461 kg/m3 is to be clarified by
centrifugation. The solution density is 801 kg/m3 and its viscosity is 100cp. The centrifuge has a
bowl has a bowl with r2 =0.02225 m, r1=0.00715 m, and height b=0.1970 m. calculate the critical
particle diameter of the largest particles in the exit stream if N=23000 rev/min and flow rate of
0.002832 m3/h.
Answer = 0.746 𝜇𝑚
2. What is the capacity in cubic meters per hour of a clarifying centrifuge operating under these
conditions?
Diameter of the bowl = 600 mm
Thickness of liquid layer = 75 mm
Depth of bowl = 400 mm
Speed = 1200 rpm
SG of liquid = 1.2
SG of solid = 1.6
Viscosity of liquid = 2 cP
Cut-size of particles = 30 𝜇𝑚
Answer = 210 m3/h
3. A centrifuge with a phosphor bronze basket, 380 mm in diameter, is to be run at 67 Hz with a 75 mm layer
of liquid of density 1200 kg/m3 in the basket. What thickness of walls are required in the basket? The
density of phosphor bronze is 8900 kg/m3 and the maximum safe stress for phosphor bronze is 87.6
MN/m2.
Answer = 15.1 mm
4. An aqueous suspension consisting of particles of density 2500 kg/m 3 in the size range 1-10 𝜇𝑚 is
introduced into a centrifuge with a basket rotating at 80 Hz. If the suspension forms a layer of 75
mm thick in the basket, approximately how long will it take for the smallest particle to settle out?
Answer = 19.3 s
5. A centrifuge basket 600 mm long and 100 mm internal diameter has a discharge weir 25 mm
diameter. What is the maximum volumetric flow of liquid through the centrifuge such that, when the
basket is rotated at 200 Hz, all particles of diameter greater than 1 µm are retained on the
centrifuge wall? The retarding force on a particle moving liquid may be taken as 3πµdu, where u is
the particle velocity relative to the liquid µ is the liquid viscosity, and d is the particle diameter. The
density of the liquid is 1000 kg/m3, the density of the solid is 2000 kg/m3 and the viscosity of the
liquid is 1.0 mNs /m2. The inertia of the particle may be neglected.
Answer = 0.00103 m3/s
6. If a centrifuge is 3 ft diameter and rotates at 1000 rpm, what must be the speed of a laboratory
centrifuge of 6in diameter be ran if it is duplicate plant conditions?
Given:
𝜃1 = 3𝑓𝑡
𝑁1 = 1000 𝑟𝑝𝑚
𝜃2 = 6𝑚
Required: 𝑁2 =?
Solution:
𝑣𝑡1 = 𝑣𝑡2
2
2
𝐷
2𝜋𝑁2
2𝜋𝑁1 2 𝐷1
( 60 ) ( 2 ) (𝐷𝑝2 )(𝜌𝑝 − 𝜌) (𝑤2 ) ( 2 ) (𝜌𝑝 − 𝜌) ( 60 )
=
18𝜇
18𝜇
𝑁12 𝐷1 = 𝑁22 𝐷2
𝑁2 = 1000√
3
𝑟𝑒𝑣
= 2449
0.5
𝑚𝑖𝑛
Flotation
Concept
1. It involves phenomena related to the relative buoyancy of objects.
a) Flotation
b) Filtration
c) Collector
d) Washer
2. An example of collector for flotation of metallic sulfides and native metals is
a) Xanthates
b) Sodium silicate
c) Sodium sulphide
d) Sphalerite
3. Which of the following is an example of deflocculant?
a) Sulfuric acid
b) Lignin sulforate
c) Dithiophosphate
d) Molybderite
4. These are used to make a mineral surface amenable to collector coating
a) Modifiers
b) Activators
c) Regulators
d) Collectors
5. Added to strengthen temporarily covering film of the air bubbles
a) Frothers
b) Collectors
c) Modifiers
d) Promoters
6. A flotation modifier which assists in the selectivity or stop unwanted minerals from floating
a) Depressants
b) Activators
c) Alkalinity regulators
d) Promoters
7. Froth Flotation is most suitable for treating
a) iron ores
b) sulfide ores
c) quartz
d) metal ores
8. In Froth Flotation, chemical agent added to cause air adherence is called
a) collector
b) frother
c) modifier
d) promoter
9. Pine oil used in forth flotation technique acts as a
a) collector
b) modifier
c) frother
d) activator
10. Which of the following is the most suitable for cleaning of fine coal dust (< 0.5 m)?
a) Through washer
b) Baum Jig Washer
c) Spiral separator
d) Froth Flotation
Problem Solving
1. Ground lead ore is to be concentrated by a single flotation process using 1.5 oz of reagent per ton
of ore. The feed concentrate and tailings have the following composition by weight on a dry basis
Feed %
Concentrate %
Tailings %
PbS
30
90
0.9
ZnS
25
3
35.6
SiO2
45
7
63.5
Water is fed to the cell at the rate of 1000 gallons per ton of wet concentrate with 99% of the water
leaving with the tailings and 1% with the concentrate.
The mass of wet concentrate produced per hour when ten tons of ore are fed to the cell per 24
hours is?
Answer: 3.4
2. Total water required in pounds per hour is?
Answer: 1185 lb/hr
3. A flotation plant produces 3,000 tons per day of CuFeS2 (chalcopyrite). It produces 80 tons of Cu
concentrate assaying 25% Cu. If the analyzes 0.7%, calculate the percentage recovery.
Answer: 95.24%
4. Laboratory experiments indicated that the water to solids ratio, L/S=2 and the contact time is 10
min in the rougher; L/S=4, contact time=18 minutes in the scavenger. On the basis of 300 tons per
day of ore treated. Calculate the volume of the rougher needed.
% CuS
%SiO2
Feed
5
95
Concentrate
85
15
Rougher tailings
1
99
Scavenger
concentrate
Final tailings
10
90
0.3
99.7
Answer: 169.8 ft3
5. A flotation machine has the following specifications:
Number of cells = 4
Flotation time = 12 min
Cell volume = 60 cu. ft
Hp per cell = 10 Hp
The material treated has the following specifications:
Pulp = 40% solids
Specific gravity of ore = 3
The capacity of the machine in tons of dry ore per 24 hours is?
Solution:
0.4 =
2000
2000 + 𝑥
𝑥 = 3000 𝐻2𝑂
𝐹 = 3000 + 2000 = 5000
𝑑=
2000
3000
+
= 58.76
3𝑥62.4 62.4
4=
12(𝑥)(58.76)
60(1440)
𝑐𝑎𝑝 = 490.13
𝑡𝑜𝑛𝑠
24ℎ𝑟
Screening
Concept
1. A screen is said to be blinded when
a) Oversize are present in undersize fraction
b) Undersize are retained in oversize fraction
c) The screen is plugged with solid particles
d) Its capacity is abruptly increased
2. Size measurement of ultrafine particles can be best expressed in terms of
a) Centimetre
b) Screen size
c) Micron
d) Surface area per unit area
3. Trommels separate a mixture of particles depending on their
a) Size
b) Weetability
c) Micron
d) Screen size
4. Increasing the capacity of screen
a) Decreases the screen effectiveness
b) Increases the screen effectiveness
c) Does not affect the screen effectiveness
d) None of the given
5. The screen effectiveness is
a) Recovery + rejection
b) Recovery
c) Rejection
d) None of the given
6. As the particle size is reduced
a) Screening becomes progressively more difficult
b) Screening becomes progressively easier
c) Capacity and effectiveness of the screen is increased
d) None of the given
7. The material passing one screening surface and retained on a subsequent surface is called
a) Intermediate material
b) Minus material
c) Plus material
d) None of the given
8. Box-like machines, either round or square with a series of screen clothes nested atop one another.
a) Reciprocating screen
b) Oscillating screen
c) Electricity vibrated screen
d) Gyratory screen
9. The minimum clear space between the edges of the opening in the screening surface and is
usually given in inches and millimetres.
a) Sieve
b) Aperture
c) Mesh number
d) Holes
10. The screen used in making size separation smaller than 4 mesh and larger than 48 mesh
a) Grizzly screen
b) Gyratory screen
c) Oscillating screen
d) Vibrating screen
Problem Solving
1. It is desired to remove small particles from a crushed stone mixture by screening through a 10mesh screen. The screen analysis of feed, overflow and underflow are given in the table.
Calculate the mass ratios of the overflow and underflow to feed
Mesh
Dp(mm)
Feed
Overflow
Underflow
4
4.699
0
0
6
3.327
0.025
0.071
8
2.362
0.125
0.43
1
10
1.651
0.32
0.85
0.905
14
1.168
0.26
0.97
0.42
20
0.833
0.155
0.99
0.17
28
0.589
0.055
1
0.09
35
0.417
0.02
0.06
65
0.208
0.02
0.025
0.02
0
pan
Answer: a) xf=0.47, xp=0.85, xr=0.095;
2. Find the effectiveness of the screen.
Answer: 0.77
3. It is desired to separate 1000 kg of a mixture of crushed solids into three fractions, a coarse
fraction retained on an 20 mesh screen; a middle fraction passing through a 20 mesh screen and
retained on a 65 mesh screen; and a fine fraction passing through a 65 mesh screen. Two tyler
standard screens are used to remove particles 20/65. Screen analysis of the feed, coarse, medium
and fine fractions are given.
What is the effectiveness of the 20 screen?
Mesh
%feed
%P
%M
%Fine
+4-6
2.51
3.26
0.5
0
-6+8
12.5
13.59
11.27
0
-8+10
32.07
38.04
18.03
1.66
-10+14
25.7
27.17
22.53
16.63
-14+20
15.9
16.3
13.52
21.38
-20+28
5.38
1.36
18.03
9.03
-28+35
2.1
0.27
8.11
2.38
-35+48
1.02
0
4.51
0.48
-48+65
0.77
0
3.42
0.24
-65+100
0.58
0
0.09
13.3
-100+150
0.41
0
0
9.74
-150+200
0.31
0
0
7.36
Pan
0.75
0
0
17.81
100
100
100
100
Answer: E20 = 0.73
4. Given the following screen analysis, calculate the effectiveness of the 65 screen on problem 3.
Answer: E65 = 0.902
5. If the total percentage of particles larger than the screen opening in the feed, product, and
undersize are 36%, 89% and 3% respectively, the effectiveness of the screen is?
Answer: 88.5
6. It is desired to separate a mixture of sugar crystals into two fractions, a coarse fration returned on
an 8 mesh screen, and a fine fraction passing through it, screen analysis of feed, coarse and fine
fractions shows
Mass fraction of 18 particles
In feed=0.46
In coarse=0.88
In fine=0.32
The overall E of the screen per 100 kg of feed is?
Solution:
𝐸=
𝑋𝑝 𝑋𝑓 − 𝑋𝑟
1 − 𝑋𝑝 𝑋𝑓 − 𝑋𝑟
.
(1 −
.
)
𝑋𝑓 𝑋𝑝 − 𝑋𝑟
1 − 𝑋𝑓 𝑋𝑝 − 𝑋𝑟
0.88 0.46−0.32
1−0.88 0.46−0.32
𝐸 = 0.46 . 0.88−0.32 (1 − 1−0.46 . 0.88−0.32) 𝑥 100
𝐸 = 45.17%
Technological Institute Of The Philippines
363 P. Casal St.,Quiapo, Manila
Introduction To Particle Technology
Problem Set
(Concepts and Solved Problems)
Screening
Size Reduction
Sedimentation
Centrifugation
Flotation
Submitted by:
Sheila B. Bautista
Submitted to:
Engr. Robert Delfin
March 24, 2015
Screening
11. Increasing the capacity of screen
a. decreases the screen effectiveness
b. increases the screen effectiveness
c. does not affect the screen effectiveness
d. none of these
12. Screen efficiency is
a. recovery rejection
c. rejection
b. recovery
d. none of these
13. As particle size is reduced
a. screening becomes progressively more difficult
b. screening becomes progressively easier
c. capacity and effectiveness of the screen is increased
d. none of these
14. A screen is said to be blinded when
e. oversizes are present in undersize fraction
f. undersizes are retained in oversize fraction
g. the screen is plugged with solid particles
h. its capacity is abruptly increased
15. Size measurement of ultrafine particles can be best expressed in terms of
a. centimeter
c. micron
b. screen size
d. surface area per unit mass
16. Trommels separate a mixture of particles depending on their
c. size
c. screen size
d. wet ability
d. electrical and magnetic
17. Screen capacity is expressed in terms of
a. tons/h
b. tons/ft2
c. both a and b
18. Which of the following screens has the maximum capacity?
a. grizzlies
c. shaking screen
b. trommels
d. vibrating screen
19. For sizing of fine materials, the most suitable equipment is a
c. trommel
c. shaking screen
d. grizzly
d. vibrating screen
20. Mesh number indicates the number of holes per
a. square inch
c. square foot
b. linear inch
d. linear foot
d. tons/h-ft2
Problem Solving:
1. A screen with an aperture of 6 mesh BSS is treating a feed with 66% of +6 mesh and producing an
oversize fraction containing 89% of +6 mesh particles. If the undersize fraction contains 2% of +6 mesh
particles, calculate the effectiveness of the screen.
Answer: 75.59%
2. One ton per hour of dolomite is produced by a ball mill in a closed circuit grinding with a 100 mesh
screen. The screen analysis is given below. Calculate the screen efficiency.
Data from screen analysis:
Mesh
Feed
Oversize
Undersize
35
7.07
13.67
0.00
48
16.60
32.09
0.00
65
14.02
27.12
0.00
100
11.82
20.70
2.32
150
9.07
4.35
14.32
200
7.62
2.07
13.34
-200
33.80
0.00
70.02
100.00
100.00
100.00
Answer: 91.32%
3. Table salt is being fed to a vibrating screen at the rate of 150 kg/hour. Thedesired product is -39 +20
mesh fraction. A 30 mesh and 20 mesh screen are therefore used,the feed being introduced on the 30
mesh screen. During the operation it was observed that the average proportion of oversize (from 30 mesh
screen) : oversize (from 20 mesh screen): undersize (from 20 mesh screen) is 2:15:1. Calculate the
effectiveness of the screener from the following data.
Mass Fraction
Mesh
Feed
Oversize from 30
mesh screen
Oversize from 20
mesh screen
Undersize from 20
mesh scree
-85 + 60
0.097
0.197
0.026
0.0005
-60 + 40
0.186
0.389
0.039
0.0009
-40 + 30
0.258
0.337
0.322
0.0036
-30 + 20
0.281
0.066
0.526
0.349
-20 + 15
0.091
0.005
0.061
0.299
-15 + 10
0.087
0.006
0.026
0.347
Answer: 63.39%
4. Material is fed to a nominal 100 µm screen and separated into oversize and undersize streams. Size
distribution for the feed and two product streams are shown below. Calculate the effectiveness of the sieve
if the desired fraction of the material is the material, which is smaller than 100 µm.
Size Range µm
Feed kg/h
Oversize kg/h
Undersize kg/h
160-180
5
5
0
140-160
10
10
0
120-140
15
9
1
100-120
20
16
4
80-100
15
4
11
60-80
15
2
13
40-60
10
0
10
<60
15
0
15
Answer: 79.2%
5. A sand mixture was screened through a standard 10-mesh screen. The mass fraction of the oversize
material in feed, overflow and underflow were found to be 0.38, 0.79 and 0.22 respectively. The screen
effectiveness based on the oversize is?
Answer: 0.50
6. It is desired to separate a mixture of sugar crystals into two fraction retained on an 8mesh screen
and fine fraction passing through it. Screen analysis of feed,coarse and fine fractions show
Mass fraction of +8 particles in feed = 0.46
Product in coarse material = 0.88
Reject in the fraction = 0.32
The overall E of the screen per 100 kg of feed is?
Given:
Xf= 0.46
Xp=0.32
Xr=0.88
Req’d:
E=?
Sol’n:
OMB: F=P+R
100=P+R
Eq. 1
SB:
100(0.46)= 0.32(P) + 0.88(R)
P= 75 kg
R= 25 kg
25(0.88)
(1−0.88)(25)
E= 100(0.46)[ 1- (1−0.46)(100)
E= 45.17
Size Reduction
11. Equivalent diameter of a particle is the diameter of the sphere having the same
a. ratio of surface to volume as the actual volume
b. ratio of volume to surface as the particle
c. volume as the particle
d. none of these
12. Crushing efficiency is the ratio of
a. surface energy created by the crushing to the energy absorbed by the solid
b. the energy absorbed by the solid to that fed to the machine
c. the energy fed to the machine to the surface energy created by the crushing
d. the energy absorbed by the solid to the surface energy created by the crushing.
13. Rittinger’s crushing law states that
a. work required to form a particle of any size is proportional to the square of the surface to volume
ratio of the product.
b. work required to form a particle of a particular size is proportional to the square root of the surface
to volume ratio of the product
c. work required in crushing is proportional to the new surface created
d. for a given machine and feed, crushing efficiency is dependent on the sizes of feed and product
14. Bond crushing law
a. calls for relatively less energy for the smaller product particle than does the Rittinger law
b. is less realistic in estimating the power requirements of commercial crushes
c. states that the work required to form particle of any size from very large feed is proportional to the
square root of the volume to surface ratio of the product
d. states that the work required for the crushing is proportion
15. Kick’s law relates to
a. energy consumption
b. final particle size
c. feed size
d. none of these
16. Which of the following gives the crushing energy required to create new surface?
a. Taggart’s rule
c. Rittinger’s law
a. Fick’s Law
d. none of these
17. Size reduction mechanism used in Jaw crushers is
c. attrition
c. cutting
d. compression
d. impact
18. To get ultra fine particles, the equipment used is
a. ball mill
c. hammer crusher
b. rod mill
d. fluid energy mill
19. The material is crushed in a gyratory crusher by the action of
a. impact
c. compression
b. attrition
d. cutting
20. To get fine talc powder from its granules, the equipment used is
a. roller crusher
c. jaw crusher
b. ball mill
d. gyratory crusher
Problem Solving:
1. A batch grinding mill is charged with material of the composition. The grinding-rate function Su is
assumed to be 0.001/s for the 4/6 mesh particles. Breakage function B u is given with b=1.3 both Su and
Bu are assumed to be independent of time. How long will it take for the fraction of 4/6 mesh material to
diminish by 10 %?
Answer: 103.3s
2. What is the power required to crush 100 ton/h of limestone if 80 percent of the feed passes a 2-in.
screen and 80 percent of the product a 1/8-in. screen?
Answer: 227 hp
3.
A crusher reducing limestone of crushing strength 70 MN/m2 from 6 mm diameter average size to 0.1
mm diameter average size, requires 9 kW. The same machine is used to crush dolomite at the same
output from 6 mm diameter average size to a product consisting of 20 per cent with an average
diameter of 0.25 mm, 60 per cent with an average diameter of 0.125 mm and a balance having an
average diameter of 0.085 mm. Estimate the power required, assuming that the crushing strength of
the dolomite is 100 MN/m2 and that crushing follows Rittinger’s Law.
Answer: 5.9 kW
4. From measurements on a uniformly sized material from a dryer, it is inferred that the surface area of
the material is 1200 m2. If the density of the material is 1450 kg m-3 and the total weight is 360 kg
calculate the equivalent diameter of the particles if their value of l is 1.75.
Answer: 2200 microns
5. If it is desired to reduce the separation time for milk to at least one week (before cream will rise to the
top), what maximum diameter of cream droplet would Stokes' Law predict to be necessary for the
homogenization to achieve? Assume the depth is 10 cm.
Answer : 0.0567 microns
6. A dispersion of oil in water is to be separated using a centrifuge. Assume that the oil is dispersed in the
form of spherical globules 5.1 x 10-5 m diameter and that its density is 894 kg m-3. If the centrifuge rotates
at 1500 rev/min and the effective radius at which the separation occurs is 3.8 cm, calculate the velocity of
the oil through the water. Take the density of water to be 1000 kg m-3 and its viscosity to be
0.7 x 10-3 N s m-2
Solution:
Vm = D2N2r(rp - rf)/1640m
Vm = (5.1 x 10-5)2 x (1500)2 x 0.038 x (1000 - 894)/(1.64 x 103 x 0.7 x 10-3)
𝑽𝒎 = 𝟎. 𝟎𝟐
𝒎
𝒔
Sedimentation
11. Drag is defined as the force exerted by the
a. fluid on the solid in a direction opposite to flow
b. the fluid on the solid in the direction of flow
c. the solid on the fluid
d. none of these
12. Drag coefficient for flow of past immersed body is the ratio of
a. shear stress to the product of velocity head and density
b. shear force to the product of velocity head and density
c. average drag per unit projected area to the product of the velocity head and density
d. none of these
13. Stoke’s law is valid when the particle Reynolds number is
a. <1
c. >1
b. <5
d. none of these
14. Drag coefficient CD is given by (in Stoke’s law range)
a. CD =
16
Re p
c.
CD =
24
Re p
b. CD =
18.4
Re p
d.
CD =
0.079
Re p 0.23
15. Terminal velocity is
e. constant velocity with no acceleration
f. a fluctuating velocity
g. attained after moving one-half of total distance
h. none of these
16. In hindered settling, particles are
c. placed farther from the wall
d. not affected by other particles and the wall
c. near each other
d. none of these
17. Sedimentation on commercial scale occurs in
a. classifiers
c. thickeners
b. rotary drum filters
d. cyclones
18. At low Reynolds number
a. viscous forces are unimportant
b. viscous forces control
c. viscous forces control and inertial forces are unimportant
d. gravity forces control
19. At high Reynolds number
a. inertial forces control and viscous forces are unimportant
b. viscous forces predominate
c. inertial forces are unimportant and viscous forces control
d. none of these
20. Forces acting on a particle settling in fluid are
a. gravitational and buoyant forces
b. centrifugal and drag forces
c. gravitational or centrifugal, buoyant and drag forces
d. external, drag and viscous forces
Problem Solving:
1. Calculate the maximum velocity at which a spherical particle of galena 0.15 cm in diameter will fall in
water.
Data:
Special gravity of galena = 7.5
Special gravity of water = 1.0
Viscosity of water = 0.82 Cp
Drag Coefficient = 0.45
Answer: 53 cm/sec
2. Calculate the maximum velocity at which spherical particles of silica of 0.05 mm in diameter will fall
through still water of temperature 75 degree celcius.
Data:
Special gravity of silica = 2.70
Special gravity of water = 1.0
Viscosity of water at 75 degrees Celcius = 0.30 Cp
Answer: 0.77 cm/sec
3. Calculate the terminal velocity of a spherical particle of diameter 1mm and density 3000 kg/m3, falling
through atmospheric air.
Answer: 8.64 m/s
4. In a mixture of quartz (sp. Gr. = 2.65) and galena (sp. Gr. = 7.5) the size of the particle range from
0.0002 cm to 0.001 cm. On separation in a hydraulic classifier using water under free settling conditions,
what are the size ranges of quartz and galena in the pure products? (Viscosity of water = 0.001 kg/m-s;
Density= 1000 kg/m3)
Answer : 0.0004 – 0.0005 cm
5. A mixture of coal and sand particles having sizes smaller than 1x10-4 m in diameter is to be separated by
screeningand subsequent elutriation with water. Recommend a screen aperture such that the oversize from
the screen can be separated completely into sand and coal particles by elutriation. Calculate also the
required water velocity. Assume that Stokes law is applicable. Density of sand = 2650 kg/m3; density of
coal = 1350 kg/m3; density of water= 1000 kg/m3; viscosity of water = 1 x10 -3 kg/m-s; g= 9.812 m/s2.
Answer: Diameter of sand = 4.60 x 10-5 m
Required water velocity = 1.9079 x 10-3 m/s
6. The damage to blueberries and other fruits during handling immediately after harvest is closely related to
the terminal velocity in air. Compute the terminal velocity of a blueberry with a diameter of 0.60 cm density
of 1120kg/m3 in air at 21celcius and atmospheric pressure
K = Dp [
gρ( ρp− ρ) 1/3
]
µ2
Sol’n:
Density of air = 1000(29)/ (8314)21+273)= 1201 g/m3
9.81(1.201)( 1170− 1.201) 1/3
]
(1.828 𝑥 10−5)2
k = 0.6 x 10-2 [
k = 200.25 N
V = 1.75 √𝑔𝐷𝑝 (ρp – ρ) / ρ
V = 1.75 √
9.81( 0.6 𝑥 10−2)(1120−12.1)
1.201
V = 12.9 m/s
Centrifugation
7. For separation of sugar solution from settled out mud we use
c. sparkler filter
c. centrifugal filter
d. plate & frame filter
d. rotary drum vacuum filter
8. Moisture can be removed from lubricating oil using
c. tubular centrifuge
c. sparkler filter
d. clarifier
d. vacuum leaf filter
9. Which of the following can be most effectively used for clarification of tube oil and printing ink?
c. sparkler filter
c. disc-bowl centrifuge
d. precoat filter
d. sharpless supercentrifuge
10. If the radius of a basket centrifuge is halved and the rpm is doubled, then
e. linear speed of the basket is doubled
f. linear speed of the basket is halved
g. centrifugal force is doubled
h. capacity of centrifuge is increased
11. Where the difference in density of the two liquid phases to be separated is very small (as in milk cream
separator), the most suitable separator is
a. disc bond centrifuge
c. batch basket centrifuge
b. sharpless supercentrifuge
d. sparkler filter
6. It is a widely used piece of process equipment that can separate either liquid-solid or liquid-liquid
systems.
a. centrifuge
c. funnel
b. filter
d. sparkler
7. It is the means used to effect separation.
a. centripetal force
c. electrostatic force
b. centrifugal force
d. gravitational force
8. It is the force applied to the moving mass in the direction toward the center of curvature which causes
the mass to travel in a curved path.
a. centripetal force
c. electrostatic force
b. centrifugal force
d. gravitational force
9. This centrifugation equipment operates at lower speed and have bowls of larger diameter.
a. tubular bowl centrifuge
c. continous
b. batch centrifugals
d. disk bowl centrifuges
10. The terminal falling velocity of spherical particles of diameter at radius in a centrifugal field rotating at
rate.
a. VR
c. Vz
b. VX
d. VY
Problem Solving:
1. If a cream separator has discharge radii of 5 cm and 7.5 cm and if the density of skim milk is 1032 kg m3 and that of cream is 915 kg m-3, calculate the radius of the neutral zone so that the feed inlet can be
designed. For skim milk, r1 = 0.075m, A = 1032 kg m-3, cream r2 = 0.05 m, = 915 kg m-3
Answer : 17 cm
2. A dilute slurry containes small solid food particles having a diameter of 5 x 10 -2 mm, which are to be
removed by centrifuging. The particle density is 1050 kg/m 3 and the solution density is 1000 kg/m3. The
viscosity of the liquid is 1.2 x 10-3 Pa-s. A centrifuge at 3000 rev/min is to be used. The bowl dimensions
are b = 100.1 mm, r1 = 5.00 mm, and r2 = 30.0 mm. Calculate the expected flow rate in m 3/s just to remove
these particles.
Answer : 8.76 x 10-5 m3/
3. A centrifuge of 0.2m in a pilot plant operates at a speed of 50 Hz in order to achieve effective
separation. If this centrifuge is scaled up to a diameter of 1m in the chemical plant and the same operation
factor is to be achieved, what is the rotational speed of the scaled up centrifuge?
Answer : 22.36 Hz
4. If a centrifuge is 0.9 m diameter and rotates at 20 Hz, at what speed should a laboratory centrifuge of
150 mm diameter be run if it is to duplicate the performance of the large unit?
Answer : 49 Hz
5. An aqueous suspension consisting of particles of density 2500 kg/m3 in the size range 1–10 µm is
introduced into a centrifuge with a basket 450 mm diameter rotating at 80 Hz. If the suspension forms a
layer 75 mm thick in the basket, approximately how long will it take for the smallest particle to settle out?
Answer : 19.3 s
6. How many gravitational acceleration can be obtained in a centrifuge which can spin a liquid at 2000
rev/min at a maximum radius of 10 cm?
Sol’n:
Fc = ma = mrw2
a = rw2
w = 2πN/ 60
Fc = mrw2
= m (0.1 m) (2π x 2000 / 60 )2
= 4386.49 m
𝐹𝑐
𝐹𝑔
=
4386.49
9.81
= 447.14 g
FLOTATION
11. Froth Flotation is most suitable for treating
c. iron ores
c. quartz
d. sulfide ores
d. metal ores
12. In Froth Flotation, chemical agent added to cause air adherence is called
c. collector
c. modifier
d. frother
d. promoter
13. Pine oil used in forth flotation technique acts as a
c. collector
c. frother
d. modifier
d. activator
14. Which of the following is the most suitable for cleaning of fine coal dust (< 0.5 m)?
c. Through washer
c. Spiral separator
d. Baum Jig Washer
d. Froth Flotation
5.
A mechanical separation that includes any operation in which one type of solid is separated from
another type by floating one of them on the surface of the fluid.
a. flotation
b. sedimentation
c. centrifugation
d. size reduction
6. It is called a concentrate in the minerals industry.
a. concentrate
c. froth overflow stream
b. tailings
d. promoter
7. It is termed as tailings.
a. concentrate
b. feed
c. froth overflow stream
d. slurry underflow
8. Alters the surface of the mineral in order that it will become air- avid (cause it to adhere to air
bubbles)
a. modifiers
c. frothers
b. activators
d. promoter
9. This assist in selectivity or stop unwanted minerals from floating.
a. modifiers
c. frothers
b. depressants
d. promoter
10. Important for the control of slimes that sometimes interfere with the selectivity and increase reagent
consumption.
a. modifiers
c. frothers
b. activators
d. dispersants
Problem Solving:
1. Ground lead ore is to be concentrated by a single flotation process using 1.5 oz of reagent per ton of ore.
The feed concentrate and tailings have the following composition by weight on a dry basis
Feed %
Concentrate %
Tailings %
PbS
30
90
0.9
ZnS
25
3
35.6
SiO2
45
7
63.5
Water is fed to the cell at the rate of 1000 gallons per ton of wet concentrate with 99% of the water leaving
with the tailings and 1% with the concentrate. Find the mass of wet concentrate produced per hour when
ten tons of ore are fed to the cell / 24 hr. is?
Answer: 3.4
2. From problem number 3 find the total water required in pounds per hour
Answer: 1185
3. A typical flotation machine has the following specifications:
Number of cells = 4
Flotation time = 12min.
Cell Volume = 60 ft3
Hp per cell = 10hp
The material treated has the following specifications:
Pulp (mixture ore and water ) = 40% solids
Specific gravity of ore = 3
𝑇 𝑥 𝐶𝑎𝑝 𝑥 𝑑
𝑛=
𝑉 𝑥 1440
Where n= number of cells; V = volume in cu. Ft per cell; Cap = tons of dry ore / 24 hrs.; d= cu. Ft of pulp
(ore and water) containing one ton of solids.
Answer : 490.13
4. A copper ore initially contains 2.09% Cu. After carrying out a froth flotation separation, the products are
as shown in Table 1.
Using this data, calculate the ratio of concentration.
Answer: 10
5. From Problem number 4, calculate the weight recovery.
Answer: 10%
6. A flotation plant processes 3000 tons/day of CuFeS 2. It produces 80 tons Cu concentrate assaying 25%
Cu. If ore analyzes 0.7% Cu, the percent recovery is?
Solution:
% Cu Recovery = [(c x C)/(f·F]·100
=
(0.25𝑥80)𝑡𝑜𝑛𝑠
0.007𝑥3000
= 𝟗𝟓. 𝟐%
TECHNOLOGICAL INSTITUTE OF THE PHILIPPINES
363 P. CASAL ST QUIAPO MANILA
PROBLEM SET IN PARTICLE TECHNOLOGY
Submitted by:
Briones, Erica Louise D.
Submitted to:
Engr. Robert Delfin
March 24, 2015
SCREENING
1.) The screen used in making size separation smaller than 4 mesh and larger than 48 mesh.
a. Grizzly screen
b. Gyratory screen
c. Oscillating screen
d. Vibrating screen
Answer: d
2.) Box like machines, either around or square with a series of screen clothes nested atop one
another.
a. Reciprocating screen
b. Oscillating screen
c. Electricity vibrated screen
d. Gyratory screen
Answer: d
3.) 200 mesh screen means 200 openings per
a. cm2
b. cm
c. inch
d. inch2
Answer: c
4.) Trommels separate a mixture of particles depending on their
a. Size
b. Density
c. Wettability
d. electrical & magnetic properties
Answer: a
5.) Screen capacity is not a function of
a. its openings size.
b. screening mechanism.
c. screening surface.
d. atmospheric humidity.
Answer: d
6.) Materials which remains on a screen surface is called
a. Fines
b. Undersize
c. Intermediate material
d. Oversize
Answer: d
7.) Making a size smaller than 48 mesh is called
a. Course separation
b. Fine separation
c. Ultrafine separation
d. Scalping
Answer: c
8.) Increasing the capacity of a screen __________ the screen effectiveness.
a. decreases.
b. Increases
c. does not effect
d. none of these
Answer: a
9.) Wet sieving is employed, when the product contains __________ materials.
a. Abrasive
b. large quantity of very fine
c. coarse
d. non-sticky
Answer: b
10.) Vibrating screens have capacity in the range of __________ tons/ft 2 .mm mesh size.
a. 0.2 to 0.8
b. 5 to 25
c. 50 to 100
d. 100 to 250
Answer: a
Problem Solving
1.) A sand mixture was screened through a standard 10-mesh screen. The mass fraction of the
oversize material in feed, overflow and underflow were found to be 0.38, 0.79 and 0.22
respectively. The screen effectiveness based on the oversize is
Answer: 0.50
2.) Shape factors are important in characterizing particulate solids such as those encountered in
various unit operations. What is the specific surface (surface area per unit mass of material) in
cm2/gm for quartz with a diameter of 2inches, density of 165lb/ft 3 and a shape factor of 10?
Answer: 4.47
3.) If the total percentage of particles larger than the screen opening in the feed, product, and
undersize are 25%, 73%, and 6%, respectively, the effectiveness of the screen is ______%.
Answer: 75.03%
4.) It is desired to separate limestone into two fractions, a coarse fraction and a fine fraction. Screen
analysis of feed, coarse to fine fractions show:
Mass fraction of particles in feed = 0.52
Mass fraction of particles in coarse fraction = 0.84
Mass fraction of particles in fine fraction = 0.26
The overall effectiveness of the screen used for the separation purpose per 100kg of feed is
Answer: 61.59%
5.) A sponge-iron industry uses a reciprocating screen of 5mm aperture to separate oversize from
undersize fines which is then recycled to the furnace. The screen analysis of the furnace output
was found to contain 25% fines. The screen efficiency was known to be 50%. The underflow from
the screen contains around 95% fines. If the furnace production rate is 100 ton/h, find the product
rate and the amount of fines present in it.
Answer: Amount of fines in the product = 14.45%
Product rate= 86.60 ton/hr
6.) A quartz mixture is screened through a 8-mesh screen. The cumulative screen analysis of feed,
overflow and underflow are given in the table. Calculate the mass ratios of the overflow and
underflow to feed and the overall effectiveness of the screen.
Mesh
4
Dp (mm)
Feed
Overflow
Underflow
4.699
0
0
0
6
3.327
0.025
0.071
0
8
2.362
0.15
0.43
0
10
1.651
0.47
0.85
0.195
14
1.168
0.73
0.97
0.58
20
0.833
0.885
0.99
0.83
28
0.589
0.94
1.0
0.91
35
0.417
0.96
0.94
65
0.208
0.98
0.975
1.0
1.0
Pan
Solution:
From table: xF= 0.15, xD= 0.43, xB= 0
𝐷 𝑥𝐹 − 𝑥𝐵 0.15 − 0
=
=
= 0.35
𝐹 𝑥𝐷 − 𝑥𝐵 0.43 − 0
𝐵
𝐷
= 1 − = 1 − 0.35 = 0.65
𝐹
𝐹
𝐸=
(𝑥𝐹 − 𝑥𝐵 )(𝑥𝐷 − 𝑥𝐹 )𝑥𝐷 (1 − 𝑥𝐵 )
(𝑥𝐷 − 𝑥𝐵 )2 (1 − 𝑥𝐹 )𝑥𝐹
𝐸=
(0.15 − 0)(0.43 − 0.15)(0.43)(1 − 0)
(0.43 − 0)2 (1 − 0.15)(0.15)
𝑬 = 𝟎. 𝟕𝟔𝟔𝟏 ≈ 𝟕𝟔. 𝟔𝟏%
SIZE REDUCTION
1.) Which of the following terminology is not used for size reduction of materials to fine sizes or
powders?
a. Comminution
b. Dispersion
c. Pulverisation
d. Compression
Answer: d
2.) A tube mill as compared to a ball mill
a. employs smaller balls.
b. gives finer size reduction but consumes more power.
c. has larger length/diameter ratio (>2 as compared to 1 for ball mill).
d. all (a), (b) and (c).
Answer: d
3.) Kick's law assumes that the energy required for size reduction is proportional to the logarithm
of the ratio between the initial and the final diameters. The unit of Kick's constant is
a. kW. sec/kg
b. kWh/kg
c. kWh/sec. kg
d. kg/sec
Answer: a
4.) To get ultrafine particles, the equipment used is a
a. ball mill
b. rod mill
c. hammer crusher
d. fluid energy mill
Answer: d
5.) The capacity of a gyratory crusher is __________ that of a jaw crusher with the same gape,
handling the same feed & for the same product size range.
a. same as
b. 2.5 times
c. 5 times
d. 10 times
Answer: b
6.) States that the energy required for crushing is proportional to the new surface created
a. Rittinger’s Law
b. Kick’s Law
c. Bond Law
d. Energy Law
Answer: a
7.) Size reduction is important in chemical engineering since
a. It prevents chemical engineers from becoming overweight
b. It makes products to become uniform in size
c. It prepares raw materials of the desired sizes prior to processing
d. None of the above
Answer: c
8.) Rittinger number which designates the new surface prduced per unit of mechanical energy
absorbed by the material being crushed, depends on the
a. state or manner of application of the crushing force.
b. ultimate strength of the material.
c. elastic constant of the material.
d. all (a), (b) and (c).
Answer: d
9.) Grinding efficiency of a ball mill is of the order of __________ percent.
a. 1-5
b. 40-50
c. 75-80
d. 90-95
Answer: a
10.) Work index is defined as the
a.
b.
c.
d.
gross energy (kWh/ton of feed) needed to reduce very large feed to such a size that 80%
of the product passes through a 100 micron screen.
energy needed to crush one tonne of feed to 200 microns.
energy (kWh/ton of feed) needed to crush small feed to such a size that 80% of the
product passes a 200 mesh screen.
energy needed to crush one ton of feed to 100 microns.
Answer: a
Problem Solving
1.) A ball mill is to grind 250 tons/day of calcite having a specific surface of 90.34 sq. cm./g. No
recrushing of material is to be done. Overall energy effectiveness of the ball mill is 7.5%.
Mesh
Mass Fraction Product
10/14
0.02
14/20
0.07
20/28
0.18
28/35
0.25
35/48
0.14
48/65
0.11
65/100
0.08
100/150
0.05
150/200
0.04
200/270
0.325
270/400
0.0275
What is the theoretical horsepower requirements to grind the calcite?
Answer: 0.4033Hp
2.) A certain crusher accepts the feed rock having a volume surface mean diameter of 20mm and
discharges a product having a volume surface mean diameter of 5mm. the power required to crush
1200kg/h of material is 9.5kW. What would be the power consumption of the capacity is reduced to
100kg/h and the product size to 4mm? Assume mechanical efficiency to be same in both cases.
Answer: 1.054kW
3.) A continuous grinder obeying the Bond crushing law grinds a solid at the rate of 800kg/h form the
initial diameter of 12mm to the final diameter of 2mm. if it is required to produce particles of 1mm
sizes, what would be the output rate of the grinder (in kg/h) for the same power input?
Answer: 470.32 kg/h
4.) What will be the power required to crush 150ton/h of limestone (work index for limestone is 12.74)
if 80% of the feed passes a 60mesh screen and 80% of the product passes a 6mm screen?
Answer: 168.68kW
5.) If crushing rolls, 1 m in diameter, are set so that the crushing surfaces are 12.5 mm apart and the
angle of nip is 31◦, what is the maximum size of particle which should be fed to the rolls? If the
actual capacity of the machine is 12 per cent of the theoretical, calculate the throughput in kg/s
when running at 2.0 Hz if the working face of the rolls is 0.4 m long and the bulk density of the feed
is 2500kg/m3.
Answer: 25mm
6.) What rotational speed in rpm would you recommend for a ball mill that is 1000mm in diameter
charged with 70mm balls?
Solution:
1
𝑔
The critical speed equation 𝑁𝑐 = 2𝜋 √(𝑅−𝑟)
D= 1000mm
d= 70mm
𝑅=
𝐷 1000
=
= 500𝑚𝑚 ≈ 0.5𝑚
2
2
𝑟=
𝑑 70
=
= 35𝑚𝑚 ≈ 0.035𝑚
2
2
𝑁𝑐 =
1
9.81
√
2𝜋 (0.5 − 0.035)
𝑵𝒄 = 𝟎. 𝟕𝟑𝟎𝒓𝒑𝒔 ≈ 𝟒𝟖. 𝟑𝟓𝒓𝒑𝒎
SEDIMENTATION
1.) Two particles are called to be equal settling, if they are having the same.
a. size.
b. specific gravity.
c. terminal velocities in the same fluid & in the same field of force.
d. none of these.
Answer: c
2.) The operation by which solids are separated from liquids due to difference in their respective
densities is
a. screening
b. adsorption
c. sedimentation
d. absorption
Answer: c
3.) the separation of solid particles into several size fractions based upon the settling velocities in a
medium is called
a. settling
b. filtration
c. flotation
d. classification
Answer: d
4.) Solid particles separation based on the difference in their flow velocities through fluids is termed as
the
a. Clarification
b. Classification
c. Elutriation
d. Sedimentation
Answer: b
5.) Gravity settling process is not involved in the working of a
a. Hydrocyclone
b. Classifier
c. dorr-thickener
d. sedimentation tank
Answer: a
6.) Which of the following is a batch sedimentation equipment?
a. Dust catcher
b.
c.
d.
Filter thickener
Dry cyclone separator
Rotary sprayer scrubber.
Answer: b
7.) Separation of a suspension or slurry into a supernatant clear liquid (free from particles) and a thick
sludge containing a high concentration of solid is called
a. Classification
b. Sedimentation
c. Clarification
d. Decantation
Answer: b
8.) Sedimentation on commercial scale occurs in
a. Classifiers
b. Thickeners
c. rotary drum filters
d. cyclones
Answer: c
9.) If the motion of a particle is impeded by other particles, which will happen when the particles are
near each other even though they are not actually colliding, the process is called
a. Free settling
b. Unhindered settling
c. Gravity settling
d. Hindered settling
Answer: c
10.) Which of the following is not a wet classifier?
a. Sharpies super-centrifuge
b. Hydrocyclones
c. Dorr Oliver rake classifier
d. None of these
Answer: d
Problem Solving
1.) Square mica plates, 1/32 in thick and 0.01 sq. in area are falling randomly through oil with a density
of 55 lb/cu ft and with viscosity of 15 centipoise. The specific gravity of the mica is 3.0 the settling
velocity is
Answer: 7.6 cm/s
2.) The terminal settling velocity of a 6mm diameter glass sphere (density: 2500kg/m 3) in a viscous
Newtonian liquid (density: 1500kg/m3) is 100µm/s. If the particle Reynolds number is small and the
value of acceleration due to gravity is 9.81m/s2, then the viscosity of the liquid (in Pa.s) is
Answer: 196.2 Pa.s
3.) Many animal cells can be cultivated on the external surface of dextran beads. These cell-laden
beads or “microcarriers” have a density of 1.02 g/cm3 and a diameter of 150 µm. A 50-liter stirred
tank is used to cultivate cells grown on microcarriers to produce a viral vaccine. After growth, the
stirring is stopped and the microcarriers are allowed to settle. The microcarrierfree fluid is then
withdrawn to isolate the vaccine. The tank has a liquid height to diameter ratio of 1.5; the carrierfree fluid has a density of 1.00 g/cm3 and a viscosity of 1.1 cP. Estimate the settling time by
assuming that these beads quickly reach their maximum terminal velocity.
Answer: 2379 seconds
4.) A particle settles through a suspension containing 35% solids by weight. Given that the velocity
under free settling motion is ut = 0.00669 m/s. Find us. [density of particle = 2800 kg/m3, Dp = 200
mesh, density of fluid = 996.5 kg/m3, viscosity of fluid = 0.8Cp]
Answer: 0.00306 m/s
5.) Spherical particles with diameter d = 0.5 mm = 0.0005 m, and true particle densitare suspended in
water: density of water ρF= 1000 kg/m3; viscosity of water ηF =The concentration of the particles =
5 vol%. Acceleration of gravity g = 9.81 m/s2 Calculate the settling velocity.
Answer: 2.3mm/s
6.) Oil droplets having a diameter of 20 µm (0.020 mm) are to be settled from air at temperature of
37.8°C and 101.3 kPa pressure. The density of the oil is 900 kg/m3. Calculate the terminal settling
velocity of the droplets. Air at 37.8°C: ρ = 1.137 kg/m3 , µ = 1.90 X 10-5 Pa.s
Solution:
Dp= 20x10-6 m
ρoil= 900kg/m3
ρair=1.137kg/m3
µ=1.90 X 10-5 Pa.s
1
3
𝑔𝜌(𝜌𝑝 − 𝜌)
𝑘 = 𝐷𝑝 [
]
𝜇2
1
(9.81)(1.137)(900 − 1.137) 3
𝑘 = (20𝑥10−6 ) [
]
2
1.90𝑥10−5
𝑘 = 0.606 ∴ 𝑆𝑡𝑜𝑘𝑒 ′ 𝑠 𝐿𝑎𝑤
𝑣𝑡 = [
𝑔𝐷2𝑝 (𝜌𝑝 − 𝜌)
18𝜇
]
2
(9.81) (20𝑥10−6 ) (900 − 1.137)
𝑣𝑡 = [
]
−5
18(1.90𝑥10 )
𝒗𝒕 = 𝟎. 𝟎𝟏𝟎𝟑𝒎/𝒔
CENTRIFUGATION
1.) Tabular bowl centrifuges as compared to disk bowl centrifuges
a. operate at higher speed.
b. employ bowl of larger diameter.
c. can not be operated under pressure/vacuum.
d. can't be used for separation of fine suspended solids from a liquid.
Answer: a
2.) __________ centrifuge is normally used in sugar mills.
a. Tubular bowl
b. Disc-bowl
c. Suspended batch basket
d. Perforated horizontal basket continuous
Answer: c
3.) If radius of a batch basket centrifuge is halved & the r.p.m. is doubled, then the
a. linear speed of the basket is doubled.
b. linear speed of the basket is halved.
c. centrifugal force is doubled.
d. capacity of centrifuge is increased.
Answer: c
4.) Ultracentrifuges running at speeds upto 100000 rpm is normally used for the
a. separation of isotopes based on their density or molecular weights difference.
b. concentration of rubber latex.
c. separation of cream from milk.
d. dewaxing of lubricating oil.
Answer: a
5.) Flocculation method will improve the
a. Centrifugation
b. Filtration
c. Lyophilization
d. Drying
Answer: a
6.) Separation of isotopes is generally done using a/an __________ centrifuge.
a. Ultra
b. disk-bowl
c. both (a) & (b)
d. neither (a) nor (b)
Answer: a
7.) Ultra centrifuges are used for the separation of __________ solid particles.
a. Coarse
b. Fine
c. Colloidal
d. Dissolved
Answer: c
8.) Moisture can be removed from lubricating oil using
a. tubular centrifuge
b. clarifier
c. sparkler filter
d. vacuum leaf filter
Answer: a
9.) Where the density difference of the two liquid phase to be separated is very small (as in milk cream
separator), the most suitable separator is a
a. disc bowl centrifuge.
b. sharpies supercentrifuge.
c. batch basket centrifuge.
d. sparkler filter.
Answer: a
10.) is a separation process which uses the action of centrifugal force to promote accelerated settling of
particles in a solid-liquid mixture.
a. Sedimentation
b. Filtration
c. Membrane separation
d. Centrifugation
Answer: d
Problem Solving
1.)
A viscous solution containing particles with a density of 1461 kg/m3 is to be clarified by
centrifugation. The solution density is 801 kg/m3 and its viscosity is 100 cp. The centrifuge has a
bowl with r2 = 0.02225 m, r1 = 0.00715 m, and height b = 0.1970 m. Calculate the critical particle
diameter of the largest particles in the exit stream if N = 23,000 rev/min and flow rate of 0.002832
m3/h.
Answer: 0.746 µm
2.)
In a test conducted using laboratory centrifuge, it was found that the optimum recovery of protein
from coconut oil was achieved with an RPM of 2500. Industrial size centrifuge come in 2.5, 3.0,
3.5, 4.0 and 5.0 ft. diameter with warranty if the centrifuge is operated not over 1000 RPM. For
optimum commercial operations, the centrifuge size you will recommend is
Data: Laboratory Centrifuge
Height= 9in.
Diameter=5in.
RPM= 2,000-3,000
Answer: 3.0ft.
3.)
The capacity of Sharples Centrifuge is estimated to increase/decrease by____% if its speed is
doubled and the cut size of the particles is reduced by 20%.
Answer: 156% increase
4.)
In a bowl centrifugal classifier operating at 60rp with water (µ=0.001kg/m.s), the time taken for a
particle (Dp=0.0001m, s.g=2.5) in seconds to traverse a distance of 0.05m from the liquid surface
is
Answer: 7.8 seconds
5.)
In a vegetable oil refining process, an aqueous phase is being separated from the oil phase in a
centrifuge. The density of the oil is 919.5kg/m3 and that of the aqueous phase is 980.35kg/m3. The
radius r1 for overflow of the light liquid has been set at 10.160mm and rhe outlet for the heavy liquid
at 10.414mm. Calculate the location of the interface in the centrifuge.
Answer: r2=13.75mm
6.)
A centrifuge of diameter of 0.2m in a pilot plant rotates at a speed of 50Hz in order to achieve
effective separation. If this centrifuge is scaled up to a diameter of 1m in the chemical plant, and
the same separation factor is to be achieved, what is the rotational speed of the scaled up
centrifuge?
Solution:
Separation factor, S=
𝜔2 𝑟
𝑔
To maintain the same separation factor,
𝜔21 𝑟1 𝜔22 𝑟2
=
𝑔
𝑔
𝜔2 = 𝜔1 √
𝑟1
𝑟2
0.2𝑚
𝜔2 = 50𝑥√
1𝑚
𝝎𝟐 = 𝟐𝟐. 𝟑𝟔𝑯𝒛
FLOTATION
1.) is a method of separation widely used in the wastewater treatment and mineral
processing industries.
a. Sedimentation
b. Flotation
c. Centrifugation
d. Size reduction
Answer: b
2.) Dispersants are important for the control of slimes which sometimes interfere with the selectivity
and increase reagent consumption. Another term for dispersant is
a. Deflocculant
b. Depressants
c. Frothers
d. Regulators
Answer: a
3.) Which one of the following ores are best concentrated by froth flotation method?
a. Cassiterite
b. Galena
c. Malachite
d. Magnetite
Answer: b
4.) Which of the following is the most suitable for cleaning of fine coal dust (<0.5 mm) ?
a. Trough washer
b. Baum jig washer
c. Spiral separator
d. Froth floatation
Answer: d
5.) Xanthates are used in the froth floatation process as a/an
a. Conditioner
b. Frother
c. Collector
d. Activator
Answer: c
6.) An example of a collector for flotation of metallic sulphides and native metals is
a. Xanthates
b. Sodium solicate
c. Sodium sulphide
d. Sphalerite
Answer: a
7.) Which of the following is an example of deflocculant?
a. Sulphuric acid
b. Lignin sulforate
c. Dithiophosphate
d. Molybderite
Answer: b
8.) In froth floatation, chemical agent added to cause air adherence is called
a. Collector
b. Frother
c. Modifier
d. Activator
Answer: a
9.) The flotation agent that prevents coalescence of air bubbles as they travel to the surface of the
water is/are
a.) Collectors
b.) Promoters
c.) Frothing agent
d.) Modifying agent
Answer: c
10.) is a water treatment process that clarifies wastewaters (or other waters) by the removal of
suspended matter such as oil or solids.
a. Dissolved air flotation
b. Induced gas flotation
c. Froth flotation
d. None of these
Answer: a
Problem Solving
1.) A typical flotation machine has the following specifications:
Number of cells=4
Cell Volume=60 cu.ft.
Flotation time=12min
Hp per cell=10Hp
The material treated as the following specifications:
Pulp (mixture of ore and water)=40% solids
s.g of ore= 3
𝑇𝑥𝐶𝑎𝑝𝑥𝑑
Equation: 𝑛 = 𝑉𝑥1440
Where:
n= number of cells
V=volume in cu ft per cell
Cap= tons of dry ore per 24hrs
d= cu ft of pulp (ore and water) containing one ton (2000lb) of solids
The capacity of the machine in tons of dry ore per 24 hour is
Answer: 490.1 tons/hr
2.) Calculate the specific flotation rate constant for 100um galena particles on 0.5mm diameter
bubbles in a continuous floatation cell that is aerated at 0.013m 3air/m pump. The bubble residence
time is 5s. Consider the case where the average induction time for the particles is 100ms.
Calculate the fraction of the non-floatable component for these particles. Consider the case when
the bubble is carrying 5 galena particles.
Answer: 0.0255
3.) A flotation plant processes 5000tons/day of PbS. It produces 60tons of Pb concentrate assaying
30% Pb. If ore analyses 0.7% Pb, the percent recovery
Answer: 51.43%
4.) Ground lead ore is to be concentrated by a single flotation process using 1.5oz. of reagent per ton
of ore. The feed concentrate and tailings have the following composition by weight on a dry basis:
Feed %
Concentrate%
Tailings%
PbS
30
90
0.9
ZnS
25
3
35.6
SiO2
45
7
63.5
Water is fed to the cell at the rate of 1000gallons per ton of wet concentrate with 99% of the water
leaving with the tailings and 1% with the concentrate. Mass of wet concentrate produced per hour
when ten tons of ore are fed into the cell per 24hrs is
Answer: 3.4 tons/hr
5.) In the previous problem, the total water required in pounds per hour is
Answer: 1185 lbs/hr
6.) A flotation section is extracting CuS. The ore consists of 8% CuS and 92% gangue, which may be
assumed to be SiO2. The following data are given:
% CuS
Feed (a)
Concentrate (b)
Rougher Tailings (c)
Scavenger Concentrate (d)
Final Tailings (e)
%SiO2
3
80
90
20
4
13
0.1
96
87
99.99
Laboratory experiments indicated that water to solid ratio, L/S=4 at the contact time of 15min in the
rougher; L/S=6 at t=18min in the scavenger. On the basis of 500tons/day ore treated. Calculate the
Volume of the rougher and scavenger needed.
Solution:
Data: ρSiO2= 2.65g/cc
ΡCuS=4.6g/cc
Basis: 100 lbs/day
OMB: 𝑎 = 𝑏 + 𝑒
100 = 𝑏 + 𝑒
SB: (100)(0.03) = (𝑏)(0.80) + (𝑒)(0.001)
3.63𝑙𝑏𝑠
𝑏=
𝑑𝑎𝑦
96.37𝑙𝑏𝑠
𝑒=
𝑑𝑎𝑦
Scavenger OMB: 𝑐 = 𝑑 + 𝑒
𝑒 =𝑐−𝑑
96.37 = 𝑐 − 𝑑
SB:(96.37)(0.001) = (𝑐)(0.04) − (𝑑)(0.13)
138.13𝑙𝑏𝑠
𝑐=
𝑑𝑎𝑦
41.76𝑙𝑏𝑠
𝑑=
𝑑𝑎𝑦
𝜌𝑎 =
100
171.16𝑙𝑏
𝑥62.4 =
8
92
𝑓𝑡3
+
4.6 2.65
𝜌𝑐 =
100
168.21𝑙𝑏
𝑥62.4 =
4
96
𝑓𝑡3
+
4.6 2.65
𝜌𝑑 =
100
175.00𝑙𝑏
𝑥62.4 =
13
87
𝑓𝑡3
+
4.6 2.65
Volume:
Rougher
𝑤𝑡 = 𝑎 + 𝑑
𝑤𝑡 = 100 + 41.76
𝑤𝑡. = 141.76𝑙𝑏𝑠
𝜌𝑎𝑑 =
141.76
172.27𝑙𝑏
𝑥62.4 =
100
41.76
𝑓𝑡3
171.16 + 175
𝑣𝑜𝑙𝑢𝑚𝑒 =
100
41.76
+
= 0.8229𝑓𝑡 3
171.16
175
𝐿
=4
𝑆
𝐿 = (4)(141.76𝑙𝑏𝑠) = 567.04𝑙𝑏𝑠 = 9.087𝑓𝑡 3
𝑣𝑜𝑙. 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑜𝑙𝑖𝑑 =
(0.0830) (
𝑣𝑜𝑙𝑢𝑚𝑒 =
𝑤𝑡.𝑎𝑑 =
0.8229
= 0.0830
9.087 + 0.8229
172.27𝑙𝑏
1𝑡𝑜𝑛
60𝑚𝑖𝑛
)(
)
(24ℎ𝑟𝑠)(
3
2000𝑙𝑏𝑠
ℎ𝑟 ) 0.6863𝑡𝑜𝑛
𝑓𝑡
=
15𝑚𝑖𝑛
𝑓𝑡 3
141.76
𝑥500 = 708.8𝑡𝑜𝑛𝑠
100
708.8𝑡𝑜𝑛𝑠
0.6863𝑡𝑜𝑛𝑠
𝑓𝑡 3
𝒗𝒐𝒍. 𝒓𝒐𝒖𝒈𝒉𝒆𝒓 = 𝟏𝟎𝟑𝟐. 𝟕𝟓𝒇𝒕𝟑
𝑣𝑜𝑙. 𝑟𝑜𝑢𝑔ℎ𝑒𝑟 =
Scavenger:
𝑤𝑡.𝑐 = 100 + 138.13 = 238.13𝑙𝑏𝑠
238.13
𝑣𝑜𝑙. 𝐶 =
= 1.416𝑓𝑡 3
168.21
𝐿
=6
𝑆
𝐿 = (6)(238.13) = 1428.78𝑙𝑏𝑠 = 22.90𝑓𝑡 3
𝑣𝑜𝑙. 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑜𝑙𝑖𝑑𝑠 =
(0.0582) (
𝑣𝑜𝑙𝑢𝑚𝑒 =
𝑤𝑡. 𝑐 =
1.416
= 0.0582
22.90 + 1.416
168.21𝑙𝑏
1𝑡𝑜𝑛
60𝑚𝑖𝑛
)(
) (24ℎ𝑟𝑠)( ℎ𝑟 ) 0.3916𝑡𝑜𝑛
3
2000𝑙𝑏𝑠
𝑓𝑡
=
18𝑚𝑖𝑛
𝑓𝑡 3
238.13
𝑥500 = 1190.65𝑡𝑜𝑛𝑠
100
1190.65𝑡𝑜𝑛𝑠
0.3916𝑡𝑜𝑛
𝑓𝑡 3
𝒗𝒐𝒍𝒖𝒎𝒆 𝒔𝒄𝒂𝒗𝒆𝒏𝒈𝒆𝒓 = 𝟑, 𝟎𝟒𝟎. 𝟒𝟕𝒇𝒕𝟑
𝑣𝑜𝑙𝑢𝑚𝑒 𝑠𝑐𝑎𝑣𝑒𝑛𝑔𝑒𝑟 =
Technological Institute of the Philippines
College of Engineering and Architecture
Chemical Engineering Department
Introduction to Particle Technology
Submitted by:
John Anthony L. Cruz
Submitted to:
Engr. Robert Delfin
March 16, 2015
SEDIMENTATION
21. Which of the following is true for the experiment
In an experiment, a sphere of density 1 and radius r is dropped in a tank of oil of viscosity 1 and density
2 . The time of descent for the sphere through the first section of height d is recorded as t1 and through the
second section of the same height as t2, 0 < t2 – t1 << 1.
Sphere
d
d
a.
b.
c.
d.
V
(1)
2
The drag force exerted on the sphere increases during it descent through the second section.
The sphere never reaches its terminal velocity while falling through both sections
The sphere reaches its terminal velocity while falling through the first section
The drag force exerted on the sphere decreases during the descent through the second section
22. In hindered settling, particles are
e. placed farther from the wall
f. not affected by other particles and the wall
c. near each other
d. none of these
23. Drag is defined as the force exerted by the
e. fluid on the solid in a direction opposite to flow
f. the fluid on the solid in the direction of flow
g. the solid on the fluid
h. none of these
24. Drag coefficient for flow of past immersed body is the ratio of
e. shear stress to the product of velocity head and density
f. shear force to the product of velocity head and density
g. average drag per unit projected area to the product of the velocity head and density
h. none of these
25. Drag coefficient in hindered settling is
e. less than in free settling
f. equal to that in free settling
g. not necessarily quarter than in free settling
h. greater than in free settling
26. The terminal velocity of a small sphere settling in a viscous fluid varies as the
e. first power of its diameter
f. inverse of the fluid viscosity
g. inverse square of the diameter
h. square of the difference in specific weights of solid and fluid
27. At low Reynolds number
e.
f.
g.
h.
viscous forces are unimportant
viscous forces control
viscous forces control and inertial forces are unimportant
gravity forces control
28. At high Reynolds number
e. inertial forces control and viscous forces are unimportant
f. viscous forces predominate
g. inertial forces are unimportant and viscous forces control
h. none of these
29. Forces acting on a particle settling in fluid are
e. gravitational and buoyant forces
f. centrifugal and drag forces
g. gravitational or centrifugal, buoyant and drag forces
h. external, drag and viscous forces
30. Terminal velocity is
i. constant velocity with no acceleration
j. a fluctuating velocity
k. attained after moving one-half of total distance
l. none of these
1. A slurry containing 5 kg of water/kg of solids is to be thickened to a sludge containing 1.5 kg of water/kg of solids in
a continuous operation. Laboratory tests using five different concentrations of the slurry yielded the following data:
Concentration (kg water/kg solid)
Rate of sedimentation (mm/s)
5.0
4.2
3.7
3.1
2.5
0.20
0.12
0.094
0.070
0.050
Calculate the minimum area of a thickener required to effect the separation of a flow of 1.33 kg/s of solids.
Answer: 31.2 sq m
2. For the sedimentation of a suspension of uniform particles in a liquid, the relation between observed sedimentation
velocity uc and fractional volumetric concentration C is given by:
Uc/Uo = (1−C)4.8
where u0 is the free falling velocity of an individual particle. Calculate the concentration at which the rate of
deposition of particles per unit area will be a maximum, and determine this maximum flux for 0.1 mm spheres of
glass (density 2600 kg/m3) settling in water (density 1000 kg/m3, viscosity 1 mNs/m2).
Answer: 6.06×10−4 m3/m2-s
3. Determine the terminal settling velocity of dust particles having a diameter of 60 microns at 294.3 K and 101.32
kPa. The dust particles can be considered spherical with a density of 1280 kg per cubic meter.
Answer: 0.14 m/s
4. Calculate the minimum area and diameter of a thickener with a circular basin to treat 0.1 m3/s of a slurry of a
solids concentration of 150 kg/m3. The results of batch settling tests are:
A value of 1290 kg/m3 for underflow concentration was selected from a retention time test. Estimate the underflow
volumetric flow rate assuming total separation of all solids and that a clear overflow is obtained.
Answer: 0.0116 m3/s
5. When a suspension of uniform coarse particles settles under the action of gravity, the relation between the
sedimentation velocity uc and the fractional volumetric concentration C is given by:
Uc/Uo = (1−C)n,
where n =2.3 and Uo is the free falling velocity of the particles. Draw the curve of solids flux ψ against concentration
and determine the value of C at which ψ is a maximum and where the curve has a point of inflexion. What is implied
about the settling characteristics of such a suspension from the Kynch theory? Comment on the validity of the Kynch
theory for such a suspension
Answer: C=0.61
6. A slurry containing 5 kg of water/kg of solids is to be thickened to a sludge containing 1.5 kg of water/kg of solids in
a continuous operation. Laboratory tests using five different concentrations of the slurry yielded the following results:
Calculate the minimum area of a thickener to effect the separation of 0.6 kg/s of solids.
Solution:
CENTRIFUGATION
12. For separation of sugar solution from settled out mud we use
e. sparkler filter
c. centrifugal filter
f. plate & frame filter
d. rotary drum vacuum filter
13. Moisture can be removed from lubricating oil using
e. tubular centrifuge
c. sparkler filter
f. clarifier
d. vacuum leaf filter
14. Which of the following can be most effectively used for clarification of tube oil and printing ink?
e. sparkler filter
c. disc-bowl centrifuge
f. precoat filter
d. sharpless supercentrifuge
15. If the radius of a basket centrifuge is halved and the rpm is doubled, then
i. linear speed of the basket is doubled
j. linear speed of the basket is halved
k. centrifugal force is doubled
l. capacity of centrifuge is increased
16. Where the difference in density of the two liquid phases to be separated is very small (as in milk cream
separator), the most suitable separator is
c. disc bond centrifuge
c. batch basket centrifuge
d. sharpless supercentrifuge
d. sparkler filter
17. A piece of equipment that puts an object in rotation around a fixed axis (spins it in a circle), applying a potentially
strong force perpendicular to the axis of spin (outward).
a. Centrifuge
b. Filter
c. Settler
d. Mixer
18. A filtering or screen centrifuge which is also known as worm screen or conveyor discharge centrifuge.
a. Screen/scroll centrifuges
b. Pusher centrifuges
c. Peeler centrifuges
d. Decanter centrifuges
19. A device that performs by rotating filtration basket in an axis.
a. Screen/scroll centrifuges
b. Pusher centrifuges
c. Peeler centrifuges
d. Decanter centrifuges
20. A type of filtration technique that offers continuous operation to de-water and wash materials such as relatively
in-compressible feed solids, free-draining crystalline, polymers and fibrous substances.
a. Pusher centrifuges
b. Peeler centrifuges
e. Decanter centrifuges
f.
Screen/scroll centrifuges
21. A device, which employs a high rotational speed to separate components of different densities.
a. Pusher centrifuges
b. Peeler centrifuges
g. Decanter centrifuges
h. Screen/scroll centrifuges
1. If a centrifuge is 0.9 m diameter and rotates at 20 Hz, at what speed should a laboratory centrifuge of 150 mm
diameter be run if it is to duplicate the performance of the large unit?
Answer: 2940 rpm
2. An aqueous suspension consisting of particles of density 2500 kg/m3 in the size range 1–10 µm is introduced into
a centrifuge with a basket 450 mm diameter rotating at 80 Hz. If the suspension forms a layer 75 mm thick in the
basket, approximately how long will it take for the smallest particle to settle out?
Answer: 19.3 s
3. A centrifuge basket 600 mm long and 100 mm internal diameter has a discharge weir 25 mm diameter. What is the
maximum volumetric flow of liquid through the centrifuge such that, when the basket is rotated at 200 Hz, all particles
of diameter greater than 1 µm are retained on the centrifuge wall? The retarding force on a particle moving liquid
may be taken as 3πµdu, whereu is the particle velocity relative to the liquid µ is the liquid viscosity, and d is the
particle diameter. The density of the liquid is 1000 kg/m3, the density of the solid is 2000 kg/m3 and the viscosity of
the liquid is 1.0 mNs/m2. The inertia of the particle may be neglected
Answer: 0.00103 m3/s
4. In a test on a centrifuge all particles of a mineral of density 2800 kg/m3 and of size 5 µm, equivalent spherical
diameter, were separated from suspension in water fed at a volumetric throughput rate of 0.25 m3/s. Calculate the
value of the capacity factor. What will be the corresponding size cut for a suspension of coal particles in oil fed at the
rate of 0.04 m3/s? The density of coal is 1300 kg/m3 and the density of the oil is 850 kg/m3 and its viscosity is 0.01
Ns/m2. It may be assumed that Stokes’ law is applicable.
Answer: 0.000004 m
5. When an aqueous slurry is filtered in a plate and frame press, fitted with two 50 mm thick frames each 150 mm
square, operating with a pressure difference of 350 kN/m2, the frames are filled in 3600 s (1 h). How long will it take
to produce the same volume of filtrate as is obtained from a single cycle when using a centrifuge with a perforated
basket, 300 mm diameter and 200 mm deep? The radius of the inner surface of the slurry is maintained constant at
75 mm and the speed of rotation is 65 Hz (3900 rpm). It may be assumed that the filter cake is incompressible, that
the resistance of the cloth is equivalent to 3 mm of cake in both cases, and that the liquid in the slurry has the same
density as water.
Answer: 4.25 min
6. A centrifuge with a phosphor bronze basket, 380 mm in diameter, is to be run at 67 Hz with a 75 mm layer of liquid
of density 1200 kg/m3 in the basket. What thickness of walls are required in the basket? The density of phosphor
bronze is 8900 kg/m3 and the maximum safe stress for phosphor bronze is 87.6 MN/m2.
FLOTATION
15. Froth Flotation is most suitable for treating
e. iron ores
c. quartz
f. sulfide ores
d. metal ores
16. In Froth Flotation, chemical agent added to cause air adherence is called
e. collector
c. modifier
f. frother
d. promoter
17. Pine oil used in forth flotation technique acts as a
e. collector
c. frother
f. modifier
d. activator
18. Which of the following is the most suitable for cleaning of fine coal dust (< 0.5 m)?
e. Through washer
c. Spiral separator
f. Baum Jig Washer
d. Froth Flotation
19. It involves phenomena related to the relative buoyancy of objects.
i. Flotation
j. Filtration
k. Collector
l. Washer
20. An example of collector for flotation of metallic sulfides and native metals is
22. Xanthates
23. Sodium silicate
24. Sodium sulfide
25. Sphalerite
21. Which of the following is an example of deflocculant?
g. Sulfuric acid
h. Lignin sulforate
i. Dithiophosphate
j. Molybderite
