Lin S.D. Water and Wastewater Calculations Manual

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solution:

Step 1. UC ⫽ d

⫽ 0.74 mm/0.54 mm

⫽ 1.37

Step 2. Find d

using Eq. (5.72)

⫽ d

(10

1.67 log UC

)

⫽ 0.54 mm(10

1.67 log 1.37

)

⫽ 0.54 mm(10

0.228

)

⫽ 0.91 mm

11.2 Mixed media

Mixed media are popular for ﬁltration units. For the improvement

process performance, activated carbon or anthracite is added on the top

of the sand bed. The approximate speciﬁc gravity (s) of ilmenite sand,

silica sand, anthracite, and water are 4.2, 2.6, 1.5, and 1.0, respectively.

For equal settling velocities, the particle sizes for media of different

speciﬁc gravity can be computed by

(5.73)

where d

, d

⫽ diameter of particles 1 and 2, respectively

, s ⫽ speciﬁc gravity of particles 1, 2, and water,

respectively

Example: Estimate the particle sizes of ilmenite (speciﬁc gravity ⫽ 4.2) and

anthracite (speciﬁc gravity ⫽ 1.5) which have same settling velocity of silica

sand 0.60 mm in diameter (speciﬁc gravity ⫽ 2.6)

solution:

Step 1. Find the diameter of anthracite by Eq. (5.73)

Step 2. Determine diameter of ilmenite sand

5 0.38 mm

d 5 s0.6 mmda

2.6 2 1

4.2 2 1

2/3

5 1.30 mm

d 5 s0.6 mmda

2.6 2 1

1.5 2 1

2/3

5 a

2 s

2/3

Public Water Supply 395

11.3 Hydraulics of ﬁlter

Head loss for ﬁxed bed ﬂow. The conventional ﬁxed-bed ﬁlters use a

granular medium of 0.5 to 1.0 mm size with a loading rate or ﬁltration

velocity of 4.9 to 12.2 m/h (2 to 5 gpm/ft

). When the clean water ﬂows

through a clean granular (sand) ﬁlter, the loss of head (pressure drop)

can be estimated by the Kozeny equation (Fair et al., 1968):

(5.74)

where h ⫽ head loss in ﬁlter depth L, m, or ft

k ⫽ dimensionless Kozeny constant, 5 for sieve

openings, 6 for size of separation

g ⫽ acceleration of gravity, 9.81 m/s or 32.2 ft/s

m ⫽ absolute viscosity of water, N ⋅ s/m

or lb ⭈ s/ft

␳ ⫽ density of water, kg/m

or lb/ft

␧ ⫽ porosity, dimensionless

A/V ⫽ grain surface area per unit volume of grain

⫽ speciﬁc surface S (or shape factor ⫽ 6.0 to 7.7)

⫽ 6/d for spheres

⫽ 6/␺d

for irregular grains

␺ ⫽ grain sphericity or shape factor

⫽ grain diameter of spheres of equal volume

v ⫽ ﬁltration (superﬁcial) velocity, m/s or fps

The Kozeny (or Carmen–Kozeny) equation is derived from the fun-

damental Darcy–Waeisback equation for head loss in circular pipes

(Eq. (4.17)). The Rose equation is also used to determine the head loss

resulting from the water passing through the ﬁlter medium.

The Rose equation for estimating the head loss through ﬁlter medium

was developed experimentally by Rose in 1949 (Rose, 1951). It is appli-

cable to rapid sand ﬁlters with a uniform near spherical or spherical

medium. The Rose equation is

(5.75)

where h ⫽ head loss, m or ft

␸ ⫽ shape factor (Ottawa sand 0.95, round sand 0.82,

angular sand 0.73, pulverized coal 0.73)

⫽ coefﬁcient of drag (Eq. (5.64))

Other variables are deﬁned previously in Eq. (5.74)

h 5

1.067 C

␸gde

kms1 2 ed

gre

396 Chapter 5

Applying the medium diameter to the area to volume ratio for homo-

geneous mixed beds the equation is

(5.76)

For a stratiﬁed ﬁlter bed, the equation is

(5.77)

where x ⫽ percent of particles within adjacent sizes

⫽ geometric mean diameter of adjacent sizes

Example 1: A dual medium ﬁlter is composed of 0.3 m (1 ft) anthracite

(mean size of 2.0 mm) that is placed over a 0.6 m (2 ft) layer of sand (mean

size 0.7 mm) with a ﬁltration rate of 9.78 m/h (4.0 gpm/ft

). Assume the grain

sphericity is

␺ ⫽ 0.75 and a porosity for both is 0.40. Estimate the head loss

of the ﬁlter at 15°C.

