MOMENTUM LINKAGE PROBLEM NOS. 1 & 2: 1. Calculate the equivalent volume sphere diameter o f a cuboid particle of side length 1,2,4 mm a.
1.71 mm
b. 1.14 mm
c. 2.48 mm
d. 3.50 mm
GIVEN: Sides: 1, 2, 4 mm REQUIRED: Equivalent Volume Sphere diameter SOLUTION:
V cubcuboid = 1x2x 1x2x8484== 8m8mmm V sphe spherere== 43 πr 8 = 43 πr r = 1.2407 mm ; d = .
2. Calculate the surface-volume equivalent sphere diameter of the same cuboid particle (side length 1, 2, and 4 mm).
a. 1.71 mm
b. 1.14 mm
c. 2.48 mm
d. 3.50 mm
Given:
1 mm
4 mm 2 mm
Required: Dsv Solution:
D = D = (V); V =LWH=1mm =LWH=1mm2mm 2mm4mm 4mm =8mm D = 8mm = 2.4814814 mm D = ; S =2LW+2LH+2WH=2 =2LW+2LH+2WH=21 × 2 + 21 × 4 + 22 × 4 =28mm
D = =2.9854 mm = .. =. LINKAGE PROBLEM NOS. 3 & 4: 3. A suspension in water of uniformly sized spheres of diameter 100µm and density 1200 kg/m3 has a solids volume fraction of 0.2. The suspension settles to a bed of solids volume fraction of 0.5. The single particle terminal velocity of the spheres in water may be taken as 1.1 mm/s Calculate the velocity at which the clear water/suspension interface settles. a. b. c. d.
0.42 mm/s 0.31 mm/s 0.25 mm/s 0.39 mm/s
Given: D = 100µm
= 1200 kg/m3
Solids concentration of initial suspension, C B= 0.20 ut = 1.1 mm/s Reqd: velocity at which the clear water/suspension interface settles Soln: Uint,AB =
AA−BB − ; =0→ ,=
UpB is the hindered settling velocity of particles relative to the vessel wall in batch settling and is given by the equation: Up = UTεn Check whether Stoke’s Law applies or not by solving for the Reynold’s number (limiting value is 0.3): Re =
µ =0. 1 1→ , ℎ =4.65
Calculate for the voidage of the initial suspension: εB = 1 – CB = 1 – 0.2 = 0.8 Up = UTεn = 1.1 x 10 -3 m/s (0.80)4.65
Up = 0.39 mm/s
4. Calculate the velocity at which the sediment/suspension interface rises. a. 0.21 mm/s b. 0.36 mm/s c. 0.26 mm/s d. 0.39 mm/s The velocity of the interface between initial suspension (B) and sediment (S) istherefore:U int,
nat sediment i d s conc′ n at i n i t i a l suspensi o nvel o ci t y at sedi m entsol i d s conc U at interface= Velocity at initial suspensiSoloinSol ds concnat initial suspensionsolids concentration at sediment
o n0. 2 00 u at interface= velocity at initi0.al2suspensi = 0.6667velocity at initial suspension 00.5
From No.3 velocity at initial suspension is 0.3897 mm/s.
Velocity at interface= (0.3897)(-0.6667) = - 0 .2598 mm/s (Negative sign signifies that interface is moving upward) ANS: 0.26 mm/s
5. Calculate the velocity at which the sediment / suspension interface rises. a. 0.21 mm/s b. 0.36 mm/s c. 0.26 mm/s d. 0.39 mm/s
uint= −− − uint= .. .−. Uint = 0.2598 mm/s
LINKAGE PROBLEM NOS. 5-6 & 7-11: Spherical particles of uniform diameter 40 μm and particle density 2000 kg/m3 form a suspension of solids volume fraction 0.32 in a liquid of density 880 kg/m 3 and viscosity 0.0008 Pa-s. Assuming Stoke’s Law applies 5. Calculate the sedimentation velocity for this suspension a. b. c. d.
0.345 mm/s 0.30 mm/s 0.203 mm/s 0.033 mm/s
Given: Dp = 40 µm
ρp = 2000 kg/m3
vol. fraction = 0.32
ρL = 880 kg/m3
Required: sedimentation velocity Solution: Assuming Stoke’s Law Applies
U = UACCAA CUBBCB CA =0; U =UB =Uε U−w = gDp 18μρ ρ 2000880 U = 9.8140x10180.−0008 −m/s U =1.2208x10 UB =1.2208x10− −10.32. UB = 2.0315x10s m =. .
