Lin S.D. Water and Wastewater Calculations Manual

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Manhole spacing varies with available sanitary sewer maintenance

methods. Typical manhole spacings range from 90 to 150 m (300 to 500

ft) in straight lines. For sewers larger than 1.5 m (5 ft), spacings of 150

to 300 m (500 to 1000 ft) may be used (ASCE and WPCF, 1982).

Where the elevation difference between inflow and outflow sewers

exceeds about 0.5 m (1.5 ft), sewer inflow that is dropped to the eleva-

tion of the outflow sewer by an inside or outside connection is called a

drop manhole (or drop inlet). Its purpose is to protect workers from the

splashing of wastewater, objectionable gases, and odors.

10.3 Inverted siphons (depressed sewers)

A sewer that drops below the hydraulic gradient to pass under an

obstruction, such as a railroad cut, subway, highway, conduit, or stream,

is often called an inverted siphon. More properly, it should be called a

depressed sewer. Because a depressed sewer acts as a trap, the veloc-

ity of sewer flow should be greater than 0.9 m/s (3 ft/s) or more for

domestic wastewater, and 1.25 to 1.5 m/s (4 to 5 ft/s) for stormwater, to

prevent deposition of solids (Metcalf and Eddy, Inc. 1991). Thus, some-

times, two or more siphons are needed with an inlet splitter box.

In practice, minimum diameters for depressed sewers are usually the

same as for ordinary sewers: 150 or 200 mm (6 or 8 in) in sanitary

sewers, and about 300 mm (12 in) in storm sewers (Metcalf and Eddy,

Inc. 1991).

The determination of the pipe size for depressed sewers is the same

as for water and wastewater mains. The size depends upon the maxi-

mum wastewater flow and the hydraulic gradient.

Due to high velocities in depressed sewers, several pipes in parallel

are commonly used. For example, it may be that a small pipe may be

designed large enough to carry the minimum flow; a second pipe carries

the difference between the minimum and average flow (or maximum

dry-weather flow); and a third pipe carries peak flow above the average

flow. Depressed sewers can be constructed of ductile iron, concrete, PVC,

or tile.

Example: Design a depressed sewer system using the following given

conditions:

■

Diameter of gravity sewer to be connected by depressed sewer ⫽ 910 mm

(36 in)

■

Slope of incoming sewer, S ⫽ 0.0016 m/m (ft/ft)

■

Minimum flow velocity in depressed sewer ⫽ 0.9 m/s (3 ft/s)

■

Length of depressed sewer ⫽ 100 m (328 ft)

■

Maximum sewer deression ⫽ 2.44 m (8 ft)

Wastewater Engineering 565

■

Design flows:

minimum flow ⫽ 0.079 m

/s (2.8 ft

/s)

average flow ⫽ 0.303 m

/s (10.7 ft

/s) ⫽ maximum dry-weather flow

full (maximum) flow ⫽ capacity of gravity sanitary sewer

■

Design three inverted siphons from the inlet chamber

(1) to carry minimum flow

(2) to carry flows from minimum to average

(3) to carry all flows above the average flow

■

Available fall from invert to invert ⫽ 1.0 m (3.3 ft)

■

Available head loss at inlet ⫽ 125 mm (0.5 ft)

■

Available head loss for friction in depressed sewer ⫽ 1.0 m (3.3 ft)

■

Available hydraulic grade line ⫽ 1 m/100 m ⫽ 0.01 m/m

■

n ⫽ 0.015 (ductile-iron pipe)

Note: The above information is required to design a depressed sewer system.

solution:

Step 1. Design the depressed sewer

(a) Calculate velocity and flow of the 910 mm (D) sewer for full flow

by Eq. (4.21) R ⫽ D/4 ⫽ 910 mm/4 ⫽ 227.5 mm

⫽ 0.2275 m

by Eq. (4.22) V ⫽ (1/n)R

2/3

1/2

⫽ (1/0.015) (0.2275)

