Post on 14-Mar-2020
Equation for one-dimensional motion of
particle through fluidβ’ Expression for acceleration of a particle settling in a fluid:
πππ’
ππ‘= πΉπ β πΉπ β πΉπ·
Where , πΉπ= πππβ’ acceleration in external field = ππ=g in gravity settling, and Ο2πin centrifugal field
β’ πΉπ =Buoyant force =π
πππ
β’ πΉπ·= Drag force, πΉπ· =πΆπ·ππ’
2π΄π
2β’ CD = Drag coefficient, Ap=Projected areaβ’ πΉπ· always increases with velocity, and soon acceleration
becomes 0.β’ Terminal velocity is the constant velocity the particle attains
when acceleration becomes 0.
DRAG COEFFICIENT &TERMINAL VELOCITY
β’ πͺπ« = ππ« π¨π
ππππ π
β’ ππ π
π π= ππ β ππ β ππ« = πππ β
π
πππππ β
πͺπ«ππππ¨π
π
β’π π
π π= ππ β
π
πππππ β
πͺπ«ππππ¨π
ππ
β’ ππ π βπ
ππ=πͺπ«ππ
ππ¨π
ππ
β’ ππ =πππ ππβπ
πͺπ«ππππ¨π=ππ
ππ«ππππ π ππβπ
πͺπ«ππππ π«ππ
π
=ππ«ππ ππβπ
ππͺπ«π
β’ π =ππ«ππ ππβπ
ππͺπ«π
Trial and error method for determination of terminal velocity
π =ππ«ππ ππ β π
ππͺπ«πβ¦ . (π)
1. Assume a value of u
2. Calculate Reynolds number of particle, ππ π,π
3. ππ π,π=π·ππ’π
π
β’ π·π = π·πππππ‘ππ ππ ππππ‘ππππ, π’ = π£ππππππ‘π¦ ππ ππππ‘ππππ,β’ π = density of fluid, π= viscosity of fluid
4. Determine πͺπ« from chart of πͺπ« π£π ππ π,π5. Calculate u from equation 16. Compare Calculated u with Assumed u, if error is not within limit
restart from step 1 for second trial
Terminal velocity in Stokes Law range and Newtons law range
π’π‘ =4π ππ β π π·π
3πΆπ·π
β’ Stokes Law range, particle Reynolds number less than 1.0
πΆπ· =24
ππ π,π, whereππ π,π =
π·ππ’ππ
π
π’π‘ =ππ·π
2 ππ β π
18π
β’ Newtons Law Range: 1000<ππ π,π<200,000, πΆπ· = 0.44
π’π‘ = 1.75ππ·π ππ β π
π
7
β’ π’π‘ =4π ππβπ π·π
3πΆπ·π
β’ π’π‘ =4π ππβπ π·π
324π·ππ’ππ
π
π=π·π2π ππβπ π’π‘
18π
β’ π’π‘ =π·π2π ππβπ
18π
Determination of Range of settling
β’ Determine the value of K
πΎ = π·πππ ππ β π
π2
1 3
β’ Stokes Law range K<2.6
β’ Newtons Law range K>68.9
β’ Intermediate range K greater than 2.6 and less than 68.9
β’ For Stokes Law range:
β’ π ππ =π·ππ’π‘π
π=π·ππ
π
π·π2π ππβπ
18π=π·π3ππ ππβπ
18π2< 1
β’ Let πΎ = π·πππ ππβπ
π2
1/3
β’1
18πΎ3 < 1, ππ πΎ < 181/3 = 2.6
β’ For Newtons law range
β’ π ππ =π·ππ’π‘π
π=π·ππ
π1.75
ππ·π ππβπ
π= 1.75π·π
1.5 ππ ππβπ
π2
3
2.1
3=
1.75πΎ1.5
β’ 1.75πΎ1.5 > 1000
β’ πΎ > 68.9
β’ Estimate the terminal velocity of 80/100 mesh particles of limestone (density 2800kg/m3) settling in water at 30oC. Viscosity =0.801cP
β’ Determine K, to find range of settling,β If K is not much larger than 2.6 Stokes law can be
assumed and recheckedβ If K is not much less than 68.9 Newtons law can be
assumedβ If K is in Intermediate range, trial and erro method to
be followed
Hindered settling velocity
β’ π’π = π’π‘ ππ where π = πππππ ππ‘π¦
β’ Particles of sphalerite (sp. Gr. 4.00) are settling under the force of gravity in the carbon tetrachloride (CCl4) at 20oC(sp.gr. 1.594). The diameter of the sphalerite particles is 0.1 mm. The free settling terminal velocity is 0.015m/s. The volume fraction of sphalerite in carbon tetrachloride is 0.2. What is the settling velocity?
