Week 5

NASSERELDEEN AHMED KABASHI

DEPARTMENT OF BIOTECHNOLOGY

ENGINEERING

[email protected]

Sem II, 2014-2015

BTE 4217 Water Treatment Plant Design

WEEK 5

Flocculation process

Sem II, 2014-2015 Prof Kabbashi BTE 4217

Typical layout of a water treatment plant


Key Words Coagulation: adding and rapid mixing of chemicals to remove

particles from water. (flash mixing)

Flocculation: adding and slow mixing of chemicals and particles to create flocs that settle out of water.

Turbidity: suspended, dissolved, and colloidal particles in pretreated water that need to be removed to optimize treatment efficiency.

Suspended Solids: particles held in suspension by the natural action of

flowing waters.

Colloidal Solids: fine silt that does not settle out of water but remain

in suspension.

Dissolved Solids: organic or inorganic molecules that are dissolved

into the aqueous phase. Sem II, 2014-2015 Prof Kabbashi BTE 4217

Flocculation is the gentle mixing phase that follows

the rapid dispersion of coagulant by the flash mixing

unit. Its purpose is to accelerate the rate of particle

collisions, causing the agglomeration of colloids by

collisions to form separable flocs

Examples - milk, blood, seawater

Mechanisms - perikinetic, collisions from Brownian

motion

- orthokinetic, induced collisions through stirring

Orthokinetic flocculation

Velocity gradient, relative movement between colloids

in a fluid body RMS velocity gradient

Camp No. Gt Typical 2x 104 - 105


FLOCCULATION

A slow stirring process that causes the gathering together of small

particles into larger settleable ones.

Controlled by the rate of collisions between particles and the

effectiveness promoting attachment


FLOCCULATION PROCESS

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.

Flocculation Purpose


Importance of Flocculator Speed

If the speed of the stirring process is to great then the floc particles will

be sheared or broken apart causing

an increase in turbidity.

If flocculator speed is to slow then short-circuiting may occur


Desirable Floc Quality

The best floc are smooth circular particles these tend to settle quicker.

Irregular shaped particles settle slower.

Large clumps (popcorn floc) settle fast but are caused by chemical

overdosing.


Considerations

For designing a proper flocculation process must

consider:

Raw water quality and flocculation characteristics Treatment process and finished water quality goals

Available hydraulic headloss and plant flow variations

Choice of flocculation tank inlets Local conditions Cost Relation to existing treatment facilities Miscellaneous items


This is the first step, design engineer must have access to data for a period of five years or longer.

The 7 important water quality are Turbidity Total organic compounds pH Alkalinity Color Algae counts Temperature

The nature of colloids (organic), size distribution of the turbidity are preferably known.

Flocculation characteristics is evaluated by jar test.

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.

Raw water quality


Jar Tests

Determination of optimum pH

The jar test a laboratory procedure to determine the optimum pH and the optimum coagulant dose

A jar test simulates the coagulation and flocculation processes

Fill the jars with raw water sample (500 or 1000 mL) usually 6 jars

Adjust pH of the jars while mixing using H2SO4 or NaOH/lime

(pH: 5.0; 5.5; 6.0; 6.5; 7.0; 7.5)

Add same dose of the selected coagulant (alum or iron) to each jar

(Coagulant dose: 5 or 10 mg/L) Jar Test


Jar Test set-up

Rapid mix each jar at 100 to 150 rpm for 1 minute. The rapid mix

helps to disperse the coagulant throughout each container

Reduce the stirring speed to 25 to 30 rpm and continue mixing for 15 to 20 mins

This slower mixing speed helps

promote floc formation by

enhancing particle collisions,

which lead to larger flocs

Turn off the mixers and allow flocs to settle for 30 to 45 mins

Measure the final residual turbidity in each jar

Plot residual turbidity against pH

Jar Tests determining optimum pH


Optimum pH: 6.3

Jar Tests optimum pH


The overall treatment process and finished water goals are second important issue.

In direct filtration process, the floccculation tanks should not produce large settable flocs since sedimentation is not

involve.

