Lin S.D. Water and Wastewater Calculations Manual

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Wastewater Engineering 535

Fixed and volatile solids. The residue from TS, TSS, or TDS tests is

ignited to constant weight at 550⬚C. The weight lost on ignition is called

volatile solids, whereas the remaining solids represent the fixed total,

suspended, or dissolved solids. The portions of volatile and fixed solids

are computed by

(6.4)

(6.5)

where G ⫽ weight of residue plus crucible before ignition, mg

H ⫽ weight of residue plus crucible or filter after ignition, mg

I ⫽ weight of dish or filter, mg

The determination of the volatile portion of solids is useful in control-

ling wastewater treatment plant operations because it gives a rough esti-

mation of the amount of organic matter present in the solid fraction of

wastewater, activated sludge, and in industrial waste.

Determination of volatile and fixed solids does not distinguish pre-

cisely between organic and inorganic matter. Because the loss on igni-

tion is not confined only to organic matter, it includes losses due to

decomposition or volatilization of some mineral salts. The determina-

tion of organic matter can be made by tests for biochemical oxygen

demand (BOD), chemical oxygen demand (COD), and total organic

carbon (TOC).

Settleable solids. Settleable solids is the term applied to material set-

tling out of suspension within a defined time. It may include floating

material, depending on the technique. Settled solids may be expressed

on either a volume (mL/L) or a weight (mg/L) basis.

The volumetric method for determining settleable solids is as follows.

Fill an Imhoff cone to the 1-L mark with a well-mixed sample. Settle for

45 min, gently agitate the sample near the sides of the Imhoff cone with

a rod or by spinning, then continue to settle for an additional 15 min and

record the volume of settleable solids in the cones as mL/L.

Another test to determine settleable solids is the gravimetric

method. First, determine total suspended solids as stated above.

Second, determine nonsettleable suspended solids from the super-

natant of the same sample which has settled for 1 h. Then determine

TSS (mg/L) of this supernatant liquor; this gives the nonsettleable

solids. The settleable solids can be calculated as

mg settleable solids/L ⫽ mg TSS/L ⫺ mg nonsettleable solids/L (6.6)

mg fixed solids/L 5

sH 2 Id 3 1000

sample volume, mL

mg volatile solids/L 5

sG 2 H d 3 1000

sample volume, mL

Example: A well-mixed 25 mL of raw wastewater is used for TS analyses.

A well-mixed 50 mL of raw wastewater is used for suspended solids analy-

ses. Weights (wt.) of evaporating dish with and without the sample either

dried, evaporated, or ignited were determined to constant weight according

to Standard Methods (APHA et al., 1998). The laboratory results are

Tare wt. of evaporating dish ⫽ 42.4723 g

Wt. of dish plus residue after evaporation at 105°C ⫽ 42.4986 g

Wt. of dish plus residue after ignition at 550°C ⫽ 42.4863 g

Tare wt. of filter plus Gooch crucible ⫽ 21.5308 g

Wt. of residue and filter plus crucible after drying at 105°C ⫽ 21.5447 g

Wt. of residue and filter plus crucible after ignition at 550°C ⫽ 21.5349 g

Compute the concentrations of total solids, volatile solids, fixed solids, total

suspended solids, volatile suspended solids, and fixed suspended solids.

solution:

Step 1. Determine total solids by Eq. (6.1)

Step 2. Determine total volatile solids by Eq. (6.4)

G ⫽ 42,498.6 mg

H ⫽ 42,486.3 mg

VS ⫽ (42,498.6 ⫺ 42,486.3) ⫻ 1000/25

⫽ 492 mg/L

Step 3. Determine total fixed solids

FS ⫽ TS ⫺ VS ⫽ (1052 ⫺ 492) mg/L ⫽ 560 mg/L

Step 4. Determine total suspended solids by Eq. (6.2)

C ⫽ 21,544.7 mg

D ⫽ 21,530.8 mg

Sample size ⫽ 50 mL

TSS ⫽ (C – D) ⫻ 1000/50

⫽ (21,544.7 ⫺ 21,530.8) ⫻ 20

⫽ 278 mg/L

5 1052 mg/L

5 s42,498.6 2 42,472.3d 3 40 mg/L

TS 5

sA 2 Bd mg 3 1000 mL/L

25 mL

B 5 42,472.3 mg

A 5 42,498.6 mg

Sample size 5 25 mL

536 Chapter 6

Step 5. Determine volatile suspended solids by Eq. (6.4)