22. These are used to make a mineral surface amenable to collector coating
a. Modifiers
b. Activators
c. Regulators
d. Collectors
23. Added to strengthen temporarily covering film of the air bubbles
a. Frothers
b. Collectors
c. Modifiers
d. Promoters
24. A flotation modifier which assists in the selectivity or stop unwanted minerals from floating
a. Depressants
b. Activators
c. Alkalinity regulators
d. Promoters
m. A flotation plant produces 3,000 tons per day of CuFeS 2 (chalcopyrite). It produces 80 tons of Cu concentrate
assaying 25% Cu. If the analyzes 0.7%, calculate the percentage recovery.
Answer: 95.24%
Link Problem: Ground lead ore is to be concentrated by a single flotation process using 1.5 oz of reagent per ton of
ore. The feed concentrate and tailings have the following composition by weight on a dry basis
Feed %
Concentrate %
Tailings %
PbS
30
90
0.9
ZnS
25
3
35.6
SiO2
45
7
63.5
Water is fed to the cell at the rate of 1000 gallons per ton of wet concentrate with 99% of the water leaving with the
tailings and 1% with the concentrate.
n. Mass of wet concentrate produced per hour when ten tons of ore are fed to the cell per 24 hours is __________
Answer: 3.4
o. Total water required in pounds per hour is ________
Answer: 1185 lb/hr
p. A flotation machine has the following specifications:
Number of cells = 4
Flotation time = 12 min
Cell volume = 60 cu. ft
Hp per cell = 10 Hp
The material treated has the following specifications:
Pulp = 40% solids
Specific gravity of ore = 3
The capacity of the machine in tons of dry ore per 24 hours is ________
Answer: 490
q. Laboratory experiments indicated that the water to solids ratio, L/S=2 and the contact time is 10 min in the
rougher; L/S=4, contact time=18 minutes in the scavenger. On the basis of 300 tons per day of ore treated.
Calculate the volume of the rougher needed.
% CuS
%SiO2
Feed
5
95
Concentrate
85
15
Rougher tailings
1
99
Scavenger concentrate
10
90
Final tailings
0.3
99.7
Answer: 169.8 ft3
FILTRATION
1. For a classification of potable water, we use
a.) gravity sand filter
b.) plate and frame filter
c.) vacuum leaf filter
2. Vacuum filter is most suitable for
a.) removal of lines from liquid
c.) liquids of very high viscosity
3. Filter aid is used
a.) to increase the rate of filtration
c.) to increase the porosity of the cake
4. During the washing of cake
a.) all resistance are constant
c.) filter medium resistance
d.) rotary vacuum filter
b.) liquid having high vapour pressure
d.) none of these
b.) to decrease the pressure drop
d.) as a support base for the septum
b.) filter medium resistance increases
d.) change resistance decreases
5. The porosity of a compressible cake is
a.) minimum at the filter medium
b.) minimum at the upstream face
c.) maximum at the filter medium
d.) same throughout the thickness
6. The specific cake resistance is
a.) gm / cm2
c.) cm / gm2
7. The unit of filter medium resistance is
a.) cm-1
c.) cm / gm-1
8. The medium resistance is controlled by
a.) the pressure drop along
b.) the flow rate alone
b.) cm / gm
d.) gm / gm.
b.) gm / cm-1
d.) gm -1
c.) both pressure drop and flow rate
d.) the cake thickness
9. Compressibility coefficient for an absolutely compressible cake is
a.) 0
b.) 1
c.) 0 to 1
d.) none of these
10. In continuous filtration (at a constant pressure drop) filtrate flow rate varies inversely as the
a.) square root of the velocity
b.) square root of the viscosity
c.) filtration time only
d.) washing time only
k. A plate and frame filter press containing 20 frames is used to filter a slurry made up to 10 lb dry solids
per 100 lb of liquid and solid mixture. The inside dimensions of each frame are 2 ft by 2 ft by 1 in thick.
The caked formed in the filtration is non compressible and contains 0.7 lb of dry solids per pound of
cake. How many pounds of solid free filtrate can be delivered before the press is filled with a wet cake
having a density of lb/ft3?
l. A rotary filter turns at the rate of 2 rpm. The fraction of total filtering area immersed in the slurry is
0.20. It has been observed that 1.5ft3 of filtration is delivered per minute per square foot of submerged
area with a given slurry under these operating conditions. If 6 ft 3 of the filtrate is delivered per
revolution by this rotary filter, what is the total filtering area of the filter cloth on the drum?
m. Experimental filtration data for a compressible cake indicate that the specific cake resistance is 3.1 lb f
– h/lb-ft when the pressure difference driving force is 2lb/in 2 and 3.9 lbf –h/lb-ft when the pressure
difference is 5lbf /in2. Estimate the value of the compressibility exponent for the cake.
n. A filtration is carried out in a plate and frame filter press at a constant pressure difference of 3lbf/in 2.
The total filtering area is 80ft2 and the specific cake resistance is 3.5 lbf –h/lb-ft. How many cubic feet
of filtrate will be obtained in 1 hour if the filter cake contains10 lb of dry solids per cubic foot of filtrate?
The resistance of the frame and the cloth may be assumed as negligible and the cake may be
considered as non compressible.
o. Drop of oil 15 microns in diameter are to be settled from their mixture with air. The specific gravity of
the oil is 0.9and the air is at 70  F and 1 atm. A settling of 1 min. is available. How high should the
chamber be to follow settling of these particles? The density of air at 70  F and 1 atm is 0.075 lb/ft3.
The viscosity of the air is 0.018 Cp.
TECHNOLOGICAL INSTITUTE OF THE PHILIPPINES
363 P. Casal St., Quiapo, Manila
Introduction to Particle Technology
Problem Set for Finals
Submitted by:
Dangat, Jennifer G.
Submitted to:
Engr. Robert Delfin
Date Submitted:
March 24, 2015
Screening
1. if the total percentage of the particle larger that the screen opening in the feed,
product and under size are 30%, 85% and 5%, respectively, the effectiveness of the
screen is___%
Ans. 92.3
2. Power of 3 kW is supplied to a machine crushing material at the rate of 0.3 kg/s from
12.5 mm cubes to a product having the following sizes: 80 per cent 3.175 mm, 10
per cent 2.5 mm and 10 per cent 2.25 mm. What power should be supplied to this
machine to crush 0.3 kg/s of the same material from 7.5 mm cube to 2.0 mm cube?
Ans. 3.6kw
3. A screen an aperture of 6 mesh BSS in treating is feed with 66% of +6mesh and
producing an oversize fraction containing 89% of +6mesh particles. If the undersize
fraction contains 2% of +6mesh particles, calculate the effectiveness of the screen..
Ans. 75.59%
4. Table salt is being fed to a vibrating screen at the rate of 150 kg/hr. The desired
product is -39 +20 mesh fraction. A 30 mesh and 20 mesh screen are therefore used
(double deck), the feed being introduced on the 30 mesh screen. During the
operation it was observed that the average proportions of oversize are (from 30
mesh screen): oversize (from 20 mesh screen): undersize (from 20 screen) is
2:1.5:1. Cal culate the effectiveness of the screener from the following data:
Ans. 63. 39 %
5. One ton per hour of dolomite is produced by a ball mill operating in a closed circuit
grinding with a 100 mesh screen. The screen analysis (weight %) is given below.
Calculate the screen efficiency. Data from screen analysis :
Ans. 91.32 %
6. One ton per hour of dolomite is produced by crushing and then screening through a
14 mesh screen. According to the screen analysis (in weight percent) given below,
calculate
a. The total load to the crusher
b. The effectiveness of the screen
XF = 0.543
XD = 0.76
XB = 0
𝐵
𝐹
𝑋𝐷−𝑋𝐹
= 𝑋𝐷−𝑋𝐵
B = 1ton/hr
1
0.76 − 0.543
=
𝐹
0.76 − 0
F = 3.5 ton/hr
E=
E=
(𝑋𝐹−𝑋𝐵)(𝑋𝑃−𝑋𝐹)𝑋𝑃(1−𝑋𝐵)
(𝑋𝐷−𝑋𝐵)2 (1−𝑋𝐹)𝑋𝐹
(0.543−0)(0.76−0.543)(0.76)(1−0)
(0.76−0)2 (1−0.543)(0.543)
E = 62.48 %
7. Removal of free water from a solid-water mixture and generally limited to 4mesh and
above
a. Scalping
c. scalping
b. Trash removal
d. dewatering
Ans. D. dewatering
8. The removal of small amount of oversize from feed w/c are predominantly fines
called
a. Scalping
c. scalping
b. Trash removal
d. dewatering
Ans. C Scalping
9. Another name for revolving screen is
a. Shaking screen
b. Vibrating screen
c. Mechanical vibrated screen
d. Trammel screen
Ans. D trammel sceen
10. With increase in the capacity of screens, the screen effectiveness
a. Decreases
b. Increases
Ans. A. Decrease
c. remains the same
d. increase exponentially
11. As particle size is reduced
a. screening becomes progressively more difficult
b. screening becomes progressively easier
c. capacity and effectiveness of the screen is increased
d. capacity and effectiveness of the screen is decreased
Ans. A
12. Pick out the wrong statement.
a. The capacity and the effectiveness of a screen are the different,
b. The capacity and the effectiveness of screen are opposing factors.
c. The screening surface of a 'reel' (a revolving screen used in flour mills) is made
of silk bolting cloth supported by wire mesh
d. Both b and c
Ans. A
13. Cumulative Analysis for determining surface is more precise than differential
analysis because of the
a. assumption that all particles in single fraction are equal in size
b. fact that screening is more effective
c. assumption that all particles in a single fraction are equal in size is not needed
d. none of these
Ans. C
14. Increasing the capacity of screen
c. decreases the screen effectiveness
effectiveness
d. increases the screen effectiveness
Ans. A
15. Screen efficiency is
a. recovery rejection
b. recovery
c.does not affect the screen
d. none of these
c. rejection
d. none of these
Ans. D
16. As particle size is reduced
e. screening becomes progressively more difficult
f. screening becomes progressively easier
g. capacity and effectiveness of the screen is increased
h. none of these
Ans. A
17. A screen is said to be blinded when
i. oversizes are present in undersize fraction
j. undersizes are retained in oversize fraction
k. the screen is plugged with solid particles
l. its capacity is abruptly increased
Ans. C
Size Reduction
6. If crushing rolls, 1 m in diameter, are set so that the crushing surfaces are 12.5 mm
apart and the angle of nip is 31◦, what is the maximum size of particle which should
be fed to the rolls? If the actual capacity of the machine is 12 per cent of the
theoretical, calculate the throughput in kg/s when running at 2.0 Hz if the working
face of the rolls is 0.4 m long and the bulk density of the feed is 2500kg/m 3.
Ans. 25mm
7. A material is crushed in a Blake jaw crusher such that the average size of particle is
reduced from 50 mm to 10 mm with the consumption of energy of 13.0 kW/(kg/s).
What would be the consumption of energy needed to crush the same material of
average size 75 mm to an average size of 25 mm: assuming Rittinger’s law applies?
Ans. 4.33 kJ/kg
8. A batch grinding mill is charged with material of the composition. The grinding-rate
function Su is assumed to be 0.001/s for the 4/6 mesh particles. Breakage function
Bu is given with b=1.3 both Su and Bu are assumed to be independent of time. How
long will it take for the fraction of 4/6 mesh material to diminish by 10 %?
Ans. 103.3s
9. From the previous problem how long will the values Xn vary with the time during 6 h
of operation? Use a time interval of 30 s in the calculations.
Ans. 0.12285
10. What is the power required to crush 100 ton/h of limestone if 80% of the feed pass a
2-in screen and 80% of the product a 1/8 in screen? The work index for limestone is
12.74.
Ans. 101.9
11. A material is crushed in a jaw crusher and the average size of the particle reduced
from 5cm to 1 cm, with the consumption of energy 1.32 x 10 4 J/kg. What will be the
consumption of energy to crush the same material of an average size of 7.5 cm to
2.5 cm, assuming
a. Rittinger’s Law
b. Kick’s Law
a. Rittinger’s Law
P
1
1
= Kr( − )
M
DP Df
kr =
p/m
=
1.32 x104
1 1
(1 − 5 )
1
1
(Dp) − (Df)
m
Kr = 1.65x104 j.
kg
P
1
1
= 1.65 x104 (
−
)
M
2.5 7.5
𝐏
𝐣
= 𝟒. 𝟒𝐱𝟏𝟎𝟑
𝐌
𝐤𝐠
b. Kick’s Law
P
Df
= Kkln ( )
M
DP
P
2x104
m
Kk =
= 1.
Df
5
ln (DP)
ln (1)
Kk = 8.202x103 j/kg
P
7.5
= 8.202x103 ln ( )
M
2.5
𝐏
𝑱
= 𝟗. 𝟎𝟏𝒙𝟏𝟎𝟑
𝐌
𝒌𝒈
12. Pick out the wrong statement.
a. Size enlargement (opposite of size reduction) is not a mechanical operation.
b. Recycled coarse material to the grinder by a classifier is termed as circulating
load.
c. Wear and tear in wet crushing is more than that in dry crushing of materials.
d. A 'dust catcher' is simply an enlargement in a pipeline which permits the solids to
settle down due to reduction in velocity of the dust laden gas.
Ans. A
13. Size reduction of __________ is accomplished in steam heated rollers and roll
crushers.
a.
Hard rubber
b. Resins
Ans. A hard rubber
c. gums
d. waxes
14. During size reduction by a jaw crusher, the energy consumed decreases with the
a. Increasing size of feed at constant reduction ratio.
b. Decreasing size of product at constant size of feed.
c. decreasing machine capacity
d. increasing machine capacity
Ans . A.
15. A fluid energy mill used
a. Cutting
b. Grinding
Ans. Ultragrinding
c. ultragrinding
d. crushing
16. For course reduction of hard solids, use
a. Impact
b. Attrition
Ans. C. Compression
c. compression
d. cutting
17. Soft and non- abrasive materials can be made into fines by
a. Impact
c. compression
b. Attrition
d. cutting
Ans. B. attrition
18. Crushing efficiency is the ratio of
e. surface energy created by the crushing to the energy absorbed by the solid
f. the energy absorbed by the solid to that fed to the machine
g. the energy fed to the machine to the surface energy created by the crushing
h. the energy absorbed by the solid to the surface energy created by the crushing.
Ans. A
19. Rittinger’s crushing law states that
e. work required to form a particle of any size is proportional to the square of the
surface to volume ratio of the product.
f. work required to form a particle of a particular size is proportional to the square
root of the surface to volume ratio of the product
g. work required in crushing is proportional to the new surface created
h. for a given machine and feed, crushing efficiency is dependent on the sizes of
feed and product
Ans. C
20. Bond crushing law
e. calls for relatively less energy for the smaller product particle than does the
Rittinger law
f. is less realistic in estimating the power requirements of commercial crushes
g. states that the work required to form particle of any size from very large feed is
proportional to the square root of the volume to surface ratio of the product
h. states that the work required for the crushing is proportion
Ans. A
21. The operating speed of a ball mill should be
a. less than the critical speed
speed
b. much more than the critical speed
Ans. A
Flotation
c.
at least equal to the critical
d. none of these
1. A copper ore initially contains 2.09% Cu. After carrying out a froth flotation
separation, the products are as shown in Table 1. Using this data, calculate:
a. Ratio of concentration
.
Ans. 10 tons of feed produce 1 ton of concentrate.
b. % Metal Recovery
Ans. . 95.7%
2. Sample in no. 1 calculate % Metal Loss and % Weight Recovery, or % Yield
a. % Metal Loss
Ans. 4.3%
b. % Weight Recovery
Ans. 10%
3. A flotation plant processes 4000tons/day of CuFeS2. It produces 70 ton of Cu
concentration assaying 37% Cu if ore analysis 0.9 Cu, what is the percent recovery?
Ans. 68.1 %
4. Calculate the specific flotation rate constant for 100um galena particles on 0.5mm
diameter bubbles in a continuous floatation cell that is aerated at 0.013m 3air/m
pump. The bubble residence time is 5s. Consider the case where the average
induction time for the particles is 100ms. Calculate the fraction of the non-floatable
coponent for these particles. Consider the case when the bubble is carrying 5 galena
particles.
Ans. C. 0.0255
5. For a laboratory flotation of an iron ore in water, it was observed that 2 mg was
collected while traversing 2 m of the flotation column. The concentration of the ore in
water was 0.5 kg/m3 . The average diameter of the bubbles was 2 mm and the
average diameter of the particles was 0.1 mm. Compute flotation recovery.
Ans. A. 0. 578
6. Flotation agent that prevents coalescence of air bubles as they travel to the surface
of the water is/are
a. Collector
b. Frothing agent
c. Promoter
d. Modifying agent
Ans. B. Frothing agent
7. A flotation modifier which assist in the selectivity (sharpness of separation) or stop
unwanted minerals from floating
a. Activator
b. Alkalinity regulators
c. Depressant
d. Regulator
Ans. C. depressant
8. Which of the following is an example of an deflocculant
a. Sulfuric acid
b. Lignin sulforate
c. C. dithiophosphate
d. Molybderite
Ans. B. Lignin sulforate
9. The following is a typical anionic collector used in flotation
a. Potassium ethyl xanthate
b. Ethyl dixanthogen
c. Trimethyl cetyl ammonium bromide.
d. lAuryl amine hydrochloride.
Ans. A Potassium ethyl xanthate
10. froth floatation process may be used to increase the concentration of the mineral in
a. chalcopyrites
c. haematie
b. Bauxite
d. calmine
Ans. C. haematie
11. which one of the ff ores is best concentrated by froth floatation method
a. cassiterite
c. malachite
b. Galena
d. magnetite
Ans. B Galena
12. flotation is also called
a. flotage
c. froth flotation
b.
floters
ans. A. flotage
d. floater
13. In Froth Flotation, chemical agent added to cause air adherence is called
g. collector
c. modifier
h. frother
d. promoter
Ans. A Collector
14. Pine oil used in forth flotation technique acts as a
g. collector
c. frother
h. modifier
d. activator
Ans. C frother
15. Which of the following is the most suitable for cleaning of fine coal dust (< 0.5 m)?
g. Through washer
c. Spiral separator
h. Baum Jig Washer
d. Froth Flotation
Ans. D. Froth Flotation
Sedimentation
1. A mixture of coal and sand particles having sizes smaller than 1x10 -4 m in diameter is
to be separated by screening and subsequent elutriation with water. Recommend a
screen aperture such that the oversize from the screen can be separated completely
into sand and coal particles by elutriation. Calculate also the reqiured water velocity.
Assume that Stokes law is applicable.
Density of sand = 2650 kg/m3 Density of coal = 1350 kg/m3
Density of water = 1000 kg/m3
Viscosity of water = 1x10-3kg/m.s
Recommend a screen aperture such that the oversize from the screen can be
separated completely into sand and coal particles by elutriation
Ans. 4.61 x10-5 m
2. The water required water velocity
Ans. 1.91 x10-3 m/s
3. Slurry containing 0.2 kg of solids per kg of water is to be thickened to a sludge
containing 0.70 kg solids per kg of water in a continuous settling process. With five diff.
concentration of the slurry the following test results were obtained:
What would be the minimum area of thickener to effect a separation at the rate of 0.625
kg of solids per second?
Ans. 14. 57 m2
4. Solid spherical particles of coffee extract from a dryer having a diameter of 400µm
are falling through air at a temperature of 422k. The density of the particles is
1030kg/m3 . Calculate the terminal settling velocity and the distance of fall in 5s. The
pressure is 101.32 kPa.
Terminal settling velocity
Ans. 1.49 m/s
5.Distance of fall in 5s
Ans. 7.4m/s
6. Calculate the maximum velocity at which spherical particles of silica of 0.05mm in
diameter will fall through still water of temperature 75C. Assume stokesian condition
Data given: special gravity of silica = 2.70
Special gravity of water = 1.0
Viscosity at 75C = 0,30 Cp
gd2 (𝑃𝑠 − 𝑃𝑙)
Vm =
18µ
960(5x10−3 )2 (2.7 − 1.0)
Vm =
18(0.0030)
𝐕𝐦 = 𝟎. 𝟕𝟕𝐜𝐦/𝐬
7. Pick up the incorrect statement from the following :
a. Detention period for plain sedimentation tanks ranges between 6 to 10 hours
b. Detention period for sedimentation tanks, using coagulants usually ranges
between 2 to 4 hours
c. The horizontally flow velocity in sedimentation tanks, is generally limited to 0.3
m/minute
d. All the above.
Ans. A
8. In sewage treatment, its sedimentation is speeded up by commonly adding
a. Lime
c. copper sulphate
b. Hydrochloric acid
d. sodium sulphate
Ans. lime
9. Dorr thickener is an equipment used for
a. Sedimentation
b. Clarification
Ans. Sedimentation
c. leaching
d. drying
10. Depositon within a meandering stream ussually occurs on the inside of the curves
because the
a. Water velocity decreases
b. Stream gradient increases
c. Water is deeper
d. Stream is narrower
Ans. A. Water velocity decreases
11. The most sandstone bedrock is composed of sediment that was
a. Sorted by size and not layered
b. Sorted by size and layered
c. Unsorted and not layered
d. Unsorted and layered
Ans. B. Sorted by size and layered
12. What is the largest sediment that can be transported by a stream that has a velocity
of 125cm/s.
a. cobbles
b. pebbles
c. sand
d. clay
Ans. B. Pebbles
13. Forces acting on a particle settling in fluid are
i. gravitational and buoyant forces
j. centrifugal and drag forces
k. gravitational or centrifugal, buoyant and drag forces
l. external, drag and viscous forces
Ans. C.
14. Terminal velocity is
m. constant velocity with no acceleration
n. a fluctuating velocity
o. attained after moving one-half of total distance
p. none of these
Ans. A
15. In hindered settling, particles are
g. placed farther from the wall
h. not affected by other particles and the wall
Ans. C near each other
c. near each other
d. none of these
16. Drag coefficient in hindered settling is
i. less than in free settling
j. equal to that in free settling
k. not necessarily quarter than in free settling
l. greater than in free settling
Ans. D greater than in free settling
Centrifugation
1. In a test on a centrifuge all particles of a mineral of density 2800 kg/m3 and of size 5
μm, equivalent spherical diameter, were separated from suspension in water fed at a
volumetric throughput rate of 0.25 m3/s. Calculate the value of the capacity factor.
What will be the corresponding size cut for a suspension of coal particles in oil fed at
the rate of 0.04 m3/s? The density of coal is 1300 kg/m3 and the density of the oil is
850 kg/m3 and its viscosity is 0.01 Ns/m2.
Ans. 4 μm.
2. When an aqueous slurry is filtered in a plate and frame press, fitted with two 50 mm
thick frames each 150 mm square, operating with a pressure difference of 350
kN/m2, the frames are filled in 3600 s (1 h). How long will it take to produce the
same volume of filtrate as is obtained from a single cycle when using a centrifuge
with a perforated basket, 300 mm diameter and 200 mm deep? The radius of the
inner surface of the slurry is maintained constant at 75 mm and the speed of rotation
is 65 Hz (3900 rpm). It may be assumed that the filter cake is incompressible, that
the resistance of the cloth is equivalent to 3 mm of cake in both cases, and that the
liquid in the slurry has the same density as water.
Ans. 4.25 min.
3. Two centrifuges rotates at the same peripheral velocity of 53.34 m/s. the first bowl
has a radius of r1 = 76.2 mm and the second r2 = 305mm. Calculate the revolution
per min in each bowl.
Ans. 1670 rev/min
4. How fast (in rpm's) must a centrifuge rotate if a particle 7.49 cm from the axis of
rotation is to experience an acceleration of 86,000 g's?
Ans. 320312 rpm
5. If a centrifuge is 0.9 m diameter and rotates at 20 Hz, at what speed should a
laboratory centrifuge of 150 mm diameter be run if it is to duplicate the performance
of the large unit?
Ans. 49 Hz
6. A centrifuge of diameter 0.2 m in a pilot plant rotates at a speed of 50 Hz in order to
achieve effective separation. If this centrifuge is scaled up to a diameter of 1 m in the
chemical plant, and the same separation factor is to be achieved, what is the
rotational speed of the scaled up centrifuge?
Solution:
Separation factor
S=
𝜔2 𝑟
𝑔
ω1 2 r1
g
=
ω2 2 r2
g
𝑟
0.2
𝑟2
1
𝜔2 = 𝜔1 √ 1 = 50 x √
𝝎𝟐 = 22.36 Hz
7. The yeast generated during the fermentation of beer is generally separated by
a. Centrifugation
c. sedimentation
b. Filtration
d. extraction
Ans. A. Centrifugation
8. Viruses can be purified based on their size and density by using
a. Gradient centrifugation
c. precipitation
b. Differential centrifugation
d. coagulation
Ans. A. Gradient centrifugation
9. If a force greater than that of gravity is used to separate solids & fluids of different
densities, the process is termed as the
a. Centrifugation
c. dispersion
b. Sedimentation
d. coagulation
Ans. A centrifugation
10. Which of the following separations cannot be carried out using a centrifuge? The
separation of .
a. water from wet clothes
d. red blood cells from plasma
b. salt from sea water
c. cream from milk
Ans. salt from sea water
11. For separation of sugar solution from settled out mud we use
g. sparkler filter
c. centrifugal filter
h. plate & frame filter
d. rotary drum vacuum filter
Ans. C. centrifugal filter
12. Moisture can be removed from lubricating oil using
g. tubular centrifuge
c. sparkler filter
h. clarifier
d. vacuum leaf filter
Ans. A. tubular centrifuge
13. Which of the following can be most effectively used for clarification of tube oil and
printing ink?
r. sparkler filter
c. disc-bowl centrifuge
s. precoat filter
d. sharpless supercentrifuge
Ans. D. sharpless supercentrifuge
14. If the radius of a basket centrifuge is halved and the rpm is doubled, then
m. linear speed of the basket is doubled
n. linear speed of the basket is halved
o. centrifugal force is doubled
p. capacity of centrifuge is increased
Ans. D. centrifugal force is doubled
15. Where the difference in density of the two liquid phases to be separated is very
small (as in milk cream separator), the most suitable separator is
p. disc bond centrifuge
c. batch basket centrifuge
q. sharpless supercentrifuge
d. sparkler filter
Ans. A disc bond centrifuge
16. which of the ff separation process method is suited for a protein sample with large
differences in molecular mass
a. dialysis
b. salting out process
c. density gradient centrifugation
d. rate zonal centrifugation
Ans. D
Technological institute of the Philippines
363 p. Casal st. Quiapo Manila
PROBLEM SET AND CONCEPT
IN
PARTICLE TECHNOLOGY
Submitted to:
Engr. Robert Delfin
Submitted by:
De leon, Jerome C.
MARCH 25, 2015
Size reduction concept
1) Pick out the wrong statement.
a. Recycled coarse material to the grinder by a classifier is termed as circulating load.
b. Wear and tear in wet crushing is more than that in dry crushing of materials.
c. Size enlargement (opposite of size reduction) is not a mechanical operation.
d. A 'dust catcher' is simply an enlargement in a pipeline which permits the solids to settle down
due to reduction in velocity of the dust laden gas.
2) Which of the following relationships between co-efficient of friction (μ) between rock & roll and a (half of
the angle of nip) of the particle to be crushed is correct?
a. μ > tan α
b. μ ≥ tan α
c. μ > tan 2α
d. μ ≤ tan α
3) The distribution given by microscopic analysis of powder is
a. number
b. length
c. area
d. volume
4) For a non-spherical particle, the sphericity
a. is defined as the ratio of surface area of a sphere having the same volume as the particle
to the actual surface area of the particle.
b. has the dimension of length.
c. is always less than 1.
d. is the ratio of volume of a sphere having the same surface area as the particle to the actual
volume of the particle.
5) Rittinger's number designates the new surface created per unit mechanical energy absorbed by the
material being crushed. Larger value of Rittinger's number of a material indicates its
a. easier grindability
b. poor grindability
c. high power consumption in grinding
d. none of these
6) Bond crushing law
a. Calls for relatively less energy for the smaller product particles, than does the Rittinger
law.
b. is less realistic in estimating the power requirements of commercial crushers.
c. states that the work required to form particle of any size from very large feed is proportional to
the square root of the volume to surface ratio of the product.
d. states that the work required for the crushing is proportional to the new surface created.
7) The unit of specific cake resistance is
a. gm/cm2
b. cm/gm
c. cm/gm2
d. gm/gm
8) Which of the following is the hardest material?
a. Calcite
b. Quartz
c. Corundum
d. Gypsum
9) Size measurement of ultrafine particles can be best expressed in terms of
a. centimetre
b. screen size
c. micron
d. surface area per unit mass
10) rpm of a trommel at critical speed is given by (where, D = Diameter of trommel in ft)
a.
b.
c.
d. 76.75 D
11) Maximum size reduction in a fluid energy mill is achieved by
a. compression
b. interparticle attrition
c. cutting
d. impact
12) Which of the following comes in the category of primary crusher for hard and tough stone ?
a. Jaw crusher
b. Cone crusher
c. Gyratory crusher
d. None of these
13) During size reduction by a jaw crusher, the energy consumed decreases with the
a. decreasing size of product at constant size of feed.
b. decreasing machine capacity.
c. increasing size of feed at constant reduction ratio.
d. None of these
14) Reciprocal of sphericity is termed as the
a. specific surface ratio
b. shape factor
c. sauter diameter
d. surface area per unit mass
15) The optimum moisture content in solids to be crushed/ground ranges from __________ percent.
a. 3 to 4
b. 8 to 10
c. 10 to 15
d. 15 to 20
Screening Concept
1) Pick out the wrong statement
a. Hammer crushers operate by impact action.
b. Standard screens have circular opening.
c. With increase in mesh number of screens, their diameter in microns decreases.
d. 200 mesh screen has 200 openings per linear cm.
2) The ratio of the actual mesh dimension of Taylor series to that of the next smaller screen is
a. 1
b. 2
c. 1.5
d. 3
3) Screen capacity is proportional to (where, S = screen aperture)
a. S
b 1/S
c. S2
d. 1x S
4) Which of the following is not industrial screening equipment?
a. Sharpies centrifuge
b Vibrating screen
c. Grizzly
d. Trommel
5) The ratio of the area of openings in one screen (Taylor series) to that of the openings in the next smaller
screen is
a. 1.5
b 1
c. 2
d. none of these
6) 200 mesh seive size corresponds to __________ microns.
a. 24
b 74
c. 154
d. 200
7) Vibrating screens have capacity in the range of __________ tons/ft 2 .mm mesh size
a. 0.2 to 0.8
b. 5 to 25
c. 50 to 100
d. 100 to 250
8) Wet sieving is employed, when the product contains __________ materials.
a. abrasive
b. large quantity of very fine
c. coarse
d.non-sticky
9) The opening of a 200 mesh screen (Taylor series) is
a. 0.0074 cm
b. 0.0074 mm
c. 0.0047 cm
d. 74 mili-microns
10) Trommels are revolving screens which normally operate in the range of __________ rpm
a. 1 - 2
b. 15 - 20
c. 40 - 50
d. 60 - 75
11) In screen analysis, the notation +5 mm/-10 mm means particles passing through
a. 10 mm screen and retained on 5 mm screen.
b. 5 mm screen and retained on 10 mm screen.
c. both 5 mm and 10 mm screens.
d. neither 5 mm nor 10 mm screen.
12) As particle size is reduced
a. screening becomes progressively more difficult.
b. screening becomes progressively easier.
c. capacity and effectiveness of the screen is increased.
d. none of these.
13) Mesh indicates the number of holes per
a. square inch
b. linear inch
c. square foot
d. linear foot
14) increasing the capacity of a screen __________ the screen effectiveness.
a. decreases.
b increases
c. does not effect
d. none of these
15) Screen capacity is not a function of
a. its openings size.
b screening mechanism.
c. screening surface.
d. atmospheric humidity.
Sedimentation concept
1) Coal washing waste water containing about 3% suspended solids (comprising of clay, slate, stone etc.)
is treated for solid particles removal
a. by chemical coagulation
b in sedimentation tanks equipped with mechanical scrapper.
c. in vacuum filter.
d. in clarifiers.
2) Suspended solid present in the waste water generated in blast furnace gas cooling and cleaning plant is
removed by
a. biological oxygen pond.
b radial settling tank (thickener) using coagulant (lime & ferrous sulphate).
c. lagoons.
d. filtration
3) The amount of chemical coagulant added for treatment of polluted water __________ with increase in
temperature of the polluted water to be treated.
a. decreases.
b. increases.
c. remains constant.
d. may increase or decrease ; depends on the chemical characteristics of polluted water.
4) Particles having diameter greater than 75 μm (micrometer = 10-6 mm) are called
a. grit
b. dust
c. powder
d. smoke
5) In a sedimentation tank, the detention period for water ranges from __________ hours.
a. 2 to 4
b. 8 to 12
c. 16 to 20
d. 24 to 32
6) Pick out the one which is not a chemical coagulant.
a. Aluminium sulphate
b. Ferrous sulphate
c. Hydrated lime
d. Chloramine
7) Which of the following is the most efficient for removal of very finely divided suspended solids and
colloidal matter from the polluted water stream ?
a. Sedimentation tank
b. Circular clarifier
c. Mechanical flocculation
d. Chemical coagulation
8) in continuous thickeners, separation of solid particles can be achieved if the settling velocity of the solid
is _____ compared to the velocity of the displaced liquid
a. Aluminium sulphate
b. Ferrous sulphate
c. Hydrated lime
d. Chloramine
9) in order for a particle to move through a fluid under the influence of gravity, there must be
a. velocity difference
b. pressure difference
c. density difference
d. temperature difference
10) The ratio of drug force per unit area to the product of fluid density and the velocity head is called
a. buoyant coefficient
b. drag coefficient
c. density difference
d. temperature difference
Flotation concept
1) pine oil used in forth flotation technique acts as a
a. collector
b. modifier
c. frother
d. activator
2) in froth flotation, the chemical agent added to cause air adherence is called
a. collector
b. modifier
c. frother
d. activator
3) froth flotation is most suitable for treating
a. iron ores
b. sulfides ores
c. quartz
d. metal ores
4) which of the following is an example of a deflocculant ?
a. sulfuric acid
b. lignin sulforate
c. dithiophosphate
d. molybderite
5) an example of a collector for flotation of metallic sulfides and native metals is
a. xanthates
b. sodium silicates
c. sodium sulfides
d. sphalerite
6) dispersants are important for the control of slimes which sometimes interfere with the selectivity and
increase reagent consumption. Another term for dispersants is
a. deflocculant
b. depressants
c. frothers
d. regulators
7) a flotation modifier which assist in the selectivity (sharpness of separation ) or stop unwanted minerals
from floating
a. collector
b. modifier
c. frother
d. activator
8) the flotation agent that prevents coalescence of air bubbles as they travel to the surface of the water
is/are
a. collector
b. frothing agent
c. promoters
d. modifying agent
9) Any operation in which one solid is separated from another by floating one of them at or on the surface
of a fluid
a. coagulation
b. flotation
c. centrifugation
d. sedimentation
10) What is the selectivity index, if the grade of tailings & concentrate is the same ?
a. 0
b. ∞
c. 1
d. 0.5
Centrifugation concept
1) Tabular bowl centrifuges as compared to disk bowl centrifuges
a. operate at higher speed
b. employ bowl of larger diameter.
c. cannot be operated under pressure/vacuum
d. can't be used for separation of fine suspended solids from a liquid
2) __________ centrifuge is normally used in sugar mills.
a. Tubular bowl
b. Disc-bowl
c. Suspended batch basket
d. Perforated horizontal basket continuous
3) Ultra centrifuges are used for the separation of __________ solid particles.
a. coarse
b. fine
c. colloidal
d. dissolved
4) Where the density difference of the two liquid phase to be separated is very small (as in milk cream
separator), the most suitable separator is a
a. disc bowl centrifuge.
b. sharpies supercentrifuge.
c. batch basket centrifuge
d. sparkler filter
5) Moisture can be removed from lubricating oil using
a. tubular centrifuge
b. clarifier
c. sparkler filter
d. vacuum leaf filter
6) Separation of isotopes is generally done using a/an __________ centrifuge
a. ultra
b. disk-bowl
c. both (a) & (b)
d. neither (a) nor (b)
7) Paddle agitator
a. is suitable for mixing low viscosity liquids
b. produces axial flow
c. moves at very high speed.
d. none of these.
8) Which of the following can be most effectively used for clarification of lube oil and printing ink?
a. Sparkler filter
b. Precoat filter
c. Disc-bowl centrifuge
d. Sharpies supercentrifuge
9) Ultracentrifuges running at speeds upto 100000 rpm is normally used for the
a. separation of isotopes based on their density or molecular weights difference.
b. concentration of rubber latex
c. separation of cream from milk.
d. dewaxing of lubricating oil.
10) if the radius of a basket centrifuge is halved and the rpm is doubled, then the
a. linear speed of the basket is doubled
b. linear speed of the basket is halved
c. batch basket centrifuge
d. sparkler filter
Problem solving
1) Two very small silica particles are settling at their respective terminal velocities through a highly viscous
oil column. If one particle is twice as large as the other, the larger particle will take ………the time taken by
the smallest particle to fall through the same height.
a. ¼ t
b. ½ t
c. 2/3 t
d. ¾ t
2) Sugar is ground from crystals of which it is acceptable that 80% pass a 500 mm sieve(US Standard
Sieve No.35), down to a size in which it is acceptable that 80% passes a 88 mm (No.170) sieve, and a 5horsepower motor is found just sufficient for the required throughput. If the requirements are changed such
that the grinding is only down to 80% through a 125 mm (No.120) sieve but the throughput is to be
increased by 50% would the existing motor have sufficient power to operate the grinder? Assume Bond's
equation.
a) 5.4 hp.
b) 3hp
c) 2hp
d) 1hp
3) If a cream separator has discharge radii of 5 cm and 7.5 cm and if the density of skim milk is 1032 kg m 3 and that of cream is 915 kg m-3, calculate the radius of the neutral zone so that the feed inlet can be
designed.
For skim milk, r1 = 0.075m,rA = 1032 kg m-3, cream r2 = 0.05 m,rB= 915 kg m-3
a) 17cm
b) 5cm
c) 8cm
d) 15cm
4) It is desired to remove particles from a crushed stone mixture by screening through 10-mesh. The
screen analyses of the feed, oversize and undersize are given.
a) Calculate the mass ratio of the overflow and the underflow to feed (ans. 0.58)
b) Find the effectiveness of the screen (ans. 0.669)
mesh
feed
4
0
6
0.025
oversize
undersize
0
0
0.071
0
8
0.125
0.43
0
10
0.32
0.85
0.805
14
0.26
0.97
0.92
20
0.155
0.99
0.17
28
0.055
1.00
0.09
35
0.02
0
0.06
65
0.02
0
0.025
0
0.02
0
0
5) a viscous solution containing particles with a density of 1461 kg/m 3 is to be clarified by centrifugation.
The solution density is 801 kg/m3 and its viscosity is 100 cP. The centrifuge has bowl with r2 =
0.02225m, r1 = 0.00715 and height b = 0.1970m. calculate the critical particle diameter of the
largest particles in the exit stream if N= 23,000 rev/min and flowrate of 0.002832 m 3/h.
ANS. Dpc = 0.746μm
6) if a centrifuge is 3-ft diameter & rotates at 1000 rpm, what must be the speed of a laboratory centrifuge
of 6-in diameter be ran if its duplicate plant condition?
Given :
@ 1 lab centrifuge
D1 = 3 ft
N1 = 1000 rpm
1
@2 lab centrifuge 6(2)ft
N2 =?
SOLUTION:
Ut1 = Ut2
(
2𝜋𝑁1
𝐷1
) 2𝑥
𝑥
60
2
Dp2 x (ρp −ρ)
18𝜇
=
(
2𝜋𝑁2
𝐷2
) 2𝑥
𝑥 Dp2 x (ρp −ρ)
60
2
18𝜇
N12 D1 = N22 D2
D1
N2 = N1 √D2
= 1000 rpm (√
3 ft
1
2
6( )ft
N2 = 2449.49 Ans.
)
TECHNOLOGICAL INSTITUTE OF THE PHILIPPINES
363 P. CASAL ST., QUIAPO, MANILA
INTRODUCTION TO PARTICLE TECHNOLOGY
PROBLEM SET
SUBMITTED BY:
DELA TORRE, JUDY ANN T.
SUBMITTED TO:
ENGR. ROBERT DELFIN
MARCH 24, 2015
SCREENING
1.
Making a size separation smaller than 48-mesh is called
a. coarse separation
c. ultrafine separation
b. fine separation
d. scalping
2. The mesh number of a screen denotes
a. the area of the screen in square inch
b. the number of openings per linear inch of screen
c. the number of layers in a screen system
d. the number of layers in a screen system
3. Materials which remain on a screen surface are called
a. Fines
c. Intermediate material
b. Undersize
d. oversize
4. Increasing the capacity of a screen
a. Decreases the screen effectiveness
b. Increases the screen effectiveness
c. Does not affect the screen effectiveness
d. None of these
5. As the particle size is reduced
a. Screening becomes progressively more difficult
b. Screening becomes progressively easier
c. Capacity and effectiveness of the screen is increased
d. None of these
6. The screen is said to be blinded when
a. Oversize are present in undersize fraction
b. Undersize are retained in oversize fraction
c. The screen is plugged with solid particles
d. Its capacity is abruptly increased
7. Trommels separate a mixture of particles depending on their
a. Size
c. Screen size
b. Wettability
d. Electrical and magazine
8. The minimum clear space between the edges of the opening in the screening surface and is usually
given in inches or millimeters.
a. Sieve
c. Mesh number
b. Aperture
d. Vibrating screen
9. Box-like machines, either round or square with a series of screen clothes nested atop one another.
a. Reciprocating screen
c. Electricity vibrated screen
b. Oscillating screen
d. Gyratory screen
10. The material passing one screening surface and retained on a subsequent surface is called
a. Intermediate material
c. Plus material
b. Minus material
d. None of these
11. The size distribution of a dust as measured by a microscope is as follows. Convert these data to
obtain the distribution on a mass basis, and calculate the specific surface, assuming spherical
particles of density 2650 kg/m3.
Size range (µm)
Number of particles in range (−)
0–2
2000
2–4
600
4–8
8–12
12–16
16–20
20–24
140
40
15
5
2
Ans 0.732 x 106 m2/m3
12. A sample of dust from the air in a factory is collected on a glass slide. If dust on the slide was
deposited from one cubic centimeter of air, estimate the mass of dust in g/m3 of air in the factory,
given the number of particles in the various size ranges to be as follows:
Size range (µm)
Number of particles (−)
0–1
2000
1–2
1000
2–4
500
4–6
200
6–10
100
10–14
40
It may be assumed that the density of the dust is 2600 kg/m3 , and an appropriate allowance should
be made for particle shape. Ans 0.25 g/m3
13. 1800 lbs of dolomite per hour is produced by crushing and then screening through a 14-mesh screen.
The screen analysis is as follows:
Tyler Mesh
Feed to Screen
Undersize Product
4 on
8 on
14 on
28 on
48 on
100 on
100 through
14.3
20.0
20.0
28.5
8.6
5.7
2.86
40
30
20
10
Screen Oversize
Circulating load
20
28
24
24
0 through 24 mesh
(a) The total load to the crusher is
(b) The effectiveness of the screen
Ans. (a) 6320 lbs/hr ; (b) 62.5%
14. Fine silica is fed at 1500 lbs/hr to a double deck vibrating screen combination to obtain a 48/85 mesh
(Tyler) product. The silica feed is introduced into the upper screen of the 48-mesh and the product is
discharged off the surface of the lower screen of 65 mesh. During the screening operation, the ratio of
1
oversize to product to undersize is 2:1: .
2
Laboratory Analyis of the different fractions:
Screen Mesh
Feed Mass
Oversize Mass
Fraction
Fraction
10/14 to 28/35
0.2821
0.5855
35 /48
0.2580
0.3370
48/65
0.2810
0.0660
65/100
0.0910
0.0050
100/150 to
0.0870
0.0060
150/200
Product Mass
0.3385
0.3220
0.5260
0.0670
0.0260
Undersize Mass
Fraction
0.00453
0.00360
0.34400
0.29900
0.35300
(a) The effectiveness of the screening equipment is
(b) If the screen measures 5ft x 8ft each, the capacity in MT/day-ft2-mm of the 65 mesh screen on the
basis of a perfectly functioning 48 mesh screen is
(c) The capacity in MT/day-ft2-mm on the basis of actual performance of the 48 mesh screen
Ans. (a) 48.7% ; (b) 0.901 ; (c) 1.09
15. A sand mixture was screened through a standard 10-mesh screen. The mass fraction of the oversize
material in feed, overflow and underflow were found to be 0.38, 0.79 and 0.22 respectively. The
screen effectiveness based on the oversize is. Ans. 0.58
16. If the total percentage of particles larger than the screen opening in the feed, product and undersize is
25%, 90%, and 5% respectively, calculate the effectiveness of the screen.
Given:
XF = 0.25
XP = 0.90
XR = 0.05
Required:
Effectiveness of the screen
Solution:
(𝑋 −𝑋 )(𝑋 )
(𝑋 −𝑋 )(1−𝑋 )
𝐸 = (𝑋𝐹 −𝑋𝑅 )(𝑋𝑃 ) [1 − (𝑋𝐹 −𝑋𝑅 )(1−𝑋𝑃 )]
𝑃
𝐸=
𝑅
𝐹
𝑃
𝑅
𝐹
(0.25 − 0.05)(0.90)
(0.25 − 0.05)(1 − 0.90)
[1 −
(0.90 − 0.05)(0.25)
(0.90 − 0.05)(1 − 0.25)
𝑬 =0.8205 or 82.05%
SIZE REDUCTION
1. _________ states that the energy required for crushing is proportional to the new surface created.
a. Rittinger’s Law
c. Bond Law
b. Kick’s Law
d. Energy Law
2. It is defined as the efficiency of technical grinding compared with that of laboratory crushing experiments.
a. Grinding efficiency
c. Practical Energy Efficiency
b. Bond Work Index
d. None of these
3. In comminution, the energy requirement is determined theoretically by
a. The initial and final sizes of the particles
b. The type of equipment
c. The change in shape of the particle
d. None of these
4. The operating speed of a ball mill should be
a. Less than the critical speed
b. Much more than the critical speed
c. At least equal to the critical speed
d. None of these
5. The term applied to all ways in which particles of solids are cut broken into smaller pieces.
a. Size reduction
c. Comminution
b. Screening
d. Crushing
6. Ideal grinder/ crusher
a. Have a large capacity
b. Require a small power input per unit of product
c. Yield a product of the single size or the size distillation required
d. All of these
7. Size reduction is important in chemical engineering since
a. It prevents chemical engineers from becoming overweight
b. It makes products to become uniform in size
c. It prepares raw materials of the desired sizes prior to processing
d. None of these
8. For coarse reduction of hard solids, use
a. Impact
c. Compression
b. Attrition
d. cutting
9. Based on Bond’s Crushing Law, the power required to crush a certain material will change by ______ %
if the diameter of the product is made smaller by 50%.
a. 50%
c. 25%
b. 41%
d. 75%
10. Soft and non-abrasive materials can be made into fines by
a. Attrition
c. Cutting
b. Compression
d. None of these
11. In crushing a certain ore, the feed is such that 80% is less than 50.8 mm size and the product size is
such that 80% is less than 6.35 mm. The power required is 89.5 kW. Based on the Bond equation the
power required using the same feed so that 80% is less than 3.18 mm is______. Ans. 146.7 kW
12. If 20% of pulverized limestone with a work index of 12.74 kW-h/ton is retained by a 150-mesh screen
from an original uniform size of 35 mesh, the energy required in kW to process 1 ton/hr is __. Ans. 6.25 kW
13. A material is crushed in a Blake Jaw Crusher and the average size of particles reduced from 50 mm to
10 mm with the consumption of energy needed to crush the same material of an average size 75 mm to an
average size 25 mm assuming Kick’s Law applies is ____. Ans. 8.88 kW/(kg/s)
14. If it is desired to reduce the separation time for milk to at least one week (before cream will rise to the
top), what maximum diameter of cream droplet would Stokes' Law predict to be necessary for the
homogenization to achieve? Assume the depth is 10 cm. Ans. 0.0567 µm
15. It is found that the energy required to reduce particles from a mean diameter of 1 cm to 0.3 cm is 11 kJ
kg-1. Estimate the energy requirement to reduce the same particles from a diameter of 0.1 cm to 0.01 cm
assuming:
(a) Kick's Law,
(b) Rittinger's Law,
(c) Bond's Equation.
Ans. a) 21 kJ/kg
b) 423 kJ/kg
c) 91 kJ/kg
16. It is desired to crush 100 ton/hr of phosphate rock from a feed size where 80% is less than 4 in. to a
product where 80% is less than 1/8 in. The work index is 10.13 (P1).
a. calculate the power required.
b. calculate the power required to crush the product further where 80% is less than 1000 µm.
given:
T = 100 ton/hr
X1 = 4 in = 0.3333
X2 = 1/8 in = 0.0104 ft
Ei = 10.13
@80% , X2 = 1000 µm = 1000 x 10—6 m = 3.281 x 10 -3 ft
required:
a) Power
b) Power at X2 =1000 µm
solution: (a)
𝑃
1
1
= 1.46 𝐸𝑖 (
−
)
𝑇
√𝑥2 √𝑥1
𝑃
1
1
= 1.46 (10.13)(
−
)
𝑔𝑎𝑙 1 𝑚𝑖𝑛
𝑓𝑡
√0.0104
√0.3333𝑓𝑡
100 𝑚𝑖𝑛 𝑥 60 𝑠𝑒𝑐
𝑷 = 𝟏𝟗𝟖. 𝟗𝟗 𝒉𝒑
(b)
𝑃
1
1
= 1.46 𝐸𝑖 (
−
)
𝑇
√𝑥2 √𝑥1
𝑃
1
1
= 1.46 (10.13)(
−
)
𝑔𝑎𝑙 1 𝑚𝑖𝑛
−3 𝑓𝑡
𝑥
10
√3.281
√0.3333𝑓𝑡
100 𝑚𝑖𝑛 𝑥 60 𝑠𝑒𝑐
𝑷 = 𝟑𝟖𝟕. 𝟔𝟒 𝒉𝒑
FLOTATION
1. Unit operation of flotation is based on major step(s)
a. Conditioning
c. Both a and b
b. Separation
d. Neither a nor b
2. Type of chemical reagent(s) used during the froth flotation process
a. Collectors
c. Modifiers
b. Frothers
d. All of these
3. ______ are added to strengthen temporarily covering film of the air bubbles.
a. Collectors
c. Modifiers
b. Frothers
d. None of these
4. Any operation in which one solid is separated from another by floating one of them at or the surface of a
fluid.
a. Coagulation
c. Centrifugation
b. Flotation
d. sedimentation
5. Froth flotation is most suitable for treating
a. Iron ores
c. Quartz
b. Sulfide ores
d. Metal ores
6. Pine oil used in a flotation process acts as a
a. Collector
c. Frother
b. Modifier
d. activator
7. A flotation modifier which assists in the selectivity or stop unwanted minerals from floating
a. Depressants
c. Alkalinity regulators
b. Activators
d. promoters
8. The flotation agent that prevents coalescence of air bubbles as they travel to the surface of the water is
a. Collectors
c. Frothing agent
b. Promoters
d. Modifying agent
9. An example of a collector for flotation of metallic sulfides and native metals is
a. Xanthates
c. Sodium sulfide
b. Sodium silicate
d. sphalerite
10. Which of the following is an example of a deflocculant?
a. Sulfuric acid
c. Dithiophosphate
b. Lignin sulforate
d. molybderite
11. A flotation plant processes 3000 tons/day of CuFeS2. It produces 80 tons Cu concentrate assaying
25% Cu. If ore analyzes 0.7% Cu, the percent recovery is? Ans. 94%
12. Ground lead ore is to be concentrated by a single flotation process using 1.5 oz of reagent per ton of
ore. The feed concentrate and tailings have the following composition by weight on a dry basis
Feed
Concentrate
Tailings
PbS
30
90
0.9
ZnS
35
3
35.6
SiO2
45
7
63.5
Water is fed to the cell at the rate of 1000 gallons per ton of wet concentrate with 99% of the water leaving
with the tailings and 1% with the concentrate. Find the mass of wet concentrate produced per hour when
ten tons of ore are fed to the cell / 24 hr. is? Ans. 32.4 kg/hr
13. For a laboratory flotation of an iron ore in water, it was observed that 2 mg was collected while
traversing 2 m of the flotation column. The concentration of the ore in water was 0.5 kg/m3. The average
diameter of the bubbles was 2 mm and the average diameter of the particles was 0.1 mm. Compute
flotation recovery. Ans. 0.578
14. For a flotation operation, the net feed is 2000 tons/day to the rougher. From the following data, compute
the capacity and number of flotation cells in each unit and the power consumption. Compute for the volume
of the rougher and the scavenger.
SiO2
CuS
Feed (a)
98
2
Tailings from Rougher. (b)
Rougher Concentrate (c)
99
60
1
40
Tailings from scavenger (d)
Scavenger Concentrate (e)
Tailings from cleaner (f)
Final Concentrate (g)
Specific Gravity
SiO2
2.65
CuS
4.60
H2O
1.0
99.6
50
80
1
Water to Solid Ratios
Rougher
2/1
Scavenger
4/1
Cleaner
6/1
0.4
50
20
99
Contact time, min
Rougher
8
Scavenger
12
Cleaner
10
Ans. Vrougher= 855.45 ft3 ; Vscavenger= 2328.19 ft3
15. Use Denver No. 24 cells whose cubic capacity is 50 ft 3 and whose power consumption is 4.2 hp per cell.
Compute the capacity and number of flotation cells in each unit and the power consumption.
Ans: Rougher
Scavenger
Ncells = 18 cells
Ncells= 47 cells
Power = 75.6 hp
Power= 197.4 hp
16. A typical flotation machine has the following specifications:
Number of cells = 4
Flotation time = 12min.
Cell Volume = 60 ft3
Hp per cell = 10hp
The material treated has the following specifications:
Pulp (mixture ore and water) = 40% solids
Specific gravity of ore = 3
𝑛=
𝑇 𝑥 𝐶𝑎𝑝 𝑥 𝑑
𝑉 𝑥 1440
Where n= number of cells; V = volume in cu. ft per cell; Cap = tons of dry ore / 24 hrs; d= cu. ft of pulp (ore
and water) containing one ton of solids.
The capacity of the machine in tons of dry ore per 24 hours is
Solution:
0.4 =
2000
2000 + 𝑥
𝑥 = 3000 𝐻2 𝑂
𝐹 = 3000 + 2000 = 5000
𝑑=
2000
3000
+
3 𝑥 62.4 63.4
𝑑 = 58.76
4=
(12)(𝑥 )(58.76)
(60)(1440)
𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚 = 𝟒𝟗𝟎
𝒕𝒐𝒏𝒔 𝒐𝒇 𝒅𝒓𝒚 𝒐𝒓𝒆
𝟐𝟒 𝒉𝒐𝒖𝒓𝒔
SEDIMENTATION
1. ________ is the separation of dilute slurry into a clear fluid and dense slurry by gravity settling.
a. Sedimentation
c. Centrifugation
b. Flotation
d. None of these
2. Drag is defined as the force exerted by
a. The fluid on the solid in a direction opposite to the motion of the solid
b. Shear force to the product of velocity head and density
c. Average drag per unit projected area to the product of the velocity head and density
d. None of these
3. Stoke’s law is valid when the particle Reynolds Number is
a. <1
c. <2
b. >1
d. none of the
4. Newton’s law is valid when the particle Reynolds Number is
a. >500
c. <500
b. 0.44
d. None of these
5. Intermediate Law is valid when the particle Reynolds Number is
a. 18.5
c. 0.6
b. 2 to 500
d. None of these
6. At low Reynolds Number
a. Viscous forces are unimportant
b. Viscous forces equal the inertial forces are unimportant
c. Viscous forces control and inertial forces are unimportant
d. Gravity forces control
7. At high Reynolds Number
a. Inertial forces control and viscous forces are unimportant
b. Viscous forces predominate
c. Inertial forces are unimportant and viscous forces control
d. None of these
8. An apparatus in which particles settle in a liquid by gravitational or centrifugal forces and are removed as
a concentrated slurry.
a. Classifier
c. Elutriator
b. Thickener
d. agitator
9. In continuous thickeners, separation of solid particles can be achieved if the settling velocity of the solids
is ______ compared to the velocity of the displaced liquid.
a. Equal
c. Greater
b. Less
d. None of these
10. In a motion of a particle through fluids, forces act on a particle moving through a fluid. The force which
appears whenever there is a relative motion between the particle and the fluid is called__.
a. Gravitational force
c. Drag force
b. Centrifugal force
d. Buoyant force
11. Free settling of sludge is 0.25 cm/min. Using an original height of 25cm, the sludge settled to a height of
18cm after the free-settling period. The sludge was found to settle to a height of 10cm after 110min. This
particular sludge was found to settle completely to a height of 4cm. The time to settle to a height of 1/5 of
its original height in a cylindrical tank whose diameter is 85% of its depth if it is 85% full and consider 1000
cu ft of sludge in the tank is ___. Ans 62.4 hours
12. Square mica plates, 1/32 in thick and 0.01 sq in. in area are falling randomly through oil with a density
of 55 lb/cu ft and with viscosity of 15 centipoise. The specific gravity of the mica is 3.0, the settling velocity
is __. Ans. 7.2 cm/s
13. Solid spherical particles of coffee extract from a dryer having a diameter of 400 μm are falling through
air at a temperature of 422 K. The density of the particles is 1030 kg/m3. Calculate the terminal settling
velocity and the distance of fall in 5 sec. The pressure is 101.32 kPa. Ans. vt = 1.49 m/s ; 7.45 m fall
14. Calculate the terminal settling velocity of the pyrite particles which may be assumed to be cubic in
shape. Ans. 0.022 m/s
15. Solid spherical particles having a diameter of 0.090 mm and a solid density of 2002 kg/m 3 are settling
in a solution of water at 26.7°C. The volume fraction of the solids in water is 0.45. Calculate the settling
velocity and the Reynolds number. Ans. Vt = 2.369x10-4 m/s ; NRe = 9.89x10-3
16. Calculate the terminal settling velocity of dust particles having a diameter of 60μm in air at 294.3 K and
101.32 kPa. The dust particles can be considered spherical with a density of 1280 kg/m 3.
Given:
Dpi = 60μm = 60x10-6 m
T = 293.5 K
P = 101.32 kPa
ρ = 1280 kg/m3
Required:
Vt
Solution:
@T=293.5 K
ρwater = 1.202 kg/m3
µ = 0.01828x10-3 Pa.s
P = 101.32 kPa
𝑉𝑡 =
𝑔𝐷𝑝2 (𝜌𝑝−𝜌)
18µ
(9.81)(60𝑥10−6 )2 (1280 − 1.202)
𝑉𝑡 =
18(0.01828𝑥10−3 )
𝑽𝒕 = 𝟎. 𝟏𝟑𝟕𝟑𝒎/𝒔
CENTRIFUGATION
1. A mechanical process of separating multi-phase mixture via the use of centrifugal force.
a. Centrifugation
c. Sedimentation
b. Flotation
d. None of these
2. If the radius of a basket centrifuge is halved and the rpm is doubled, then the
a. Linear speed of the basket is doubled
b. Linear speed of the basket is halved
c. Centrifugal force is doubled
d. Capacity of centrifuge is increased
3. For the separation of sugar solution from settled out mud, we use
a. Sparkler filter
c. Centrifugal filter
b. Plate and frame filter
d. Vacuum leaf filter
4. The capacity of a Sharples centrifuge is estimated to increase/decrease by ___% if its speed is doubled
and the cut size of the particles is reduced by 20%.
a. 20% increase
c. 156% increase
b. 80% decrease
d. None of these
5. Moisture can be removed from lubricating oil using
a. Tubular centrifuge
c. Sparkler filter
b. Clarifier
d. Rotary drum vacuum filter
6. Which of the following can be most effectively used for clarification of lube oil and printing ink?
a. Sparkler filter
c. Disc-bowl centrifuge
b. Precoat filter
d. Sharpless supercentrifuge
7. Removing solids from liquids by causing particles to settle through the liquid radially toward or away from
the center of rotation by use of a centrifuge.
a. Centrifugal sedimentation
b. Centrifugal filtration
c. Centrifugal decantation
d. None of these
8. The removal of a liquid from a slurry by introducing the slurry into a rapidly rotating basket, where the
solids are retained on a porous screen and the liquid is forced out of the cake by centrifugal action.
a. Centrifugal sedimentation
b. Centrifugal filtration
c. Centrifugal decantation
d. None of these
9. ____ is a process for the separation of mixtures, by removing a layer of liquid generally one from which a
precipitate has settled.
a. Centrifugal sedimentation
b. Centrifugal filtration
c. Centrifugal decantation
d. None of these
10. It is used when higher centrifugal fields are required for separation.
a. Tubular centrifuge
c. Centrifugal sedimentation
b. Disk centrifuge
d. None of these
11. A centrifuge with a radius of 76.2 mm rotates at a peripheral velocity of 53.34 m/s. The centrifugal force
developed compared to gravitational force in the bowl centrifuge is __.
a. 2800 g
c. 3800 g
b. 3100 g
d. 4250 g
12. A centrifuge with a phosphor bronze basket, 380 mm in diameter, is to be run at 67 Hz with a 75 mm
layer of liquid of density 2300 kg/m3 in the basket. What thickness of wall is required in the basket? The
density of phosphor bronze is 8900 kg/m3 and of the maximum safe stress for phosphor bronze is 87.6
MN/m2. Ans. 15.1mm
13. The capacity in cubic meters per hour of a clarifying centrifuge operating under these conditions is
Diameter of the bowl
= 600 mm
Specific gravity of liquid
= 1.2
Thickness of liquid layer
= 75 mm
Specific gravity of solid
= 1.6
Depth of bowl
= 400 mm
Viscosity of liquid
= 2 cp
Speed
= 1200 rpm
Cut size of particles
=210 m3/h
3
Ans. 210 m /h
14. In a test conducted using a laboratory centrifuge, it was found that the optimum recovery of protein from
coconut oil was achieved with an rpm of 2,500. Industrial size centrifuge comes in 2.5, 3.0, 3.5, 4.0, and 5.0
ft diameter with warranty if the centrifuge is operated not over 1000 rpm. For optimum commercial
operations, the centrifuge size you will recommend is __
Data: Laboratory centrifuge : height = 9 in ; diameter = 5in.
rpm = 2,000 to 3,000
Ans. 3.0 ft
15. A centrifuge with a bowl which is 500 mm long and has an inside radius of 50.5 mm is to be used to
separate crystals from a dilute aqueous mother liquor. The optimum speed of rotation for the centrifuge is
60,000 rpm, and the discharge weir is adjusted so that the depth of liquid at the bowl wall is 38.5 mm. The
crystals are approximately spherical and none are smaller that 2 x 10-6 min diameter. The maximum
volumetric flow rate in m3/s of the mother liquor that can be processed by this centrifuge if all the crystals
have to be removed is ___.Ans. 0.054
6. A dilute slurry contains small solid food particles having a diameter of 5x10-2 mm which are to be
removed by centrifuging. The particle density is 1050 kg/m 3 and the solution density is 1000 kg/m3. The
viscosity of the liquid is 1.2x10-3 Pa.s. A centrifuge at 3000 rev/min is to be used. The bowl dimensions are
b= 100.1mm, r1= 5.00 mm, r2= 30.0mm. Calculate the expected flow rate in m 3/s just to remove these
particles.
Given: Dpi = 5x10-2 mm
ρp = 1050 kg/m3
μ = 1.2x10-3 Pa.s
N = 3000 rev/min
Required: q (flow rate in m3/s)
Solution:
b =100.1 mm
r1= 5.00 mm
r2= 30.0mm
𝜔=
2𝜋𝑁
60
𝑞=
=
2𝜋(3000)
60
= 314.159 𝑟𝑎𝑑/𝑠
𝜋𝑏𝜔2 (𝜌𝑝 − 𝜌) 𝑟22 − 𝑟12
[
]
2𝑟
18𝜇
ln 𝑟 +2𝑟
2
1
100.1
𝜋( 1000 𝑚)(314.159)2 (0.03)2 − (0.005)2
[
]
𝑞=
2(0.03)
18(1.2𝑥10−3 )2
ln 0.03 + 0.005
𝒒 = 𝟏𝟗𝟒𝟑𝟖𝟖𝟐. 𝟕𝟕𝟖 𝒎𝟑 /𝒔
TECHNOLOGICAL INSTITUTE OF THE PHILIPPINES
#363 PASCAUL ST. QUIAPO MANILA
INTRODUCTION TO PARTICLE TECHNOLOGY
COMPILATION OF PROBLEMS
Submitted by:
GUEVARRA. Hazel Lynn G.
Submitted to:
Engr. Robert Delfin
Size reduction
1. Shape factor for a cylinder whose length equals its diameter
A. 1.5
C. 1
B. 0.5
D. 5
2. Work index is defined as the gross energy requirement in kWh/ton of feed needed to
reduce very large feed to such a size that 80% of the product passes a 100 microns
screen. What is the work index of gypsum rock?
A. 8.16
C. 13.11
B. 25.17
Dqw
3. A 6,000 lb of a material goes through a crusher and grinder per hour in succession ( om
the same power drive). Screen analysis from the crusher shows a surface area of
product of 500 ft2 per lb. screen analysis of the grinder product indicates a surface area
of 4,200 ft2 per lb. the rittinger number is estimated to be 25% while that of the grinder
is 30%. The total power to be delivered to the equipment is ____
A. 34.8 hp
C. 38.4 hp
B. 35.4 hp
D. 40.4 hp
4. The hardness of a mineral is a criterion of its resistance to crushing. Which of the
following is an example of a harp material?
A. Talc
C. sapphire
B. Calcite
D. feldspar
5. States that the energy required for crushing is proportional to the new surface created.
A. Rittinger’s law
C. bond law
B. Kick’s law
D. energy law
6. It is defined as the efficiency of technical grinding compared with that of laboratory
crushing experiments.
A. Grinding Efficiency
C. practical energy efficiency
B. Bond work Index
D. energy efficiency
7. The term applied to all ways in which particles of solids are cut or broken into smaller
pieces
A. Size reduction
C. comminution
B. Screening
D. crushing
