PART 2 PARTICLE MECHANICS II SAMPLE QUESTIONS - most

0
PART 2
PARTICLE MECHANICS II
SAMPLE QUESTIONS
1
Chapter - 1
Sedimentation and Thickening
1.*
A single batch-settling test was made on a limestone slurry. The interface
between clear liquid and suspended solids was observed as a function of time, and the
results are tabulated below. The test was made using 236 gm of limestone per liter of
slurry. Prepare a curve showing the relation ship between settling rate and solids
concentration.
Test Data
Time, hr
Height of Interface ,cm
0
0.25
0.50
1.00
1.75
3.0
4.75
12.0
20.0
36.0
32.4
28.6
21.0
14.7
12.3
11.55
9.8
8.8
2.*
A limestone-water slurry equivalent to that of problem 1 is fed to a thickener
at a rate of 50 tons of dry solids/hr to produce a thickened sludge of 550 gm limestone
per liter. For an initial slurry concentration of 236 gm limestone per liter of slurry
concentration of 236 gm limestone per liter of slurry specify the thickener area
required.
3.*
A slurry of calcium carbonate in water, containing 45 g of calcium carbonate
per liter, was allowed to settle in a 6.0 cm I.D glass cylinder. The height of the line
between clear liquid and zone B-the zone of relative constant solid concentration, was
measured as a function of time. The results are given below. Prepare a curve of
settling rate Vs concentration.
2
Time, min
0
100
200
300
400
500
Height, in
44
34
25
17
10
7
4.*
A batch sedimentation test was made with 5 micron silica particles in water at
86Вє F. A 4.5 cm I.D cylinder was used, and the initial slurry concentration was 0.125 g
of silica per cu cm of slurry. The data are given below. Prepare the corresponding
settling rate vs concentration curve.
Final height = 5.53 cm
Time, min
Height, cm
0
34
10.8
25
16.8
20
26.4
15
43.1
10
65.8
7.5
89.5
6.4
100
6.1
5.*
The data given below were obtained from a single batch sedimentation test on
an ore slurry. The true density of the solids in the slurry was 2.5 g/cu-cm, and the
density of the liquid was 1 g/cu-cm. Determine the diameter required for a thickener
to handle 100 tons of solids per day from a feed concentration of 64.5 g/liter to an
underflow concentration of 485 g/liter.
Concentration (g of solids/liter of slurry)
64.5
70.9
94.3
111.7
139.9
173.9
222.0
331.0
Settling rate (cm/hr)
139.9
103.6
71.9
49.4
27.1
16.8
10.0
6.4
3
6.** Estimate the depth of the thickener required to perform the operation of
Example 1.2. The batch-settling test indicated a value of z в€ћ = 7.7 cm. The specific
gravity of the limestone is 2.09.
7.*** Waste water from a drinking plant is to be clarified by continuous
sedimentation. Feed to the thickener is one million gal per day containing 1.2 % by
weight solids. The underflow from the unit analyzes 8 % solids. Specify the depth and
diameter of the thickener.
A single batch - settling test on the feed material gave the following
information:
Specific gravity of solids
2.00
Specific gravity of solution
1.00
Concentration of solids in test
1.2 %
Time, min
0
5
Height of
31
21
Liquid-solid interface, cm
10
10
20
3.2
40
2.2
60
2.1
180
2.0
240
1.96
О±
1.94
4
Chapter 2
Classification
8.*
Calculate the terminal velocity for sphericity droplets of coffee extract, 400
microns in diameter, falling through air. The specific gravity of the coffee
extract is 1.03, and the air is at a temperature of 300F.
9.*
A mixture of silica and galena is to be separated by hydraulic classification.
The mixture has a size range between 0.008 cm and 0.07 cm. The density of
galena is 7.5 gm/cu cm, and the density of the silica is 2.65 gm/cu cm.
Assume the sphericity, П€=0.806. (a) What water velocity is necessary for a
pure galena product? Assume unhindered settling of the particles and a water
temperature of 65 В°F.(b) What is the maximum size range of the galena
product?
10.*
Drops of oil 15 microns in diameter are to be settle from their mixture with air.
The specific gravity of the oil is 0.9, and the air is at 21ЛљC and 1 atm. A
settling time of 1 min is available. How high should the chamber be to allow
settling of these particles? Assume that the motion of the particle lies in the
StokesвЂ™-Law range.
11.*
Air is being dried by being bubbled (in very small bubbles) through
concentrated sulfuric acid (specific gravity, 1.84, viscosity, 15 cp, temperature
100ЛљF). The sulfuric acid fills a 24 in tall, 2 in ID glass tube to a depth of 6 in.
The dry air above the acid is at a pressure of 0.8 atm and at 100ЛљF. If the dry
air rate is 3.5 cfm, what is the maximum diameter of a sulfuric acid spray
droplet which might be carried out of the apparatus by entrainment in the air
stream.
12.*
A thin water suspension of soil is prepared at 10 A.M. At noon, a sample is
drawn from the suspension at a depth of 5 cm. What is the largest particle
probably removed by a pipette at this depth (5 cm) if the specific gravity of the
soil is 2.5?
13.** Particles of sphalerite (specific gravity = 4.0) are settling under the force of
gravity through a slurry consisting of 25% by volume of quartz particles
(specific gravity = 2.65) and water. The diameter of the sphalerite particles is
0.006 in. The volumetric ratio of sphalerite to slurry is 0.25. The temperature
is 50ЛљF. What is the terminal velocity of the sphalerite? What is the density of
slurry?
5
14.** Calculate the settling velocity for the hindered settling of glass sphere in water
at 68ЛљF when the suspension contains 1206 g of glass spheres in 1140 cm3 of
total volume. The average diameter of the spheres, as determined from
photomicrographs, was 0.0061 in and the true density of the spheres was 154
lb/ft3.
15.*** Urea pellets are made by spraying drops of molten urea into cold gas at the top
of a tall tower and allowing the material to solidify as it falls. Pellets 1/4 in. in
diameter are to be made in a tower 80 ft high containing air at 70ЛљF. The
density of urea is 83 lb/cu-ft.
(a) What would be the terminal velocity of the pellets, assuming free-settling
conditions?
(b) Would the pellets attain 99 % of this velocity before they reached the
bottom of the tower?
16.*** Quartz and pyrites are separated by continuous hydraulic classifications. The
feed to the classifier ranges in size between 10 microns and 300 microns.
Three fractions are obtained: a pure-quartz product, a pure-pyrites product, and
a mixture of quartz and pyrites. The specific gravity of quartz is 2.65 and that
of pyrites is 5.1. What is the size range of the two materials in the mixed
fraction for each of the following cases.
(a) The bottoms product to contain the maximum amount of pure pyrites.
(b) The overhead product to contain the maximum amount of pure quartz.
17.*** A mixture of coal and sand in particle size smaller than 20 mesh is to be
completely separated by screening and then elutriating each of the cuts from
the screening operation with water as the elutriating fluid. Recommend a
screen size such that the oversize cut can be completely separated into coal and
sand fractions by water elutriation. What water velocity will be required? The
specific gravity for sand and coal is 2.65 and 1.35 respectively.
