Lin S.D. Water and Wastewater Calculations Manual

Подождите немного. Документ загружается.

Wastewater Engineering 655

where X

⫽ return activated-sludge suspended solids, mg/L

⫽ flow of RAS, m

/s or Mgal/d

X ⫽ mixed liquor suspended solids (MLSS), mg/L

Q ⫽ flow of secondary influent (or plant flow),

/s or Mgal/d

The values of X and X

include both the volatile and nonvolatile (inert)

fractions. Equations (6.106) and (6.110) assume no loss of suspended

solids in the basin. Equation (6.109) is essentially the same as Eq. (6.107).

Example 1: Determine the return activated-sludge flow as a percentage of

the influent flow 10.0 Mgal/d (37,850 m

/d). The sludge settling volume in

30 min is 255 mL.

solution:

Step 1. Compute RAS flow Q

in %

Using Eq. (6.107)

Step 2. Compute RAS flow rate in Mgal/d

⫽ 0.34 Q ⫽ 0.34 ⫻ 10 Mgal/d

⫽ 3.40 Mgal/d

⫽ 12,900 m

Example 2: The MLSS concentration in the aeration tank is 2800 mg/L. The

sludge settleability test showed that the sludge volume, settled for 30 min

in a 1-L graduated cylinder, is 285 mL. Calculate the sludge volume index

and estimate the SS concentration in the RAS and the required return sludge

ratio.

solution:

Step 1. Calculate SVI

Using Eq. (6.102)

⫽ 102 mL/g

This is in the typical range of 80 to 150 mL/g

SVI 5

SV 3 1000 mg/g

MLSS

285 mL/L 3 1000 mg/g

2800 mg/L

5 34%

, % 5

1000 2 SV

255 mL/L 3 100%

1000 mL/L 2 255 mL/L

656 Chapter 6

Step 2. Calculate SS in RAS

Using Eq. (6.105)

⫽ 9804 mg/L

⫽ 0.98%

Step 3. Calculate Q

Using Eq. (6.107)

⫽ 0.40

⫽ 40% return

Example 3: Compute the return activated-sludge flow rate in m

/d and as

a percentage of the influent flow of 37,850 m

/d (10 Mgal/d). The laboratory

results show that the SVI is 110 mg/L and the MLSS is 2500 mg/L.

solution:

Step 1. Compute the suspended solids in RAS based on the SVI

Using Eq. (6.105)

⫽ 9090 mg/L

Step 2. Compute RAS flow rate Q

, based on SVI

Using Eq. (6.109)

⫽ 14,360 m

⫽ 3.79 Mgal/d

Step 3. Compute RAS flow as percentage of influent flow

⫽ 37.9%

Say ⫽ 38% of influent flow

RAS flow, % 5

3 100%

3.79 Mgal/d 3 100%

10.0 Mgal/d

2 X

Q 5

2500 mg/L 3 37,850 m

9090 mg/L 2 2500 mg/L

in RAS 5

1,000,000 mg/L

SVI

1,000,000 mg/L

110

1000 2 SV

285 mg/L

s1000 2 285d mg/L

1,000,000

SVI

1,000,000 smL/Ldsmg/gd

102 mL/g

Wastewater Engineering 657

Waste activated sludge. The excess of activated sludge generated from

the secondary clarifier must be wasted, usually from the return sludge

line. The waste activated sludge (WAS) is discharged to either the pri-

mary clarifiers or to sludge thickeners. Withdrawing mixed liquor

directly from the aeration basin or from the basin effluent line is an

alternative method of sludge wasting. The amount of WAS removed

affects mixed liquor settleability, oxygen consumption, growth rate of the

microorganisms, nutrient quantities, input occurrence of foaming and

sludge bulking effluent quality, and possibility of nitrification.

The purpose of WAS is to maintain a given food-to-microorganisms

ratio or mean cell residence time of the system. It should remove just

the amount of microorganisms that grow in excess of the microorganism

death rate.