6. Calculate its sedimentation volumetric flux.
a. b. c. d.
0.065 mm/s 0.076 mm/s 0.065 cm/s 0.076 cm/s
SOLUTION: Ups=Up(1−ε) Up = 0.203 mm/s (FROM #5)
ε = 1-CB = 1 – 0.32 = 0.68 Ups = (0.203 mm/s)(1-0.68) = 0.0650 mm/s
A height-time curve for the sedimentation of a suspension in a cylindrical vessel is shown below. The initial concentration of the suspension for this test is 0.12 m 3/m3. 7. Calculate the velocity of the interface between clear liquid and a suspension of concentration, 0.12 m3/m3. a. 1.11 cm/s b. 0.30 cm/s c. 1.27 cm/s d. 0.52 cm/s
Solution: U – Velocity of the interface (slope of the straight line portion)
cm U= 22.2550 = 1. 1 1 50 s
8. Calculate the velocity of the interface between clear liquid and a suspension of concentration,0.2 m3/m3. a. 0.506 cm/s b. 0.514 cm/s c. 0.345 cm/s d. 0.122 cm/s Given: C = 0.2 Cb = 0.12 Ho = 50 cm3 Req’d: velocity of the interface between clear liquid and a suspension Sol’n:
1= 0.120.2×50 =30
*drawing tangent line through t=0, H1 = 30 cm3 to the curve locates the point on the curve corresponding to 0.2 m 3/ m3. Coordinates of the point: t = 35 sec and h = 18 cm 3
Slope = U1 = 1830 = 0. 3 429 350
(-) moving downwards
Ans: C. 0.345 cm/s downwards 9. Calculate the velocity at which a layer of concentration, 0.2m3/m3 propagates upwards from the base of the vessel.
a. b. c. d.
0.514 cm/s 0.122 cm/s 0.506 cm/s 0.345 cm/s upward propagation velocity of this layer = h/t = 18/35 = 0.514 cm/s
10. Calculate the concentration of the final sediment. a. b. c. d.
0.2 0.4 0.6 0.8
***Draw a tangent to the part of the curve corresponding to t he final sediment and projecting it to the h-axis h1s = 15 cm C0h0 = C1sh1s C1s = (0.12* 50)/ 15 = 0.4 m3/m3 Answer: 0.4 11. Calculate the velocity at which the sediment propagates upwards from the base. (Given: Hei ghttime curve) a. b. c. d.
0.30 cm/s downwards 0.30 cm/s upwards 0.41 cm/s downwards 0.41 cm/s upwards
Given: Height-time curve
Required: Velocity @ w/c sediment propagates upwards from t he base
Solution: Settling distance at 25s from the base = 22.5 cm – 15 cm = 7.5 cm
V= dt V= 7.25s5 cm
V = 0.3 cm/s upward 12. A slurry with a density of 2000 kg/m3, a yield stress of 0.5 N/m2, and a plastic viscosity of 0.3 Pa is flowing in a 1.0 cm diameter pipe which is 5m long. A pressure driving force o f 4 kPa is being used. Calculate the flow rate of the slurry. a. b. c. d.
5.60 x 10-4 m3/s 4.11 x 10-4 m3/s 4.37 x 10-7 m3/s 4.05 x 10-7 m3/s
Given: density of slurry = 2000 kg/m 3 Yield stress = 0.5 N/m 2 Viscosity = 0.3 Pa D= 1 cm
, L= 5 m
P=4kPA Required: flow rate of the slurry Solution:
Ʈ Ʈ Ʈ + Ʈ Ʈ ∆ .. =2.0 /2 . .. / . . + . . .
Uav=
Ʈ0=
(1-
=
Uav=
(1-
Q= Uav
= (.0056 m/s) (
= .0056 m/s
= 4.37 x
−/
13. A packed bed of solid particles of density 2500 kg/m3 occupies a depth of 1 meter in a vessel of cross-sectional area 0.04m2. The mass of solids in the bed is 50 kg and the surface-volume mean diameter of the particles is 1mm. A liquid of density 800 kg/m3 and viscosity 0.002 Pa-s flows upward through the bed, which is restr ained at its upper surface. Calculate the voidage of t he bed. a. b. c. d.