8. It is defined as the efficiency of technical grinding compared with that of laboratory
crushing experiments.
C. Grinding Efficiency
D. Bond work Index
9. For coarse reduction of hard solids, use
C. practical energy efficiency
D. none of these
A. Impact
C. compression
B. Attrition
D. cutting
10. Soft and non-abrasive materials can be made into fines by
A. Attrition
B. Compression
11. Cement clinker is reduced to fine size
C. cutting
D. cutting
A. Roll crusher
B. Ball mill
12. Wet grinding in a revolving mill
C. Tube mill
D. hammer mill
A. Gives less wear on chamber walls than dry grinding
B. Requires more energy than for dry grinding
C. Increases capacity compared to dry grinding
D. Complicates handling of the product compared to dry grinding.
13. A fluid energy mill is used for
A. Cutting
B. Grinding
14. The operating speed of a ball mill should be
A. Less than the critical speed
critical speed
B. Much more than the critical speed
C. ultragrinding
D. crushing
C. at least equal to the
D. none of these
15. The critical speed of a ball mill in rpm whose diameter is 12 in. with grinding balls of
diameter ½ in is approximately _____ rpm.
A. 60
C. 90
B. 50
D. 80
16. A crushing mill reduces limestones from a mean particle size of 45mm to the following
product
Size (mm)
amount of product (percent)
12.5
0.5
7.5
7.5
5.0
45.0
2.5
19.0
1.5
16.0
0.75
8.0
0.40
3.0
0.20
1.0
It requires 21 kJ/kg of material crushed. Calculate the power required to crush the same
material at the same rate, from a feed having a mean size of 25mmto a product with a
mean size of 1 mm.
Solution:
The mean size product may be obtained thus
N1
0.5
7.5
45.0
19.0
16.0
8.0
3.0
1.0
Total
Dv
D1
12.5
7.5
5.0
2.5
1.5
0.75
0.40
0.20
13,049
= ∑ n1d4 1 / ∑n1d3 1
= (101,510/13,049) = 7.78 mm
N1d13
3906
3164
5625
2969
54.0
3.375
0.192
0.008
Total
N1d4
48.828
23.731
28.125
742.2
81.0
2.531
0.0768
0.0016
101,510
Kick’s law is used as the present case may be regarded as coarse crushing
Case 1:
E = 21 kJ/kg, L1 = 45 mm and L2 = 7.8 mm
21 = KKfc ln(45/7.8)
and:
KKfc = 11.98 kJ/kg
Case 2:
L1 = 25 mm and L2 = 1.0 mm.
Thus:
E = 11.98 ln(25/1.0)
= 38.6 kJ/kg
Sedimentation
1. A gravity settling tank is to be used to clean waste water from an oil refinery. The waste
water contains 1% oil by volume as small droplets ranging in size from 100 to 1000
microns which will be removed from the water before the latter is to be discharged into
the river. The tank is of rectangular section 2 ft wide by 4 ft deep with provisions for
smooth continuous discharge of clean water is to be cleaned of oil droplets, specific
gravity of oil 0.87, the length of the settling tank is
A. 4500 ft
C. 5000 ft
B. 4850 ft
D. 4000 ft
2. Square mica plate, 1/32 in. thick and 0.01 sq. in area are falling randomly through oil
with a density of 55 lb.cu ft and with viscosity of 15 cp. The specific gravity of mica is 3.0,
the settling velocity is
A. 6.2 cm/s
C. 5.2 cm/s
B. 7.2 cm/s
D. 4.2 cm/s
3. Solid spherical particles having a diameter of 0.09 mm and a solid density of 2,002 kg/m 3
are settling in a solution of water at 26.7°C. The volume fraction of the solids in the
solution is 0.45. the settling velocity is
A. 5.3 x 10-3 m/s
C. 2.369 x 10-4 m/s
B. 3.269 x 10-3 m/s
D. 3 x 10-5 m/s
4. A slurry containing 5 kg of water/kg of solids is to be thickened to a sludge containing
1.5 kg of water/kg of solids in a continuous operation. Laboratory tests using five
different concentrations of the slurry yielded the following results:
Concentration Y (kg water/kg solid)
5.0
4.2
3.7
3.1
2.5
Rate of sedimentation uc (mm/s)
0.17
0.10
0.08
0.06
0.042
A. 13.6 m2
C. 11.3 m2
B. 16.5 m2
D. 16.2 m2
Problem 5 and 6.
Calculate the minimum area and diameter of a thickener with a circular basin to treat
0.1 m3 /s of a slurry of a solids concentration of 150 kg/m3 . The results of batch settling
tests are:
Solid concentration
Settling velocity
100
148
200
91
300
55.33
400
33.25
500
21.40
600
14.50
700
10.29
800
7.38
900
5.56
1000
4.20
1100
3.27
A value of 1290 kg/m3 for underflow concentration was selected from a retention time
test. Estimate the underflow volumetric flow rate assuming total separation of all solids
and that a clear overflow is obtained.
5. Find the area
A. 974 m2
C. 500 m2
B. 450 m2
D. 950 m2
6. Volumetric flowrate
Solution:
d = [(4 × 974)/π] 0.5 = 35.2 m
The volumetric flow rate of underflow, obtained from a mass balance, is:
[(0.1 × 150)/1290] = 0.0116 m3/s
7. The drag coefficient in hindered settling is _______________ compared to free settling
A. Greater than
C. constant
B. Less than
D. varying
8. The operation by which solids are separated from liquids due to difference in their
respective densities is
A. Screening
B. Adsorption
C. sedimentation
D. absorption
9. The separation of solid particles into several size fractions based upon the settling
velocities in a medium is called
A. Settling
C. flotation
B. Filtration
D. classification
10. Device in which a current of air separates particles with different sedimentation
velocities
A. Agitator
C. classifier
B. Air elutriator
D. air conveyor
11. The constant velocity with which a body ,=moves relative ti the surrounding fluid when
the forces acting on it are equal to the friction force acting against the movement.
A. Terminal velocity
C. maximum velocity
B. Settling velocity
D. all of these
12. Stoke’s law is valid when the particle Reynolds Number is
A. <1
C. >1
B. <5
D. none of these
13. In order for a particle to move through a fluid under the influence of gravity, there must
be,
A. Velocity difference
C. density difference
B. Pressure difference
D. temperature difference
14. At Reynolds number
A. Viscous forces are unimportant
C. viscous forces predominate
B. Viscous forces control
D. none of these
15. Separation of dilute slurry
A. Sedimentation
C. flotation
B. Centrifugation
D. size reduction
16. Range motion Newton’s law for n
A. 0
C. 2
B. 1
D.4
Centrifugation
1. If a centrifuge is 3 ft diameter and rotates at 1,000 rpm, the speed of a laboratory
centrifuge of 6 in. diameter be ran if it is to duplicate plant conditions is
A. 2500 rpm
C.2469 rpm
B. 2449 rpm
D. 2000 rpm
2. Determine the filtration rate that can be expected from a basket centrifugal filter using
the data below :
Basket height
= 12 in
Inside basket diameter
= 26 in
Rotation rate
= 2,000 rpm
Material to be filtered:
Gypsum slurry, α
Ε
= 2.52 x 1011 ft/lb
= 0.5
Specific
Gravity of CaSO4.2H2O
= 2.65
Assume that the cake is incompressible, filter medium resistance is negligible and the
liquid surface corresponds to the filter cake surface and the thickness of the cake is 1 in
A. 0.60 gpm
C. 0.85 gpm
B. 0.95 gpm
D. 0.43 gpm
3. The terminal velocity of 10 micron particle (specific gravity = 1.2) clarified of water in
the laboratory centrifuge in Problem 1 is
A. 560 ft/s
C. 5564.5 ft/s
B. 565 ft/s
D. 556 ft/s
4. An aqueous suspension consisting of particles of density 2500 kg/m3 in the size range 1–
10 µm is introduced into a centrifuge with a basket 450 mm diameter rotating at 80 Hz.
If the suspension forms a layer 75 mm thick in the basket, approximately how long will it
take for the smallest particle to settle out?
A. 20.3 s
C. 18.3 s
B. 19.3 s
D. 17.3 s
5. A centrifuge basket 600 mm long and 100 mm internal diameter has a discharge weir 25
mm diameter. What is the maximum volumetric flow of liquid through the centrifuge
such that, when the basket is rotated at 200 Hz, all particles of diameter greater than 1
µm are retained on the centrifuge wall? The retarding force on a particle moving liquid
may be taken as 3πµdu, where u is the particle velocity relative to the liquid µ is the
liquid viscosity, and d is the particle diameter. The density of the liquid is 1000 kg/m3 ,
the density of the solid is 2000 kg/m3 and the viscosity of the liquid is 1.0 mN s/m2 . The
inertia of the particle may be neglected.
A. 2.0 m3/s
C. 0.00002 m3/s
B. 1.03 m3/s
D. 1.03 × 10−3 m3/s
6. If a centrifuge is 0.9 m diameter and rotates at 20 Hz, at what speed should a laboratory
centrifuge of 150 mm diameter be run if it is to duplicate the performance of the large
unit?
Solution:
x1 = 0.45 m,
ω1 = (20 × 2π ) = 40π rad/s,
x2 = 0.075 m
Thus,
0.45(40π )2 /g = 0.075ω2 2/g
ω2 = [6(40π )2 ] = (2.45 × 40π ) = 98π rad/s
and the speed of rotation = (98π/2π ) = 49 Hz (2940 rpm)
7. Moisture can be removed from lubricating oil using
A. Tubular centrifuge
C. sparkler filter
B. Clarifier
D. vacuum leaf filter
8. Which of the following can be the most effectively used for clarification of tube oil and
printing ink?
A. Sparkle filter
C. disc-bowl centrifuge
B. Precoat filter
D. sharpless supercentrifuge
9. Mechanical process of separating multiphase mixture via the use of centrifugal force
A. Centrifugation
B. Flotation
C. screening
D. size reduction
10. Where the difference in density of the two liquid phases to be separated is very small (
as in milk cream separator), the most suitable separator is
A. Disc bond centrifuge
C. batch basket centrifuge
B. Sharpless supercentrifuge
D. sparkler filter
11. If the radius of a basket centrifuge is halved and the rpm is doubled, then the
A. Linear speed of the basket is doubled
C. centrifugal force is doubled
B. Linear speed of the basket is halved
D. capacity of centrifuge is
increased
12. Ultra centrifuges are used for the separation of __________ solid particles.
A.
coarse
C. fine
B.
colloidal
D. dissolved
13. Is created by moving a mass in a curved path and is exerted in the direction away from
the center of curvature of the path.
A
force
C. settling
B.
centrifugal force
D. raising
14. Is the force applied to the moving mass in the direction toward the center of curvature
which causes the mass in the direction toward the center of curvature which causes the
mass to travel in a curved path.
A
centripetal force
C. rotational force
B
centrifugal force
D. filtration
15. Separation of isotopes is generally done using a/an __________ centrifuge.
A.
ultra
C. both a & b
B.
disk-bowl
D. neither a nor b
16. Uses the concept that an object whirled about an axis at a constant radial distance from
the point is acted on by a force.
A
filtration
C. centrifugal separation
B.
sedimentation
D. none of these
Flotation
1. A flotation plant processes 3000 lbs/day of CuFeS 2(chalcopyrite). It produces 80 tons
copper concentrate assaying 25% copper. If ore analyzes 0.7 copper, the percent recovery is
A. 90.5%
B. 98%
C. 92.54%
D. 95.24%
For problems 2 and 3:
Ground lead ore is to be concentrated by a single flotation process using 1.5 oz of reagent per
ton of ore. The feed concentrate and tailings have the following composition by weight on a dry
basis:
Feed%
Concentrate%
Tailings%
PbS
30
90
0.9
ZnS
25
3
35.6
SiO2
45
7
63.5
Water is fed to the cell at the rate of 1,000 gallons per ton of wet concentrate with 99% of the
water leaving with the tailings and 1% with the concentrate.
2. The mass of wet concentrate produced per hour when tons of ore are fed to the cell per
24 hours is
A. 3.4
B. 4.3
C. 14.23
D. 4.7
3. The total water required in pounds per hour is
A. 1150
B. 1200
C. 1185
D. 1285.8
4. A flotation modifier which is assists in the selectivity (sharpness of separation) or stop
unwanted minerals from floating.
A. Depressants
B. Activators
C. alkalinity regulators
D. promoters
5. Dispersants are important for the control of slimes which sometimes with the selectivity
and increase reagent consumption. Another term for dispersant is
A. Deflocculant
B. depressants
C. frothers
D. regulators
6. An example of a collector for flotation of metallic sulfides and native metals is
A. xanthates
B. sodium silicate
C. sodium sulfide
D. sphalerite
7. Which of the following is an example of a deflocculant?
A. Sulfuric acid
B. Lignin sulforate
C. dithiophosphate
D. molybderite
8. Substances added to maintain the proper pH
A. Regulators
C. activators
B. Depressors
D. promoters
9. Prevents the absorption of a collector by a mineral particle and thereof inhibit the
flotation of mineral
A. Regulators
C. activators
B. Depressors
D. promoters
10. Enhance the absorption of a collector by mineral particles
A. Regulators
C. activators
B. Depressors
D. promoters
11. Alter the surface of the mineral in order that it will become air-avid( to cause it to
adhere to air bubbles)
A. Regulators
B. Depressors
C. activators
D. promoters
12. Froth Flotation is most suitable for treating
g. iron ores
c. quartz
h. sulfide ores
d. metal ores
13. In Froth Flotation, chemical agent added to cause air adherence is called
i. collector
c. modifier
j. frother
d. promoter
14. Pine oil used in forth flotation technique acts as a
i. collector
c. frother
j. modifier
d. activator
15. Which of the following is the most suitable for cleaning of fine coal dust (< 0.5 m)?
i. Through washer
c. Spiral separator
j. Baum Jig Washer
d. Froth Flotation
16. Any operation in which one solid is separated from another by floating one of them at
or on the surface of a fluid.
a. coagulation
c. centrifugation
b. flotation
d. sedimentation
C.
Screening
Linked problem 1-3
Fine silica is fed at 1500 lbs/hour to a double-deck vibrating screen combination to obtain a
48/65 mesh (Tyler ) product. The silica feed is introduced into the upper screen of the 48 mesh
and the product is discharged off the surface of the lower screen of 65 mesh. During the
screening operation, the ratio of oversize to product to undersize is 2:1 1⁄2.
Laboratory analysis of the different fractions:
Screen Mesh
Feed Mass
Oversize Mass
Product Mass
Undersize Mass
Fraction
Fraction
10/14 to 28/35
0.2821
0.5855
0.3385
0.00453
35/48
0.2580
0.3370
0.3220
0.00360
48/65
0.2810
0.0660
0.5260
0.34400
65/100
0.0910
0.0050
0.0670
0.39900
100/150 to
0.0870
0.0060
0.0260
0.35300
Fraction
150/200
1. The effectiveness of the screening equipment is
A. 58.7%
B. 48.7%
C. 68.6%
D. 45.6%
2. If the screens measure 5ft x 8ft each, the capacity in MT/day-ft2 –mm of the 65 mesh
screen on the basis of a perfectly functioning 48 mesh screen is
A. 0.901
B. 1.09
C. 0.801
D. 0.75
3. The capacity in MT/day-ft2-mm on the basis of the actual performance of the 48 mesh
screen
A. 1.09
B. 0.901
C. 1.29
D. 1.49
4. The fineness characteristic of a powder on a cumulative basis is represented by a
straight line from the origin to 100 per cent undersize at a particle size of 50 µm. If the
powder is initially dispersed uniformly in a column of liquid, calculate the proportion by
mass which remains in suspension in the time from commencement of settling to that at
which a 40 µm particle falls the total height of the column. It may be assumed that
Stokes’ law is applicable to the settling of the particles over the whole size range.
Solution:
For settling in the Stokes’ law region, the velocity is proportional to the diameter squared
and hence the time taken for a 40 µm particle to fall a height h m is:
t = h/402k
k where k a constant.
During this time, a particle of diameter d µm has fallen a distance equal to:
kd2 h/402 k = hd2 /402
The proportion of particles of size d which are still in suspension is:
= 1 − (d2 /402 )
and the fraction by mass of particles which are still in suspension is:
40
= ∫0 1 − (d2 /402 )dw
Since dw/dd = 1/50, the mass fraction is:
= (1/50) 40 0 [1 − (d2 /402 )]dd
= (1/50)[d − (d3 /4800)] 400
= 0.533 or 53.3 per cent of the particles remain in suspension.
5. In a mixture of quartz of density 2650 kg/m3 and galena of density 7500 kg/m 3 , the
sizes of the particles range from 0.0052 to 0.025 mm. On separation in a hydraulic
classifier under free settling conditions, three fractions are obtained, one consisting of
quartz only, one a mixture of quartz and galena, and one of galena only. What are the
ranges of sizes of particles of the two substances in the original mixture?
A. 0.0103–0.0126 mm
C. 0.0012-0.0015 mm
B. 0.0112- 0.0114 mm
D. 0.011-0.12 mm
6. A sample of dust from the air in a factory is collected on a glass slide. If dust on the slide
was deposited from one cubic centimeter of air, estimate the mass of dust in g/m3 of air
in the factory, given the number of particles in the various size ranges to be as follows:
Size range (µm)
0–1
1–2
2–4
4–6
6–10 10–14
Number of particles (-)
2000
1000 500
200
100
40
It assumed that the density of the dust is 2600 kg/m3 , and an appropriate allowance
should be made for particle shape.
A. 0.75 g/m3
C. 0.25 g/m3
B. 0.65 g/m3
D. 0.30 g/m3
7. The removal of a small amount of oversize from a feed which are predominantly fines is
called
A. Scalping
C. coarse separation
B. Desliming
D. dewatering
8. Removal of free water from a solids-water mixture and is generally limited to a 4-mesh
and above
A. Scalping
C. trash removal
B. Dewatering
D. separation
9. Making a size separation smaller than 48-mesh is called
A. Coarse separation
C. ultrafine separation
B. Fine separation
D. scapling
10. The material passing one screening surface and retained on a subsequent surface is
called
A. Intermediate material
B. Minus material
C. plus material
D. none of these
11. The minimum clear space between the edges of the opening in the screening surface
and is usually given in inches or millimeters.
A. Sieve
C. mesh number
B. Aperture
D. holes
12. The screen used in making size separation smaller than 4-mesh and larger than 48-mesh
A. Grizzly screen
C. oscillating screen
B. Gyratory screen
D. vibrating screen
13. Box-like machines either round or square with a series of screen clothes nested stop one
another
A. Reciprocating screen
C. electricity vibrated screen
B. Oscillating screen
D. gyratory screen
14. Another name for revolving screen is
A. Shaking screen
C. mechanically vibrated screen
B. Vibrating screen
D. trommel screen
15. Materials which remain on a screen surface are called
A. Fines
C. intermediate material
B. Undersize
D. oversize
16. Screen capacity is expressed in terms of
A. Tons/h
B. Tons/ft2
C. both a and b
D. tons/h-ft2
TECHNOLOGICAL INSTITUTE OF THE PHILIPPINES
363 P. Casal St., Quiapo, Manila
College of Engineering and Architecture
Chemical Engineering Department
Problem Set in Particle Technology
Joaquico, Jeffrey C.
5th Yr Che Student
Engr. Robert Delfin
Instructor
March 25, 2015
CONCEPTS
SIZE REDUCTION
1. Equivalent diameter of a particle is the diameter of the sphere having the same
Answer: ratio of surface to volume as the actual volume.
2. If dp is the equivalent diameter of a non-spherical particle, Vp its volume and sp its surface area,
then its sphericity is Φs is defined by
Answer: Φs = 6 Vp/dp sp
3. Pick out the wrong statement.
A.
Recycled coarse material to the grinder by a classifier is termed as circulating load.
B.
Wear and tear in wet crushing is more than that in dry crushing of materials.
C.
Size enlargement (opposite of size reduction) is not a mechanical operation.
D.
A 'dust catcher' is simply an enlargement in a pipeline which permits the solids to settle
down due to reduction in velocity of the dust laden gas.
Answer: C
4. For sizing of fine materials, the most suitable equipment is a
Answer: vibrating screen
5. Pebble mills are tumbling mills widely used for grinding in the manufacture of paints & pigments
and cosmetic industries, where iron contamination in the product is highly objectionable. Pebbles
used in pebble mill are made of
Answer: flint or porcelain
6. Which of the following relationships between co-efficinet of friction (μ) between rock & roll and a
(half of the angle of nip) of the particle to be crushed is correct ?
A.
μ > tan α
B.
μ ≥ tan α
C.
μ > tan 2α
D.
μ ≤ tan α
Answer: B
7. Pick out the wrong statement pertaining to the roll crushers.
A.
Maximum feed size determines the required roll diameter.
B.
For hard material's crushing, the reduction ratio should not exceed 4.
C.
Both the rolls run necessarily at the same speed.
D.
Reduction ratio and differential roll speed affect production rate & energy consumed per
unit of surface produced.
Answer: C
8. Which of the following is not an ultrafine grinder (colloid mill)?
A.
Micronizers
B.
Agitated mills and fluid energy mills
C.
Toothed roll crusher
D.
Hammer mills with internal classification
Answer: C
9. Two particles are called to be equal settling, if they are having the same.
Answer: terminal velocities in the same fluid & in the same field of force.
10. Number of particles in a crushed solid sample is given by (where, m = mass of particles in a
sample, Vp = volume of one particle, ρ= density of particles)
Answer: m/ρ . Vp
SCREENING
1. Pick out the correct statement:
A.
Removal of iron from ceramic material is necessitated (by magnetic separation method) so
as to avoid discolouration of ceramic products.
B.
The operating cost of shaking screen is more than that of a vibrating screen.
C.
Screen capacity does not depend upon the specific gravity of the minerals.
D.
Asphalt is best crushed using toothed roll crusher.
Answer: A
2. Jigging is a technique by which different particles can be
Answer: separated by particle size.
3. 200 mesh screen means 200 openings per
Answer: inch
4. It is a separation method used to separate solids based on their particle size alone.
Answer: Screening
5. It is a series of testing sieves having openings in a fixed succession.
Answer: Sieve Scales
6. Screening equipment which consists of a Set of parallel metal bars in an inclined stationary
frame.
7. Screening equipment in which the frequency of the screen is mainly controlled by an
electromagnetic vibrator which is mounted above and directly connected to the screening
surface.
Answer: Vibrating Screens
8. Screening equipment which is Composed of a rotating perforated drum set in an inclined
position
Answer: Rotary Screen
9. It is calculated by the product of recovery of desired material in the product and recovery of
undesired material in the reject
Answer: Screen Effectiveness
10. Important characteristics of an individual particle are the following except:
A. Composition
B. Size
C. Shape
D. None of the Above
Answer: D
SEDIMENTATION
1. Solid particles separation based on the difference in their flow velocities through fluids is
termed as the
Answer: classification
2. Pick out the wrong statement.
A.
Close circuit grinding is more economical than open circuit grinding.
B.
Cod oil, beef tallow or aluminum stearates are used as grinding aids in cement 'industries'.
C.
The equipment used for the removal of traces of solids from a liquid is called a classifier.
D.
Size enlargement is a mechanical operation exemplified by medicinal tablet making.
Answer: C
3. Gravity settling process is not involved in the working of a
Answer: hydrocyclone
4. Its main purpose is to remove the particle from the fluid stream so that the fluid is free of
particle contaminants.
Answer: Sedimentation
5. When a particle is at sufficient distance from the walls of the container and from other particles
so that its fall is not affected by them, the process is called _______.
Answer: Free Settling
6. When the particles are crowded, they settle at a lower rate and the process is called _______.
Answer: Hindered Settling
7. It is the period of constant velocity fall.
Answer: Free Settling Velocity
8. It is the random motion imparted to the particle by collisions between the molecules of the fluid
surrounding the particle.
Answer: Brownian movement
9. It is a device known for the separation of solid particles into several fractions based upon their
rates of flow or settling through fluids.
Answer: Classifiers
10. The separation of solid particles into several size fractions based upon the settling velocities in
a medium is called ______.
Answer: Differential settling
FLOTATION
1. Froth floatation is the most suitable for treating
Answer: sulfide ores
2. Which of the following is the most suitable for cleaning of fine coal dust (<0.5 mm) ?
A.
Trough washer
B.
Baum jig washer
C.
Spiral separator
D.
Froth floatation
Answer: D
3. It is is a method of separation widely used in the wastewater treatment and mineral processing
industries.
Answer: Flotation
4. It is a water treatment process that clarifies wastewaters (or other waters) by the removal of
suspended matter such as oil or solids. The removal is achieved by dissolving air in the water
or wastewater under pressure and then releasing the air at atmospheric pressure in a flotation
tank or basin.
Answer: Dissolved air flotation
5. It is a process wherein colloids come out of suspension in the form of floc or flake; either
spontaneously or due to the addition of a clarifying agent.
Answer: Flocculation
6. They are used to remove suspended solids from liquids by inducing flocculation.
Answer: Clarifying agents
7. It is the creation of a solid in a solution or inside another solid during a chemical reaction or by
diffusion in a solid.
Answer: Precipitation
8. It is a chemical additive to prevent a colloid from coming out of suspension or to thin
suspensions or slurries. It is used to reduce viscosity or prevent flocculation and is sometimes
incorrectly called a "dispersant."
Answer: deflocculant
9. It is a process for treating sewage and industrial wastewaters using air and a biological floc
composed of bacteria and protozoa.
Answer: Activated sludge
10. They are settling tanks built with mechanical means for continuous removal of solids being
deposited by sedimentation.
Answer: Clarifiers
CENTRIFUGATION
1. Tabular bowl centrifuges as compared to disk bowl centrifuges
Answer: operate at higher speed.
2. Which of the following is the most suitable for handling fibrous and dense slurries?
A.
Propeller agitator
B.
Cone type agitator
C.
Turbine agitator
D.
Radial propeller agitator
Answer: B
3. __________ centrifuge is normally used in sugar mills.
Answer: Suspended batch basket
4. If a force greater than that of gravity is used to separate solids & fluids of different densities,
the process is termed as the
Answer: centrifugation
5. If the object being rotated is a cylindrical container, the contents of the fluid and solids exert an
equal and opposite force called ________.
Answer: Centrifugal forces
6. Equipment for centrifugation in which the bowl is tall and has a narrow diameter, 100 to 150
mm. It develops a force about 13000 times the force of gravity.
Answer: Supercentrifuges
7. Centrifuge having a diameter of 75mm and very high speeds of 60,000 or so rev/min are
known as _______.
Answer: Ultracentrifuges
8. It is a centrifuge that is often used in liquid-liquid separations. The feed enters the actual
compartment at the bottom and travels upward through vertically spaced feed holes, filling the
spaces between the disks.
Answer: Disk bowl centrifuge
9. A kind of filter in which the slurry is fed continuously to a rotating basket which has a perforated
wall is covered with a filter cloth.
Answer: Centrifugal filter
10. _______ make use of the common principle that an object whirled about an axis or center point
at a constant radial distance from the point is acted on by a force.
Answer: Centrifugal separators
PROBLEMS
SIZE REDUCTION
1. Calculate the surface volume mean diameter for the following particulate material.
Size Range (um)
Mass of particles in the range (gm)
-704 + 352
25
-352 +176
37.5
-176 +88
62.5
-88 +44
75
PAN
50
Answer: 61 um
2. Calculate the sphericity of a solid particle of cubical shape.
𝝅 𝟏⁄
𝟑
Answer: ( 𝟔)
3. The size distribution function of a sample of freshly ground material is given by
f(d) = 1.37𝑥 10−7 𝑑 3 𝑒 −0.03𝑑
For the integral values of m.
𝑑2
∫
𝑑1
𝑚
𝑚
𝑗=1
𝑗=1
(𝑏𝑑1)𝑗
(𝑏𝑑2)𝑗
𝑎𝑑 𝑚 𝑒 −𝑏𝑑 𝑑 (𝑑 ) = 𝑒 −𝑏𝑑 (1 + ∑
) − 𝑒 −𝑏𝑑2 ( 1 + ∑
)
𝑗!
𝑗!
Where d is the particle diameter in um.
Answer:
Most common particle size = m/b = 3/0.003 = 100 um
Length mean particle size = 133.3 um
Fraction of particles below 75 um diameter = 0.190
Fraction of particles between 125 and 175 um in size = 0.2521
4. What is the power required to crush 100 ton/h of limestone if 80% of the feed passes a 2 in screen
and 80% of the product a 1/8 in screen?
Answer: 169.6Kw
5. A batch grinding mill is charged with material of the composition shown in Table 29.2. The griningrate function Su is assumed to be 0.001/s for the 4/6 mesh particles. Breakage function Bv is given
by Eq(29.13) with B=1.3. Both Su and Bv are assumed to be independent of time. How long will it
take for the fraction of 4/6 mesh material to diminish by 10 percent?
Mesh
n
Dpn
Xno
Sn
4/6
1
3.327
0.0251
10
6/8
2
2.362
0.1250
3.578
8/10
3
1.651
0.3207
1.222
10/14
4
1.168
0.2570
0.4326
14/20
5
0.833
0.1550
0.1569
20/28
6
0.589
0.0538
0.0554
28/35
7
0.417
0.0210
0.0196
Answer: 105.3 s
6. It is desired to crush 10 ton/h of iron ore hematite. The size of the feed is such that 80% passes a 3
in screen and 80% of the product is to pass a 1/8 inch screen. Calculate the gross power required.
Use a work index E1 for iron ore hematite of 12.68
Solution:
3
Feed Size = 12 = 0.25𝑓𝑡
Product Size =
Feed rate = 10
1/8⁄
12 = 0.104𝑓𝑡
𝑡𝑜𝑛
ℎ𝑟
1ℎ𝑟
𝑡𝑜𝑛
𝑥 60 𝑚𝑖𝑛𝑠 = 0.167 𝑚𝑖𝑛
𝑃
1
1
= (1.46)(12.68)(
−
)
0.167
√0.104 √0.250
𝑷 = 𝟐𝟒. 𝟏𝒉𝒑
SCREENING
1. The size analysis of a powdered material on a weight basis is represented by a straight line from
0% weight at 1 um particle size to 100% weight at 101 um particle size. Calculate the mean
surface diameter of particles constituting the system.
Answer: 21.7 um
2. The table below gives the results of microscopic size analysis of a particulate solid sample. The
lower limit of the resolution of the instrument was 40 um. By the plotting the data logarithmically,
find whether the size distribution corresponds to the gamma distribution and if so, of what order.
Estimate the percentage of the particles in the 20 – 40 um range.
Size range (um)
Number of % of particles
40-60
0.996
60-80
2.025
80-100
3.189
100-120
4.314
120-140
5.275
140-160
6.003
160-180
6.474
180-200
6.695
200-220
6.697
Answer: It corresponds to the gamma distribution. The gradient of the straight part of the
line is 2.04, so the value of m can be assumed to be 2. The percent of the particles in the 2040 um range is 0.325%
3. A sieve analysis of a sample of sand yielded the following results:
BSS No.
Aperture
Weight
Sieve
Size (um)
Retained (%)
100
152
0.41
120
125
2.36
150
104
48.30
170
89
40.00
200
76
5.93
Pan
3.00
In a separate experiment the density of the sad was determined as follows: A 25 ml density bottle
was weighed, carefully filled with the sand and reweighed. Benzene was then poured into the bottle
until no more could be accommodated and the bottle was weighed again. The weights recorded
were:
Weight of empty bottle
15.3600g
Weight of bottle and sand
50.0835g
Weight of bottle, sand and benzene
58.0806g
Find the linear, surface area, volume and surface mean diameters of the sand as well as its bulk
and true density.
Answer:
Weight of sand in bottle = 34.7235g
Weight of benzene = 7.9941g
Volume of benzene = 9.1668ml
Volume of sand = 15.8322ml
True density of sand = 2.1931g/ml
4. A quartz mixture having the screen analysis shown in Table is screened through a standard 10mesh screen. The cumulative screen analysis of overflow and underflow are given in Table.
Calculate the mass ratios of the overflow and underflow to feed.
Mesh
Dpi
Feed
Overflow
4
4.699
0
0
6
3.327
0.025
0.071
8
2.362
0.15
0.43
0
10
1.1651
0.47
0.85
0.195
14
1.168
0.73
0.97
0.58
20
0.833
0.885
0.99
0.83
28
0.589
0.94
1.00
0.91
35
0.417
0.96
0.94
65
0.208
0.98
0.975
1.00
1.00
Pan
Underflow
Answer:
Overflow/ Feed = 0.420
Underflow/Feed = 0.58
5. A quartz mixture is screened through a 10-mesh screen. The cumulative screen analysis of feed,
overflow and underflow are given in the table in number 4. Calculate the overall effectiveness of
the screen.
Answer: 0.669
6.
Mesh
Screen
Mass
Average
Cumulative
Opening
Fraction
particle
Fraction
Xi/Dpi
Xi Dpi
Xi/ Dpi^3
diameter in
smaller than
(Xi)
increment
Dpi
4
4.699
0.000
-
1.000
0.006255 0.100726 0.000388
6
3.327
0.0251
4.013
0.9749
0.043937 0.355625 0.005428
8
2.362
0.1250
2.845
0.8499
0.159791 0.643645
10
1.651
0.3207
2.007
0.5292
0.183453 0.360031 0.093478
14
1.168
0.2570
1.4009
0.2722
0.158841 0.159159 0.158524
20
0.833
0.1590
1.001
0.1132
0.075668 0.038252 0.149683
28
0.589
0.0538
0.711
0.0594
0.04175 0.010563 0.165012
35
0.417
0.0210
0.503
0.0384
0.028652 0.003631 0.226074
48
0.295
0.0102
0.356
0.0282
0.030556
0.00194
65
0.208
0.0077
0.252
0.0205
0.038202
0.00121 1.205727
100
0.147
0.0068
0.178
0.0147
0.03254 0.000517 2.049615
150
0.104
0.0041
0.126
0.0106
0.034871 0.000276 4.412212
200
0.074
0.0031
0.0889
0.0075
0.202703 0.000278 148.0663
0.0075
0.037
0.000
0.006255 0.100726 0.000388
PAN
•
Retained
Calculate Volume Surface Mean Diameter
𝐷𝑠 =
𝐷𝑠 =
•
1
∑𝑁
𝑖=1
𝑋𝑖
𝐷𝑝𝑖
1
= 0.9641
1.0372
Calculate Mass Mean Diameter
𝑁
𝐷𝑛 = ∑ 𝑋𝑖 𝐷𝑝𝑖 = 1.6759
𝑖=1
•
Calculate Volume Mean Diameter
0.03967
0.48116
𝐷𝑣 =
1
∑𝑁
𝑖=1
𝑋𝑖
𝐷𝑝𝑖 3
=
1
= 0.006367
157.0532
SEDIMENTATION
1. Oil droplets having a diameter of 20 um are to be settled from air at an air temperature of 37.8 C at
101.3Kpa pressure. The density of the oil is 900 kg/m3. Calculate the terminal settling velocity of
the drops.
Answer: 0.0103 m/s
2. Calculate the settling velocity of glass spheres having a diameter of 1.554𝑥10−4 m in water at
293.2 K. The slurry contains 60% wt solids. The density of the glass sphere is 2467 kg/m3.
Answer: 𝟓. 𝟎𝟑𝒙𝟏𝟎−𝟑
3.
𝒇𝒕⁄
𝒔
A mixture of silica (B) and galena (A) solid particles having a size range of 5.21𝑥10−6 m to
2.50𝑥10−5 𝑚 is to be separated by hydraulic classification using free settling conditions in water
at 293.2K. The specific gravity of silica is 2.65 and that of galena is 7.5. Calculate the size range of
the various fractions obtained in the settling. If the settling is in the laminar region, the drag
coefficients will be reasonably close to that for spheres.
Answer:
Pure Galena:
DpA3 =1.260𝑥10−5 𝑚 to
DpA4 = 2.5𝑥10−5 𝑚
Mixed Fraction size range
DpB2 = 1.033𝑥10−5m to
DpB4 = 2.5𝑥10−5 m
DpA1 = 5.21𝑥10−6 𝑚 to
DpA3 = 1.260𝑥10−5 𝑚
Size range of the third fraction of pure silica
DpB1 = 5.21𝑥10−6 𝑚 to
DpB2 = 1.033𝑥10−5 𝑚
4. A protein/RNA complex (v = 0.71 cm3 g-1) gives a sedimentation coefficient of 12.7 S in 10%
sucrose, 50 mM Tris buffer, pH 7.4 at 4oC. What will be the velocity of sedimentation of the
complex under these conditions when the complex is found 6.0 cm from the central axis of a rotor
spinning at 40,000 rpm?
Answer: 1.337 x 10-4 cm/ s
5. A slurry containing 5 kg of water/kg of solids is to be thickened to a sludge containing 1.5 kg of
water/kg of solids in a continuous operation. Laboratory tests using five different concentrations of
the slurry yielded the following data:
Concentration (kg water/kg solid)
Rate of Sedimentation (mm/s)
5.0
0.20
4.2
0.12
3.7
3.1
2.5
0.094 0.070 0.052
Calculate the minimum area of a thickener to effect the separation of 1.33 kg/s of solids.
Answer: 31.2 𝑚2
6. A continuous separating tank is to be designed to follow after a water washing plant for liquid oil.
Estimate the necessary area for the tank if the oil, on leaving the washer, is in the form of globules
5.1 x 10-5 m diameter, the feed concentration is 4 kg water to 1 kg oil, and the leaving water is
effectively oil free. The feed rate is 1000 kg h-1, the density of the oil is 894 kg m-3 and the
temperature of the oil and of the water is 38°C. Assume Stokes' Law.
Viscosity of water = 0.7 x 10-3 N s m-2.
Density of water = 1000 kg m-3.
Diameter of globules = 5.1 x 10-5 m
Solution
𝑉𝑡 =
𝐷𝑝2 𝑔(𝜌𝑝 − 𝜌)
18 𝜇
𝑘𝑔
(5.1 x 10 − 5 m)2 (9.81 𝑚⁄𝑠 2 ) (1000 − 894) 3
𝑚
𝑉𝑡 =
𝑘𝑔
18 (0.7𝑥10−3 𝑚 − 𝑠)
𝑚
𝑉𝑡 = 2.15𝑥10−4
𝑠
𝑘𝑔
Since F = 4 and L = 0, and dw/dt = flow of minor component = 1000/5 = 200 ℎ𝑟
𝐴=
FLOTATION
4𝑥200
= 1.0𝑚2
(0.77𝑥1000)
1. A mixture of coal and sand particles having sizes smaller than 1 x 10-4 m in diameter is to be
separated by screening and subsequent elutriation with water. Recommend a screen aperture
such that the oversize from the screen can be separated completely into sand and coal particles by
elutriation. Calculate also the required water velocity. Assume that Stokes law is applicable.
Density of sand = 2650 kg/m3; density of coal = 1350 kg/m3; density of water = 1000 kg/m3;
viscosity of water = 1 x 10-3 kg/m.s; g = 9.812 m/s2.
Answer:
The size of screen aperture needed so that the oversize particles can be separated completely into
sand and coal by elutriation = 4.6056 x 10-5 m
The required water velocity = 1.9079 x 10-3 m/sec
2. A copper ore initially contains 2.09% Cu. After carrying out a froth flotation separation, the products
are as shown in Table 1. Using this data, calculate: (a) Ratio of concentration (b) % Metal
Recovery (c) % Metal Loss (d) % Weight Recovery, or % Yield
Product
% Weight
%Cu Assay
Feed
100
2.09
Concentrate
10
20.0
Tailings
90
0.1
Answer:
a. 10
b. 95.7%
c. 4.3%
d. 10%
3. Compute the force ratios given in table for an aqueous flotation system for which R = 1 um, ∆𝜌 =
𝑘𝑔
3000 𝑚3 , 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 = 100𝑚𝑉 and Ah =1𝑥10−20 𝐽. Take h = 1nm
Answer:
Gravitational force/Electrical Force = 4.2𝑥10−6
Gravitational force/van der Waals force = 2.9𝑥10−6
Inertial Force/Electrical Force = 3.7𝑥10−10
Inertial Force/van def Waals force = 2.6𝑥10−10
4. Calculate the flotation recovery of an ore in water if the velocity of bubble is 20 mm/s and the
settling velocity of particle is 10 mm/s. The probability of adhesion is 0.7 and the probability of
detachment is 0.3. The diameter of the bubble is 1 mm and the same of the particle is 100 um.
Answer: 65.2%
5. For a laboratory flotation of an iron ore in water, it was observed that 2 mg was collected while
traversing 2 m of the flotation column. The concentration of the ore in water was 0.5 kg/m3. The
average diameter of the bubbles was 2 mm and the average diameter of the particles was 0.1 mm.
Compute flotation recovery.
Given:
𝐷𝑝 = 1𝑥10−4 𝑚
𝐷𝑏 = 2𝑥10−3 𝑚
H = 2m
𝑐 = 0.5
𝑘𝑔⁄
𝑚3
𝑁𝑐 = 2𝑥10−6 𝑘𝑔
Solution:
2𝑥10−6
2𝑥10−6
𝑅=𝜋
=
−4
−3 )2 𝑥 (2 𝑥 0.5)
3.46𝑥10−6
(
4 1𝑥10 + 2𝑥10
𝑹 = 𝟎. 𝟓𝟕𝟖 = 𝟓𝟕. 𝟖%
CENTRIFUGATION
1. A centrifuge rotor is spinning at 25,000. The ‘top’ of the cell is 5.5 cm from the rotor’s central axis,
and the ‘bottom’ of the cell is 9.5 cm from the central axis. What are the gforces on a particle found
at the top and at the bottom of the tube?
Answer:
At r = 5.5 cm, the centrifugal force is equivalent to 38,400 x g
At r = 9.5 cm, the centrifugal force is equivalent to 66,400 x g
2. A pure protein (M = 67.0 kDa, vp = 0.72 cm3 g-1) is centrifuged at 10,000 rpm in an analytical
centrifuge, at 5oC and in the same buffer solution as was used in the last problem. At equilibrium, if
the concentration of the protein at r = 6.5 cm is 0.8 mg ml-1, what will be the protein concentration
at r = 6.0 cm?
Answer: r = 6.0 cm will be 0.050 mg/ ml
3. A centrifuge having a radius of the bowl of 0.1016 m is rotating at N = 1000 rev/min. Calculate the
centrifugal force developed in terms of gravity forces.
Answer: 113.6 g
4. A viscous solution containing particles with a density 1461 kg/m3 is to be clarified by centrifugation.
The solution density is 801 kg/m3 and its viscosity is 100cp. The centrifuge has a bowl with r2 =
0.02225 m, r1 = 0.00716 m and height b = 0.1970m. Calculate the critical particle diameter of the
largest particles in the exit stream if N = 23, 000 rev/min and the flow rate q = 0.002832 m3/h.
Answer: 0.746 um
5. In a vegetable-oil process and aqueous phase is being separated from the oil phase in a
centrifuge. The density of the oil is 919.5 kg/m3 and that of the aqueous phase is 980.3 kg/m3. The
radius r1 for overflow of the light liquid has been set at 10.160 mm and the outlet for the heavy
liquid at 10.414 mm. Calculate the location of the interface in the centrifuge.
Answer: 13.75 mm
6. A viscous solution contains particles with a density 𝜌𝑝 = 1200
centrifugation. The solution density is 𝜌 = 850
𝑘𝑔⁄
𝑚3 is to be clarified by
𝑘𝑔⁄
𝑚3 and its viscosity is 80 cp. The centrifuge
has a bowl r2 = 0.02m and r1 = 0.01m and height b =0.25m. Calculate the critical particle diameter
3
of the largest particles in the exit stream if N = 15,000 rpm and flow rate 𝑞 = 0.002 𝑚 ⁄ℎ𝑟.
Solution:
𝜔=
2𝜋𝑁 2𝜋 (15,000)
=
= 1570 𝑟𝑎𝑑⁄𝑠
60
60
Bowl Volume:
𝑉 = 𝜋𝑏(𝑟 2 2 − 𝑟 21 )
𝜋(0.25𝑚)(0.02𝑚2 − 0.01𝑚2 )
𝑉 = 2.355𝑥10−4 𝑚3
0.002 𝑚3
3
⁄𝑠 = 5.56𝑥10−7 𝑚 ⁄𝑠
3600
𝑘𝑔
15702 (1200 − 850) ⁄𝑚3 𝐷𝑝2
−7 𝑚3⁄
5.56𝑥10
= 84.16𝑥103 𝐷𝑝2
𝑠=
2𝑥0.02
18(80𝑥10−3 )ln (0.01 + 0.02)
𝑄=
𝑫𝒑 = 𝟐. 𝟓𝟕𝝁𝒎
TECHNOLOGICAL INSTITUTE OF THE PHILIPPINES
363 P. CASAL ST., QUIAPO, MANILA
PROBLEM SET
INTRODUCTION TO PARTICLE TECHNOLOGY
(FINALS)
SUBMITTED BY:
LANGAS, JAN CARLO V.
BSChE – IV
1430087
SUBMITTED TO:
ENGR. ROBERT DELFIN
DATE:
MARCH 25,2015
SIZE REDUCTION
PROBLEM 1
Sugar is ground from crystals of which it is acceptable that 80% pass a 500 mm sieve(US Standard Sieve
No.35), down to a size in which it is acceptable that 80% passes a 88 mm (No.170) sieve, and a 5horsepower motor is found just sufficient for the required throughput. If the requirements are changed such
that the grinding is only down to 80% through a 125 mm (No.120) sieve but the throughput is to be
increased by 50% would the existing motor have sufficient power to operate the grinder? Assume Bond's
equation.
SOLUTION
𝐸1 =
𝐸2 =
1
5
100
88 1
= 𝐸𝑖 (
𝑥 10−6 )2 [1 − (
)2 ]
𝑚
88
500
1
𝑥
100
125 1
= 𝐸𝑖 (
𝑥 10−6 )2 [1 − (
)2 ]
1.5𝑚
125
500
𝑥
0.500
= 0.84 𝑥 (
)
7.5
0.58
𝑥 = 0.74
Answer:
5.4 hp
PROBLEM 2
Calculate the shape factors ~ for model systems in which the particles are:
(a) cylinders with L = 2D,
(b) tetrahedra with their sides being equilateral triangles (the volume of a tetrahedron being the area of
the base multiplied by 1/3 the vertical height)
Estimate the specific surface area of a powder consisting of equal numbers of the above two shapes in
which there are 4 x 103 particles kg-1. The cylinders have a density of 1330 kg m-3 and the tetrahedra a
density of 1500 kg m-3.
Answer
(a) 0.83
(b) 2.4
(c) 0.81
PROBLEM 3
It is found that the energy required to reduce particles from a mean diameter of 1 cm to 0.3 cm is 11 kJ kg1. Estimate the energy requirement to reduce the same particles from a diameter of 0.1 cm to 0.01 cm
assuming:
(a) Kick's Law,
(b) Rittinger's Law,
(c) Bond's Equation.
Answer:
(a) 21 KJ/kg
(b) 423 KJ/kg
(c) 91 KJ/kg
PROBLEM 4
It is suspected that for a product of interest the oxidation reactions, which create off-flavours, are surface
reactions which proceed at a rate which is uniform with time, and if the shelf life of the product is directly
related to the percentage of the off-flavours that have been produced, estimate the percentage reduction in
shelf life consequent upon the size reductions of example 3, that is from 1 cm to 0.3 cm and from 0.1 cm to
0.01 cm in diameter, assuming l = 1.5.
Answer:
(a) 10:1
(b) 100:1
PROBLEM 5
If it is desired to reduce the separation time for milk to at least one week (before cream will rise to the top),
what maximum diameter of cream droplet would Stokes' Law predict to be necessary for the
homogenization to achieve? Assume the depth is 10 cm.
Answer:
0.0567 microns
PROBLEM 6
From measurements on a uniformly sized material from a dryer, it is inferred that the surface area of the
material is 1200 m2. If the density of the material is 1450 kg m-3 and the total weight is 360 kg calculate the
equivalent diameter of the particles if their value of l is 1.75.
Answer:
2200 microns
SCREENING
PROBLEM 1
In an analysis of ground salt using Tyler sieves, it was found that 38% of the total salt passed through a 7
mesh sieve and was caught on a 9 mesh sieve. For one of the finer fractions, 5% passed an 80 mesh
sieve but was retained on a 115 mesh sieve. Estimate the surface areas of these two fractions in a 5 kg
sample of the salt, if the density of salt is 1050 kg m -3 and the shape factor () is 1.75.
Aperture of Tyler sieves, 7 mesh = 2.83 mm, 9 mesh = 2.00 mm, 80 mesh = 0.177 mm, 115 mesh =
0.125 mm.
Mean aperture 7 and 9 mesh
= 2.41 mm = 2.4 x 10-3m
Mean aperture 80 and 115 mesh = 0.151 mm = 0.151 x 10-3m
Answer:
16.6 m2
PROBLEM 2
A sand mixture was screened through a standard 10-mesh screen. The mass fraction of the oversize
material in feed, overflow and underflow were found to be 0.38, 0.79 and 0.22 respectively. The screen
effectiveness based on the oversize is.
Answer:
0.58
PROBLEM 3
The collection efficiency of a cyclone is 45% over the size range 0-5μm, 80% over the size range 5-10μm,
and 96% for particles exceeding 10 μm. Calculate the efficiency of collection for a dust with a mass
distribution of 50% 0-5 μm, 30% 5-10μm and 20% above 10μm.
Size (micron)
0-5
5 - 10
> 10
% Efficiency
45
80
96
30
20
Dust Particle
% mass
Answer:
PROBLEM 4
50
65.70%
The screen with a total feed to bottom of 75.1 tons/h are shaking screens with a capacity of 4 metric
tons/m2-h-mm mesh size. How many square meters of screen are needed for each of the screens 8 mesh
to 14 mesh, if the feed to the first screen is 100tons/h?
Answer:
16.08 m2
FLOTATION
PROBLEM 1
A flotation plant processes 3000 tons/day of CuFeS2. It produces 80 tons Cu concentrate assaying 25%
Cu. If ore analyzes 0.7% Cu, the percent recovery is?
Answer:
92.24%
PROBLEM 2
Ground lead ore is to be concentrated by a single flotation process using 1.5 oz of reagent per ton of ore.
The feed concentrate and tailings have the following composition by weight on a dry basis
Feed
Concentrate
Tailings
PbS
30
90
0.9
ZnS
25
3
35.6
SiO2
45
7
63.5
Water is fed to the cell at the rate of 1000 gallons per ton of wet concentrate with 99% of the water leaving
with the tailings and 1% with the concentrate. Find the mass of wet concentrate produced per hour when
ten tons of ore are fed to the cell / 24 hr. is?
Answer:
3.4 kg/hr
CENTRIFUGATION
PROBLEM 1
A dispersion of oil in water is to be separated using a centrifuge. Assume that the oil is dispersed in the
form of spherical globules 5.1 x 10-5 m diameter and that its density is 894 kg m-3. If the centrifuge rotates
at 1500 rev/min and the effective radius at which the separation occurs is 3.8 cm, calculate the velocity of
the oil through the water. Take the density of water to be 1000 kg m-3 and its viscosity to be 0.7 x 10-3 N s
m-2.
SOLUTION
𝑉𝑡 =
𝐷 2𝑟(2𝜋
𝑁 2
) (𝜌𝑝 − 𝜌)
𝐷 2 𝑁 2 𝑟(𝜌𝑝 − 𝜌)
60
=
18𝜇
164𝜇
(5.1 𝑥 10−5 )2 (1500)2 (0.038)(1000 − 894)
𝑉𝑡 =
(164)(0.7 𝑥 10−3 )
𝑨𝑵𝑺𝑾𝑬𝑹: 𝑉𝑡 = 0.02 𝑚/𝑠
PROBLEM 2
If a cream separator has discharge radii of 5 cm and 7.5 cm and if the density of skim milk is 1032 kg m-3
and that of cream is 915 kg m-3, calculate the radius of the neutral zone so that the feed inlet can be
designed.
For skim milk, r1 = 0.075m, rA = 1032 kg m-3, cream r2 = 0.05 m, rB= 915 kg m-3
Answer:
17cm
PROBLEM 3
A viscous solution containing particles with a density of 1461 kg/m3 is to be clarified by centrifugation. The
solution density is 801 kg/m3 and its viscosity is 100 cp. The centrifuge has a bowl with r2 = 0.02225 m, r1
= 0.00715 m, and height b = 0.1970 m. Calculate the critical particle diameter of the largest particles in the
exit stream if N = 23,000 rev/min and flow rate of 0.002832 m3/h.
Answer:
0.746 micron
PROBLEM 4
A centrifuge bowl is spinning at a constant speed of 2000 rpm. The radius of bowl in cm needed to create a
force of 455 g is ?
Answer: 10.2 m
PROBLEM 5
If a centrifuge is 3-ft diameter and rotates at 1,000 rpm, what must be the speed of a laboratory centrifuge
of 6-in diameter be ran if it is duplicate plant conditions?
Answer:
2449 rev/min
PROBLEM 6
5. If a particle of mass m is rotating at radius x with an angular velocity ω, it is subjected to a centrifugal
force mxω2 in a radial direction and a gravitational force mg in a vertical direction. The ratio of the
centrifugal to gravitational forces, xω2/g, is a measure of the separating power of the machine, and in order
to duplicate conditions this must be the same in both machines.
Answer
49 Hz
SEDIMENTATION
PROBLEM 1
A mixture of coal and sand particles having sizes smaller than 1 x 10-4 m in diameter is to be separated by
screening and subsequent elutriation with water. Recommend a screen aperture such that the oversize
from the screen can be separated completely into sand and coal particles by elutriation. Calculate also the
required water velocity. Assume that Stokes law is applicable. Density of sand = 2650 kg/m3; density of
coal = 1350 kg/m3; density of water = 1000 kg/m3; viscosity of water = 1 x 10-3 kg/m.s; g = 9.812 m/s2.
SOLUTION
For laminar settling regimes, terminal-settling velocity is given by Stokes law: (valid for NRe,p < 0.1)
𝐷 2 (𝜌𝑝 − 𝜌)𝑔
𝑉𝑡 =
18𝜇
Settling velocity for larger coal
(1𝑋104 )2 (1350 − 1000)(9.81)
𝑉𝑡 =
18(.0010)
𝑨𝑵𝑺𝑾𝑬𝑹: 𝑉𝑡 = 1.9079 𝑋 10−3 𝑚/𝑠
Diameter of the sand corresponding to the settling velocity of larger coal
1.9079 𝑋 10−3 =
(𝐷)2 (2650 − 1000)(9.81)
18(.0010)
𝑨𝑵𝑺𝑾𝑬𝑹 ∶ 𝐷 = 4.6056 𝑋 10−5 𝑚
PROBLEM 2
Calculate the settling velocity of dust particles of (a) 60 mm and (b)10 mm diameter in air at 21°C and 100
kPa pressure. Assume that the particles are spherical and of density 1280 kg m-3, and that the viscosity of
air = 1.8 x 10-5 N s m-2 and density of air = 1.2 kg m-3.
Answer:
(a) 0.14 m/s
(b) 0.0039 m/s
PROBLEM 3
A continuous separating tank is to be designed to follow after a water washing plant for liquid oil. Estimate
the necessary area for the tank if the oil, on leaving the washer, is in the form of globules 5.1 x 10-5 m
diameter, the feed concentration is 4 kg water to 1 kg oil, and the leaving water is effectively oil free. The
feed rate is 1000 kg h-1, the density of the oil is 894 kg m-3 and the temperature of the oil and of the water
is 38°C. Assume Stokes' Law.
Viscosity of water = 0.7 x 10-3 N s m-2.
Density of water = 1000 kg m-3.
Diameter of globules = 5.1 x 10-5 m
Answer:
1.0 m2
PROBLEM 4
A particle settles through a suspension containing 35% solids by weight. Given that the velocity under free
settling motion is Vt = 0.00669 m/s. Find Vs.[density of particle = 2800 kg/m3, Dp = 200 mesh, density of
fluid = 996.5 kg/m3, viscosity of fluid = 0.8Cp]
Answer:
0.003 m/s
PROBLEM 5
A suspension of uniform particles in water at a concentration of 500 kg of solids per cubic meter of slurry is
settling in a tank. Density of the particles is 2500 kg/m3 and the terminal velocity of a single particle is 20
cm/s. What will be thee settling velocity of suspension? Richardson and Zaki index is 4.6
Answer:
7.16 cm/s
PROBLEM 6
Oil droplets having a diameter of 20 μm (0.020 mm) are to be settled from air at temperature of 37.8°C and
101.3 kPa pressure. The density of the oil is 900 kg/m3. Calculate the terminal settling velocity of the
droplets. Air at 37.8°C: ρ = 1.137 kg/m3, μ = 1.90 X 10-5 Pa•s
Answer:
0.0103 m/s
PART II
CONCEPTS
(MULTIPLE CHOICE)
SIZE REDUCTION
[1] Power required to drive a ball mill with a particular ball load is proportional to (where, D = diameter of
ball mill)
A.
D
B.
1/D
C.
D2.5
D.
1/D2.5
[2] Pick out the wrong statement
A.
Recycled coarse material to the grinder by a classifier is termed as circulating load.
B.
Wear and tear in wet crushing is more than that in dry crushing of materials.
C.
Size enlargement (opposite of size reduction) is not a mechanical operation
D.
A 'dust catcher' is simply an enlargement in a pipeline which permits the solids to settle down due
to reduction in velocity of the dust laden gas.
[3] For sizing of fine materials, the most suitable equipment is a
A.
Trommel
B.
Grizzly
C.
Shaking Screen
D.
Vibrating Screen
[4] Pebble mills are tumbling mills widely used for grinding in the manufacture of paints & pigments and
cosmetic industries, where iron contamination in the product is highly objectionable. Pebbles used in
pebble mill are made of
A.
bronze
B.
stainless steel
C.
flint or porcelain
D.
concrete
[5]
Which of the following relationships between co-efficinet of friction (μ) between rock & roll and a
(half of the angle of nip) of the particle to be crushed is correct ?
A.
μ > tan α
B.
μ ≥ tan α
C.
μ > tan 2α
D.
μ ≤ tan α
[6] Pick out the wrong statement pertaining to the roll crushers
A.
Maximum feed size determines the required roll diameter.
B.
For hard material's crushing, the reduction ratio should not exceed 4
C.
Both the rolls run necessarily at the same speed
D.
Reduction ratio and differential roll speed affect production rate & energy consumed per unit of
surface produced.
[7] Which of the following is not an ultrafine grinder (colloid mill)?
A.
Micronizers
B.
Agitated mills and fluid energy mills
C.
Toothed roll crusher
D.
Hammer mills with internal classification
[8] Number of particles in a crushed solid sample is given by (where, m = mass of particles in a sample, Vp
= volume of one particle, ρ= density of particles)
A.
m/ρ . Vp
B.
m . ρ/Vp
C.
m . Vp/ρ
D.
Vp/m . ρ
[9] Ball mills and tube mills with flint or porcelain balls are used for size reduction of
A.
asbestos
B.
rubber
C.
non-metallic ores
D.
limestone
[10] Size reduction of __________ is accomplished in steam heated rollers and roll crushers
A.
resins
B.
gums
C.
hard rubber
D.
wax
SCREENING
[1] Equivalent diameter of a particle is the diameter of the sphere having the same
A.
ratio of surface to volume as the actual volume.
B.
ratio of volume to surface as the particle.
C.
volume as the particle.
D.
none of these.
[2] If dp is the equivalent diameter of a non-spherical particle, Vp its volume and sp its surface area, then its
sphericity is Φs is defined by
A.
Φs = 6 Vp/dp sp
B.
Φs = Vp/dp sp
C.
Φs = Vp/dp sp
D.
Φs = dp Sp/Vp
[3] Filter aids like asbestos, kieselguhr, diatomaceous earth etc. are used to increase the porosity of the
final filter cake & reducing the cake resistance during filtration. Filter aid is
A.
added to the feed slurry
B.
precoated on the filter medium prior to filtration.
C.
separated from the cake by dissolving solids or by burning it off
D.
all 'a', 'b'&'c'
[4] Pick out the correct statement:
A.
Removal of iron from ceramic material is necessitated (by magnetic separation method) so as to
avoid discolouration of ceramic products.
B.
The operating cost of shaking screen is more than that of a vibrating screen.
C.
Screen capacity does not depend upon the specific gravity of the minerals.
D.
Asphalt is best crushed using toothed roll crusher.
[5] Jigging is a technique by which different particles can be
A.
separated by particle size
B.
separated by particle density.
C.
separated by particle shape
D.
mixed
[6] 200 mesh screen means 200 openings per
A.
cm2
B.
cm
C.
inch
D.
inch2
[7] Screen capacity is not a function of
A.
its openings size
B.
screening mechanism
C.
screening surface
D.
atmospheric humidity
[8] Increasing the capacity of a screen __________ the screen effectiveness
A.
decreases
B.
increases
C.
does not affect
D.
none of these
[9] The removal of a small amount of oversize from a feed which are predominantly fines is called
A.
Scalping
B.
Coarse Separation
C.
Desliming
D.
Dewatering
[10] Making a size separation smaller than 48-mesh is called
A.
Ultrafine separation
B.
Coarse Separation
C.
Scalping
D.
Fine Separation
FLOTATION
[1] Froth floatation is the most suitable for treating
A.
iron ore
B.
sulphide ore
C.
quartzite
D.
none of these
[2] What is the selectivity index, if the grade of tailings & concentrate is the same?
A.
0
B.
1
C.
infinite
D.
1/2
[3] Any operation in which one solid is separated from another by floating one of them at or on the surface
of a fluid.
A.
coagulation
B.
flotation
C.
centrifugation
D.
sedimentation
[4] Pine oil used in forth flotation technique acts as a
A.
collector
B.
frother
C.
modifier
D.
activator
[5] Which of the following is the most suitable for cleaning of fine coal dust (< 0.5 m)?
A.
though washer
B.
spiral separator
C.
baum jig washer
D.
froth flotation
[6] In Froth Flotation, chemical agent added to cause air adherence is called
A.
collector
B.
modifier
C.
frother
D.
promoter
[7] Alter the surface of the mineral in order that it will become air-avoid (causing it to adhere to air bubbles)
A.
regulators
B.
activators
C.
depressor
D.
promoter
[8] Enhance the absorption of a collector by mineral particles
A.
promoter
B.
regulator
C.
depressor
D.
activator
[9] Prevents the absorption of a collector by a mineral particle and thereof inhibit the flotation of mineral
A.
depressor
B.
regulator
C.
activator
D.
promoter
[10] A flotation modifier which is assists in the selectivity (sharpness of separation) or stop unwanted
minerals from floating.
A.
depressants
B.
activators
C.
alkaline regulator
D.
promoter
SEDIMENTATION
[1] Two particles are called to be equal settling, if they are having the same.
A.
Size
B.
Specific Gravity
C.
Terminal velocities in the same fluid & in the same field of force.
D.
None of these
[2] Gravity settling process is not involved in the working of a
A.
hydrocyclone
B.
classifier
C.
dorr-thickener
D.
sedimentation tank
[3] Pine oil used in froth floatation technique acts as a/an
A.
collector
B.
modifier
C.
frother
D.
activator
[4] __________ are used for the separation of coarse particles from a slurry of fine particles
A.
thickeners
B.
classifiers
C.
hydrocyclones
D.
decanters
[5] A suspension of glass beads in ethylene glycol has a hindered settling velocity of 1.7 mm/s, while the
terminal settling velocity of a single glass bead in ethylene glycol is 17 mm/s. If the Richardson-Zaki
hindered settling index is 4.5, the volume fraction of solids in the suspension is
A.
0.1
B.
0.4
C.
0.6
D.
none of these
[6] For a cyclone of diameter 0.2 m with a tangential velocity of 15 m/s at the wall, the separation factor is
A.
2250
B.
1125
C.
460
D.
230
[7] Ore concentration by jigging is based on the difference in the __________ of the particles.
A.
specific gravities
B.
wettability
C.
shape
D.
none of these
[8] Cyclones are used primarily for separating
A.
solids
B.
solids from liquids
C.
liquids
D.
solids from solids
[9] Agglomeration of individual particles into clusters (flocs) is called flocculation. To prevent flocculation,
the most commonly used dispersing agents are
A.
carbonates
B.
sulphates
C.
silicates & phosphates
D.
bicarbonates
[10] Device in which a current of air separates particles with different sedimentation velocities
A.
agitator
B.
air elutriator
C.
classifier
D.
air conveyor
CENTRFUGATION
[1] Tabular bowl centrifuges as compared to disk bowl centrifuges
A.
operate at higher speed.
B.
employ bowl of larger diameter
C.
can not be operated under pressure/vacuum
D.
can't be used for separation of fine suspended solids from a liquid
[2] __________ centrifuge is normally used in sugar mills
A.
Tubular bowl
B.
Disc-bowl
C.
Suspended batch basket
D.
Perforated horizontal basket continuous
[3] If a force greater than that of gravity is used to separate solids & fluids of different densities, the process
is termed as the
A.
sedimentation
B.
flocculation
C.
dispersion
D.
centrifugation
[4] Where the density difference of the two liquid phase to be separated is very small (as in milk cream
separator), the most suitable separator is a
A.
disc bowl centrifuge
B.
sharpies supercentrifuge
C.
batch basket centrifuge
D.
sparkler filter
[5] For separation of sugar solution from settled out mud, we use a __________ filter.
A.
sparkler
B.
plate and frame
C.
centrifugal
D.
rotary drum vacuum
[6] The speed of a rotary drum vacuum filter may be about __________ rpm.
A.
1
B.
50
C.
100
D.
500
[7] Ultra centrifuges are used for the separation of __________ solid particles
A.
coarse
B.
fine
C.
colloidal
D.
dissolved
[8] Driving force in case of filtration by a centrifuge is the
A.
speed of the centrifuge
B.
centrifugal pressure exerted by the liquid
C.
narrow diameter of the vessel
D.
formation of highly porous cake
[9] Is created by moving a mass in a curved path and is exerted in the direction away from the center of
curvature of the path.
A.
force
B.
settling
C.
centrifugal force
D.
raising
[10] Is the force applied to the moving mass in the direction toward the center of curvature which causes
the mass in the direction toward the center of curvature which causes the mass to travel in a curved path.
A.
centripetal force
B.
rotational force
C.
centrifugal force
D.
filtration
Technological Institute of the Philippines – Manila
363 P. Casal Street, Quiapo, Manila
College of Engineering
Chemical Engineering Department
INTRODUCTION TO PARTICLE TECHNOLOGY
(Compilation of Worked Problems)
Submitted by:
LIWANAG, MARY CHRISTINE B.
Submitted to:
ENGR. ROBERT E. DELFIN
Date Submitted:
25 MARCH 2014
Screening
CONCEPT QUESTIONS
1.
In Tyler series, the ratio of the aperture size of a screen to that of the next smaller screen is:
A.
1/√2
B.
√2
C.
1.5
D.
2
Answer:
2.
B
In the Tyler standard screen scale series, when the mesh number increases from 3 mesh to 10
mesh, then
A.
the clear opening decreases
B.
the clear opening increases
C.
the clear opening is unchanged
D.
the wire diameter increases
Answer:
A
3.
The material passing one screening surface and retained on a subsequent surface is called
A.
Intermediate material
B.
Minus material
C.
Plus material
D.
None of these
Answer:
4.
A
The minimum clear space between the edges of the opening in the screening surface and is
usually given in inches or millimeters
A.
Sieve
B.
Aperture
C.
Mesh number
D.
Holes
Answer:
B
5.
The screen used in making size separation smaller than 4-mesh and larger than 48-mesh
A.
Grizzly screen
B.
Gyratory screen
C.
Oscillating screen
D.
Vibrating screen
Answer:
6.
D
When the object is used to recover an oversize product from the screen, efficiency may be
expressed as
A.
Ratio of the amount of undersize obtained to the amount of undersize in the feed
B.
Using the Taggart Formula, E =
100(e - v)
×100 , where e = percentage of undersize in the feed;
e(100 - v)
v = percentage of undersize in the screen oversize
C.
Ratio of the amount of oversize obtained to the amount of true oversize
D.
None of these
Answer:
C
7.
Box-like machines, either round or square with a series of screen clothes nested atop one another
A.
Reciprocating screen
B.
Oscillating screen
C.
Electricity vibrated screen
D.
Gyratory screen
Answer:
D
8.
The removal of a small amount of oversize from a feed which are predominantly fines is called
A.
Scalping
B.
Desliming
C.
Coarse separation
D.
Dewatering
Answer:
A
9.
Removal of free water from a solids-water mixture and is generally limited to 4 mesh and above
A.
Scalping
B.
Dewatering
C.
Trash removal
D.
Separation
Answer:
B
10.
Making a size separation smaller than 48-mesh is called
A.
Coarse separation
B.
Fine separation
C.
Ultrafine separation
D.
Scalping
Answer:
C
PROBLEM-SOLVING
1.
A sand mixture was screened through a standard 10-mesh screen. The mass fraction of the
oversize material in feed, overflow and underflow were found to be 0.38, 0.79 and 0.22 respectively. The
screen effectiveness based on the oversize is _____.
F=D+B
Fx f = Dx d + Bx b
D x f - xb
=
F xd - xb
Screen effectiveness
oversize material in the overflow
=
oversize material in the feed
Dx d
=
Fx f
 x - x  x 
=  f b  d 
 xd - xb   x f 
 0.38 - 0.22   0.79 
=