Sludge wasting can be accomplished on an intermittent or continu-

ous basis. The parameters used for control guidelines are F/M ratio,

mean cell residence time, MLVSS, and sludge age.

Secondary clariﬁer mass balance. Although it assumes the sludge blan-

ket is level in the secondary clarifier, the secondary clarifier mass bal-

ance approach is a useful tool to determine the RAS flow rate. The

calculations are based on the secondary clarifier (Fig. 6.15). Assuming

the sludge blanket in the clarifier is not changed and the effluent sus-

pended solids are negligible, the SS entering the clarifier is equal to the

SS leaving. The relationship can be written as

(Q ⫹ Q

) (MLSS) ⫽ Q

(WAS) ⫹ Q

(RAS)

Q(MLSS) ⫺ Q

(WAS) ⫽ Q

(RAS ⫺ MLSS)

(6.111)

where Q ⫽ flow of tank influent, m

⫽ return activated-sludge flow, m

⫽ waste activated-sludge flow rate, m

MLSS ⫽ mixed liquor suspended solids, mg/L

WAS ⫽ SS of waste activated sludge, mg/L

RAS ⫽ SS of return activated sludge, mg/L

Note: WAS is equal to RAS (Fig. 6.15a).

Sludge age. Sludge age is a measure of the length of time a particle of

suspended solids has been retained in the activated-sludge process. It

is also referred to as mean cell residence time and is an operational

parameter related to the F/M ratio. Sludge age is defined as that of the

QsMLSSd 2 Q

sWASd

RAS 2 MLSS

658 Chapter 6

suspended solids under aeration (kg or lb) divided by the suspended

solids added (kg/d or lb/d) and is calculated by the following equation

which relates MLSS in the system to the influent suspended solids

(6.112)

where Sludge age ⫽ mean cell residence time, day

MLSS ⫽ mixed liquor suspended solids, mg/L

V ⫽ volume of aeration basin, m

or Mgal

⫽ suspended solids in waste sludge, mg/L

⫽ flow of waste sludge, m

/d or Mgal/d

⫽ suspended solids in wastewater effluent, mg/L

⫽ flow of wastewater effluent, m

/d or Mgal/d

Note: Equations (6.112) and (6.72) are the same.

Sludge age can also be expressed in terms of the volatile portion of

suspended solids, which is more representative of biological mass.

Solids retention time in an activated-sludge system is measured in

days, whereas the liquid aeration period is in hours. In most activated-

sludge plants, sludge age ranges from 3 to 8 days (US EPA, 1979). The

liquid aeration periods vary from 3 to 30 h (Hammer, 1986). Wastewater

passes through the aeration tank only once and rather quickly, whereas

the resultant biological growths and extracted waste organics are

repeatedly recycled from the secondary clarifier back to the aeration

basin.

Example 1: An activated-sludge system has an influent flow of 22,700 m

(6.0 Mgal/d) with suspended solids of 96 mg/L. Three aeration tanks hold

1500 m

(53,000 ft

) each with MLSS of 2600 mg/L. Calculate the sludge age

for the system.

solution:

Step 1. Calculate the SS under aeration, MLSS ⫻ V

MLSS ⫽ 2600 mg/L ⫽ 2600 g/m

⫽ 2.6 kg/m

MLSS ⫻ V ⫽ 2.6 kg/m

⫻ 1500 m

⫻ 3

⫽ 11,700 kg

⫽ 25,800 lb

Sludge age 5

MLSS 3 V

3 Q

1 SS

3 Q

Sludge age 5

SS under aeration, kg or lb

SS added, kg/d or lb/d

Wastewater Engineering 659

Step 2. Calculate the SS added

SS added ⫽ Influent flow ⫻ SS conc.

