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 86º 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 ∞ = 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
α 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, ψ=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 °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 21˚C 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 Stokes’-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 100˚F). 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 100˚F. 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 50˚F. 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 68˚F 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 70˚F. 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 55˚F 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 µ 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 45° 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 66˚F 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 (α), the cake porsity (ε),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
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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 ρ of CaCO3 ρ 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 α (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. µ of water = 1 cp ρ of water = 62.3 lb/ft3 ρ 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 (-∆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 αo for this sludge.
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-∆ 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.
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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 °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 (θ) (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 θ = 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/dθ = υ = 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 θ,hr
Zi, cm
υ 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.
υ , cm/hr
cL, gm/l
1 cL
1 1 cL cυ
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 ∞
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 LL c L = 4730 cm / hr S 1 / gm Mm
corresponding to υ = 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
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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
θL θ, min
θL, 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 − z L
θL
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 86º 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
θL
θL , min
zL, cm
θ , min zi , zna
    Â
    Â
    Â
vL =
zi − z L
θL
    Â
, cm/min
cL =
co z o , g/lit zL
    Â
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
ρs ρL = = = =?
= = 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 − c L cu
1 1 − c L cu
       Â
       Â
VL / ( 1/ CL – 1 / CU )
cm/hr / lit/g
VL / ( 1/ CL – 1 / CU ) min
VL, cm/hr
vL 1 1 c −c u L
A = 100
π 4
= Lo co = A min min
ton lit / g × day cm / hr
D2 =
Â
D =
Â
ft
Â
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 ∞ = 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 − z∞ zo − z∞
(z-z ∞ )
0 0.25 0.50 1.00 1.75 3.0 4.75 12.0 20.0 ∞
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-Z∞ / ZO - Z∞
ZO
0.6
0.2 ZO’
0
θC
Time , hr
14
Figure : Settling Characteristics
Equation 1.18 is used to evaluate the compression zone volume. Lc Lc W V = o o (θ − θ ) + o o ∫ θθ c 1 dθ 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 (θ - θc) = 2.6 hr Therefore, V =
100,000 100,000 × 2.6 + × 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 ≅ 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
α 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 Zα = 1.99 cm Zo -Zα = 29.06 cm
z,cm
0
θL, min
zL, cm
zi, cm
vL , cm/min
θ, min
3
cL, g/cm
VL / ( 1/ CL – 1 / CU )
cm/ min / cm3/g
VL / ( 1/ CL – 1 / CU ) min
VL, cm/ min
1/cL
vL 1 1 − c L cu
From fig vL 1 1 − cL cu
l o co = A A=
= A= D =
= min
Â
Â
cm / min cm 3 / g
cm / min cm 3 / g
g 105 × 0.012 gal min 1day 1 ft 3 × 3× × × day cm cm 24 × 60 min 7.48 gal ft2
π 4
D2
Â
θ, 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- z∞ , cm
z − zα zo − zα
3
1/CL , cm /gal
ln (Z-Z∞)/ (Z0- Z∞)
0
θc = θu =
 Â
θC θu θ, min
θC θ, min
min
volume of compression zone =
loco (θ u − θ c ) + l o c o ∫ θθcu 1 dθ ys cL
= 105
gal g × 0.012 3 (− ) day cm
min ×
1 day cm 3 1 ft 3 × × 2 g 24 × 60 min 7.48 gal
+ 105
gal g cm 3 × 0.012 3 × − − − − × min× day cm g
1 day 24 × 60 min
×
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 )/ 3µ2] with slop of -2. 400 × 10 −4 = 1.31 x 10-3 ft 30.48 µair = 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 × = 0.0524 lb/cu ft pair = 359 760
Dp =
CD = CD =
4 gDp 3 p ( ps − p ) 3µ 2
(4)(32.2)(0.00131)3(0.0524)(64.1)(0.0524) (3)(1.747 × 10 −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.ψ = 1.0, NRe =14. Therefore, Dp υ tp / µ = 14, and
υt =
9.*
14 µ 14(1.747 × 105 ) = = 3.57 ft / sec D p P (1.31 × 10 −3 )(5.24 × 10 −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, ψ=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 °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.
