Flocculation

PHEH401 for San4

Shalasai Hungprasert

Dept Env H Sc, Public Health,Mahidol U.

1. Purpose

 

·        Flocculation is the gentle mixing phase that follows the rapid dispersion of coagulant by the flash mixing unit.

·        The purpose of flocculation is to accelerate the pace at which the particles collide, causing the agglomeration of electrolytically destabilized particles into setteable and filterable sizes

·        The aggregation of particulate matter is a 2-step process

            - The coagulant is added and reduces the interparticle forces; this is coagulation.

            - The particles then collide and enmesh into larger particles, floc; this is flocculation.

2. Considerations

·        Raw Water Quality and Flocculation Characteristics. Flocculation characteristics can be evaluated by a jar test. Repeatability of the results must be demonstrated. F3.2.4-1. P.106. The residual turbidity is directly proportional to alum dosage, but not always. Note the effects of alum on kaolinite, topsoil and diatomaceous earth. The initial reaction is a decrease in turbidity, but if too much alum is added, the turbidity increases.

·        Finished Water Quality Goals. Less than .5TU.

·        Example:

Given: A bentonite soil.

Find: What dosage of alum is effective and what is the resulting turbidity.

From F3.2.4-1, p.106

An alum dosage of 45 mg/l appears effective. Lesser dosage give markedly poorer results; increased dosages do not markedly effect the turbidity removal

The residual turbidity at an alum dosage is about 5 NTUs.

 

3. Type and Selection Guide

·        Flocculation can be provided by either mechanical mixers or baffles

·        F3.2.4-3, p. 111 show the most common types of units. The mixing system include:

Mechanical Mixing.

            - Vertical shaft with turbine or propeller type blades

            - Paddle type with either horizontal or vertical shafts

            - Proprietary units such as Walking Beam

 

Baffled Channel Basins.

            - Horizontally baffled channels

            - Vertically baffled channels

Others including: Reactor Clarifier Proprietary Systems, Contact Flocculation and Diffused Air Agitation

·        Selection Criteria: Selection of the flocculation process should be based on the following criteria:

            - treatment process: conventional, direct, softening or sludge conditioning

            - raw water: turbidity, color, TDS

            - Flocculation characteristics in response to mixing characteristics. For example: if the floc is hard, more rigorous mixing intensity can be applied.

·        The order of preference is:

            - vertical shaft flocculators in properly compartmentalized, horizontal tanks.

            - paddle flocculators in properly compartmentalized, horizontal tanks.

            - baffled channel flocculation for constant flow rate plants

 

4. Discussion of Alternatives

 

·        Mechanical Mixing System. The mixer should have the following characteristics:

            - Must delivered the required G value which may vary by compartment.

            - The shear must be low at the edge of the blades.

            - Low maintenance and operation.

Regardless of the type of flocculator, tapered mixing across the tank is important. The initial mixing is rigorous, but as the floc grows in size, a more gentle, less disruptive regime is in order.

Vertical shaft flocculators are usually the first choice for the following reasons:

            - minimal maintenance

            - operational flexibility

            - very little head loss across the tank

            - easy control of mixing intensity

·        Baffled Channel System. Requires a moderate head loss across the tank. Suitable for developing countries that may not be able to afford a mechanical system.

 

5. Design Criteria(T 3.2.4-2, p.121)

 

·        The  general design criteria for a basic rectangular flocculation tank are as follows:

Energy input: Gt=10,000 to 100,000, t =5x10⁴ s average, G=30 s^-1 average, 10-70 range

DT: 20-30 minutes at Qmax.

Depth: 10-15’

Stages: 3-4 common, 2-6 range

 

 

·        Among the first considerations are the selection of the mode of mixing and the physical relationship between the flocculators and clarifiers. Subsequent decisions include: the number of tanks, number of mixing stages and their energy level and baffling type

·        Design usually based on:

            - DT

            - mixing energy level

·        The energy level is the G value or velocity gradient as defined by Camp:

G = [ ]^.5 , units p.117

Given: P= 850J/s, and the space influenced by the flocculator 4x6x6m. The temperature is 15°C.

Find:

1.)G

2.)What size motor is 850J/s, assume e=70%

 

1.) G

From App.6, p.604 @15°C, m=1.17x10^-3 N.s/m², N=Newton = kg.m/s², the units of m are kg/m.s

V = 4x6x6m

V = 144m³

G = []^.5 =  []^.5

G = 71s^-1

 

2.)How many hp is 850J/s

1kW=1000J/s

850J/s x (1kW / 1000J/s) = .85kw

1hp = .746kw

.85kw x (1hp/.746kW) = 1.14hp

Motor size = 1.14/e = 1.14/.7

Motor size = 1.63hp

The velocity gradient indicates the contacts that are being made. The gradients are produced by hydraulic or mechanical mixing.