18.*** A mixture of spherical particles of silica contains particles ranging in size from
14 mesh to 200 mesh. This mixture is to be divided into two fractions by
elutriation, utilizing the upward velocity of a stream of water at 55ЛљF rising
through a tube 4 in in diameter.
(a) What quantity of water, in gpm, will probably be needed to divide the
mixture at a size equal to the aperture of a 48-mesh screen?
(b) When the water flow through the tube is 1.5 gpm, what is the smallest size
of silica particle which will probably settle through the stream? The specific
gravity for silica is 2.65.
6
Chapter 3
Centrifugation
1.*
A Tubular bowl centrifuge is to be used to separate nitrobenzene with the
density of 75 lb/ft3 , from an aqueous wash liquid having a density of 64 lb/ft3. The
centrifuge has a bow 1.4 in inside dia & rotates at 15,000 rpm. The radius of the dam
overwhich the light phase flow is 1 in. If the centrifuge bowl is to contain equal
volumes of the two liquids, what should be the radial distance from the rotation axis
to the dam overwhich the heavy phase flow?
2.*
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 Вµ m
3.** A liquid-detergent solution of 100 centipoises viscosity and 0.8 gm/cu cm
density is to be clarified of fine Na2SO4 crystals (Ps= 1.46 gm/cu cm)by
centrifugation. Pilot runs in a laboratory super centrifuge operating at 23,000 rpm
indicate that satisfactory clarification is obtained at a throughput of 5 lb/hr of solution.
This centrifuge has a bow 734 in. long internally with r2 =7/8 in, and(r2-r1)=19/32 in.
(a) Determine the critical particle diameter for this separation.
(b) If the separation is to done in the plant using a No,2 disk centrifuge with
50 dicks at 45В° half angle, what production rate could be expected?
4.** In the primary refining of vegetable oils, the crude oil is partially saporified
with caustic and the refined oil separated immediately from the resulting soap stock in
a centrifuge. In such a process, the oil has a density of 0.92 gm/cu cm and a viscosity
of 20 centipoises, and the soap phase has a density of 0.98 gm/cu cm and a viscosity
of 300 centipoises. It is proposes to separate these phases in a tubular-bow1 centrifuge
with a bow1 30 in long and 2 in. I.D. rotating at 18.000 rpm. The radius of the dam
over which the light phase flows is 0.500 in, whereas that over which the heavy phase
flows is 0.510 in.
(a) Determine the location of the liquid-liquid interface within the centrifuge.
7
(b) If this centrifuge is fed at a rate of 50 gal/hr with feed containing 10
volume percent soap phase, what is the critical droplet diameter of oil held in
the soap?
8
Chapter 4
Filteration
1.*
Ruth and Kempe report the results of laboratory filtration tests on a precipitate
of CaCO3 suspended in water A specially designed plate-frame with a single frame
was used. The frame had a filtening area of 0.283 sq ft and a thickenss of 1.18 in. All
tests were conducted at 66ЛљF and with a slurry containing 0.0723 weight fraction
CaCO3. The density of the dried cake was 100 lb/cu ft. Test results for one run are
given below:
P = 40 psi = constant
Volume of Filtrate, 1.
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
22
24
26
28
Time, sec
1.8
4.2
7.5
11.2
15.4
20.5
26.7
33.4
41.0
48.8
57.7
67.2
77.3
88.7
Determine the filtrate volume equivalent in resistance to the filter medium and piping
(Ve), the specific cake resistance (О±), the cake porsity (Оµ),and the cake specific
surface(So).
2.*
The results of laboratory tests on 6 in plate and frame filter press using two
frames, each 2 in thick and having a total active filter area of 1 sq-ft are given below.
A slurry of calcium carbonate in water was used.
Experimental data for constant pressure filtration
9
Time of filtration(sec)
0
26
98
211
361
555
788
1083
pressure difference across the press
wt ratio of wet cake to dry cake
p of dry cake
wt fraction of CaCO3 in slurry
viscosity of filtrate
ПЃ of CaCO3
ПЃ of filtrate
wt of filtrate (lb)
0
5
10
15
20
25
30
35
30 lb.f/in2
1.473
73.8 lb/ft3
0.139
2.07 lb/ft-hr
264 lb/ft3
62.2 lb/ft3
Determine (a) the value of Ve
(b) the value of the cake specific resistance О±
(c) the value of the porosity of the cake
(d) determine the time required to wash the cake if the washing is
carried out at 30 psi and the volume of wash water used is equal to
three times the void volume of the cake.
3.*
A homogeneous sludge forming a uniform noncompressible cake is filtered
through a batch leaf filter at a constant difference in pressure of 40 psi forming a 3/4 in
cake in 1 hr with a filtrate volume of 1500 gal. Three minutes are required to drain
liquor from the filter. Two minutes are required to fill the filter with water. Washing
proceeds exactly as filtration. using 300 gal. Opening dumping and closing take 6
min. Assume the filtrate to have the same properties as wash water, and neglect the
resistances of the filter cloth and flow lines.
(a) How many gallons of filtrate are produced on the average per 24 hr?
(b) How many gallons of filtrate would be produced if a cake of 1/2 in
thickness were formed, using the same ratio of wash water to filtrate with other
conditions the same?
4.** An open sand filter uses a 3-ft-deep bed of -20 +28 mesh sand as primary filter
bed. The sand particles used have an estimated spheroid of 0.9 . If the slurry being
filtered is essentially water and stands 2 ft deep over the top of the sand, determine the
maximum flow rate through the bed which occurs immediately after backwashing.
10
5.** A rotary drum filter with 30 % submergence is to be used to filter a
concentrated aqueous slurry of calcium carbonate containing 14.7 lb of solid cu-ft of
slurry. The pressure drop is to be 20 in Hg. If the filter cake contains 50 % moisture
(wet basis), calculate the filter area required to filter 10 gal/min of slurry when the
filter cycle time is 5 min. Assume that specific cake resistance is 18 x 1011 ft/lb.
Вµ of water
= 1 cp
ПЃ of water
= 62.3 lb/ft3
ПЃ of solid
= 164 lb/ft3
6.** A certain filter press when tested on a homogeneous slurry at constant rate
yielded the following data.
Filtrate, lb
Time, min
P1-P2, psi
0
0
1
6
10
3
8
30
5
10
50
6
11
60
10
15
100
20
25
200
30
35
300
If this press is to be operated at a constant pressure drop of 10 psi and the time
for cleaning and washing between cycles is 20 min, compute the optimum cycle time
(time for filtration, cleaning and washing) for 400 lb of filtrate.
7.*** A 30 by 30 in plate-and frame filter press with twenty frames 2.50 in, thick is
to be used to filter the CaCO3 slurry which was used in the test of problem 1. The
effective filtering area per frame is 9.4 sq ft, and Ve may be assumed to be the same as
that found in the test run. If filtration is carried out at constant pressure with (-в€†P)=40
psi, determine the volume of slurry that will be handled until the frames are full, and
the time required for this filtration.