⫽ 22,700 m

/d ⫻ 0.096 kg/m

⫽ 2180 kg/d

⫽ 4800 lb/d

Step 3. Calculate sludge age

Using Eq. (6.112)

⫽ 5.4 days

Note: It is in the typical range of 3 to 8 days.

Example 2: In an activated-sludge system, the solids under aeration are

13,000 kg (28,700 lb); the solids added are 2200 kg/d (4850 lb/d); the return

activated-sludge SS concentration is 6600 mg/L; the desired sludge age is 5.5

days; and the current waste activated sludge (WAS) is 2100 kg/d (4630 lb/d).

Calculate the WAS flow rate using the sludge age control technique.

solution:

Step 1. Calculate the desired SS under aeration for the desired sludge age

of 5.5 days

SS ⫽ solid added ⫻ sludge age

⫽ 2200 kg/d ⫻ 5.5 days

⫽ 12,100 kg

⫽ 26,800 lb

Step 2. Calculate the additional suspended solids removed per day

SS aerated – SS desired ⫽ 13,000 kg – 12,100 kg ⫽ 900 kg

Step 3. Calculate the additional WAS flow (q) to maintain the desired sludge age

q (SS in RAS) ⫽ 900 kg/d

q ⫽ (900 kg/d)/(6.6 kg/m

)

⫽ 136.4 m

/d ⫽ 0.095 m

/min

⫽ 25.0 gal/min

Step 4. Calculate total WAS flow

Current flow ⫽ (2100 kg/d)/(6.6 kg/m

)

⫽ 318.2 m

/d ⫽ 0.221 m

/min

⫽ 58.4 gal/min

Sludge age 5

SS under aeration

SS added

11,700 kg

2180 kg/d

660 Chapter 6

Total WAS flow ⫽ (0.095 ⫹ 0.221) m

/min

⫽ 0.316 m

/min

⫽ 83.4 gal/min

Example 3: The aeration basin volume is 6600 m

(1.74 Mgal). The influ-

ent flow to the basin is 37,850 m

/d (10.0 Mgal/d) with BOD of 140 mg/L. The

MLSS is 3200 mg/L with 80% volatile portion. The SS concentration in the

return activated sludge is 6600 mg/L. The current waste activated-sludge flow

rate is 340 m

/d (0.090 Mgal/d). Determine the desired WAS flow rate using

the F/M ratio control technique with a desired F/M ratio of 0.32.

solution:

Step 1. Calculate BOD loading

BOD ⫽ 140 mg/L ⫽ 140 g/m

⫽ 0.14 kg/m

Loading ⫽ BOD ⫻ flow

⫽ 0.14 kg/m

⫻ 37,850 m

⫽ 5299 kg/d

⫽ 11,680 lb/d

Step 2. Calculate the desired MLVSS with the desired F/M ⫽ 0.32

Desired MLVSS ⫽ BOD loading/(F/M) ratio

⫽ (5299 kg/d)/(0.32 kg/(d ⭈ kg))

⫽ 16,560 kg

Step 3. Calculate the desired MLSS

Desired MLSS ⫽ MLVSS/0.80 ⫽ 16,560 kg/0.80

⫽ 20,700 kg

Step 4. Calculate actual MLSS under aeration

Actual MLSS ⫽ concentration ⫻ basin volume

⫽ 3.2 kg/m

⫻ 6600 m

⫽ 21,120 kg

Step 5. Calculate the additional solids to be removed daily

(actual ⫺ desired) MLSS ⫽ 21,120 kg/d ⫺ 20,700 kg/d

⫽ 420 kg/d

Step 6. Calculate the additional WAS flow q required

q ⫽ additional SS removed/SS in RAS

⫽ (420 kg/d)/(6.6 kg/m

)