solution:

Step 1. Determine head loss through anthracite layer

Using the Kozeny equation (Eq. (5.74))

where k ⫽ 6

g ⫽ 9.81 m/s

m/r ⫽ ␯ ⫽ 1.131 ⫻ 10

⫺6

⋅ s (Table 4.1a) at 15°C

␻ ⫽ 0.40

A/V ⫽ 6/0.75d ⫽ 8/d ⫽ 8/0.002

v ⫽ 9.78 m/h ⫽ 0.00272 m/s

L ⫽ 0.3 m

then

5 0.0508 smd

h 5 6 3

1.131 3 10

9.81

s1 2 0.4d

0.4

3 a

0.002

s0.00272ds0.3d

kms1 2 ed

gre

␷

h 5

1.067Lv

␸ge

⌺

h 5

1.067C

␸ge

⌺

Public Water Supply 397

Step 2. Compute the head loss passing through the sand

Most input data are the same as in Step 1, except

k ⫽ 5

d ⫽ 0.0007 m

L ⫽ 0.6 m

then

Step 3. Compute total head loss

h ⫽ 0.0508 m ⫹ 0.6918 m

⫽ 0.743 m

Example 2: Using the same data given in Example 1, except the average size

of sand is not given. From sieve analysis, d

(10-percentile of size diameter) ⫽

0.53 mm, d

⫽ 0.67 mm, d

⫽ 0.73 mm, d

⫽ 0.80 mm, and d

⫽ 0.86 mm,

estimate the head of a 0.6 m sand ﬁlter at 15°C.

solution:

Step 1. Calculate head loss for each size of sand in the same manner as

Step 2 of Example 1

⫽ 0.3387 ⫻ 10

⫺6

⫽ 0.3387 ⫻ 10

⫺6

/(0.00053)

⫽ 1.206 (m)

⫽ 0.3387 ⫻ 10

⫺6

/(0.00067)

⫽ 0.755 (m)

Similarly h

⫽ 0.635 m

⫽ 0.529 m

⫽ 0.458 m

Step 2. Taking the average of the head losses given above

h ⫽ (1.206 ⫹ 0.755 ⫹ 0.635 ⫹ 0.529 ⫹ 0.458)Ⲑ5

⫽ 0.717 (m)

h 5 5 3

1.131 3 10

9.81

0.6

0.4

s0.00272ds0.6d

5 0.3387 3 10

/s0.0007d

5 0.6918 smd

398 Chapter 5

Note: This way of estimation gives slightly higher values than

Example 1.

Head loss for a fluidized bed. When a filter is subject to back wash-

ing, the upward flow of water travels through the granular bed at a

sufficient velocity to suspend the filter medium in the water. This is

fluidization. During normal filter operation, the uniform particles of

sand occupy the depth L. During backwashing the bed expands to a

depth of L

. When the critical velocity (␷

) is reached, the pressure

drops (⌬p ⫽ h␳g ⫽ ␥c) and is equal to the buoyant force of the grain.

It can be expressed as

h␥ ⫽ (␥

⫺ ␥) (1 ⫺ ␧

) L

h ⫽ L

(1 ⫺ ␧

)(␥

⫺ ␥)/␥ (5.78)

The porosity of the expanded bed can be determined by using (Fair

et al., 1963)

␧

⫽ (u/v

)

0.22

(5.79)

where u ⫽ upﬂow (face) velocity of the water

⫽ terminal settling velocity of the particles

A uniform bed of particles will expand when

(5.80)

The depth relationship of the unexpanded and expanded bed is

(5.81)

The minimum ﬂuidizing velocity (u

) is the superﬁcial ﬂuid velocity

needed to start ﬂuidization. The minimum ﬂuidization velocity is impor-

tant in determining the required minimum backwashing ﬂow rate. Wen

and Yu (1966) proposed the U

equation excluding shape factor and

porosity of ﬂuidization:

(5.82)

s1135.69 1 0.0408G

0.5

33.7 m

5 La

1 2 e

u 5 v

4.5

Public Water Supply 399

where

(5.83)

Other variables used are expressed in Eq. (5.74).

In practice, the grain diameter of spheres of equal volume d

is not

available. Thus the d

sieve size is used instead of d

. A safety factor

of 1.3 is used to ensure adequate movement of the grains (Cleasby and

Fan, 1981).