2/3

(0.0016)

1/2

⫽ 0.994 (m/s)

Flow Q ⫽ AV ⫽ 3.14 (0.455 m)

(0.994 m/s)

⫽ 0.646 m

(b) Determine the size of the small depressed sewer pipe to carry the mini-

mumflow (d ⫽ diameter of the pipe)

Q ⫽ ␲(d/2)

(1/n)(d/4)

2/3

1/2

⫽ (0.3115/n) d

8/3

1/2

⫽ 0.079 (m

/s)

and

S ⫽ 0.01 m/m

then

0.079 ⫽ (0.3115/0.015) d

8/3

(0.01)

1/2

0.079 ⫽ 2.077d

8/3

566 Chapter 6

8/3

⫽ 0.038

d ⫽ 0.293 (m)

> 300 mm

⫽ 12 in

Using a 12-in (300 mm) pipe will just carry the 0.079 m

/s flow Check velocity

V ⫽ (0.397/n) d

2/3

1/2

⫽ (0.397/0.015) (0.304)

2/3

(0.01)

1/2

⫽ 2.647 (0.304)

2/3

⫽ 1.20 (m/s) (verified, ⬎0.9 m/s)

Note: A nomograph for the Manning equation can be used without calculation.

0.397 ⫽1/4

2/3

weather flow above the minimum flow

Q ⫽ (0.303 ⫺ 0.079) m

⫽ 0.224 m

0.224 ⫽ 2.077 d

8/3

(refer to step (b))

8/3

⫽ 0.1078

d ⫽ 0.434 (m)

⫽ 17.1 in

A standard 18-in (460 mm) pipe would be used. Check velocity of 460 mm pipe,

from Step 1b

V ⫽ 2.647 (0.460)

2/3

⫽ 1.58 (m/s)

The capacity of the 460 mm pipe would be (refer to step (b))

Q ⫽ 2.077 (0.46)

8/3

⫽ 0.262 (m

/s)

(d) Determine the size of the third pipe to carry the peak flow, from steps (a)

and (c). The third pipe must carry (0.646 ⫺ 0.079 ⫺ 0.262) m

⫽ 0.305 m

The size (d) required would be

8/3

⫽ 0.305/2.077

8/3

⫽ 0.1468

d ⫽ 0.487 (m)

Wastewater Engineering 567

The size of 500-mm (20-in) diameter standard is chosen.

The capacity and velocity of a 500-mm pipe with 0.01 hydraulic slope is

Q ⫽ 2.077 (0.50)

8/3

⫽ 0.327 (m

/s)

V ⫽ 2.647 (0.50)

2/3

⫽ 1.67 (m/s)

(e) Calculate total capacity of the three pipes (300, 460, and 500 mm)

Q ⫽ (0.079 ⫹ 0.262 ⫹ 0.327) m

⫽ 0.668 m

/s (0.646 m

/s is needed)

Step 2. Design the inlet and outlet chambers

These depressed sewer pipes are connected from the inlet chamber and

outlet chamber. Weirs (2 m in length) are installed to divide the chamber into

three portions. The design detail can be found elsewhere (Metcalf and Eddy,

Inc. 1981).

11 Pumping Stations

The pumping station must be able to adjust to the variations of waste-

water flows. The smallest capacity pump should be able to pump from

the wet well and discharge at a self-cleansing velocity of about 0.6 m/s

(2 ft/s). It should be connected to a 100-mm (4-in) force main which

would have a capacity of approximately 280 L/min or 74 gal/min. The

wet well capacity should contain sufficient wastewater to permit the

pump to run for at least 2 min and restart not more than once in 5 min

(Steel and McGhee, 1979). The pump running time (t

) and the filling

time (t

) in the wet well are computed as

(6.21)

and

⫽ (6.22)

where t

⫽ pump running time, min

V ⫽ storage volume of wet well, L or gal

D ⫽ pump discharge, at peak flow, L/min (m

/s) or gpm, L/min

or gpm

Q ⫽ average daily, L/min (m

/s) or gpm inflow, m

/s or gpd

⫽ filling time with the pump off, min

D 2 Q

568 Chapter 6

and the total cycle time (t)is

(6.23)

The starting limitations on pump motors usually dictate the minimum

size of a well. The wet well should be large enough to prevent pump

motors from overheating due to extensive cycling, but small enough to

accommodate cycling times that will reduce septicity and odor problems.