⫽ 63.6 m

Wastewater Engineering 661

Step 7. Calculate the total WAS flow Q

Q ⫽ (340 ⫹ 63.6) m

⫽ 403.6 m

/d ⫽ 0.28 m

/min

⫽ 74 gal/min

Example 4: Calculate the waste activated-sludge flow rate using the mean

cell residence time (MCRT) method. The given conditions are the same as

those in Example 3. In addition, the desired MCRT is 7.5 days and the SS level

in the effluent is 12 mg/L.

solution: (use English system)

Step 1. Calculate SS concentration in the aeration tank

MLSS ⫽ 3200 mg/L

V ⫽ 1.74 Mgal

SS ⫽ 1.74 Mgal ⫻ 3200 mg/L ⫻ 8.34 lb/(Mgal ⭈ mg/L)

⫽ 46,440 lb ⫽ 21,070 kg

Step 2. Determine SS lost in the effluent

⫽ 12 mg/L

⫽ 10 Mgal/d

⫻ Q

⫽ 10 Mgal/d ⫻ 12 mg/L ⫻ 8.34 lb/(Mgal ⭈ mg/L)

⫽ 1000 lb/d ⫽ 453 kg/d

Step 3. Calculate the desired SS in waste sludge

Using Eq. (6.113)

SS wasted ⫽ (46,440 lb – 7500 lb)/7.5 days

⫽ 5192 lb/d ⫽ 2355 kg/d

Step 4. Calculate the WAS flow rate Q

⫽ 6600 mg/L

SS wasted ⫽ Q

⫻ SS

⫻ 8.34

⫽ 0.0943 Mgal/d

SS wasted

3 8.34

5192 lb/d

6600 mg/L 3 8.34 lb/sMgal

mg/Ld

7.5 days 5

46,440 lb

SS wasted, lb/d 1 1000 lb/d

MCRT, day 5

SS in aerator, lb

SS wasted, lb/d 1 SS in effluent, lb/d

662 Chapter 6

⫽ 0.0943 Mgal/d ⫻ 694(gal/min)/(Mgal/d)

⫽ 65.4 gal/min

⫽ 356 m

Example 5: The given operating conditions are the same as in Example 3.

The desired MLVSS is 16,500 kg (36,400 lb). The SS concentration in RAS is

6600 mg/L. Using MLVSS as a control technique, calculate the desired WAS

flow rate. The current WAS flow rate is 420 m

/d (0.11 Mgal/d)

solution:

Step 1. Calculate actual MLVSS

Tank volume ⫽ 6600 m

VSS/TSS ⫽ 0.80

MLSS ⫽ 3200 mg/L ⫽ 3.2 kg/m

Actual MLVSS ⫽ tank volume ⫻ volatiles ⫻ MLSS

⫽ 6600 m

⫻ 0.8 ⫻ 3.2 kg/m

⫽ 16,900 kg

Step 2. Calculate additional VSS to be wasted daily

Wasted ⫽ actual ⫺ desired (Step 3 of Example 3)

⫽ (16,900 ⫺ 16,500) kg/d

⫽ 400 kg/d

Step 3. Calculate additional WAS flow, q

VSS in RAS ⫽ 6600 mg/L ⫻ 0.8 ⫽ 5280 mg/L

⫽ 5.28 kg/m

q ⫽ (400 kg/d)/(5.28 kg/m

)

⫽ 75.8 m

/d ⫽ 13.9 gal/min

Step 4. Calculate the desired WAS flow rate

Q ⫽ current ⫹ additional flow

⫽ (420 ⫹ 75.8) m

⫽ 495.8 m

⫽ 91 gal/min

Sludge bulking.

A desirable activated sludge is one which settles rap-

idly leaving a clear, odorless, and stable supernatant. The efficiency of

treatment achieved in an activated-sludge process depends directly on

settleability of the sludge in the secondary clarifier. At times, poorly

Wastewater Engineering 663

settling sludge and foam formation cause the most common operational

problems of the activated-sludge process. Sludge with poorly flocculated

(pin) particles or buoyant filamentous growths increases its volume and

does not settle well in the clarifier. Light sludge in the clarifier then spills

over the weirs and is carried away in the effluent. The concentrations

of BOD and suspended solids increase in the effluent. This phenomenon

is called sludge bulking and it frequently occurs unexpectedly.

