Lin S.D. Water and Wastewater Calculations Manual

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Public Water Supply 365

Step 5. Compute the height of tower z by Eq. (5.38)

z ⫽ (HTU) (NTU) ⫽ 1.82 m ⫻ 2.42

⫽ 4.4 m

6.4 Nozzles

There are numerous commercially available spray nozzles. Nozzles and

tray aerators are air-water contact devices. They are used for iron and

manganese oxidation, removal of carbon dioxide and hydrogen sulﬁde

from water, and removal of taste and odor causing materials.

Manufacturers may occasionally have mass transfer data such as K

and a values. For an open atmosphere spray fountain, the speciﬁc inter-

facial area a is (Calderbrook and Moo-Young, 1961):

(5.43)

where d is the droplet diameter which ranges from 2 to 10,000 µm.

Under open atmosphere conditions, C

remains constant because there

is an inﬁnite air-to-water ratio and the Lewis and Whitman equation

can express the mass transfer for nozzle spray (Fair et al., 1968;

Cornwell, 1990)

⫺ C

⫽ (C

⫺ C

)[1 ⫺ exp(⫺K

at)] (5.44)

where c

, c

⫽ solute concentrations in bulk water and in

droplets, respectively, ␮g/L

t ⫽ time of contact between water droplets and air

⫽ twice the rise time, t

(5.45a)

or t

⫽ V Sin␾/g (5.45b)

where ␾ ⫽ angle of spray measured from horizontal

V ⫽ velocity of the droplet from the nozzle

(5.46)

where C

⫽ velocity coefﬁcient, 0.40 to 0.95, obtainable from

the manufacturer

g ⫽ gravitational acceleration, 9.81 m/s

V 5 C

22 gh

2V sin f

a 5

366 Chapter 5

Combine Eq. (5.45b) and Eq. (5.46)

The driving head h is

(5.47)

Neglecting wind effect, the radius of the spray circle is

r ⫽ Vt

cos ␾

⫽ V(V sin ␾/g) cos ␾

⫽ (V

/g) sin ␾ cos ␾

⫽ C

⋅ 2gh/g ⋅ sin 2␾

⫽ 2C

h sin 2␾ (5.48)

and the vertical rise of the spray is

From Eq. (5.47) (5.49)

Example 1: Determine the removal percentage of trichloroethylene from a

nozzle under an operating pressure of 33 psi (22.8 kPa).

The following data is given:

d ⫽ 0.05 cm

⫽ 0.50

␾ ⫽ 30°

⫽ 0.005 cm/s

⫽ 131 ␮g/L

solution:

Step 1. Determine the volumetric interfacial area a by Eq. (5.43):

Step 2. Compute the velocity of the droplet V by Eq. (5.46)

The pressure head

h 5 33

psi 3

70.3 cm of waterhead

1 psi

5 2320 cm

a 5

0.05 cm

5 120 cm

hs2C

sin

fd 5 C

h sin

h 5 gt

/s2C

sin

5 C

22 gh

sinf

Using Eq. (5.46)

Step 3. Compute the time of contact by Eq. (5.45a)

Step 4. Compute the mass transfer C

⫺ C

by Eq. (5.44)

⫽ 68.1 (␮g/L) ⫽ 52% remained

⫽ 48% removal

Example 2: Calculate the driving head, radius of the spray circle, and ver-

tical rise of spray for (a) a vertical jet and (b) a jet at 45° angle. Given the

expose time for water droplet is 2.2 s and C

is 0.90.

solution:

Step 1. For question (a), using Eq. (5.45b)

␾ ⫽ 90º

sin ␾ ⫽ sin 90 ⫽ 1

sin 2 ␾ ⫽ sin 180 ⫽ 0

Using Eq. (5.47)

⫽ 9.81 m/s

(1.1s)

/(2 ⫻ 0.9

⫻ 1

)

⫽ 7.33 m

Using Eq. (5.48)

5 0

r 5 2C

h sin 2f

h 5 gt

/s2C

sin

3 2.2 s 5 1.1 s

5 131 2 131 3 0.48

2 131 5 s0 2 131d[1 2 exp s 2 0.005 3 120 3 1.09d]

2 C

5 sC

2 C

d[1 2 exp s 2K

atd]

5 1.09 s

2 3 1067 3 0.5

981

t 5

2V sin f

2 3 1067 cm/s sin 30

981 cm/s

5 1067 cm/s

V 5 C

22 gh

5 0.5 22 3 981 3 2320

Public Water Supply 367

Since sin 2␾ ⫽ 0, and using Eq. (5.49)