G ⫽ 21,544.7 mg

H ⫽ 21,534.9 mg

VSS ⫽ (G ⫺ H) ⫻ 1000/50

⫽ (21,544.7 – 21,534.9) ⫻ 20

⫽ 196 mg/L

Step 6. Determine fixed suspended solids by Eq. (6.5)

H ⫽ 21,534.9 mg

I ⫽ 21,530.8 mg

FSS ⫽ (H ⫺ I ) ⫻ 1000/50

⫽ (21,534.9 ⫺ 21,530.8) ⫻ 20

⫽ 82 mg/L

FSS ⫽ TSS – VSS ⫽ (278 – 196) mg/L

⫽ 82 mg/L

2.2 Chemical constituents of wastewater

The dissolved and suspended solids in wastewater contain organic and

inorganic material. Organic matter may include carbohydrates, fats,

oils, grease, surfactants, proteins, pesticides and other agricultural

chemicals, volatile organic compounds, and other toxic chemicals (house-

hold and industrial). Inorganics may include heavy metals, nutrients

(nitrogen and phosphorus), pH, alkalinity, chlorides, sulfur, and other

inorganic pollutants. Gases such as carbon dioxide, nitrogen, oxygen,

hydrogen sulfide, and methane may be present in a wastewater.

Normal ranges of nitrogen levels in domestic raw wastewater are

25 to 85 mg/L for total nitrogen (the sum of ammonia, nitrate, nitrite,

and organic nitrogen); 12 to 50 mg/L ammonia nitrogen; and 8 to

35 mg/L organic nitrogen (WEF 1996a). The organic nitrogen con-

centration is determined by a total kjeldahl nitrogen (TKN) analysis,

which measures the sum of organic and ammonia nitrogen. Organic

nitrogen is then calculated by subtracting ammonia nitrogen from

the TKN measurement.

Typical total phosphorus concentrations of raw wastewater range

from 2 to 20 mg/L, which includes 1 to 5 mg/L of organic phosphorus and

1 to 15 mg/L of inorganic phosphorus (WEF 1996a). Both nitrogen and

phosphorus in wastewater serve as essential elements for biological

growth and reproduction during wastewater treatment processes and

in the natural water.

Wastewater Engineering 537

The strength (organic content) of a wastewater is usually measured as

5-days biochemical oxygen demand (BOD

), chemical oxygen demand,

and total organic carbon. The BOD

test measures the amount of oxygen

required to oxidize the organic matter in the sample during 5 days of bio-

logical stabilization at 20°C. This is usually referred to as the first stage

of carbonaceous BOD (CBOD), not nitrification (second phase). Secondary

wastewater treatment plants are typically designed to remove CBOD, not

for nitrogenous BOD (except for advanced treatment). The BOD

of raw

domestic wastewater in the United States is normally between 100 and

250 mg/L. It is higher in other countries. In this chapter, the term BOD

represents 5-days BOD, unless stated otherwise.

The ratio of carbon, nitrogen, and phosphorus in wastewater is very

important for biological treatment processes, where there is normally

a surplus of nutrients. The commonly accepted BOD/N/P weight ratio

for biological treatment is 100/5/1; i.e. 100 mg/L BOD to 5 mg/L nitro-

gen to 1 mg/L phosphorus. The ratios for raw sanitary wastewater and

settled (primary) effluent are 100/17/5 and 100/23/7, respectively.

Example 1: Calculate the pounds of BOD

and TSS produced per capita per

day. Assume average domestic wastewater flow is 100 gallons (378 L) per

capita per day (gpcpd), containing BOD

and TSS concentrations in waste-

water of 200 and 240 mg/L, respectively.

solution:

Step 1. For BOD

BOD ⫽ 200 mg/L ⫻ 1 L/10

mg ⫻ 100 gal/(c ⭈ d) ⫻ 8.34 lb/gal

⫽ 0.17 lb/(c ⭈ d)

⫽ 77 g/(c ⭈ d)

Step 2. For TSS

TSS ⫽ 240 mg/L ⫻ 10

–6

L/mg ⫻ 100 gal/(c ⭈ d) ⫻ 8.34 lb/gal

⫽ 0.20 lb/(c ⭈ d)

⫽ 90 g/(c ⭈ d)

Note: 100 gal/(c ⭈ d) of flow and 0.17 lb BOD

per capita per day are commonly

used for calculation of the population equivalent of hydraulic and BOD load-

ings for other wastewaters.