Typically, submersible pumps can cycle four to ten times per hour. Amax-

imum storage volume for a cycling time should be no more than 30 min.

If the selected pumps have a capacity equal to the peak (maximum)

flow rate, the volume of a wet well is calculated as (WEF, 1993a)

V ⫽ TQ/4 (6.24)

where V ⫽ storage volume of wet well, gal

T ⫽ pump cycle time, min

Q ⫽ peak flow, gal/min

Example 1: A subdivision generates an average daily wastewater flow of

144,000 L/d (38,000 gal/d). The minimum hourly flow rate is 20,000 L/d (5300

gal/d) and the peak flow is 500,000 L/d (132,000 gal/d). Determine the pump-

ing conditions and the size of a wet well.

solution:

Step 1. Determine pump capacity for peak flow

D ⫽ 500,000 L/day ⫻ 1 day/1440 min

⫽ 347 L/min

Step 2. Calculate the minimum volume (V

) for 2-min running time

⫽ 347 L/min ⫻ 2 min

⫽ 694 L

Step 3. Calculate volume (V

) for 5-min cycle using Eq. (6.23)

Average flow Q ⫽ 144,000 L/d ⫽ 100 L/min

5 356 L

100V

1 247V

5 5 3 247 3 100

5 min 5

s347 2 100dL/min

100L/min

t 5

D 2 Q

t 5 t

1 t

D 2 Q

Wastewater Engineering 569

Step 4. Determine the control factor

Since V

⬍ V

, therefore the pump running time is the control factor. Say V

⫽

700 L for design.

Step 5. Calculate the actual time of the pumping cycle

Step 6. Determine size of wet well

A submergence of 0.3 m (1 ft) above the top of the suction pipe is required for

an intake velocity of 0.6 m/s (2 ft/s). The depth between the well bottom and

the top of submergence is 0.5 m (1.6 ft). If a 1.2-m (4-ft) diameter of wet well

is chosen, surface area is 1.13 m

(12.2 ft

For storage

⫽ 700 L ⫽ 0.7 m

the depth would be

d ⫽ 0.7/1.13 ⫽ 0.62 (m)

Typically, 0.6 m (2 ft) of freeboard is required

Thus total depth of the wet well ⫽ (0.50 ⫹ 0.62 ⫹ 0.60) m

⫽ 1.72 m

⫽ 5.6 ft

Example 2: Wastewater is collected from a subdivision of 98.8 acre (40.0 ha)

area that consists of 480 residential units and 2.2 acres (0.89 ha) of commercial

center. Each of the two pumps will be cycled, alternately, four times per hour.

Determine the volume of wet well needed.

solution:

Step 1. Determine domestic sewer flow q

Assume the residential units have 3.5 persons (United States) and each

produces 100 gal/d (378 L/d)

⫽ 100 gal/(c ⭈ d) ⫻ 3.5 person/unit ⫻ 480 unit

⫽ 168,000 gal/d

Step 2. Estimate commercial area contribution q

Assume 1500 gal/d ⭈ per acre (468 m

/d)

⫽ 1500 gal/(d ⭈ a) ⫻ 2.2 a

⫽ 3300 gpd

t 5

700 L

s347 2 100dL/min

700 L

100 L/min

5 9.83 min

570 Chapter 6

Step 3. Estimate infiltration/inflow (I/I) q

Assume I/I is 1000 gal/(d ⭈ a) or (9.35 m

/(d ⭈ ha))