0.45 0.71 0.50 0.80
Solution:
=1 . =0.5 14. Calculate the frictional pressure drop across t he bed when the volume flow rate of liquid is 1.44 m3/hr. Use Ergun Equation
a. b. c. d.
6560 Pa 7070 Pa 6650 Pa 7700 Pa
Solution
= 0.5
U=
. . . . . +1. 7 5−} h{150 µ− ∅ ∅ . −. . −. . 1 m150 . . +1.75 . . = 0.01
Re =
∆ ∆ ∆
= 4
P =
P =
P = 6560 Pa
15. A mixture of quartz and galena of a size range from 0 .015mm to 0.065mm 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 7500 kg/m3 and the density of quartz is 2650 kg/m3.
a. b. c. d.
2377 kg/m3 1960 kg/m3 3100 kg/m3 1190 kg/m3
Given: Ρgalena = 7500 kg/m3 Ρquartz = 2650 kg/m3 Size range = 0.015 mm – 0.065 mm
Required: minimum apparent density of the fluid to give separation Solution: let A-galena; B-quartz
= 0.0.006515 = 2650 7500 =77772. 8 125 / ℎ 0.0.006515 = 7500 2650 =. / larger particles as heavier
16. The capacity (in tons/h) of the flight conveyor of 10 by 24 inches travelling at 100 fpm and handling the crushed limestone is a. b. c. d.
300 350 400 450
Given:
D = 10 in B = 24 in
S = 100 fpm
Required:
Capacity (T) in tons /hr
Solution: Crushed limestone bulk density from 95 to 103 lb/ft3 from table 2 -120 and 2-326 Pb = 995+103)/2 = 99 lb/ft 3 T= T=
∗∗∗
= 396 tons/hr
Final answer = 400 tons/hr 17. What is the horsepower requirement of a 45 ft length scr ew conveyor which will handle 20 tons/hr of a material with average density of 50 lb/ft 3. a.
3.64
b. 6.82
c. 5.23 d. 7.5
GIVEN: Length= 45 ft Capacity=20 tons/hr ρ = 50 lb/ft3 Required: Horsepower requirement SOLUTION: HP=
, , , ,
HP=
HP= 3.6364 18. Find the horsepower requirement fo r a continuous bucket elevator with loading leg which will lift solids at a rate of 50 tons per hour at a vertical distance of 22 ft. a. 2 b. 3 c. 4 d. 5 SOLN:
25022 HP= 2TΔZ = 1000 1000 =. ≈ 19. A glass sphere having a diameter of 1.554 x 10-4 m in water at 293.2 K and the slurry contains 60 wt% solids. The density of the glass spheres is 2467 kg/m3. The settling velocity of glass sphere in m/ s. a. b. c. d.
3.1x10-2 1.5x10-3 5.1x10-4 0.265
Given: D = 1.554 x 10-4 m T = 293.2K Density = 2467 kg/m 3 Required: Ut
40 ∈= 1000 401000+ 2467 60 =0.622 1 φ = =0. 2 05 . −. 10 − 9. 8 11. 5 54x10 246710000. 6 22 x0. 2 05 − U = =. / 180.001
Solution:
20. What is the porosity of a solid if tis bulk density is 1125 kg/m 3 and its true density is 1500 kg/m3?
a. b. c. d.
0.25 0.33 0.75 1.33
Solution: True Density =
−
1125 =. 1500 m3kg = 1
21. If the total percentage of particles larger than the screen opening in the feed, oversize, undersize are 36%, 89%, and 3% respectively, the effectiveness of the screen if the undersize is the product is a. b. c. d.
0.98 0.76 0.65 0.89
Given: XF = 0.36 XP = 0.89 XR = 0.03 Required: Effectiveness Solution:
1XpXfXr E= XpXfXr [ XfXpXr 1XfXpXr]
8 9. 3 6. 0 3 E= ..839.6.386.9.00331. 1. 3 6.89.03 E =.8861
E = 0.89 22. The wire diameter of a 10-mesh screen whose aperture is 0.065 inch is a. 0.025’’
c.0.045’’
b. 0.035’’
d. 0.05’’
Given: Mesh = 10
Aperture= 0.065 inch
Solution: 1 = Mesh (Aperture + Wire Diameter) 1 = 10(0.065 + W d)
Wd = 0.035’’ 23. If 20% of pulverized limestone (work index= 12.74) is retained by a 150 mesh scr een from an original uniform size of 35 mesh, the e nergy required in kW to process 1 ton/h. a. 2.65
b. 3.80
d. 8.30
c. 6.25
Mesh 35= .0176 inch = 0.417 mm Mesh 150= 0.104 mm
√ . √ .
