Lin S.D. Water and Wastewater Calculations Manual

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Peripheral weirs shall be placed at least 300 cm (1 ft) from the well (Ten

States Standards, GLUMRB, 1996).

Example 1: Design circular clarifiers using English system units with the

same given information as in the example for rectangular clarifiers design.

solution:

Step 1. Calculate surface area A

Wastewater Engineering 615

Inlet zone

Settling zone

Outlet

zone

Figure 6.12 Typical circular basin design-plan view and cross section.

Design 2 circular clarifiers, each treating 1.0 Mgal/d with an overflow rate

of 700 gal/(d ⭈ ft

A ⫽ 1,000,000 (gal/d)/700 gal/(d ⭈ ft

)

⫽ 1429 ft

⫽ 132.8 m

Step 2. Determine the tank radius r

⫽ 1429

⫽ 1429/3.14 ⫽ 455

r ⫽ 21.3 ft

Use d ⫽ 44 ft (13.4 m) diameter tank

Surface area ⫽ 1520 ft

(141 m

)

Step 3. Check overflow rate

Overflow rate ⫽ 1,000,000 (gal/d)/1520 ft

⫽ 658 gal/(d ⭈ ft

)

⫽ 26.8 m

/(m

⭈ d)

From Fig. 6.11, BOD

removal for an overflow rate of 658 gal/(d ⭈ ft

)

is 35%.

Step 4. Determine detention time t

Use a wastewater side wall depth of 10 ft and, commonly, plus 2-ft freeboard

for wind protection

Step 5. Calculate weir loading rate

Use an inboard weir with diameter of 40 ft (12.2 m)

Length of periphery ⫽ p ⫻ 40 ft ⫽ 125.6 ft

Weir loading ⫽ 1,000,000 (gal/d)/125.6 ft

⫽ 7960 gal/(d ⭈ ft)⫽98.8 m

/m ⭈ d

Step 6. Calculate the number (n) of V notches

5 2.73 h

5 0.114 day

t 5

1520 ft

3 10 ft 3 7.48 gal/ft

1,000,000 gal

616 Chapter 6

Use 90⬚ standard V notches at rate of 8 in center-to-center of the launders.

n ⫽ 125.6 ft ⫻ 12 in/ft ⫼ 8 in

⫽ 188

Step 7. Calculate average discharge per notch at average design flow q

q ⫽ 1,000,000 gal/d ⫻ (1 d/1440 min) ⫼ 188

⫽ 3.7 gal/min ⫽ 14 L/min

Example 2: If the surface overflow rate is 40 m

/(m

⭈ d) [982 gal/(d ⭈ ft

)] and

the weir overflow rate is 360 m

/(d ⭈ m) [29,000 gal/(d ⭈ ft)], determine the max-

imum radius for a circular primary clarifier with a single peripheral weir.

solution

Step 1. Compute the area (A) required with a flow (Q)

Let r ⫽ the radius of the clarifier

A ⫽ pr

⫽ Q/40 m

/(m

⭈ d) or Q ⫽ 40 pr

Step 2. Compute the weir length (⫽ 2pr)

2pr ⫽ Q/360 m

/(d ⭈ m)

Q ⫽ 720pr m

/(d ⭈ m)

Step 3. Compute r by solving the equations in Steps 1 and 2

⫽ 720pr [m

/(d ⭈ m)]/40 [m

/(m

⭈ d)]

r ⫽ 18 m (59 ft)

20 Biological (Secondary) Treatment

Systems

The purpose of primary treatment is to remove suspended solids and

floating material. In many situations in some countries, primary treat-

ment with the resulting removal of approximately 40% to 60% of the sus-

pended solids and 25% to 35% of BOD

, together with removal of

material from the wastewater, is adequate to meet the requirement of

the receiving water body. If primary treatment is not sufficient to meet

the regulatory effluent standards, secondary treatment using a biolog-

ical process is mostly used for further treatment due to its greater

removal efficiency and less cost than chemical coagulation. Secondary

treatment processes are intended to remove the soluble and colloidal

organics (BOD) which remain after primary treatment and to achieve

further removal of suspended solids and, in some cases, also to remove

Wastewater Engineering 617

nutrients such as phosphorus and nitrogen. Biological treatment

processes provide similar biological activities to waste assimilation,

which would take place in the receiving waters, but in a reasonably

shorter time. Secondary treatment may remove more than 85% of BOD

and suspended matter, but is really not effective for removing nutrients,

nonbiodegradable organics, heavy metals, and microorganisms.