8.*** The following table shows data obtained in a constant rate filtraion of a sludge
consisting of magnesium carbonate and water. The rate was 0.1 ft/sec, the viscosity of
the filtrate was 0.92 cp, and the concentration of solids in the slurry was 1.08 lb/ft3 of
filtrate. Evalurate , in foot , lbf , second units, the constants Rm, S , K1 and О±o for this
sludge.
11
-в€† P, psi
1, sec
4.4
5.0
6.4
7.5
8.7
10.2
11.8
13.5
15.2
17.6
20.0
10
20
30
40
50
60
70
80
90
100
110
9.*** A leaf filter press with 10 sq ft of filtering area operating at a constant
pressure of 40 psig gave the following results:
Time, min
Filtrate volume,ft3
10
141
20
215
30
270
45
340
60
400
The original slurry contained 10 % by weight of solid calcium carbonate and a
small amount of dissolved alkalinity in the water.
Determine the time required to wash the cake formed at the end of 70min of
filterring at the same pressure, using 100 ft3 of wash water.
12
Chapter - 5
Handling of Solids
1.*
A belt conveyor is required to deliver crushed limestone having a bulk density
of 75 lb/cu ft at the rate of 200 tons/hr. The conveyor is to be 200 ft between
centers of pulleys with a rise of 25ft. The largest lumps are 4 in an and
constitute 15% of the total. The conveyor will discharge over the end. For a
belt speed of 200 fpm, what is the minimum width of belt that can be used?
Calculate the horsepower for the drive motor.
2.*
What is the capacity of a fight conveyor of 12 by 24 in traveling at 100 fpm
and handling the crushed limestone. These materials are to be moved
horizontally a distance of 100 ft. Weight of light conveyor is 1.0 lb/in of width
per running foot . 10 x 24 in flight conveyor calculate the h.p required.
3.*
A screw conveyors is to be installed to convey 800 bushels is of wheat per
hour over a distance of 80 ft Determine the size (diameter speed ( revolutions
per minute) and horsepower requirements for the installation ( 1 bushel = 8
gallons, bulk density of wheat = 48 lb/ft3 )
13
Chapter - 6
Agitation and Mixing
1.*
A flat-blade turbine with six blades is installed centrally in a vertical tank. The
tank is 6 ft (1.83 m ) in diameter: the turbine is 2 ft (0.61 m ) in diameter and is
positioned 2 ft (0.61 m ) from the bottom of the tank. The turbine blades are 5
in (127 mm) wide. The tank is filled to a depth of 6 ft (1.83m) with a solution
of 50 percent caustic soda. at 150 F (65.6 В°C) which has a viscosity of 12 cP
and a density of 93.5 lb/ft3 (1498 kg/m3). The turbine is operated at 90r/min.
The tank is baffled. What power will be required to operate the mixer?
2.
What would the power requirement be in the vessel described in problem 1 if
the tank were unbaffled?
3.
The mixer of Problem 1 is to used to mix a rubber-later compound having a
viscosity of 1200 P and a density of 70lb/ft3 (1120 kg/m3). What power will be
requied?
WORKED OUT
EXAMPLES
Chapter - 1
Sedimentation and Thickening
1.*
A single batch-settling test was made on a limestone slurry. The interface
between clear liquid and suspended solids was observed as a function of time, and the
results are tabulated below. The test was made using 236 gm of limestone per liter of
slurry. Prepare a curve showing the relation ship between settling rate and solids
concentration.
Test Data
Time, hr
Height of Interface ,cm
0
0.25
0.50
1.00
1.75
3.0
4.75
12.0
20.0
36.0
32.4
28.6
21.0
14.7
12.3
11.55
9.8
8.8
Using the test data. The height of the interface (z) is plotted as a function of
time (Оё) (Figure)
Height of interface, cm
403020100Figure: Height of Interface as a Function of Time
From the solid concentration of the initial slurry
co zo = 236 x 36 = 8500 gm cm/l
From Equation 1.8,
c = 8500/Zi gm/ml
V, settling velocity y ,
cm/hr
The tangent to the curve at Оё = 2 hr, is found to have an intercept of z1 = 20
cm. The settling velocity at that time is the slope of the curve, dz/dОё = П… = 2.78 cm/hr,
and c =425 gm/l. Other points are obtained in the same way, tabulated in Table, and
plotted in Figure.
0
800
Solid conc, g/lit
Figure: Settling-Rate-Concentration Relationship
Оё,hr
Zi, cm
П… i,cm/hr
C,gm/l
0.5
1.0
1.5
2.0
3.0
4.0
8.0
36
36
23.8
20
16.2
14.2
11.9
15.65
15.65
5.00
2.78
1.27
0.646
0.158
236
236
358
425
525
600
714
2.*
A limestone-water slurry equivalent to that of problem 1 is fed to a thickener
at a rate of 50 tons of dry solids/hr to produce a thickened sludge of 550 gm limestone
per liter. For an initial slurry concentration of 236 gm limestone per liter of slurry
concentration of 236 gm limestone per liter of slurry specify the thickener area
required.
П… , cm/hr
cL, gm/l
1
cL
1
1
cL
cП…
LL cl
S
10
8
6
3
2
1
265
285
325
415
465
550
0.00377
0.00351
0.00307
0.00241
0.00215
0.00182
0.00195
0.000169
0.00125
0.00059
0.00033
0
5140
4740
4800
5090
6060
в€ћ
To determine the minimum value of LL cL / S, the data of Table 1.2 are plotted
in Figure 1.9 This plot yields a minimum value of
пЈ« LL c L пЈ¶
пЈ¬
пЈ· = 4730 cm / hr
пЈ¬ S пЈ·
1 / gm
пЈ
пЈё Mm
corresponding to П… = 6.9 cm/hr, and, from Figure 1.7, cL = 310 gm /l. Since
no solids leave in the overflow, a solids material balance (Equation 1.14) gives
LLcL= 50 tons/hr = 100,000 lb solids /hr
Now, 4730
S=
cm / hr
ft / hr
= 9.68
1 / gm
cu ft / lb
and
100,00
= 10,32.sq ft
9.68
LLCL/S
6200
4600
V, settling velocity
10
3.*
A slurry of calcium carbonate in water, containing 45 g of calcium carbonate
per liter, was allowed to settle in a 6.0 cm I.D glass cylinder. The height of the line
between clear liquid and zone B-the zone of relative constant solid concentration, was
measured as a function of time. The results are given below. Prepare a curve of
settling rate Vs concentration.
Time, min
0
100
200
300
400
500
Height, in
44
34
25
17
10
7
ZO
Co = 45
g/lit
Zo = 44
Z, in
Zi
ZL
in
0
ОёL
Оё, min
ОёL, min
100
200
300
400
450
ZL , in
34
25
17
10
Zi, in
43.5
43
38
27.8
20.5
VL, jn / min
CL , g/Lit
VL =
zi в€’ z L
ОёL
0.095
0.09
, in/min
cL=
co zo
, g / lit
zi
45.517
46.046
4.*
A batch sedimentation test was made with 5 micron silica particles in water at
86Вє F. A 4.5 cm I.D cylinder was used, and the initial slurry concentration was 0.125 g
of silica per cu cm of slurry. The data are given below. Prepare the corresponding
settling rate vs concentration curve.