Types of sedimentation tanks
β’ Circular Radial flow
β’ Conical Vertical flow
β’ Rectangular horizontal flow
17
Circular Radial Flow Gravity Thickener
β’ 1 = Raw wastewater inlet pipe2 = Inlet stilling well (baffle)3 = Clarified water overflow weir4 = Primary sludge outlet pipe
β’
20
β’ After the removal of inorganic screenings and grit, the next step in wastewater treatment is the removal of the grosser suspended solids from the raw sewage. This is achieved via the process of primary settling or primary sedimentation. The wastewater flow is directed into one or more settling tanks, also known as primary clarifiers. Large mechanically-raked circular tanks are generally used at the larger works,.
β’ The overall volume of the primary settling tank is designed to ensure the incoming wastewater will take a certain amount of time to flow completely through the structure. This is known as the retention time, and must be sufficiently long enough to allow suitable settling of the solids to take place, yet short enough to prevent the anaerobic breakdown of the settled solids from occurring (i.e. the settled sludge turning septic).
β’ The retention time can vary from 1 to 3 hours, depending on the plant loading and incoming wastewater characteristics. In a tank with a retention time of 2 hours, 50 to 70% of the suspended solids may be removed by settling and flotation (scum removal). Removal of these solids will also reduce the BOD by 30 to 35%.
β’ Of equal importance in the design of the primary settling tank is the surface loading rate. This will dictate the overall diameter of a circular settling tank. This is effectively the upward velocity of the incoming flow from the base of the tank to the top of the overflow weirs. Upflowvelocities of around 1.0m per hour are generally desired.
21
VERTICAL SETTLING TANKDortmund Type
β’ For small plants conical Dortmund-type settling tanks are preferred. Occasionally rectangular structures are used, with complex sludge and scum scraper mechanisms.
β’ Upward flow tank suited small treatment plants
β’ Steep floor slope, large lower hopper volume enable large sludge storage
β’ No need for scarping.β’ Depth of hopper at least
equal to the top dimension which is less than 6m
22
RECTANGULAR HORIZONTAL FLOW
β’ .
β’ Rectangular are for large primary settling tank
β’ Sensitive depth to width /length ratio
β’ Occasionally rectangular structures are used, with complex sludge and scum scraper mechanisms.
β’ Rectangular are for large primary settling tank
β’ Sensitive depth to width /length ratio
23
THICKENERS AND CLARIFIERSSETTLING OF FLOCCULATED PARTICLES
β’ Fine particles form agglomerates entrapping water within
β’ Flocculating agents β strong electrolytes, polymeric -polyacrylamide
β’ Inexpensive water treatment materials lime alumina, sodium silicates, ferrous sulphate, ferric chloride etc. form loose agglomerates and removes fine particles.
Batch Sedimentation test
β’ A= Clear liquid
β’ B = Uniform initial conc.
β’ C=Transition zone
β’ D= Settled solid
24
gffd
β’ Plot settling rate curve from the following settling experimental data, and report (a)Hindered settling velocity (b) settling in compression zone (c) critical composition.
Time min Interface height m
0 2.5
20 1.56
40 0.97
60 0.75
80 0.63
100 0.53
120 0.44
140 0.39
160 0.36
26
SETTLING TANK DESIGNRef: McCabe Smith
β’ There are two ways the solids can move to the bottom:
1. Under the influence of their settling velocity, downward
2. Due to the continuous removal of sludge at the bottom as underflow
Total downward flux of solids, πΎπ π2βπ, consists of two parts the 1. Transport Flux, πΊπ‘2. Settling flux πΊπ
Feed overflowππ, πΆπ ππ
Underflow
ππ , ππ, ππ’=Flowrate of inlet, overflow and underflow
πΆπ , πΆπ’=Concentration of solids in inlet and underflow
27
β’ Settling flux, Gs, also called the batch settling flux, this is due to the settlement of solids and as expected is a function of solids concentration and the settling velocity of particles.
πΊπ =ππ»
ππ‘π
πΆπ
ππ»
ππ‘ π, πΆπ= velocity and concentration at a layer
"i" in the thickener. It passes through a maxima, as at very low concentration settling rate is constant, and at high concentration the rate decreases rapidly.
β’ Transport flux, Gt,Fluxof solids carried by liquid, is independent of the solids settling in the thickener, whether the solids settle or not, there pumping of the sludge from the bottom of the tank.
πΊπ‘ = π’πΆπwhere, u= the velocity created by the underflow
28
SETTLING TANKAREA
β’ Total flux curve has a minimum value between the influent solids concentration (Co) and underflow solids concentration (Cu). This minimum flux is the maximum allowable solids loading for the thickener to work successfully. When flux is at minimum, we calculate for the maximum area required. If this limit is exceeded, the solids will overflow in the effluent. So the design of a thickener is thus reduced to the point of determining this flux.