In newly developed communities, generally designed wastewater flow is

120 gpcpd. With the same concentrations, the loading rate for design purpose

would be

BOD ⫽ 0.20 lb/(c ⭈ d)

⫽ 90 g/(c ⭈ d)

TSS ⫽ 0.24 lb/(c ⭈ d)

⫽ 110 g/(c ⭈ d)

538 Chapter 6

Example 2: An example of industrial waste has an average flow of 1,230,000

gpd (4656 m

/d) with BOD

of 9850 lb/d (4468 kg/d). Calculate BOD concen-

tration and the equivalent populations of hydraulic and BOD loadings.

solution:

Step 1. Determine BOD concentration

Note: MGD ⫽ million gallons per day

Mgal ⫽ million gallons

Step 2. Calculate equivalent populations, E. P.

For the SI units calculations:

Step 3. Compute BOD concentration

Step 4. Compute E.P., refer to Example 1

Chemical oxygen demand. Concept and measurement of BOD are fully

covered in Chapter 1. Since the BOD

test is time consuming, chemical

oxygen demand is routinely performed for wastewater treatment opera-

tions after the relationship between BOD

and COD has been developed

for a wastewater treatment plant. Many regulatory agencies accept the

COD test as a tool of the wastewater treatment operation. The total

5 12,300 persons

Hydraulic E.P. 5

4656 m

/d 3 1000 L/m

378.5/c/d

> 58,000 capitasspersonsd

BOD E.P. 5

4468 kg/d 3 1000 g/kg

77 g/c/d

BOD 5

4468 kg/d 3 10

mg/kg

4656 m

/d 3 10

L/m

5 960 mg/L

5 12,300 persons

Hydraulic E.P. 5

1,230,000 gpd

100 g/c/d

5 57,940 persons scapitasd

BOD E.P. 5

9850 lb/d

0.17 lb/c/d

Q 5 1.23 MGD

BOD 5

9850 lb/d

1.23 Mgal/d 3 8.34 lb/sMgal ? mg/Ld

5 960 mg/L

Wastewater Engineering 539

organic carbon test is also a means for defining the organic content in

wastewater.

The chemical oxygen demand is a measurement of the oxygen equiv-

alent of the organic matter content of a sample that is susceptible to oxi-

dation by a strong chemical oxidant, such as potassium dichromate.

For a sample, COD can be related empirically to BOD, organic carbon,

or organic matter.

After the correlation of BOD

and COD has been established, the

COD test is useful for controlling and monitoring wastewater treat-

ment processes. The COD test takes 3 to 4 h rather than 5 days for BOD

data. The COD results are typically higher than the BOD values. The

correlation between COD and BOD varies from plant to plant. The BOD:

COD ratio also varies across the plant from influent to process units to

effluent. The ratio is typically 0.5:1 for raw wastewater and may drop

to as low as 0.1:1 for well-stabilized secondary effluent. The normal

COD range for raw wastewater is 200 to 600 mg/L (WEF 1996a).

COD test. Most types of organic matter in water or wastewater are oxi-

dized by a boiling mixture of sulfuric and chromic acids with excess of

potassium dichromate (K

). After 2-h refluxing, the remaining

unreduced K

is titrated with ferrous ammonium sulfate to meas-

ure the amount of K

consumed and the oxidizable organic matter

is calculated in terms of oxygen equivalent. The COD test can be per-

formed by either the open reflux or the closed reflux method. The open

reflux method is suitable for a wide range of wastewaters, whereas the

closed reflux method is more economical but requires homogenization

of samples containing suspended solids to obtain reproducible results.