·        The number of particle contacts is:

 

N = n₁n₂(G/6)(d₁ + d₂)³ units p.117

 

N is the number of contacts between n₁and n₂ particles.  Therefore, the rate of flocculation increases with the number and size of the particles and with the power input but decreases with the viscosity of the fluid.

·        The mean velocity gradient for baffled systems is:

G = ()^.5 =  ()^.5  , units p.117

in which d is the specific weight, 62.4lb/ft³ and

m=absolute viscosity, 2.73x10^-5 lb.s/ft² @50°F

The gradient increases with the head loss across the tank and decrease when the viscosity and time increase. In plane English, the more turns and curves the more the mixing; the thicker the liquid and the longer it takes, the less the mixing.

 

Example:

Given: Rectangular basin, L=W=4’, D=4.5’, Q=8.7MGD, Water @50°F, m=absolute

viscosity, 2.73x10^-5 lb.s/ft²

 

 

Find: G

t=V/Q=4’ x 4’ x 4.5’ / 8.7MGDx1.547cfs/MGD = 72/13.46

t= 5.35s

h total  = no of baffles x loss per baffle

v=Q/A = 13.46/(4x1) Note the entire distance is 4’, but there are 4 compartments, therefore each compartment is 4/4=1’wide.

v=3.37fps

h = head loss per baffle, see graphic =.5v²/2g = .5(3.37)²/(2x32.2)

h = head loss per baffle =  .0882ft

head total = 3 baffles x .0882ft

head total = .264ft

G = ()^.5 = ()^.5

G = 336.3 fps/ft 

 

·        For mechanical systems with paddles:

 

G = (C_DAv³/2nV)^.5 units p.117

 

C_D is a shape factor, use 1.8 (p.117). A is the cross sectional area of the paddles. G is increases as the area of the paddle increases, the velocity of the paddle increase and G decreases and the fluid gets thicker or the volume of the tank increases. A very important consideration is that v, the relative velocity of the paddle with respect to the fluid is from .5-.75 of the peripheral velocity of the paddle.

 

Given: If G = (C_DAv³/2nV)^.5 and G = []^.5

Find: A power equation

 (C_DAv³/2nV)^.5 = []^.5 square both sides

P = mV  x C_DAv³/2nV, mn=r

P = C_DArv³/2

Given: The outside blade of a paddle wheel is rotating at 4rpm. The distance from the center of the shaft to the outside of the paddle element (a piece of wood, perhaps a 6”x10’) is 7’.

Find:

1.) What is the peripheral speed of the outside blade

2.) Estimate the blade velocity relative to the water.

 

1.) What is the peripheral speed of the outside blade

v = rps x pD/revolution = 4rev/min x 1min/60s x p(7ftx2), 7’ is a radius

v = 2.93fps

 

2.) Estimate the blade velocity relative to the water.

v, the relative velocity of the paddle with respect to the fluid is from .5-.75 of the peripheral velocity of the paddle.

v’= 2.93fps x .5

v’ = 1.47fps

v’= 2.93fps x .75

v’ = 2.20fps, use .75 unless otherwise stated

 

·        The number of stages used in design is determined by:

            - Type of subsequent treatment unit.

            - Raw water.

            - Short circuiting and types of baffles

            - Local conditions

·        Vertical Shaft Flocculators. D/T>35, where D is the diameter of the blade and T is the tank diameter.  The maximum flow induced by the mixing blade should be less than 8fps in the first stage and less than 2fps in the last stage. When properly produced, the floc should have the characteristics and appearance of snow flakes.

·        Horizontal Shaft Flocculators. The advantage of the horizontal shaft is that one shaft can operate a number of agitators and thereby reduce the number of drive units but if a drive fails, more agitators will be put out of business. Basic design criteria include:

            - The total paddle area should be 10-25% of the tank cross-sectional area. If the paddles are too big, they will rotate the water causing a reduction in eddies and turbulence.

            - Each arm should have a minimum of 3 paddles, so that dead space especially near the shaft will be eliminated.

            - The peripheral speed of the paddles should be between .5 and 3.3fps.

            - The G value should be 50s^-1 and then be reduced to 10 or 5s^-1 in the last stage

·        Baffled Walls. Each baffle should have orifices that 4-6 inches in diameter uniformly distributed across the vertical surface and a velocity of 1.2-1.8 fps should be produced in each hole at maximum flow. The baffle is placed perpendicular to the flow. The top of the baffle is slightly submerged, 1/2 inch, to allow scum to flow over the top. The orifice formula is Q=CA(2gh)^.5, C=.8, see p.118.

