Lin S.D. Water and Wastewater Calculations Manual

Подождите немного. Документ загружается.

Public Water Supply 375

8.1 Jar test

For the jar test, chemical (coagulant) is added to raw water sample for

mixing in the laboratory to simulate treatment-plant mixing conditions.

Jar tests may provide overall process effectiveness, particularly to

mixing intensity and duration as it affects ﬂoc size and density. It can

also be used for evaluating chemical feed sequence, feed intervals, and

chemical dilution ratios. A basic description of the jar test procedure and

calculations is presented elsewhere (APHA 1995).

It is common to use six 2 L Gator jars with various dosages of chemi-

cal (alum, lime, etc); and one jar as the control without coagulant.

Appropriate coagulant dosages are added to the 2 L samples before the

rapid mixing at 100 revolution per minute (rpm) for 2 min. Then the sam-

ples and the control were ﬂocculated at 20 rpm for 20 or more minutes

and allow to settle. Water temperature, ﬂoc size, settling characteristics

(velocity, etc.), color of supernatant, pH, etc. should be recorded.

The following examples illustrate calculations involved in the jar

tests; prepare stock solution, jar test solution mixture.

Example 1: Given that liquid alum is used as a coagulant. Speciﬁc gravity

of alum is 1.33. One gallon of alum weighs 11.09 pounds (5.03 kg) and contains

5.34 pounds (5.42 kg) of dry alum. Determine: (a) the alum concentration,

(b) mL of liquid alum required to prepare a 100 mL solution of 20,000 mg/L

alum concentration, (c) the dosage concentration of 1 mL of stock solution in

a 2000 mL Gator jar sample.

solution:

Step 1. Determine alum concentration in mg/mL

Step 2. Prepare 100 mL stock solution having a 20,000 mg/L alum

concentration

Let x ⫽ mg of alum required to prepare 100 mL stock solution

Step 3. Calculate mL (y) of liquid alum to give 2000 mg

y 5 3.125 mL

y mL

1 mL

2000 mg

640 mg

x 5 2000 mg

100 mL

20,000 mg

1000 mL

5 640 mg/mL

Alum smg/Ld 5

s5.34 lbd s453,600 mg/lbd

s1 galds3785 mL/gald

Note: Or add 6.25 mL of liquid alum into the 200 mL stock solution, since

3.125 mL is difﬁcult to accurately measure.

Step 4. Find 1 mL of alum concentration (z) in 2000 mL sample (jar)

Note: Actual ﬁnal volume is 2001 mL; using 2000 mL is still reasonable.

Example 2: Assuming that 1 mL of potassium permanganate stock solu-

tion provides l mg/L dosage in a 2 L jar, what is the weight of potassium

permanganate required to prepare 100 mL of stock solution? Assume

100% purity.

solution:

Step 1. Find concentration (x mg/L) of stock solution needed

Step 2. Calculate weight ( y mg) to prepare 100 mL (as 2000 mg/L)

Add 200 mg of potassium permanganate into 100 mL of deionized water to

give 100 mL stock solution of 2000 mg/L concentration.

Example 3: Liquid polymer is used in a 2 L Gator jar test. It weighs

8.35 lb/gal. One mL of polymer stock solution provides 1 mg/L dosage.

Compute how many mL of polymer are required to prepare 100 mL of stock

solution.

solution:

Step 1. Calculate (x) mg of polymer in l mL

5 1000 mg/mL

x 5

s8.35 lbds453,600 mg/lbd

s1 galds3785 mL/gald

y 5 200 mg

y mg

100 mL

2000 mg

1000 mL

x 5 2000 mg/L

sx mg/Ld s1 mLd 5 s1 mg/Ld sas 2000 mLd

5 10 mg/L

z 5

20,000

2000

smg/Ld

sz mg/Ld s2000 mLd 5 s20,000 mg/Lds1 mLd

376 Chapter 5

Step 2. As in Example 2, ﬁnd the weight (y mg) to prepare 100 mL of 2000

mg/L stock solution

Step 3. Compute volume (z mL) of liquid polymer to provide 200 mg

Note: Use 0.20 mL of polymer in 100 mL of distilled water. This stock solu-

tion has 2000 mg/L polymer concentration. One mL stock solution added to

a 2 L jar gives 1 mg/L dosage.