β’ πΊ = πΊπ‘ + πΊπ
= π’πΆπ +ππ»
ππ‘π
πΆπ
As Solid introduced = solid removedππΆπ = πππ‘ππ πππ’π₯ π₯ ππππ
π΄ =ππΆπ
πΊπ‘ + πΊπ πππ
29
KYNCH Method to obtain settling flux from one batch settling data
β’ Locate any time, ti
β’ Draw tangent at the point to cut y axis t Hiβ and x axis at tiβ
β’ Determine Ci: π»0πΆ0 = π»πβ²πΆπ πππ πΆπ =π»0πΆ0
π»πβ²[conc. at top of settling zone]
β’ Determine settling velocity, ππ»
ππ‘ π=π»πβ²
π‘πβ²
β’ Determine settling Flux : πΊπ π =ππ»
ππ‘ ππΆπ
β’ Plot Gsi vs Ci
ππ ππβ² π―πβ² πΆπ ππ»
ππ‘π
πΊπ π
30
β’ Following is the results of a settling experiment, with initial concentration of 60 g/l of calcium carbonate. Data is to be used to design a settling tank to handle 0.03m3/second of slurry, with an underflow velocity of 0.05m/h.
β’ Plot (a) Settling flux vs concentration, (b) transport flux vs concentration,(c) total flux vs concentration
β’ Determine minimum total flux, and cross sectional area of the tank.
Time,min Height of interface, mm
0 250
10 175
20 123
30 103
40 86
50 75
60 65
70 57
80 52
31
Tubular Centrifuge β’ Centrifugal liquid liquid
separation
β’ Removal of solids from lubricating oil, ink, beverages etc.
β’ 10 to 15 cm dia
β’ 15,000rpm
β’ Solid/dirt accumulate inside the bow, and periodically removed
33
Disk Centrifugeβ’ Liquid liquid separationβ’ Removal of solids from lubricating
oil, ink., beverages etc.β’ Short wide bowl 20to 50cmdiaβ’ Bowl filled with discs- cones of
sheet metal set one above the other, with matching holes in the middle, which form the liquid passage
β’ Heavy liquid moves outwards and light liquid inwards.
β’ Soon Heavy Liquid touches lower side of a cone and separates out
β’ Shearing helps break emulsions β’ Solid/dirt accumulate inside the
bowl, and periodically removed
34
Nozzle discharge Centrifuge
β’ Modified disk centrifuge
β’ 3 mm dia holes at the periphery
β’ Dilute slurry leaves through hole
β’ Alternatively the holes are plugged most of the time and occasionally opened and thick slurry removed.
35
HELICAL CONVEYER CENTRIFUGEβ’ Bowl diameter β 10cm to 140 cm
β’ Solid removal rate β 1 to 2 tons/h to 50tons/h
β’ May not clarify 100% may require subsequent clarifier
β’ Separate free particles from water, like crystals, dewater etc
36
Sedimenting Centrifuge
β’ π2 = radius of the bowlβ’ π1 = radius at liquid surfaceβ’ π =depth of bowl
β’ π·π = ππππ‘ππππ π ππ§π
β’ ππ΄= position of particle at inletβ’ ππ΅ = position of particle at outlet
β’ π’π‘ =π2ππ·π
2 ππβπ
18π
β’ Since π’π‘ = ππ ππ‘
Time required for a particle of size Dp to travel from radius rA to rB.
β’ 0π‘π ππ‘ = ππ΄
ππ΅ 18π
π2 ππβπ π·π2
ππ
π
β’ π‘π =18π
π2 ππβπ π·π2 ππππ΅
ππ΄
β’ Residence time
β’ π‘π =ππππ’ππ
π=ππ π2
2βπ12
π
β’ π =πππ2 ππβπ π·π
2
18π
π22βπ1
2
ππππ΅ππ΄
37
CUT DIAMETER
β’ Cut point diameter particles that travels half the distance between π2 πππ π1in the available time
β’ For the particles to be removed ππ΄ =π1+π2
2and
ππ΅ = π2
β’ π =πππ2 ππβπ π·π
2
18π
π22βπ1
2
ππππ΅ππ΄
β’ ππ =πππ2 ππβπ π·ππ
2
18π
π22βπ1
2
ππ2π2π1+π2
38
Special case: very thin liquid layer π1 β π2
β’ π’π‘ =π2π2π·π
2 ππβπ
18π
β’ Let the thickness = s
β’ If Cut dia is π·ππ the particles have to travel a distance s/2
β’ π’π‘ =π
2π‘π
β’ Residence time π‘π =π
ππ
ππ =π
π‘π=2ππ’π‘π =2ππ2π2π·π
2 ππ β π
18ππ
39
Sigma Value: a parameter for scale up
ππ =2ππ2π2π·π
2 ππ β π
18ππ Let re and se be average value of radius and liquid layer thickness
ππ =2ππ2πππ·π
2 ππ β π
18ππ π
=2ππ2ππππ π
π·π2π ππ β π
18π= 2 ππΊ
πππππ π£πππ’π, =ππ2ππ
ππ π, is a characteristics of centrifuge .
It is the cross sectional area of a gravity settling tank of the same separation capacity as the centrifuge.
40
β’ What is the capacity in cubic meters per hour of a clarifying centrifuge operating under the following conditions?
β Diameter of bowl, 600mm
β Thickness of liquid layer, 75mm
β Sp.gr of liquid 1.2 sp.gr of solid 1.6
β Depth of bow 400 mm Viscosity of liquid 2 cP
β Speed, 1,200rpm Cut size 30ππ
42