Final height = 5.53 cm
Time, min
Height, cm
0
34
10.8
25
16.8
20
26.4
15
43.1
10
65.8
7.5
89.5
6.4
100
6.1
co = 0.125
zo = 34
g / cm3
cm
ZO
Zi
Z , cm
ZL
0
ОёL
ОёL , min
zL, cm
Оё , min
zi , zna
Г‚
Г‚
Г‚
Г‚
Г‚
Г‚
Г‚
Г‚
Г‚
Г‚
Г‚
Г‚
Г‚
Г‚
Г‚
vL =
zi в€’ z L
ОёL
Г‚
Г‚
Г‚
Г‚
Г‚
, cm/min
cL =
co z o
, g/lit
zL
Г‚
Г‚
Г‚
Г‚
Г‚
vL , cm/min
cL , g/lit
5.*
The data given below were obtained from a single batch sedimentation test on
an ore slurry. The true density of the solids in the slurry was 2.5 g/cu-cm, and the
density of the liquid was 1 g/cu-cm. Determine the diameter required for a thickener
to handle 100 tons of solids per day from a feed concentration of 64.5 g/liter to an
underflow concentration of 485 g/liter.
Concentration (g of solids/liter of slurry)
64.5
70.9
94.3
111.7
139.9
173.9
222.0
331.0
Lo co
co
cu
D
ПЃs
ПЃL
=
=
=
=?
=
=
100
64.5
485
2.5
g / cm3
1
g / cm3
tons / day
g / lit
g / lit
Settling rate (cm/hr)
139.9
103.6
71.9
49.4
27.1
16.8
10.0
6.4
cL , g / lit
vL (cm/hr)
64.5
139.9
70.9
103.6
94.3
71.9
111.7
49.4
139.9
27.1
173.9
16.8
222.0
10.0
331.0
6.4
vL
1
1
в€’
c L cu
1
1
в€’
c L cu
Г‚
Г‚
Г‚
Г‚
Г‚
Г‚
Г‚
Г‚
Г‚
Г‚
Г‚
Г‚
Г‚
Г‚
Г‚
Г‚
VL / ( 1/ CL вЂ“ 1 / CU )
cm/hr / lit/g
VL / ( 1/ CL вЂ“ 1 / CU )
min
VL, cm/hr
пЈ«
пЈ¬
пЈ¬ vL
1
пЈ¬ 1
пЈ¬c в€’c
u
пЈ L
A = 100
ПЂ
4
пЈ¶
пЈ·
пЈ· = пЈ«пЈ¬ Lo co пЈ¶пЈ· =
пЈ·
пЈ A пЈё min
пЈ·
пЈё min
ton lit / g
Г—
day cm / hr
D2 =
Г‚
D =
Г‚
ft
Г‚
cm / hr
lit / q
6.** Estimate the depth of the thickener required to perform the operation of
Example 1.2. The batch-settling test indicated a value of z в€ћ = 7.7 cm. The specific
gravity of the limestone is 2.09.
From the information of Problem 1 (must know), the following Table may be
prepared.
Time hr
z в€’ zв€ћ
zo в€’ zв€ћ
(z-z в€ћ )
0
0.25
0.50
1.00
1.75
3.0
4.75
12.0
20.0
в€ћ
28.3
24.7
20.9
13.3
7.0
4.6
3.85
2.1
1.1
0
1.0
0.871
0.739
0.470
0.247
0.162
0.136
0.0743
0.0389
0.0
Wg/WL
Wl /Wg
gm solid / 1, water
1/gm solid
236
236
236
236
392
525
644
850
0.000444
0.000444
0.000444
0.000444
0.00255
0.00191
0.00155
0.00118
The critical time is evaluated from Figure by determining determining the time
at a value of (zo + zo' ) /2. This time is found to be 0.80 hr. The final concentration is
specified by Problem 2 (must know) as 550 gm/l. From Table, a time of about 3.4 hr
is required to produce a concentration of solids equal to 550 gm/l. Thus the retention
time in the compression zone is (3.4-0.8)or 2.6 hr.
Z-Zв€ћ / ZO - Zв€ћ
ZO
0.6
0.2
ZOвЂ™
0
ОёC
Time , hr
14
Figure : Settling Characteristics
Equation 1.18 is used to evaluate the compression zone volume.
Lc
Lc
W
V = o o (Оё в€’ Оё ) + o o в€« ОёОё c 1 dОё
Ps
P
Ws
Using the data of Table, the integral of Equation 1.18 is graphically evaluated
and found to be 6.89 hr.
Loco = 100,000 lb/hr (from Example 1.2)
Ps
= 2.09 x 62.3 = 130 lb /cu ft
(Оё - Оёc) = 2.6 hr
Therefore, V =
100,000
100,000
Г— 2.6 +
Г— 6.89
130
62.3
V = 2000 + 11,000 = 13,000 cu ft
The thickener area is 10,320 sq ft from Problem2.
Depth of thickener.
Compression zone 13,000/10,320 в‰… 1.3 ft
Bottom pitch
2.0 ft
Storage capacity
2.0 ft
Submergence of feed
2.0 ft
Total depth
=7.3 ft
7.*** Waste water from a drinking plant is to be clarified by continuous
sedimentation. Feed to the thickener is one million gal per day containing 1.2 % by
weight solids. The underflow from the unit analyzes 8 % solids. Specify the depth and
diameter of the thickener.
A single batch - settling test on the feed material gave the following
information:
Specific gravity of solids
2.00
Specific gravity of solution
1.00
Concentration of solids in test
1.2 %
Time, min
0
5
Height of
31
21
Liquid-solid interface, cm
10
10
20
3.2
40
2.2
60
2.1
180
2.0
240
1.96
О±
1.94
LO
V
CO
Lu
Cu
Lo
= 106 gal / day
Co = 1.2 % solid =
Cu = 8 % solid =
1 .2 g
100 g / 1 g / cm
3
= 0.012 g/cm3
8
= 0.08 g/cm3
100 / 1
Zo = 31 cm
ZО± = 1.99 cm
Zo -ZО± = 29.06 cm
z,cm
0
ОёL, min
zL, cm
zi, cm
vL , cm/min
Оё, min
3
cL, g/cm
VL / ( 1/ CL вЂ“ 1 / CU )
cm/ min / cm3/g
VL / ( 1/ CL вЂ“ 1 / CU )
min
VL, cm/ min
1/cL
vL
1
1
в€’
c L cu
From fig
пЈ«
пЈ¬
пЈ¬ vL
1
пЈ¬ 1
в€’
пЈ¬
пЈ cL cu
l o co
=
A
A=
=
A=
D =
пЈ¶
пЈ·
пЈ·
=
пЈ·
пЈ·
пЈё min
Г‚
Г‚
cm / min
cm 3 / g
cm / min
cm 3 / g
g
105 Г— 0.012 gal
min
1day
1 ft 3
Г— 3Г—
Г—
Г—
day cm
cm 24 Г— 60 min 7.48 gal
ft2
ПЂ
4
D2
Г‚
Оё, min
z, cm
0
5
10
20
40
60
180
240
31
21
10
3.2
2.2
2.1
2.0
1.96
ft
z- zв€ћ , cm
z в€’ zО±
zo в€’ zО±
3
1/CL , cm /gal
ln (Z-Zв€ћ)/ (Z0- Zв€ћ)
0
Оёc =
Оёu =
Г‚
Г‚
ОёC
Оёu
Оё, min
ОёC
Оё, min
min
volume of compression zone =
loco
(Оё u в€’ Оё c ) + l o c o в€« ОёОёcu 1 dОё
ys
cL
= 105
gal
g
Г— 0.012 3 (в€’ )
day
cm
min Г—
1 day
cm 3
1 ft 3
Г—
Г—
2 g 24 Г— 60 min 7.48 gal
+ 105
gal
g
cm 3
Г— 0.012 3 Г— в€’ в€’ в€’ в€’
Г— minГ—
day
cm
g
1 day
24 Г— 60 min
Г—
1ft 3
gal =
7.48
ft 3
Chapter 2
Classification
8.*
Calculate the terminal velocity for sphericity droplets of coffee extract, 400
microns in diameter, falling through air. The specific gravity of the coffee
extract is 1.03, and the air is at a temperature of 300F.