Given: The total orifice area available is 20ft². Q=50MGD

Find:

1.) total number of orifices

2.) orifice velocity

3.) headloss through the orifice

 

1.) total number of orifices

4-6” in diameter, use 5”

A = D² = .785 x (5/12)²

A = .136ft²

 

number or orifices = total area/orifice area = 20ft² / .136ft²/orifice

number or orifices = 147 orifices

 

2.) orifice velocity

v=Q/A = 50MGD x 1.547cfs/MGD / 20ft²

v=3.87 fps, Note: this is too high, should be from 1.2-1.8fps

 

3.) headloss through the orifice

Q=CA(2gh)^.5

50MGD x 1.547cfs/MGD = .8 x  .136ft² (2 x 32.2 x h)^.5

4.84 = (64.4h)^.5

23.39 = 64.4h

h = .36ft

6. Operation and Maintenance

 

·        Three basic operational procedures:

            - check the size of the floc

            - removal of the scum from the water surface

            - algae control

7. Example

Given: A flocculation basin. Q=12MGD, horizontal shaft, paddle wheel. The mean  G=25s^-1 @ 50°F. t=45min. The Gt must be between 50,000-100,000. Use 3 stages of equal depth in which the G’s decrease: 45,20,10. L=.5W, L=3H. The paddles are to be made of redwood, 10’x6”. The outside blade is to be 1.5’ from the floor of the tank as well as from the top of the water surface. Use 6 blades/wheel and maintain a clear spacing of 12” between blades. Adjacent wheels are to maintain a clear spacing of 24-36” between blades. The wall clearance is to be between 12-18”. C_D=1.50 for the paddles. Use the power equations P=.97C_DAv³ and P=mVG².

 

 

Find:

1.) Basin dimensions

2.) Design the paddle wheel

3.) Calculate the P for each compartment and the rotational speed of the paddles, provide for a 1:4 turndown.

 

1.) Basin dimensions

V = Qt = 12MGD x 45 minutes x 1.547cfs/MGD x 60s/minute

V = 50,122.8ft³

L=.5W, L=3H, W=2L=6H

V = WLH =  3H x 6H x H

V = 18H³

18H³ = 50,122.8ft³

H = 14.07’ use 14’-3”

W = 6H = 6(14.07)

W = 84.4’ use 85’-0”

L = 3H = 3(14’-3”)

L = 42’-9”

new V = WLH = (14.25)(85.0)(42.75)

V = 51,780.9 ft³

H =  14’-3”

W =  85’-0”

L = 42’-9”

 

2.)Paddle wheel

Calculate the numbers of individual 10’ wheels across the 85’ width.

Since the blades are 10’x6”, the width is 10’. The width of the tank is 85’, 85/10»8 wheels, but there are clearances involved, so try 7 wheels.

Let the clear space between the wheels be s, s/2 at the walls, therefore, let s=clearance.

7s + 7(10) = 85

s = 2.14’ which is between 24-36”, so 7 wheels is OK.

 

Check maximum blade area;

Max blade area = 20% x cross-sectional area = .20HW = .20(85)(14.25)

Max blade area = 242.25ft²

Calculate actual blade area

Try 6 blades per wheel, given but typical.

A = wheels/tank width x blades/wheel x blade area

A = 7 wheels x 6 blades/wheel x (6”/12 x 10’)

A = 210ft² < Max blade area = 242.25ft² \ OK

 

 

3.) P, power

t = V/Q = 51,780.9ft³/12x10⁶gpd x x 1440 minutes/day

t = 46.48 minutes

Gt = 25s^-1 x 46.48 minutes x 60s/minute

Gt = 69,720 between 50,000-100,000 \ OK

 

velocity of the water,v = 75% of the maximum peripheral velocity

The distance traveled is pD or 2pr per revolution, rev/s x pD/rev = pD/sec

v = .75 x 2pr x R(revolutions per second)

v₁(first compartment) = .75 x 2p(5.25’) x R

v₁(first compartment) = 24.74R

v₂(second compartment) = .75 x 2p(3.75) x R

v₂(second compartment) = 17.67R

v₃(third compartment) = .75 x 2p(2.25’) x R

v₃(third compartment) = 10.60R

 

P=.97C_DAv³ =.97C_DA₁v₁³  +.97C_DA₂v₂³  + .97C_DA₃v₃³  =  .97C_DA(v₁³  + v₂³  + v₃³), A₁=A₂=A₃