Step 2. For question (b)

⫽ 1.1 s

␾ ⫽ 45º

sin 2 ␾ ⫽ sin 90 ⫽ 1

Using Eq. (5.47)

⫽ 9.81 m/s

⫻ (1.1 s)

/(2 ⫻ 0.9

⫻ 0.707

)

⫽ 14.66 m

Using Eq. (5.48)

and using Eq. (5.49)

7 Solubility Equilibrium

Chemical precipitation is one of the most commonly employed methods

for drinking water treatment. Coagulation with alum, ferric sulfate, or

ferrous sulfate, and lime softening involve chemical precipitation. Most

chemical reactions are reversible to some degree. A general chemical

reaction which has reached equilibrium is commonly written by the law

of mass action as

aA ⫹ bB ↔ cC ⫹ dD (5.50a)

where A, B ⫽ reactants

C, D ⫽ products

a, b, c, d ⫽ stoichiometric coefﬁcients for A, B, C, D, respectively

5 C

h sin

5 s0.9d

3 14.66 m 3 s0.707d

5 5.93 m

r 5 2C

h sin 2f

5 2s0.9d

s14.66 mds1d

5 23.7 m

h 5 gt

/s2C

sin fd

sin f 5 sin 45 5 1/ 22

5 0.707

5 5.93 m

3 9.81 m/s

3 s1.2 sd

368 Chapter 5

The equilibrium constant K

for the above reaction is deﬁned as

(5.50b)

where K

is a true constant, called the equilibrium constant, and the

square brackets signify the molar concentration of the species within

the brackets. For a given chemical reaction, the value of equilibrium

constant will change with temperature and the ionic strength of the

solution.

For an equilibrium to exist between a solid substance and its solu-

tion, the solution must be saturated and in contact with undissolved

solids. For example, at pH greater than 10, solid calcium carbonate

in water reaches equilibrium with the calcium and carbonate ions in

solution: consider a saturated solution of CaCO

that is in contact with

solid CaCO

. The chemical equation for the relevant equilibrium can

be expressed as

CaCO

(s) ↔ Ca

2⫹

(aq) ⫹ CO

⫺

(aq) (5.50c)

The equilibrium constant expression for the dissolution of CaCO

can

be written as

(5.50d)

Concentration of a solid substance is treated as a constant called K

mass-action equilibrium, thus [CaCO

] is equal to k

. Then

(5.50e)

The constant K

is called the solubility product constant. The rules

for writing the solubility product expression are the same as those

for the writing of any equilibrium constant expression. The solubil-

ity product is equal to the product of the concentrations of the ions

involved in the equilibrium, each raised to the power of its coefficient

in the equilibrium equation. For the dissolution of a slightly soluble

compound when the (brackets) concentration is denoted in moles,

the equilibrium constant is called the solubility product constant.

The general solubility product expression can be derived from the

general dissolution reaction

(5.50f)A

ssd 4 xA

1 yB

5 [Ca

][Co

] 5 K

[Ca

][CO

]

[CaCo

]

[C]

[D]

[A]

[B]

Public Water Supply 369

and is expressed as

(5.50g)

Solubility product constants for various solutions at or near room tem-

perature are presented in Appendix C.

If the product of the ionic molar concentration [A

y⫹

]

x⫺

]

is less than

the K

value, the solution is unsaturated, and no precipitation will occur.

In contrast, if the product of the concentration of ions in solution is greater

than K

value, precipitation will occur under supersaturated condition.

Example 1: Calculate the concentration of OH

⫺

in a 0.20 mole (M) solution

of NH

at 25°C. K

⫽ 1.8 ⫻ 10

⫺5

solution:

Step 1. Write the equilibrium expression

Step 2. Find the equilibrium constant

Step 3. Tabulate the equilibrium concentrations involved in the equilibrium:

Let x ⫽ concentration (M) of OH

⫺

(aq) ⫹ H

O (1) ↔ NH

⫹

(aq) ⫹ OH

⫺

(aq)

Initial: 0.2M 0 M 0 M

Equilibrium: (0.20 ⫺ x)M xM x M

Step 4. Inserting these quantities into Step 2 to solve for x

Neglecting x term, then approximately

Example 2: Write the expression for the solubility product constant for (a)