 0.79 - 0.22   0.38 
= 0.58
2.
Answer:
3.
Answer:
4.
The wire diameter of a 14-mesh screen (Tyler Standard) is _____.
0.025 in.
The wire diameter of a 10-mesh screen whose aperture is 0.065 in. is _____.
0.035 in.
It is desired to separate 1000 kg of a mixture of crushed solids into three fractions, a coarse
fraction retained on a 20 mesh screen; a middle fraction passing through a 20 mesh screen and retained on
a 65 mesh screen; and a fine fraction passing through a 65 mesh screen. Two Tyler Standard Screens are
used to remove particles 20/65.
Screen analysis of the feed, coarse, medium and fine fractions are given. What are the
effectiveness of the 20 and 65 screens, given the following screen analysis?
Mesh
% Feed
%P
%Medium
%Fine
+4 - 6
2.51
3.26
0.5
0
-6 + 8
12.5
13.59
11.27
0
-8 + 10
32.07
38.04
18.03
1.66
-10 + 14
25.7
27.12
22.53
16.63
-14 + 20
15.9
16.3
13.52
21.38
-20 + 28
5.38
1.36
18.03
9.03
-28 + 35
2.1
0.27
8.11
2.38
-35 + 48
1.02
0
4.51
0.48
-48 + 65
0.77
0
3.42
0.24
-65 + 100
0.58
0
0.09
13.3
-100 + 150
0.41
0
0
9.74
-150 + 200
0.31
0
0
7.36
PAN
0.75
0
0
17.81
100
100
100
100
Answer:
65-mesh: 96.6%, 20-mesh: 71.8%
Powdered coal with screen analysis given below is fed to a vibrating 48-mesh screen in an attempt
to remove the undesired fine material. The oversize product comes at a rate of 80 MT/hr.
Screen Analyses – Mass Fractions
Mesh
Feed
Oversize
-3 + 4
0.010
0.014
-4 + 6
0.022
0.031
-6 + 8
0.063
0.088
-8 + 10
0.081
0.112
-10 + 14
0.102
0.142
Undersize
-14 + 20
0.165
0.229
-20 + 28
0.131
0.182
-28 + 35
0.101
0.104
0.093
-35 + 48
0.095
0.065
0.171
-48 + 65
0.070
0.025
0.186
-65 + 100
0.047
0.008
0.146
-100 + 150
0.031
0.111
-150 + 200
0.020
0.071
-200
0.062
0.222
5.
What is the effectiveness of the screen?
Answer:
6.
81.06%
If the capacity of the screen is 2 MT/m2-hr-mm mesh size, how many square meters of screen are
needed?
Answer:
188.39 m2
Size Reduction
CONCEPT QUESTIONS
1.
In a gyratory crusher size reduction is effected primarily by:
A.
Compression
B.
Impact
C.
Attrition
D.
Cutting action
Answer:
2.
A
How is work index defined?
Answer:
Work index is defined as the gross energy (expressed in kWH) per ton of feed required to
convert a very large feed to a size such that 80% of the product passes through a 100mm (micron) screen.
3.
A fluid energy mill is used for
A.
Cutting
B.
Grinding
C.
Ultra grinding
D.
Crushing
Answer:
C
4.
To produce talcum powder, use:
A.
Ball mill
B.
Hammer mill
C.
Jet mill
D.
Pin mill
Answer:
A
5.
The work index in Bond’s law for crushing of solids has the following dimensions:
A.
No units (dimensionless)
B.
kWh/ton
C.
kW/ton
D.
kWh-m1/2/ton
Answer:
6.
B
Arrange the following size reduction equipment in the decreasing order of the average particle size
produced by each of them.
A.
Jaw crusher, ball mills, fluid energy mills
B.
Ball mills, jaw crusher, fluid energy mills
C.
Fluid energy mills, jaw crusher, ball mills
D.
Fluid energy mills, ball mills, jaw crusher
Answer:
A
7.
Energy requirement (per unit mass of material crushed/ground) is highest for
A.
Jaw crusher
B.
Rod mill
C.
Ball mill
D.
Fluid energy mill
Answer:
D
8.
The critical speed of the ball mill of radius R, which contains balls of radius r, is proportional to
A.
(R – r)-0.5
B.
(R – r)-1
C.
(R – r)
D.
(R – r)2
Answer:
A
9.
Size reduction of coarse hard solids using a crusher is accomplished by
A.
Attrition
B.
Compression
C.
Cutting
D.
Impact
Answer:
B
10.
The power required for size reduction in crushing is
A.
1/surface energy of the material
B.
1/(surface energy of the material)1/2
C.
Proportional to the surface energy of the material
D.
Independent of the surface energy of the material
Answer:
C
PROBLEM-SOLVING
1.
A continuous grinder obeying the Bond crushing law grinds a solid at the rate of 1000 kg/hr from
the initial diameter of 10 mm to the final diameter of 1 mm. If the market now demands particles of size 0.5
mm, the output rate of the grinder (in kg/hr) for the same power input would be reduced to _____.
Answer:
62.3%
 1
P
1 