P= 1 ton/hr (.3162)(12.74)(
)
P= 6.2532 kW= C 24. Calculate the power needed to reduce 100 ton/h of quartz (work index = 17.5 kW-hr/ton) in a grinding ball mill from an 80% passing size of 2 inches to an 80% passing size of 1/8 inch
a. b. c. d.
11.4 hP 17.8 hP 23.1 hP 27.5 hP
Given: 100 ton/hr Wi = 17.5 kW-hr/ton F = 2 inches = 50800 P = 1/8 inches = 3175
Required: W Solution: Using Bond’s law
=10 √ 1 √ 1 1 1 =10 17.5 ℎ √ 3175 √ 50800 1 ℎ ℎ ℎ =2.3293 ℎ =3. 1 237 100 0.7457 ℎ =312.3657 ℎ 25. 6000 lb of a material goes through a crusher and grinder per hour in succession (on 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 ft per lb. the Rittingers number of the material processed is 163 in2 per ft-lb. the efficiency of thecrusher is estimated to be 25%, while that of the grinder is 30%. Estimate the total power to be delivered to the equipment. a. b. c. d.
18.5 hP 26.4 hP 30.5 hP 38.4 hP
Given: m = 6000lb a = 500ft2/lb product = 4200 ft/lb Required: P Solution: Grinder P =
Eff grinder = 30% Eff crusher = 25% Rf = 163 in2/ft-lb
144 =33.02 hp − 6000 . 6000 . 144=5.35 hp
Crusher P =
Total Power delivered = 38.3 HP
26.
Soft and non- abrasive materials can be made into fines by a. Attrition b. Cutting c. Compression d. Impact - Attrition- the action or process of gradually reducing the strength or effectiveness of someone or something through sustained attack or pressure. 26. In a plate and frame filter press, the rate of washing is equal to the ________ of the final filtration rate. a. 1/4 b. ½ c. 1 d. Sqrt of 2 - A plate and frame filter press is the most fundamental design, and many now refer it as a "membrane filter plate". This type of filter press consists of many plates and frames assembled alternately with the supports of a pair of rails.
27. A solid handling equipment used for moving powdered or granular materials to and from storage or between reaction vessels as in moving bed catalytic. a. Screw conveyor b. Bucket elevator c. Belt conveyor d. Pneumatic conveyor - Bucket elevator- A bucket elevator, also called a grain leg, is a mechanism for hauling flowable bulk materials (most often grain or fertilizer) vertically. 28. It is a portable platform on which package d materials can be handled and stored. a. Hopper trucks b. Baler bags c. Pallets d. Steel drums - Pallet- a flat wooden structure that heavy goods are put onto so that they can be moved using a fork-lift truck (a small vehicle with two strongbars of metal on the front that is used for lifting heavy goods).
29. A type of pneumatic conveyor system characterized by a material moving in air stream or pressure less than ambient is a. Fluidizing b. Pressure c. Pressure-vacuum d. Vacuum -Vacuum- a space without any gas or other matter in it, or a space from which most of the air or gas has been removed.
30.
It is the angle at which a material will rest on a pile a. Angle of inclination b. Angle of repose c. Banking angle d. Contact angle -Angle of repose- the steepest angle at which a sloping surface formed of a particular loose material is stable.
31.
Which of the following conveyors can be used in conveying mate rials up and down an incline? a. Screw conveyor b. Bucket elevator c. Belt conveyor d. Pneumatic conveyor - Bucket elevator- A bucket elevator, also called a grain leg, is a mechanism for hauling flowable bulk materials (most often grain or fertilizer) vertically.
32.
A flotation modifier which assists in selectivity or stop unwanted minerals from floating is a. Activators b. Depressants c. Promoters d. Regulators - The function of depressants is to prevent, temporarily, or sometimes permanently, the flotation of certain minerals without preventing the desired mineral from being readily floated.