8.2 Mixing

Mixing is an important operation for the coagulation process. In prac-

tice, rapid mixing provides complete and uniform dispersion of a chem-

ical added to the water. Then follows a slow mixing for ﬂocculation

(particle aggregation). The time required for rapid mixing is usually 10

to 20 s. However, recent studies indicate the optimum time of rapid

mixing is a few minutes.

Types of mixing include propeller, turbine, paddle, pneumatic, and

hydraulic mixers. For a water treatment plant, mixing is used for coag-

ulation and ﬂocculation, and chlorine disinfection. Mixing is also used

for biological treatment processes for wastewater. Rapid mixing for

coagulant in raw water and activated-sludge process in wastewater

treatment are complete mixing. Flocculation basins after rapid mixing

are designed based on an ideal plug ﬂow using ﬁrst-order kinetics. It is

very difﬁcult to achieve an ideal plug ﬂow. In practice, bafﬂes are

installed to reduce short-circuiting. The time of contact or detention

time in the basin can be determined by:

For complete mixing

(5.52a)

For plug ﬂow

(5.52b)t 5

aln

t 5

2 C

5 0.20 mL

z 5

200 mg

200 mg 3 1 mL

1000 mg

y 5 200 mg sof polymer in 100 mL distilled waterd

y mg

100 mL

2000 mg

1000 mL

Public Water Supply 377

where t ⫽ detention time of the basin, min

V ⫽ volume of basin, m

or ft

Q ⫽ ﬂow rate, m

Ⲑs or cfs

K ⫽ rate constant

⫽ inﬂuent reactant concentration, mgⲐL

⫽ efﬂuent reactant concentration, mgⲐL

L ⫽ length of rectangular basin, m or ft

v ⫽ horizontal velocity of ﬂow, m/s or ftⲐs

Example 1: Alum dosage is 50 mgⲐL, 379 ⫽ 90 per day based on laboratory

tests. Compute the detention times for complete mixing and plug ﬂow reac-

tor for 90% reduction.

solution:

Step 1. Find C

Step 2. Calculate t for complete mixing

Using Eq. (5.52a)

Step 3. Calculate t for plug ﬂow

Using Eq. (5.52b)

Power requirements.

Power required for turbulent mixing is tradition-

ally based on the velocity gradient or G values proposed by Camp and

Stein (1943). The mean velocity gradient G for mechanical mixing is

(5.53)G 5 a

␮V

1/2

5 36.8 min

t 5

aln

b 5

1440 min

aln

5 144 min

1day

1440 min

1 day

3 9

t 5

2 C

b 5

90/day

50 mg/L 2 5 mg/L

5 mg/L

5 5 mg/L

5 s1 2 0.9dC

5 0.1 3 C

5 0.1 3 50 mg/L

378 Chapter 5

where G ⫽ mean velocity gradient; velocity (ft/s)/distance (ft) is equal

to per second

P ⫽ power dissipated, ft ⭈ lb/s or N ⭈ m/s (W)

m ⫽ absolute viscosity, lb ⭈ s/ft

or N ⭈ s/m

V ⫽ volume of basin, ft

or m

The equation is used to calculate the mechanical power required to

facilitate rapid mixing. If a chemical is injected through oriﬁces with

mixing times of approximately 1.0 s, the G value is in the range of 700

to 1000/s. In practice, G values of 3000 to 5000/s are preferable for rapid

mixing (ASCE and AWWA, 1990).

Eq. (5.53) can be expressed in terms of horsepower (hp) as

(5.54a)

For SI units the velocity gradient per second is:

(5.54b)

where kW ⫽ energy input, kW

V ⫽ effective volume, m

m ⫽ absolute viscosity, centipoise, cp

⫽ 1 cp at 20°C (see Table 4.1a, 1cp ⫽ 0.001 N ⋅ s/m

)

The equation is the standard design guideline used to calculate the

mechanical power required to facilitate rapid mixing. Camp (1968)

claimed that rapid mixing at G values of 500 to 1000/s for 1 to 2 min pro-

duced essentially complete ﬂocculation and no further beneﬁt for pro-

longed rapid mixing. For rapid mixing the product of Gt should be 30,000

to 60,000 with t (time) generally 60 to 120 s.

Example 2: A rapid mixing tank is l m ⫻ l m ⫻ 1.2 m. The power input is

746 W (1 hp). Find the G value at a temperature of 15°C.

solution: At 10°C, m ⫽ 0.00113 N ⭈ s/m

(from Table 4.1a)

V ⫽ 1m ⫻ 1m ⫻ 1.2 m ⫽ 1.2 m

P ⫽ 746 W ⫽ 746 N ⋅ m/s

G 5 a

kW 3 10

1/2

G 5 a

550 hp

1/2

Public Water Supply 379

Using Eq. (5.53):

9 Flocculation

After rapid mixing, the water is passed through the ﬂocculation basin.