P = .97(1.50)(.5’x10’board dim.)(2 boards,1up,1down)[24.74³+17.67³+10.61³]R³

P=317,976R³

 

first compartment

P=mVG² = m= 2.73x10^-5 lb.s/ft² x  51,780.9 ft³ /3(3 compartments)x 45²

P=950.7 ft.lb/s  x  1hp/550ft.lb/s

P₁=1.73hp

950.7 ft.lb/s  / 7wheels = 317,976R³

R = .075 rps

RPM(max) = .075 rps x 60s/min

RPM(max) = 4.50rpm

RPM(min @ 1:4 turndown) = 4.50rpm/4

RPM(min @ 1:4 turndown) = 1.13rpm

 

Peripheral speed of outside blade

v = circumference x RPM

v₁ (actual v as opposed to 75%) = R x 2pr

v₁ = .075 x 2p(5.25)

 v₁ = 2.47fps

 

second compartment

P=mVG² = m= 2.73x10^-5 lb.s/ft² x  51,780.9 ft³ /3(3 compartments)x 20²

P=187.8 ft.lb/s  x  1hp/550ft.lb/s

P₂=.34hp

187.8 ft.lb/s  / 7wheels = 317,976R³

R = .044 rps

RPM(max) = .044 rps x 60s/min

RPM(max) = 2.64rpm

RPM(min @ 1:4 turndown) = 2.64rpm/4

RPM(min @ 1:4 turndown) = .66rpm

 

third compartment

P=mVG² = m= 2.73x10^-5 lb.s/ft² x  51,780.9 ft³ /3(3 compartments)x 10²

P=46.95 ft.lb/s  x  1hp/550ft.lb/s

P₃=.085 hp

46.95 ft.lb/s  / 7wheels = 317,976R³

R = .0276 rps

RPM(max) = .0276 rps x 60s/min

RPM(max) = 1.66 rpm

RPM(min @ 1:4 turndown) = 1.66 rpm/4

RPM(min @ 1:4 turndown) = .42 rpm

 

 

HOMEWORK No. 5, Flocculation

Read Chapter 3 pp. 104-139

Problems:

 

5A. Given: Diatomaceous earth

Find: What dosage of alum is effective and what is the resulting turbidity.

 

5B. Given: The power delivered by a tank is 1.5hp, G=35s^-1, T=30°C.

Find: What are the required tank dimensions if L=W and the depth =15’, 1hp=550ft.lb/s

 

5C. Given:  See F.3.2.4-10. The boards are 6”x10’s except the diagonals which are 11’long and the vertical members which are 7’ long. The channel x-section is 11’ wide by 12’ deep.

Find: The total paddle area should be 10-25% of the tank cross-sectional area.

Has the criterion been met?

 

5D. Given: A baffled tank. G=60s^-1, the kinematic viscosity,n,  is .898x10^-6 m²/s, 25°C. and Q=3m³/s. There are 30 turns. The head loss at the turn is h=Kv²/2g, where K=1.5, p.125. t=10min

Find:

1.) total head loss

2.) The velocity at each turn, slit.

 

5E. Given: A cross-flow, horizontal shaft, paddle-wheel flocculation basis is to be designed for a flow of 25,000 m³/day with a mean G of 30s^-1 @ 10°C(m=1.307 centipoise, 1centipoise=10^-3N-s²/kg-m)  and a t=50 minutes. Assume L=3W and W=D. The Gt should be between 50,000-100,00. Tapered flocculation is to be provided and three compartments of equal depth in series are to be used. The G values determined from laboratory tests are: G₁=50s^-1, G₂=25s^-1 and G₃=15s^-1. These give an average of 30s^-1. The compartments are to be separated by slotted, redwood baffle fences and the floor of the basin is level. The basin should be 15m in width to adjoin the settling basin. The speed of the blades relative to the water is 3/4 of the peripheral blade speed. Assume 6 pieces of wood per paddle, 3 up, 3 down: D₁=1.70m, D₂=2.60m, D₃=3.50m, four assemblies (groups of 6 pieced of wood) per shaft. Each piece of wood is 15cm wide and 3m long.

Use the power equations: P=C_DArv³/2 and P=mVG², in which C_D=1.50, r=999.7kg/m³.

Find:

1.) Gt and V

2.) Basin dimensions

3.) Design the paddle wheels

4.) P in each compartment

5.) Rotational speed of each horizontal shaft in rpm

6.) The peripheral speed of the outside paddle blades in m/s

 

 

5F. Given: Design the flocculators for your project.