⫽ 1000 gal/(d ⭈ a) ⫻ 98.8 a

⫽ 98,800 gal/d

Step 4. Determine total sewer flow q

This average daily flow

q ⫽ q

⫹ q

⫹q

⫽ (168,000 ⫹ 3300 ⫹ 98,800) gal/d

⫽ 270,100 gal/d

⫽ 188 gal/min

Step 5. Estimate population equivalent (PE) and peak flow (Q)

PE ⫽ 270,100 gal/d/100 gal/(c ⭈ d)

⫽ 2700 persons

Take

Q ⫽ 3.5q for the peak flow

Q ⫽ 188 gal/min ⫻ 3.5

⫽ 658 gal/min

Note: The selected pump should be able to deliver 658 gal/min. The pipe

diameter that will carry the flow and maintain at least 2.0 ft/s (0.6 m/s) of

velocity should be selected by the manufacturer’s specification. For this exam-

ple a 10-in (254-mm) pressure class ductile iron pipe will be used.

Step 6. Determine the volume of the wet well, V

The selected pumps can cycle four times per hour. Alternating each pump

between starts gives 8 cycles per hour. The time between starts, T, is

T ⫽ 60 min/8 ⫽ 7.5 min

It means that one pump is capable of starting every 7.5 min. Using Eq. (6.24)

V ⫽ TQ/4

⫽ 7.5 min ⫻ 658 (gal/min)/4

⫽ 1230 gal (⫽4662 L)

⫽ 164 ft

Wet wells are typically available in cylindrical sections of various sizes. In this

example, a 6-ft (1.8-m) diameter gives 28.3 ft

of surface area.

Wastewater Engineering 571

The depth D of the wet well is

D ⫽ 164/28.3

⫽ 5.8 ft

Note: One foot of freeboard should be added. Thus the well is 6 ft in

diameter and 6.8 ft in depth.

12 Sewer Construction

Conduit material for sewer construction consists of two types: rigid pipe

and flexible pipe. Specified rigid materials include asbestos–cement,

cast iron, concrete, and vitrified clay. Flexible materials include ductile

iron, fabricated steel, corrugated aluminum, thermoset plastic (rein-

forced plastic mortar and reinforced thermosetting resin), and thermo-

plastic. Thermoplastic consists of acrylonitrile–butadiene–styrene (ABS),

ABS composite, polyethylene (PE), and polyvinyl chloride (PVC). Their

advantages, disadvantages, and applications are discussed in detail

elsewhere (ASCE and WPCF, 1982, WEF, and ASCE, 1993a).

Nonpressure sewer pipe is commercially available in the size range

from 4 to 42 in (102 to 1067 mm) in diameter and 13 ft (4.0 m) in length.

Half-length sections of 6.5 ft (2 m) are available for smaller size pipes.

12.1 Loads and buried sewers

Loads on sewer lines are affected by conditions of flow, groundwater,

adjacent earth, and superimposed situation. Loads include hydraulic

loads, earth loads, groundwater loads, and superimposed loads (weight

and impact of vehicles or other structure). Crushing strength of the

sewer material, type of bedding, and backfill load are important factors.

Marston’s equation. Figure 6.2 illustrates common cuts used for sewer

pipe installations. Marston’s equation is widely used to determine the

vertical load on buried conduits caused by earth forces in all of the most

572 Chapter 6

Figure 6.2 Common trench cuts for sewer pipes.

commonly encountered construction conditions (Marston, 1930). The

general form of Marston’s formula is

W ⫽ CwB

(6.25)

where W ⫽ vertical load on pipe as a result of backfill, lb per linear ft

C ⫽ dimensionless load coefficient based on the backfill and

ratio of trench depth to width; a nomograph can be used

w ⫽ unit weight of backfill, lb/ft

B ⫽ width of trench at top of sewer pipe, ft (see Fig. 6.2)