The COD test procedures (open reflux) are (APHA et al., 1998):

■

Place appropriate size of sample in a 500-mL refluxing flask use

50 mL if COD ⬍ 900 mg/L

use less sample and dilute to 50 mL if COD ⬎900 mg/L

use large size of sample if COD ⬍50 mL

■

Add 1 g of HgSO

and several glass beads

■

Slowly add 5 mL sulfuric acid reagent and mix to dissolve HgSO

■

Cool the mixture

■

Add 25.0 mL of 0.0417 mole K

solution and mix

■

Attach the refluxing flask to condenser and turn on cooling water

■

Add 70 mL sulfuric acid reagent through open end of condenser, then

swirl and mix

■

Cover open end of condenser with a small beaker

■

Reflux for 2 h (usually 1 blank with several samples)

540 Chapter 6

■

Cool and wash down condenser with distilled water to about twice

volume

■

Cool to room temperature

■

Add 2–3 drops ferroin indicator

■

Titrate excess K

with standard ferrous ammonium sulfate (FAS)

■

Take as the end point of the titration of the first sharp color change

from blue-green to reddish brown

■

Calculate COD result as (APHA et al., 1998)

(6.7)

where A ⫽ mL FAS used for blank

B ⫽ mL FAS used for sample

M ⫽ molarity of FAS to be determined daily against standard

solution ⬵ 0.25 mole

Example: The results of a COD test for raw wastewater (50 mL used) are

given. Volumes of FAS used for blank and the sample are 24.53 and 12.88 mL,

respectively. The molarity of FAS is 0.242. Calculate the COD concentration

for the sample.

solution: Using Eq. (6.7)

2.3 Biological characteristics

of wastewater

The principal groups of microorganisms found in wastewater are bac-

teria, fungi, protozoa, microscopic plants and animals, and viruses. Most

microorganisms (bacteria, protozoa) are responsible and are beneficial

for biological treatment processes of wastewater. However, some path-

ogenic bacteria, fungi, protozoa, and viruses found in wastewater are of

public concern.

Indicator bacteria. Pathogenic organisms are usually excreted by

humans from the gastrointestinal tract and discharge to wastewater.

Water-borne diseases include cholera, typhoid, paratyphoid fever,

5 451

5 s24.53 2 12.88d 3 0.242 3 8000/50

COD as mg O

/L 5

sA 2 Bd 3 M 3 8000

mL sample

COD as mg O

/L 5

sA 2 Bd 3 M 3 8000

mL sample

Wastewater Engineering 541

diarrhea, and dysentery. The number of pathogenic organisms in waste-

waters is generally low in density and they are difficult to isolate and

identify. Therefore, indicator bacteria such as total coliform (TC), fecal

coliform (FC), and fecal streptococcus (FS) are used as indicator organ-

isms. The concept of indicator bacteria and enumeration of bacterial den-

sity are discussed in Chapter 1.

Tests for enumeration of TC, FC, and FS can be performed by multiple-

tube fermentation (most probable number, MPN) or membrane filter

methods; the test media used are different for these three groups of

indicators. Most regulatory agencies have adopted fecal coliform density

as an effluent standard because FC is mostly from fecal material.

3 Sewer Systems

Sewers are underground conduits to convey wastewater and storm-

water to a treatment plant or to carry stormwater to the point of dis-

posal. Sewers can be classified into three categories: sanitary, storm, and

combined. Community sewer systems, according to their discharging

types, can be divided into separated and combined sewer systems.

3.1 Separated sewer system

Separated sewers consist of sanitary sewers and stormwater sewer

networks separately. Sanitary sewers carry a mixture of household and

commercial sewage, industrial wastewater, water from groundwater

infiltration/inflow, basement and foundation drainage connections, and

cross-connections between sanitary sewers and stormwater drainage.

Separated sanitary sewers should be free of stormwater, but they

seldom are.

Storm sewers are commonly buried pipes that convey storm drainage.

They may include open channel elements and culverts, particularly

when drainage areas are large.

Storm sewer networks convey mainly surface storm runoff from roofs,

streets, parking lots, and other surfaces toward the nearest receiving

water body. An urban drainage system with separated sewers is more

expensive than a combined sewer system because it uses two parallel

conduits. Sanitary and storm sewers are usually designed to operate

under gravity flow conditions. Pressure or vacuum sewers are rare.

Storm sewers are dry much of the time. When rainfalls are gentle, the

surface runoffs are usually clear and low flows present no serious prob-

lem. However, flooding runoffs wash and erode unprotected areas and

create siltation.

Illicit connections from roofs, yards, and foundations drain flow to

sanitary sewers through manhole covers that are not tight. The flow

542 Chapter 6

rates vary and are as high as 70 gal/(c ⭈ d) and average 30 gal/(c ⭈ d). A

rainfall of 1 in (2.54 cm)/ h on 1200 ft

(111 m

) of roof produces a flow of

12.5 gal/min (0.788 L/s) or 18,000 gal/d. Spreading over an acre (0.4047 ha),

1 in/h equals 1.008 ft

/s (28.5 L/s) of runoff. Leakage through manhole

covers can add 20 to 70 gal/min (1.26 to 4.42 L/s) to the sewer when

there is as much as 1 in of water on the streets (Fair et al., 1966).