In this problem, the terminal velocity can be calculated using Equation 2.12;
but since the velocity is unknown, CD cannot be directly evaluated. This
method would require a trial-and-error solution, but the problem is easily
solved using Equation 2.21. This equation will be plotted on Figure 2.1 using
the specified data. It will pass through the point NRc = 1.0 and
CD =[4gD3 p(ps-p )/ 3Вµ2] with slop of -2.
400 Г— 10 в€’4
= 1.31 x 10-3 ft
30.48
Вµair = 0.026 x 6.72 x 10-4 = 1.747 x 10-5 lb/ft sec
ps = 1.03 x 62.3 = 64.1 lb/cu ft
29 492
Г—
= 0.0524 lb/cu ft
pair =
359 760
Dp =
CD =
CD =
4 gDp 3 p ( ps в€’ p )
3Вµ 2
(4)(32.2)(0.00131)3(0.0524)(64.1)(0.0524)
(3)(1.747 Г— 10 в€’5 )2
CD = 611
On Figure 2.1a at (CD =611, NRe = 1.0), draw a line of slope of -2 At its intersection
with.П€ = 1.0, NRe =14.
Therefore, Dp П… tp / Вµ = 14,
and
П…t =
9.*
14 Вµ
14(1.747 Г— 105 )
=
= 3.57 ft / sec
D p P (1.31 Г— 10 в€’3 )(5.24 Г— 10 в€’2 )
A mixture of silica and galena is to be separated by hydraulic classification.
The mixture has a size range between 0.008 cm and 0.07 cm. The density of
galena is 7.5 gm/cu cm, and the density of the silica is 2.65 gm/cu cm.
Assume the sphericity, П€=0.806. (a) What water velocity is necessary for a
pure galena product? Assume unhindered settling of the particles and a water
temperature of 65 В°F.(b) What is the maximum size range of the galena
product?
(a) For equal-sized silica and galena particles, the heavier galena will settle
faster. Therefore, the settling velocity of the largest silica particle will
determine the water velocity.A water velocity equal to this settling velocity
should give a pure galena product . By Equation 2.21, using the metric system
of units.
пЈ« 4 gD p 3 ( p sil в€’ p ) пЈ¶
пЈ·
log CD = -2 log NRe + log пЈ¬
2
пЈ¬
пЈ·
3
Вµ
пЈ
пЈё
The viscosity of water at 65ЛљF = 0.01 poise
пЈ« (4 )(981)(0.07 )3 (1.0 )(2.65 в€’ 1) пЈ¶
пЈ·
Log CD = -2 log NRe - log пЈ¬пЈ¬
2
пЈ·
3(0.01)
пЈ
пЈё
Log CD = -2 log NRe + log 7400
This is a straight line on Figure 2.1 passing through (NRe = 1, CD = 7400) with a slope
of -2.
This line intersects the П€ = 0.806 curve at NRe = 28.
A Reynolds number of 28 corresponds to a settling velocity of
N Вµ (28)(0.01)
П… t = Re =
= 4.0 cm /sec
D p p (0.07 )(1.0 )
This velocity must also be the water velocity to a ensure a clean galena product, since
it will carry all silica overhead.
(b) calculation of the size of a galena particle that settles at a velocity of 4.0 cm/sec
fixes the smallest galena particle in the galena product. By Equation 2.22.
пЈ« 4 g ( p gat в€’ p )Вµ пЈ¶
пЈ·пЈ·
log CD = log NRe + log пЈ¬пЈ¬
3 p 2П… 3
пЈ
пЈё
пЈ« (4)(981)(7.5 в€’ 1.0)0.01пЈ¶
log CD = log NRe + log пЈ¬
пЈ·
пЈ
пЈё
(3)(1)2 (4.0)3
log CD = log NRe + log 1.33
This is a straight line on Figure 2.1 passing through (NRe = 1.0, CD= 1.33) with a slop
of + 1. This line intersects the П€ = 0.806 curve at NRe = 9.0. This Reynolds number
corresponds to a diameter of
Dp =
N Re Вµ
П…p
=
(9(0.01)) = 0.0225cm
(4.0)(1.0)
Thus, The galena product size ranges between 0.0225 cm and 0.07 cm. Galena
particles smaller than 0.0225 cm are carried overhead along with all silica.
10.*
Drops of oil 15 microns in diameter are to be settle from their mixture with air.
The specific gravity of the oil is 0.9, and the air is at 21ЛљC and 1 atm. A
settling time of 1 min is available. How high should the chamber be to allow
settling of these particles? Assume that the motion of the particle lies in the
StokesвЂ™-Law range.
oil
Dp = 15 Вµ
sp.gr of oil = 0.9
air , 21В°C , 1 atm
time = 1 min
StokeвЂ™s-law , range
D p2 g (ПЃ s в€’ ПЃ )
vt =
18Вµ
Dp = 15 x 10-4 cm
ПЃs = 0.9 g/cm3
Вµ = 0.018 x 10-2 g/cm-5
pv = nRT
R=
pv 1atm Г— 22.4 Г— 10 3 cm 3
=
nt
1 g.mol Г— 273K
ПЃair =
pM
RT
=
1atm Г— 29 g / g в€’ mol
1 Г— 22.4 Г— 10 3
Г— 294K
273
= 0.0012 g/ cm3
(15 Г— 10 )
vt =
Г— 980 Г— (0.9 в€’ 0.0012 )
= 0.611 cm/s
18 Г— 0.018 Г— 10 в€’.2
в€’4 2
height of chamber = 0.611 x 60 s = 36.66 cm = 1.2 ft
11.*
Air is being dried by being bubbled (in very small bubbles) through
concentrated sulfuric acid (specific gravity, 1.84, viscosity, 15 cp, temperature
100ЛљF). The sulfuric acid fills a 24 in tall, 2 in ID glass tube to a depth of 6 in.