The finished water quality goals also influence the degree of flocculation

Excessive amounts of the following substances in the raw water may effectively be removed by improving the

flocculation and sedimentation

Color as aprecursor to DBPs Algae as source of taste and odor Asbestos fibers Certain toxic metals and compounds

Treatment Process and Finished Water Quality


it is another consideration in the design of flocculation process

If the allowable headloss across is limited, hydraulic flocculation methods are ruled out.

Thus, the design engineer is forced to use mechanical methods

Also, plant flow rate flocculation is an important consideration.

If the flocculation is relatively minor year-round, that is, 50% variation from the daily average flow

rate, hydraulc flocculation is applicable and will

perform effectively.

Available Hydraulic Headloss and plant flow variations


Almost all reactor-clarifiers are designed with a single pipe inlet to the flocculation zone.

Most municipal water treatment plants have a rectangular flocculation tank located upstream of the sedimentation tanks or

flotation tanks (DAF system)

Most simple and cost effective design is a single inlet to each flocculation tank.

Engineers have two basic choices when designing the tank inlet A series of identical individual weirs set the same elevation A series of submerged orifices each fitted with an isolation valve or gate

4 basic problems with weir inlet choices The inlet channel must be large enough to maintain nearly equal water surface elevation

Heavy suspended solids accumulate in the channel due to the large size of the channel and flow velocity

Each tank is isolated by means of stop logs Slight changes in water depth result in significant changes in flow rate

Choice of Flocculation Tank Inlets


The second choice of submerged orifices fitted with isolation valves operators or gates

The drawbacks are 1. higher capital cost, 2. more loss of head (see table 3.2.4-1)

Advantages over weir inlets: The flow rate is NOT significantly affected by slight change in water level from inlet effluent to the other

point

Influent size may be much smaller and a velocity of 2-3 ft/ (0.6-0.9 m/s) may be allowed

There is min deposition of suspended solids in the influent due to higher velocity and also the flow

exists near the bottom of the channel

The isolation valves provide virtually driptight closured

Choice of Flocculation Tank Inlets


When selecting an appropriate type of flocculation process, local conditions should be

analyzed

5 major factors should be evaluated are: Site topography Climate conditions Availability of services Capability of the operational personnel Level of the local technology

Local condition


Is always important and must be evaluated Both the initial cost (Capital) The operation and maintenance expenses

Cost


In case of plant expansion, the relation of the new flocculation process to the existing process is a realistic

problem.

Make all the flocculation units identical so that uniform performance and operation and

maintenance procedure can be maintained

Relation to Existing Treatment Facilities


Miscellaneous considerations such as the hydraulic characteristics of the flocculation tank and scum and

sludge removal, must be incorporated during the design

phase.

Prevention of flow short circuiting, the creation of effective eddies and turbulence with the tank, and

minimization of high shear forces by the mixing blades

Important operational problems include the ease of scum removal from water surface and of sludge and silt

from the basin

Some chemists also emphasize on ion concentration

Miscellaneous Items


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

Type and Selection Guide


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

Type and Selection Guide


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

Discussion of Alternatives


Type of treatment process: conventional, dissolved air flotation, direct filtration, softening, or sludge

conditioning

Raw water quality: turbidity, color, and temperature

Flocculation characteristics in response to change in mixing intensity and mixing times

The following criteria when selecting mixing type: Local condition as high wind etc Avaiilable headoss across the plant Shape and depth of the basin Capital and operation and maintenance (O&M) costs

Selection Criteria


The general design criteria for a basic rectangular

flocculation tank are as follows:

Energy input: Gt=10,000-100,000, 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

Design Criteria


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: , units p.117

Given: P= 850J/s, and the space influenced by the

flocculator 4x6x6m. The temperature is 15C.

Find:

1.)G

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

Design Criteria

5.0

V

PG


1.) G

From App.6, p.632 @15C, =1.17x10-3 N.s/m2,

N=Newton = kg.m/s2, the units of are kg/m.s

V = 4x6x6m= 144m3

=

G = 71s-1

2.)How many hp is 850J/s

1kW=1000J/s

850J/s x (1kW / 1000J/s) = 0.85kw

1hp = 0.746kw

0.85kw x (1hp/0.746kW) = 1.14hp

Motor size = 1.14/e = 1.14/0.7

Motor size = 1.63hp

Design Criteria

5.0

V

PG

33 144./1017.1

/850

msmKg

sJG


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.