It is intended to mix the water to permit agglomeration of turbidity set-

tled particles (solid capture) into larger ﬂoes which would have a mean

velocity gradient ranging 20 to 70 s

⫺1

for a contact time of 20 to 30 min

taking place in the ﬂocculation basin. A basin is usually designed in four

compartments (ASCE and AWWA 1990).

The conduits between the rapid mixing tank and the ﬂocculation

basin should maintain G values of 100 to 150 s

⫺1

before entering the

basin.

For bafﬂed basin, the G value is

(5.55)

where G ⫽ mean velocity gradient, s

⫺1

Q ⫽ ﬂow rate, ft

␥ ⫽ speciﬁc weight of water, 62.4 lb/ft

H ⫽ head loss due to friction, ft

␮ ⫽ absolute viscosity, lb ⋅ s/ft

V ⫽ volume of ﬂocculator, ft

t ⫽ detention time, s

For paddle ﬂocculators, the useful power input of an impeller is

directly related to the drag force of the paddles (F). The drag force is the

product of the coefﬁcient of drag (C

) and the impeller force (F

). The drag

force can be expressed as (Fair et al., 1968):

(5.56a)

(5.57)

then

(5.56b)F 5 0.5 C

rAv

5 rA

F 5 C

G 5 a

QgH

0.5

5 a

62.4H

0.5

5 742 s

5 a

746 N

m/s

0.00113 N

s/m

3 1.2 m

0.5

G 5 sP/mVd

0.5

380 Chapter 5

where F ⫽ drag force, lb

⫽ dimensionless coefﬁicient of drag

r ⫽ mass density, lb ⋅ s

/ft

A ⫽ area of the paddles, ft

v ⫽ velocity difference between paddles and water, fps

The velocity of paddle blades (v

) can be determined by

(5.58)

where v

⫽ velocity of paddles, fps

n ⫽ number of revolutions per minute, rpm

r ⫽ distance from shaft to center line of the paddle, ft

The useful power input is computed as the product of drag force and

velocity difference as below:

(5.59)

Example 1: In a bafﬂed basin with detention time of 25 min. Estimate head

loss if G is 30/s, m ⫽ 2.359 ⫻ 10

⫺5

lb ⭈ s/ft

at T ⫽ 60°F (Table 4.1b).

solution: Using Eq. (5.55)

Example 2: A bafﬂed ﬂocculation basin is divided into 16 channels by 15

around-the-end bafﬂes. The velocities at the channels and at the slots are

0.6 and 2.0 fps (0.18 and 0.6 m/s), respectively. The ﬂow rate is 12.0 cfs (0.34 m

/s).

Find (a) the total head loss neglecting channel friction; (b) the power dis-

sipated; (c) the mean velocity gradient at 60°F (15.6°C); the basin size is

16 ⫻ 15 ⫻ 80 ft

; (d) the Gt value, if the detention (displacement) time is

20 min, and (e) loading rate in gpd/ft

solution:

Step 1. Estimate loss of head H

Loss of head in a slot 5

64.4

5 0.0621 sftd

Loss of head in a channel 5

0.6

2 3 32.2

5 0.00559 sftd

5 0.51 ft

H 5

62.4

s30s

s2.359 3 10

s/ft

d 3 s25 3 60 sd

62.4 lb/ft

G 5 a

62.4H

1/2

P 5 Fv 5 0.5 C

r Av

2p r n

Public Water Supply 381

(a)

Step 2. Compute power input P

(b)

Note: g ⫽ 62.37 lb/ft

at 60°F (Table 5.1b)

Step 3. Compute G

From Table 4.1b

At 60°F,

Using Eq. (5.53)

(c)

Step 4. Compute Gt

(d)

Step 5. Compute loading rate

Q ⫽ 12 cfs ⫽ 12 cfs ⫻ 0.646 MGD/cfs

⫽ 7.75 MGD

⫽ 7.75 ⫻ 10

gpd

(e) Loading rate ⫽ Q/V ⫽ 7.76 ⫻ 10

gpd/19,200 ft

⫽ 404 gpd/ft

Example 3: A ﬂocculator is 16 ft (4.88 m) deep, 40 ft (12.2 m) wide, and 80 ft

(24.4 m) long. The ﬂow of the water plant is 13 MGD (20 cfs, 0.57 m

/s).