=K
T
Df 
 Dp
where:
P = power required
T = feed rate
K = constant (depends on the nature of particle and the system of units used)
Dp = diameter of product
Df = diameter of feed
P
1 
 1
=K
 = K × 0.684
T1
10 mm 
 1 mm
P
1
1 

=K
 = K × 1.098
T2
10 mm 
 0.5 mm
T2
0.684
=
= 0.623
T1 1.098
2.
The power required to crush 100 tons/hr of a material is 179.8 kW, if 80% of the feed passes
through a 51 mm screen and 80% of the product passes through a 3.2 mm screen. What is the work index
of the material?
Answer:
3.
13.571
What will be the power required for the same feed at 100 tons/hr to be crushed to a product such
that 80% is to pass through a 1.6 mm screen?
Answer:
4.
279.157 kW
Particles of average feed size 25 x 10-4 m are crushed to an average product size of 5 x 10-4 m at
the rate of 15 tons per hour. At this rate the crusher consumes 32 kW of power of which 2 kW are required
for running the mill empty. What would be the power consumption if 10 tons per hour of this product is
further crushed 1 x 10-4 m size in the same mill? Assume that Rittinger’s law is applicable.
Answer:
5.
P = 100 kW
The energy required per unit mass to grind limestone particles of very large size to 100 Mm is 12.7
kWh/ton. An estimate (using Bond’s Law) of the energy to grind the particles from a very large size to 50
mm is _____.
Answer:
6.
18 kWh/ton
What is the critical rotational speed, in revolutions per second, for a ball mill of 1.2 m diameter
charged with 70 mm diameter balls?
Answer:
0.66
Flotation
CONCEPT QUESTIONS
1.
Froth floatation is the most suitable for treating
A.
Iron ores
B.
Sulphide ores
C.
Quartzite
D.
None of these
Answer:
B
2.
Pine oil used in froth floatation technique acts as a/an
A.
Collector
B.
Modifier
C.
Frother
D.
Activator
Answer:
C
3.
What is the selectivity index, if the grade of tailings & concentrate is the same?
A.
0
B.
∞
C.
1
D.
0.5
Answer:
4.
C
There is practically no alternative/competitor to __________ in the beneficiation treatment of
sulphide ores.
A.
Classification
B.
Tabling
C.
Jigging
D.
Froth floatation
Answer:
5.
D
For benefication of iron ore, the most commonly used method is
A.
Flocculation
B.
Froth floatation
C.
Jigging & tabling
D.
None of these
Answer:
6.
C
In coal washeries, three products namely the valuable product (i.e. clean/washed coal), discarded
product (i.e. mineral matter) and an additional concentrated product called __________ is produced.
A.
Concentrate
B.
Tailing
C.
Middling
D.
None of these
Answer:
C
7.
Xanthates are used in the froth floatation process as a/an
A.
Conditioner
B.
Frother
C.
Collector
D.
Activator
Answer:
C
8.
Which of the following is the most suitable for cleaning of fine coal dust (<0.5 mm)?
A.
Trough washer
B.
Baum jig washer
C.
Spiral separator
D.
Froth floatation
Answer:
9.
D
Activators are those chemicals which help buoying up one mineral in preference to the other in the
froth floatation process. Which of the following is an activator?
A.
Cresylic acid
B.
Copper sulphate
C.
Calcium carbonate
D.
Sodium carbonate
Answer:
10.
B
A flotation modifier which assists in the selectivity (sharpness of separation) or stop unwanted
minerals from floating
A.
Depressants
B.
Activators
C.
Alkalinity regulators
D.
Promoters
Answer:
A
PROBLEM-SOLVING
Ground lead ore is to be concentrated by a single flotation process using 1.5 oz of reagent per ton
of ore. The feed concentrate and tailings have the following concentration by weight on a dry basis:
Feed %
Concentrate %
Tailings %
PbS
30
90
0.9
ZnS
25
3
35.6
SiO2
45
7
63.5
Water is fed to the cell at the rate of 1000 gallons per ton of wet concentrate with 99% of the water leaving
with the tailings and 1% with the concentrate.
1.
Mass of wet concentrate produced per hour when ten tons of ore are fed to the cell per 24 hours is
_____.
F=C+T
10 tons/24 h = C + T  (1)
FX F = CX C + TX T
(10 tons/24 h) × 0.30 = 0.90C + 0.009T
 (2)
Solving, C = 3.266 tons/24-h, T = 6.734 tons/24-h
 1000 gal

1 ft 3
62.43 lb
1 ton
WC = 3.266 tons/24-h + WC × 
×
×
×
× 0.01
3
7.481 gal
1 ft
2000 lb
 ton WC

WC = 3.4 tons/24-h
2.
Answer:
3.
Total water required in pounds per hour is _____.
1200
A flotation plant processes 3000 tons/day of CuFeS 2 (chalcopyrite). It produces 80 tons Cu
concentrate assaying 25% Cu. If ore analyzes 0.7% Cu, the percent recovery is _____.
Answer:
95.24%
4.
A typical flotation machine has the following specifications:
Number of cells = 4
Flotation time = 12 min
Cell volume = 60 ft.3
Hp per cell = 10 hp
The material treated has the following specifications:
Pulp (mixture of ore and water) = 40% solids
SG of ore = 3
Equation: n = T × cap × d where n = number of cells; V = volume in ft. 3 per cell; cap = tons of dry ore per
V × 1440
24 hours; d = ft.3 of pulp (ore and water) containing one to (2000 lbs. of solids)
The capacity of the machine in tons of dry ore per 24 hours is _____.
Answer:
490
A flotation section of a mining company is extracting CuS from covellite ores. The ore consists of
5% CuS and 95% gangue, which may be assume to be SiO 2. The following data are given:
% CuS
% SiO2
Feed
5
95
Concentrate
85
15
Rougher tailings
1
99
Scavenger concentrate
10
90
Final tailings
0.3
99.7
Laboratory experiments indicated that the water to solids ratio L/S =2 and the contact time is 10 min. in the
rougher; L/S = 4, contact time = 18 min. in the scavenger. On the basis of 300 tons per day of ore treated.
Data:
ρSiO2 = 2.65 g/cc, ρCuS = 4.6 g/cc
5.
The volume of the rougher needed is _____.
Answer:
6.
Answer:
169.8 ft.3
The volume of the scavenger needed is _____.
535 ft.3
Sedimentation
CONCEPT QUESTIONS
1.
A particle ‘A’ of diameter 10 microns settles in an oil of specific gravity 0.9 and viscosity 10 poise
under Stokes law. A particle ‘B’ with diameter 20 microns settling in the same oil will have a settling velocity
A.
Same as that of ‘A’
B.
One-fourth as that of ‘A’
C.
Twice as that of ‘A’
D.
Four times as that of ‘A’
Answer:
2.
D
For separating particles of different densities, the differential settling method uses a liquid sorting
medium of density
A.
Intermediate between those of the light and the heavy one
B.
Less than that of either one
C.
Greater than that of either one
D.
Of any arbitrary value
Answer:
A
3.
A particle attains its terminal settling velocity when
A.
Gravity force + drag force = buoyancy force
B.
Gravity force – drag force = buoyancy force
C.
Buoyancy force + drag force = gravity force
D.
Drag force = buoyancy force
Answer:
B
4.
Stokes equation is valid in the Reynolds number range
A.
0.01 to 0.1
B.
0.1 to 2
C.
2 to 10
D.
10 to 100
Answer:
A
5.
For a sphere falling in the constant drag coefficient regime its terminal velocity depends on its
diameter (d) as
A.
d
B.
√d
C.
d2
D.
1/d
Answer:
6.
C
Two identically sized spherical particles A and B having densities ρA and ρB respectively; are
settling in a fluid of density ρ. Assuming free settling under turbulent flow conditions, the ratio of the
terminal settling velocity of particle A to that of particle B is given by
A.
(ρA - ρ )
( ρB - ρ )
B.
( ρB - ρ )
(ρA - ρ )
C.
(ρA
( ρB
D.
( ρB
(ρA
- ρ)
- ρ)
- ρ)
- ρ)
Answer:
A
7.
Two particles are called to be equal settling, if they are having the same
A.
Size
B.
Specific gravity
C.
Terminal velocities in the same fluid and in the same field of force
D.
None of these
Answer:
8.
C
For a sphere falling in the constant drag co-efficient regime, its terminal velocity depends on its
diameter (d) as
A.
d
B.
√d
C.
d2
D.
1/d
Answer:
9.
C
Separation of solid suspended in liquid into a supernatant clear liquid and a denser slurry employs
a process termed as the
A.
Coagulation
B.
Flocculation
C.
Sedimentation
D.
Clarification
Answer:
C
10.
The process by which fine solids is removed from liquids is termed as
A.
Decantation
B.
Flocculation
C.
Sedimentation
D.
Classification
Answer:
C
PROBLEM-SOLVING
1.
Quartz and pyrites (FeS2) are separated by continuous hydraulic classifications using free settling
conditions in water at 28 0C. The feed to the classifier ranges in size between 10 microns and 55 microns.
The specific gravity of quartz is 2.65 and that of pyrites is 5.1. Calculate the size range of the various
fractions obtained in the settling.
H2O at 28o C:
ρ = 996.2 kg/m3
μ = 8.392×10-4
kg
m-s
Range: 10μm-55μm
Let: A = quartz, B = FeS 2
SG A = 2.65, SGB = 5.1
 gρ(ρPA - ρ) 
k = Dp A 