The dry air above the acid is at a pressure of 0.8 atm and at 100ЛљF. If the dry
air rate is 3.5 cfm, what is the maximum diameter of a sulfuric acid spray
droplet which might be carried out of the apparatus by entrainment in the air
stream.
air
dry air
6 in
0.8 atm, 100ЛљF
S/A conc
sp.gr = 1.84
Вµ = 15 cp, 100ЛљF
QЛљ = 3.5 ft3 /min
vt - A = 3.5 ft3 / min
ПЂ
2
3.5 3
пЈ«2пЈ¶
ft / s
vt Г— Г— пЈ¬ пЈ· =
4 пЈ 12 пЈё
60
vt =
ft/s
ПЃp = 1.84 x 62.4 lb / ft3
PM
ПЃt =
=Г‚
RT
Вµf = Вµair
Dp
s/^
=?
log CD = log NRe + log
(
)
4g ПЃ p в€’ ПЃ f Вµ
3v ПЃ
3
t
f
2
f
at NRe = 1
DD =
(
)
4g ПЃ p в€’ ПЃ f Вµ f
3v ПЃ
3
t
2
f
=
Г‚
fig 2.1
NRe =
D p vПЃ
Вµ
Dp =
Г‚
=
Г‚
Г‚
12.* A thin water suspension of soil is prepared at 10 A.M. At noon, a sample is
drawn from the suspension at a depth of 5 cm. What is the largest particle
probably removed by a pipette at this depth (5 cm) if the specific gravity of the
soil is 2.5?
5cm,10A,
at noon
Sp.gr of soil =2.5
Dp=?
water
vt = 5 cm /2 hr
ПЃp = 2.5 g/ cm3
ПЃf = 62.4 lb/ft3
Вµf = 1 cp
Dp can be calculated same as problem 4.
13.** Particles of sphalerite (specific gravity = 4.0) are settling under the force of
gravity through a slurry consisting of 25% by volume of quartz particles
(specific gravity = 2.65) and water. The diameter of the sphalerite particles is
0.006 in. The volumetric ratio of sphalerite to slurry is 0.25. The temperature
is 50ЛљF. What is the terminal velocity of the sphalerite? What is the density of
slurry?
sphalerite
sp gr = 4
Dp = 0.006 in
50В°F
vol of sphalerite
vol of slurry
= 0.25
vt = ?
ПЃslurry = ?
ПЃslurry = 2.65 x 0.25 + 1 x 0.75 =
vH =
Fs =
slurry
quarty+ water
25% vol
sp gr=2.65
(
D 2p g ПЃ p в€’ ПЃ f
18Вµ f
)=F
Г‚
g/cm3
s
X2
101в€’82 (1в€’ X )
x = 0.75
Dp = 0.006 in x
2.54cm
= 0.006 x 2.54cm
1in
ПЃp = 4 g/cm3
Вµf = 1.31 cp = 1.31 x 10 -2 g/cm.s
ПЃf = 62.42 lb/ft3 x
Fs =
0.752
101.82 (1в€’0.75)
1g / cm 3
= 1 g/cm3
62.37lb / ft 3
= 0.197
(0.006 Г— 2.54) 2 Г— 980( 4 в€’ 1)
Г— 0.197
18 Г— 1.31 Г— 10 в€’2
= 0.57 cm/s
vH =
14.** Calculate the settling velocity for the hindered settling of glass sphere in water
at 68ЛљF when the suspension contains 1206 g of glass spheres in 1140 cm3 of
total volume. The average diameter of the spheres, as determined from
photomicrographs, was 0.0061 in and the true density of the spheres was 154
lb/ft3.
gass sphere
Dp =0.0061 in
ПЃp = 154 lb /ft3
vH = ?
68ЛљF
water
ПЃt = 62.31 lb/ft3
Вµt = 0.982 cp
1206
154 / 62.4 = 0.571
1140
1140 в€’
x=
Fs =
vH =
x2
101в€’82 (1в€’ x )
(
==
D 2p g ПЃ p в€’ ПЃ f
18Вµ f
0.5712
=
101.82 (1в€’0571)
Г‚
) .F
s
(0.0061 / 12) 2 Г— 32.2(154 в€’ 62.32)
18 Г— 0.982 Г— 6.72 Г— 10 в€’ 4
= 0.0034 ft /s
=
x
Г‚
15.*** Urea pellets are made by spraying drops of molten urea into cold gas at the top
of a tall tower and allowing the material to solidify as it falls. Pellets 1/4 in. in
diameter are to be made in a tower 80 ft high containing air at 70ЛљF. The
density of urea is 83 lb/cu-ft.
(a) What would be the terminal velocity of the pellets, assuming free-settling
conditions?
(b) Would the pellets attain 99 % of this velocity before they reached the
bottom of the tower?
molten urea
80 ft
air
70ЛљF
(a)
urea pellets
Dp = 1/4 in = 1/48 ft
ПЃurea = 83 lb/ft3
vt = ?
Dp = 1/48 ft
Вµf =
Г‚
ПЃp = 83 lb/ft3
PM
RT
ПЃf =
log cD = -2 log NRe + log
4D 3p g (ПЃ p в€’ ПЃ f )ПЃ f
3Вµ f2
at NRe = 1
4D 3p g (ПЃ p в€’ ПЃ f )ПЃ f
cD =
3Вµ f2
cD
slope = -2
П€ =1
cD
1
NRe = Г‚
NRe
Fig 2.1
NRe =
D p vt ПЃ
Вµ
vt
(b)
=
Г‚
=
Г‚
80 ft vp
vt
> 0.99 would attain,
not attain
v = 0.99 vt
0 ft
a=
vb
dv
dt
dПЃ . a =
dv
dz
dt
dПЃ . a = v
0
80
в€« dПЃ =
Vb
0
dv
в€«
v
a
dv = -80
в€‘F = mpa = FE - FB - FR
C D A pПЃf v 2
= mp g - mf g -
2
c D ПЂ / 4 D 2p ПЃ f v 2
mp
mpa = mpg ПЃt g в€’
2
ПЃp
mp =
a =
ПЃp в€’ ПЃf
ПЃp
ПЂ
6
.g в€’
D p3 ПЃ p
3c D ПЃ f v 2
4D p ПЃ p
3 Г— 0.075c D v 2
83 в€’ 0.075
Г— 62.4 в€’
1
83
4 Г— Г— 83
48
DpvПЃ f в€’ в€’ в€’ Г— v
NRe , p =
=
в€’в€’в€’
Вµf
a=
Assume
v c ft/s
NRe
0
cD
0
a
v
a
0
Г‚
0
vt
v
a
в€ґ vb =
Г‚
- 80
0
v
vb vt
vb
=
vt
Г‚
16.*** Quartz and pyrites are separated by continuous hydraulic classifications. The
feed to the classifier ranges in size between 10 microns and 300 microns.
Three fractions are obtained: a pure-quartz product, a pure-pyrites product, and
a mixture of quartz and pyrites. The specific gravity of quartz is 2.65 and that
of pyrites is 5.1. What is the size range of the two materials in the mixed
fraction for each of the following cases.
(a) The bottoms product to contain the maximum amount of pure pyrites.
(b) The overhead product to contain the maximum amount of pure quartz.