A1(OH)

and (b) Ca

(PO

)

x 5 1.9 3 10

0.2 3 1.8 3 10

[NH

][OH

]

[NH

]

sxdsxd

s0.20 2 xd

5 1.8 3 10

5 s0.20 2 xds1.8 3 10

[NH

][OH

]

[NH

]

5 1.8 3 10

saqd 1 H

Os1d 4 NH

saqd 1 OH

saqd

5 [A

]

370 Chapter 5

solution:

Step 1. Write the equation for the solubility equilibrium

(a) A1(OH)

(s) ↔ Al

3⫹

(aq) ⫹ 3OH

⫺

(aq)

(b) Ca

(PO

)

(s) ↔ 3Ca

2⫹

(aq) ⫹ 2PO

⫺

(aq)

Step 2. Write K

by using Eq. (5.50g) and from Appendix C

(a) K

⫽ [Al

3⫹

] [OH

⫺

]

⫽ 2 ⫻ 10

⫺32

(b) K

⫽ [Ca

2⫹

]

[PO

3⫺

]

⫽ 2.0 ⫻ 10

⫺29

Example 3: The K

for CaCO

is 8.7 ⫻ 10

⫺9

(Appendix C). What is the sol-

ubility of CaCO

in water in g/L?

solution:

Step 1. Write the equilibrium equation

CaCO

(s) ↔ Ca

2⫹

(aq) ⫹ CO

2⫺

(aq)

Step 2. Set molar concentrations

For each mole of CaCO

that dissolves, 1 mole of Ca

2⫹

and 1 mole of CO

2⫺

enter the solution.

Let x be the solubility of CaCO

in mol/L. The molar concentrations of Ca

2⫹

and CO

2⫺

are

[Ca

2⫹

] ⫽ x and [CO

2⫺

] ⫽ x

Step 3. Solve solubility x in M/L

1 mole of CaCO

⫽ 100 g/L of CaCO

then

8 Coagulation

Coagulation is a chemical process to remove turbidity and color pro-

ducing material that is mostly colloidal particles (1 to 200 millimi-

crons, mm) such as algae, bacteria, organic and inorganic substances,

x 5 9.3 3 10

s100 g/Ld

5 0.0093 g/L

5 9.3 mg/L

5 [Ca

][CO

] 5 8.7 3 10

sxdsxd 5 8.7 3 10

x 5 9.3 3 10

M/L

Public Water Supply 371

and clay particles. Most colloidal solids in water and wastewater are

negatively charged. The mechanisms of chemical coagulation involve

the zeta potential derived from double-layer compression, neutral-

ization by opposite charge, interparticle bridging, and precipitation.

Destabilization of colloid particles is inﬂuenced by the Van der Waals

force of attraction and Brownian movement. Detailed discussion of the

theory of coagulation can be found elsewhere (American Society of Civil

Engineers and American Water Works Association, 1990). Coagulation

and ﬂocculation processes were discussed in detail by Amirtharajah and

O’Melia (1990).

Coagulation of water and wastewater generally add either aluminum,

or iron salt, with and without polymers and coagulant aids. The process

is complex and involves dissolution, hydrolysis, and polymerization. pH

values play an important role in chemical coagulation depending on

alkalinity. The chemical coagulation can be simpliﬁed as the following

reaction equations:

Aluminum sulfate (alum):

(SO

) ⋅ 18H

O ⫹ 3Ca(HCO

)

→ 2A1(OH)

↓ ⫹ 3CaSO

⫹

6CO

⫹ 18H

O (5.51a)

When water does not have sufﬁcient total alkalinity to react with

alum, lime, or soda ash is usually also dosed to provide the required alka-

linity. The coagulation equations can be written as below:

(SO

)

⋅ 18H

O ⫹ 3Ca(OH)

→ 2A1(OH)

↓

⫹ 3CaSO

⫹ 18H

O (5.51b)

(SO

)

⋅ 18H

O ⫹ 3Na

⫹ 3H

O → 2A1(OH)

↓

⫹ 3Na

⫹ 3CO

⫹ 18H

O (5.51c)

Ferric chloride:

2FeCl

⫹ 3Ca(HCO

)

→ 2Fe(OH)

↓ ⫹ 3CaCl

⫹ 6CO

(5.51d)

Ferric sulfate:

Fe(SO

)

⫹ 3Ca(HCO

)

→ 2Fe(OH)