Design Criteria


The mean velocity gradient for baffled systems is:

, units p.117

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

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

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.

Design Criteria

5.05.0

t

h

Vt

ghG


Example:

Given: Rectangular basin, L=W=4, D=4.5,

Q=8.7MGD, Water @50F, =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=1wide.

Design Criteria

Loss = .5V2/2g (typical)


v=3.37fps

h = head loss per baffle, see graphic =0.5v2/2g =

0.5(3.37)2/(2x32.2)

h = head loss per baffle = 0.0882ft

head total = 3 baffles x 0.0882ft

head total = 0.264ft

G = 336.3 fps/ft

Design Criteria

Loss = .5V2/2g (typical)

5.0

25

5.0

35.5/.1073.2

264.04.62

stftsIbt

hG


For mechanical systems with paddles:

G = (CDAv3/2V)0.5 units p.117

CD 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 0.5-0.75 of

the peripheral velocity of the paddle

Design Criteria


Given: If G = (CDAv3/2V).5 and

Find: A power equation

(CDAv3/2V).5 = square both sides

P = V x CDAv3/2V, =

P = CDAv3/2

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 6x10) 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 D/revolution = 4rev/min x 1min/60s x (7ftx2), 7 is

a radius, v = 2.93fps

Design Criteria 5.0

V

PG

5.0

V

PG


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

v, the relative velocity of the paddle with respect to the fluid is

from 0.5-0.75 of the peripheral velocity of the paddle.

v= 2.93fps x 0.5

v = 1.47fps

v= 2.93fps x 0.75

v = 2.20fps, use 0.75 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

Design Criteria


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 0.5 and 3.3fps.

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

last stage

Design Criteria


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)0.5, C=0.8, 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 =/4 D2 = 0.785 x (5/12)2 A = 0.136ft2

number or orifices = total area/orifice area = 20ft2 / 0.136ft2/orifice

number or orifices = 147 orifices

Design Criteria


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)0.5

50MGD x 1.547cfs/MGD = 0.8 x 0.136ft2 (2 x 32.2 x h).5

4.84 = (64.4h)0.5

23.39 = 64.4h

h = 0.36ft

Design Criteria


2/5/2015 water treatment 42

Slide 13 of 27


Transport Mechanisms

Brownian motion: for relatively small particles

which follow random motion and collide with

other particles (perikinetic motion)

Differential settling: Particles with different

settling velocities in the vertical alignment

collide when one overtakes the other (orthokinetic motion)

Design of Flocculator (Slow & Gentle mixing)

Flocculators are designed mainly to provide enough interparticle contacts to

achieve particles agglomeration so that they can be effectively removed by

sedimentation or flotation


Mechanical Mixing Systems

Selection of a proper flocculation unit depends on the

overall unit process that is selected

If the raw water quality is good and direct filtration is feasible, the filter bed will be a reverse-graded high rate

filter bed as dual media type

Floc should be small, physically strong to resist the high shearing within the bed.

Vertical shaft high energy flocculators produce a floc that meets these requirements.

When the raw water quality is poor the suspended matter is removed from raw water through the

production of good floc and effective sedimentation


Mixer Characteristics

1. It must deliver the G values

2. Must provide the sufficient eddies and turbulence for

required velocity gradient

3. Provide low shear forces

4. Low M&O cost


Vertical-Shaft Flocculator

It is the first choice for the following reasons

Minimal maintenance Operational flexibility Very little head loss across the tank Easy control of mixing intensity Effectiveness Minimal impact to the overall performance if one unit

malfunction


Cross flow Flocculator (sectional view)

Plan (top view)

L

H

W

Mechanical Flocculator


Mechanical flocculators


Mecahnical flocculators


Mechanical flocculators


Another mechanical flocculator


of 27

Differential settling flocculation


Flocculators integrated with settling


Flocculators both sides of settling


Flocculator perforated wall (in background)