μ2


1/3


-6  9.81 × 996.2 × (5100 - 2650) 
k = 5.5 × 10
2
-4


8.392
×
10


(
k = 2.12, Stoke's Law
)
1/3
Dp A
DpB
3
ρpB - ρ
=
ρpA - ρ
4
DpA = 5.5 × 10-6 ×
3
2650 - 996.2
5100 - 996.2
DpA = 3.49 × 10-5 m
3
Dp A
1
DpB
ρpB - ρ
=
ρpA - ρ
2
5100 - 996.2
2650 - 996.2
-5
= 1.58 × 10 m
DpB = 1 × 10-5 ×
2
DpB
2
Size Range:
Pure A: Dp3 = 3.49 × 10-5 m, Dp4 = 5.5 × 10 -5 m
Pure B: Dp1 = 1 × 10-5 m, D p2 = 1.58 × 10 -5 m
Mixture:
Dp1A = 1 × 10-5 m to Dp3A = 3.49 × 10 -5 m
Dp2B = 1.58 × 10-5 m to D p4B = 5.5 × 10 -5 m
2.
The maximum particle of a spherical sand particle (ρ = 2850 kg/m3) that will settle in the Stoke’s
law region in water (ρ = 1000 kg/m3, μ = 0.001 kg/m-s) is _____.
Answer:
3.
32.64 μm
In a mixture of quartz (SG = 2.85) and galena (SG = 7.5), the size of the particles range from
0.0002 cm to 0.001 cm. On separation in a hydraulic classifier using water under free settling conditions,
what is the maximum size of quartz and minimum of galena in pure products? (μ = 0.001 kg/m-s, ρ = 1000
kg/m3)
Answer:
cm
Largest quartz particle in galena: 0.00053 cm, Smallest galena particle in quartz: 0.000374
4.
What is the terminal velocity in m/s, calculated from Stokes law, for a particle diameter 0.1 x 10 -3 m,
density 21800 kg/m3 settling in water of density 1000 kg/m3 and viscosity = 10-3 kg/m-s? Assume g = 10
m/s2
Answer:
5.
10-2 m/s
Two spherical particles have the same outer diameter but are made of different materials. The first
one (with material density ρ1) is solid, whereas the second (with material density ρ2) is a hollow sphere with
the inner shell diameter equal to half the outer diameter. If both the spheres have the same terminal
velocity in any fluid, then the ratio of their material densities ρ1/ρ2 is _____
Answer:
6.
8/7
The terminal settling velocity of a 6 mm diameter glass sphere (density: 2500 kg/m 3) in a viscous
Newtonian liquid (density: 1500 kg/m3) is 100 μm/s. If the particle Reynolds number is small and the value
of acceleration due to gravity is 9.81 m/s 2, then the viscosity of the liquid (in Pa-s) is
Answer:
196.2
Centrifugation
CONCEPT QUESTIONS
1.
Tabular bowl centrifuges as compared to disk bowl centrifuges
A.
Operate at higher speed
B.
Employ bowl of larger diameter
C.
Cannot be operated under pressure/vacuum
D.
Cannot be used for separation of fine suspended solids from a liquid
Answer:
2.
A
If a force greater than that of gravity is used to separate solids & fluids of different densities, the
process is termed as the
A.
Sedimentation
B.
Flocculation
C.
Dispersion
D.
Centrifugation
Answer:
3.
D
Where the density difference of the two liquid phase to be separated is very small (as in milk cream
separator), the most suitable separator is a
A.
Disc bowl centrifuge
B.
Sharpies supercentrifuge
C.
Batch basket centrifuge
D.
Sparkler filter
Answer:
A
4.
Ultra centrifuges are used for the separation of __________ solid particles.
A.
Coarse
B.
Fine
C.
Colloidal
D.
Dissolved
Answer:
C
5.
Driving force in case of filtration by a centrifuge is the
A.
Speed of the centrifuge
B.
Centrifugal pressure exerted by the liquid
C.
Narrow diameter of the vessel
D.
Formation of highly porous cake
Answer:
B
6.
Moisture can be removed from lubricating oil using
A.
Tubular centrifuge
B.
Clarifier
C.
Sparkler filter
D.
Vacuum leaf filter
Answer:
A
7.
Which of the following can be most effectively used for clarification of lube oil and printing ink?
A.
Sparkler filter
B.
Precoat filter
C.
Disc-bowl centrifuge
D.
Sharpies supercentrifuge
Answer:
D
8.
Ultracentrifuges running at speeds up to 100000 rpm is normally used for the
A.
Separation of isotopes based on their density or molecular weights difference
B.
Concentration of rubber latex
C.
Separation of cream from milk
D.
Dewaxing of lubricating oil
Answer:
A
9.
If radius of a batch basket centrifuge is halved & the r.p.m. is doubled, then the
A.
Linear speed of the basket is doubled
B.
Linear speed of the basket is halved
C.
Centrifugal force is doubled
D.
Capacity of centrifuge is increased
Answer:
C
10.
__________ centrifuge is the most suitable for separation of non-friable crystals.
A.
Tubular bowl
B.
Disc-bowl
C.
Perforated horizontal basket continuous
D.
Suspended batch basket
Answer:
C
PROBLEM-SOLVING
1.
A centrifuge of diameter 0.2 m in a pilot plant rotates at a speed of 50 Hz in order to achieve
effective separation. If this centrifuge is scaled up to a diameter of 1 m in the chemical plant, and the same
separation factor is to be achieved, what is the rotational speed of the scaled up centrifuge?
Separation Factor =
ω 2r
g
To maintain the same separation factor:
ω12r1 ω 2 2r2
=
g
g
ω 2 = ω1
r1
0.2 m
= 50 ×
r2
1m
ω 2 = 50 Hz
2.
Ut =
In the Stokes regime, the terminal velocity of particles for centrifugal sedimentation is given by:
ω 2r ( ρp - ρ ) dp 2
18μ
where ω = angular velocity; r = distance of the particle from the axis of rotation; ρp =
density of the particle; ρ = density of the fluid; dp = diameter of the particle; μ = viscosity of the fluid.
In a Bowl centrifugal classifier operating at 60 rpm with water (μ = 0.001 kg/m-s), the time taken for
a particle (dp = 0.001 m, SG = 2.5) in seconds to traverse to a distance of 0.05 m from the liquid surface is
_____.
Answer:
3.
7.8
A centrifuge bowl is spinning at a constant 2000 rev/min. What radius bowl is needed for a force of
455 g’s?
Answer:
r = 0.1017 m
4.
A cream-separator centrifuge has an outlet discharge radius r1 = 50.8 mm and outlet radius r4 =
76.2 mm. The density of the skim milk is 1032 kg/m 3 and that of the cream is 865 kg/m3. Calculate the
radius of the interface neutral zone.
Answer:
5.
r2 = 150 mm
A batch centrifugal filter similar to Fig. 14.4-5 has a bowl height b = 0.0457 m and r2 = 0.381 m and
operates at 33.33 rev/s at 25.0 0C. The filtrate is essentially water. At a given time in the cycle, the slurry
and cake formed have the following properties: c s = 60.0 kg solids/m3 filtrate, ε = 0.82, ρp = 2002 kg
solids/m3, cake thickness = 0.152 m, α = 6.28 x 1010 m/kg, Rm = 8.53 x 1010 m-1, r1 = 0.2032 m. Calculate
the rate of filtrate flow.
Answer:
6.
q = 6.11 x 10-4 m3/s
A laboratory tubular-bowl centrifuge has the following dimensions, with respect to Figure 19.28,
and operating conditions: bowl speed 800 rps, R 0 = 0.875 inch, R1 = 0.65 in., and bowl length = L = 4.5
inches. When used to remove E. coli cells from the following fermentation broth, a satisfactory volumetric
feed capacity of the centrifuge Q, of 0.011 gpm is achieved.
Broth: ρf = 1.01 g/cm3 and μ = 1.02 x 10-3 kg/m-s
E. coli: Smallest diameter = dp min = 0.7 μm and ρp = 1.04 g/cm3
Assuming the applicability of Stokes’ Law, estimate the feed capacity of the centrifuge.
Answer:
0.0860 gpm
REFERENCES
(n.d.). Retrieved March 23, 2015, from http://unitoperation.com/
Chemical
Engineering Questions
and
Answers.
(n.d.).
Retrieved March
23,
2015,
from
http://www.indiabix.com/chemical-engineering/questions-and-answers/
Geankoplis, C. (2003). Principles of Transport Processes and Separation Processes (4th ed.). Upper
Saddle River, NJ: Prentice Hall Professional Technical Reference.
McCabe, W., & Smith, J. (1976). Unit operations of chemical engineering (3rd ed.). New York: McGraw-Hill.
Olano, S., Bungay, V., Centeno, C., Medina, L., & Salazar, C. (2012). Reviewer for Chemical Engineering
Licensure Examination (3rd ed.). Manila: Manila Review Institute.
Seader, J., & Henley, E. (2011). Separation process principles (3rd ed.). New York: Wiley.
TECHNOLOGICAL INSTITUTE OF THE PHILIPPINES
363 P.CASAL ST. QUIAPO MANILA
INTRODUCTION TO PARTICLE TECHNOLOGY
PROBLEM SET
SUBMITTED BY:
LLANETA, JIZELLE LHEA B.
SUBMITTED TO:
ENGR. ROBERT DELFIN
Screening
1.
It is removing a small amount of oversize from a feed which is predominantly fines
A. Scalding
C. Classifier
B. Separation
D. Desliming
2. It is making a size separation smaller than mesh no.4 and larger than mesh no.48
A. Scalding
C. Desliming
B. Separation
D. Dewatering
3. It is the removal of free water from a solids-water mixture
A. Desliming
C. Scalding
B. Classifier
D. Dewatering
4. It is the removal of extremely fine particles from a wet material by passing it over a screening surface.
A. Desliming
C. Scalding
B. Separation
D. Dewatering
5. It is a method of separation of particles according to size alone.
A. Size Reduction
C. Crystallization
B. Screening
D. NOTG
6. It is the ratio of surface of a spherical particle to actual surface area of non-spherical particle whose both
volume are equal
A. Sphericity
C. Screening
B. Particle Shape
D. Size Reduction
A. Particle of different sizes or not equi- dimensional are characterized based on its equivalent
B. Screening
D. Particle Size
C. Shape
E. Mean diameter
7. The most predominant used as mesh
A. Mesh number
C. Screening
B. Standard sieve
D. Tyler Series
8. For sizing of fine materials, the most suitable equipment is a
A. Trommel
B. Grizzly
C. Shaking screen
D. Vibrating screen
9. 200 mesh seive size corresponds to __________ microns.
A. 24
C. 154
B. 74
D. 200
10. Increasing the capacity of a screen ________ the screen effectiveness.
A. Decreases
B. Increases
C. No effect
D. Not on the given
11. Powdered coal with the screen analysis given below as feed is fed to a vibrating 48- mesh screen in an
attempt to remove the undesired fine materials. When the screen was new the oversize and undersize
analysis are listed under column headed ”NEW”. After 3 months operation, the analysis are headed “OLD”.
OVERSIZE
UNDERSIZE
MESH
FEED
NEW
OLD
NEW
OLD
-3+4
0.010
0.012
0.014
-----
-----
-4+6
0.022
0.027
0.031
-----
-----
-6+8
0.063
0.078
0.088
-----
-----
-8+10
0.081
0.100
0.112
-----
-----
-10+14
0.102
0.126
0.142
-----
-----
-14+20
0.165
0.204
0.229
-----
-----
-20+28
0.131
0.162
0.182
-----
-----
-28+35
0.101
0.125
0.104
-----
0.093
-35+48
0.095
0.117
0.065
-----
0.171
-48+65
0.070
0.029
0.025
0.246
0.186
-65+100
0.047
0.015
0.008
0.183
0.146
-100+150
0.031
0.005
-----
0.141
0.111
-150+200
0.020
-----
-----
0.105
0.071
200
0.062
-----
-----
0.325
0.222
What is the effectiveness of the screen if new?
ANSWER: 82.75%
12. What is the effectiveness of the screen if old?
ANSWER: 81.06%
13. Table salt is being fed to a vibrating screen at the rate of 3,000 lb/hr. The desired product is the 48/65 mesh
fraction. A 48 and 65 mesh screen therefore use (double deck), the feed being introduced on the 48 mesh
screen, the product being discharged from the mesh 65 mesh screen. During the operation it was observed
that the average proportion of oversize: product: undersize was 2:1 1/2: 1
SCREEN MESH
FEEDMASS
OVERSIZE MASS PRODUCT MASS UNDERSIZED
FRACTION
FRACTION
FRACTION
MASS FRACTION
-10+14
0.000356
0.0008
-----
-----
-14+20
0.00373
0.0008
0.0005
0.00003
-20+28
0.089
0.189
0.016
0.00012
-28+35
0.186
0.389
0.039
0.0009
-35+48
0.258
0.337
0.322
0.0036
-48+65
0.281
0.066
0.526
0.344
-65+100
0.091
0.005
0.067
0.299
-100+150
0.062
0.005
0.024
0.237
-150_200
0.025
0.001
0.002
0.110
What is the effectiveness of the screener?
ANSWER: 48.68%
14. What will be the overall effectiveness of mesh no. 65 on screen
ANSWER: 65.26%
15. What will be the overall effectiveness of mesh no. 48 on screen
ANSWER: 70.68%
Size Reduction
It is also known as “comminution”
A.
Screening
B.
Agitation
C.
Size Reduction
D.
Floatation
The typical feed sizes of the product are ½ to 3 inches while product sizes are ½ inch to about 20 mesh size.
A.
Jaw Crusher
B.
Crushing Rolls
C.
Gyratory
D.
Revolving mills
It yields very fine product from soft non-abrasive materials
A.
Impact
B.
Compression
C.
Attrition
D.
Shearing
Pick out the wrong statement
A.
Recycled coarse material to the grinder by a classifier is termed as circulating load.
B.
Wear and tear in wet crushing is more than that in dry crushing of materials.
C.
Size enlargement is not a mechanical operation
D.
A 'dust catcher' is simply an enlargement in a pipeline which permits the solids to settle down due to
reduction in velocity of the dust laden gas.
The term applied to all ways in which particles of solids are cut or broken into smaller pieces
A. Size reduction
C. Comminution
B. Screening
D. Crushing
Size reduction of the __________ can be suitably done by ball mills, crushing rolls and rod mills.
A. metalliferrous ores
B. non-metallic ores
C. basic slags
D. asbestos & mica
If it is desired to reduce the separation time for milk to at least one week (before cream will rise to the top), what
maximum diameter of cream droplet would Stokes' Law predict to be necessary for the homogenization to achieve?
Assume the depth is 10 cm.
ANSWER: 0.0567 microns
It is found that the energy required to reduce particles from a mean diameter of 1 cm to 0.3 cm is 11 kJ/kg. What will
be the energy requirement to reduce the same particles from a diameter of 0.1 cm to 0.01 cm assuming if used;
Kick's Law
ANSWER: 21 kJ/kg
Rittinger's Law
ANSWER: 423 kJ/kg
Bond's Equation
ANSWER: 91 kJ/kg
From measurements on a uniformly sized material from a dryer, it is inferred that the surface area of the material is
1200 m2. If the density of the material is 1450 kg m-3 and the total weight is 360 kg, what will be the equivalent
diameter of the particles if their value of is 1.75.
ANSWER: 2200 microns
Sedimentation
When a dilute slurry is settled by gravity into a clear fluid and a slurry of higher solids concentration
A.
Centrifugation
B.
Filtration
C.
Sedimentation
D.
Crystallization
A centrifugal force is used instead of a pressure diffenrence to cause a flow of slurry in a filter where a cake of solids
builds up on a creen.
A.
Centrifugal filtration
B.
Filtration
C.
Sedimentation
D.
Crystallization
If the object being rotated is a cylindrical container, the contents of fluid and solids exert an equal and opposite force,
called _____.
A.
Force
B.
Centrifugal Force
C.
Gravity
D.
None of the above
Device in which a current of air separates particles with different sedimentation velocities
A. Agitator
B. classifier
C. Air elutriator
D. air conveyor
Industrially, sedimentation operations are often carried out continuously in equipment called?
A. Thickener
B. Simple gravity settling
C. Sedimentation thickener
D. None of the these
The operation by which solids are separated from liquids due to difference in their respective densities is
C. Screening
c. sedimentation
D. Adsorption
d. absorption
Oil droplets having a diameter of 20 μm are to be settled from air at an air temperature of 37.8 OC at 101.3 kPa
pressure. The density of the oil is 900 kg/m 3. What will be the terminal settling velocity of the drops?
A.
0.8906 m/s
B.
0.3489 m/s
C.
0.0103 m/s
D.
0.0012 m/s
What will be the settling velocity of glass spheres having a diameter of 1.554 x 10 -4 m in water at 293.2 K. The slurry
contains 60 wt% solids. The density of the glass spheres is ρ p = 2467 kg/m3.
A.
5.03 x 10-3 ft/s
B.
5.03 x 103 ft/hr
C.
5.03 x 10-4 ft/s
D.
5.03 x 104 ft/hr
What is the terminal settling velocity of dust particles having a diameter of 60 μm in air at 294.3 K and 101.32 kPa.
The dust particles can be considered spherical with a density of 1280 kg/m 3.
A.
0.4565 m/s
B.
0.1372 m/s
C.
0.2228 m/s
D.
0.3489 m/s
A mixture of Galena (ρ = 7500 kg/m3) and Silica (ρ = 2650 kg/m3) has size range between 0.08 mm to 0.7mm. a).
What is the velocity of water needed to obtain a pure galena product? b) What is the maximum size range of the
galena product?
Answers: (a.) 0.44 m/s, (b.) 0.35 mm < Dp galena ≤ 0.70 mm
Centrifugation
It removes or concentrate particles of solids by causing the particles to migrate through the fluid radially towards or
away from the axis of rotation.
A.
Centrifugal sedimentation
B.
Sedimentation
C.
Centrifugation
D.
Filtration
It causes the liquid to flow through the increasing thickness of solids with as being hold on a filter medium.
A.
Centrifugation
B.
Sedimentation
C.
Centrifugal filters
D.
Extractors
This process is used to separate two immiscible liquids
A.
Centrifugation
B.
Sedimentation
C.
Crystallization
D.
None of the above
What will be the filtration rate that can be expected from basket centrifugal filters using data given below:
Basket height – 30 cm
Sp. Cake resistance – 1.71 x 1011 m/kg
Inside Basket diameter – 66 cm
Porosity – 0.5
Rotation Rate – 2000 rpm
Specific gravity of CaSO4. 2H2O = 2.65
Material to be filtered – gypsum slurry
A.
4.11 x 10-5 m3/s
B.
4.11 x 10-5 m3/min
C.
4.11 x 10-5 m3/hr
D.
4.11 x 105 m3/s
What is the capacity of a clarifying centrifuge operating under these condition:
Diameter of the bowl = 60 cm
Thickness of the liquid layer = 8 cm
Depth of the bowl = 40 cm
Speed = 1000 rpm
A.
B.
C.
D.
0.001 m3/s
0.02 m3/s
0.78 m3/s
0.98 m3/s
A centrifuge having a radius of the bowl of 0.1016 m is rotating at N= 1000 rev/min. What will be the centrifugal force
developed in terms of gravity forces? In English unit
A.
115.7
B.
123.45
C.
113.6
D.
45.67
Calculate the settling velocity of dust particles of (a) 60 mm and (b)10 mm diameter in air at 21°C and 100 kPa
pressure. Assume that the particles are spherical and of density 1280 kg m -3, and that the viscosity of air = 1.8 x 105 N s m-2 and density of air = 1.2 kg m -3.
Answer: (a) 0.14m/s (b) 3.9 x 10-3m/s
Flotation
It is a highly versatile method for physically separating particles based on differences in the ability of air bubbles to
selectively adhere to specific mineral surfaces in a minera/water slurry.
A. Froth flotation
B. Sedimentation
C. Crystallization
D. None of the Above
It is the inverse of the ratio of concentration
A. % Metal recovery
B. % Metal loss
C. Enrichment Ratio
D. % Weight Recovery
It is calculated directly from assays as c/f, weights are not involved in the calculation.
A. % Metal recovery
B. % Metal loss
C. Enrichment Ratio
D. % Weight Recovery
Which of the following can be most effectively used for clarification of tube oil and printing ink?
t.
u.
sparkler filter
precoat filter
C. disc-bowl centrifuge
D. sharpless supercentrifuge
Solid particles separation based on the difference in their flow velocities through fluids is termed as the
A.
Clarification
B.
Classification
C.
Elutriation
D.
Sedimentation
If a force greater than that of gravity is used to separate solids & fluids of different densities, the process is
termed as the
A. Sedimentation
B. Flocculation
C. Dispersion
D. Centrifugation
A copper ore initially contains 2.09% Cu. After carrying out a froth flotation separation, the products are shown:
Product
%Weight
%Cu Assay
Feed
100
2.09
Concentrate
10
20.0
Tailings
90
0.1
What is the ratio of concentration?
A. 10
B. 1
C. 2
D. None of the choices
What is the % metal loss?
A. 5.7 %
B. 6.2 %
C. 7.8 %
D. 4.3%
What will be the enrichment ratio?
A. 6.78
B. 4.56
C. 9.57
D. 6.78
If a centrifuge is 3-ft diameter and rotates at 1,000 rpm, what must be the speed of a laboratory centrifuge of 6-in
diameter be ran if it is duplicate plant conditions?
Answer: 2449 rev/min
TECHNOLOGICAL INSTITUTE OF THE PHILIPPINES
363 P. Casal St., Quiapo, Manila
Introduction to Particle Technology
Problem Set for Finals:
Screening
Size Reduction
Flotation
Sedimentation
Centrifugation
Submitted by:
Mandap, Melvin Christopher C.
1220118
Submitted to:
Engr. Robert Delfin
March 24, 2015
I.
Size reduction
1. Rittinger's number designates the new surface created per unit mechanical energy absorbed
by the material being crushed. Larger value of Rittinger's number of a material indicates its
a. poor grindability
b. easier grindability
c. high power consumption in
grinding
d. none of these
2. For a non-spherical particle, the sphericity
a. is defined as the ratio of surface area of a sphere having the same volume as
the particle to the actual surface area of the particle.
b. has the dimension of length.
c. is always less than 1.
d. is the ratio of volume of a sphere having the same surface area as the particle
to the actual volume of the particle.
3. Which of the following is the hardest material?
a. Calcite
b. Quartz
c. Corundum
d. Gypsum
4. Size reduction mechanism used in Jaw crushers is
e. attrition
c. cutting
f. compression
d. impact
5. The main size reduction operation in ultra-fine grinders is
c. cutting
c. compression
d. attrition
d. impact
6. Wet grinding in a revolving mill
e. gives less wear on chamber walls than dry grinding
f. requires more energy than for dry grinding
g. increases capacity compared to dry grinding
h. complicates handling of the product compared to dry grinding
7. Reciprocal of sphericity is termed as the
a. specific surface ratio
b. shape factor
8.
The reduction ratio for grinders is defined as
c. Df/Dp
c. Df – Dp
d. Dp/Df
d. Dp – Df
c. sauter diameter
d. surface area per unit mass
9.
A fluid energy mill is used for
c. cutting
d. grinding
c. ultragrinding
d. crushing
10.
Cement clinker is reduced to fine size
a. Roll crusher
c. Tube mill
b. Ball mill
d. Hammer mill
11. From measurements on a uniformly sized material from a dryer, it is inferred that the
surface area of the material is 1200 m2. If the density of the material is 1450 kg m -3 and the
total weight is 360 kg calculate the equivalent diameter of the particles if their value of l is 1.75.
Answer: 2200 microns
12.
If it is desired to reduce the separation time for milk to at least one week (before cream
will rise to the top), what maximum diameter of cream droplet would Stokes' Law predict to be
necessary for the homogenization to achieve? Assume the depth is 10 cm.
Answer: 0.0567 microns
13.
It is desired to crush 10 ton/h of iron ore hematite. The size of the feed is such that 80%
passes a 3-in. (76.2-m screen ad 80% of the product is to pass a 1/8 – in. (3.175-mm) screen.
Calculate the gross power required. Use a work index Ei for iron ore hematite of 12.68 (P1).
Answer: 24.1 hp
14.
Sugar is ground from crystals of which it is acceptable that 80% pass a 500 mm
sieve(US Standard Sieve No.35), down to a size in which it is acceptable that 80% passes a
88 mm (No.170) sieve, and a 5-horsepower motor is found just sufficient for the required
throughput. If the requirements are changed such that the grinding is only down to 80%
through a 125 mm (No.120) sieve but the throughput is to be increased by 50% would the
existing motor have sufficient power to operate the grinder? Assume Bond's equation.
Answer: 5.4 horsepower
15.
It is suspected that for a product of interest the oxidation reactions, which create offflavours, are surface reactions which proceed at a rate which is uniform with time, and if the
shelf life of the product is directly related to the percentage of the off-flavours that have been
produced, estimate the percentage reduction in shelf life consequent upon the size reductions
of example 3, that is from 1 cm to 0.3 cm and from 0.1 cm to 0.01 cm in diameter, assuming l =
1.5.
Answers: (a) 10:1 ; (b) 100:1
16.
A crushing mill reduces limestones from a mean particle size of 45mm to the following
product
Size (mm)
amount of product (percent)
12.5
0.5
7.5
7.5
5.0
45.0
2.5
19.0
1.5
16.0
0.75
8.0
0.40
3.0
0.20
1.0
It requires 21 kJ/kg of material crushed. Calculate the power required to crush
the same material at the same rate, from a feed having a mean size of 25mmto a
product with a mean size of 1 mm.
Solution:
The mean size product may be obtained thus
N1
0.5
7.5
45.0
19.0
16.0
8.0
3.0
1.0
Total
D1
12.5
7.5
5.0
2.5
1.5
0.75
0.40
0.20
13,049
Dv
N1d13
3906
3164
5625
2969
54.0
3.375
0.192
0.008
Total
101,510
N1d4
48.828
23.731
28.125
742.2
81.0
2.531
0.0768
0.0016
= ∑ n1d4 1 / ∑n1d3 1
= (101,510/13,049) = 7.78 mm
Kick’s law is used as the present case may be regarded as coarse crushing
Case 1:
E = 21 kJ/kg, L1 = 45 mm and L2 = 7.8 mm
21 = KKfc ln(45/7.8)
and:
KKfc = 11.98 kJ/kg
Case 2:
L1 = 25 mm and L2 = 1.0 mm.
Thus:
E = 11.98 ln(25/1.0)
= 38.6 kJ/kg
II.
Screening
11. With increase in the capacity of screens, the screen
.effectiveness
a. remains unchanged
b. increases
c. decreases
d. decreases exponentially
2. In screen analysis, the notation +5 mm/-10 mm means
particles passing through
3
.a. 10 mm screen and retained on 5 mm screen
b. 5 mm screen retained n 10 mm screen
c. both 5 and 10 mm screen
d. neither 5 nor 10 mm screen
3. For sizing of fine materials, the most suitable equipment is a
6 a. Trommel
. b. Grizzly
c. Shaking screen
d. Vibrating screen
4. 200 mesh screen means 200 openings
per
a.
b.
c.
d.
Cm2
Inch
Cm
Inch2
5. increasing the capacity of a screen
__________ the screen effectiveness.
a. decreases
b. increases
c. no effect
d. none of these
6. The ratio of the actual mesh dimension of Taylor series to that of the next smaller screen is
a.
b.
c.
d.
1
2
1.5
3
7. Which of the following is not industrial screening equipment?
a. Sharpies centrifuge
b Vibrating screen
c. Grizzly
d. Trommel
8. 200 mesh seive size corresponds to __________ microns.
a. 24
b 74
c. 154
d. 200
9.
The ratio of the area of opening in one screen (Taylor series) to that of openings in the
next smaller screen is
a. 1.5
b. 1
c. 2
d. none of these
10.
Screen capacity is expressed in terms of
a. tons/h
b. tons/ft2
c. both a and b
d. tons/h-ft2
11. The particle size distribution of the feed and collected solids for a gas cyclone are given
below
Size range
1-5
5-10
10-15
15-20
Wt. of feed in the 2.0
3.0
5.0
6.0
size range
Wt. of collected
0.1
0.7
3.6
5.5
solids in the size
range
What is the collection efficiency (%) of the gas cyclone?
20-25
25-30
3.0
1.0
2.9
1.0
Answer: 69%
12.A quartz mixture is screened through a 10-mesh screen. The cumulative screen analysis of
feed, overflow and underflow are given in the table. Calculate the mass ratios of the overflow
and underflow to feed and the overflow effectiveness of the screen.
Mesh
Dp (mm)
Feed
Overflow
Underflow
4
4.699
0
0
0
6
3.37
0.025
0.071
0
8
2.362
0.15
0.43
0
10
1.651
0.47
0.35
0.195
14
1.168
0.73
0.97
0.58
20
0.833
0.885
0.99
0.83
28
0.589
0.94
0.91
35
0.417
0.96
0.94
65
0.208
Pan
0.98
0.975
1
1
Answer: 66.91%
13-15. Fine silica is fed at 1500 lbs/hour to a double-deck vibrating screen combination to
obtain a 48/65 mesh (Tyler ) product. The silica feed is introduced into the upper screen of the
48 mesh and the product is discharged off the surface of the lower screen of 65 mesh. During
the screening operation, the ratio of oversize to product to undersize is 2:11⁄2.
Laboratory analysis of the different fractions:
Screen Mesh
Feed Mass
Oversize Mass
Product Mass
Undersize
Fraction
Fraction
10/14 to 28/35
0.2821
0.5855
0.3385
0.00453
35/48
0.2580
0.3370
0.3220
0.00360
48/65
0.2810
0.0660
0.5260
0.34400
65/100
0.0910
0.0050
0.0670
0.39900
100/150 to
0.0870
0.0060
0.0260
0.35300
Mass Fraction
150/200
17. The effectiveness of the screening equipment is
a. 58.7%
b. 48.7%
c. 68.6%
d. 45.6%
18. If the screens measure 5ft x 8ft each, the capacity in MT/day-ft2 –mm of the 65
mesh screen on the basis of a perfectly functioning 48 mesh screen is
E. 0.901
F. 1.09
G. 0.801
H. 0.75
19. The capacity in MT/day-ft2-mm on the basis of the actual performance of the 48
mesh screen
E. 1.09
F. 0.901
G. 1.29
H. 1.49
16. A sand mixture was screened through a standard 10-mesh screen. The mass fraction of
the oversize material in feed, overflow and underflow were found to be 0.38, 0.79 and 0.22
respectively. The screen effectiveness based on the oversize is?
Given:
10-mesh screen
Xf = 0.38
Xb = 0.22
Xd = 0.79
Solution:
𝐸=
=
(𝑋𝑓 − 𝑋𝑏 ) 𝑋𝑑
( )
(𝑋𝑑 − 𝑋𝑏 ) 𝑋𝑓
(0.38 − 0.22) 0.79
(
)
(0.79 − 0.22) 0.38
= 𝟎. 𝟓𝟖
III. Flotation
1.
Pine oil used in forth flotation technique acts as a
k.
collector
c. frother
l.
modifier
d. activator
2.
The flotation agent that prevents coalescence of air bubbles as they travel to the
surface of the water is/are
a. collectors
c. frothing agent
b. promoters
d. modifying agent
3.
An example of a collector for floatation of metallic sulfides and native metals is
a.
b.
xanthates
sodium silicate
d.
c. sodium sulfide
sphalerite
4.
Dispersants are important for the control of limes which sometimes interfere with the
selectivity and increase reagent consumption. Another term for dispersant is
a. deflocculant
c. frothers
b. depressants
d. regulators
5.
The flotation agent that prevents coalescence of air bubbles as they travel to the
surface of the water is/are
a. collectors
c. frothing agent
b. promoters
d. modifying agent
6. pine oil used in forth flotation technique acts as a
a. collector
c. frother
b. modifier
d. activator
7. dispersants are important for the control of slimes which sometimes interfere with the
selectivity and increase reagent consumption. Another term for dispersants is
a. deflocculant
c. frothers
b. depressants
d. regulators
8. the flotation agent that prevents coalescence of air bubbles as they travel to the surface of
the water is/are
a. collector
c. promoters
b. frothing agent
d. modifying agent
9. What is the selectivity index, if the grade of tailings & concentrate is the same ?
a. 0
c. 1
b. ∞
d. 0.5
10. froth flotation is most suitable for treating
a. iron ores
c. quartz
b. sulfides ores
d. metal ores
11.
A flotation plant processes 3000 tons/day of CuFeS2. It produces 80 tons Cu
concentrate assaying 25% Cu. If ore analyzes 0.7% Cu, the percent recovery is?
Answer: 95.24%
12.
Ground lead ore is to be concentrated by a single flotation process using 1.5 oz of
reagent per ton of ore. The feed concentrate and tailings have the following composition by
weight on a dry basis
Feed %
Concentrate %
Tailings %
13.
PbS
30
90
0.9
ZnS
25
3
35.6
SiO2
45
7
63.5
Water is fed to the cell at the rate of 1000 gallons per ton of wet concentrate with
99% of the water leaving with the tailings and 1% with the concentrate. Find the
mass of wet concentrate produced per hour when ten tons of ore are fed to the
cell / 24 hr. is?
Answer: 3.4
From problem number 3 find the total water required in pounds per hour
Answer: 1185
14. A flotation plant processes 3000 lbs/day of CuFeS2(chalcopyrite). It produces 80 tons
copper concentrate assaying 25% copper. If ore analyzes 0.7 copper, the percent recovery is
a. 90.5%
b. 98%
c. 92.54%
d. 95.24%
15-16. Ground lead ore is to be concentrated by a single flotation process using 1.5 oz of
reagent per ton of ore. The feed concentrate and tailings have the following composition by
weight on a dry basis:
Feed%
Concentrate%
Tailings%
PbS
30
90
0.9
ZnS
25
3
35.6
SiO2
45
7
63.5
Water is fed to the cell at the rate of 1,000 gallons per ton of wet concentrate with 99%
of the water leaving with the tailings and 1% with the concentrate.
The mass of wet concentrate produced per hour when tons of ore are fed to the cell per
24 hours is
a. 3.4
b. 4.3
c. 14.23
d. 4.7
The total water required in pounds per hour is
a. 1150
b. 1200
c. 1185
d. 1285.8
IV.
17.
18.
Sedimentation
The drag coefficient in hindered settling is _______________ compared to free settling
C. Greater than
c. constant
D. Less than
d. varying
The separation of solid particles into several size fractions based upon the settling
velocities in a medium is called
C. Settling
c. flotation
D. Filtration
d. classification
19. The operation by which solids are separated from liquids due to difference in their
respective densities is
E. Screening
c. sedimentation
F. Adsorption
d. absorption
20. Stoke’s law is valid when the particle Reynolds Number is
C. <1
c. >1
D. <5
d. none of these
21. Range motion Newton’s law for n
C. 0
c. 2
D. 1
d.4
6. Coal washing waste water containing about 3% suspended solids (comprising of clay, slate,
stone etc.) is treated for solid particles removal
a. by chemical coagulation
b in sedimentation tanks equipped with mechanical scrapper.
c. in vacuum filter.
d. in clarifiers.
5. Particles having diameter greater than 75 μm (micrometer = 10-6 mm) are called
a. grit
b. dust
c. powder
d. smoke
7. Pick out the one which is not a chemical coagulant.
8.
9.
is
a. Aluminium sulphate
b. Ferrous sulphate
c. Hydrated lime
d. Chloramine
Terminal velocity is
q. constant velocity with no acceleration
r. a fluctuating velocity
s. attained after moving one-half of total distance
t. none of these
For the free settling of a spherical particle through a fluid, the slope of C D vs log NRe plot
a. 1
b. 0.5
d.
c. – 1
– 0.5
10. in order for a particle to move through a fluid under the influence of gravity, there must be
a. velocity difference
c. density difference
b. pressure difference
d. temperature difference
11. A mixture of Galena (ρ = 7500 kg/m3) and Silica (ρ = 2650 kg/m3) has size range between
0.08 mm to 0.7mm. a). What is the velocity of water needed to obtain a pure galena product?
b) What is the maximum size range of the galena product?
Answers: (a.) 0.44 m/s, (b.) 0.35 mm < Dp galena ≤ 0.70 mm
12. A particle settles through a suspension containing 35% solids by weight. Given that the
velocity under free settling motion is ut = 0.00669 m/s. Find us.
Given:
[density of particle = 2800 kg/m3, Dp = 200 mesh, density of fluid = 996.5 kg/m3,
viscosity of fluid = 0.8Cp]
Answer: 0.003036m/s
13. Calculate the settling velocity of dust particles of (a) 60 mm and (b)10 mm diameter in air
at 21°C and 100 kPa pressure. Assume that the particles are spherical and of density 1280 kg
m-3, and that the viscosity of air = 1.8 x 10-5 N s m-2 and density of air = 1.2 kg m-3.
Answer: (a) 0.14m/s (b) 3.9 x 10-3m/s
V. Centrifugation
17.
Uses the concept that an object whirled about an axis at a constant radial distance from
the point is acted on by a force.
a. filtration
c. centrifugal separation
b. sedimentation
d. none of these
18.
Is the force applied to the moving mass in the direction toward the center of curvature
which causes the mass in the direction toward the center of curvature which causes the mass
to travel in a curved path.
a.
centripetal force
c. rotational force
b.
centrifugal force
d. filtration
19.
Is created by moving a mass in a curved path and is exerted in the direction away from
the center of curvature of the path.
a. Force
b. Centrifugal force
c. Settling
d. raising
20.
If the radius of a basket centrifuge is halved and the rpm is doubled, then the
C.
Linear speed of the basket is doubled
D.
Linear speed of the basket is halved
increased
c. centrifugal force is doubled
d. capacity of centrifuge is
21.
Mechanical process of separating multiphase mixture via the use of centrifugal force
C.
D.
Centrifugation
Flotation
22.
Moisture can be removed from lubricating oil using
C.
D.
Tubular centrifuge
Clarifier
23.
Moisture can be removed from lubricating oil using
i.
j.
tubular centrifuge
clarifier
c. screening
d. size reduction
c. sparkler filter
d. vacuum leaf filter
c. sparkler filter
d. vacuum leaf filter
24.
Which of the following can be most effectively used for clarification of tube oil and
printing ink?
v.
w.
25.
sparkler filter
precoat filter
c. disc-bowl centrifuge
d. sharpless supercentrifuge
Tabular bowl centrifuges as compared to disk bowl centrifuges
a. operate at higher speed
b. employ bowl of larger diameter.
c. cannot be operated under pressure/vacuum
d. can't be used for separation of fine suspended solids from a liquid
26.
Where the density difference of the two liquid phase to be separated is very small (as in
milk cream separator), the most suitable separator is a
a. disc bowl centrifuge.
b. sharpies supercentrifuge.
c. batch basket centrifuge
d. sparkler filter
11. A centrifuge bowl is spinning at a constant speed of 2000 rpm. The radius of bowl in cm
needed to create a force of 455 g is.
Answer: 10.2
12. How many "g" can be obtained in a centrifuge which can spin a liquid at 2000 rev/min at a
maximum radius of 10 cm?
Answer: 450
13.
If a centrifuge is 3-ft diameter and rotates at 1,000 rpm, what must be the speed of a
laboratory centrifuge of 6-in diameter be ran if it is duplicate plant conditions?
Answer: 2449 rev/min
TECHNOLOGICAL INSTITUTE OF THE PHILIPPINES
363 P. Casal St., Quiapo, Manila
Introduction to Particle Technology
Problem Set for Finals
Submitted by:
Mejia, Ferowen Joy D.
BSChE – 5th year
Submitted to:
Engr. Robert Delfin
Date Submitted:
March 24, 2015
SCREENING
1) Screening is a method of separating particles according to
a) size alone
b) size and shape
c) size, shape and texture
d) All of the above
2) In the screening process, characteristics of an individual particle are considered. Which of
the following is not included?
a) Composition
b) Size
c) Shape
d) None of these
3) In screening, particles are separated into two fractions, one is obtained from passing feed
into a single screen and either upper or lower size limit can be indicated. This particle is
called
a) Unsized particles
b) Sized particles
c) Sieved particles
d) Both b and c
4) Material passed through a series of screens of different sizes is separated into sized
fractions, i.e. fractions in which _____________ particle sizes are known.
a) maximum
b) minimum
c) both a and b
d) neither a and b
5) The capacity of a screen is measured by the mass of material that can be fed per unit time
to a unit area of the screen. To obtain maximum effectiveness, the capacity must be
_______ and large capacity is obtainable only at the expense of a reduction in
effectiveness.
a) large
c) cannot be determined
b) small
d both a and b
6) Mesh screens are arranged with __________ mesh number, thus _______ size of
opening, from top to bottom.
a) increasing, decreasing
c) constant, increasing
b) decreasing, increasing
d) increasing, constant
7) It is the data that provides the Mesh number versus the cumulative fraction larger than
Dp.
a) differential screen analysis
c) screen analysis
b) cumulative screen analysis
d) particle size distribution
8) Which of the following is not screening equipment?
a) Grizzlies
c) Rotary
b) Trammels
d) None of these
9) It is a type of screening equipment where the set of parallel metal bars in an inclined
stationary frames.
a) Grizzlies
c) Rotary
b) Trammels
d) Vibrating
10) It is calculated by the product of recovery of desired material in the product and recovery
of undesired material in the reject.
a) Screen analysis
c) Screen Effectiveness
b) Mesh Analysis
d) Particle Size
SIZE REDUCTION
11) Size reduction can be divided into two major categories depending on whether the
material is a solid or a liquid.
a) True
c) Both a and b
b) False
d) None of these
12) In size reduction, there are _____ commonly used methods.
a) five
c) three
b) four
d) two
13) In size reduction, crusher must
a) have a large capacity
c) yield uniform distribution desired
b) require a small power input per unit of product d) all of the above
14) Grinding and cutting reduce size of the materials by _______ action dividing them into
smaller particles.
a) mechanical
c) physical
b) chemical
d) no action
15) The important factors to be studied in the grinding process.
a) amount of energy used
b) the amount of new surface formed by grinding
c) the amount of raw materials used for grinding
d) both a and b
16) The energy required to reduce a material size was directly proportional was directly
proportional to the size reduction ratio dL/L.
a) Rittinger’s Law
c) Bond’s Law
b) Kick’s Law
d) None of these
17) The energy required for size reduction is directly proportional, not to the change in length
dimensions, but to the change in surface area.
a) Rittinger’s Law
c) Bond’s Law
b) Kick’s Law
d) None of these
18) Bond defines the quantity Ei by this equation: L is measured in and so Ei is the amount of
energy required to reduce unit mass of the material from an infinitely large particle size
down to a particle size of 100 mm. It is expressed in terms of q, the reduction ratio where
a) q = (L1)(L2)
c) q = L1 = L2
b) q = L1 + L2
d) q = L1/L2
19) All are size reduction equipment except
a) Hammer Mill
b) Cutters
c) Crushers
d) All of the above
20) It is a type of size reduction that is used to break down large pieces of food into smaller
pieces suitable for further processing, such as in the preparation of meat for retail sales
and in the preparation of processed meats and processed vegetables.
a) Crushing
c) Grinding
b) Cutting
d) Miller
FLOTATION
21) In flotation process, separation of mixtures involves substances will be based on whether
they
a) sink
c) sink or float
b) float
d) none of these
22) Flotation process involves _______ treatment of the ore pulp to create conditions
favorable for the attachment of certain mineral particles to the air bubbles.
a) physical
c) both a and b
b) chemical
d) either a or b
23) All are part of the flotation process except:
a) Centrifugation
b) Conditioning
c) Cleaning
d) both b and c
24) The overflow from the conditioner is fed into flotation cell is termed as
a) Tailings
c) Rougher
b) Scavenger
d) None of these
25) The material which floated off in the first separation in flotation process is called
a) Tailings
c) Scavenger
b) Concentrate
d) All of the above
26) The other material which sinks in the water and is removed from the bottom is called
a) Tailings
c) Scavenger
b) Concentrate
d) both a and b
27) It is the equipment in which the material is actually separated of floated from the residual
tailings.
a) flotation unit
c) both a and b
b) cell
d) either a or b
28) The floated product from the scavenger is returned or cycled to the rougher with the feed.
a) True
b) False
29) The tailings from the scavenger are called
a) Final tailings
b) Scavenger
c) Rougher
d) Concentrate
30) Flotation
a) depends upon the relative adsorption
b) depends on wetting of solid surface
c) both a and b
d) none of these
SEDIMENTATION
31) Sedimentation is the separation of the dilute slurry (solid suspended in fluid) into a clear
dense slurry by gravity settling.
a) True
b) False
32) Assuming that the concentration of solids is small,
a) FcL =LCL = UCu
b) FcL =LCL + UCu
c) FcL +LCL + UCu
d) FcL +LCL = UCu
33) The drag coefficient in hindered settling is _______ compared to free settling.
a) greater than
c) constant
b) less than
d) varying
34) If the external force in sedimentation process is due to gravity,
a) ae = rw2
c) ae = g
b) ae = 0
d) none of these
35) If the external force in sedimentation process is due to centrifugal action,
c) ae = rw2
c) ae = g
d) ae = 0
d) none of these
36) When the velocity of the particle attains the maximum (or terminal) velocity,
a) dV/dt is constant
c) dV/dt is not applicable
b) dV/dt is zero
d) both a and b
37) If the external force in sedimentation process is due to gravity,
e) ae = rw2
c) ae = g
f) ae = 0
d) none of these
38) The constant velocity with which a body moves relative to the surrounding fluid when the
forces acting on it are equal to the friction force acting against the movement.
a) settling velocity
c) maximum velocity
b) terminal velocity
d) all of these
39) In a motion of particle through fluids, forces act on a particle moving through a fluid. The
force which appears whenever there is a relative motion between the particle and the fluid
is called
a) drag force
c) centrifugal force
b) gravitational force
d) buoyant force
40) The ration of drag force per unit area to the product of fluid density and the velocity head
is called
a) buoyant coefficient
c) friction factor
b) drag coefficient
d) shear coefficient
CENTRIFUGATION
41) In centrifugation process, a _________ process of separating multi-phase mixture,
usually fluid-solid system with the use of centrifugal force.
a) physical
c) mechanical
b) chemical
d) thermal
42) Which of the following is not a centrifugation process
a) Clarification
b) Classification
c) Thickening
d) None of these
43) Centrifugation can be used for solid-liquid separation provided that the solids are ______
liquids.
a) lighter than
c) same as
b) heavier than
d) none of these
44) A centrifuge is used for centrifugation process in which is a piece of equipment,
generally driven by an electric motor, that puts an object in rotation around a fixed axis,
applying a force _____________ to the axis to separate substances of different densities.
45) a) perpendicular
c) constant
b) tangential
d) varies
46) Much ________ forces can be obtained by introducing centrifugal action, in a centrifuge.
a) greater
c) equal
b) lesser
d) constant
47) Gravity still acts and the net force is a combination of the centrifugal force with gravity
as in the cyclone. Because in most industrial centrifuges, the centrifugal forces imposed
are so much _____ than gravity, the effects of gravity can usually be _______ in the
analysis of the separation.
a) greater, equal
c) greater, neglected
b) Lesser, neglected
d) lesser, equal
48) Centrifuges are devices used in a variety of scientific and technical applications which
spin carrier vessels (centrifuge tubes) _____ at rotation speeds and very high centrifugal
force.
a) low
c) both a and b
b) high
d) none of these
49) The centrifugal force (expressed as gravities) generated is _________ to the rotation rate
of the rotor (in rpm) and the distance between the rotor center and the centrifuge tube.
a) equal
c) constant
b) proportional
d) varying
50) Centrifuges generally work under vacuum and are refrigerated to reduce heating caused
by frictional forces as the rotor spins.
a) True
b) False
SCREENING
It is desired to separate a mixture of crystals into three fractions, a coarse fraction retained on a
8- mesh screen, a middle fraction passing on 8-mesh but retained in a 14-mesh and a fine fraction
passing a 14-mesh screen. Two screens in series are used on 8-mesh and a 14-mesh conforming
to the Tyler standard. Screen analysis of feed, coarse, medium, and fine fractions are given
below. Assuming the analysis are accurate. What do they show as to the ratio be weight of each
of the three fractions actually obtained? What is the efficiency of each screen?
Screen
Feed
3/4
4/6
6/8
8/10
10/11
14/20
20/28
28/35
35/44
3.5
15
27.5
23.5
16.0
9.1
3.4
1.3
0.7
Solution:
Basis: F = 100 lb
F = F1 + F2 + F3
M = M1 + M2 + M3
R = R1 + R 2 + R 3
100 = M + P + R → (1)
F1 = M1 + P1 + R1
46 = (0.4)M + 0.88P1 → (2)
F2 = M2 + P2 + R2
39.5 = 0.491M + 0.12P + 0.467R → (3)
M = 60.1926
P = 24.9125
R = 14.8949
At 8-mesh:
Coarse fraction: 0.88P
Middle fraction: 0.491M
Fine fraction: 46.7R
Coarse Fraction
(P)
14
50
24
8
4
Middle Fraction
(M)
42
35.8
30.8
14.3
10.2
0.7
Fine Fraction (R)
20.0
26.7
20.2
19.6
8.9
4.6
At 14-mesh:
Coarse fraction: Middle fraction: 0.109 M
Fine fraction: 0.533 R
Ratio of Coarse:
24.91
= 0.2491
100
Middle:
60.19
= 0.6019
100
Fine:
14.90
= 0.149
100
Effectiveness:
@ first screen:
𝐸=
𝐸=
𝑃1
𝑃2
(1 −
)
𝐹1
𝐹2
0.88(24.91)
0.12 (24.91)
(1 −
)
46
39.5
𝑬 = 𝟎. 𝟒𝟒𝟎𝟏 ≈ 𝟒𝟒. 𝟎𝟏 %
@ second screen:
𝐸=
𝑀2
𝑀3
(1 −
)
𝐹2 − 𝑃2
𝐹2 − 𝑃3
0.491
0.109 (60.19)
(1 −
)
14.5 − 0
(39.5 − (0.12 𝑥 24.91))
𝐸=
𝑬 = 𝟎. 𝟒𝟒𝟑𝟐 ≈ 𝟒𝟒. 𝟑𝟐 %
1)
Limestone is produced by crushing and then screening through a 14-mesh screen.
One thousand eight hundred pounds have been screened.
Tyler
Feed to Screen
4 on
8 on
14 on
28 on
48 on
100 on
100 through
14.3
20.0
20.0
28.5
8.6
5.7
2.86
The total load to the crusher is
Undersized Product Oversized Product
40
30
20
10
20
28
24
24
0 through 24 mesh
Answer: 6320 lbs/hour
2) The effectiveness of the screen
Answer: 62.5 %
3) Laboratory Analysis of the different fractions:
Screen Mesh
Feed Mass
Fraction
Oversize
Mass
Fraction
Product
Mass
Undersize
Mass
Fraction
0.2821
0.5855
0.3385
0.00453
0.2580
0.3370
0.3220
0.00360
0.2810
0.0660
0.5260
0.3440
0.0910
0.0050
0.0670
0.29900
0.0870
0.0060
0.0260
0.35300
10/14 to
28/35
35/48
48/65
65/100
100/150 to
150/200
What is the effectiveness of the screen?
Answer: 48.7 %
4) The capacity in MT/day-ft2-mm on the basis of the actual performance of the 48 mesh
screen
Answer: 1.09
5) A sand mixture was screened through a standard 10-mesh screen. The mass fraction of
the oversize material in feed, overflow and underflow were found to be 0.38, 0.79 and
0.22 respectively. The screen effectiveness based on the oversize is?
Answer: E = 58
SIZE REDUCTION
Sugar is ground from crystals of which it is acceptable that 80% pass a 500 m sieve (US
Standard Sieve No.35), down to a size in which it is acceptable that 80% passes a 88 m
(No.170) sieve, and a 5-horsepower motor is found just sufficient for the required throughput. If
the requirements are changed such that the grinding is only down to 80% through a 125 m
(No.120) sieve but the throughput is to be increased by 50% would the existing motor have
sufficient power to operate the grinder? Assume Bond's equation.
Given:
@ mesh 35: 500 µm ≈ 50 x 10-3 mm
@ mesh 170: 88 µm ≈ 8.8 x 10-6 mm
@ mesh 120: 125 µm = 125 x 10-3 mm
Required: P2 = ?
Solution:
1
1
P1
)
0.3162 𝐸𝑖 (
−
T2
√𝑥2 √𝑥1
=
P2
1
1
)
0.3162 𝐸𝑖 (
−
T2
√𝑥2 √𝑥1
1
1
5 hp
)
0.3162 𝐸𝑖 (
−
T2
√88 √500
=
P2
1
1
)
0.3162 𝐸𝑖 (
−
T2
√125 √500
P2 = 5.4 horsepower.
In an analysis of ground salt using Tyler sieves, it was found that 38% of the total salt
passed through a 7 mesh sieve and was caught on a 9 mesh sieve. For one of the finer fractions,
5% passed an 80 mesh sieve but was retained on a 115 mesh sieve. Estimate the surface areas of
these two fractions in a 5 kg sample of the salt, if the density of salt is 1050 kg m -3 and the shape
factor is 1.75.
Answer: A1 = 7.88 m2
A2 = 16.6 m2
1)
What is the power required to crush 100 ton/h of limestone if 80% of the feed pass a 2-in
screen and 80% of the product a 1/8 in screen? The work index for limestone is 12.74.
Answer: 169.6 kW
2)
Particles of average feed size 25 x 10^-4 m are crushed to an average product size of 5
x10^-4 m
at the rate of 15 tons per hour. At this rate the crusher consumes 32 kW of power of which 2 kW
are required for running the mill empty. What would be the power consumption if 10 tons per
hour of this product is further crushed 1 x10^-4 m size in the same mill? Assume the Rittinger’s
law is applicable.
3)
Answer: P = 100 kW
It is desired to crush 10 ton/h of iron ore hematite. The size of the feed is such that 80%
passes a 3-in. (76.2-m screen ad 80% of the product is to pass a 1/8 – in. (3.175-mm) screen.
Calculate the gross power required. Use a work index Ei for iron ore hematite of 12.68 (P1).
4)
Answer: P = 24.1 hp(17.96 kW)
It is suspected that for a product of interest the oxidation reactions, which create offflavours, are surface reactions which proceed at a rate which is uniform with time, and if the
shelf life of the product is directly related to the percentage of the off-flavors that have been
produced, estimate the percentage reduction in shelf life consequent upon the size reductions of
example 3, that is from 1 cm to 0.3 cm and from 0.1 cm to 0.01 cm in diameter, assuming l = 1.5.
5)
Answers: (a) 10:1 ; (b) 100:1
FLOTATION
A typical flotation machine has the following specifications the number of 4 cells, having a
flotation time of 12min. The cell Volume is 60 ft3 and Hp per cell has 10hp. The material treated
has the following specifications:
Pulp (mixture ore and water ) = 40% solids
Specific gravity of ore = 3
𝑛=
𝑇 𝑥 𝐶𝑎𝑝 𝑥 𝑑
𝑉 𝑥 1440
n= number of cells; V = volume in cu. Ft per cell; Cap = tons of dry ore / 24 hrs.; d= cu. Ft of
pulp (ore and water) containing one ton of solids.
Solution:
2000
2000 + 𝑥
𝑥 = 3000 𝐻2𝑂
𝐹 = 3000 + 2000 = 5000
2000
3000
𝑑=
+
= 58.76
3𝑥62.4 62.4
12(𝑥)(58.76)
4=
60(1440)
𝑻 = 𝟒𝟗𝟎. 𝟏𝟑
0.4 =
1) A copper ore initially contains 2.09% Cu. After carrying out a froth flotation separation,
the products are as shown.
Feed
100
2.09
Concentrate
10
20
Tailings
90
0.1
Using this data, calculate:
(a) Ratio of concentration
(b) % Weight Recovery, or % Yield
Answers: (a) 10, (d) 10%
2) A crushing mill reduces limestone from a mean particle size of 45mm to the following
product
Size (mm)
Amount of Product
12.5
0.5
7.5
75.5
5.0
45.0
2.5
19.0
1.5
16.0
0.75
8.0
0.4
3.0
0.2
1.0
It requires 21 kJ/kg of material crushed. Calculate the power required to crush the same material
at the same rate, from a feed having a mean size of 25mmto a product with a mean size of 1 mm.
Answer = 38.6 kJ/kg
3)
A flotation plant processes 3000 lbs/day of CuFeS2(chalcopyrite). It produces 80 tons
copper concentrate assaying 25% copper. If ore analyzes 0.7 copper, the percent recovery is
Answer: 95.24%
4-5)
Ground lead ore is to be concentrated by a single flotation process using 1.5
ounce of reagent per ton of ore. The feed concentrate and tailings have the following
composition by weight on a dry basis:
Feed%
Concentrate%
Tailings%
PbS
30
90
0.9
ZnS
25
3
35.6
SiO2
45
7
63.5
Water is fed to the cell at the rate of 1,000 gallons per ton of wet concentrate with 99% of the water leaving
with the tailings and 1% with the concentrate.
What is the mass of wet concentrate produced per hour when tons of ore are fed to the cell per 24 hours?
Answer: 3.4 kg/hr
What is the total water required in pounds per hour is?
Answer: 1185 lbs/hr
SEDIMENTATION
Many animal cells can be cultivated on the external surface of dextran beads. These cell-laden
beads or “microcarriers” have a density of 1.02 g/cm 3 and a diameter of 150 mm. A 50-liter
stirred tank is used to cultivate cells grown on microcarriers to produce a viral vaccine. After
growth, the stirring is stopped and the microcarriers are allowed to settle. The microcarrier-free
fluid is then withdrawn to isolate the vaccine. The tank has a liquid height to diameter ratio of
1.5; the carrier-free fluid has a density of 1.00 g/cm3 and a viscosity of 1.1 cP. (a) Estimate the
settling time by assuming that these beads quickly reach their maximum terminal velocity. (b)
Estimate the time to reach this velocity.
Solution:
Data:
d = 150 mm = 0.015 cm;
µ = 1.1
cP = 0.011 g/cm-s;
rs = 1.02 g/cm3
r = 1.00 g/cm3
g = 980 cm/s2
a)
vg =
d2
(  s −  ) g → vg = 0.022 cm/s
18 
Check: N Re =
vd 1 0.022  0.015
=
= 0.03  1

0.011
 d t2 
 h 
  h =   h = 50 L
Liquid volume, V = 
4  1.5 
 4 
2
 h = 52.3 cm  Settling time =
h
52 .3 cm
=
= 2379 s
vg 0.22 cm/s
b)
Force balance: m
m=
d 3  s
6
dv
mg
v 2
= mg −
− CD A
dt
s
2
v 2
 24   d 2  v 2 