Example: Estimate the minimum ﬂuidization velocity and backwash rate

for the sand ﬁlter at 15°C. The d

size of sand is 0.88 mm. The density of sand

is 2.65 g/cm

solution:

Step 1. Compute the Galileo number

From Table 4.1a, at 15°C

␳ ⫽ 0.999 g/cm

m ⫽ 0.00113 N ⋅ s/m

⫽ 0.00113 kg/m ⋅ s ⫽ 0.0113 g/cm ⋅ s

m/r ⫽ 0.0113 cm

g ⫽ 981 cm/s

d ⫽ 0.088 cm

␳

⫽ 2.65 g/cm

Using Eq. (5.83)

Step 2. Compute U

by Eq. (5.82)

5 0.627 scm/sd

0.0113

0.999 3 0.088

s1135.69 1 0.0408 3 8635d

0.5

33.7 3 0.0113

0.999 3 0.088

5 8635

5 s0.088d

s0.999ds2.65 2 0.999ds981d/s0.0113d

5 d

rsr

2 rdg/m

5 d

rsr

2 rdg/m

5 Galileo number

400 Chapter 5

Step 3. Compute backwash rate

Apply a safety factor of 1.3 to U

as backwash rate

Backwash rate ⫽ 1.3 ⫻ 0.627 cm/s ⫽ 0.815 cm/s

⫽ 0.00815 m/s

⫽ 0.00815 m/s ⫻ 86,400 s/d

⫽ 704.16 m/d or (m

⋅ d)

⫽ 704 m/d/(0.01705 gpm/ft

) ⫻ l (d/m)

⫽ 12.0 gpm/ft

11.4 Washwater troughs

In the United States, in practice washwater troughs are installed at even

spaced intervals (5 to 7 ft apart) above the gravity ﬁlters. The washwater

troughs are employed to collect spent washwater. The total rate of dis-

charge in a rectangular trough with free ﬂow can be calculated by (Fair

et al., 1968; ASCE and AWWA, 1990)

Q ⫽ Cwh

1.5

(5.84a)

where Q ⫽ ﬂow rate, cfs

C ⫽ constant (2.49)

w ⫽ trough width, ft

h ⫽ maximum water depth in trough, ft

For rectangular horizontal troughs of such a short length friction

losses are negligible. Thus the theoretical value of the constant C is

2.49. The value of C may be as low as 1.72 (ASCE and AWWA, 1990).

In European practice, backwash water is generally discharged to the

side and no troughs are installed above the ﬁlter. For SI units, the rela-

tionship is

Q ⫽ 0.808 wh

1.5

(5.84b)

where Q ⫽ ﬂow rate, m

/(m

⋅ s)

w ⫽ trough width, m

h ⫽ water depth in trough, m

Example 1: Troughs are 20 ft (6.1 m) long, 18-in (0.46 m) wide, and 8 ft

(2.44 m) to the center with a horizontal ﬂat bottom. The backwash rate is

24 in/min (0.61 m/min). Estimate: (1) the water depth of the troughs with free

ﬂow into the gullet, and (2) the distance between the top of the troughs and

the 30-in sand bed. Assuming 40% expansion and 6 in of freeboard in the

troughs and 8 in of thickness.

Public Water Supply 401

solution:

Step 1. Estimate the maximum water depth ( h) in trough

v ⫽ 24 (in/min) ⫽ 2 ft/60 s ⫽ 1/30 fps

A ⫽ 20 ft ⫻ 8 ft ⫽ 160 ft

Q ⫽ VA ⫽ 160/30 cfs

⫽ 5.33 cfs

Using Eq. (5.84a)

Q ⫽ 2.49wh

1.5

, w ⫽ 1.5 ft

h ⫽ (Q/2.49w)

2/3

⫽ [5.33/(2.49 ⫻ 1.5)]

2/3

⫽ 1.27 (ft)

Say h ⫽ 16 in ⫽ 1.33 ft

Step 2. Determine the distance ( y) between the sand bed surface and the top

troughs

Depth of sand bed ⫽ 30 in ⫽ 2.5 ft

Height of expansion ⫽ 2.5 ft ⫻ 0.4

Freeboard ⫽ 6 in ⫽ 0.5 ft

Thickness ⫽ 8 in ⫽ 0.67 ft (the bottom of the trough)

y ⫽ 2.5 ft ⫻ 0.4 ⫹ 1.33 ft ⫹ 0.5 ft ⫹ 0.67 ft

⫽ 3.5 ft

Example 2: A ﬁlter unit has surface area of 16 ft (4.88 m) wide and 30 ft

(9.144 m) long. After ﬁltering 2.88 Mgal (10,900 m

) for 50 h, the ﬁlter is back-

washed at a rate of 16 gpm/ft

(0.65 m/min) for 15 min. Find: (a) the average

ﬁltration rate, (b) the quantity of washwater, (c) percent of washwater to

treated water, and the ﬂow rate to each of the four troughs.