Velocity Gradient: relative velocity of the two fluid particles/distance

G = dv/dy = 1.0/0.1 = 10 s-1

Mixing and Power

The degree of mixing is measured by Velocity Gradient (G)

Higher G value, intenser mixing

0.1

m

1 m/s

In mixer design, the following equation is useful

G= velocity gradient, s-1;

P = Power input, W

V = Tank volume, m3;

= Dynamic viscosity, (Pa.s)


G value for coagulation: 700 to 1000 S-1; 3000 to 5000 S-1 for Mixing time: 30 to 60 S in-line blender; 1-2 sec

G value for flocculation: 20 to 80 S-1; Mixing time: 20 to 60 min

In the flocculator design, Gt (also known Camp No.); a product of G

and t is commonly used as a design parameter

Typical Gt for flocculation is 2 x 104 - 105

Large G and small T gives small but dense floc

Small G and large T gives big but light flocs

We need big as well as dense flocs

which can be obtained by designing

flocculator with different G values

1 2 3 G1:40 G2:30 G3:20


Power Calculation

What horsepower level do we need to supply to a flocculation basin to

provide a G value of 100s-1 and a Gt of 100,000 for 10 MGD flow? (Given:

= 0.89 x 10-3 Pa.s; 1 hp = 745.7 watts)

Solution:

Retention time, t = Gt/G = 100,000/100 = 1000 secs

Volume of Flocculation basin, V = (0.438 m3/sec) x (1000 sec)

= 438 m3

P = G2 V x

= 1002 x 438 x 0.89 x10-3 = 3900 W

= 3900/746 = 5.2 hp


WATER TREATMENT ENERGY

CALCULATIONS

F = ma. In a gravity field, F = mg Force in N, where a N is the force to accelerate 1kg @1m/s2

Force to move h, Potential energy = Fh =

mgh Dimensions MLT-2L, kgm2s-2 = Nm or J

Force moving at a certain speed, introduces time dimension

Dimensions here are MT-1, L/s (1L=1kg)

Rate of energy usage, or power, P = mgh/t Dimensions are now ML2T-3, or kgm2s-3 = J/s or W.

Power (W) to pump water to h, flow rate in L/s (or kg/s)

W = kg/s x h x 9.8 m/s2

kW, divide by 1000 HP, divide by 746


Viscosity of water is a measure of its resistance

to flow

The cgs unit is the Poise, 1 gcm-1s-1.

Water viscosity is c. 1cP = 0.01P = 0.001 Pa.s

Pa = N/m2 or kgms-2m-2, so Pa.s = kgms-2m-2s =

kgm-1s-1

This could also have been derived from going

from gcm-1s-1, multiplying by 100/1000.

Therefore 1cP = 0.001kgm-1s-1

VISCOSITY MEASUREMENT


Calculation of Velocity Gradient

Calculate the velocity gradient in a flocculator, where

the required energy is 1 J/L. Flow rate is 4ML/d,

retention time = 20 min

Volume, V = 4000/(24 x 60/20) = 55.5 m3

Flow rate = 4000 x 1000 = 46.3 L/s

24 x 60 x 60

_____ _________________

G = P/V = 1 x 46.3/0.001x55.5

= 28 s-1


Calculate height required for hydraulic

flocculator

Calculate the head difference in water through a

hydraulic flocculator, where the required energy

input is 1 J/L and the flow rate is 4 ML/d.

Power = energy/time

1 J x L/s = kg/s x 9.8 x h

Therefore, h = 1/9.8 m

= 0.102m


Calculate Camp No

Calculate the Camp No for the hydraulic

flocculator in the previous example

Camp No = G.t

= 28 x 20 x 60

= 33,000 (within the boundaries of 20,000 200,000)


Where F = drag force, N

CD = dimensionless drag coefficient for plates moving faces normal to direction of

motion

A = cross-sectional area of the paddles, m2

= relative velocity between paddles and fluid, m/s

= density, 1000 kg/m3

The power input can be computed as the product of drag force and velocity:

P = F = CDA3/2

If this is substituted in the equation for G, the mean velocity gradient G becomes