 = 3dva

; CD A
= 
2

vd
4
2




dv
g 18v
=g−
−
dt
s d 2 s

dv 18 

+ 2 v = 1 −  g (I.C.: t = 0, v = 0)
dt d  s
 s 
v=

 − 18  
d2
(  s −  ) g 1 − exp  2 t 
18 
 d  s 

d2
(s −  ) g
At steady state (t → ), v = vg =
18 
 When
18t
 1 , v = vg
d 2 s
 When t >> 1.16  10-3 s, v = vg
For v = 0.99vg, t = 5.34  10-3 s
1) A mixture of Galena (ρ = 7500 kg/m3) and Silica (ρ = 2650 kg/m3) has size range
between 0.08 mm to 0.7mm. a). What is the velocity of water needed to obtain a pure galena
product? b) What is the maximum size range of the galena product?
Answers: (a.) 0.44 m/s, (b.) 0.35 mm < Dp galena ≤ 0.70 mm
2) A continuous separating tank is to be designed to follow after a water washing plant for
liquid oil. Estimate the necessary area for the tank if the oil, on leaving the washer, is in the form
of globules 5.1 x 10-5 m diameter, the feed concentration is 4 kg water to 1 kg oil, and the
leaving water is effectively oil free. The feed rate is 1000 kg h -1, the density of the oil is 894 kg
m-3 and the temperature of the oil and of the water is 38°C. Assume Stokes' Law.
Answer: 1.0 m2
3)
Solid spherical particles having a diameter of 0.09 mm and a solid density of 2,002
kg/m3 are settling in a solution of water at 26.7°C. The volume fraction of the solids in the
solution is 0.45. the settling velocity is
Answer: 2.369 x 10-4 m/s
4)
A gravity settling tank is to be used to clean waste water from an oil refinery. The
waste water contains 1% oil by volume as small droplets ranging in size from 100 to 1000
microns which will be removed from the water before the latter is to be discharged into the river.
The tank is of rectangular section 2 ft wide by 4 ft deep with provisions for smooth continuous
discharge of clean water is to be cleaned of oil droplets, specific gravity of oil 0.87, the length of
the settling tank is
Answer: 4850 ft.
5) A particle settles through a suspension containing 35% solids by weight. Given that the
velocity under free settling motion is µt = 0.00669 m/s. Find us.
Data:
Density of particle = 2800 kg/m3
Dp = 200 mesh
Density of fluid = 996.5 kg/m3
Viscosity of fluid = 0.8Cp
Answer: 0.003036m/s
CENTRIFUGATION
A centrifuge of diameter 0.2 m in a pilot plant rotates at a speed of 50 Hz in order to achieve
effective separation. If this centrifuge is scaled up to a diameter of 1 m in the chemical plant, and
the same separation factor is to be achieved, what is the rotational speed of the scaled up
centrifuge?
Solution:
Separation factor S =
𝜔1 2 𝑟1
𝑔
=
𝜔2 𝑟
𝑔
𝜔2 2 𝑟2
𝑔
r1
0.2
ω2 = ω1 √r = 50 x √ 1
2
𝛚𝟐 = 𝟐𝟐. 𝟒 𝑯𝒛
1)
What is the capacity in cubic meters per hour of a clarifying centrifuge operating
under these conditions?
Diameter of the bowl = 600
mm
Thickness of liquid layer = 75
mm
Depth of bowl = 400 mm
Speed = 1200 rpm
SG of liquid = 1.2
SG of solid = 1.6
Viscosity of liquid = 2 cp
Cut-size of particles = 30 μm
Answer: 210 m3/h
2)
If a particle of mass m is rotating
at radius x with an angular velocity ω, it is
subjected to a centrifugal force mxω2 in a radial
direction and a gravitational force mg in a
vertical direction. The ratio of the centrifugal to
gravitational forces, xω2/g, is a measure of the
separating power of the machine, and in order to
duplicate conditions this must be the same in
both machines.
Answer: speed of rotation = 49 Hz
3) If a centrifuge is 3 ft diameter and
rotates at 1,000 rpm, the speed of a laboratory
centrifuge of 6 in. diameter be ran if it is to
duplicate plant conditions is
Answer: 2449 rpm
4) An aqueous suspension consisting of
particles of density 2500 kg/m3 in the size range
1–10 µm is introduced into a centrifuge with a
basket 450 mm diameter rotating at 80 Hz. If the
suspension forms a layer 75 mm thick in the
basket, approximately how long will it take for
the smallest particle to settle out?
Answer: 19.3 s
5) If a centrifuge is 0.9 m diameter and
rotates at 20 Hz, at what speed should a laboratory
centrifuge of 150 mm diameter be run if it is to
duplicate the performance of the large unit?
Answer:
2940 rpm
TECHNOLOGICAL INSTITUTE OF THE PHILIPPINES
363 P.CASAL ST. QUIAPO MANILA
Screening
INTRODUCTION TO PARTICLE TECHNOLOGY
PROBLEM SET
SUBMITTED BY:
PABILLON, JESTA R..
SUBMITTED TO:
ENGR. ROBERT DELFIN
1. For sizing of fine materials, the most suitable
equipment is a
a. Trommel
c.
Shaking screen
b. Grizzly
d.
Vibrating screen
2. Trommels separate a mixture of particles
depending on their
a. Size
c.
wettability
b. Density
d.
electrical and magnetic properties
3. Screen capacity is not a function of
a. Its operating size
c.
screening surface
b. Screening mechanism
d.
atmospheric humidity
4. Ratio of oversize material A that is actually in
the overflow to the amount of A entering with
the feed
a. Screen mechanism
c.
Screen reduction
b. Screen effectiveness
d.
Screen capacity
5. It is the measure of the mass of material that
can be fed per unit of time to a unit area of the
screen
a. Screen effectiveness
c.
Screen reduction
b. Screen mechanism
d.
Screen capacity
6. A type of screen that are inclined slightly.
Speeds are at the range of 30-1000
strokes/min
7.
8.
9.
10.
a. Flat Screens
c.
Rotaty sifters
b. Reciprocating screens
d.
Trommel
It is either a gyratory or reciprocating type of
screening equipment that operate at 500-600
rpm.
a. Flat Screens
c. Mesh
Rotaty sifters
b. Reciprocating screens
d. 4
Trommel
These screens are vibrated or shaken to force 6
circulation of the bed of particles and to
8
prevent binding of the opening by oversize
particles and usually operates at 600-7000 10
strokes/min.
a. Flat Screens
c. 14
Rotaty sifters
b. Reciprocating screens
d. 20
Trommel
It is a material defined by mesh size, which 28
can be made by any type o material such as
steel, stainless, rubber, polyurethance, brass. 35
a. Shaker
c.
65
Screen mesh
b. Screen handle
d.
Pan
Screen media
Refers to the number of open slots per linear
inch.
a. Mesh
c.
slots
b. Scalps
d.
deck
11. A quartz mixture is screened through a
10-mesh screen. The cumulative screen
analysis of feed, overflow and underfolw are
given in the table. Calculate the mass ratios of
the overflow and underflow to feed and the
overall effectiveness of the screen.
Dp (mm)
Feed
Overflow
Underflow
4.699
0
0
0
3.327
0.025
0.071
0
2.362
0.15
0.43
0
1.651
0.47
0.85
0.195
1.168
0.73
0.97
0.58
0.833
0.885
0.99
0.83
0.589
0.94
1.0
0.91
0.417
0.96
0.94
0.208
0.98
0.975
1.0
1.0
Solution:
From the table, xF=0.47, xD=0.85, xB=0.195
12. A san mixture was screened through a
standard 12 mesh screen. The mass fraction
of the oversize material in feed, overflow and
underflow were found to be 0.4, -.8 and -.2
respectively. Calculate the screen
effectiveness based on the oversize materials.
Answer: 59.26%
13. A sponge-iron industry uses a reciprocating
screen of 5-mm aperture to separate oversize
from undersize fines which is then recycled to
the furnace. The screen analysis of the
furnace output was found to contain 25%
fines. The screen efficiency was known to be
50%. The underflow from the screen contains
95% fines. If the furnace production rate is
100 ton/h, find the product rate and the
amount of fines present. Answer: 85.55%,
86.6 ton/hr
14. If the screen dimensions in problem 59 were 2
ft by 4 ft, calculate the capacity of the 65mesh screen on the basis of a perfectly
functioning 48-mesh screen and also on the
basis of the actual performance of the screen.
Answer: 408.2 kg/m2-mm-hr
15. From measurements on a uniformly sized
material from a dryer, it is inferred that the
surface area of the material is 1200 m2. If the
density of the material is 1450 kg m-3 and the
total weight is 360 kg calculate the equivalent
diameter of the particles if their value of l is
1.75. Answer: 2200 microns
16. The capacity in MT/day-ft2-mm on the basis of
the actual performance of the 48 mesh
screen. Answer: 1.09
Size Reduction
1. Ball mills and tube mills or porcelain balls are
used for size reduction of
a. Asbestos
c.
non-metallic ores
b. Rubber
d.
limestone
2. Size reduction of _______ is accomplished in
steam heated rollers and roll crushers.
a. Resins
c.
hard rubber
b. Gums
d.
waxes
3. Size reduction of the _______ can be suitably
done by ball mills, crushing rolls and rod mills.
a. Metalliferrous ores
c.
Non-metallic ores
b. Basic slags
d.
Asbestos and mica
4. The main size reduction operation in ultrafine
grinders
a. Cutting
b.
attrition
b. Compressions
d.
impact
5. Balls for ball mills are never made of
a. Forged/cast steel
c.
Lead
b. Cast iron
d.
Alloy steel
6. Rittinger’s number designates the new surface
created per unit mechanical energy absorbed
by the material being crushed. Larger value of
Rittinger’s number of a material indicates
a. Easier grindability
c.
high power consumption in grinding
b. Poor grindability
d.
None of these
7. The particle is broken by two force and the
size reduction is done between two surfaces,
with the work being done by one of both
surfaces.
a. Compression
c.
Impact
b. Attrition
d.
Cutting
8. It is a method of size reduction that is done by
rubbing or scrubbing the materials between
two hard surfaces.
a. Compression
c.
Impact
b. Attrition
d.
Cutting
9. It is the resistance of a material to impact.
a. Toughness
c.
Hardness
b. Friability
d.
Brittleness
10. It is the tendency to fracture during normal
handling
a. Toughness
c.
Hardness
b. Friability
d.
Soapiness
11. A set of coarse-sized screens was constructed
from steel rods and used to evaluate the
efficiency of a 60 cm and 40 cm blake jaw
crusher. The standard tyler relationship was
maintained between the screen apertures in
th3 above set of 30hp motor. The screen
analysis of feed and product is given below:
Datas:
Sg of calcite = 2.71
Sg of quartz = 2.65
Rittinger’s number of calcite = 76.05 cm2/kgcm
Rittinger’s number of quartz = 17.51 cm2/kgcm
Calcuate the efficiency of the crusher
assuming the motor to be operated at an
average or 1/6th of the normal rating. Answer:
96%
12. From the above problem, calculate the ton.hr of
quartz that can be fed to the crusher and
reduced on the same size range with the same
power. Answer: 10.22 ton/hr
13. Particles of the average feed size of 50x10^-4
m are crushed to an average product size of
10x10^-4 m at the rate of 20 tons/hr. At this
rate, the crusher consumes 40KW of power f
which 5 KW are required for running the mill
empty. Calculate the power consumption if 12
tons/hr of this product is further crushed to
5X10^-4 m sie in the same mill. Use rittinger’s
law
Solution:
35
1
1
= 𝐾𝑟 (
−
)
20
10 × 10−4 50 × 10−4
𝑘𝑊ℎ − 𝑚
𝑡𝑜𝑛𝑛𝑒
𝑃
1
= 2.1875 × 10−3 (
12
5 × 10−4
1
−
)
10 × 10−4
𝑷 = 𝟐𝟔. 𝟐𝟓 𝒌𝑾
14. Find the power required for crushing 5ton/hr of
liestonne (Rittinger’s number = 0.0765 m2/J) if
the specific surface areas of the feed and the
product are 100 and 200 m2/kg respectively. If
the machine consumes a power of 4 hp,
calculate its efficiency. Answer: 60.75%
15. A sample of materials is crushed in a blake jaw
crusher such that the average size of the
particles is reduced from 50 mm to 10 mm with
the energy consumption of 13kW. Determine
the consumption of energy to crush the same
material of 75 mm average size to an average
size of 23 mm using Rittinger’s law. Answer:
1.218 kWh/ton
16. For the same problem use Kick’s law. Answer:
2.24 kWh/ton
𝐾𝑟 = 2.1875 × 10−3
Centrifugation
1. Tubular bowl centrifuges as compared to disk
bowl centrifuges
a. Operate at higher speed
c.
Cannot be operated under
pressure/vacuum
b. Employ bowl of larger diameter
d.
Can’t be used for separation of fine
suspended solids
from liquid
2. __________ centrifuge is normally used in
sugar mills.
a. Tubular bowl
c.
Disc-bowl centrifuge
b. Suspended batch basket
d.
perforated horizontal basket continuous
3. If a force greater than that of gravity is used to
separate solids and fluids of different
densities, the process is termed as the
a. Sedimentation
c.
dispersion
b. Flocculation
d.
centrifugation
4. Where the density difference of the two liquid
phase to be separated is very small (as in milk
cream separator), the most suitable separator
is a _____.
a. Disc bowl centrifuge
c.
Sharpies supercentrifuge
b. Batch basket centrifuge
d.
Sparkler filter
5. Classified as a centrifuge that is separation is
dependent on a difference of density between
solid and liquid phases.
a. Sedimentation centrifuges
c.
Classifying centrifuges
b. Filtration centrifuges
d.
Thickening centrifuges
6. Use centrifugal force to drive the filtrate
through the filter cake.
a. Classifying filter
c.
thickening filter
b. Centrifugal filter
d.
Filtration
7. A type of centrifuge that is considered high
speed, vertical axis, that are used for the
separation of immiscible liquids, such as water
and oil and for the separation of fine solids.
a. Tubular bowl
c.
solid-bowl
b. Disc-bowl
d.
scroll discharge
8. A type of centrifuge which has conical disc
that split the liquid flow into number of thin
layers, which greatly increases the separating
efficiency.
a. Tubular bowl
c.
solid-bowl
b. Disc-bowl
d.
scroll discharge
9. In this type of centrifuge, the solids are
deposited on the wall of the bowl and are
reoved by a helical screw conveyor which
revolves at a slightly different speed from the
bowl.
a. Tubular bowl
c.
solid-bowl
b. Disc-bowl
d.
scroll discharge
10. The simplest type of centrifuge.
a. Tubular bowl
c.
solid-bowl
b. Disc-bowl
d.
scroll discharge
11. If a centrifuge is 0.9 m diameter and rotates at
20Hz, at what speed should a laboratory
centrifuge of 150 mm diameter be run if its is
to duplicate the performance of a large unit?
Solution:
0.45 (40𝜋)2 0.075 (𝜛)2
=
𝑔
𝑔
𝜛2 = √6(40𝜋)2
𝑟𝑎𝑑
𝜛2 = 98𝜋
𝑠
𝒓𝒂𝒅
𝟗𝟖𝝅 𝒔
𝑵=
= 𝟒𝟗 𝑯𝒛 = 𝟐𝟗𝟒𝟎 𝐫𝐩𝐦
𝟐𝝅
12. An aqueous suspension consisting of particles
of density 2500 kg/m3 in the size range of 110 micrometer is introduced into a centrifuge
with a basket of 450 mm diameter rotating at
80Hz. If the suspension forms a layer of 75 m
thick in the basket, approximately how long
will it take for the smallest particle to settle
out? Answer: 19 s
13. A centrifuge basket 600 mm ong and 100 mm
internal diameter has a discharge weir 25 mm
diameter. What is the maximum volumetric
flow of liquid through the centrifuge that, when
the basket is rotated at 200 Hz, all particles of
diameter greater than 1 micrometer are
retained on the centrifuge wall? The retarding
force on a particle moving liquid may be taken
as 3𝜋𝜇 𝑑𝜇 where u is the particle velocity
relative to the liquid e is the liquid viscosity
and d is the particle diameter. The density of
the liquid is 1000kg/m3, the density of solid is
2000kg/m3 and viscosity of the liquid is 1 mNs/m2. The inertia of the particle may be
neglected. Answer: 1cm3/s
14. A centrifuge with a phosphor bronze basket,
380 mm in diameter, is to be run at 67 Hz with
a 75 mm layer of liquid density 1200 kg/m3 in
the basket. What thickness of walls are
required in the basket? The density of
phosphor bronze is 8900 kg/m3 and the
maximum safe stress for the bronze is 876
MN/m2. Answer: 15.5 mm
15. A dispersion of oil in water is to be separated
using a centrifuge. Assume that the oil is
dispersed in the form of spherical globules
5.1x10^-5 m diameter and that its density if
894 kg.m3. If the centrifuge rotates at 1500
rpm and the effective radius at which
separation occurs is 3.8 cm, calculate the
velocity of the oil through water. Take the
density of water to be 1000kg/m3 and its
viscosity to be 0.7x10^-3 Ns/m2. Answer: 0.2
m/s
16. If a centrifuge is 3-ft diameter and rotates at
1,000 rpm, what must be the speed of a
laboratory centrifuge of 6-in diameter be ran if
it is duplicate plant conditions? Answer: 2449
rev/min
Flotation
1. Froth flotation is the most suitable for treating
a. Iron ores
b.
sulphide ores
b. Quartzite
d.
none of these
2. The products obtained from a floatation
operation is known as
a. Concentrate
c.
Conditioners
b. Tailings
d.
Froths
3. The underflow slurry containing the gangue
a. Concentrate
c.
Conditioners
b. Tailings
d.
Froths
4. Particles which are easily wetted by water and
tends to remain in suspernsion
a. Hydrophobic
c.
Hydraneous
b. Hydrophilis
d.
Hydrophilic
5. The conditioning step is known as
a. The feed slurry is then prepared by
adding water
b. Made free of any agglomerates by adding
dispersants
c. Agitating them in tank
d. All of these
6. A type of mechanical floatation equipment in
which air is dispersed by mechanical means
a. Launcher
c. Frother
b. Floatation cell
d. Thickener
7. The floatation technique is used for
a. Recovery of fine coal
b. Concentration of barite, iron oxide, mica,
talk, pyrite, feldspar and many more
minerals
c. Application of waste water treatment
d. All of these
8. Methyl isobutyl carbinol is used in floatation as
a
a. Depressor
c.
Collector
b. Frother
d.
pH-regulator
9. A depressor which can depress pyrites from
lead ore to float galena is
a. Calcium cynamide
b. Pine oil
c. Polypropylene glycol
d. Oleic acid
10. Floatation is carried out generally at a pH
value of
a. Less than 7
b. Greater than 7
c. 0-14
d. Nothing in particular
11. A flotation plant processes 3000 tons/day of
CuFeS2 (chalcopyrite). It produces 80 tons of
Cu concentrate assaying 25%Cu. If ore
analyzes 0.7% Cu, what is the percent
recovery? Answer: 95.24%
12. A copper ore initially contains 2.09% Cu. After
carrying out a froth flotation separation, the
products are shown, using this data, calculate
the metal recovery. Answer: 95.7%
Product
%Weight
%Cu Assay
Feed
100
2.09
Concentrate
10
20.0
Tailings
90
0.1
13. Water is fed to the cell at the rate of 1,000
gallons per ton of wet concentrate with 99% of
the water leaving with the tailings and 1% with
the concentrate.
14. What is the mass of wet concentrate produced
per hour when tons of ore are fed to the cell
per 24 hours? Answer: 3.4 kg/hr
15. A copper ore initially contains 2.09% Cu. After
carrying out a froth flotation separation, the
products are as shown.
Feed
100
2.09
Concentrate
10
20
Tailings
90
0.1
Using this data, calculate:
(a) Ratio of concentration Answer: 10
16. A typical flotation machine has the following
specifications:
Number of cells = 4
Flotation time = 12min.
Cell Volume = 60 ft3
Hp per cell = 10hp
Pulp (mixture ore and water ) = 40% solids
Specific gravity of ore = 3
Solution
𝑇 𝑥 𝐶𝑎𝑝 𝑥 𝑑
𝑛=
𝑉 𝑥 1440
2000
2000 + 𝑥
𝑥 = 3000 𝐻2𝑂
𝐹 = 3000 + 2000 = 5000
𝟐𝟎𝟎𝟎
𝟑𝟎𝟎𝟎
𝒅=
+
= 𝟓𝟖. 𝟕𝟔
𝟑𝒙𝟔𝟐. 𝟒 𝟔𝟐. 𝟒
0.4 =
Sedimentation
1. Two particles are called to be equal settling, if
they are having the same ____.
a. Size
c. terminal
velocities in the same fluid and field of force
b. Specific gravity
d. none of
these
c. Its purpose is to increase the
concentration of solids.
a. Thickeners
b.
Separators
b. Clarifiers
d.
Filters
d. Its purpose is to remove a relatively small
quantity of suspended solids to get a clear
liquid.
a. Thickeners
b.
Separators
b. Clarifiers
d.
Filters
e. It is a physical process used to separate
the suspended solids from a liquid under
the influence of gravity.
a. Filtration
c.
Sedimentation
b. Centrifugation
d.
Settling
f. When particles fall in a gravitational field
though a stationary fluid and their fall is
not affected by walls of the container and
by other particles, the settling process is
a. Velocity settling
c.
Hindered settling
b. Flocculant settting
d.
Free settling
g. When the particle through a stationary
fluid is impeded by the presence of other
particles, the process is called
a. Velocity settling
c.
Hindered settling
b. Flocculant settting
d.
Free settling
h. In free settling, the falling velocity of the
particle increases until the resisting and
accelerating forces becomes equal. When
this condition is reached, the particles
settle at a constant velocity known as
a. Free settling velocity
c.
Terminal settling velocity
b. Hindered settling velocity
d.
Speed
i. The law where in the relationship between
the draw coefficient and Reynolds number
is given by a straight line of slope.
a. Newton’s Law
c.
Free settling
b. Intermediate Law
d.
Stoke’s Law
j. When two solid particles having different
densities but the same size are separated
using a fluid, the method is known
a. Floatation
c.
Clarification
b. Filtration
d.
Elutriation
k. If there is a wide difference in densities of
two materials, the separation technique is
a. Floatation
c.
Clarification
b. Filtration
d.
Elutriation
l. Find the terminal settling velocity for
particles of 40 micron size having a
specific gravity of 2.6 falling though still
water is the settling one is laminar. All
particles may be assumed spherical and
the wall effect may be neglected.
Viscosity of water may be taken as 1 cP.
Solution:
𝐷𝑝2 (𝜌𝑝 − 𝜌𝑓 )𝑔
𝑣𝑡 =
18𝜇𝑓
𝑣𝑡
=
(40 × 10−6 )2 (2600 − 1000)(9.8)
18(1 × 10−3 )
𝒎
𝒔
A sample of bauxite ore is to be cleaned
using water in a classifier. The ore
particles have a size range of 10-500
microns. The mixture is being separated
into 3 parts: pure bauxite (sg=2.2), pure
silica (sg = 2.8) and the third fraction is
the middling which is recycled. Assuming
the flow to be laminar and neglecting any
wall effect, estimate the size range of
bauxite. Answer: 408.25 microns
A concentrated suspension of galena
particles in water settle under gravity in a
settling tank. If the density of particles is
7500 kg/m3 and the particle diameter is
0.0002 m, find the upwared velocity of
water in the tank. The porosity suspension
is 0.5 and the index is 4.5. Answer:
6.26x10^-3 m/s
A glass sphere of diameter 6 mm and
density 2600 kg.m3, falls through a layer
of oil of density 900 kgm3 into water. If the
oil layer is sufficiently deep for the particle
to have reached its free falling velocity in
the oil, how far will it have penetrated into
the water before its velocity is on 1%
above its free falling velocity in water. It
may be assumed that the force on that
particle is given by Newton’s law and that
the drag coefficient is 0.22. Answer:
0.096 s, 48 mm
Two spherical particles, one density 3000
kg/m3 and diameter 20 microns and the
other density 2000kg/m3 and diameter is
30 microns start settling from rest at the
same horizontal level in a liquid density
900kg/m3 and of viscosity of 3mN-s/m2.
After what period of settling will the
𝒗𝒕 = 𝟎. 𝟎𝟎𝟏𝟒
m.
n.
o.
p.
particles be again at the same horizontal
level. It may be assumed that Stoke’s law
is applicable and the effect of added mass
of the liquid moved with each sphere may
be ignored. Answer: 7.81x10^-5 s
.
Particle
Technology
Mardo Lex
FLOTATION
1. Regulators such as lime, caustic soda, soda
ash, and sulfuric acid are used to control or
adjust pH
A. Activators
B. pH Regulator
C. Depressants
D. Flocculants
2. Any operation in which one solid is
separated from another by floating one of
them at or on the surface of a fluid.
A. Coagulation
B. Flotation
C. Centrifugation
D. Sedimentation
3. Pine oil is used in flotation process acts as a
A. Collector
B. Modifier
C. Frother
D. Activator
4. In Froth Flotation, the chemical agent added
to cause air adherence is called
A. Collector
B. Modifier
C. Frother
D. Activator
5. Froth Flotation is most suitable for treating
A. Iron ores
B. Sulfide ores
C. Quartz
D. Metal ores
6. Which of the following is an example of a
deflocculant?
A. Sulfuric Acid
B. Lignin sulforate
C. Dithiophosphate
D. Molybderite
7. An example of a collector for flotation of
metallic sulfides and native metals is
A. Xanthates
B. Sodium Silicate
C. Sodium sulfide
D. Sphalerite
8. A flotation modifier which assists in the
selectivity (sharpness of separation) or stop
unwanted minerals from floating
A. Depressants
B. Activators
C. Alkalinity Regulators
D. Promoters
9. Dispersants are important for the control of
slimes which sometimes interfere with the
selectivity
and
increase
reagent
consumption. Another term for dispersant is
A. Deflocculant
B. Activators
C. Alkalinity regulators
D. Promoters
10. The flotation agent that prevents
coalescence of air bubbles as they travel to
the surface of the water is/are
A. Collectors
B. Promoters
C. Frothing Agent
D. Modifying Agent
SEDIMENTATION
1. In continuous thickeners, separation of solid
particles can be achieved if the settling
velocity of the solids is _________ compared
to the velocity of the displaced liquid
A. Equal
B. Less
C. Greater
D. None of these
2. In order for a particle to move through a
fluid under the influence of gravity, there
must be
A. Velocity Difference
B. Pressure Difference
C. Density Difference
D. Temperature Difference
3. The ratio of drag force per unit area to the
product of fluid density and the velocity
head is called
A. Buoyant Coefficient
B. Drag Coefficient
C. Friction Factor
D. Shear Coefficient
4. At high Reynolds Number
A. Inertial force control viscous forces are
unimportant
B. Viscous forces predominate
C. Inertial forces are unimportant and
viscous forces control
D. None of these
5. At low Reynolds Number
A. Viscous forces are unimportant
B. Viscous forces equal the inertial forces
C. Viscous forces control and inertial forces
are unimportant
D. Gravity forces control
6. Stoke’s Law is valid when the particle
Reynolds Number is
A. <1
B. >1
C. <2
7. Drag coefficient for flow past an immersed
body is the ratio of
A. Shear stress to the product of velocity
head and density
B. Shear force to the product of velocity
head and density
C. Average drag per unit projected area to
the product of the velocity head and
density
D. None of these
8. Drag is defined as the force exerted by
A. The fluid on the solid in a direction
opposite to the motion of the solid
B. The fluid on the solid in the direction of
motion of the solid
C. The solid on the fluid
D. None of these
9. The drag coefficient in hindered settling
is_________ compared to free settling
A. Greater than
B. Less than
C. Constant
D. Varying
10. The operation by which solids are separated
from liquids due to difference in their
respective densities is
A. Screening
B. Adsorption
C. Sedimentation
D. Absorption
CENTRIFUGATION
1. If the thickness of the fluid in the centrifuge
is small, terminal velocity is
A. Equals zero
B. Constant
C. Increases
D. Decreases
2. Which of these is not a type of centrifugation
A. Centrifugal sedimentation
B. Centrifugal filtration
C. Centrifugal Decantation
D. Centrifugal distillation
3. The most common type of equipment for
clarifying centrifuge.
A. Sharples centrifuge
B. Plate and frame filter
C. Sparkler filter
D. Rotary drum vacuum filter
4. A mechanical process of separating multiphase (usually fluid-solid system) via the use
of centrifugal force.
A. Screening
B. Sedimentation
C. Centrifugation
D. Leaching
5. Settling process due to the difference in
densities of the solid and fluid media.
A. Clarifying centrifuge
B. Disc bond centrifuge
C. Batch basket centrifuge
D. Sparkler filter
6. Where the difference in density of the two
liquid phases to be separated is very small
(as in milk cream separator), the most
suitable separator is
A. Disc bond centrifuge
B. Sharpless supercentrifuge
C. Batch basket centrifuge
D. Sparkler filter
7. If the radius of a basket centrifuge is halved
and the rpm is doubled, then the
A. Linear speed of the basket is doubled
B. Linear speed of the basket is halved
C. Centrifugal Force is doubled
D. Capacity of centrifuge is increased
8. Which of the following can be most
effectively used for clarification of lube oil
and printing ink?
A. Sparker filter
B. Precoat filter
C. Disc-bowl centrifuge
D. Sharpless supercentrifuge
9. Moisture can be removed from lubricating
oil using
A. Tubular Centrifuge
B. Clarifier
C. Sparkler filter
D. Vacuum leaf filter
10. For the separation of sugar solution from
settled out mud, we use
A. Sparkler filter
B. Plate and frame filter
C. Centrifugal filter
D. Rotary drum vacuum filter
SCREENING
1. Trommels separate a mixture of particles
depending on their
2.
3.
4.
5.
6.
A. Size
B. Wettability
C. Screen size
D. Electrical and magazine
Size measurement of ultrafine particles
can best expressed in terms of
A. Centimeter
B. Screen size
C. Micron
D. Surface area per unit mass
A screen is said to be blinded when
A. Oversize are present in undersize
fraction
B. Undersize are retained in oversize
fraction
C. The screen is plugged with solid
particles
D. Its capacity is abruptly increased
As the particle size is reduced
A. Screening becomes progressively
more difficult
B. Screening becomes progressively
easier
C. Capacity and effectiveness of the
scree increased
D. None of these
The screen efficiency is
A. Recovery + rejection
B. Recovery
C. Rejection
D. None of these
Increasing the capacity of screen
A. Decrease the screen effectiveness
B. Increase the screen effectiveness
C. Does not affect the screen
effectiveness
D. None of these
7. The ratio of the area of the opening in
one screen (Tyler Series) to the opening
of the next smaller screen is
A. 1.5
B. 1
C. Square root of 2
D. None of these
8. The clear opening of a 200-mesh screen
in the Tyler Standard scree series
A. 0.0074 cm
B. 0.0074 mm
C. 0.0047 cm
D. 0.0047 mm
9. Materials which remain on a screen
surface are called
A. Fines
B. Undersize
C. Intermediate material
D. Oversize
10. Another name for revolving screen is
A. Shaking screen
B. Vibrating screen
C. Mechanically vibrated screen
D. Trommel screen
SIZE REDUCTION
1. Cement clinker is commonly reduced to fine
size using a
A. Roll Crusher
B. Ball mill
C. Tube Mill
D. Hammer Mill
2. Wet grinding in a revolving mill
A. Gives less wear on chamber walls than
dry grinding
B. Requires more energy than for dry
grinding
C. Increase capacity compared to dry
grinding
D. Complicates handling of the product
compared to dry grinding
3. A fluid mill is used for
A. Cutting
B. Grinding
C. Ultragrinding
D. Crushing
4. The operation speed of a ball mill should be
A. Less than the critical speed
B. Much more than the critical speed
C. At least equal to the critical speed
D. None of these
5. Soft and non-abrasive materials can be made
into fines by
A. Attrition
B. Compression
C. Cutting
D. None of these
6. For coarse reduction of hard solids, use
A. Impact
B. Attrition
C. Compression
D. Cutting
7. Equivalent diameter of a particle is the
diameter of the sphere having the same
A. Ratio of surface to volume as the actual
volume
B. Ratio of volume to surface as the particle
C. Volume as the particle
D. None of these
8. Sphericity for a cylinder whose length equals
its diameter is
A. 1.5
B. 0.5
C. 0.87
D. 5
9. Based on the Bond’s Crushing Law, the
power required to crush a certain material
will change by ______% if the diameter of
the product is made smaller by 50 %
A. 50%
B. 41%
C. 25%
D. 75%
10. Size reduction is important to chemical
engineering since
A. It prevents the chemical engineers from
becoming overweight
B. It makes products to become uniform in
size
C. It prepares raw materials of the desired
sizes prior to processing
D. None of these
Problems With Answers
RACHO, CARLO ANGELO E.
FLOTATION
1. A typical flotation machine has the following specifications:
Number of cells = 4
Flotation time = 12 min
Cell volume
= 60 cu. Ft
Hp per cell
= 10 Hp
The material treated has the following specifications:
Pulp (mixture of ore and water) = 40% solids
Specific gravity of ore
=3
Equation: 𝑛 =
𝑇𝑥 𝐶𝑎𝑝𝑥𝑑
𝑉𝑥1440
Where n= number of cells; V= volume in cu. ft. per cell;
Cap = tons of dry ore per 24 hours; d = cu. ft. of pulp
(ore and water) containing one ton (2000 lbs) of solids.
What is the capacity of the machine in tons of dry ore per 24 hours?
Ans. 490
2. A flotation plant produces 3000 tons/day of CuFeS₂. It produces 80 tons of Cu concentrate
assaying 25% Cu. If ore analyses 0.7% Cu, what is the percent recovery?
Ans. 95.2%
3. Ground lead ore is to be concentrated by a single flotation process using 1.5 oz of reagent per
ton of ore. The feed concentrate and tailings have the following composition by weight on a dry
basis:
Feed %
Concentrate %
Tailings %
PbS
30
90
0.9
ZnS
25
3
35.6
SiO₂
45
7
63.5
Water is fed to the cell at the rate of 1,000 gallons per ton of wet concentrate with 99% of the
water leaving with the tailings and 1% with the concentrate. What mass of wet concentrate
produced per hour when ten tons of ore are fed to the cell per 24 hours? What is the total water
required in pounds per hour?
Ans. 3.4 , 1185
4. A flotation section of a mining company is extracting CuS from covellite ores. The ore consists of
5% CuS and 95% gangue, which may be assumed to be SiO₂. The following data are given:
%CuS
%SiO₂
Feed
5
95
Concentrate
85
15
Rougher Tailings
1
99
Scavenger Tailings
10
90
Final Tailings
0.3
99.7
Laboratory experiments indicated that the water to solids ratio, L/S = 2 and the contact time is
10 min. in the rougher; L/S = 4, contact time = 18 min. in the scavenger. On the basis of 300 tons
per day of ore treated.
Data: Density of SiO₂ = 2.65 g/cc
Density of CuS = 4.6 g/cc
What is the volume of the rougher needed and what is the volume of the scavenger needed?
Ans. 169.8 ft³, 535 ft³
5. A flotation plant processes 3,000 tons/day of CuFeS₂ (chalcopyrite). It produces 80 tons Cu
concentrate assaying 25% Cu. If ore analyzes 0.7% Cu, the percent recovery is?
Ans. 95.24%
SEDIMENTATION
1. In an experiment, a sphere of density ρ₁ and radius 𝑟 is dropped in a tank of oil of viscosity µ₁
and density ρ₂. The time of descent of the sphere through the first section of height 𝑑 is
recorded as 𝑡₁ and through the second section of the same height as 𝑡₂, 0<𝑡₂-𝑡₁<<1.
Sphere
r
ρ₂
𝑑
𝑑
Derived the drag force exerted on the sphere during its descent through the second section.
4
3
Ans. (𝜌1 − 𝜌2 )𝑔 𝜋𝑟 3
2. A continuous thickener is required to concentrate a slurry of calcium carbonate in water from a
solids content of 50 kg/m³ to 130 kg/m³, and to produce a clear overflow containing no calcium
carbonate. The density of the dry calcium carbonate is 2300 kg/m³. A single batch sedimentation
test produced the data below:
Time, hr
0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
Z, m
1.1
0.98
0.82
0.68
0.54
0.42
0.35
0.31
0.28
0.27
If the thickener is fed with slurry at a rate of 0.06 m³/s. What is the minimum thickener area in
square meter required? What is the flow rate of clarified water in m³/s?
Ans. 1005, 0.037
0.27
3. An intimate mixture of two different ores of densities 2600 kg/m³ (A) and 1500 kg/m³ (B) is
finely ground and given a screen analysis. The various fractions collected on the screens are then
analyzed chemically to determine the amount of each ore in each size fraction, and in this way
the following particles size analysis for each component of the original mixture is obtained.
Diameter, microns
15
25
35
45
55
65
75
85
95
% undersize, A
0
22
35
47
59
68
75
81
100
% undersize, B
0
0
21
33.5
46
57.5
67
75
100
The ratio of the heavy ore to the light ore in the mixture was by analysis, 1:5 by weight. If the
ores are to be separated from the original mixture by elutriation using water at a velocity of 4.00
x 10−3 m/s and a temperature of 18℃. What is the composition of the bottom product in terms
of A? What is the liquid velocity in m/s required to produce a bottoms product that does not
contain any lighter particle?
Ans. 67%, 7.4x10−3
4. Square mica plates, 1/32 in. thick and 0.01 sq. in. in area are falling randomly through oil with a
density of 55 lb/cu ft and with viscosity of 15 centipoise. The specific gravity of the mica is 3.0,
what is the settling velocity?
Ans. 7.2 cm/s
5. Free settling of sludge is 0.25 cm/min. Using an original height of 25cm, the sludge settled to a
height of 18 cm after the free-settling period. The sludge was found to settle to a height of 10cm
after 110 min. This particular sludge was found to settle completely to a height of 4 cm. What is
the time to settle to a height of 1/5 of its original height in a cylindrical tank whose diameter is
85% of its depth if it is 85% full and consider 1000 cu ft of sludge in the tank.
Ans. 62.4 hours
CENTRIFUGATION
1. A centrifugal bowl is spinning at a constant speed of 2000 rpm. What is the radius of bowl in cm
needed to create a force of 455 g?
Ans. 10.2
2. A centrifuge with a radius of 76.2 mm rotates at a peripheral velocity of 53.34 m/s. What is the
centrifugal force developed compared to gravitational force in the bowl centrifuge?
Ans. 3800 g
3. Two centrifuges rotate at the same peripheral velocity of 53.34 m/s. The first bowl has a radius
of r₁= 305 mm. Calculate the rev/min and the centrifugal forces developed in each bowl.
Ans. N₁= 6684 rev/min
N₂= 1670 rev/min, 3806 g’s in bowl 1,951 g’s in bowl 2
4. A centrifuge bowl is spinning at a constant 2000 rev/min. What radius bowl is needed for the
following?
(a) A force of 455 g’s
(b) A force four times that in part (a)
Ans. (a) r = 0.1017 m
5. A cream separator centrifuge has an outlet discharge radius r₁ = 50.8 mm and outlet radius r₄ =
76.2 mm. The density of the skim milk is 1032 kg/m³ and that of the cream is 863 kg/m³.
Calculate the radius of the interface neutral zone.
Ans. r₂ = 150 mm
SCREENING
1. Rock used as gravel is crushed after quarrying. It is crushed by six units operating in parallel and
the products separated by six 35 mesh screen also in parallel, into two fractions. The effective
dimensions of each screen is 6ft by 20ft. The common undersize from the screen comes out at
the rate of 50 tons/hr. Assume no losses.
Mesh size
Feed size
Undersize
Oversize
6/8
0.075
0.050
0.200
8/10
0.125
0.145
0.055
10/20
0.100
0.170
0.020
20/28
0.125
0.150
0.085
28/35
0.125
0.280
0.100
35/48
0.175
0.175
0.150
48/65
0.225
0.150
65/100
0.050
0.250
100/150
0.100
Calculate the efficiency of the screening operation?
Ans. 52.82%
2. If the total percentage of particles larger that the screen opening in the feed, product and
undersize are 36%, 89% and 3% respectively, calculate the effectiveness of the screen.
Ans. 88.61%
3. It is a desired to separate a mixture of table salt crystals into two fractions, a course fraction
retained on a 10 mesh screen and a fine fraction passing through it. Screen analysis of feed,
course and fine fractions show
Mass fraction of +10 feed = 0.46
Mass fraction of +10 particles in course fraction = 0.88
Mass fraction of +10 articles in fine fractions = 0.32
Calculate the overall effectiveness of the screen used for the separation purpose per 100kg of
feed.
Ans. 45.17%
4. 1800 kgs of calcite per hour is produced by crushing and then screening through a 14 mesh. The
screen analysis is as follows
Tyler Mesh
Feed to screen
Undersize product
Screen oversize
4 on
14.3
20
8 on
20
28
14 on
20
24
28 on
28.5
40
48 on
8.6
30
100 on
5.7
20
1000 through
2.86
10
Calculate the effectiveness of the screen.
Ans. 62.42%
5. If the total percentage of particles larger that the screen opening in the feed, product and
undersize are 31%, 93% and 11% respectively, calculate the effectiveness of the screen.
Ans. 71.36%
SIZE REDUCTION
1. The energy required per unit mass to grind limestone particles of very large size to 100 μm is
12.7 kWh/ton. An estimate (using Bond's law) of the energy to grind the particles from a very
large size to 50 μm.
Ans. 18 kWh/ton
2. Capacity (in tons/hr) of jaw/gyratory crusher is equal to (where, L = length of the receiving
opening, cm S = greater width of the discharge opening, cm)
Ans. 0.087 LS
3. What is the critical rotation speed in revolutions per second, for a ball mill of 1.2 m diameter
charged with 70 mm diameter balls?
Ans. 0.66
4. A ball mill, 1.2 m in diameter, is run at 0.80 Hz and it is found that the mill is not working
properly. Should any modification in the conditions of operation be suggested?
Ans. 2.42 rad/s
0.39 Hz
the speed of rotation should be halved
5. A material is crushed in a Blake jaw crusher such that the average size of particle is reduced
from 50 mm to 10 mm with the consumption of energy of 13.0 kW/(kg/s). What would be the
consumption of energy needed to crush the same material of average size 75 mm to an average
size of 25 mm:
a) Assuming Rittinger’s law applies?
b) Assuming Kick’s law applies?
Which of these results would be regarded as being more reliable and why?
Ans. (a) 4.33 kJ/kg
(b) 8.88 kJ/kg
Problems With Solutions
RACHO, CARLO ANGELO E.
FLOTATION
A copper ore initially contains 2.09% Cu. After carrying out a froth flotation separation, the products are
as shown in Table 1. Using this data, calculate:
(a) Ratio of concentration
(b) % Metal Recovery
(c) % Metal Loss
(d) % Weight Recovery, or % Yield
(e) Enrichment Ratio
Solution
%𝐶𝑢 𝑅𝑒𝑐𝑜𝑣𝑒𝑟𝑦 =
(10)(20)
𝑥100 = 95.7%
(2.09)(100)
%𝐶𝑢 𝐿𝑜𝑠𝑠 = 100 − 95.7 = 4.3%
2.09 − 0.1
𝑥100 = 10%
20 − 0.1
20.0
𝐸𝑛𝑟𝑖𝑐ℎ𝑚𝑒𝑛𝑡 𝑅𝑎𝑡𝑖𝑜 =
= 9.57
2.09
%𝑊𝑒𝑖𝑔ℎ𝑡 𝑅𝑒𝑐𝑜𝑣𝑒𝑟𝑦 =
SEDIMENTATION
A mixture of silica (B) and galena (A) solid particles having a size range of 5.21𝑥10−6 m to 2.50𝑥10−5 m
is to be separated by hydraulic classification using free settling conditions in water at 293.2 K. The
specific gravity of silica is 2.65 and that of galena is 7.5. Calculate the size range of the various fractions
obtained in the settling. If the settling is in the laminar region, the drag coefficient will be reasonably
close for that for spheres.
Solution
𝑣𝑡𝐴 =
2
𝑔𝐷𝑝𝐴
(𝜌𝑝𝐴 − 𝜌)
18𝜇
𝑣𝑡𝐴 =
9.807(2.50𝑥10−5 )2 (7500 − 998)
= 2.203𝑥10−3 𝑚⁄𝑠
18(1.005𝑥10−3 )
𝑁𝑅𝑒 =
𝐷𝑝𝐴 𝑣𝑡𝐴 𝜌
𝜇
𝑁𝑅𝑒 =
(2.50𝑥10−5 )(2.203𝑥10−3 )998
= 0.0547
1.005𝑥10−3
𝐷𝑝𝐴3
2650 − 998 1⁄2
=
(
)
2.50𝑥10−5
7500 − 998
𝐷𝑝𝐴3 = 1.260𝑥10−5 𝑚
5.21𝑥10−6
2650 − 998 1⁄2
=(
)
𝐷𝑝𝐴3
7500 − 998
𝐷𝑝𝐵2 = 1.033𝑥10−5 𝑚
CENTRIFUGATION
In a test on a centrifuge all particles of a mineral of density 2800 kg/m³ and of size 5 μm, equivalent
spherical diameter, were separated from suspension in water fed at a volumetric throughput rate of
0.25 m³/s. Calculate the value of the capacity factor.
What will be the corresponding size cut for a suspension of coal particles in oil fed at the rate
of 0.04 m³/s? The density of coal is 1300 kg/m³ and the density of the oil is 850 kg/m³ and its
viscosity is 0.01 Ns/m².
It may be assumed that Stokes’ law is applicable.
Solution
𝑢𝑜 =
𝑑 2 (𝜌𝑠 − 𝜌)𝑔 25𝑥10−12 𝑥(12800 − 1000)𝑥9.81
=
18𝜇
18𝑥10−3
𝑢𝑜 = 2.45𝑥10−5 𝑚⁄𝑠
𝑄 = 𝑢𝑜 𝛴
𝛴=
0.25
= 1.02𝑥104 𝑚2
(2.45𝑥10−5 )
𝑢𝑜 =
𝑄
0.04
=
= 3.92𝑥10−6 𝑚⁄𝑠
4)
(1.02𝑥10
𝛴
𝑑2 =
18𝜇𝑢𝑜
(𝜌𝑠 − 𝜌)𝑔
𝑑2 =
18𝑥10−2 𝑥3.92𝑥10−6
(1300 − 850)𝑥9.81
𝑑 = 4.0𝑥10−6 𝑚 𝑜𝑟 4 𝜇𝑚
SIZE REDUCTION
If crushing rolls, 1 m in diameter, are set so that the crushing surfaces are 12.5 mm apart and the angle
of nip is 31◦, what is the maximum size of particle which should be fed to the rolls? If the actual capacity
of the machine is 12 per cent of the theoretical, calculate the throughput in kg/s when running at 2.0 Hz
if the working face of the rolls is 0.4 m long and the bulk density of the feed is 2500kg/m³.
Solution
The particle size may be obtained from:
cos 𝑎 =
(𝑟1 + 𝑏)
(𝑟1 + 𝑟2 )
2𝑎 = 31°
cos 𝑎 = 0.964
𝑏=
𝑟1 =
12.5
2
= 6.25 𝑚𝑚 𝑜𝑟 0.00625𝑚
1
= 0.5𝑚
2
0.964 =
(0.5 + 0.00625)
(0.5 + 𝑟2 )
r2 = 0.025 m or 25 mm
The cross sectional area for flow = (0.0125 × 0.4) = 0.005 m²
SCREENING
A quartz mixture is screened through a 10-mesh screen. The cumulative screen analysis of feed,
overflow and underflow are given in the table. Calculate the mass ratios of the overflow and underflow
to feed and the overall effectiveness of the screen.
Mesh
Dp (mm)
Overflow
Underflow
4
4.699
0
0
0
6
3.327
0.025
0.071
0
8
2.362
0.15
0.43
0
10
1.651
0.47
0.85
0.195
14
1.168
0.73
0.97
0.58
20
0.833
0.885
0.99
0.83
28
0.589
0.94
1.0
0.91
35
0.417
0.96
0.94
65
0.208
0.98
0.975
1.0
1.0
Pan
Solution:
𝑥𝑓 = 0.47, 𝑥𝐷 = 0.85, 𝑥𝐵 = 0.195
𝑥𝑓 − 𝑥𝐵 0.47 − 0.195
𝐷
=
=
= 0.42
𝐹
𝑥𝐷 − 𝑥𝐵 0.85 − 0.195
𝐵
𝐵
= 1 − = 0.58
𝐹
𝐹
𝐸=
Feed
(𝑥𝑓 − 𝑥𝐵 )(𝑥𝐷 − 𝑥𝑓 )𝑥𝐷 (1 − 𝑥𝐵 )
(𝑥𝐷 − 𝑥𝐵 )2 (1 − 𝑥𝑓 )𝑥𝑓
= 0.669
Technological Institute of the Philippines
363 P. Casal St. Quiapo, Manila
Chemical Engineering Department
Introduction to Particle Technology
PROBLEM SET
(Finals)
Submitted by:
RILLERA, Regina C.
BS ChE
CHEP592 – CH51FB1
Submitted to:
Engr. Robert Delfin
24 March 2015
SCREENING
1. The material passing one screening surface and retained on a subsequent surface is
called
A. intermediate material B. minus material C. plus material
D. none of these
2. The minimum clear space between the edges of the opening in the screening surface
and is usually given in inches or millimeters.
A. sieve
B. aperture
C. mesh number
D. holes
3. The screen used in making size separation smaller than 4-mesh and larger than 48mesh.
A. grizzly screen
B. gyratory screen C. oscillating screen
D.vibrating screen
4. The removal of a small amount of oversize from a feed which are predominantly fines
is called
A. scalping
B. desliming
C. coarse separation
D. dewatering
5. Removal of free water from a solids-water mixture and is generally limited to 4 mesh
and above.
A. scalping
B. dewatering
C. trash removal
D. separation
6. Making a size separation smaller than 48-mesh is called
A. coarse separation
B. fine separation C. ultrafine separation
7. Another name for revolving screen is
A. shaking screen
B. vibrating screen
D. scalping
C. mechanically vibrated screen
D. trommel screen
8. The mesh number of a screen denotes
A. the area of the screen in square inch
B. the number of openings per linear inch of screen
C. the number of layers in a screen system
D. the number of screens needed to obtain the required fines
9. Materials which remain on a screen surface are called
A. fines
B. undersize
C. intermediate material
D. oversize
10. The wire diameter of a 10-mesh screen whose aperture is 0.065 in. is
A. 0.045 in.
B. 0.025 in.
C. 0.035 in.
D. 0.050 in.
12. Fine silica is fed at 1500 lbs/hour to a double-deck vibrating screen combination to
obtain a 48/65 mesh (Tyler) product. The silica feed is introduced into the upper screen
of the 48 mesh and the product is discharged off the surface of the lower screen of 65
mesh. During the screening operation, the ratio of oversize to product to undersize is
2:1:1/2.
Laboratory analysis of the different fractions:
Feed Mass
Oversize Mass
Screen Mesh
Fraction
Fraction
Product
Mass
Undersize Mass
Fraction
10/14 to 28/35
0.2821
0.5855
0.3385
0.00453
35/48
0.2580
0.3370
0.3220
0.00360
48/65
0.2810
0.0660
0.5260
0.34400
65/100
0.0910
0.0050
0.0670
0.29900
100/150 to
150/200
0.0870
0.0060
0.0260
0. 35300
a. The effectiveness of the screening equipment is(Ans. 48.7%)
b. If the screens measure 5ft x 8ft each, the capacity in MT/day-ft2-mm of the 65
mesh screen on the basis of a perfectly functioning 48 mesh screen is
(Ans. 0.901)
13. The fineness characteristic of a powder on a cumulative basis is represented by a
straight line from the origin to 100 per cent undersize at a particle size of 50 m. If the
powder is initially dispersed uniformly in a column of liquid, calculate the proportion by
mass which remains in suspension in the time from commencement of settling to that at
which a 40 m particle falls the total height of the column. It may be assumed that
Stokes’ law is applicable to the settling of the particles over the whole size range.
(Ans. 53.3%)
14. In a mixture of quartz of density 2650 kg/m3 and galena of density 7500 kg/m3, the
sizes of the particles range from 0.0052 to 0.025 mm. On separation in a hydraulic
classifier under free settling conditions, three fractions are obtained, one consisting of
quartz only, one a mixture of quartz and galena, and one of galena only. What are the
ranges of sizes of particles of the two substances in the original mixture?
(Ans. 0.0103–0.0126 mm)
15. The size distribution of a dust as measured by a microscope is as follows. Convert
these data to obtain the distribution on a mass basis, and calculate the specific surface,
assuming spherical particles of density 2650 kg/m3.(Ans. 0.731 × 106 m2/m3)
Size range (μm)
Number of particles in range (−)
0–2
2000
2–4
600
4–8
140
8–12
40
12–16
15
16–20
5
20–24
2
16. It is desired to separate a mixture of salt crystals into two fractions, a course fraction
retained on an 8-mesh screen, and a fine fraction passing through it. Screen analysis of
feed, coarse and fine fractions show
Mass fraction of +8 particles in feed
Mass fraction of +8 particles in course fraction
Mass fraction of +8 particles in fine fraction
= 0.51
= 0.90
= 0.42
The overall effectiveness of the screen used for the separation purpose per 100 kg of
feed is (Ans. 31.8%)
Solution:
𝜺=
(𝒙𝒇 − 𝒙𝑹 )(𝒙𝑷)
(𝒙𝑷 − 𝒙𝑹 )(𝒙𝒇)
[𝟏 −
(𝟎.𝟓𝟏−𝟎.𝟒𝟐)(𝟎.𝟗𝟎)
(𝒙𝒇 − 𝒙𝑹 )(𝟏 − 𝒙𝑷 )
(𝒙𝑷 − 𝒙𝑹 )(𝟏 − 𝒙𝒇)
= (𝟎.𝟗𝟎−𝟎.𝟒𝟐)(𝟎.𝟓𝟏) [𝟏 −
𝜺 = 𝟑𝟏. 𝟖%
] × 𝟏𝟎𝟎
(𝟎.𝟓𝟏−𝟎.𝟒𝟐)(𝟏−𝟎.𝟗𝟎)
(𝟎.𝟗𝟎−𝟎.𝟒𝟐)(𝟏−.𝟓𝟏)
] × 𝟏𝟎𝟎
SIZE REDUCTION
1. The term applied to all ways in which particles of solids are cut or broken into
smaller pieces
A. Size reduction
C. comminution
B. Screening
D. crushing
2. The hardness of a mineral is a criterion of its resistance to crushing. Which of the
following is an example of a hard material?
A. Talc
C. sapphire
B. Calcite
D. feldspar
3. States that the energy required for crushing is proportional to the new surface
created.
A. Rittinger’s Law
C. Bond Law
B. Kick’s Law
D. Energy Law
4. It is defined as the efficiency of technical grinding compared with that of
laboratory crushing experiments.
A. Grinding Efficiency
C. Practical Energy Efficiency
B. Bond Work Index
D. none of these
5. In comminution, the energy requirement is determined theoretically by
A. The initial and final sizes of the particles
B. The type of equipment
C. The change in shape of the particle
D. None of these
6. Size reduction is important in chemical engineering since
A. It prevents chemical engineers from becoming overweight
B. It makes products to become uniform in size
C. It prepares raw materials of the desired sizes prior to processing
D. None of these
7. Equivalent diameter of a particle is the diameter of the sphere having the same
A. Ratio of surface to volume as the actual volume
B. Ratio of volume to surface as the particle
C. Volume as the particle
D. None of these
8. For coarse reduction of hard solids, we use
A. Impact
C. compression
B. Attrition
D. none of these
9. Soft and non-abrasive materials can be made into fines by
A. Attrition
C. cutting
B. Compression
D. none of these
10. Crushing efficiency is the ratio of
A. Surface energy created by the crushing to the energy absorbed by the
solid
B. Energy absorbed by the solid to that fed to the machine
C. Energy fed to the machine to the surface energy created by the crushing
D. Energy absorbed by the solid to the surface energy created by the crushing
11. The energy required per unit mass to grind limestone particles of very large size
to 100 μm is 12.7 kWh/ton. An estimate (using Bond's law) of the energy to grind
the particles from a very large size to 50 μrn is(Ans. 18 kWh/ton)
12. A crusher was used to crush a material with a compressive strength of 22.5
MN/m2.The size of the feed was minus 50 mm, plus 40 mm and the power
required was13.0 kW/(kg/s). The screen analysis of the product was:
Size of aperture (mm)
Amount of product (per cent)
through 6.0
all
on 4.0
26
on 2.0
18
on 0.75
23
on 0.50
8
on 0.25
17
on 0.125
3
through 0.125
5
What power would be required to crush 1 kg/s of a material of compressive
strength45 MN/m2 from a feed of minus 45 mm, plus 40 mm to a product of 0.50
mm averagesize?(Ans. 47.8 kW)
13. A crusher reducing limestone of crushing strength 70 MN/m2 from 6 mm
diameter averagesize to 0.1 mm diameter average size, requires 9 kW. The
same machine is used to crushdolomite at the same output from 6 mm diameter
average size to a product consisting of20 per cent with an average diameter of
0.25 mm, 60 per cent with an average diameterof 0.125 mm and a balance
having an average diameter of 0.085 mm. Estimate the powerrequired, assuming
that the crushing strength of the dolomite is 100 MN/m2 and thatcrushing follows
Rittinger’s Law.(Ans. 5.9 kW)
14. A crushing mill reduces limestone from a mean particle size of 45 mm to the
followingproduct:
Size (mm)
Amount of product (per cent)
12.5
0.5
7.5
7.5
5.0
45.0
2.5
19.0
1.5
16.0
0.75
8.0
0.40
3.0
0.20
1.0
It requires 21 kJ/kg of material crushed. Calculate the power required to crush
the samematerial at the same rate, from a feed having a mean size of 25 mm to
a product with amean size of 1 mm. (Ans. 38.6 kJ/kg)
15. Power of 3 kW is supplied to a machine crushing material at the rate of 0.3 kg/s
from12.5 mm cubes to a product having the following sizes: 80 per cent 3.175
mm, 10 per cent 2.5 mm and 10 per cent 2.25 mm. What power should be
supplied to this machineto crush 0.3 kg/s of the same material from 7.5 mm cube
to 2.0 mm cube?(Ans. 3.6 kW)
16. A material is crushed in a Blake jaw crusher such that the average size of particle
is reducedfrom 50 mm to 10 mm with the consumption of energy of 13.0
kW/(kg/s). What would be theconsumption of energy needed to crush the same
material of average size 75 mm to an averagesize of 25 mm:
a) assumingRittinger’s law applies?
b) assuming Kick’s law applies?
Which of these results would be regarded as being more reliable and why?
Solution:
a) Rittinger’s law.
This is given by:
E = KRfc[(1/L2) - (1/L1)]
Thus:
13.0 KRfc[(1/10) - (1/50)]
and:
KRfc = (13.0 × 50/4) = 162.5 kW/(kg mm)
Thus the energy required to crush 75 mm material to 25 mm is:
E = 162.5[(1/25) − (1/75)] = 4.33 kJ/kg
b) Kick’s law.
This is given by:
E = KKfcln(L1/L2)
Thus:
13.0 = KKfcln(50/10)
and:
KKfc = (13.0/1.609) = 8/08 kW/(kg/s)
Thus the energy required to crush 75 mm material to 25 mm is given by:
E = 8.08 ln(75/25) = 8.88 kJ/kg
FLOTATION
1. Any operation in which one solid is separated from another by floating one of them at
or on the surface of a fluid.
A. coagulation
B. flotation
C. centrifugation
D. sedimentation
2. The flotation agent that prevents coalescence of air bubbles as they travel to the
surface of the water is/are
A. collectors
B. promoters
C. frothing agent
D. modifying agent
3. A flotation modifier which assists in the selectivity (sharpness of separation) or stop
unwanted minerals from floating.
A. depressants
B. activators
C. alkalinity regulators
D. promoters
4. Dispersants are important for the control of slimes which sometimes interfere with the
selectivity and increase reagent consumption. Another term for dispersant is
A. deflocculant
B. depressants
C. frothers
D. regulators
5. An example of a collector for flotation of metallic sulfides and native metals is
A. xanthates
B. sodium silicate C. sodium sulfide
D. sphalerite
6. Which of the following is an example of a deflocculant?
A. sulfuric acid
B. lignin sulforate C. dithiophosphate
D. molybderite
7. Froth Flotation is most suitable for treating
A. iron ores
B. quartz
C. sulfide ores
D. metal ores
8. In Froth Flotation, the chemical agent added to cause air adherence is called
A. collector
B. frother
C. modifier
D. promoter
9. Pine oil used in a flotation process acts as a
A. collector
B. modifier
C. frother
D. activator
10. The flotation agent that prevents coalescence of air bubbles as they travel to the
surface of the water is/are
A. frothing agent B. promoters
C. collectors
D. modifying agent
11. Ground lead ore is to be concentrated by a single flotation process using 1.5 oz of
reagent per ton of ore. The feed concentrate and tailings have the following composition
by weight on a dry basis:
Feed %
Concentrate %
Tailings %
PbS
30
90
0.9
ZnS
25
3
35.6
SiO₂
45
7
63.5
Water is fed to the cell at the rate of 1,000 gallons per ton of wet concentrate with
99% of the water leaving with the tailings and 1% with the concentrate. What
mass of wet concentrate produced per hour when ten tons of ore are fed to the
cell per 24 hours? What is the total water required in pounds per hour?
(Ans. 3.4 , 1185)
12. A flotation plant processes 3,000 tons/day of CuFeS₂ (chalcopyrite). It produces 80
tons Cu concentrate assaying 25% Cu. If ore analyzes 0.7% Cu, the percent recovery
is? (Ans. 95.24%)
13. A typical flotation machine has the following specifications:
Number of cells = 4
Flotation time
= 12 min
Cell volume
= 60 cu. Ft
Hp per cell
= 10Hp
The material treated has the following specifications:
Pulp (mixture of ore and water) = 40% solids
Specific gravity of ore
=3
Equation: 𝑛 =
𝑇𝑥 𝐶𝑎𝑝𝑥𝑑
𝑉𝑥1440
Where n= number of cells; V= volume in cu. ft. per cell;
Cap = tons of dry ore per 24 hours; d = cu. ft. of pulp
(ore and water) containing one ton (2000 lbs) of solids.
What is the capacity of the machine in tons of dry ore per 24 hours? (Ans. 490)
14. A flotation plant produces 3000 tons/day of CuFeS₂. It produces 80 tons of Cu
concentrate assaying 25% Cu. If ore analyses 0.7% Cu, what is the percent recovery?
(Ans. 95.2%)
15. A flotation section of a mining company is extracting CuS from covellite ores. The
ore consists of 5% CuS and 95% gangue, which may be assumed to be SiO₂. The
following data are given:
%CuS
%SiO₂
Feed
5
95
Concentrate
85
15
Rougher Tailings
1
99
Scavenger Tailings
10
90
Final Tailings
0.3
99.7
Laboratory experiments indicated that the water to solids ratio, L/S = 2 and the contact
time is 10 min. in the rougher; L/S = 4, contact time = 18 min. in the scavenger. On the
basis of 300 tons per day of ore treated.
Data: Density of SiO₂ = 2.65 g/cc
Density of CuS = 4.6 g/cc
What is the volume of the rougher needed and what is the volume of the scavenger
needed?
(Ans. 169.8 ft³, 535 ft³)
16. A copper ore initially contains 2.09% Cu. After carrying out a froth flotation
separation, the products are as shown in Table 1. Using this data, calculate:
(a) Ratio of concentration
(b) % Metal Recovery
(c) % Metal Loss
(d) % Weight Recovery, or % Yield
(e) Enrichment Ratio
Solution:
%𝐶𝑢 𝑅𝑒𝑐𝑜𝑣𝑒𝑟𝑦 =
(10)(20)
𝑥100 = 95.7%
(2.09)(100)
%𝐶𝑢 𝐿𝑜𝑠𝑠 = 100 − 95.7 = 4.3%
2.09 − 0.1
𝑥100 = 10%
20 − 0.1
𝟐𝟎. 𝟎
𝑬𝒏𝒓𝒊𝒄𝒉𝒎𝒆𝒏𝒕 𝑹𝒂𝒕𝒊𝒐 =
= 𝟗. 𝟓𝟕
𝟐. 𝟎𝟗
%𝑊𝑒𝑖𝑔ℎ𝑡 𝑅𝑒𝑐𝑜𝑣𝑒𝑟𝑦 =
SEDIMENTATION
1. The motion of a particle is impeded by other particles, which will happen when the
particles are near each other even though they are not actually colliding, the process is
called
A. free settling
B. unhindered settling
C. gravity settling D.
hindered settling
2. The drag coefficient in hindered settling is ______ compared to free settling.
A. greater than
B. less than
C. constant
D. varying
3. The operation by which solids are separated from liquids due to difference in their
respective densities is
A. screening
B. adsorption
C. sedimentation D. absorption
4. The separation of solid particles into several size fractions based upon the settling
velocities in a medium is called
A. settling
B. filtration
C. flotation
D. classification
5. Device in which a current of air separates particles with different sedimentation
velocities.
A. agitator
B. air elutriator
C. classifier
D. air
conveyor
6. The constant velocity with which a body moves relative to the surrounding fluid when
the forces acting on it (gravitational or centrifugal or electrostatic forces) are equal to the
friction force acting against the motion.
A. terminal velocity
B. settling velocity
C. maximum velocity
D. all
of these
7. An apparatus in which particles settle in a liquid by gravitational or centrifugal force
and are removed as a concentrated slurry.
A. classifier
B. thickener
C. elutriator
D. agitator
8. In a motion of a particle through fluids, forces act on a particle moving through a fluid.
The force which appears whenever there is a relative motion between the particle and
the fluid is called
A. gravitational force
buoyant force
B. centrifugal force
C. drag force
D.
9. Which of the following is true for the experiment?
A. the drag force exerted on the sphere increases during its descent through the
second section.
B. the sphere never reaches its terminal velocity while falling through both
sections.
C. the sphere reaches its terminal velocity while falling through the first section.
D. the drag force exerted on the sphere decreases during the descent
through the second section.
10. The drag is defined as the force exerted by
A. the fluid on the solid in a direction opposite to the motion of the solid
B. the fluid on the solid in the direction of motion of the solid
C. the solid on the fluid
D. none of these
11. A slurry containing 5 kg of water/kg of solids is to be thickened to a sludge
containing 1.5 kg of water/kg of solids in a continuous operation. Laboratory tests using
five different concentrations of the slurry yielded the following results: concentration Y
(kg water/kg solid)
5.0 4.2
3.7
3.1
2.5
rate of sedimentation uc (mm/s) 0.17 0.10 0.08 0.06 0.042
Calculate the minimum area of a thickener to effect the separation of 0.6 kg/s of solids.
(Ans. 16.5 m2)
12. When a suspension of uniform coarse particles settles under the action of gravity,
therelation between the sedimentation velocity 𝜐𝑐 and the fractional volumetric
concentration
C is given by:
𝜐𝑐
= (1 − 𝐶 )𝑛
𝜐0
wheren = 2.3 and u0 is the free falling velocity of the particles. Draw the curve of solids
flux  against concentration and determine the value of C at which  is a maximum and
where the curve has a point of inflexion. What is implied about the settling
characteristics of such a suspension from the Kynch theory? Comment on the validity of
the Kynchtheory for such a suspension.(Ans. C = 0.30)
13. Calculate the minimum area and diameter of a thickener with a circular basin to treat
0.1 m3/s of a slurry of a solids concentration of 150 kg/m3. The results of batch settling
tests are:
Solids concentration (kg/m3)
Settling velocity (μm/s)
100
148
200
91
300
55.33
400
33.25
500
21.40
600
14.50
700
10.29
800
7.38
900
5.56
1000
4.20
1100
3.27
A value of 1290 kg/m3 for underflow concentration was selected from a retention time
test. Estimate the underflow volumetric flow rate assuming total separation of all solids
and that a clear overflow is obtained.(Ans. 974 m2, 0.0116 m3/s)
14. Solid spherical particles having a diameter of 0.090 mm and a solid density of 2002
kg/m3 are settling in a solution of water at 26.7 °C. The volume fraction of the solids in
the water is 0.45. Calculate the settling velocity and the Reynolds number.
(Ans. 2.369x10-4 m/s, Re=9.89x10-3)
15. For the sedimentation of a suspension of uniform fine particles in a liquid, the
relation between observed sedimentation velocity 𝜐𝑐 and fractional volumetric
concentration C is given by:
𝜐𝑐
= (1 − 𝐶 )𝑛
𝜐0
where𝜐0 is the free falling velocity of an individual particle. Calculate the concentration
at which the rate of deposition of particles per unit area is a maximum and determine 41
this maximum flux for 0.1 mm spheres of glass of density 2600 kg/m3 settling in water
of density 1000 kg/m3 and viscosity 1 mNs/m2. It may be assumed that the resistance
force F on an isolated sphere is given by
Stokes’ Law.(Ans. 6.06 × 10-4 m3/m2s)
16. A mixture of silica (B) and galena (A) solid particles having a size range of 1.27x10 -2
mm to 5.08x10-2 mm is to be separated by hydraulic classification using free settling
conditions in water at 293.2K. The specific gravity of silica is 2.65 and that of galena is
7.5. Calculate the size range of the various fractions obtained using free settling
conditions. Also calculate the value of the largest Reynolds number occurring.
Solution:
0.5
𝐷𝑝𝐴3
𝜌𝑝𝑩 − 𝜌
=(
)
𝐷𝑝𝐵4
𝜌𝑝𝐴 − 𝜌
2650−998 0.5
𝐷𝑝𝐴3
= (7500−998)
5.08x10−2
DpA3=2.5606x10-5 m
𝐷𝑝𝐴1 1.27 × 10−5
2650 − 998
)
=
=(
𝐷𝑝𝐵2
𝐷𝑝𝐵2
7500 − 998
𝐷𝑝𝐴3
5.08x10−2
0.5
=
DpB2=2.520x10-5 m
Pure A
Mixed A and B
Pure B
DpA3=2.5606x10-5 m to DpA4=5.08x10-5 m
DpB2=2.520x10-5 m to DpB4=5.08x10-5 m
DpB1=1.27x10-5 m to DpB2=2.520x10-5 m
CENTRIFUGATION
1. For the separation of sugar solution from settled out mud, we use
A. Sparkler filter
C. centrifugal filter
B. Plate and frame filter
D. rotary drum vacuum filter
2. Moisture can be removed from lubricating oil using
A. Tubular centrifuge
C. sparkler filter
B. Clarifier
D. vacuum leaf filter
3. Which of the following can be most effectively used for clarification of lube oil and
printing ink?
A. Sparkler filter
C. disc-bowl centrifuge
B. Precoat filter
D. sharplesssupercentrifuge
4. If the radius of a basket centrifuge is halved and the rpm is doubled, then the
A. Linear speed of the basket is doubled
B. Linear speed of the basket is halved’
C. Centrifugal force is doubled
D. Capacity of centrifuge is increased
5. Where the difference in density of the two liquid phases to be separated is very
small (as in milk cream separator), the most suitable separator is
A. Disc bowl centrifuge
C. batch basket centrifuge
B. Sharplesssupercentrifuge
D. sparkler filter
6. __________ centrifuge is normally used in sugar mills.
A.
B.
C.
D.
Tubular bowl
Disc-bowl
Suspended batch basket
Perforated horizontal basket continuous
7. Ultra centrifuges are used for the separation of __________ solid particles.
A.
coarse
B.
fine
C.
colloidal
D.
dissolved
8. Driving force in case of filtration by a centrifuge is the
A. speed of the centrifuge.
B. centrifugal pressure exerted by the liquid.
C. narrow diameter of the vessel.
D. formation of highly porous cake.
9. Cyclones are used primarily for separating
A. solids
B. solids from fluids
C. liquids
D. solids from solids
10. Separation of isotopes is generally done using a/an __________ centrifuge.
A. ultra
B. disk-bowl
C. both (a) & (b)
D. neither (a) nor (b)
11. A centrifuge bowl is spinning at a constant 2000 rev/min. What radius bowl is
needed for the following?
(a)
A force of 455 g’s
(b)
A force four times that in part (a)
Ans. (a) r = 0.1017 m
12. A centrifugal bowl is spinning at a constant speed of 2000 rpm. What is the
radius of bowl in cm needed to create a force of 455 g? (Ans. 10.2)
13. Two centrifuges rotate at the same peripheral velocity of 53.34 m/s. The first
bowl has a radius of r₁= 305 mm. Calculate the rev/min and the centrifugal forces
developed in each bowl.
(Ans. N₁= 6684 rev/min; N₂= 1670 rev/min, 3806g’s in bowl 1,951 g’s in bowl
2)
14. A centrifuge with a radius of 76.2 mm rotates at a peripheral velocity of 53.34
m/s. What is the centrifugal force developed compared to gravitational force in
the bowl centrifuge? (Ans. 3800 g)
15. A cream separator centrifuge has an outlet discharge radius r₁ = 50.8 mm and
outlet radius r₄ = 76.2 mm. The density of the skim milk is 1032 kg/m³ and that of
the cream is 863 kg/m³. Calculate the radius of the interface neutral zone.
(Ans. r₂ = 150 mm)
16. In a test on a centrifuge all particles of a mineral of density 2800 kg/m³ and of
size 5 μm, equivalent spherical diameter, were separated from suspension in
water fed at a volumetric throughput rate of 0.25 m³/s. Calculate the value of the
capacity factor.
What will be the corresponding size cut for a suspension of coal
particles in oil fed at the rate of 0.04 m³/s? The density of coal is 1300 kg/m³ and
the density of the oil is 850 kg/m³ and its viscosity is 0.01 Ns/m². It may be
assumed that Stokes’ law is applicable.
Solution:
𝑢𝑜 =
𝑑 2 (𝜌𝑠 − 𝜌)𝑔 25𝑥10−12 𝑥(12800 − 1000)𝑥9.81
=
18𝜇
18𝑥10−3
𝑢𝑜 = 2.45𝑥10−5 𝑚⁄𝑠
𝑄 = 𝑢𝑜 𝛴
𝛴=
0.25
= 1.02𝑥104 𝑚2
(2.45𝑥10−5 )
𝑢𝑜 =
𝑄
0.04
=
= 3.92𝑥10−6 𝑚⁄𝑠
𝛴 (1.02𝑥104 )
𝑑2 =
18𝜇𝑢𝑜
(𝜌𝑠 − 𝜌)𝑔
𝑑2 =
18𝑥10−2 𝑥3.92𝑥10−6
(1300 − 850)𝑥9.81
𝒅 = 𝟒. 𝟎𝒙𝟏𝟎−𝟔 𝒎 𝒐𝒓 𝟒 𝝁𝒎
Technological Institute of the Philippines
363 P. Casal St. Quiapo, Manila
College of Engineering and Technology
Chemical Engineering Department
CHEP 592 Intro to Particle Technology
PROBLEM SET
(Screening, Size Reduction, Flotation, Sedimentation,
& Centrifugation)
Submitted by:
Kharren Mae C. Rosario
ChE – 1320071
Submitted to:
Engr. Robert Delfin
25 March 2015
SCREENING
Concepts
1. Screen capacity is not a function of
A. its opening size
B. screening mechanism
C. screening surface
D. atmospheric humidity
2. Increasing the capacity of a screen __________ the screen effectiveness.
A. decreases
C. does not affect
B. increases
D. none of these
3. Pulverised coal passing through 200 mesh screen has a diameter of 0.074 mm (74 micron). The same
passing through 50 mesh screen will have a diameter of __________ mm.
A. 0.007
B. 0.30
C. 50
D. 0.014
4. Mesh indicates the number of holes per
A. square inch
B. linear inch
C. square foot
D. linear foot
5. Trommels employ __________ for screening of materials.
A. fibrous cloth
C. punched plate
B. woven wire screen
D. none of these
6. As particle size is reduced
A. screening becomes progressively easier
B. capacity and effectiveness of the screen is increased
C. screening becomes progressively more difficult
D. capacity and effectiveness of the screen is decreased
7. Pick out the wrong statement.
A. The capacity and the effectiveness of screen are opposing factors.
B. The capacity and the effectiveness of a screen are the different,
C. The screening surface of a 'reel' (a revolving screen used in flour mills) is made of silk bolting cloth
supported by wire mesh
D. Both b and c
8. The material passing one screening surface and retained on a subsequent surface is called
A. Intermediate material
B. Minus material
C. Plus material
D. None of the these
9. The screen used in making size separation smaller than 4-mesh and larger than 48-mesh
A. Grizzly screen
B. oscillating screen
C. Gyratory screen
D. vibrating screen
10. Another name for revolving screen is
A. Shaking screen
C. mechanically vibrated screen
C. Vibrating screen
D. trommel screen
Problem Solving
1. A sand mixture was screened through a standard 10-mesh screen. The mass fraction of the oversize
material in feed, overflow and underflow were found to be 0.38, 0.79 and 0.22 respectively. The screen
effectiveness based on the oversize is _______. 0.50
2. It is desired to separate a mixture of crystals into three fractions, a coarse fraction retained on an 8mesh screen, a middle fraction passing an 8-mesh but retained on a 14-mesh screen, and a fine
fraction passing a 14-mesh. Two screens in series are used, an 8-mesh and a 14-mesh, conforming to
the Tyler standard. Screen analyses of feed, coarse, medium, and fine fractions are given in the table
below. Assuming the analyses are accurate, what is the overall effectiveness of each screen? E1 =
0.4405 ; E2 = 0.4432
3. The screens used in the previous problem are shaking screens with a capacity of 4 MT/m 2-h-mm mesh
size. How many square meters of screen are needed for each of the screens in the previous problem if
the feed to the first screen is 100 ton/h? 16.08 m2
4. A quartz mixture having the screen analysis shown in the table below is screened through a standard
10-mesh screen. The cumulative screen analysis of overflow and underflow are given below. Calculate
the mass ratios of the overflow and underflow to feed and the overall effectiveness of the screen. (O/F)
= 0.420 ; (U/F) = 0.58 ; E = 0.669
5. Limestone is crushed by six units operating in parallel and the products separated by six 35-mesh
screens, also in parallel, into two fractions. The effective dimensions of each screen is 6 ft x 20 ft. The
common undersize comes out at the rate of 50 tons/hr. Assume no losses. The efficiency of screening
operation is _____. 0.5282
6. The capacity of each screen in the previous problem is ________lbf/ft 2.
𝑭= 𝑷+𝑹
𝐹𝑋𝐹 = 𝑃𝑋𝑃 + 𝑅𝑋𝑅
𝐹 = 50 + 𝑅
0.45𝐹 = 50(0.65) + 0.175𝑅
𝐹 = 86.36
𝑡𝑜𝑛𝑠
ℎ𝑟
𝑅 = 36.36
𝑡𝑜𝑛𝑠
ℎ𝑟
𝑡𝑜𝑛𝑠
86.36
ℎ𝑟 × 24 ℎ𝑟
𝑙𝑏
𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 = 6 𝑢𝑛𝑖𝑡𝑠
× 2000
2
(6 × 20)𝑓𝑡
𝑡𝑜𝑛
𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 = 5800
𝑙𝑏
𝑓𝑡 3
SIZE REDUCTION
Concepts
1. Pick out the wrong statement.
A.
Recycled coarse material to the grinder by a classifier is termed as circulating load.
B.
Wear and tear in wet crushing is more than that in dry crushing of materials.
C.
Size enlargement (opposite of size reduction) is not a mechanical operation.
D.
A 'dust catcher' is simply an enlargement in a pipeline which permits the solids to settle down
due to reduction in velocity of the dust laden gas.
2. States that the energy required for crushing is proportional to the new surface created.
A. Rittinger’s law
B. Kick’s law
C. Bonds law
D. Energy law
3. Ball mills and tube mills with flint or porcelain balls are used for size reduction of
A. asbestos
B. rubber
C. non-metallic ores
D. limestone
4. Size reduction of __________ is accomplished in steam heated rollers and roll crushers.
A. resins
C. hard rubber
B. gums
D. waxes
5. Soft and non—abrasive materials can be made into fines by
A. Attrition
B. Compression
C. Cutting
D. None of the given
6. Which of the following is not a part of the Blake jaw crusher?
A. Hanger
B. Check plates
C. Toggles
D. Pitman
7. The term applied to all ways in which particles of solids are cut or broken into smaller pieces
A. Size reduction
C. Comminution
B. Screening
D. Crushing
8. The hardness of a mineral is a criterion of its resistance to crushing. Which of the following is an
example of a harp material?
A. Talc
B. sapphire
C. Calcite
D. feldspar
9. The energy required per unit mass to grind limestone particles of very large size to 100 μm is 12.7
kWh/ton. An estimate (using Bond's law) of the energy to grind the particles from a very large size to 50
μrn is
A. 6.35 kWh/ton
B. 9.0 kWh/ton
C. 18 kWh/ton
D. 25.4 kWh/ton
10. Size reduction of the __________ can be suitably done by ball mills, crushing rolls and rod mills.
A. metalliferrous ores
B. non-metallic ores
C. basic slags
D. asbestos & mica
Problem Solving
1. A crusher was used to crush a material with a compressive strength of 22.5 MN/m2. The size of the
feed was minus 50 mm, plus 40 mm and the power required was 13.0 kW/(kg/s). The screen analysis
of the product was:
What power would be required to crush 1 kg/s of a material of compressive strength 45 MN/m2 from a
feed of minus 45 mm, plus 40 mm to a product of 0.50 mm average size? 47.8 kW
2. A crusher reducing limestone of crushing strength 70 MN/m2 from 6 mm diameter average size to 0.1
mm diameter average size, requires 9 kW. The same machine is used to crush dolomite at the same
output from 6 mm diameter average size to a product consisting of 20 per cent with an average
diameter of 0.25 mm, 60 per cent with an average diameter of 0.125 mm and a balance having an
average diameter of 0.085 mm. Estimate the power required, assuming that the crushing strength of the
dolomite is 100 MN/m2 and that crushing follows Rittinger’s Law. 5.9 kW
3. A crushing mill reduces limestone from a mean particle size of 45 mm to the following product:
It requires 21 kJ/kg of material crushed. Calculate the power required to crush the same material at the
same rate, from a feed having a mean size of 25 mm to a product with a mean size of 1 mm. 38.6 kJ/kg
4. What is the power required to crush 100 ton/h of limestone if 80 percent of the feed passes a 2-in.
screen and 80 percent of the product a 1/8-in. screen? 227 hp
5. A batch grinding mill is charged with material of the composition. The grinding-rate function Su is
assumed to be 0.001/s for the 4/6 mesh particles. Breakage function B u is given with b=1.3
both Su and Bu are assumed to be independent of time. How long will it take for the fraction of 4/6 mesh
material to diminish by 10 %? 105.3 s
6. Power of 3 kW is supplied to a machine crushing material at the rate of 0.3 kg/s from 12.5 mm cubes to
a product having the following sizes: 80 per cent 3.175 mm, 10 percent 2.5 mm and 10 per cent 2.25
mm. What power should be supplied to this machine to crush 0.3 kg/s of the same material from 7.5
mm cube to 2.0 mm cube?
Solution
*Calculate the mass mean diameter.
*Crushing is considered intermediate and Bond’s law will be used.
𝑷
𝟏
𝟏
= 𝟎. 𝟑𝟏𝟔𝟐 𝑬𝒊 (
−
)
𝑻
√𝑿𝟐 √𝑿𝟏
0.3
𝑘𝑔
𝑡𝑜𝑛𝑠
= 1.189
𝑠
ℎ𝑟
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑚𝑒𝑎𝑛 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 = 3.175
𝑃
1
1
= 0.3162 𝐸𝑖 (
−
)
1.189 𝑡𝑜𝑛𝑠/ℎ𝑟
√2.0 𝑚𝑚 √7.5 𝑚𝑚
3 𝑘𝑊
1
1
= 0.3162 𝐸𝑖 (
−
)
1.189 𝑡𝑜𝑛𝑠/ℎ𝑟
𝑚𝑚
√3.175
√12.5 𝑚𝑚
𝑃 = 3.69 𝑘𝑊
FLOTATION
Concepts
1. Which of the following is the most suitable for cleaning of fine coal dust (<0.5 mm) ?
A. Trough washer
C. Spiral separator
B. Baum jig washer
D. Froth flotation
2. Which of the following mineral dressing operations is termed as 'comminution'?
A. Panning
B. Spiralling
C. Tabling
D. None of these
3. Froth floatation is the most suitable for treating
A. iron ores
B. sulphide ores
C. quartzite
D. none of these
4. Pine oil used in froth floatation technique acts as a/an
A. collector
B. modifier
C. frother
D. activator
5. It involves phenomena related to the relative buoyancy of objects.
A. Flotation
C. Centrifugation
B. Sedimentation
D. Filtration
6. The crushed material received for separation is called feed or
A. tailing
C. concentrate
B. heading
D. middling
7. What is the selectivity index, if the grade of tailings & concentrate is the same ?
A. 0
B. ∞
C. 1
D. 0.5
8. The following is a typical anionic collector used in flotation
A. Potassium ethyl xanthate
B. Ethyl dixanthogen
C. Trimethyl cetyl ammonium bromide.
D. lauryl amine hydrochloride.
9. Which of the following is an example of deflocculant?
A. sulfuric acid
B. lignin sulforate
C. dithiophosphate
D. Molybderite
10. Froth flotation is most suitable for
A. quartz
B. iron ores
C. sulfide ores
D. metal ores
Problem Solving
1. A flotation plant processes 3000 tons/day of CuFeS2 (chalcopyrite). It produces 80 tons of Cu
concentrate assaying 25%Cu. If ore analyzes 0.7% Cu, what is the percent recovery? 95.24%
2. For a laboratory flotation of an iron ore in water, it was observed that 2 mg was collected while
traversing 2 m of the flotation column. The concentration of the ore in water was 0.5 kg/m3 . The
average diameter of the bubbles was 2 mm and the average diameter of the particles was 0.1 mm.
Compute flotation recovery. 0.578
3. A copper ore initially contains 2.09% Cu. After carrying out a froth flotation separation, the products are
as shown in Table 1. Using this data, calculate: (a) Ratio of concentration (b) % Metal recovery (c) %
Metal Loss (d) % Weight Recovery, or % Yield (e) Enrichment Ratio. (a) 10; (b) 95.7%; (c) 4.3% (d)
10% (e) 9.57
(http://www.chem.mtu.edu/chem_eng/faculty/kawatra/Flotation_Fundamentals.pdf
Solution
(a) From Table 1, the Ratio of Concentration can be calculated as
F/C = 100/10 = 10.
If only assays are available, the ratio of concentration equals (20 – 0.1)/(2.09 – 0.1) = 10
So, for each 10 tons of feed, the plant would produce 1 ton of concentrate.
(b) Using the example data from Table 1, the % Cu recovery calculated from weights and assays is:
% Cu Recovery = [(10·20)/(2.09·100)]·100 = 95.7%
The calculation using assays alone is % Cu Recovery = 100(20/2.09)(2.09 – 0.1)/(20 – 0.1) = 95.7%
This means that 95.7% of the copper present in the ore was recovered in the concentrate, while the rest
was lost in the tailings.
(c) The % Cu Loss can be calculated by subtracting the % Cu Recovery from 100%:
% Cu Loss = 100 – 95.7 = 4.3%
This means that 4.3% of the copper present in the ore was lost in the tailings.
(d) The % Weight Recovery is equal to the % Weight of the concentrate in Table 1. It can also be
calculated from the assay values given in the table, as follows:
% Weight Recovery = 100·(2.09 - 0.1)(20 – 0.1) = 10%
(e) The Enrichment Ratio is calculated by dividing the concentrate assay in Table 1 by the feed assay:
Enrichment Ratio = 20.0/2.09 = 9.57
This tells us that the concentrate has 9.57 times the copper concentration of the feed.
SEDIMENTATION
Concepts
1. Two particles are called to be equal settling, if they are having the same.
A.
Size
B.
Specific gravity
C.
Terminal velocities in the same fluid & in the same field of force
D.
None of these
2. Gravity settling process is not involved in the working of a
A. hydrocyclone
B. classifier
C.dorr-thickener
D. sedimentation tank
3. Device in which a current of air separates particles with different sedimentation velocities
A. Agitator
B. classifier
C. Air elutriator
D. air conveyor
4. Pick up the incorrect statement from the following :
A. Detention period for plain sedimentation tanks ranges between 6 to 10 hours
B. Detention period for sedimentation tanks, using coagulants usually ranges between 2 to 4 hours
C. The horizontally flow velocity in sedimentation tanks, is generally limited to 0.3 m/minute
D. All the above.
5. In sewage treatment, its sedimentation is speeded up by commonly adding
A. sodium sulphate
B. copper sulphate
C. Hydrochloric acid
D. Lime
6. Industrially, sedimentation operations are often carried out continuously in equipment called?
A. Thickener
B. Simple gravity settling
C. Sedimentation thickener
D. None of the these
7. Dorr thickener is an equipment used for
A. clarification
B.leaching
C. drying
D. sedimentation
8. A suspension of glass beads in ethylene glycol has a hindered settling velocity of 1.7 mm/s, while the
terminal settling velocity of a single glass bead in ethylene glycol is 17 mm/s. If the Richardson-Zaki
hindered settling index is 4.5, the volume fraction of solids in the suspension is
A. 0.1
B. 0.4
C. 0.6
D. none of these
9. When particles are crowded, they settle at a lower rate and the process is called?
A. Hindered Settling
B. Free Settling
C. Compression Settling
D. Terminal Settling
10. To remove dirt from the flowing fluid, we use a
A. coagulant
B. gravity settler
C. strains
D. clarifier
Problem Solving
1. A mixture of quartz and galena of a size range from 0.015 mm to 0.065 mm is to be separated into two
pure fractions using a hindered settling process. What is the minimum apparent density of the fluid that
will give this separation? The density of galena is 7500 kg/m3 and the density of quartz is 2650 kg/m3.
1196 – 2377 kg/m3
2. Calculate the terminal velocity of a steel ball, 2 mm diameter and of density 7870 kg/m 3 in an oil of
density 900 kg/m3 and viscosity 50 mNs/m2. 0.189 m/s
3. What is the terminal velocity of a spherical steel particle, 0.40 mm in diameter, settling in an oil of
density 820 kg/m3 and viscosity 10 mN s/m2? The density of steel is 7870 kg/m3. 51 mm/s
4. What will be the terminal velocities of mica plates, 1 mm thick and ranging in area from 6 to 600 mm2,
settling in an oil of density 820 kg/m3 and viscosity 10 mN s/m2? The density of mica is 3000 kg/m3.
0.154 m/s – 0.159 m/s
5. A spherical glass particle is allowed to settle freely in water. If the particle starts initially from rest and if
the value of the Reynolds number with respect to the particle is 0.1 when it has attained its terminal
falling velocity, calculate: (a) the distance travelled before the particle reaches 90 per cent of its terminal
falling velocity, (b) the time elapsed when the acceleration of the particle is one hundredth of its initial
value. x = 1.05 mm ; t = 0.0016 s
6. Solid spherical particles of coffee extract from a dryer having a diameter of 400 microns are falling
through air at a temperature of 422 K. The density of the particles is 1030 kg/m 3. Calculate the terminal
settling velocity and the distance of fall in 5 s. The pressure is 101.32 kPa.
𝑎𝑡 𝑇 = 422 𝐾; 𝜌𝑎𝑖𝑟 = 0.8378
𝑘𝑔
; 𝜇 = 0.0237 × 10−3 𝑘𝑔/𝑚 ∙ 𝑠
𝑚3
𝟏
1
(9.81)(0.8378)(1030 − 0.8378) 3
𝒈𝝆(𝝆𝑷 − 𝝆) 𝟑
𝑲 = 𝑫𝒑 [
] = 400 × 10−6 [
]
𝟐
(0.0237 × 10−3 )2
𝝁
= 9.86 ≈ 𝐼𝑛𝑡𝑒𝑟𝑚𝑒𝑑𝑖𝑎𝑡𝑒 𝐿𝑎𝑤
𝒗𝒕 =
=
𝟎. 𝟏𝟓𝟑𝒈𝟎.𝟕𝟏 𝑫𝒑 𝟏.𝟏𝟒 (𝝆𝑷 − 𝝆)𝟎.𝟕𝟏
𝝁𝟎.𝟒𝟑 𝝆𝟎.𝟐𝟗
0.153(9.81)0.71 (400 × 10−6 )1.14 (1030 − 0.8378)0.71
(0.0237 × 10−3 )0.43 (0.8378)0.29
𝑣𝑡 = 1.46
𝑥 = 1.46
𝑚
𝑠
𝑚
× 5 𝑠 = 7.31 𝑚
𝑠
CENTRIFUGATION
Concepts
1. Where the density difference of the two liquid phase to be separated is very small (as in milk cream
separator), the most suitable separator is a
A. disc bowl centrifuge
C. batch basket centrifuge
B. sharpies supercentrifuge
D. sparkler filter
2. Dust laden air can be purified using a
A. cyclone separator
B. bag filter
C. gravity settler
D. tubular centrifuge
3. Tabular bowl centrifuges as compared to disk bowl centrifuges
A.
operate at higher speed.
B.
employ bowl of larger diameter.
C.
can not be operated under pressure/vacuum.
D.
an't be used for separation of fine suspended solids from a liquid.
4. __________ are used for the separation of coarse particles from a slurry of fine particles.
A. Thickeners
C. Hydrocyclones
B. Classifiers
D. Decanters
5. This centrifuge is normally used in sugar mills.
A.
Tubular bowl
B.
Disc-bowl
C.
Suspended batch basket D.
Perforated horizontal basket continuous
6. Solid particles separation based on the difference in their flow velocities through fluids is termed as the
A.
Clarification
B.
Classification
C.
Elutriation
D.
Sedimentation
7. If a force greater than that of gravity is used to separate solids & fluids of different densities, the
process is termed as the
A. Sedimentation
B. Flocculation
C. Dispersion
D. Centrifugation
8. For separation of sugar solution from settled out mud, we use a __________ filter.
A. sparkler
B. plate and frame
C. centrifugal
D. rotary drum vacuum
9. Traces of solids are removed from, liquid in a
A. classifier
B. clarifier
C. sparkler filter
D. rotary vacuum filter
10. The yeast generated during the fermentation of beer is generally separated by
A. Centrifugation
C. Filtration
B. Sedimentation
D. Extraction
Problem Solving
1. Two centrifuges rotate at the same peripheral velocity of 53.34 m/s. The first bowl has a radius of 76.2