Biological treatment systems are designed to maintain a large active

mass and a variety of microorganisms, principally bacteria (and fungi,

protozoa, rotifers, algae, etc.), within the confined system under favor-

able environmental conditions, such as dissolved oxygen, nutrient, etc.

Biological treatment processes are generally classified mainly as sus-

pended growth processes (activated sludge, Fig. 6.13), attached (film)

growth processes (trickling filter and rotating biological contactor,

RBC), and dual-process systems (combined). Other biological waste-

water treatment processes include the stabilization pond, aerated

lagoon, contaminant pond, oxidation ditch, high-purity oxygen acti-

vated sludge, biological nitrification, denitrification, and phosphorus

removal units.

In the suspended biological treatment process, under continuous

supply of air or oxygen, living aerobic microorganisms are mixed thor-

oughly with the organics in the wastewater and use the organics as

food for their growth. As they grow, they clump or flocculate to form an

active mass of microbes. This is so-called biologic floc or activated sludge.

618 Chapter 6

Figure 6.13 Conventional activated sludge process.

20.1 Cell growth

Each unicellular bacterium grows and, after reaching a certain size,

divides to produce two complete individuals by binary fission. The period

of time required for a growing population to double is called the gener-

ation time. With each generation, the total number increases as the

power (exponent) of 2, i.e. 2

, 2

. . . The exponent of 2 correspon-

ding to any given number is the logarithm of that number to the base

2 (log

). Therefore, in an exponentially growing culture, the log

of the

number of cells increasing in proportion to time, is often referred to as

logarithmic growth.

The growth rate of microorganisms is affected by environmental con-

ditions, such as DO levels, temperature, nutrient levels, pH, micro-bial

community, etc. Exponential growth does not normally continue for a

long period of time. A general growth pattern for fission-reproduction

bacteria in a batch culture is sketched in Fig. 6.14. They are the lag

phase, the exponential (logarithmic) growth phase, the maximum

(stationary) phase, and the death phase.

When a small number of bacteria is inoculated into a fixed volume of

vessel with culture medium, bacteria generally require time to accli-

matize to their environmental condition. For this period of time, called

the lag phase, the bacterial density is almost unchanged. Under excess

Wastewater Engineering 619

Figure 6.14 Bacterial density with growth time.

food supply, a rapid increase in number and mass of bacteria occurs in

the log growth phase. During this maximum rate of growth a maxi-

mum rate of substrate removal occurs. Either some nutrients become

exhausted, or toxic metabolic products may accumulate. Subsequently,

the growth rate decreases and then ceases. For this period, the cell

number remains stationary at the stationary phase. As bacterial den-

sity increases and food runs short, cell growth will decline. The total

mass of protoplasm exceeds the mass of viable cells, because bacteria

form resistant structures such as endospores.

In the endogenous growth period, the microorganisms compete for

the limiting substrate even to metabolize their own protoplasm. The

rate of metabolism decreases and starvation occurs. The rate of death

exceeds the rate of reproduction; cells also become old, die exponen-

tially, and lysis. Lysis can occur in which the nutrients from the dead

cells diffuse out to furnish the remaining living cells as food. The

results of cell lysis decrease both the population and the mass of

microorganisms.

In the activated-sludge process, the balance of food to microorganisms

is very important. Metabolism of the organic matter results in an

increased mass of microorganisms in the process. The excess mass of

microorganisms must be removed (waste sludge) from the system to

maintain a proper balance between mass of microorganisms and sub-

strate concentration in the aeration basin.

In both batch and continuous culture systems the growth rate of bac-

terial cells can be expressed as

⫽ mX (6.57)

where r

⫽ growth rate of bacteria, mg/(L ⭈ d)

m ⫽ specific growth rate, per day

X ⫽ mass of microorganism, mg/L

since

(6.58)

Therefore

(6.59)

Substrate limited growth. Under substrate limited growth conditions,

or for an equilibrium system, the quantity of solids produced is equal

5 mX

5 r

620 Chapter 6

to that lost, and specific growth rate (the quantity produced per day) can

be expressed as the well-known Monod equation

(6.60)

where m ⫽ specific growth rate, per day

⫽ maximum specific growth rate, per day

S ⫽ concentration for substrate in solution, mg/L

⫽ half velocity constant, substrate concentration at

one-half the maximum growth rate, mg/L

⫽ cell decay coefficient reflecting the endogenous burn-up

of cell mass, mg/(mg ⭈ d)

Substituting the value of ␮ from Eq. (6.60) into Eq. (6.57), the resulting

equation for the cell growth rate is

(6.61)

21 Activated-Sludge Process

The activated-sludge process was first used in Manchester, England.