G2 = P/V = CDA3/ 2V

CDA2

2

F = CDA2/2

PADDLE FLOCCULATORS


What you need to know

How to determine the velocity gradient and

volume, chemical and energy requirements

for flocculation

Be able to size settling tanks on the basis of

particle settling rates and identify important

zones in the settling tank

Softening calculations


Operation and Maintenance

O&M Three basic operational procedures:

check the size of the floc

removal of the scum from the water surface

algae control

Example

Given: A flocculation basin. Q=12MGD, horizontal shaft, paddle wheel. The

mean G=25s-1 @ 50F. t=45min. The Gt must be between 50,000-

100,000. Use 3 stages of equal depth in which the Gs decrease: 45,20,10.

L=.5W, L=3H. The paddles are to be made of redwood, 10x6. 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.

CD=1.50 for the paddles. Use the power equations P=0.97CDAv3 and

P=VG2.


Operation and Maintenance

O&M

Profile

24-36" typ

Flow in and out ofthe paper, 6 blades

1.5'

1.5'

12" typ6" typ

10' typ

Section

Plan

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.



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

V = 50,122.8ft3

L=0.5W, 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

Solution


2.)Paddle wheel

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

Since the blades are 10x6, the width is 10. The width of the tank is 85, 85/108 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 = 0.20HW = 0.20(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

Solution


Solution

1.5'

1.5'

12" typ6" typ

10' typConsider the shaft 2' indiameter

Starting from shaft at the center and working down1'(shaft) + 1'(clearance) +6/12 /2(1/2board) = 2.25'2.25' + 6/12 /2(1/2 board) + 1' (clearance) +6/12 /2 =3.75'3.75 +6/12 /2(1/2 board) + 1' (clearance) +6/12 /2 =5.25'

5.25'

3.75'

2.25'

H=14.25'

1.875'left over> 1,5', 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 D or 2r per revolution, rev/s x D/rev = D/sec

v = 0.75 x 2r x R(revolutions per second)

v1(first compartment) = 0.75 x 2(5.25) x R

v1(first compartment) = 24.74R

v2(second compartment) = 0.75 x 2(3.75) x R

v2(second compartment) = 17.67R

v3(third compartment) = 0.75 x 2(2.25) x R

v3(third compartment) = 10.60R

Solution


P=.97CDAv3 =0.97CDA1v1

3 +0.97CDA2v23 + 0.97CDA3v3

3 = 0.97CDA(v13 + v2

3 + v33), A1=A2=A3

P = 0.97(1.50)(0.5x10board dim.)(2 boards,1up,1down)[24.743+17.673+10.613]R3

P=317,976R3

first compartment

P=VG2 = = 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 =0.075 rps

RPM(max) = 0.075 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 2r

v1 = 0.075 x 2(5.25)

v1 = 2.47fps

Solution


second compartment

P=VG2 = = 2.73x10-5 lb.s/ft2 x 51,780.9 ft3 /3(3

compartments)x 202


P2=0.34hp

187.8 ft.lb/s / 7wheels = 317,976R3

R = 0.044 rps


RPM(max) = 2.64rpm

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

RPM(min @ 1:4 turndown) = 0.66rpm

Solution


third compartment

P=VG2 = = 2.73x10-5 lb.s/ft2 x 51,780.9 ft3 /3(3

compartments)x 102


P3=0.085 hp

46.95 ft.lb/s / 7wheels = 317,976R3

R = 0.0276 rps


RPM(max) = 1.66 rpm

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

RPM(min @ 1:4 turndown) = 0.42 rpm

Solution

Go over example problems in book, p.123


Performance Monitoring

Flow measurement is important to accurately

establish chemical feed rates, wash water

rates, and unit loadings

Mixing needs to be adequate

Chemical feed systems need dosage control

Monitor pH for optimum conditions

Jar Test at the Start of Every Shift or more!


Homework No. 3: 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=30C.

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 6x10s except the diagonals which are 11long 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,, is .898x10-6 m2/s, 25C. 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 @ 10C(=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=CDAv3/2 and P=VG2, in which CD=1.50, =999.7kg/m

3.

Find:

1.) Gt and V


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.


Thank You

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Week 5