4 gD p 3 ( p sil − p ) log CD = -2 log NRe + log 2 3 µ
The viscosity of water at 65˚F = 0.01 poise (4 )(981)(0.07 )3 (1.0 )(2.65 − 1) Log CD = -2 log NRe - log 2 3(0.01) 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 ψ = 0.806 curve at NRe = 28. A Reynolds number of 28 corresponds to a settling velocity of N µ (28)(0.01) υ 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. 4 g ( p gat − p )µ log CD = log NRe + log 3 p 2υ 3 (4)(981)(7.5 − 1.0)0.01 log CD = log NRe + log (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 ψ = 0.806 curve at NRe = 9.0. This Reynolds number corresponds to a diameter of Dp =
N Re µ
υp
=
(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 21˚C 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 Stokes’-Law range.
oil
Dp = 15 µ sp.gr of oil = 0.9
air , 21°C , 1 atm time = 1 min Stoke’s-law , range
D p2 g (ρ s − ρ )
vt =
18µ
Dp = 15 x 10-4 cm ρs = 0.9 g/cm3 µ = 0.018 x 10-2 g/cm-5 pv = nRT R=
pv 1atm × 22.4 × 10 3 cm 3 = nt 1 g.mol × 273K
ρair =
pM RT
=
1atm × 29 g / g − mol 1 × 22.4 × 10 3 × 294K 273
= 0.0012 g/ cm3
(15 × 10 ) vt =
× 980 × (0.9 − 0.0012 ) = 0.611 cm/s 18 × 0.018 × 10 −.2
−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 100˚F). 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 100˚F. 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, 100˚F
S/A conc
sp.gr = 1.84 µ = 15 cp, 100˚F Q˚ = 3.5 ft3 /min vt - A = 3.5 ft3 / min
π
2
3.5 3 2 ft / s vt × × = 4 12 60 vt = ft/s ρp = 1.84 x 62.4 lb / ft3 PM ρt = =Â RT µf = µair
Dp
s/^
=?
log CD = log NRe + log
(
)
4g ρ p − ρ f µ 3v ρ 3 t
f
2 f
at NRe = 1 DD =
(
)
4g ρ p − ρ f µ f 3v ρ 3 t
2 f
=
Â
fig 2.1 NRe = D p vρ
µ Dp =
 =
Â
Â
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 ρp = 2.5 g/ cm3 ρf = 62.4 lb/ft3 µf = 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 50˚F. What is the terminal velocity of the sphalerite? What is the density of slurry?
sphalerite sp gr = 4 Dp = 0.006 in 50°F
vol of sphalerite vol of slurry
= 0.25
vt = ?
ρslurry = ?
ρslurry = 2.65 x 0.25 + 1 x 0.75 = vH = Fs =
slurry quarty+ water 25% vol sp gr=2.65
(
D 2p g ρ p − ρ f 18µ f
)=F
Â
g/cm3
s
X2
101−82 (1− X ) x = 0.75 Dp = 0.006 in x
2.54cm = 0.006 x 2.54cm 1in
ρp = 4 g/cm3 µf = 1.31 cp = 1.31 x 10 -2 g/cm.s ρf = 62.42 lb/ft3 x Fs =
0.752 101.82 (1−0.75)
1g / cm 3 = 1 g/cm3 62.37lb / ft 3
= 0.197
(0.006 × 2.54) 2 × 980( 4 − 1) × 0.197 18 × 1.31 × 10 −2 = 0.57 cm/s vH =