This process is perhaps the most widely used process for secondary

treatment of wastewaters. Recently, the activated-sludge process has

been applied as a nitrification and denitrification process, and modified

with an anoxic and anaerobic section for phosphorus removal.

A basic suspended-growth (activated-sludge) process is depicted in

Fig. 6.13. The wastewater continuously enters an aeration tank where

previously developed biological flocs are brought into contact with the

organic material of the wastewater. Air or oxygen-enriched air is con-

tinuously injected into the aeration tank as an oxygen source to keep the

system aerobic and the activated sludge in suspension. Approximately

8 m

of air is required for each m

of wastewater.

The microbes in the activated sludge consist of a gelatinous matrix of

filamentous and unicellular bacteria which are fed on protozoa. The

predominant bacteria, such as Pseudomonas, utilize carbohydrate and

hydrocarbon wastes, whereas Bacillus, Flavobacterium, and Alcaligenes

consume protein wastes. When a new activated-sludge system is first

started, it should be seeded with activated sludge from an existing oper-

ating plant. If seed sludge is not available, activated sludge can be pre-

pared by simply continuously aerating, settling, and returning settled

solids to the wastewater for a few weeks (4 to 6 weeks) (Alessi et al, 1978;

Cheremisinoff, 1995).

1 S

2 k

m 5 m

1 S

2 k

Wastewater Engineering 621

The microorganisms utilize the absorbed organic matter as a carbon

and energy source for cell growth and convert it to cell tissue, water, and

oxidized products (mainly carbon dioxide, CO

). Some bacteria attack

the original complex substance to produce simple compounds as their

waste products. Other bacteria then use these waste products to produce

simpler compounds until the food is used up.

The mixture of wastewater and activated sludge in the aeration basis

is called mixed liquor. The biological mass (biomass) in the mixed liquor

is called the mixed liquor suspended solids (MLSS) or mixed liquor

volatile suspended solids (MLVSS). The MLSS consists mostly of

microorganisms, nonbiodegradable suspended organic matter, and other

inert suspended matter. The microorganisms in MLSS are composed of

70% to 90% organic and 10% to 30% inorganic matter (Okun, 1949;

WEF and ASCE, 1996a). The types of bacterial cell vary, depending on

the chemical characteristics of the influent waste-water tank conditions

and the specific characteristics of the microorganisms in the flocs.

Microbial growth in the mixed liquor is maintained in the declining or

endogenous growth phase to insure good settling properties.

After a certain reaction time (4 to 8 h), the mixed liquor is discharged

from the aeration tank to a secondary sedimentation basin (settling

tank, clarifier) where the suspended solids are settled out from the

treated wastewater by gravity. However, in a sequencing batch reactor

(SBR), mixing and aeration in the aeration tank are stopped for a time

interval to allow MLSS to settle and to decant the treated wastewater;

thus a secondary clarifier is not needed in an SBR system. Most con-

centrated biological settled sludge is recycled back to the aeration tank

(so-called return activated sludge, RAS) to maintain a high population

of microorganisms to achieve rapid breakdown of the organics in the

wastewater. The volume of RAS is typically 20% to 30% of the wastewater

flow. Usually more activated sludge is produced than return sludge.

The treated wastewater is commonly chlorinated and dechlorinated,

then discharged to receiving water or to a tertiary treatment system

(Fig. 6.3). The preliminary, primary, and activated-sludge (biological)

processes are included in the so-called secondary treatment process.

21.1 Aeration periods and BOD loadings

The empirical design of activated sludge is based on BOD loading, food-

to-microorganism ratio (F/M), sludge age, and aeration period. Empirical

design concepts are still acceptable. The Ten States Standards states

that when activated-sludge process design calculations are not submit-

ted, the aeration tank capacities and permissible loadings for the sev-

eral adoptions of the processes shown in Table 6.11 are used as simple

design criteria. Those values apply to plants receiving diurnal load

622 Chapter 6

ratios of design peak hourly BOD

to design BOD

ranging from 2:1 to

4:1 (GLUMRB, 1996).