solution:

Step 1. Determine ﬂow rate Q

say ⫽ 120 m/d

Q 5

2,880,000 gal

50 h 3 60 min/h 3 16 ft 3 30 ft

5 2.0 gpm/ft

2 gal 3 0.0037854 m

3 1440 min 3 10.764 ft

min 3 ft

3 1 gal 3 1 day 3 1 m

5 117.3 m/d

402 Chapter 5

Public Water Supply 403

Step 2. Determine quantity of washwater q

q ⫽ (16 gal/min ft

) ⫻ 15 min ⫻ 16 ft ⫻ 30 ft

⫽ 115,200 gal

Step 3. Determine percentage (%) of washwater to treated water

% ⫽ 115,200 gal ⫻ 100%/2,880,000 gal

⫽ 4.0%

Step 4. Compute ﬂow rate (r) in each trough

r ⫽ 115,200 gal/(15 min ⫻ 4 troughs)

⫽ 1920 gpm

11.5 Filter efﬁciency

The ﬁlter efﬁciency is deﬁned as the effective ﬁlter rate divided by the

operation ﬁltration rate as follows (AWWA and ASCE, 1998):

(5.84c)

where E ⫽ ﬁlter efﬁciency, %

⫽ effective ﬁltration rate, gpm/ft

or m

/h (m/h)

⫽ operating ﬁltration rate, gpm/ft

or m

/h (m/h)

UFRV ⫽ unit ﬁlter run volume, gal/ft

or L/m

UBWV ⫽ unit backwash volume, gal/ft

or L/m

Example: A rapid sand ﬁlter operating at 3.7gpm/ft

(9.0 m/h) for 46 h. After

this ﬁlter run, 295 gal/ft

(12,000 L/m

) of backwash water is used. Find the

ﬁlter efﬁciency.

solution:

Step 1. Calculate operating ﬁltration rate, R

⫽ 3.7 gal/min/ft

⫻ 60 min/h ⫻ 46 h

⫽ 10,212 gal/ft

Step 2. Calculate effective ﬁltration rate, R

⫽ (10212 ⫺ 295) gal/ft

⫽ 9917 gal/ft

Step 3. Calculate ﬁlter efﬁciency, E, using Eq (5.84c)

E ⫽ 9917/10212 ⫽ 0.97

⫽ 97%

E 5

UFRV 2 UBWV

UFRV

404 Chapter 5

12 Water Softening

Hardness in water is mainly caused by the ions of calcium and magne-

sium. It may also be caused by the presence of metallic cations of iron,

sodium, manganese, and strontium. These cations are present with

anions such as Cl

–

and The carbonates and

bicarbonates of calcium, magnesium, and sodium are called carbonate

hardness or temporary hardness since it can be removed and settled by

boiling of water. Noncarbonate hardness is caused by the chloride and

sulfate slots of divalent cations. The total hardness is the sum of car-

bonate and noncarbonate hardness.

The classiﬁcation of hardness in water supply is shown in Table 5.5.

Although hard water has no health effects, using hard water would

increase the amount of soap needed and would produce scale on bath

ﬁxtures, cooking utensils. Hardness also causes scale and corrosion

in hot-water heaters, boilers, and pipelines. Moderate hard water

with 60 to 120 mg/L as CaCO

is generally publicly acceptable (Clark

et al., 1977).

12.1 Lime-soda softening

The hardness in water can be removed by precipitation with lime,

Ca(OH)

and soda ash, Na

and by an ion exchange process. Ion

exchange is discussed in another section. The reactions of lime-soda

ash precipitation for hardness removal in water and recarbonation are

shown in the following equations:

Removal of free carbon dioxide with lime

(5.85)

Removal of carbonate hardness with lime

(5.86)Ca

1 2HCO

1 CasOHd

2CaCO

T 1 2H

d 1 CasOHd

CaCO

T 1 H

SiO

.NO

,HCO

TABLE 5.5 Classiﬁcation of Hard water

mg/L as CaCO

Hardness classiﬁcation US International

Soft 0⫺60 0–50

Moderate soft 51–100

Slightly hard 101–150

Moderate hard 61–120 151–200

Hard 121–180 201–300

Very hard >180 >300