14.** Calculate the settling velocity for the hindered settling of glass sphere in water at 68˚F 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 ρp = 154 lb /ft3
vH = ? 68˚F
water ρt = 62.31 lb/ft3 µt = 0.982 cp
1206 154 / 62.4 = 0.571 1140
1140 − x=
Fs =
vH =
x2
101−82 (1− x )
(
==
D 2p g ρ p − ρ f 18µ f
0.5712 = 101.82 (1−0571)
Â
) .F
s
(0.0061 / 12) 2 × 32.2(154 − 62.32) 18 × 0.982 × 6.72 × 10 − 4 = 0.0034 ft /s =
x
Â
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 70˚F. 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 70˚F
(a)
urea pellets Dp = 1/4 in = 1/48 ft ρurea = 83 lb/ft3 vt = ? Dp = 1/48 ft
µf =
Â
ρp = 83 lb/ft3
PM RT
ρf =
log cD = -2 log NRe + log
4D 3p g (ρ p − ρ f )ρ f 3µ f2
at NRe = 1 4D 3p g (ρ p − ρ f )ρ f
cD =
3µ f2
cD
slope = -2 ψ =1
cD 1
NRe = Â
NRe Fig 2.1 NRe =
D p vt ρ
µ vt
(b)
=
Â
=
Â
80 ft vp vt
> 0.99 would attain, not attain
v = 0.99 vt 0 ft a=
vb dv dt
dρ . a =
dv dz dt
dρ . a = v 0 80
∫ dρ =
Vb 0
dv
∫
v a
dv = -80
∑F = mpa = FE - FB - FR C D A pρf v 2
= mp g - mf g -
2
c D π / 4 D 2p ρ f v 2 mp mpa = mpg ρt g − 2 ρp mp = a =
ρp − ρf ρp
π 6
.g −
D p3 ρ p
3c D ρ f v 2 4D p ρ p
3 × 0.075c D v 2 83 − 0.075 × 62.4 − 1 83 4 × × 83 48 Dpvρ f − − − × v NRe , p = = −−− µf
a=
Assume
v c ft/s
NRe 0
cD 0
a
v a
0
Â
0
vt
v a
∴ vb =
Â
- 80 0
v
vb vt
vb = vt
Â
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 300µ
- 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 ρp = 2.56 g/cm3 ρf =
g/cm3
1
µf = 1 x 10-2 gl cm.s vt = ? log cD = - 2 log NRe + log
4 D 3p g (ρ p − ρ )ρ
at NRe = 1
cD =
4 D p3 g (ρ p − ρ )ρ 3µ 2
=
Â
3µ 2
slope = -2 cD cD ρ=1 1
NRe =
Â
NRe NRe =
ρp
µ vt =
For pyrite vt =
ρ
vt
=
Â
Â
Dp = ? cm/s
Â
g/cm3
ρp = 5.1
log cD = log NRe + log
4 g ( ρ p − ρ )µ 3vt3 ρ 2
at NRe = 1 cD =
4 g ( ρ p − ρ )µ 3vt3 ρ 2 slope = 1
cD
cD ρ = 1 NRe = 1
NRe =
Dp
vt
µ
ρ
=
Â
Dp = Â 10 to 300 µ for quart 2 10 to _ Â _ _ µ for pyrite
(b)
Q& p
Q (pure) 10 to Dp
10 to 300µ
H2 O
P + Q Dp to 300 µ
For pure quartz, total pyrite in downstream For pyrite DP = 10 µ = 10-3 cm ρP = 5. 1 g/cm3 vt = ?
4 D p3 g (ρ p − ρ )ρ
log cD = - 2 log NRe + log
3µ 2
at NRe = 1 = > cD =
4 D p3 g (ρ p − ρ )ρ
=
3µ 2
 slope = -2
cD cD ψ =1 NRe =
1
Â
NRe NRe =
ρp
ρ
vt
µ vt =
Â
For quartz vt =
Â
ρp = 2.65 Dp = Â ? log cD = log NRe + log at NRe = 1
(
)
4g ρ p − ρ µ 3v ρ 3 t
2 f
=
Â
cD =
(
)
4g ρ p − ρ µ 3v 3t ρ f2 slope = 1
cD
cD ρ = 1 NRe = 1 NRe =
Â
Dp
vt
µ
ρ
=
Â
Dp = Â 10 to 300 µ for pyrite
Â
to 300 µ 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 ρp = 1.35 g/cm3 1 g/cm3 ρf = µt = 10-2 g/cm .s vt = ?