33. The flotation agent that prevents coalescence o f air bubbles as they travel to the surface of the water is/are a. Collectors b. Modifying agent c. Promoters d. Frothing Agent -Frothing agent- is a material that facilitates formation of foam such as a surfactant or a blowing agent. A surfactant, when present in small amounts, reduces surface tension of a liquid (reduces the work needed to creat e the foam) or increases its colloidal stability by inhibiting coalescence of bubbles. 34.
In sedimentation process, settling is very fast if the a. Density of the particles are big and the viscosity of the liquid is high b. Density of the particles are low and the viscosity of the liquid is high c. Density of the particles are high and the viscosity of the liquid is low d. Particle size is big and the density is low -A particle or droplet will settle in a fluid if its density is greater than that of the fluid in which it is suspended. The (laminar) settling velocity of particles whose concentration is very low, that
is when the flow of fluid around a particle does not affect the flow around neighboring particles. 35.
Cement clinker is reduced to fine size a. Roll crusher b. Ball mill c. Hammer mill d. Tube mill -Tube mill - a grinding mill that consis ts of a long revolving tube cont aining flint pebbles or steel balls or slugs and is used for pulverizing (as in cement manufacturing).
36.
Removal of activated carbon from glycerin is done by a. Plate and frame b. Rotary vacuum filter c. Batch backed centrifuge d. None of these - Plate-and-frame filter- a filter pre ss in which the spaces for the caked solid matter are formed by inserting hollow frames between each pair of plates instead of providing the plates with raised edges.
37.
Bond crushing law a. Calls for relatively less energy for the smaller product particle than does the Ri ttinger 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 product d. States that the work required for crushing is proportion - Bond crushing law: The total work input represented by a given weight of crushed product is inversely proportional to the square root of t he diameter of the product particles.
39.
In Froth Flotation, chemical agent added to cause air adherence is called a. collector b. frother c. modifier d. promoter Collectors are chemical agents used in froth flotation to cause air adherence.
40.
The operating speed of a ball mill should be a. atleast equal to the critical speed c. much more than the critical speed b. less than the critical speed d. none of the choices The rate of rotation of ball mills should be less than the speed at which the charge is held against the inside surface by centrifugal force, since no size reduction will take place unless the balls fall upon the material to be crushed.
41.
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 gr inding In wet grinding, water holds dry m aterials to be reduced in size that increases it in capacity than dry grinding. 42.
Drag coefficient CD is given by (in Stoke’s Law r ange) a.
=
43.
b.
= ..
c.
=
d.
= .
Perry’s Chemical Engineers’ Handbook 8 th Edition 6-52 Equation 6-230
Screen capacity is expressed in terms of a. tons/h b. tons/ft2 c. tons/h-ft d. tons/h-ft2 Capacity of screen is expressed in terms of mass of feed per unit area per time.
44.
With increase in the capacity of screens, the screen effectiveness a. remains unchanged b. increases c. decreases d. decreases exponentially Screen capacity and effectiveness are closely related. If a low effectiveness may be tolerated, the screen may be operated at high capacity.
45.
The material passing one screening surface and retained on a subsequent surface is called a. intermediate material b. plus material c. minus material d. overflow material When two screens are used for screening, the material that retained on the second screen is called the intermediate material.
46.
In screen analysis, notation +5 mm/-10 mm means passing through a. 5 mm screen and retained on 10 mm screen b. 10 mm screen and retained on 5 mm screen c. both 5 mm and 10 mm screens d. neither 5 mm and 10 mm screens Oversize (plus) materials are those that fail to pass through the screen and undersize (minus) materials are those that pass through.
47.
A screen is said to be blinded when a. oversize are present in undersize fraction b. undersize are retained in the over size fraction c. the screen is plugged with solid particles d. its capacity is abruptly increased Blinding occurs when materials are plugged in t he screen opening which causes the screen effectiveness to decrease.
48.
Cumulative Analysis for determining surface is more precise than differe ntial 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 single fraction are equal in size is not needed d. all of the above In cumulative analysis, the assumption of “all particles in a single fraction are equal in size” is not required.
49.
It defines the effective membrane area installed per volume of a module. a. osmotic pressure b. water flux c. solid flux d. packing density Packing density is the effective membrane are a installed per volume of a module.
50.
Trommels separate a mixture of particles de pending on their a. size b. wetability c. screen size d. AOTA Trommels or also called revolving screens. Screening is t he separation of materials on the basis of size.