The load coefficient C can be calculated as

(6.26)

where e ⫽ base of natural logarithms

k ⫽ Rankine’s ratio of lateral pressure to vertical pressure

m ⫽

⫽ coefficient of internal friction of backfill material

⫽ coefficient of friction between backfill material and the

sides of the trench ⱕ ␮

H ⫽ height of backfill above pipe, ft (see Fig. 6.2)

Load on sewer for trench condition. The load on a sewer conduit for the

trench condition is affected directly by the soil backfill. The load varies

widely over different soil types, from a minimum of approximately 100

lb/ft

(1600 kg/m

) to a maximum of about 135 lb/ft

(2160 kg/m

) (WEF

and ASCE, 1992). The unit weight (density) of backfill material is as fol-

lows (McGhee, 1991):

100 lb/ft

(1600 kg/m

) for dry sand, and sand and damp topsoil;

115 lb/ft

(1840 kg/m

) for saturated topsoil and ordinary sand;

120 lb/ft

(1920 kg/m

) for wet sand and damp clay; and

130 lb/ft

(2080 kg/m

) for saturated clay.

The average maximum unit weight of soil which will constitute the

backfill over the sewer pipe may be determined by density measure-

ments in advance of the structural design of the sewer pipe. A design

value of not less than 120 or 125 lb/ft

(1920 or 2000 kg/m

) is recom-

mended (WEF and ASCE, 1992).

mr 5 tan ⌽r

tan ⌽

k 5

1 1

2 m

1 1 1 m

1 2 sin ⌽

1 1 sin ⌽

C 5

1 2 e

22kmrsH/Bd

2kmr

Wastewater Engineering 573

The load on a sewer pipe is also influenced by the coefficient of fric-

tion between the backfill and the side of the trench (␮′) and by the coef-

ficient of internal friction of the backfill soil (␮). For most cases these

two values are considered the same for design purposes. But, if the

backfill is sharp sand and the sides of the trench are sheeted with fin-

ished lumber, ␮ may be substantially greater than ␮′. Unless specific

information to the contrary is available, values of the products k␮ and

k␮′ may be assumed to be the same and equal to 0.103. If the back-

fill soil is slippery clay, k␮′ and k␮′ are equal to 0.110 (WEF and

ASCE, 1992).

The values of the product k␮′ in Eq. (6.26) range from 0.10 to 0.16 for

most soils; specifically, 0.110 for saturated clay, 0.130 for clay, 0.150 for

saturated top soil, 0.165 for sand and gravel, and 0.192 for cohesionless

granular material (McGhee, 1991). Graphical solutions of Eq. (6.26) are

presented elsewhere (ASCE and WPCF, 1982).

Example: A 20-in (508-mm) ductile iron pipe is to be installed in an ordi-

nary trench of 10 ft (3.05 m) depth at the top of the pipe and 4 ft (1.22 m) wide.

The cut will be filled with damp clay. Determine the load on the sewer pipe.

solution:

Step 1. Compute load coefficient C by Eq. (6.26)

Step 2. Compute the load W by Eq. (6.25)

w ⫽ 120 lb/ft

⫽ 1920 kg/m

W ⫽ CwB

⫽ 1.92 ⫻ 1920 kg/m

⫻ (1.22 m)

⫽ 5487 kg/m

⫽ 3687 lb/ft

13 Wastewater Treatment Systems

As discussed in Chapters 1 and 2, the natural waters in streams, rivers,

lakes, and reservoirs have a natural waste assimilative capacity to

remove solids, organic matter, even toxic chemicals in the wastewater.

However, it is a long process.

5 1.92

C 5

1 2 e

22kmrH/B

2kmr

1 2 e

22s0.11ds2.5d

2s0.11d

H/B 5 3.05 m/1.22 m 5 2.5

kmr 5 0.11

574 Chapter 6