↓ ⫹ 3CaSO

⫹ 6CO

(5.51e)

Ferrous sulfate and lime:

FeSO

⋅ 7H

O ⫹ Ca(OH)

→ Fe(OH)

⫹ CaSO

⫹ 7H

O (5.51f)

followed by, in the presence of dissolved oxygen

4Fe(OH)

⫹ O

⫹ 2H

O → 4Fe(OH)

↓ (5.51g)

372 Chapter 5

Chlorinated copperas:

3FeSO

⋅ 7H

O ⫹ 1.5 Cl

→ Fe

(SO

)

⫹ FeCl

⫹ 21H

O (5.51h)

followed by reacting with alkalinity as above

(SO

)

⫹ 3Ca(HCO

)

→ 2Fe(OH)

↓ ⫹3CaSO

⫹ 6CO

(5.51i)

and

2FeCl

⫹ 3Ca(HCO

)

→ 2Fe(OH)

↓ ⫹3CaCl

⫹ 6CO

(5.5lj)

Each of the above reaction has an optimum pH range.

Example 1: What is the amount of natural alkalinity required for coagula-

tion of raw water with dosage of 15.0 mg/L of ferric chloride?

solution:

Step 1. Write the reactions equation (Eq. 5.51j) and calculate MW

2FeCl

⫹ 3Ca(HCO

)

→ 2Fe(OH)

⫹ 3CaCl

⫹ 6CO

2(55.85 ⫹ 3 ⫻ 35.45) 3[40.08 ⫹ 2(1 ⫹ 12 ⫹ 48)]

⫽ 324.4 ⫽ 486.3

The above equation suggests that 2 moles of ferric chloride react with

3 moles of Ca(HCO

)

Step 2. Determine the alkalinity needed for X

X ⫽ 1.50 ⫻ mg/L FeCl

⫽ 1.50 ⫻ 15.0 mg/L

⫽ 22.5 mg/L as Ca(HCO

)

Assume Ca(HCO

)

represents the total alkalinity of the natural water.

However, alkalinity concentration is usually expressed in terms of mg/L as

CaCO

. We need to convert X to a concentration of mg/L as CaCO

MW of CaCO

⫽ 40.08 ⫹ 12 ⫹ 48 ⫽ 100.1

MW of Ca(HCO

)

⫽ 162.1

Then

X 5 22.5 mg/L 3

100.1

162.1

5 13.9 mg/L as CaCO

mg/L CasHCO

mg/L FeCl

486.2

324.4

5 1.50

Public Water Supply 373

Example 2: A water with low alkalinity of 12 mg/L as CaCO

will be treated

with the alum-lime coagulation. Alum dosage is 55 mg/L. Determine the lime

dosage needed to react with alum.

solution:

Step 1. Determine the amount of alum needed to react with the natural

alkalinity

From Example 1

⫽ 19.4 mg/L as Ca(HCO

)

MW of A1

(SO

)

⋅ 18H

O ⫽ 27 ⫻ 2 ⫹ 3(32 ⫹ 16 ⫻ 4) ⫹ 18(2 ⫹ 16)

⫽ 666

MW of Ca(HCO

)

⫽ 162.1

Equation (5.51a) suggests that 1 mole of alum reacts with 3 moles of

Ca(HCO

)

Therefore, the quantity of alum to react with natural alkalinity is Y:

⫽ 26.6 mg/L as alum

Step 2. Calculate the lime required

The amount of alum remaining to react with lime is (dosage ⫺Y )

55 mg/L ⫺ 26.6 mg/L ⫽ 28.4 mg/L

MW of Ca(OH)

⫽ 40.1 ⫹ 2(16 ⫹ 1) ⫽ 74.1

MW of CaO ⫽ 40.1 ⫹ 16 ⫽ 56.1

Equation (5.51b) indicates that 1 mole of alum reacts with 3 moles of

Ca(OH)

. Ca(OH)

required:

⫽ 9.48 mg/L as Ca(OH)

Let dosage of lime required be Z

Z 5 9.48 mg/L as CasOHd

56.1 as CaO

74.1 as CasOHd

5 7.2 mg/L as CaO

28.4 mg/L alum

3 3 74.1 as CasOHd

666 alum

Y 5 19.4 mg/L as CasHCO

666 as alum

3 3 162.1 as CasHCO

Alkalinity 5 12 mg/L as CaCO

162.1 as CasHCO

100.1 as CaCO

374 Chapter 5