Q& p
Q+P
10 to 300Вµ
- pure Q
- pure P
- Q& p
H2 O
P (pure)
sp.gr of Q = 2 .65
p = 5.1
(a)
total quart 2 in upstream for pure pyrite
for quarty
Dp = 300 x 10-4 cm
ПЃp = 2.56 g/cm3
ПЃf =
g/cm3
1
Вµf = 1 x 10-2 gl cm.s
vt = ?
log cD = - 2 log NRe + log
4 D 3p g (ПЃ p в€’ ПЃ )ПЃ
at NRe = 1
cD =
4 D p3 g (ПЃ p в€’ ПЃ )ПЃ
3Вµ 2
=
Г‚
3Вµ 2
slope = -2
cD
cD
ПЃ=1
1
NRe =
Г‚
NRe
NRe =
ПЃp
Вµ
vt =
For pyrite
vt =
ПЃ
vt
=
Г‚
Г‚
Dp = ?
cm/s
Г‚
g/cm3
ПЃp = 5.1
log cD = log NRe + log
4 g ( ПЃ p в€’ ПЃ )Вµ
3vt3 ПЃ 2
at NRe = 1
cD =
4 g ( ПЃ p в€’ ПЃ )Вµ
3vt3 ПЃ 2
slope = 1
cD
cD
ПЃ = 1
NRe = 1
NRe =
Dp
vt
Вµ
ПЃ
=
Г‚
Dp = Г‚
10 to 300 Вµ for quart 2
10 to _ Г‚ _ _ Вµ for pyrite
(b)
Q& p
Q (pure) 10 to Dp
10 to 300Вµ
H2 O
P + Q Dp to 300 Вµ
For pure quartz, total pyrite in downstream
For pyrite
DP = 10 Вµ = 10-3 cm
ПЃP = 5. 1 g/cm3
vt = ?
4 D p3 g (ПЃ p в€’ ПЃ )ПЃ
log cD = - 2 log NRe + log
3Вµ 2
at NRe = 1 = >
cD =
4 D p3 g (ПЃ p в€’ ПЃ )ПЃ
=
3Вµ 2
Г‚
slope = -2
cD
cD
П€ =1
NRe =
1
Г‚
NRe
NRe =
ПЃp
ПЃ
vt
Вµ
vt =
Г‚
For quartz
vt =
Г‚
ПЃp = 2.65
Dp = Г‚ ?
log cD = log NRe + log
at NRe = 1
(
)
4g ПЃ p в€’ ПЃ Вµ
3v ПЃ
3
t
2
f
=
Г‚
cD =
(
)
4g ПЃ p в€’ ПЃ Вµ
3v 3t ПЃ f2
slope = 1
cD
cD
ПЃ = 1
NRe = 1
NRe =
Г‚
Dp
vt
Вµ
ПЃ
=
Г‚
Dp = Г‚
10 to 300 Вµ for pyrite
Г‚
to 300 Вµ for quartz
17.*** A mixture of coal and sand in particle size smaller than 20 mesh is to be
completely separated by screening and then elutriating each of the cuts from
the screening operation with water as the elutriating fluid. Recommend a
screen size such that the oversize cut can be completely separated into coal and
sand fractions by water elutriation. What water velocity will be required? The
specific gravity for sand and coal is 2.65 and 1.35 respectively.
s&s
0
Dp, max = 20 mesh
coal + sand
0
< 20 mesh
sp.gr = 1.35
u
screen size = ?
> Dps
for completely separated
vt = ?
For coal
Dp = 20 mesh = 0.0833 cm
ПЃp = 1.35 g/cm3
1
g/cm3
ПЃf =
Вµt = 10-2 g/cm .s
vt = ?
H2o
sand
sp.gr=2.65
cD =
NRe =
cD =
cD =
4D p g (ПЃ p в€’ ПЃ f )
3v 2t ПЃ
Dp
ПЃ
vt
Вµ
(
)
4D P g ПЃ p в€’ ПЃ ПЃ
3ПЃ
2
Г—
4 D p3 g (ПЃ p в€’ ПЃ )ПЃ
3Вµ
D p ПЃ2
N 2Re Г— Вµ 2
.
2
1
2
N Re
4 D p3 g (ПЃ p в€’ ПЃ )ПЃ
log cD = -2 log NRe + log
at NRe =1
cD =
=
3Вµ 2
4 D p3 g (ПЃ p в€’ ПЃ )ПЃ
3Вµ 2
4 Г— 0.08333 Г— 980(1.35 в€’ 1)1
3 Г— (10
)
в€’2 2
= 0.26 Г— 10 4
П€=1
NRe , p = 30
Dp
vt
Вµ
vt = Г‚
For sand
Dp = ?
vt =
Г‚
log cD = log NRe + log
at NRe = 1
cD =
4 g ( ПЃ p в€’ ПЃ )Вµ
3vt3
ПЃ
=
Г‚
4 g (ПЃ p в€’ ПЃ )Вµ
3vt3 ПЃ
ПЃ
=
Г‚
(From Fig 2.1)
slope = 1
cD
cD
П€= 1
NRe = Г‚
1
NRe
NRe
=
Dp
vt
Вµ
ПЃ
=
Г‚
Dp = Г‚
18.*** A mixture of spherical particles of silica contains particles ranging in size from
14 mesh to 200 mesh. This mixture is to be divided into two fractions by
elutriation, utilizing the upward velocity of a stream of water at 55ЛљF rising
through a tube 4 in in diameter.
(a) What quantity of water, in gpm, will probably be needed to divide the
mixture at a size equal to the aperture of a 48-mesh screen?
(b) When the water flow through the tube is 1.5 gpm, what is the smallest size
of silica particle which will probably settle through the stream? The specific
gravity for silica is 2.65.
silica
200-48
200-14
4 in dia
H2 O
55ЛљF
sp - gr = 2.65
48-14
(a) Q = ? g pm
Dp = 48 mesh
ПЃp = 2.65 x 62.4 lb /ft3
ПЃf =
Г‚
Вµf = Г‚
vt = ?
ПЂ
Q = vt . A = vt x
D2 =
4
4 D p3 g (ПЃ p в€’ ПЃ )ПЃ
log cD = -2 log NRe + log
3Вµ 2
at NRe = 1
cD =
4 D p3 g ( ПЃ p в€’ ПЃ ) ПЃ
=
3Вµ 2
cD
Г‚
slope = -2
cD
П€ =1
1
NRe = 1
NRe
NRe = Г‚
Dp
vt
ПЃ
=
Вµ
vt =
Г‚
Г‚
Q= vt x
ПЂ
4
D2 =
(l)
200-14
H 2O
200-Dp
Dp-14
Qв€† = 1.5 gpm
vt . A = 1.5 gal/min
1.5
1 ft 144 1
Г—
Г—
Г—
vt =
ПЂ
7
.
48
1
cos
2
Г—4
4
= 0.0383 ft/s
Г‚
Dp = ?