Example: A house has a roof of 30 ft by 40 ft (9.14 m ⫻ 12.19 m) and is occu-

pied by four people. What is the percentage of rainfall of 1 in/h, using a rate

of leakage through manholes of 60 gal/(c ⭈ d)?

solution:

Step 1. Compute stormwater runoff

Runoff ⫽ 30 ft ⫻ 40 ft ⫻ 1 in/h ⫻ (1 ft/12 in) ⫻ (24 h/d)

⫻ 7.48 gal/ft

⫼ 4 c ⫽ 4488 gal/(c ⭈ d)

Say ⫽ 4500 gal/(c ⭈ d)

Step 2. Determine percent of leakage

% ⫽ 60 gal/(c ⭈ d) ⫻ 100/4500 gal/(c ⭈ d)

⫽ 1.33

3.2 Combined sewers

Combined sewers are designed for collection and conveyance of both

sanitary sewage, including industrial wastewaters, and storm surface

runoff in one conduit. Combined sewer systems are common in old US

and European urban communities. In the United States, there has been

a trend to replace combined sewers by separate sewer systems. The

dry-weather flow (sanitary and industrial wastewaters, plus street

drainage from washing operations and lawn sprinkling and infiltra-

tion) is intercepted to the treatment facility. During storm events, when

the combined wastewater and stormwater exceed the capacity of the

treatment plant, the overflow is bypassed directly to the receiving water

body without treatment or is stored for future treatment. The overflow

is the so-called combined sewer overflow (CSO).

The average BOD

in stormwater is approximately 30 mg/L. The aver-

age BOD

in combined sewer overflows is between 60 and 120 mg/L

(Novotny and Chesters, 1981). Since the CSO bypass may result in sig-

nificant pollution, even during high river flows, it will create a hazardous

threat to downstream water users. Thus, proper management of CSO is

required (e.g. storage and later treatment). One obvious solution is to

replace combined sewers by separate sewers; however, costs and legali-

ties of shifting are major concerns.

Wastewater Engineering 543

Example: The following information is given:

Separate sanitary sewer flow ⫽ 250 L/(c ⭈ d)

Population density ⫽ 120 persons/ha

BOD

of raw wastewater ⫽ 188 mg/L

BOD

of plant effluent or stormwater ⫽ 30 mg/L

A storm intensity ⫽ 25 mm/d (1 in/d)

Impervious urban watershed ⫽ 72%

Compare flow and BOD

pollution potential produced by wastewater after

treatment and by the stormwater.

solution:

Step 1. Compute flow and BOD

loads for wastewater

Flow ⫽ 250 L/(c ⭈ d) ⫻ 120 c/ha

⫽ 30,000 L/(d ⭈ ha)

⫽ 30 m

/(d ⭈ ha)

Raw wastewater BOD

⫽ 30,000 L/(d ⭈ ha) ⫻ 188 mg/L

⫽ 5,640,000 mg/(d ⭈ ha)

⫽ 5640 g/(d ⭈ ha)

Effluent BOD

⫽ 30,000 L/(d ⭈ ha) ⫻ 30 mg/L

⫽ 900 g/(d ⭈ ha)

BOD

removed ⫽ (5640 ⫺ 900) g/(d ⭈ ha)

⫽ 4740 g/(d ⭈ ha) (84% removal)

Step 2. Compute flow and BOD

load for stormwater

Flow ⫽ 25 mm/d ⫻ 10,000 m

/ha ⫻ 0.72 ⫻ 0.001 m/mm ⫻ 1000 L/m

⫽ 180,000 L/(d ⭈ ha)

⫽ 180 m

/(d ⭈ ha)

BOD

load ⫽ 180,000 L/(d ⭈ ha) ⫻ 30 mg/L

⫽ 5,400,000 mg/(d ⭈ ha)

⫽ 5,400 g/(d ⭈ ha)

Step 3. Compare BOD

load

Storm water BOD

: effluent BOD

⫽ 5400 : 900

⫽ 6 : 1

544 Chapter 6