H2o
sand sp.gr=2.65
cD =
NRe =
cD = cD =
4D p g (ρ p − ρ f ) 3v 2t ρ Dp
ρ
vt
µ
(
)
4D P g ρ p − ρ ρ 3ρ
2
×
4 D p3 g (ρ p − ρ )ρ 3µ
D p ρ2 N 2Re × µ 2 .
2
1 2 N Re
4 D p3 g (ρ p − ρ )ρ
log cD = -2 log NRe + log at NRe =1 cD = =
3µ 2
4 D p3 g (ρ p − ρ )ρ 3µ 2 4 × 0.08333 × 980(1.35 − 1)1 3 × (10
)
−2 2
= 0.26 × 10 4
ψ=1
NRe , p = 30 Dp
vt
µ vt = Â For sand Dp = ? vt =
Â
log cD = log NRe + log at NRe = 1 cD =
4 g ( ρ p − ρ )µ 3vt3
ρ
=
Â
4 g (ρ p − ρ )µ 3vt3 ρ
ρ
=
Â
(From Fig 2.1)
slope = 1 cD
cD ψ= 1 NRe = Â
1
NRe NRe
=
Dp
vt
µ
ρ
=
Â
Dp = Â
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 55˚F 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 55˚F
sp - gr = 2.65 48-14
(a) Q = ? g pm Dp = 48 mesh ρp = 2.65 x 62.4 lb /ft3 ρf =
 µf = Â
vt = ?
π
Q = vt . A = vt x
D2 =
4
4 D p3 g (ρ p − ρ )ρ
log cD = -2 log NRe + log
3µ 2
at NRe = 1 cD =
4 D p3 g ( ρ p − ρ ) ρ
=
3µ 2
cD
Â
slope = -2
cD ψ =1 1
NRe = 1 NRe
NRe = Â Dp
vt
ρ
=
µ vt =
Â
Â
Q= vt x
π 4
D2 =
(l) 200-14
H 2O
200-Dp
Dp-14
Q∆ = 1.5 gpm vt . A = 1.5 gal/min 1.5 1 ft 144 1 × × × vt = π 7 . 48 1 cos 2 ×4 4 = 0.0383 ft/s
Â
Dp = ? φ=1 ρp = 2.65 x 62 .4 lb/ft3 log cD = log NRe + log
at NRe = 1
cD =
4 g (ρ p − ρ )µ 3vt3 ρ 2
4 g (ρ p − ρ )µ 3 vt3
=
slope = 1 cD
1 NRe =
Â
NRe Fig = 2.1 Dp
vt
µ
ρ
=
Dp = Â
Â
Â
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 66˚F 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 (α), the cake porsity (ε),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) + T∆z 990
0.03( 200 + 150)( 200 + 0.03 × 30 × 200) + 200 × 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 BDSδ b 10 × 24 × 100 × 75 T= = 300tons / hr = 6000 6000 aTL + bWLs + 10L h.p = 1000 0.6 × 300 × 100 + 0.04 × 24 × 100 × 100 + 10 × 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 × = 855.6 ft 3 /hr h 7.48gal
(coefficient )(capacity, lb / min )(length, ft ) 3300
1.3 × 85.5ft 3 / hr × 48lb / ft 3 × 1hr / 60 min × 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 °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 µ = 12 x 6.72 x 10-1 = 8.06 x 10-3 lb/ft-s ρ = 93.5 lb /ft3 , g = 32.17 ft/s2 Then
NRe =
D a2 nρ 2 2 x1.5 × 93.5 = = 69.600 µ 8.06 × 10 −3
From curve A (Fig 6.5). for NRe = 69.600. NRe = 5.8 and from Eq. (6.11) 5.8 × 93.5x1.53 × 2 3 = 1821ft − lb.s 32.17 The power requirement is 1821/550 = 3.31 hp (2.47 kw). ρ=
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 × 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 − 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 × 93.5 × 1.53 × 2 3 = 406 ft − 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 × 1.5 × 70 = 5.2 1200 × 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 × 70 × 1.53 × 2 3 = 2938 ft − 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.