Rotating paddles are supported parallel to four horizontal shafts. The rotat-

ing speed is 2.0 rpm. The center line of the paddles is 5.5 ft (1.68 m) from the

shaft (mid-depth of the basin). Each shaft equipped with six paddles. Each

paddle blade is 10 in (25 cm) wide and 38 ft (11.6 m) long. Assume the mean

velocity of the water is 28% of the velocity of the paddles and their drag coef-

ﬁcient is 1.9. Estimate:

5 49.200

Gt 5 41 s

3 20 min 3 60 s/min

5 41.0 s

G 5 a

1/2

5 a

763 f t”

lb/s

2.359 3 10

s/f t

3 19,200 f t

1/2

V 5 16 ft 3 15 ft 3 80 ft 5 19,200 ft

m 5 2.359 3 10

s/ft

5 763 ft

lb/s

P 5 Q

H 5 12 ft

/s 3 62.37 lb/ft

3 1.02 ft

5 1.02 sftd

H 5 16 3 0.00559 1 15 3 0.0621

382 Chapter 5

(a) the difference in velocity between the paddles and water

(b) the useful power input

(d) the detention time

(e) the value of G and Gt at 60°F

(f) the loading rate of the ﬂocculator

solution:

Step 1. Find velocity differential v

Using Eq. (5.58)

(a)

⫽ 0.83 fps

Step 2. Find P

A ⫽ paddle area ⫽ 4 shaft ⫻ 6 pads/shaft ⫻38 ft ⫻ 10/12 ft/pad ⫽ 760 ft

Using Eq. (5.59):

(b) P ⫽ 0.5C

␳A␯

⫽ 0.5 ⫻ 1.9 ⫻ 1.938 lb ⭈ s

/ft

⫻ 760 ft

⫻ (0.83 fps)

⫽ 800 ft ⋅ lb/s

or ⫽ (800/550) hp

⫽ 1.45 hp

or ⫽ 1.45 ⫻ 0.746 kW

⫽ 1.08 kW

Step 3. Determine energy consumption E

(c)

5 1.99 kW

h/Mgal

E 5

1.08

13 Mgal/d

24 h

5 2.68 hp

h/Mgal

E 5

1.45 hp

13 Mgal/d

24 h

r 5

62.4 lb/f t

32.2 f t/s

5 1.938 lb

/f t

v 5 v

s1 2 0.28d 5 1.15 fps 3 0.72

5 1.15 fps

2prn

2 3 3.14 3 5.5 ft 3 2

60 s

Public Water Supply 383

Step 4. Determine detention time t

Basin volume, V ⫽ 16 ft ⫻ 40 ft ⫻ 80 ft ⫽ 5.12 ⫻ 10

⫽ 5.12 ⫻ 10

⫻ 7.48 gal/ft

⫽ 3.83 ⫻ 10

gal

(d)

⫽ 0.0295 day ⫻ 1440 min/d

⫽ 42.5 min

Step 5. Compute G and Gt

Using Eq. (5.53):

(e) G ⫽ (P/mV)

0.5

⫽ (800/2.359 ⫻ 10

⫺5

⫻ 5.12 ⫻ 10

)

0.5

⫽ 25.7 (fps/ft, or s

⫺1

)

Gt ⫽ (25.7s

⫺1

) (42.5 ⫻ 60 s)

⫽ 65,530

Step 6. Compute the loading rate

(f )

⫽ 254 gpd/ft

10 Sedimentation

Sedimentation is one of the most basic processes of water treatment.

Plain sedimentation, such as the use of a presedimentation basin (grit

chamber) and sedimentation tank (or basin) following coagulation—

ﬂocculatiaon, is the most commonly used in water treatment facilities.

The grit chamber is generally installed upstream of a raw water pump-

ing station to remove larger particles or objects. It is usually a rectan-

gular horizontal-ﬂow tank with a contracted inlet and the bottom should

have at minimum a 1:100 longitudinal slope for basin draining and

cleaning purposes. A trash screen (about 2-cm opening) is usually

installed at the inlet of the grit chamber.

Loading rate 5

13 3 10

gpd

5.12 3 10

t 5

3.83 3 10

gal

13 3 10

gal/d

5 0.0295 day

384 Chapter 5