The aeration period is the retention time of the influent wastewater

flow in the aeration basin and is expressed in hours. It is computed from

the basin volume divided by the average daily flow excluding the return

sludge flow. For normal domestic sewage, the aeration period commonly

ranges from 4 to 8 h with an air supply of 0.5 to 2.0 ft

/gal (3.7 to 15.0 m

)

of wastewater. The return-activated sludge is expressed as a percentage

of the wastewater influent of the aeration tank.

The organic (BOD

) loading on an aeration basin can be computed

using the BOD in the influent wastewater without including the

return sludge flow. BOD loadings are expressed in terms of lb BOD

applied per day per 1000 ft

or [kg/(d ⭈ m

)] of liquid volume in the

aeration basin and in terms of lb BOD applied per day per lb of mixed

liquid volatile suspended solids. The latter is called the food-to-

microorganism ratio.

BOD loadings per unit volume of aeration basin vary widely from 10

to more than 100 lb/1000 ft

(0.16 to 1.6 kg/(d ⭈ m

)), while the aeration

periods correspondingly vary from 2.5 to 24 h (Clark et al., 1977). The

relationship between the two parameters is directly related to the BOD

concentration in the wastewater.

21.2 F/M ratio

The F/M ratio is used to express BOD loadings with regard to the micro-

bial mass in the process. The value of the F/M ratio can be calculated

Wastewater Engineering 623

TABLE 6.11 Permissible Aeration Tank Capacities and Loadings

Organic (BOD

) F/M ratio, lb

loading lb/(d ⭈ 1000 ft

) BOD

/d per lb

Process (kg/d ⭈ m

)

∗

MLVSS MLSS, mg/L

‡

Conventional step aeration 40(0.64) 0.2–0.5 1000–3000

complete mix

Contact stabilization

†

(0.80) 0.2–0.6 1000–3000

Extended aeration single-stage 15(0.24) 0.05–0.1 3000–5000

nitrification

Note: Loadings are based on the influent organic load to the aeration tank at plant design

average BOD

†

Total aeration capacity includes both contact and reaeration capacities; normally the

contact zone equals 30% to 35% of the total aeration capacity.

‡

The values of MLSS are dependent on the surface area provided for secondary settling

and the rate of sludge return as well as the aeration processes.

SOURCE: GLUMRB (Ten States Standards) (1996)

by the following equation:

(6.62)

(6.63)

(6.64)

where F/M ⫽ food-to-microorganism ratio, kg (lb) of BOD per day

per kg (lb) of MLSS

Q ⫽ wastewater flow, m

/d or Mgal/d

BOD ⫽ wastewater 5-day BOD, mg/L

V ⫽ liquid volume of aeration tank, m

or Mgal

MLSS ⫽ mixed liquor suspended solids in the aeration

tank, mg/L

In Eqs. (6.62) and (6.63), some authors use mixed liquor volatile sus-

pended solids instead of MLSS. The MLVSS is the volatile portion of the

MLSS and ranges from 0.75 to 0.85. Typically they are related, for

design purposes, by MLVSS ⫽ 0.80 ⫻ MLSS. The use of MLVSS may

more closely approximate the total active biological mass in the process.

The F/M ratio is also called the sludge loading ratio (SLR). The equa-

tion for the calculation of the SLR is (Cheremisinoff, 1995)

(6.65)

where SLR ⫽ sludge loading, g of BOD/d per g of MLVSS

BOD ⫽ wastewater BOD, mg/L

MLVSS ⫽ mixed liquor volatile suspended solids, mg/L

t ⫽ retention time, d

R ⫽ recyle ratio

Example 1: An activated-sludge process has a tank influent BOD con-

centration of 140 mg/L, influent flow of 5.0 Mgal/d (18,900 m

/d) and

35,500 lb (16,100 kg) of suspended solids under aeration. Calculate the

F/M ratio.

solution:

Step 1. Calculate BOD in lb/d

SLR 5

24 BOD

MLVSS stds1 1 Rd

F/M 5

BOD, kg/d

MLSS, kg

Q sMgal/dd 3 BOD smg/Ld 3 8.34 slb/gald

V sMgald 3 MLSS smg/Ld 3 8.34 slb/gald

F/M 5

BOD, lb/d

MLSS, lb

624 Chapter 6