mm and the second radius of 305 mm. Calculare the rpm and the centrifugal forces developed in each
bowl. N1= 6684 rpm; N2= 1670 rpm; F1= 3806 g’s; F2= 951 g’s
2. A centrifuge bowl is spinning at a constant 2000 rpm. What bowl radius is needed for the following? (a)
A force of 455 g’s; (b) A force four times that in part (a). r= 0.1017 m; r= 0.4068 m
3. A dilute slurry containes small solid food particles having a diameter of 5 x 10-2 mm, which are to be
removed by centrifuging. The particle density is 1050 kg/m 3 and the solution density is 1000 kg/m3. The
viscosity of the liquid is 1.2 x 10-3 Pa-s. A centrifuge at 3000 rev/min is to be used. The bowl dimensions
are b = 100.1 mm, r1 = 5.00 mm, and r2 = 30.0 mm. Calculate the expected flow rate in m 3/s just to
remove these particles. q = 8.76 x 10-5 m3/s
4. How fast (in rpm's) must a centrifuge rotate if a particle 7.49 cm from the axis of rotation is to
experience an acceleration of 86,000 g's? w= 320312 rpm
5. An aqueous suspension consisting of particles of density 2500 kg/m3 in the size range 1–10 µm is
introduced into a centrifuge with a basket 450 mm diameter rotating at 80 Hz. If the suspension forms a
layer 75 mm thick in the basket, approximately how long will it take for the smallest particle to settle
out? t = 19.3 s
6. If a centrifuge is 0.9 m diameter and rotates at 20 Hz, at what speed should a laboratory centrifuge of
150 mm diameter be run if it is to duplicate the performance of the large unit?
𝜔1 = 20 × 2𝜋 = 40𝜋
𝑎𝑡 𝑥2 =
𝑟𝑎𝑑
0.9
; 𝑥1 =
𝑚
𝑠
2
150
𝑚𝑚, 𝜔2 = ?
2
𝝎𝟏 𝟐 𝒓𝟏 = 𝝎𝟐 𝟐 𝒓𝟐
𝜔2 =
(0.45)
𝜔1 2 𝑟1
= 40𝜋 × √
= 307.81
(0.075)
𝑟2
𝜔2 =
307.81
= 49 𝐻𝑧
2𝜋
TECHNOLOGICAL INSTITUTE OF THE PHILIPPINES
363 P. CASAL ST., QUIAPO, MANILA
CHEMICAL ENGINEERING
PROBLEM SET
PARTICLE TECHNOLOGY
SUBMITTED BY:
SALADAGA, RONE MAUREEN O.
SUBMITTED TO:
ENGR. ROBERT DELFIN
Sedimentation
1. A laboratory test of a suspension of solid in a liquid gave the following information: original height
of sludge before settling is 10 inches free settling rate is 0.10 in/min, height of sludge at end of free
settling period is 6.5 inches, a height of sludge at end of 120 minutes and height of sludge when
settled completely. One thousand cubic feet of similar sludge is to be settled in a vertical cylindrical
tank, the diameter of which is to be equal to the depth of the liquid suspension in it. Calculate the
time that it would take the solid to settle to a height of 20 % of the original height of the sludge.
a. 13.56 hrs
b. 11.56 hrs
c. 12.25 hrs
d. 11.34 hrs
2. Complete the area required of a thickener to handle 20 tons/hr of slurry. Producing a clear overflow
and an underflow of sludge containing 20% by weight of solids. Assume that the constant rate of
settling in the batch sedimentation is the rate of settling in the clarification zone of the continous
thickener. Data on 2.91% by clarification zone of the continuous thickener, weight of CaCO 3 in pure
water, particle size about 5 microns are:
Time , 48.33
min
0
46.17
2
43.22
4
40.31
6
37.51
8
34.55
10
31.72
12
28.79
14
26.21
16
19.30
21
16.40
24
13.55
30
Find s.
a. 192. Ft2
b. 129 ft2
c. 124 ft2
d. 124 ft2
3. A cylindrical – bowl internal screw centrifugal is to be used to separate MgSO4.6H2O crystals from
the mother liquor which comes as a product from a vacuum crystallizer. The centrifuge has a bowl
8 inches in radius and 23 inches long and carries liquid to a depth of 4 inches. It rotates at 3000
rpm. If no crystals smaller than 5 microns in diameter exist in the slurry, what volumetric feed rate
to the centrifugal will result in complete removal of the solids? Assume that the internal screw does
not lift any solid particles back into the liquid or disturb their fall through the liquid. Density of the
solution is 1.2 g/cc, viscosity of solution is 1.5 centipoises, density of crystals is equal to 1.66 g/cc.
a. 44.65 gpm
b. 55.64 gpm
c. 67.65 gpm
d. 34.65 gpm
4. What is the settling time for the following spherical particles to settle under free settling conditions
through 5ft of H20 at 70⁰F.
Particle
Material
Sp. Gr
Diameter
1
Galena
75
0.01
2
Galena
7.5
0.001
3
Galena
7.65
0.01
4
Galena
2.65
0.001
a. 18.63 s
b. 16.63 s
c. 17.63 s
d. 19.63 s
5. Solid spherical particle of coffee extracted froma dryer having a diameter of 400µm. Ore falling
through air at a temperature of 422 K. The density of the particle is 1030 kg/m 3 . Calculate the
terminal setting viscosity
a. 1.29 m/c
b. 2.29 m/c
c. 3.29 m/c
d. 4.29 m/c
6. Solid spherical particles having a diameter of 0.09 mm and a solid density of 2002 kg/m are settling
in a solution of water at 26.7⁰C. the volume fraction of the solution is 0.45. the settling velocity is for
H20 ρ =994.7 kg/m3, µ= 0.861x10-3 pa-s.
Sol’n:
𝑔 𝐷𝑝 (𝑃𝜌− 𝜌) 2Vt =
(Ɛ φp)
18 µ
Φp = 1/ 101.82(1.053)
Φp = 0.1517
Vt =
(9.8)(0.09𝑥10)2(2002−994.7
18 (0.861𝑥10−3 )
(0.552- 0.1517)
Vt= 2.37x10-4 m/s
Centrifugation
1. Find the filtration rate that can be expected from basket centrifugal filters using data given. Basket
weight is 30 cm, inside diameter is 66 cm, rotation rate is 2000 rpm, material to be filter is gypsum
slurry, specific cake resistance is 1.71x10111 m/lg, porosity is 0.5 and specific gravity of
CaSO4.2H2O is 2.65. Assume the cake is incompressible, the filter medium resistance is negligible
and the liquid surface corresponds to the cake surface with a cake thickness of 2.5 cm.
a. 4.11x10-5 m3/s
b. 5.66 x10-5 m3/s
c. 8.90 x10-5 m3/s
d. 3.67 x10-5 m3/s
2. What is the capacity of a clarifying centrifuge operating under the following condition:
Diameter of bowl = 60 cm
Thickness of the liquid layer = 8cm
Speed = 1000 rpm
3.
4.
5.
6.
Depth of bowl = 40 cm
SG of the liquid = 1.3
SG of the solid = 1.6
Viscosity of liquid = 3cP
Cut size of particle = 30x10-6
a. 0.02 m3/s
b. 9.03 m3/s
c. 3.89 m3/s
d. 6.78 m3/s
A viscous solution containing particles with density of 1461 kg/m 3 is to be clarified by centrifugation.
The solution density P = 801 kg/m3, and its viscosity is 100 cp. The centrifuge has a bowl with r2 =
0.02225 m, r1 = 0.00718 m, and height b = 0.1970 m. Calculate the critical particle diameter of the
largest particles in the exit stream if N = 23000 rev/min and the flow rate q = 0.0022832 m 3/h.
a. 0.746x10-6 m
b. 3.789 x10-6 m
c. 2.987 x10-6 m
d. 2.987 x10-6 m
The capacity in cubic meters per hour of a clarifying centrifuge operating under these condition:
Diameter of the bowl = 600mm
Thicknenss of the liquid layer = 75mm
Depth of the bowl = 400 mm
Speed = 1200 rev/min
Sp. Gr of liquid 1.2
Sp. Gr of solid = 1.6
Viscosity of liquid = 2cp
Cut size particle = 30 µm
a. 210 m3/hr
b. 24 5 m3/hr
c. 987 m3/hr
d. 345 m3/hr
A centrifuge bowl is spinning at a constant speed of 200 rpm. The radius of the bowl in cm needed
to create a force of 455 gis
a. 12.0
b. 9.0
c. 7.0
d. 10.2
In attest conducted using a laboratyory centrifuge it was found that the optimum recovery of protein
from coconut oil was achieved with an rpm of 2500. Industrial size centrifuge comes in 2.5, 3.0,
3,5, 4.0 and 5 ft diameter with warranty. If the centrifuge is operated not over 1000 rpm. For
optimum commercial operation what is the centrifuge size you will recommend?
Sol’n:
DiNi2 = DiNi2
𝐷𝑖𝑁𝑖^2
D2 = 𝑁𝑖2 ^2
=
5𝑓𝑡 (1000)^2
(2500)2
D2 = 0.8 ft
Flotation
1. Ground lead ore is concentrated in a single flotation cell using 1.5 0z. of reagent per ton of ore.
The feed concentrate and tailings have the following analysis by weight on the dry basis: feed is
30%, concentrate is 90%, and tailings is .90% (% PbS). Water is fed to the cell at the rate of
1000 gallons per metric ton of wet concentrate with 99% of the water leaving with the tailings and
1.0 % with the concentrate. Ten metric tons of ore are fed to the cell per 24-hr-day operation.
Calculate the metric tons of wet tailings produced per hour.
a.
b.
c.
d.
Tw = 0.79 MT/hr
Tw = 0.80 MT/hr
Tw = 0.81 MT/hr
Tw = 0.82 MT/hr
2. A typical flotation machine has the following specifications: No. of cells = 5, cell volume = 0.02
cu.m., and time = 10 minutes. Material specifications: Pulp (mixture of water and ore) is 40% solid
and specific gravity of ore is 3. Calculate the capacity of the machine in metric tons of dry one per
day.
a.
b.
c.
d.
7.83 MT/day
6.98 MT/day
7.89 MT/day
7.01 MT/day
3. Ground lead ore is to be concentrated by a single flotation using 1.5 oz of reagent per ton of ore.
The feed concentrate and tailings have the following compositions by weight on dry basis:
Feed
Concentrate
Tailings
%
%
%
PbS
30
90
0.9
ZnS
20
5
27.3
CaC03
40
3
57.9
SIO2
10
2
13.9
Water is fed to the cell at the rate of 1100 gal/ton of dry concentrate with 99% of the water leaving
with the tailings and 1.0 with concentrate. Calculate the mass of wet concentrate produced per
hour when 10 tons of ore are fed to the cell per 24hr.
a.
b.
c.
d.
4.
Screening
Wet C = 0.1533
Wet C = 0.2142
Wet C = 0.1422
Wet C = 0.1322
1. One ton per hour dolomite is produced by crushing and then screening through a 14-mesh screen.
According to the screen analysis below, calculate the effectiveness of the screen.
Tyler Mesh
Feed to Screen
Undersize, Product
Oversize, circulating
load
4 on
14.3
......
20
8 on
20.0
......
28
14 on
20.0
0.0
28
28 on
28.5
40.0
24
48 on
8.6
30.0
0 through 8
100 on
5.7
20.0
Mesh
100 through
2.86
10.0
a.
b.
c.
d.
62.6%
72.3%
59.8%
75.01%
2. Powdered coal with the screen analysis given below as “feed” is fed to a vibrating 48 – mesh in an
attempt to remove the undersized fine material. When the screen was new the oversize and the
undersized analysis were as listed under columns headed “new”. After 5 month’s operation, the
analysis are as headed “Old”. What is the effectiveness of the screen when old?
Screen Analysis – Mass fractions
OVERSIZE
OLD
NEW
UNDERSIZE
OLD
MESH
FEED
NEW
-3 +4
0.010
0.012
0.014
- 4 +6
0.022
0.27
0.031
--
--
-6 + 8
0.063
0.078
0.088
--
--
-8 + 10
0.081
0.100
0.112
--
--
-10 +14
0.102
0.125
0.142
--
--
-14 +20
0.186
0.204
0.229
--
--
-20 +28
0.131
0.162
0.182
--
--
-28 +35
0.101
0.125
0.104
--
--
-35 +48
0.095
0.117
0.065
--
0.093
-48 +65
0.070
0.029
0.025
--
0.171
-65 +100
0.047
0.015
0.008
0.240
0.186
-100 +150
0.031
0.005
--
0.183
0.146
-150 +200
0.20
--
--
0.141
0.111
-200
0.062
--
--
0.105
0.071
a. 80.5 %
b. 90.3%
c. 75.7%
d. 83.02%
3. 800 kg of dolomite per hour is produced by crushing and then screening through a 14 mesh
screen. Calculate the total load to the crusher and the overall screen effectiveness. Shown below
are the screen analysis of different streams.
MESH
FEED %(F)
OVERSIZE %(R)
UNDERSIZE %P
+4
14.30
24.00
--
-4 + 8
20.00
28.00
--
-8 + 14
20.00
24.00
--
-14 + 28
28.50
24.00
40.00
-28 + 48
8.60
--
30.00
-48 + 100
5.20
--
20.00
-100 Through
2.86
--
10.00
a. 65. 53%
b. 64.35%
c. 61.98%
d. 62.42%
4. What is the spherecity of a quarter of a sphere?
a. 0.7
b. 0.25
c. 1
d. 1.25
5. The total perentage of particles larger than that of the screen opening in the feed product and
undersize are 36 %, 89% and 32 respectively. Calculate the effectiveness of the screen. Y f = 0.36,
Xp= 0.89, Xr = 0.03
a.
b.
c.
d.
88.61 %
98.76%
56.89%
54.98%
6.
Size Reduction
1. A material is crushed in a blake jaw crusher and the size of particle was reduced from 50 mm to 10
mm the consumption of energy needed to crush some material of average size of 25 mm.
Assuming rittingers apply.
a. 4.33 kw/kg/s
b. 4.44 kw/kg/s
c. 2.55 kw/kg/s
d. 5.66 kw/kg/s
2. What is the power required to crush 100 tons/ hr of limestone if 50 % of the feed passes through a
2 inch screen and 80% of the product passes on a ½ inch screen?
a. 123.4 kw
b. 126.8 kw
c. 169.6 kw
d. 134.8 kw
3. A ball mill is to grind 250 metric tons per day of limestone. Φ= 0.65 having a specific surface of
30.84 cm2/g. No recrushing of materials is to be done. Overall energy effectiveness of the ball mill
is 75%. Size range of the product are as follows:
Mesh
XI
Mesh
xI
0.02 -65+100 0.08
-10 + 14
-14 + 20
-20 + 28
-28 + 35
-35 + 48
-48 +65
0.07 100+150
0.18 150+200
0.25 200+270
0.14 270+400
0.11
0.05
0.04
0.0225
0.275
What is the specific surface of the product?
a. 64354.54 cm2/lb
b. 54987.56 cm2/lb
c. 34598.90 cm2/lb
d. 23456.87 cm2/lb
4. If crushing rolls 1 meter in diameter are set so that crushing surfaces are 12.5 mm apart and the
angle of nip is 31⁰, what is the maximum size if particle which should be fed to the rolls?
a. 25 mm
b. 0.025 mm
c. 52 mm
d. 0.52 mm
5. In crushing a certain ore, the feed is is such that 80% less than 50.8 mm and the product is such
that 80% is less than 6.35 mm. The power requirement is 89.5 wt. What will be the power required
if the same feed is sued so that 80% of the product is less than 3.18mm?
a. 146,69 w
b. 169.46 w
c. 179.89 w
d. 145.98 w
6. A material is crushed in a blake jaw crusher and the average size of particle was reduced from 50
mm to 10 mm. The consumption of energy was at the rate of 13 kw/kg/s. What will be the
consumption of energy needed to crush same material of average size 75mm to an average size of
25mm. Assuming kick’s law applies
Solution:
ρ/m = Kk ln Dp/Dp2
13 kw/kg/s = Kk ln (50/10)
Kk = 8.08 kw/kg s
ρ/m = 8.08 kw/kg s ln (75/25)
E = ρ/m 8.88 kw/kg/s
Sedimentation
1. The drag coefficient in hindered settling is
a. Varying
b. Less than
c. Greater than
d. Constant
2. Device in which a current of air separates particles with different sedimentation velocities
a. Agitator
b. Classifier
c. Air conveyor
3.
4.
5.
6.
7.
8.
9.
10.
d. Air elutriator
Have different settlings characteristics from suspensions of dispersed dense solids
a. Flocculated particles
b. Flocculation
c. Gravity separation
d. Clarifiers
Also known as settling, removal of solid particles from a suspension by settling under gravity
a. Sedimentation
b. Flotation
c. Screening
d. Size reduction
Is a similar term, which usually refers to the function of a sedimentation tank in removing
suspended matter.
a. Clarification
b. Thickening
c. Sedimentation
d. Water treatment
Is the process whereby the settled impurities are concentrated and compacted on the floor of the
tankand in the sludge collecting hoppers.
a. Concentrated impurities
b. Thickening
c. Sedimentation
d. Clarification
The drag coefficient in hindered settling is
a. Greater than
b. Constant
c. Less than
d. Varying
Device in which a current of air separates particles with different sedimentation velocities
a. Agitator
b. Air elutriator
c. Classifier
d. Air conveyor
The operation by which solids are separated from liquids due to difference in the respective
densities is
a. Screening
b. Sedimentation
c. Adsorption
d. Absorption
The separation of solid particles into several size fractions based upon the settling velocities in a
medium is called
a. Settling
b. Flotation
c. Settling
d. Classification
Centrifugation
1. Is a mechanical process of separating multi-phase mixture via the use of centrifugal force
a. Centrifugation
b. Sedimentation
c. Distillation
d. Flotation
2. If the radius of a basket centrifuge is halved and the rpm doubled, then the
a. Linear speed of the basket is doubled
b. Linear speed of the basket is halved
c. Centrifugal force is doubled
d. Capacity of centrifuge is increased
3. The most effectively used for clarification of lube oil and printing ink.
a. Disc bowl centrifuge
b. Sparkler filter
c. Precoat filter
d. Sharpless supercentrifuge
4. Alter the surface of the mineral in order that it will become air avid (to cause it to adhere to air
bubbles).
a. Collectors
b. Frothers
c. Fatty amines
d. Fuel oil
5. Added to streghthen temporarily covering film of the air bubbles.
a. Frother
b. Modifier
c. Flotation
d. Dispersants
a. Acid
6. Used in liquid liquid separation
a. Disk centrifuge
b. Sharples centrifuge
c. Centrifugal sedimentation
d. Centrifugal filtritation
7. Separates liquid liquid emulsions
a. Sharples or tubular centrifuge
b. Disk centrifuge
c. Particle trajectory
d. Centrifugal filtration
8. If the radius of a bsket centrifuge is halved and the rpm is doubled, then the
a. Linear speed of the basket is doubled
b. Linear speed of the basket is halved
c. Centrifugal force is doubled
d. Capacity of centrifuge is increased
9. For the separation of sugar solution from settled out mud, we use
a. Sparkler filter
b. Centrifugal filter
c. Plate and frame filter
d. Rotary drum
10. Moisture can be removed from lubricating oil using
a. Tubular centrifuge
b. Clarifier
c. Sparkler filter
d. Vacuum leaf filter
Flotation
1. Includes any operation in which one type of solid is separated from another type by floating one of
them on the surface of the fluid
a. Flotation
b. Separation
c. Conditioning
d. Distillation
2. It involves chemical treatment of the ore pulp to create conditions favourable for the attachment of
certain mineral particles to the air bubbles.
a. Separation process
b. Flotation process
c. Distillation process
d. Sedimentation process
3. The flotation agent that prevents coalescence of air bubbles as they travel to the surface of the
water is
a. Collectors
b. Frothing agent
c. Promoters
d. Modifying agent
4. The flotation agent that prevents coalescence of air bubbles as they travel to the surface of the
water is
a. Collectors
b. Frothing agent
c. Promoters
d. Modifying agent
5. Any operation in which one solid is separated from another by floating one of them at or on the
surface of a fluid
a. Coagulation
b. Flotation
c. Centrifugation
d. Sedimentation
6. A flotation modifier which assists in the selectivity (sharpness of separation) or stop unwanted
minerals from floating
a. Depressants
b. Activators
c. Alkalinity regulators
d. Promoters
7. Dispersants are important for the control of slimes which sometimes interfere with the selectivity
and increase reagent consumption.
a. Deflocculant
b. Depressants
c. Frothers
d. Regulators
8. An example of a collector for flotation of metallic sulphides and native metals is
a. Xanthates
b. Sodium silicate
c. Sodium sulphide
d. Sphalerite
9. Which of the following is an example of a deflocculant
a. Sulphuric acid
b. Dithiophosphate
c. Molybderite
d. Lignin sulforate
10. Operation on which solid is separated from another solid by floating them on the surface of the
liquid.
a. Flotation
b. Sedimentation
c. Centrifugation
d. Classification
Screening
1. Is a method of separating particles according to size alone
a. Screening
b. Adsorption
c. Gas absorption
d. Distillation
2. Is a grid of parallel metal bars set in an inclined stationary frame
a. Gyrating screen
b. Grizzly
c. Vibrating screen
d. Screener
3. Is measured by the mass of material that can be fed per unit time to aunit area of the screen
a. Capacity
b. Screen efficiency
c. Filtration
d. Screener
4. Number of openings per linear inch
a. Mesh number
b. Mesh screens
c. Screen analysis
d. Screen mesh
5. Frequency of the screen is mainly controlled by an electromagnetic vibrator which is mounted
above and directly connected to the screening surface.
a. Vibrating screens
b. Rotary screen
c. Screen effectiveness
d. Grizzly screens
6. Set of parallel metal bars in an inclined stationary frames
a. Vibrating screens
b. Grizzly screens
c. Rotary screens
d. Screen effectiveness
7. Composed of a rotating perforated drum set in an inclined position.
a. Vibrating screen
b. Rotary screen
c. Grizzly screen
d. Vibrating screen
8. May be due to blinding, rupture, or blockage of the screen openings and improper orientation of the
particle as it hits the screen
a. Screen effectiveness
b. Vibrating screen
c. Grizzly screen
d. Rotary screen
9. Data consust of mesh number vs. Weight fraction retained on the screen
a. Differential screen analysis
b. Cumulative screen analysis
c. Screening
d. Sedimentation
10. Data consist of mesh number vs cumulative fraction smaller than n
a. Screening
b. Sedimentation
c. Cumulative screen analysis
d. Differential screen analysis
Size Reduction
1. This term is applied to all the ways in which particles of solids are cut or broken into smaller pieces
a. Mixing effectiveness
b. Screening
c. Size Reduction
d. Filtration
2. Are slow speed machines for coarse reduction of large quantities of solids
a. Jaw crushers
b. Gyrating crushers
c. Roll crushers
d. Crushers
3. In this type of crusher the feed is admitted between two jaws, set to form a V open at the top.
a. Jaw crushers
b. Gyratory crushers
c. Roll crushers
d. Grinders
4. Use of a shape factor called spherecity
a. Particle Shape
b. Particle surface area
c. Bulk
d. Average particle size
5. The volume of a particle is proportional to its diameter cubed
6.
7.
8.
9.
10.
a. Number of particles in a mixture
b. Size reduction
c. Particle surface
d. Bulk
Common application of size reduction used in micronization of API (Active pharmaceutical
ingredients). For better dissolution rate and bioavailability; fast acting drugs
a. Pharmaceutical
b. Petrochemical
c. Mining
d. Crusher
Common application that increase in reactivity
a. Petrochemical
b. Mining
c. Highly inefficient
d. Kick’s law
Combines the two features of breaking and screening
a. Bradford beaker for coal
b. Toothed roll crusher
c. Hammer mill
d. Squirrel cage disintegrator
Consist of a swinging jaw which moves back and forth, working against a stationary surface called
the anvil jaw with which it forms a V shape chamber.
a. Crusher
b. Gyratory crusher
c. Crushing rolls
d. Cone crusher
This machine is widely used in the clay industry, but little anywhere else.
a. Edge runners
b. Cone crushers
c. Crushing rolls
d. Ball mill
TECHNOLOGICAL INSTITUTE OF THE PHILIPPINES
363 P. Casal St., Quiapo, Manila
College of Engineering and Architecture
Chemical Engineering Department
Problem Set in Particle Technology
(Finals Period)
Floatation-Centrifugation
Salmingo, Alvin N.
5th Yr Che Student
Engr. Robert Delfin
Instructor
March 24, 2015
SCREENING
1.
With increase in the capacity of screens, the screen effectiveness
A
.remains unchanged
B.
increases
C
.decreases
D.
decreases exponentially
2.
As particle size is reduced
A
.screening becomes progressively more difficult.
B
.screening becomes progressively easier.
C.
capacity and effectiveness of the screen is increased.
D
.none of these.
3.
In screen analysis, the notation +5 mm/-10 mm means particles passing through
A.
10 mm screen and retained on 5 mm screen.
B.
5 mm screen and retained on 10 mm screen.
C.
both 5 mm and 10 mm screens.
D.
neither 5 mm nor 10 mm screen.
.4.
Screen capacity is expressed in terms of
A.
tons/hr
B.
tons/ft2
C
.both (a) & (b)
D.
tons/hr-ft2
5.
A screen is said to be blinded, when the
A.
oversizes are present in undersize fraction.
B.
undersizes are retained in oversize fraction.
C.
screen is plugged with solid particles.
D.
screen capacity is abruptly increased.
6.
The distribution given by microscopic analysis of powder is
A.
number
B.
length
C
.area
D.
volume
7.
200 mesh screen means 200 openings per
A.
cm2
B.
m
C.
inch
D
.inch2
8.
increasing the capacity of a screen __________ the screen effectiveness.
A.
decreases.
B.
increases
C.
does not effect
D.
none of these
9.
Higher is the mesh number, smaller will be the aperture size of the screen. It means that the
aperture size of a 200 mesh screen will be smaller than that of 20 mesh screen. This is valid for
A
.British standard screens.
B.
German standard screens (DIN 1171) etc.
C.
American standard screens (ASTM and Tyler standard screens).
D.
all (a), (b) and (c).
10.
200 mesh screens means
a.
200 openings/cm2
c. 200 openings/cm
b.
200 openings/inch
d. 200 openings/inch2
SIZE REDUCTION
11.
Ball mills and tube mills with flint or porcelain balls are used for size reduction of
A.
asbestos
B
.rubber
C.
non-metallic ores
D.
limestone
12.
Size reduction of __________ is accomplished in steam heated rollers and roll crushers.
A.resins
B.gums
C. hard rubber
D.waxes
13.
Which of the following is not a part of the Blake jaw crusher?
A.Hanger
B.Check plates
C.Toggles
D.Pitman
14.
Size reduction of the __________ can be suitably done by ball mills, crushing rolls and rod mills.
A.metalliferrous ores
B.non-metallic ores
C.basic slags
D.asbestos & mica
15.
The main size reduction operation in ultrafine grinders is
A.cutting
B.attrition
C.compression
D.impact
16.
In case of a hammer crusher,
A.crushing takes place by impact breaking.
B.maximum acceptable feed size is 30 cms.
C.reduction ratio can be varied by adjusting the distance from cage to hammers.
D.all (a), (b) and (c).
17.
Crushing of mineral particles is accomplished in a 'cage mill', when one or more alloy steel bars are
revolved in opposite directions. It is a type of __________ mill.
A.impact
B.roll
C.vibratory
D.none of these
18.
A tube mill as compared to a ball mill
A
employs smaller balls.
B.
gives finer size reduction but consumes more power.
C
has larger length/diameter ratio (>2 as compared to 1 for ball mill).
D.
all (a), (b) and (c).
19.
In case of a hammer crusher, the
A
.feed may be highly abrasive (Moh's scale >5).
B.
minimum product size is 3 mm.
C.
maximum feed size may be 50 mm.
D.
rotor shaft carrying hammers can be vertical or horizontal.
20.
Length/diameter ratio of a ball mill is
A.
1.5
B.1
C.
<1
D.> 1
FLOATATION
21.
Froth floatation is the most suitable for treating
A.
iron ores
B.
sulphide ores
C.
quartzite
D.
none of these
22.
Pine oil used in froth floatation technique acts as a/an
A.
collector
C.
frother
B.
D.
modifier
activator
23. In froth floatation, chemical agent added to cause air adherence is called
A.
collector
B.
C.
modifier
24.
Role of a "collector" in froth floatation is to
A.
form a water repelling film on the mineral surface.
B.
create and stabilise the froth.
C.
act as a surface.
D.
collect the minerals according to their specific gravity.
D.
frother
activator
25.
Ore concentration by froth floatation utilises the __________ of ore particles.
A.
density difference
B.
wetting characteristics
C.
terminal velocities
D.
none of these
26.
Frother is added in the froth floatation cell used in ore beneficiation to stabilise the air bubbles (i.e.,
froth), which will hold the ore particles, but it does not affect the floatability of minerals. Which of the
following is not used as a frother?
A.
Cresylic acid
B.
Xanthaies
C.
Pine oil
D.
all 'a' 'b' & 'c'
27.
Activators are those chemicals which help buoying up one mineral in preference to the other in the
froth floatation process. Which of the following is an activator?
A.
Cresylic acid
B.
Copper sulphate
C.
Calcium carbonate
D.
Sodium carbonate
28.
Froth floatation is used for
A.
washing fine coal dust (< 0.5 mm size).
B.
washing lump coal (> 80 mm size).
C.
removing ash from the coal based on difference in specific gravity of coal and ash.
D.
all (a), (b)and(c).
29.
Xanthates are used in the froth floatation process as a/an
A.
conditioner
B.
frother
C.
collector
D.
activator
30.
The flotation takes place on a gas–liquid interface. ____________ particles, which may be
molecular, colloidal, or macro-particulate in size, are selectively adsorbed or attached
A.
Hydrophilic
B.
Hydrophobic
C.
Both
D.
None of the above
SEDIMENTATION
31.
Gravity settling process is not involved in the working of a
A.
hydrocyclone
B.
classifier
C.
dorr-thickener
D.
sedimentation tank
32.
Which is the most practical and economical method for removal of suspended solid matter from
polluted water ?
A.
Sedimentation
C.
Chlorination
B.
Skimming off
D.
Biological oxidation
33.
Which one of the following types of settling phenomenon can be analysed by the classic
sedimentation laws of Newton and Stokes ?
A.
Discrete settling
B.
Flocculent settling
C.
Hindered settling
D.
Compression settling
34.
A suspension of glass beads in ethylene glycol has a hindered settling velocity of 1.7 mm/s, while
the terminal settling velocity of a single glass bead in ethylene glycol is 17 mm/s. If the Richardson-Zaki
hindered settling index is 4.5, the volume fraction of solids in the suspension is
A.
0.1
B
.0.4
C
.0.6
D.
none of these
35.
Tabling process used for separating two materials of different densities by passing the dilute pulp
over a table/deck, which is inclined from the horizontal surface at an angle of about
A.
1 to 2°
C.
5 to 10°
B.
D.
2 to 5°
10 to 15°
36.
Suspended solid present in the waste water generated in blast furnace gas cooling and cleaning
plant is removed by
A.
biological oxygen pond.
B.
radial settling tank (thickener) using coagulant (lime & ferrous sulphate).
C.
lagoons.
D.
filtration.
37.
Gravity setting chambers are used in industries to remove
A.
Sox
B.
NOx
C.
suspended particulate matter
D.
CO
38.
A gravity decanter is meant for the separation of two __________ density.
A.
immiscible liquids of different
B.
miscible liquids of different
C.
immiscible liquids of same
D.
miscible liquids of same
39.
The settling velocity of the particles larger than 0.06 mm in a settling tank of depth 2.4 is 0.33 m
per sec. The detention period recommended for the tank, is
A.
30 minutes
B.
1 hour
C.
1 hour and 30 minutes
D.
2 hours
40.
For motion of spherical particles in a stationary fluid, the drag co-efficient in hindered settling
compared to that in free settling is
A.
more
B.
less
C.
equal
D.
more or less, depending on the type of particle
.
CENTRIFUGATION
41.
Drag co-efficient in hindered settling is __________ that in free settling.
A.
less than
B.
equal to
C.
not necessarily greater than
D.
always greater than
42.
If a force greater than that of gravity is used to separate solids & fluids of different densities, the
process is termed as the
A.
sedimentation
B.
flocculation
C.
dispersion
D.
centrifugation
43.
Operating principle of cyclone separator is based on the action of __________ dust particles.
A.
diffusion of
B.
centrifugal force on
C.
gravitational force on
D.
electrostatic force on
44.
Where the density difference of the two liquid phase to be separated is very small (as in milk cream
separator), the most suitable separator is a
A.
disc bowl centrifuge.
B.
sharpies supercentrifuge.
C.
batch basket centrifuge.
D.
sparkler filter.
45.
Operating principle of cyclone separator is based on the action of __________ dust particles.
A.
diffusion of
B.
centrifugal force on
C.
gravitational force on
D.
electrostatic force on
46.
Viruses can be purified based on their size and density by using
A.
gradient centrifugation
C.
precipitation
47.
The yeast generated during the fermentation of beer is generally separated by
A.
centrifugation
B.
filtration
C.
cell disruption
D.
all of these
48.
Driving force in case of filtration by a centrifuge is the
A.
speed of the centrifuge.
B.
centrifugal pressure exerted by the liquid.
C.
narrow diameter of the vessel.
B.
differential centrifugation
D.
none of these
D.
formation of highly porous cake.
49.
If radius of a batch basket centrifuge is halved & the r.p.m. is doubled, then the
A.
linear speed of the basket is doubled.
B.
linear speed of the basket is halved.
C.
centrifugal force is doubled.
D
capacity of centrifuge is increased.
50.
The disk centrifuge is the type of centrifuge used most often for bio separations due to its
A.continuous operation
C.higher speed
B.lesser cost
D.ease in operation
PROBLEM SOLVING
Screening
1. A mixture of coal and sand particles having sizes smaller than 1 x 10-4 m in diameter is to be
separated by screening and subsequent elutriation with water. Recommend a screen aperture
such that the oversize from the screen can be separated completely into sand and coal particles by
elutriation. Calculate also the required water velocity. Assume that Stokes law is applicable.
Density of sand = 2650 kg/m3; density of coal = 1350 kg/m3; density of water = 1000 kg/m3;
viscosity of water = 1 x 10-3 kg/m.s; g = 9.812 m/s2.
Calculations:
For laminar settling regimes, terminal-settling velocity is given by
Stokes law
𝑣𝑡 =
𝑔𝐷𝑝2 (𝜌𝑝 − 𝜌)
𝟏𝟖 µ
where D is the diameter of particle
𝜌𝑝 is the density of solid
vt of larger diameter coal particle
= ((1 𝑥 10−4 𝑚)2 𝑥
𝑘𝑔
)
𝑚3 𝑥9.812 𝑚
𝑘𝑔
𝑠2
18 𝑥 0.001 𝑚 𝑠
(1350 − 1000)(
= 1.9079 𝑥 10−3 𝑚/𝑠
Diameter of sand particle corresponding to the vt of larger diameter coal particle:
1.9079 𝑥 10−3 = 𝐷 2 𝑥
𝑘𝑔
)
𝑚3 𝑥9.812 𝑚
𝑘𝑔
𝑠2
18 𝑥 0.001 𝑚 𝑠
(1350 − 1000)(
𝐷 = 4.6056 𝑥 10−5
The size of screen aperture needed so that the oversize particles can be separated completely into sand
and coal by elutriation = 4.6056 x 10-5 m
The required water velocity = 1.9079 x 10-3 m/sec
By operating at the water velocity of 1.9079 x 10-3 m/sec, coal particles will be carried along with water,
and sand particles will settle down.
2. A sand mixture was screened through a standard 10-mesh screen. Mass fraction of the oversize
material is feed, overflow, and underflow were to be found to be 0.38, 0.79, and 0.22 respectively.
The screen effectiveness based on the undersize is
Solution
𝐷𝑥𝑑 𝑥𝑓 − 𝑥𝑏 𝑥𝑑
=
∗
𝐹𝑥𝑓 𝑥𝑑 − 𝑥𝑏 𝑥𝑓
xf= 0.38
xd = 0.79
xb = 0.22
Screen effectiveness = (0.38 – 0.22)/(0.79-0.22) *(0.79/0.38)
Screen effectiveness = 0.58
3. A stainless steel woven wire screen with a square aperture had an aperture 3.18 mm square. The
diameter of the wire was 1.2 mm. Determine the percent open area when the screen was operated
in an horizontal position
Solution
La = 3.18
Dw= 1.2
2
𝐿𝑎
) ∗ 100%
𝐴𝑜 = (
𝑙𝑎 + 𝑑𝑤
𝐴𝑜 = (
2
3.18
) ∗ 100%
3.18 + 1.2
𝑨𝒐 = 𝟓𝟐. 𝟕%
4. A gold ore is screened through a 30 mm screen. The average size distribution of the feed, oversize
and undersize were determined and graphed below. Determine the efficiency of the screen.
Solution (Efficiency 1) From the graph we can see that for a 30 mm separating size, mum = 46%, mu(p> in
oversize = 7.5% and mu(u) in undersize = 90%.
𝐸=[
𝑀𝑓 − 𝑀𝑢 𝑀𝑜 − 𝑀𝑓 1 − 𝑀𝑜 𝑀𝑢
][
][
][ ]
𝑀𝑜 − 𝑀𝑢 𝑀𝑜 − 𝑀𝑢 1 − 𝑀𝑢 𝑀𝑓
𝐸=[
0.46 − 0.9 0.075 − 0.46 1 − 0.075 0.46
][
][
][
]
0.075 − 0.9 0.075 − 0.9 1 − 0.46 0.9
𝑬 = 𝟎. 𝟖𝟑𝟒
5. From a crushed quartz sample the fraction less than 2 mm had to be removed by screening. The
feed sample contained 35% of minus 2 mm material. After screening the oversize fraction
contained 10% of minus 2 mm size and the undersize contained 82% of minus 2 mm size.
Solution
𝐸=[
𝑀𝑓 − 𝑀𝑢 𝑀𝑜 − 𝑀𝑓 1 − 𝑀𝑜 𝑀𝑢
][
][
][ ]
𝑀𝑜 − 𝑀𝑢 𝑀𝑜 − 𝑀𝑢 1 − 𝑀𝑢 𝑀𝑓
𝐸=[
0.32 − 0.82 0.10 − 0.35 1 − 0.1 0.82
][
][
][
]
0.1 − 0.82 0.1 − 0.82 1 − 0.35 0.35
𝑬 = 𝟎. 𝟖𝟕𝟐
6. The size fractions of a screen feed, oversize and undersize stream sample are given in the table
below. The oversize represented 62.5% of the feed mass flow rate. Draw the Tromp curve for the
separation and determine: 1. The separating size, 2. The probable error, 3. The imperfection.
In the table: Columns A and C are the analyses of the oversize and undersize streams Column B = Column
A x yield in oversize (0.625 in this example) Column D = Column C x yield in undersize (0.375 in this
example) Column E = Sum of columns B and D giving the reconstituted feed Column F = Partition
Coefficient = B/(B + D).
The separation size, d5o = 2800 µm and the d25 and d75 = 1200 µm and 6600 µm respectively
𝐸𝑝 =
𝐸𝑝 =
𝑑75 − 𝑑25
2
6600−1200
2
µm
𝑬𝒑 = 𝟐𝟕𝟎𝟎 µm
𝐼=
𝐸𝑝
𝑑50
𝐼=
2700
2800
𝑰 = 𝟎. 𝟗𝟔
Size Reduction
7.
Particles of average size 25 x10^-4 m are crushed to an average product size of 5x10^-4m at the
rate of 15 tons per hour. At this rate the crusher consumes 32 KW of power of which 2 KW are required for
running the mill empty. What would be the power consumption if 10 tons per hour of this product is further
crushed 1x10^-4 m size in the same mill? Assume Rittnger’s Law is applicable.
𝑃
1
1
= 𝐾𝑟[
−
]
𝑇
𝐷𝑢𝑠𝑏 𝐷𝑢𝑠𝑎
𝑃 = 𝑝𝑜𝑤𝑒𝑟 (𝐾𝑊 ) 𝑇 = 𝑓𝑒𝑒𝑑 𝑟𝑎𝑡𝑒 (
𝑡𝑜𝑛𝑠
)
ℎ𝑟
𝐷𝑢𝑠𝑎, 𝐷𝑢𝑠𝑏 = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑠𝑢𝑟𝑓𝑎𝑒 𝑚𝑒𝑎𝑛 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑓𝑒𝑒𝑑 𝑎𝑛𝑑 𝑝𝑟𝑜𝑑𝑢𝑐𝑡
𝐾𝑟 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑃 = 32 − 2 = 30𝑘𝑊
𝑇 = 15
𝑡𝑜𝑛
ℎ𝑟
𝐷𝑢𝑠𝑎 = 5 𝑥 10−4
𝐷𝑢𝑠𝑏 = 25 𝑥 10−4
30
1
1
= 1.25𝑥10−3 [
−
]
10
5 𝑥 10−4 25 𝑥 10−4
𝑘𝑟 = 1.25 𝑥 10−3
𝑃
1
1
= 1.25𝑥10−3 [
−
]
−4
10
1 𝑥 10
5 𝑥 10−4
P = 100kW
8.
The Power required to crush 100 tons/hr of a material is 179.8 kw, if 80% of the feed passes
through a 51 mm screen and 80% of the product passes through a 3.2 mm screen. What is the work index
of the material?
𝑃
1
1
= 𝐾𝑟[
−
]
𝑇
𝐷𝑢𝑠𝑏 𝐷𝑢𝑠𝑎
P = 179.8 Kw
T = 100 tons/hr
Dusa = 51mm
Dusb = 3.2mm
179.8 𝐾𝑤
1
1
=
0.361𝑤𝑖[
−
]
.5
𝑡𝑜𝑛𝑠
3.2
51.5
100
ℎ𝑟
Wi = 13.571
9.
The Power required to crush 100 tons/hr of a material if 80% of the feed passes through a 51 mm
screen and 80% of the product passes through a 1.6 mm screen
𝑃
1
1
= 𝐾𝑟[
−
]
𝑇
𝐷𝑢𝑠𝑏 𝐷𝑢𝑠𝑎
P = 179.8 Kw
Kr = 0.361 * 13.571
Dusa = 51mm
Dusb = 1.6 mm
𝑃
1
1
= 0.361 ∗ 13.571[ .5 − .5 ]
100
1.6
51
P = 279.157 Kw
10.
The energy required per unit mass of the grind limestone particles of very large size to 100 Mm is
12.7 kWh/ton. An estimate(using Bonds Law) of the energy to grind the particles from a very large size to
50 mm is.
Solution:
1 1/2
𝐸=𝐾 [ ]
𝑥2
𝐸2
100𝑥10−6
= [
]
𝐸1
50𝑥10−6
1/2
𝐸1 = 12.7
100𝑥10−6
𝐸2 = 12.7 [
]
50𝑥10−6
𝑬𝟐 = 𝟏𝟖
1/2
𝒌𝑾𝒉
𝒕𝒐𝒏
11.
A continuous grinder obeying the bond crushing law grinds at the rate of 1000 kg/hr from the initial
diameter of 10 mm of the final diameter of 1mm. If the market now demands particles of size 0.5 mm, the
output rate of the grinder in (kg/hr) for the same power input would be reduced to?
Solution:
𝑃
1
1
= 𝐾𝑟[
−
]
𝑇
𝐷𝑢𝑠𝑏 𝐷𝑢𝑠𝑎
Dusa1 = 1mm
Dusa2 = 0.5 mm
Dusb = 10 mm
𝑃
1 1
= 𝐾𝑟[ − ]
𝑇1
1 10
𝑃
1
1
= 𝐾𝑟[ −
]
𝑇2
1 0.5
𝑇2 0.684
=
= 𝟎. 𝟔𝟐𝟑
𝑇1 1.098
12.
In an analysis of ground salt using Tyler sieves, it was found that 38% of the total salt passed
through a 7 mesh sieve and was caught on a 9 mesh sieve. For one of the finer fractions, 5% passed an 80
mesh sieve but was retained on a 115 mesh sieve. Estimate the surface areas of these two fractions in a 5
kg sample of the salt, if the density of salt is 1050 kg m -3 and the shape factor (l) is 1.75.
Solution
Aperture of Tyler sieves, 7 mesh = 2.83 mm, 9 mesh = 2.00 mm, 80 mesh = 0.177 mm, 115 mesh = 0.125
mm.
Mean aperture 7 and 9 mesh = 2.41 mm = 2.4 x 10-3m
Mean aperture 80 and 115 mesh = 0.151 mm = 0.151 x 10-3m
Now from Eqn. (11.6)
A1 = (6 x 1.75 x 0.38 x 5)/(1050 x 2.41 x 10-3)
= 7.88 m2
A2 = (6 x 1.75 x 0.05 x 5)/(1050 x 0.151 x 10-3)
= 16.6 m2.
Floatation
13.
Compute the force ratios given in Table 6.6.2 for an aqueous flotation system for which, R = 1 μm ,
 =  3000 kg/m3 ,  = 100 mV and 20 AH 1 10− =  J. Take h = 1 nm. Comment on your results.
Solution: For the given aqueous system, we have, 3  1 10 Pa s − =   = 80
The permittivity of free space, o= 8.854 𝑥
10−12 𝐶 2
𝐽𝑚
𝑮𝒓𝒂𝒗𝒊𝒕𝒂𝒕𝒊𝒐𝒏𝒂𝒍𝒇𝒐𝒓𝒄𝒆 (𝟏𝒙𝟏𝟎−𝟗)(𝟏𝒙𝟏𝟎−𝟔)(𝟑𝟎𝟎𝟎)(𝟗. 𝟖)
=
= 𝟒. 𝟐𝒙𝟏𝟎−𝟔
𝑬𝒍𝒆𝒄𝒕𝒓𝒊𝒄 𝒇𝒐𝒓𝒄𝒆
𝟖𝟎(𝟖. 𝟖𝟓𝟒𝒙𝟏𝟎−𝟏𝟐)(𝟎. 𝟏)𝟐
(𝟏𝒙𝟏𝟎−𝟗)(𝟏𝒙𝟏𝟎−𝟔)(𝟑𝟎𝟎𝟎)(𝟗. 𝟖)
𝑮𝒓𝒂𝒗𝒊𝒕𝒂𝒕𝒊𝒐𝒏𝒂𝒍𝒇𝒐𝒓𝒄𝒆
=
= 𝟐. 𝟗𝒙𝟏𝟎−𝟔
𝑽𝒂𝒏 𝒅𝒆𝒓 𝒘𝒂𝒂𝒍𝒔 𝒇𝒐𝒓𝒄𝒆
𝟏𝒙𝟏𝟎−𝟏𝟐
(𝟏𝒙𝟏𝟎−𝟗 )(𝟏𝒙𝟏𝟎−𝟔 )(𝟑𝟎𝟎𝟎)𝟐 (𝟗. 𝟖)𝟐
𝑰𝒏𝒆𝒓𝒕𝒊𝒂𝒍 𝒇𝒐𝒓𝒄𝒆
=
= 𝟑. 𝟕𝒙𝟏𝟎−𝟏𝟎
𝑬𝒍𝒆𝒄𝒕𝒓𝒊𝒄 𝒇𝒐𝒓𝒄𝒆 𝟖𝟎(𝟖. 𝟖𝟓𝟒𝒙𝟏𝟎−𝟏𝟐 )(𝟎. 𝟏)𝟐 (𝟏𝒙𝟏𝟎−𝟑 )𝟐
(𝟏𝒙𝟏𝟎−𝟗)𝟐 (𝟏𝒙𝟏𝟎−𝟔 )𝟓 (𝟑𝟎𝟎𝟎)𝟑(𝟗. 𝟖)𝟐
𝑰𝒏𝒆𝒓𝒕𝒊𝒂𝒍 𝒇𝒐𝒓𝒄𝒆
=
= 𝟐. 𝟔𝒙𝟏𝟎−𝟏𝟎
𝑽𝒂𝒏 𝒅𝒆𝒓 𝒘𝒂𝒂𝒍𝒔 𝒇𝒐𝒓𝒄𝒆
𝟏𝒙𝟏𝟎−𝟐𝟎(𝟏𝒙𝟏𝟎−𝟑)𝟐
14.
For a laboratory flotation of an iron ore in water, it was observed that 2 mg was collected while
traversing 2 m of the flotation column. The concentration of the ore in water was 0.5 kg/m3 . The average
diameter of the bubbles was 2 mm and the average diameter of the particles was 0.1 mm. Compute
flotation recovery
Given:
𝑑𝑝 = 1𝑥10−3 , 𝑑𝑏 = 2𝑥10−3 , 𝐻 = 2𝑚, 𝑐 = 0.5
𝑘𝑔
𝑎𝑛𝑑 𝑁𝑐 = 2𝑥10−6
3
𝑚
𝑁𝑐
𝑅=𝜋
(
)
4 𝑑𝑝 + 𝑑𝑏 𝑥𝐻𝑥𝑐
(2𝑥10−6 )
𝑅=𝜋
= 𝟎. 𝟓𝟕𝟖
(0.1𝑥10−3 + 2𝑥10−3 )𝑥2𝑥0.5
4
15.
Problem: A copper ore initially contains 2.09% Cu. After carrying out a froth flotation separation,
the products are as shown in Table 1. Using this data, calculate: (a) Ratio of concentration (b) % Metal
Recovery (c) % Metal Loss (d) % Weight Recovery, or % Yield (e) Enrichment Ratio Table 1:
Grade/recovery performance of a hypothetical copper ore flotation process.
a. R= (c-t)/(f-t)
R= (20 – 0.1)/(2.09 – 0.1) = 10
b. Using the example data from Table 1,
% Cu Recovery = [(c x C)/(f·F]·100
% Cu Recovery = [(10·20)/(2.09·100)]·100 = 95.7%
c. % Cu Loss = F-R
% Cu Loss = 100 – 95.7 = 4.3%
d. % Weight Recovery = F·(f - t)(c– t) = 10%
% Weight Recovery = 100·(2.09 - 0.1)(20 – 0.1) = 10%
e. Enrichment Ratio = c/f
Enrichment Ratio = 20.0/2.09 = 9.57
16.
A typical flotation machine has the following specifications:
Number of cells = 4
Flotation time = 12min.
Cell Volume = 60 ft3
Hp per cell = 10hp
The material treated has the following specifications:
Pulp (mixture ore and water ) = 40% solids
Specific gravity of ore = 3
𝑇 𝑥 𝐶𝑎𝑝 𝑥 𝑑
𝑛=
𝑉 𝑥 1440
Where n= number of cells; V = volume in cu. Ft per cell; Cap = tons of dry ore / 24 hrs.; d= cu. Ft
of pulp (ore and water) containing one ton of solids.
Solution:
2000
2000 + 𝑥
𝑥 = 3000 𝐻2𝑂
𝐹 = 3000 + 2000 = 5000
0.4 =
𝑑=
2000
3000
+
= 58.76
3𝑥62.4 62.4
12(𝑥)(58.76)
4=
60(1440)
𝑐𝑎𝑝 = 𝟒𝟗𝟎. 𝟏𝟑
17.
A flotation plant processes 3000 tons/day of CuFeS2. It produces 80 tons Cu concentrate assaying
25% Cu. If ore analyzes 0.7% Cu, the percent recovery is?
Solution:
% Cu Recovery = [(c x C)/(f·F]·100
=
(0.25𝑥80)𝑡𝑜𝑛𝑠
0.007𝑥3000
= 𝟗𝟓. 𝟐%
Sedimentation
18.
Calculate the settling velocity of dust particles of (a) 60 m and (b)10 m diameter in air at 21°C and
100 kPa pressure. Assume that the particles are spherical and of density 1280 kg m -3, and that the
viscosity of air = 1.8 x 10-5 N s m-2 and density of air = 1.2 kg m-3.
For 60 m particle:
𝑘𝑔
)
𝑚3 𝑥9.812 𝑚
𝑉𝑡 = ((60𝑥10^ − 6)𝑚)2 𝑥
𝑘𝑔
𝑠2
18 𝑥 0.000018 𝑚 𝑠
(1280 − 1.2)(
𝑉𝑡 = 0.14
𝑚
𝑠
For 10 m particles since vm is proportional to the squares of the diameters,
𝑘𝑔
)
𝑚3 𝑥9.812 𝑚
𝑉𝑡 = ((10𝑥10^ − 6)𝑚)2 𝑥
𝑘𝑔
𝑠2
18 𝑥 0.000018 𝑚 𝑠
(1280 − 1.2)(
𝑽𝒕 = 𝟎. 𝟎𝟑𝟗
𝒎
𝒔
Checking the Reynolds number for the 60 m particles,
𝑅𝑒 =
𝐷𝑣𝜌
µ
𝑘𝑔
𝑚
60𝑥10−6 𝑚 𝑥 .14 𝑠 𝑥 1.2 3
𝑚
𝑅𝑒 =
𝑘𝑔
1.8 𝑥 10−5 𝑚 𝑠
𝑹𝒆 = 𝟎. 𝟓𝟔
19.
Oil droplets having a diameter of 20 µm (0.020 mm) are to be settled from air at temperature of
37.8°C (311 K) and 101.3 kPa pressure. The density of the oil is 900 kg/rm3. Calculate the terminal settling
velocity of the droplets.
Solution:
Dp = 2.0 X 10-5 m and ρp, = 900 kg/m3
ρ = 1.137 kg/m3, µ = 1.90 X 10^-5 Pa•s
𝜌(𝜌𝑠 − 𝜌)
]
µ2
𝑘 = 𝐷𝑝[
1
𝑘𝑔
𝑘𝑔 3
1.137 3 (900 − 1.137) 3
𝑚
𝑚
𝑘 = 20𝑥10−6 𝑚
2
𝑘𝑔
(1.90 𝑥 10−5
)
[
𝑚𝑠
]
𝑘 = 0.283 𝑠𝑡𝑟𝑜𝑘𝑒
𝑘𝑔
3)
𝑚
𝑚
𝑉𝑡 = ((20𝑥10−6 )𝑚)2 𝑥
𝑥9.812 2
𝑘𝑔
𝑠
18 𝑥 1.90 𝑥 10−5 𝑚 𝑠
(900 − 1.137)(
𝑽𝒕 = 𝟎. 𝟎𝟏
20.
𝒎
𝒔
Calculate the settling velocity of glass spheres having a diameter of 1.554 X 10 -4 m (5.10
x 10-4 ft) in water at 293.2 K (20°C). The slurry contains 60 wt % solids. The density of
the glass spheres is ρp, = 2467 kg/m3 (154 lbm/ft3).
Solution:
ρ= 998 kg/m3
µ = 1.005 x 10-3 Pa·s
𝙴=
40
)
998
40
60
( )+
998 2467
(
= 0.622
𝜌𝑚 = 𝑒𝜌 + (1 − 𝑒)𝜌𝑝 = 0.662(998) + (1 − 0.622)(2467)
𝜌𝑚 = 1553
𝑘𝑔
𝑚3
Ф𝑝 = 1/(101.82(1−𝑒)
Ф𝑝 = 1/(101.82(1−0.622)
21.
A suspension of uniform particles in water at a concentration of 500 kg of solids per cubic meter of
slurry is settling in a tank. Density of the particles is 2500 kg/m3 and terminal velocity of a single particle is
20 cm/s. what will be the settling velocity of suspension? Richardson and Zaki index is 4.6.
𝙴=
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 + 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑
500
[1 − (
)] 𝑚3
2500
𝙴=
1 𝑚3
𝙴 = 0.8
𝙴𝑛 =
(𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑙 𝑠𝑒𝑡𝑡𝑙𝑖𝑛𝑔 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 𝑠𝑢𝑠𝑝𝑒𝑛𝑠𝑖𝑜𝑛) 𝑢𝑠
=
𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑎 𝑠𝑖𝑛𝑔𝑙𝑒 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒
𝑢𝑜
𝙴𝑛 𝑢𝑜 = 𝑢𝑠
𝑢𝑠 = 20
𝑐𝑚
𝑥 0.84.6
𝑠
𝑢𝑠 = 7.16
𝑐𝑚
𝑠
22.
What is the terminal settling velocity in m/s, calculated from the strokes law for a particle diameter
of 0.1 x 10^-3 m, density 21800 kg/m^3 settling in water density 1000 kg/m^3 and viscosity of 10^-3
kg/ms?
Solution
𝑔𝐷𝑝2 (𝜌𝑝 − 𝜌)
𝑣𝑡 =
𝟏𝟖 µ
9.81
𝑣𝑡 =
𝑣𝑡 = 0.1
𝑚
𝑥 (0.1 𝑥 10−3 )2 𝑚(21800 − 1000)
𝑠2
𝒌𝒈
𝟏𝟖𝒙 𝟎. 𝟎𝟎𝟏 𝒎𝒔
𝑚
𝑠
23.
The terminal settling velocity of a 6mm diameter glass sphere (density 2500 kg/m3) in a viscous
Newtonian Liquid(density 1500 kg/m3) is 100 µm/s. if the particle Reynolds number is small in value and
the acceleration due to gravity is 9.81m/s^2, then the viscosity of the liquid in (Pa*s) is
Solution
ρ= 1500 kg/m3
ρp= 2500 kg/m3
vt = 100 x 10^-6 m/s
𝐷 = 6 𝑥 10−3 𝑚
µ=
𝐷𝑝2 (𝜌𝑝 − 𝜌)
𝟏𝟖𝒗𝒕
6 𝑥 10−3 𝑚2 (2500 − 1500)
µ=
𝒌𝒈
𝟏𝟖(𝟏𝟎𝟎 𝒙 𝟏𝟎−𝟔 ) 𝒎𝒔
µ = 𝟏𝟗𝟔. 𝟐
𝒌𝒈
𝒎𝒔
𝑘𝑔
𝑚3
Centrifugation
24.
If a cream separator has discharge radii of 5 cm and 7.5 cm and if the density of skim milk is 1032
kg m-3 and that of cream is 915 kg m-3, calculate the radius of the neutral zone so that the feed inlet can be
designed.
For skim milk, r1 = 0.075m, rA = 1032 kg m-3, cream r2 = 0.05 m, rB= 915 kg m-3
Solution
rn2 = (rAr12 - rBr22) / (rA - rB)
rn2 = [1032 x (0.075)2 - 915 x (0.05)2] / (1032 - 915)
= 0.03 m2
rn = 0.17 m
= 17 cm
25.
A dispersion of oil in water is to be separated using a centrifuge. Assume that the oil is dispersed in
the form of spherical globules 5.1 x 10-5 m diameter and that its density is 894 kg m-3. If the centrifuge
rotates at 1500 rev/min and the effective radius at which the separation occurs is 3.8 cm, calculate the
velocity of the oil through the water. Take the density of water to be 1000 kg m -3 and its viscosity to be
0.7 x 10-3 N s m-2
Solution:
Vm = D2N2r(rp - rf)/1640m
vm = (5.1 x 10-5)2 x (1500)2 x 0.038 x (1000 - 894)/(1.64 x 103 x 0.7 x 10-3)
𝑽𝒎 = 𝟎. 𝟎𝟐
𝒎
𝒔
26.
A centrifuge of 0.2m in a pilot plant operates at a speed of 50 Hz in order to achieve effective
separation. If this centrifuge is scaled up to a diameter of 1m in the chemical plant and the same operation
factor is to be achieved, what is the rotational speed of the scaled up centrifuge?
Solution
𝑟
Separation factor 𝑆 = 𝑤 2 𝑔
1
𝑟1 2
𝑤2 = 𝑤1 ∗ 𝑠𝑞𝑟𝑡 ( )
𝑟2
w1= 50 Hz
r1= 0.2m
r2= 1m
1
0.2 2
𝑤2 = 50𝐻𝑧 ∗ 𝑠𝑞𝑟𝑡 ( )
1
𝒘𝟐 = 𝟐𝟐. 𝟑𝟔 𝑯𝒛
27.
A viscous solution containing particles with density ρp=1461 Kg/m3 is clarified by centrifugation.
Solution density ρ=801 Kg/m3 ; Viscosity μ=100 cp. Centrifuge bowl with r2=0.02225 m.; r1=0.00716 m.;
height=b=0.197 m.; Calculate critical particle diameter of the largest particles in the exit stream if N=23000
rev/min, q=0.002832 m3 /hr
Solution
Ω= 2π
N
60
Ω= 2π
23000
60
Ω = 2410
rad
s
V = πb(r22 − r12 )
V = π(0.197)(0.022252 − 0.007162 )
V = 2.747 ∗ 10−4 𝑚3
µ = 10−3 ∗ 100 𝑝𝑎 ∗ 𝑠
µ = 0.1 𝑝𝑎 ∗ 𝑠
𝑞𝑐 = 0.002832𝑚3 /3600𝑠
𝑞𝑐 = 7.887 𝑥 10
−7
𝑚3
𝑠
𝑫𝒑𝒄 = 𝟎. 𝟕𝟒𝟔 µ𝒎
28.
A viscous solution contains particles with a density is to be clarified by centrifugation. The solution
density is and its viscosity is 80 cp. The centrifuge has a bowl with and 3 1200 / p ρ = kg m 3 ρ = 850 / kg
m 2r m = 0.02 1 r = 0.01 m and height b=0.25 m. Calculate the critical particle diameter of the largest
particles in the exit stream if N=15000 rpm and flow rate q=0.002 m3 /hr?
Solution
Ω= 2π
N
60
Ω= 2π
1500
60
Ω = 1570
rad
s
V = πb(r22 − r12 )
V = π(0.25)(0.022 − 0.012 )
V = 2.355 ∗ 10−4 𝑚3
𝑞𝑐 = 0.002𝑚3 /3600𝑠
𝑞𝑐 = 5.56 𝑥 10−7
𝑚3
𝑠
Ω2 (𝑝𝑝 − 𝑝)𝐷𝑝𝑐 2
𝑞𝑐 =
𝑥𝑉
2𝑥𝑟2
18 𝑥µ ln (𝑟1 + 𝑟2)
5.56 𝑥 10−7
𝑚3
15702 (1200 − 850)𝐷𝑝𝑐 2
=
𝑥2.355 ∗ 10−4
2𝑥0.02
𝑠
18 𝑥80𝑥10−3 ln (0.01 + 0.02)
𝑫𝒑𝒄𝟐 = 𝟐. 𝟓𝟕µ𝒎
29.
In the above problem, we would like to scale up the centrifuge. We have to design a centrifuge
such that it can handle 1.5 times q in Question (3). r1 and r2 remains same, find the length of the
centrifuge. Both the centrifuges have same rotational speed?
Solution
𝑞1 𝑞2
=
𝐸1 𝐸2
𝑞1 𝐸1
=
𝑞2 𝐸2
Ω2 (𝑝𝑝 − 𝑝)𝐷𝑝𝑐 2
𝑞𝑐 =
𝑥𝑉
2𝑥𝑟2
18 𝑥µ ln (𝑟1 + 𝑟2)
𝑞1
𝑏1
=
1.5𝑞1 𝑏2
𝑏2 = 1.5𝑏1 = 1.5 ∗ 0.25𝑚
𝒃𝟐 = 𝟎. 𝟑𝟕𝟓𝒎
Screening
1. These are made of punched metal plate or woven wire mesh, usually set an angle with the
horizontal up to about 60 degrees.
a.
b.
c.
d.
Vibrating Screen
Oscillating Screen
Grizzly
Stationary Screen
2. The term stating the number of openings per linear inch of screen surface
a.
b.
c.
d.
Size of Wire
Mesh
Gauze
Particle Size
3. Type of Screen that is usually characterized by a relatively low speed of about 300 or 400
oscillations per minute in a plane.
a.
b.
c.
d.
Oscillating Screen
Vibrating Screen
Trommels
Rotating Screen
4. Revolving Screen that are typically driven at relatively high speed
a.
b.
c.
d.
Vbrating Screen
Rotating Screen
Reels
Trommels
5. It is based upon both the recovery in the product of the desired material in the feed and the
exclusion or rejection from the product of the undesired material in the feed.
a.
b.
c.
d.
Recovery of the Screen
Effectiveness of the Screen
Rejection
Efficiency
6. Widely used for screening large sizes, particularly of 1 in. and over.
a. Vibrating Screen
b. Grizzlies
c. Rotating Screen
d. Reels
7. Screen Capacity is expressed in terms of
a.
b.
c.
d.
Tons/h
Tons/ft2
Both a and b
Tons/h-ft2
8. Screen Efficiency is
a.
b.
c.
d.
Recovery rejection
Recovery
Rejection
None of these
9. With increase in the capacity of screens, the screen effectiveness
a.
b.
c.
d.
remains unchanged
increases
decreases
decreases exponentially
10. Vibrating screens have capacity (tons/ft2 – hr – mm mesh size) in the range
a.
b.
c.
d.
0.2 to 0.8
5 to 25
50 to 100
100 to 250
11. Powdered coal with the screen analysis given below as feed is fed to a vibrating 48 mesh screen in
an attempt to remove the undesired fine material. When the screen was new the oversize and
undersize analyses were as listed under columns headed new. After 3 months operation, the
analyses are as headed Old. What is the effectiveness of the screen when new?
Oversize
Undersize
Mesh
Feed
New
Old
New
Old
-3 + 4
.010
0.012
0.014
--
--
-4 + 6
0.022
0.027
0.031
--
--
-6 + 8
0.063
0.078
0.088
--
--
-8 + 10
0.081
0.100
0.112
--
--
-10 + 14
0.102
0.126
0.142
--
--
-14 + 20
0.165
0.204
0.229
--
--
-20 + 28
0.131
0.162
0.182
--
--
-28 + 35
0.101
0.125
0.104
--
0.093
-35 + 48
0.095
0.117
0.065
--
0.171
-48 + 65
0.070
0.029
0.025
0.246
0.186
-65 +
100
0.047
0.015
0.008
0.183
0.146
-100 +
150
0.031
0.005
--
0.141
0.111
-150 +
200
0.020
--
--
0.105
0.071
-200
0.062
--
--
0.325
0.222
ANSWER: 82.75 %
12. From the problem above, what is the effectiveness when old?
ANSWER: 91.06 %
13. If the screen dimensions in problem 59 were 2 ft by 4 ft, calculate the capacity of the 65-mesh
screen on the basis of a perfectly functioning 48-mesh screen and also on the basis of the actual
performance of the screen.
ANSWER: 408.2 kg/m2-mm-hr
14. Table salt is being fed to a vibrating screen at the rate of 300 lb/hr. The desired product is the
48/65 mesh fraction. A 48- and a 65- mesh screen are therefore used (double deck), the feed
being introduced on the 48 mesh screen, the product being discharged from the 65-mesh screen.
During the operation it was observed that the average proportion of oversize:product:undersize
was 2:1.50:1. Calculate the effectiveness of the screener.
SOLUTION:
OMB:
F=O+P+U
F = 2 + 1.5 + 1
F = 4.5
Xf = 0.281
Xp = 0.526
USING RAP:
𝑃 = 300 (
𝑂 = 300 (
2
) = 133.33
4.5
𝑈 = 300 (
𝐸48/65 =
𝐸48/65 =
1.5
) = 100
4.5
1
) = 66.67
4.5
𝑋𝑝 𝑃 (1 − 𝑋𝑓 )𝐹 − (1 − 𝑋𝑝 )𝑃
(
)
𝑋𝑓 𝐹
(1 − 𝑋𝑓 )𝐹
0.526(100) (1 − 0.281)300 − (1 − 0.526)100
(
)
(1 − 0.281)300
0.281(300)
𝐸48/65 = 0.4868 = 48.68 %
Find E48:
P = 133.33
F = 300
Xp = 0.9238
Xf = 0.537
𝐸48 =
𝐸48 =
𝑋𝑝 𝑃 (1 − 𝑋𝑓 )𝐹 − (1 − 𝑋𝑝 )𝑃
(
)
𝑋𝑓 𝐹
(1 − 𝑋𝑓 )𝐹
0.9238(133.33) (1 − 0.537)300 − (1 − 0.9238)133.33
(
)
(1 − 0.537)300
0.537(300)
𝐸48/65 = 70.86 %
Basis: undersize of #65 is P
P = 66.67
F = 166.67
Xp = 0.646
XR = 0.093
Xf = ?
𝐹 𝑋𝑃 − 𝑋𝑅
=
𝑃 𝑋𝑓 − 𝑋𝑅
166.67 0.646 − 0.093
=
66.67
𝑋𝑓 − 0.093
𝑋𝑓 = 0.314
𝐸65 =
0.9238(66.67) (1 − 0.314)166.67 − (1 − 0.646)66.67
(
)
(1 − 0.314)166.67
0.537(166.67)
𝐸65 = 65.13 %
15. Based on problem no. 14, calculate the Capacity of the screener in 65 mesh.
ANSWER: 408.2 kg/m2-mm-hr
Size Reduction
16. One of the mechanism of size reduction wherein it is used for coarse reduction of hard solids to
give relatively few fines and the particle disintegration by two rigid forces.
a.
b.
c.
d.
Impact
Compression
Shear
Attrition
17. It generally gives coarse, medium or fine products and particle concussion by a single rigid force.
a.
b.
c.
d.
Impact
Attrition
Shear
Compression
18. Used for primary crushing which accepts anything that comes from the mine face and breaking it
into 6 to 10 inches lumps.
a.
b.
c.
d.
Ball Mill
Gyratory Crusher
Jaw Crusher
Rolling Mill
19. It gives particles of definite size and shape approximate 2 to 10 mm in length.
a.
b.
c.
d.
Shear
Grinders
Cutters
Crusher
20. It do the heavy work of breaking large pieces of solid materials into small lumps.
a.
b.
c.
d.
Rolling Mill
Crusher
Gyratory Crusher
Grinders
21. A cylindrical shell slowly turning about a horizontal axis and filled to about half its volume with a
solid grinding medium in the form of metal balls, pebbles, metal rods or length of chains.
a.
b.
c.
d.
Rotating Mill
Smooth Roll Crusher
Grinders
Revolving Mills
22. Shape factor for a cylinder whose length equals its diameter is
a.
b.
c.
d.
1.5
0.5
1
0
23. The operating speed of a ball mill should be
a.
b.
c.
d.
less than the critical speed
much more than the critical speed
at least equal to the critical speed
none of these
24. Ball mill is used for
a.
b.
c.
d.
Crushing
Coarse Grinding
Fine Grinding
Attrition
25. The Reduction Ratio for fine grinders is
a. 5 -10
b. 20 -40
c. 10 -20
d. As high as 100
26. If crushing rolls 1m diameter are set so that the crushing surfaces are 12.5 mm apart and the angle
of nip is 31o, what is the maximum size of particle which should be fed to the rolls?
ANSWER: 0.025 m
27. A crusher reducing limestone of crushing strength 70 MN/m 2 from 6 mm diameter average size to
0.1 mm diameter average size, requires 9 kW. The same machine is used to crush dolomite at the
same output from 6 mm diameter average size to a product consisting of 20 per cent with an
average diameter of 0.25 mm, 60 per cent with an average diameter of 0.125 mm and a balance
having an average diameter of 0.085 mm. Estimate the power required, assuming that the crushing
strength of the dolomite is 100 MN/m2 and that crushing follows Rittinger’s Law.
ANSWER: 5.9 kW
28. Quartz goes through two successive grinders on the same shaft which draws a total of 20 hp. The
feed averages 2 in. in diameter and has a surface ratio n of 10. The grinders running empty require
2 hp. Their capacity is 3 tons/hr. The analyses of their products are given below.
Calculate the horsepower in each grinder.
ANSWER: 2.20 hp and 15.80 hp
29. If crushing rolls 1 m diameter are set so that the crushing surfaces are 12.5 mm apart and the
angle of nip is 31o, what is the max size of particle which should be fed to the rolls? If the actual
capacity of machine is 12% of thee, calculate the throughput in kg/s when running at 2Hz (2m per
sec) if the working face of the rolls is 0.40 m long and the feed weighs 2500 kg per cu meter.
ANSWER: 3 kg/s
30. A material is crushed in ablake jaw crusher and the average size of particle was reduced from 50
mm to 10 mm. The consumption of energy was at the rate of 13 kW/kg-s. What will be the
consumption of energy needed to crush same material of average size 75 mm to an average size
of 25 mm.
ANSWER: 8.87 kW/kg-s
31. What is the power required to crush 100 tons per hour of limestone if 80% of the feed passes a 2
inch screen and 80% of the product passes a 1/8 inch screen?
SOLUTION:
Using bond’s law/ Work index for limestone, (Table 28.2, Mccabe and Smith, page 967)
𝑊𝑖 = 12.74 𝑘𝑊 − ℎ𝑟/𝑡𝑜𝑛
𝑆𝑝. 𝐺𝑟. = 2.66
𝐷𝑝𝑖 = 2 𝑖𝑛𝑐ℎ𝑒𝑠 (25.44
𝑚𝑚
) = 50.88 𝑚𝑚
𝑖𝑛𝑐ℎ
𝐷𝑝2 = 0.125 𝑖𝑛𝑐ℎ𝑒𝑠 (25.44
𝑚𝑚
) = 3.175 𝑚𝑚
𝑖𝑛𝑐ℎ
𝑃
1
1
= 0.3162𝑊𝑖 [(
−
)]
𝑚
√𝐷𝑝2 √𝐷𝑝𝑖
𝑃 = 0.3162(100
𝑡𝑜𝑛𝑠
𝑘𝑊 − ℎ𝑟
1
1
)]
)(12.74
[(
−
ℎ𝑟
𝑡𝑜𝑛
√3125 √50.8
𝑃 = 169.60 𝑘𝑊
Sedimentation
32. Stoke’s law is valid when the particle Reynolds number is
a.
b.
c.
d.
<1
<5
>1
None of these
33. Terminal velocity is
a.
b.
c.
d.
attained after moving one-half of total distance
constant velocity with no acceleration
a fluctuating velocity
none of these
34. For the free settling of a spherical particle through a fluid, the slope of C D vs log NRe plot is
a.
b.
c.
d.
0.5
-0.5
-1
1
35. Sedimentation on commercial scale occurs in
a.
b.
c.
d.
Thickeners
Cyclones
Magnetic separator
Classifiers
36. Drag is the force component exerted on an immersed
a.
b.
c.
d.
the component being normal to the flow direction
passing the centroid of the body at 60° to the direction of motion
the component being parallel to the flow direction
none of these
37. In hindered settling, particles are
a.
b.
c.
d.
not affected by other particles and the wall
near each other
placed farther from the wall
none of these
38. Drag coefficient CD is given by (in Stoke’s law range)
a. CD =
16
Re p
b. CD =
0.079
Re p 0.23
c. CD =
24
Re p
d. CD =
18.4
Re p
39. In sewage treatment, its sedimentation is speeded up by commonly adding
a.
b.
c.
d.
sodium sulphate
Hydrochloric acid
Lime
copper sulphate
40. When the particle concentration increases, particle settling velocities _____ because of
hydrodynamic interation between particles and the upward motion of displaced liquid.
a.
b.
c.
d.
Remain the same
Increases
Partially increases
Decreases
41. A particle falling under the action of gravity will accelerate until the drag force balances
gravitational force, after which it falls at constant______.
a.
b.
c.
d.
Size
Settling
Free Settling velocity
Terminal velocity
42. A suspension in water of uniformly sized sphere (diameter 150 µm, density 1140 kg/m 3 has a
solids concentration of 25% by volume. The suspension settles to a bed of solids concentration of
55% by volume. Calculate the rate at which the water/ suspension interface settles. The rate at
which the sediment/suspension interface rises (assume water properties: density = 1000 kg/m 3,
viscosity = 0.001 Pa-s).
ANSWER: 0.45X10-3 m/s, -0.375 mm/s
43. Oil droplets having a diameter of 200µm are settling from still air at 294.3 K and 101.325 kPa. The
density of the oil is 900 kg/m3. A settling chamber is 0.457 m high and the particles having a size
range of 1.27x10-2 mm to 5.08x10-2 mm. Calculate the size range of the various fractions obtained
using free settling conditions. Also calculate the value of the largest Reynolds number occurring.
ANSWER: 0.4589 NRe
44. What is the mass of a sphere of material of density 7500 kg/m 3 whose terminal velocity in a large
deep tank of water is 0.6 m/s?
ANSWER: 0.029 g
45. Two spherical particles, one of density 3000 kg/m3 and diameter 20µm, and the other of density
200 kg/m3 and diameter 30µm start settling from rest at the same horizontal level in a liquid of
density 900 kg/m3 and of viscosity 3 mN s/m2. After what period of settling will the particles be
again at the same horizontal level? It may be assumed that Stokes’ Law is applicable, and the
effect of added mass of the liquid moved with each sphere may be ignored.
ANSWER: 7.81x10-5 sec
46. A spherical glass particle is allowed to settle freely in water. If the particle starts initially from rest
and if the value of the Reynolds number with respect to the particle is 0.1 when it has attained its
terminal falling velocity, calculate the distance travelled before the particle reaches 90 percent of its
terminal falling velocity.
ANSWER: 1.05 mm
47. What will be the terminal falling velocity of a glass sphere 1 mm in diameter in water if the density
of glass is 2500 kg/m3.
SOLUTION:
For water:
Density = 1000 kg/m3
Viscosity = 0.001 Ns/m2 per water
2𝑥0.0013
(1000(2500 − 1000) ∗ 9.81) = 9810
3𝑥0.0012
log10 9810 = 3.992
From table 3.4
log10 𝑅𝑒𝑜 = 2.16
𝑅𝑒𝑜 = 144.5
144.5𝑥0.001
𝑚
𝑉=
= 0.145
1000𝑥0.001
𝑠
Centrifugation
48. It is fed to a rotating basket having a slotted or perforated wall covered with a filter medium such as
canvas or metal cloth.
a.
b.
c.
d.
Residue
Cakes
Slurry
None of these
49. Type of filtering centrifuges wherein the filter medium is usually the slotted wall of the basket itself.
a.
b.
c.
d.
Suspended Centrifuge
Automatic Centrifuge
Batch Machines
Conveyor Centrifuge
50. It resembles a heavy duty hammer mill except that it contains no grate or screen.
a.
b.
c.
d.
Collector
Impactor
Bearing
Reels
51. For separation of sugar solution from settled out mud we use
a.
b.
c.
d.
centrifugal filter
plate & frame filter
rotary drum vacuum filter
sparkler filter
52. It is used extensively in sugar refining, where they operate on short cycles of 2-3 min per load and
produce up to 5 ton/h of crystals per machine.
a.
b.
c.
d.
Tubular Centrifuge
Batch Centrifuge
Top-Suspended Centrifuge
Conveyor Centrifuge
53. Moisture can be removed from lubricating oil using
a.
b.
c.
d.
clarifier
vacuum leaf filter
tubular centrifuge
sparkler filter
54. Which of the following can be most effectively used for clarification of tube oil and printing ink?
a.
b.
c.
d.
sparkler filter
sharpless supercentrifuge
disc-bowl centrifuge
precoat filter
55. If the radius of a basket centrifuge is halved and the rpm is doubled, then
a.
b.
c.
d.
linear speed of the basket is halved
capacity of centrifuge is increased
linear speed of the basket is doubled
centrifugal force is doubled
56. Slimy or very fine solids that form a dense, impermeable cake quickly plug any filter medium that is
fine enough to retain them.
a.
b.
c.
d.
Slurry
Filter Media
Filter Aids
Cake
57. Where the difference in density of the two liquid phases to be separated is very small (as in milk
cream separator), the most suitable separator is
a.
b.
c.
d.
sparkler filter
disc bond centrifuge
batch basket centrifuge
sharpless supercentrifuge
58. What is the capacity in m3 / hr of the centrifuge operation under the following conditions?
Dia of bowl 600 mm
Thickness of liquid layer 75 mm
Speed 1000 rpm
Depth of bowl 400 mm
Sp-gr of liquid 1.3
Sp-gr of soild 1.6
Viscosity of liquid 3 cp
Critical particle dia 30 µ m
SOLUTION:
𝜋𝑏𝜔2 𝐷𝑝2 (𝜌𝑝 − 𝜌) 𝑟2 2 − 𝑟1 2
[
]
𝑄=
2𝑟
18𝜇
ln [𝑟 +2𝑟 ]
2
1
𝑄=
𝜋(.40)(104.5)2 (30𝑥10−6 )2 (1600 − 1300) 0.32 − 0.222
[
]
2(0.3)
18(0.003)
ln [0.22 + 0.3]
𝑄 = 0.02 𝑚3 /𝑠
59. A centrifuge having a radius of the bowl of 0.1016 m is rotating at N = 1000 rev/min. Calculate the
centrifugal force developed in terms of gravity forces.
ANSWER: 113.6
60. A centrifuge with a phosphor bronze basket, 380 mm in diameter, is to be run at 67 Hz with a 75
mm layer of liquid of density 1200 kg/m3 in the basket. What thickness of walls are required in the
basket? The density of phosphor bronze is 8900 kg/m3 and the maximum safe stress for phosphor
bronze is 87.6 MN/m2.
ANSWER: 15 mm
61. A batch centrifugal filter has a bowl height b = 0.457 m and r2 = 0.381 m and operates at 33.33
rev/s at 25 oC. The filtrate is essentially water. At a given time in the cycle the slurry and cake
formed have then following properties. Cs = 60 kg solids/m3 filtrate, ε = 0.82, ρp = 2002 kg
solids/m3, cake thickness = 0.152 m, α = 6.38x1010 m/kg, Rm = 8.53x1010 m-1, r1 = 0.0232 m.
Calulate the rate of filtrate flow.
ANSWER: 6.11x10-4 m3/s
62. Find the filtration rate that can be expected from basket centrifugal filter using the data below:
Basket height = 30 cm
Inside Basket diameter = 66 cm
Rotation rate = 2000 rpm
Material to be filtered = gypsum Slurry
Specific case resistance = 1.71x1011 m/kg
Porosity = 0.5
S.G. of CaSO4-2H2O = 2.65
Assume that the cake is incompressible, the filter medium resistance is negligible and the liquid
surface sorresponds to the filter cake surface with cake thickness of 2.5 m.
ANSWER: 4.11x10-5 m3/s
63. A rotary drum filter with 30 percent submergence is to be used to filter a concentrated aqueous
slurry of CaCO3 containing 14.7 lb of solids per cubic foot of water (236 kg/m3). The pressure drop
is to be 20 in. Hg. If the filter cake contains 50 percent moisture (wet basis), calculate the filter area
required to filter 10 gal.min of slurry when the filter cycle is 5 min. Assume that the filter-medium
resistance Rm is negligible. The temperature is 20oC.
ANSWER: 7.59 m2
Flotation
64. A highly versatile method for physically separating particles based on differences in the ability of air
bubbles to selectively adhere to specific mineral surfaces in a mineral/water slurry.
a.
b.
c.
d.
Sedimentation
Centrifugation
Froth flotation
NOTA
65. Reagents that are used to selectively adsorb onto the surfaces of particles. They form a monolayer
on the particle surface that essentially makes a thin film of non-polar hydrophobic hydrocarbons.
a.
b.
c.
d.
Non-inizing
Ionizing
Collector
Froth
66. Recovery of metallic silver form photographic residues is an example of,
a.
b.
c.
d.
Leaching
Flotation
Sedimentation
Centrifugation
67. Froth Flotation is most suitable for treating
i.
j.
k.
l.
iron ores
quartz
sulfide ores
metal ores
68. In Froth Flotation, chemical agent added to cause air adherence is called
k.
l.
m.
n.
Collector
modifier
frother
promoter
69. Pine oil used in forth flotation technique acts as a
m.
n.
o.
p.
collector
frother
modifier
activator
70. An example of a collector for floatation of metallic sulfides and native metals is
a. Sodium silicate
b. Sodium sulfide
c. Xanthates
d. Sphalerite
71. Which of the following is the most suitable for cleaning of fine coal dust (< 0.5 m)?
k.
l.
m.
n.
Through washer
Spiral separator
Baum Jig Washer
Froth Flotation
72. A copper ore initially contains 2.09% Cu. After carrying out a froth flotation separation, the products
are shown, using this data, calculate the Metal recovery.
Product
%Weight
%Cu Assay
Feed
100
2.09
Concentrate
10
20.0
Tailings
90
0.1
ANSWER : 95.7
73. Based on the problem no. 55, calculate the percentage of metal loss.
ANSWER: 4.3 %
74. A flotation section of a mining company is extracting CuS from corellite ores. The ore consists of
5% CuS and 45% gangue, which may be assumed to be SiO 2. The following data are given
%CuS
%SiO2
Feed
5
95
Concentrate
85
15
Rougher Tailings
1
99
Scavenger
Concentrate
10
90
Final tailings
0.3
99.7
Laboratory experiments indicate that the water to solid ratio, L/S = 2, and the contact time is 10
min in the rougher; L/S = 4 contact time = 18 min in the scavenger. On the basis of 300
tons/day of ore treated. Calculate the volume of the rougher and scavenger. Data (ρ SiO2 = 2.65
g/cc; ρCuS= 4.6 g/cc)
ANSWER: R= 170 ft3 ; S = 536 ft3
TECHNOLOGICAL INSTITUTE OF THE PHILIPPINES
363 P. CASAL ST. QUIAPO, MANILA
COLLEGE OF ENGINEERING AND ARCHITECTURE
CHEMICAL ENGINEERING DEPARTMENT
INTRODUCTION TO PARTICLE TECHNOLOGY
SUBMITTED BY:
TORDECILLAS, BIA D.
MONDAY/2:30-4:30 PM
SUBMITTED TO:
ENGR. ROBERT DELFIN
MARCH 24, 2015
Screening
1. Is a method of separating particles according to size alone.
a. Screening
b. Flotation
c. Sedimentation
d. Centrifugation
2. Are performed based on the physical difference between particles such as size, shape, or density.
a. Chemical Separation
b. Mechanical Separation
c. both a and b
d. None of the above
3. Mechanical separations are applicable to ________ mixtures, not to ___________ solutions.
a. heterogeneous, homogeneous
b. homogeneous, heterogeneous
c. homogeneous, aqueous
d. None of the above
4. Are fines, pass through the screen openings.
a. oversize
b. right size
c. undersize
d. none of the above
5. Are tails, do not pass through the screen openings.
a. oversize
b. right size
c. undersize
d. none of the above
6. _____ is the minimum free space between the edges of the opening in the screening surface.
a. Open area
b. aperture
c. mesh
d. all of the above
7. This type of equipment differs from the vibrating equipment such that the machine gyrates in a circular
motion at a near level plane at low angles. The drive is an eccentric gear box or eccentric weights.
a. Gyratory equipment
b. Vibrating Equipment
c. Centrifugal Sitter
d. all of the above
8. Does not require vibrations, instead, material is fed into a horizontal rotating drum with screen panels
around the diameter of the drum.
a. Trommel screens
b. Centrifugal Sitter
c. both a and b
d. none of the above
9. Are the most important screening machines for mineral processing applications.
a. Gyratory equipment
b. Vibrating Equipment
c. Centrifugal Sitter
d. all of the above
10. Which is not a stationary screen?
a. Gyratory equipment
b. Vibrating Equipment
c. Centrifugal Sitter
d. trammel screen
Problem Solving
A quartz mixture is screened through a 10-mesh screen. The cumulative screen analysis of feed, overflow
and underfolw are given in the table. Calculate the mass ratios of the overflow and underflow to feed and
the overall effectiveness of the screen.
Mesh
Dp (mm)
Feed
Overflow
Underflow
4
4.699
0
0
0
6
3.327
0.025
0.071
0
8
2.362
0.15
0.43
0
10
1.651
0.47
0.85
0.195
14
1.168
0.73
0.97
0.58
20
0.833
0.885
0.99
0.83
28
0.589
0.94
1.0
0.91
35
0.417
0.96
0.94
65
0.208
0.98
0.975
1.0
1.0
Pan
From the table, xF=0.47, xD=0.85, xB=0.195
1. A screen with an aperture of 6 mesh BSS is treating a feed with 66% of +6 mesh and producing an
oversize fraction containing 89% of +6 mesh particles. If the undersize fraction contains 2% of +6 mesh
particles, calculate the effectiveness of the screen. Answer: 75.59%
2. Table salt is being fed to a vibrating screen at the rate of 150 kg/hr. The desired product is -39 +20 mesh
fraction. A 30 mesh and 20 mesh screen are therefore used (double deck), the feed being introduced on
the 30 mesh screen. During the operation it was observed that the average proportions of oversize are
(from 30 mesh screen): oversize (from 20 mesh screen): undersize (from 20 screen) is 2:1.5:1. Calculate
the effectiveness of the screener from the following data: Ans. B. 63. 39 %
3. It is desired to separate a mixture of sugar crystals into two fractions, a coarse fration returned on an 8
mesh screen, and a fine fraction passing through it, screen analysis of feed, coarse and fine fractions
shows
Mass fraction of 18 particles
In feed=0.46
In coarse=0.88
In fine=0.32
The overall E of the screen per 100 kg of feed is?
Answer: E = 45.17 %
4. A sand mixture was screened through a standard 10-mesh screen. Mass fraction of the oversize material
is feed, overflow, and underflow were to be found to be 0.38, 0.79, and 0.22 respectively. The screen
effectiveness based on the undersize is Answer: 0.58
5. In an analysis of ground salt using Tyler sieves, it was found that 38% of the total salt passed through a
7 mesh sieve and was caught on a 9 mesh sieve. For one of the finer fractions, 5% passed an 80 mesh
sieve but was retained on a 115 mesh sieve. Estimate the surface areas of these two fractions in a 5 kg
sample of the salt, if the density of salt is 1050 kg m-3 and the shape factor ( ) is 1.75.
Aperture of Tyler sieves, 7 mesh = 2.83 mm, 9 mesh = 2.00 mm, 80 mesh = 0.177 mm, 115 mesh = 0.125
mm.
Mean aperture 7 and 9 mesh
= 2.41 mm = 2.4 x 10-3m
Mean aperture 80 and 115 mesh = 0.151 mm = 0.151 x 10-3m
Answer: A1= 7.88 m2
A2= 16.6 m2
Size Reduction
1. A process in which solid particles are cut or broken into smaller process.
a. Size Reduction
b. Screening
c. Flotation
d. Sedimentation
2. The generic term for size reduction
a. Compression
b. Impact
c. Cutting
d. Comminution
3. What is the criterion for size reduction?
a. have a large capacity
b. require a small power input per unit of produc
c. yield a product of the single size distribution desired
d. all of the above
4. A law in crushing efficiency wherein the work required in crushing is proportional to the new surface
created.
a. Rittinger’s Law
b. Kick’s Law
c. Bond’s Law
d. None of the above
5. In this law the work required for crushing a given mass of material is constant for the same reduction
ratio, that is the ratio of the initial particle size to the finial particle size.
a. Rittinger’s Law
b. Kick’s Law
c. Bond’s Law
d. None of the above
6. An energy requirement which assumes that the energy consumed is proportional to the newly generated
surface area.
a. Rittinger’s Law
b. Kick’s Law
c. Bond’s Law
d. None of the above
7. Is the reduction of solid materials from one average particle size to a smaller average particle size, by
crushing, grinding, cutting, vibrating, or other processes.
a. Compression
b. Impact
c. Cutting
d. Comminution
8. Equipment for coarse reduction of large amounts of solids consists of slow-speedd machines called
crushers.
a. Jaw Crushers
b. Gyratory Crushers
c. Roll Crushers
d. None of the above
9. Are devices used to reduce intermediate-sized material to small sizes or powder.
a. Revolving Grinding Mills
b. Hammer Mill Grinders
c. Roll Crushers
d. Gyratory Crushers
10. For intermediate and fine reduction of materials, _________ are often used.
a. Revolving Grinding Mills
b. Hammer Mill Grinders
c. Roll Crushers
d. Gyratory Crushers
Problem Solving
What is the power required to crush 100 ton/h of limestone if 80% of the feed pass a 2-in screen and 80%
1
of the product a 8in screen? The work index for limestone is 12.74.
Given:
x1= 2 in.
1
x2 = 8 in
Ei = 12.74
T = 100 ton/hr
Required: P?
Solution:
𝑃
= 0.3162 Ei (
𝑇
𝑃
100
1
𝑥2
√
1
√𝑥1
= 0.3162 (12.74) (
)
1
-
1
√3.175 √1
)
8
P = 169.6 kW
1. It is desired to crush 10 ton/hr of iron ore hematite. The size of the feed is such that 80% passes a 3 in.
1
screen and 80 % of the product is to pass a 8 in. screen. Calculate the gross power required. Use a work
index Ei for iron ore hematite 12.68. Answer: 24.1 hp (17.96 kW)
2. It is desired to crush 100 ton/h of phosphate rock from a feed size where 80 % is less than 4 in. to a
1
product where 80 % is less than 8 in. The work index is 10.13.
a. Calculate the power required. Answer: 198.7 hp ( 148.2 kW)
b. Calculate the power required to crush the product further where 80% is less than 100um. Answer: 387.6
hp (289.1 kW)
3. In crushing a certain ore, the feed is such that 80% is less than 67.2 mm size and the product size is
such that 80% is less than 7.53 mm. The power required is 95.8 kW. Based on the Bond equation the
power required using that same feed so that 80% is less than 4.02 mm is _______.
Answer: 148.88 kW
4. A material consisting originally of 25mm particles is crushed to an average size of 7mm and requires 20
kJ/kg for this size reduction. Determine the energy required to crush the material from 25mm to 3.5mm
assuming (a) Rittinger’s law, (b) Kick’s law and (c) Bond’s law.
Answer: a. 47:8kJ/kg, 30:9kJ/kg, 37.6kJ/kg
5. It is found that the energy required to reduce particles from a mean diameter of 1 cm to 0.3 cm is 11
kJ kg-1. Estimate the energy requirement to reduce the same particles from a diameter of 0.1 cm to 0.01
cm assuming:
(a)Kick'sLaw Answer: 21 kJ/kg
(b)Rittinger'sLaw Answer: 423 kJ/kg
(c) Bond's Equation Answer: 91 kJ/kg
Flotation
1. Is a process of separating mixtures which involves separating substances by whether they sink or float.
a. Screening
b. Flotation
c. Sedimentation
d. Centrifugation
2. Factors affecting floatation:
a. Particle size and density
b. Pulp density
c. Air bubble size
d. all of the above
3. Is a process for selectively separating hydrophobic materials from hydrophilic.
a. Flotation
b. Froth flotation
c. Sedimentation
d. Crystallization
4. Its objective is to remove the maximum amount of the valuable mineral at as coarse a particle size as
practical.
a. Conditioner
b. Rougher
c. Cleaner
d. Scavenger
5. Produced by the rougher.
a. Rougher Concentrate
b. Final Tailings
c. Final Concentrate
d. None of the above
6. The rougher concentrate is normally subjected to further stages of flotation to reject more of the
undesirable minerals that also reported to the froth, in a process known as ____________.
a. Conditioner
b. Rougher
c. Cleaner
d. Scavenger
7. The product of cleaning is known as the __________.
a. cleaner concentrate
b. final concentrate
c. both a and b
d. None of the above
8. Its objective is to produce as high a concentrate grade as possible.
a. Conditioner
b. Rougher
c. Cleaner
d. Scavenger
9. The rougher flotation step is often followed by a ________ flotation step that is applied to the rougher
tailings.
a. Conditioner
b. Rougher
c. Cleaner
d. Scavenger
10. Its objective is to recover any of the target minerals that were not recovered during the initial roughing
stage.
a. Conditioner
b. Rougher
c. Cleaner
d. Scavenger
Problem Solving
A flotation plant processes 5000 ton/hr of CuFeS 2 (chalcopyrite). It produces 70 tons of Cu concentrate
assaying 30 % Cu. If ore analyzes 0.8% cu, what is the amount of tailings and percent recovery?
Given:
F = 5000 tons/hr
C = 80 tons/hr
XF = 0.008
XC = 0.30
Required: T and % recovery?
Solution:
Over-all Material Balance:
F=C+T
5000 ton/hr = 80 tons/hr + T
T = 4,200 tons/hr
𝐶𝑋𝐶
% recovery = 𝐹𝑋𝐹 x 100
% recovery =
(70 𝑡𝑜𝑛𝑠/ℎ𝑟) (0.30)
𝑡𝑜𝑛𝑠
)(0.008)
ℎ𝑟
(5000
% recovery = 52.50 %
x 100
1. A copper ore initially contains 2.09% Cu. After carrying out a froth flotation separation, the products are
as shown in Table 1. Using this data, calculate Ratio of concentration and the metal recovery.
𝑭
Answer: 𝑪= 10, % recovery = 95.7 %
2. Calculate the percentage loss using the data from the table above. Answer: 4.3%
3. For a flotation. operation, the net feed is 2000 tons/day to the rougher. From the following data, compute
the capacity and number of flotation cells in each unit and the power consumption. Compute for the volume
of the rougher and the scavenger. Answer: Vrougher = 855.45 ft3, Vscavenger= 2328.19 ft3
SiO2
CuS
Feed (a)
98
2
Tailings from Rougher. (b)
99
1
Rougher Concentrate (c)
60
40
Tailings from scavenger (d)
99.6
0.4
Scavenger Concentrate (e)
50
50
Tailings from cleaner (f)
80
20
Final Concentrate (g)
1
99
Specific Gravity
Water to Solid Ratios
Contact time, min
SiO2
2.65
Rougher
2/1
Rougher
8
CuS
4.60
Scavenger
4/1
Scavenger
12
H2O
1.0
Cleaner
6/1
Cleaner
10
Required: Vrougher , Vscavenger
4. Use Denver No. 24 cells whose cubic capacity is 50 ft3 and whose power consumption is 4.2 hp per cell.
Compute the capacity and number of flotation cells in each unit and the power consumption.
Answer:
Rougher
Scavenger
Ncells = 18 cells ,
Ncells = 47 cells
Power = 75.6 hp
Power= 197.4 hp
5. Use air-lift cells whose cross-sectional area up to froth overflow is 9.85 ft2 and compute the lengths of
the troughs and air required at 2 psi. The amount of air may be assumed to be 75 cfm/ft in rougher and 100
cfm/ft in scavenger at 2 psi.
Answer:
Rougher
Scavenger
L = 130.27 ft
L = 354.55 ft
Amount of air = 9770.25 cfm
Amount of air = 35455 cfm
Sedimentation
1. Is the process of letting suspended materials settle by gravity.
a. Screening
b. Flotation
c. Sedimentation
d. Centrifugation
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;
c. both a and b
d. none of the above.
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;
c. both a and b;
d. none of the above.
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;
c. equal to the feed concentration
d. none of the above
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;
c. equal to the feed concentration.
d. none of the above
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;
c. equal to the concentration in the bottom section.
d. none of the above
7. When a particle reaches terminal velocity:
a. the particle acceleration is constant;
b. the particle acceleration is zero;
c. the particle acceleration equals the apparent weight of the particle;
d. none of the above.
8. Which of the following do not influence hindered settling velocity?
a. particle density;
b. particle size;
c. particle suspension concentration;
d. none of the above
9. Is the tendency for particles in suspension to settle out of the fluid in which they are entrained, and come
to rest against a barrier.
a. Screening
b. Flotation
c. Sedimentation
d. Centrifugation
10. 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;
c. increases with increasing fluid viscosity
d. none of the above.
Problem Solving:
Particles of quartz having a diameter of 0.127 mm and a specific gravity of 2.65 are settling in water at 293.
2 K. The volume fraction of the solids in the water is 0.45. Calculate the settling velocity.
Given:
For H2O @293.2 K
ρ = 998.22 kg/m3
µ = 1.0038 x 10 -3 Pa.s
Required: : Vt
Solution:
ɛ = 1-0.45 = 0.55
1
Ø = 10 ^1.82(1−ɛ)
=
1
10 ^1.82(1−0.55)
Ø = 0.1517
Vt =
𝑔 𝐷𝑝2 (𝜌𝑝− 𝜌)
18 µ
2
Vt =
(9.81)(1.27 𝑥 10−4 ) (𝐷𝑝2 (2650− 998.22)
18 (1.0038 𝑥 10−3)
Vt = 6.96 x 10-3 m/s
x 0.552 x 0.1517
1. A suspension in water of uniformly sized spheres of diameter 100 mm 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; Answer: 0.39 mm/s
(b) the velocity at which the sediment/suspension interface rises. Answer: 0.26 mm/s
2. A suspension in water of uniformly sized spheres of diameter 90 mm 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 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 0:44mm/s. Calculate:
(a) the velocity at which the clear water=suspension interface settles; Answer: 0.156 mm/s
(b) the velocity at which the sediment=suspension interface rises. Answer: 0.104 mm/s
3. A suspension in water of uniformly sized spheres of diameter 80 mm 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 is 1000 kg/m3 and viscosity is 0:001 Pa s.) The single particle terminal velocity of the spheres
under these conditions is 1:0mm/s.
Calculate:
(a) the velocity at which the clear water=suspension interface settles; Answer: 0.613 mm/s
(b) the velocity at which the sediment/suspension interface rises. Answer: 0.204 mm/s
4. Solid particles having a diameter of 0.090 mm and a solid density of 2002 kg/m3 are settling in a solution
of water at 26.7 °C. The volume fraction of the solids in the water is 0.45. Calculate the settling velocity and
the Reynolds number. Answer: Vt = 2.37 x 10-4 m/s, NRe = 0.121
5. Calculate the settling velocity of glass spheres having a diameter of 1.554 x 10 -4 m in water at 20 °C.
The slurry contains 60% wt. solids. The density of the glass spheres is 2467 kg/m 3. Answer: Vt = 1.53 x
10-3 m/s, NRe = 0.121
Centrifugation
1. Is a process by which solid particles are sediment and separated from a liquid using centrifugal force as
a driving force.
a. Screening
b. Flotation
c. Sedimentation
d. Centrifugation
2. Are sedimentation devices in which suspended solids are separated from a liquid under the action of
centrifugal forces generated by spinning the internal bowl of the centrifuge.
a. screen
b. centrifuges
c. gravity settling tank
d. all of the above
3. Centrifuges can be thought of as sedimentation vessels operating under high _______ forces.
a. gravitational
b. weight
c. settling
d. none of the alove
4. Consists of an elongated cylindrical rotating bowl having a tapered conical end.
a. decanter
b. thickener
c. centrifuge
d. all of the above
5. In order to leave the centrifuge the liquid must pass between conical plates where the solids can
separate after impacting on the plates
a. decanter
b. disk bowl centrifuge
c. thickener
d. none of the above
6. Centrifuges are also used in _______where a centrifugal force is used instead of a pressure difference to
cause the flow of slurry in a filter where a cake of solids builds up on a screen.
a. Centrifugal settling
b. sedimentation
c. centrifugal filtration
d. none of the above
7. These supercentrifuges are often used to separate liquid-liquid emulsions.
a. tubular centrifuge
b. disk bowl centrifuge
c. both a and b
d. none of the above
8. Centrifuges that are often used in liquid-liquid separation.
a. tubular centrifuge
b. disk bowl centrifuge
c. both a and b
d. none of the above
9. Gives smoother flow and better separation in liquid-liquid separation.
a. conical bowl
b. disk bowl centrifuge
c. tubular centrifuge
d. none of the above
10. They are used when a very short overall path of centrifugation is required.
a. disk bowl centrifuge
b. rotors
c. thickeners
d. all of the above
Problem Solving
A cream separator centrifuge has an outlet discharge radius of 50.8 mm and an outlet radius of 76.2 mm.
The density of the skim milk is 1032 kg/m3. Calculate the radius of the interface neutral zone.
Given:
ρH = 1032 kg/m3
ρL = 865 kg/m3
r1 = 50.8 mm
r4 = 76.2 mm
Required: r2?
Solution:
r22 =
r22 =
(ρHr42 − ρLr12)
ρH− ρL
(1032)(76.2)2−(865)(50.8)2 )
1032−865
r2 = 150.05 mm
1. Two centrifuges rotate at the same peripheral velocity of 53.34 m/s. The first bowl has a radius 76.2 mm
and the second 305 mm. Calculate the rev/min in each bowl. Answer: N1 = 6684 rev/min, N2 = 1670
rev/min
2. A centrifuge bowl is spinning at a constant 2000 rev/min. What radius bowl is needed for a force of 455
g’s? Answer: 0.1017 m.
3. Repeat problem 2 but double the throughout. Answer: Dp = 1.747 x 10 -6 m
4. A dilute slurry contains small solid food particles having a diameter of 5 x 10 -2 mm which are to be
removed by centrifuging. The particle density is 1050 kg/m 3 and the solution density is 1000 kg/m3. The
viscosity of the liquid is 1.2 x 10 -3 Pa.s. A centrifuge at 3000 rev/min is tp be used. The bowl dimensions
are b= 100.1 mm, r1 = 5.00 mm and r2= 30.0 mm. Calculate the expected flowrate in m3/s just to remove
these particles. Answer: 8.76 x 10 -5 m3/s
5. A batch centrifugal filter has a bowl height b= 0.457 mm and r 2= 0.381 m operates at 33.33 rev/s at 25.0
°C. The filtrate is essentially water. At a given time in the cycle the slurry and cake formed have the
following properties. Cs= 60 kg/m3 filtrate, ɛ= 0.82, ρp= 2002 kg solids/m3, cake thickness= 0.152 m, α=
6.38 x 10 10 m/kg, Rm = 8.53 x 10 10 /m, r1= 0.2032 m. Calculate the rate of filtrate flow. Answer: 6.11 x 10 4m3/s