П†=1
ПЃp = 2.65 x 62 .4 lb/ft3
log cD = log NRe + log
at NRe = 1
cD =
4 g (ПЃ p в€’ ПЃ )Вµ
3vt3 ПЃ 2
4 g (ПЃ p в€’ ПЃ )Вµ
3 vt3
=
slope = 1
cD
1
NRe =
Г‚
NRe
Fig = 2.1
Dp
vt
Вµ
ПЃ
=
Dp = Г‚
Г‚
Г‚
Chapter 4
Filteration
19.* Ruth and Kempe report the results of laboratory filtration tests on a precipitate
of CaCO3 suspended in water A specially designed plate-frame with a single frame
was used. The frame had a filtening area of 0.283 sq ft and a thickenss of 1.18 in. All
tests were conducted at 66ЛљF and with a slurry containing 0.0723 weight fraction
CaCO3. The density of the dried cake was 100 lb/cu ft. Test results for one run are
given below:
P = 40 psi = constant
Volume of Filtrate, 1.
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
22
24
26
28
Time, sec
1.8
4.2
7.5
11.2
15.4
20.5
26.7
33.4
41.0
48.8
57.7
67.2
77.3
88.7
Determine the filtrate volume equivalent in resistance to the filter medium and piping
(Ve), the specific cake resistance (О±), the cake porsity (Оµ),and the cake specific
surface(So).
20.* The results of laboratory tests on 6 in plate and frame filter press using two
frames, each 2 in thick and having a total active filter area of 1 sq-ft are given below.
A slurry of calcium carbonate in water was used.
Expermental data for constant pressure filtration
Time of filtration(sec)
0
wt of filtrate (lb)
0
26
5
Chapter - 5
Handling of Solids
1.*
A belt conveyor is required to deliver crushed limestone having a bulk density
of 75 lb/cu ft at the rate of 200 tons/hr. The conveyor is to be 200 ft between
centers of pulleys with a rise of 25ft. The largest lumps are 4 in an and
constitute 15% of the total. The conveyor will discharge over the end. For a
belt speed of 200 fpm, what is the minimum width of belt that can be used?
Calculate the horsepower for the drive motor.
Solution
minimum width of belt = 14 in
Data
friction factor =
0.03
Lo
= 150
wt of conveyor = 30
lb/ft
h.p =
F(L + L o )(T + t 0.03ws) + Tв€†z
990
0.03( 200 + 150)( 200 + 0.03 Г— 30 Г— 200) + 200 Г— 25
940
= 52 h.p
=
2.*
What is the capacity of a fight conveyor of 12 by 24 in traveling at 100 fpm
and handling the crushed limestone. These materials are to be moved
horizontally a distance of 100 ft. Weight of light conveyor is 1.0 lb/in of width
per running foot . 10 x 24 in flight conveyor calculate the h.p required.
Solution
Constant for material = 0.6
"
conveyor = 0.04
bulk density of wheat = 75 lb /ft3
BDSОґ b 10 Г— 24 Г— 100 Г— 75
T=
= 300tons / hr
=
6000
6000
aTL + bWLs + 10L
h.p =
1000
0.6 Г— 300 Г— 100 + 0.04 Г— 24 Г— 100 Г— 100 + 10 Г— 100
=
1000
= 28.6 h.p
3.*
A screw conveyors is to be installed to convey 800 bushels is of wheat per
hour over a distance of 80 ft Determine the size (diameter speed ( revolutions
per minute) and horsepower requirements for the installation ( 1 bushel = 8
gallons, bulk density of wheat = 48 lb/ft3 )
Solution
size = 8 in
speed = 180 rpm
capacity = 800 x8
h.p =
gal
1ft 3
Г—
= 855.6 ft 3 /hr
h 7.48gal
(coefficient )(capacity, lb / min )(length, ft )
3300
1.3 Г— 85.5ft 3 / hr Г— 48lb / ft 3 Г— 1hr / 60 min Г— 80ft
3300
= 2.16 h .p
=
Chapter - 6
Agitation and Mixing
tajccHar;cGef;rsm;(must know)
Problems
1.*
A flat-blade turbine with six blades is installed centrally in a vertical tank. The
tank is 6 ft (1.83 m ) in diameter: the turbine is 2 ft (0.61 m ) in diameter and is
positioned 2 ft (0.61 m ) from the bottom of the tank. The turbine blades are 5
in (127 mm) wide. The tank is filled to a depth of 6 ft (1.83m) with a solution
of 50 percent caustic soda. at 150 F (65.6 В°C) which has a viscosity of 12 cP
and a density of 93.5 lb/ft3 (1498 kg/m3). The turbine is operated at 90r/min.
The tank is baffled. What power will be required to operate the mixer?
Solution
1.
Curve A in Fig. 6.5 applies under the conditions of this problem. The
Reynolds number is calculated. The quantities for substitution are. in
consistent unit.
90
Da = 2 ft
n =
= 1.5 r/s
60
Вµ = 12 x 6.72 x 10-1 = 8.06 x 10-3 lb/ft-s
ПЃ = 93.5 lb /ft3 ,
g = 32.17 ft/s2
Then
NRe =
D a2 nПЃ 2 2 x1.5 Г— 93.5
=
= 69.600
Вµ
8.06 Г— 10 в€’3
From curve A (Fig 6.5). for NRe = 69.600. NRe = 5.8 and from Eq. (6.11)
5.8 Г— 93.5x1.53 Г— 2 3
= 1821ft в€’ lb.s
32.17
The power requirement is 1821/550 = 3.31 hp (2.47 kw).
ПЃ=
2.
What would the power requirement be in the vessel described in problem 1 if
the tank were unbaffled?
Solution
Curve D of Fig 6.5 now applies. Since the dashed portion of the curve must be
used. the Froude number is a factor, its effect is calculated as follows:
NRe =
nDa 1.5 2 Г— 2
=
= 0.14
g
32.17
From Table 61, the constants a and b for substitution into Eq. (6.10) are a =1.0
and b= 40.0 From Eq (6.10)
Do в€’ log10 69,600
m=
= 0.096
40.0
The power number read from Fig 6.5, curve D.for NRe = 69.600 , is 1.07 ; the
corrected value of NRe = is 1.07: 0.14 -0.096 =1.29. Thus, from Eq(6.11)
1.29 Г— 93.5 Г— 1.53 Г— 2 3
= 406 ft в€’ lb ds
32.17
The power requirement is 406/550 = 0.74 hp (0.55 KW)
It is usually not good practice to operate an unbaffled tank under these
conditions of agitation.
3.
The mixer of Problem 1 is to used to mix a rubber-later compound having a
viscosity of 1200 P and a density of 70lb/ft3 (1120 kg/m3). What power will be
requied?
Solution
The Reynolds number is now
p=
2 2 Г— 1.5 Г— 70
= 5.2
1200 Г— 0.0672
This is well within the range of laminar flow. From Table 6.3 KL=65: from
Eq.(6.12)NRe = 65/5.2=12.5 and
NRe =
1.25 Г— 70 Г— 1.53 Г— 2 3
= 2938 ft в€’ lbds
32.17
The power required is 2938 =5.34 hp (3.99kw). This power requirement is
independent of whether or not the tank is battled. There is no reason for baffles
in a mixer operated at low Reynolds numbers, as vortex formation does not
occur under such conditions.
p=
Note that a 10.000-folde increase in viscosity increases the power by only
about 60 percent over that required by the baffled tank operating on the lowviscosity liquid.

Download Report