Sludge bulking is caused by: (1) the growth of filamentous organisms,

or organisms that can grow in filamentous form under adverse conditions,

and (2) adverse environmental conditions such as excessive flow, insuf-

ficient aeration, short circulating of aeration tanks, lack of nutrients,

septic influent, presence of toxic substances, or overloading. Nocardia

amarae, Microthrix parvicella, N. amarae-like organisms, N. pinensis-like

organisms, and type 0092 were most commonly found in foam and bulk

sludge. Nocardia growth is supported by high sludge age, low F/M ratio,

and higher rather than lower wastewater temperature (Droste, 1997).

Fungi are mostly filamentous microorganisms. Some bacteria such as

Beggiatoa, Thiotrix, and Leucothrix grow in filamentous sheaths.

There are no certain rules for prevention and control of sludge bulking.

If bulking develops, the solution is to determine the cause and then either

eliminate or correct it or take compensatory steps in operation control.

Some remedial steps may be taken, e.g. changing parameters such as

wastewater characteristics, BOD loading, dissolved oxygen concentra-

tion in the aeration basin, return sludge pumping rate, microscopic exam-

ination of organisms (check for protozoa, rotifers, filamentous bacteria,

and nematodes) in clarifiers’ and other operating units. Chlorination of

RAS in the range of 2 to 3 mg/L of chlorine per 1000 mg/L of MLVSS is

suggested (Metcalf and Eddy, Inc. 1991). Hydroperoxide also can be used

as an oxidant. Reducing the suspended solids in the aeration basin,

increasing air supply rate, or increasing the BOD loading (which may

depress filamentous growth), and addition of lime to the mixed liquor for

pH adjustment are the remedial methods which have been used. The

F/M ratio should be maintained at 0.5 to 0.2.

21.7 Modiﬁed activated-sludge processes

Numerous modifications (variations) of the conventional activated-sludge

plug flow process have been developed and proved effective for removal

of BOD and/or nitrification. The modified processes include tapered aer-

ation, step-feed aeration, complete-mix extended aeration, modified aer-

ation, high-rate aeration, contact stabilization, Hatfield process, Kraus

process, sequencing batch reactor, high-purity oxygen, oxidation ditch,

deep shaft reactor, single-stage nitrification, and separate stage nitrifi-

cation. Except for a few processes, most modified processes basically

differ on the range of the F/M ratio maintained and in the introduction

664 Chapter 6

locations for air supply and wastewater. The design parameters and

operational characteristics for various activated-sludge processes are

presented elsewhere (Metcalf and Eddy, Inc. 1991, WEF and ASCE,

1991a). Standard and modified processes are discussed below.

Conventional process. The earliest activated-sludge process was con-

tained in long narrow tanks with air for oxygen supply and mixing

through diffusers at the bottom of the aeration tank. A conventional

plug-flow process consists of a primary clarifier, an aeration tank with

air diffusers for mixing the wastewater and the activated sludge in the

presence of dissolved oxygen, a secondary settling basin for solids

removal, and a sludge-return line from the clarifier bottom to the influ-

ent of the aeration tank (Fig. 6.18). The return activated sludge is mixed

with the incoming wastewater and passes through the aeration tank in

a plug-flow fashion. High organics concentration and high microbial

solids at the head end of the aeration tank lead to a high oxygen demand.

The conventional process is more susceptible to upset from shock loads

and toxic materials, and is used for low-strength domestic wastewaters.

The conventional process is designed to treat 0.3 to 0.6 kg BOD

applied/(m

⭈ d) (20 to 40 lb/(1000 cu ⭈ ft ⭈ d)) in the United States with

MLSS of 1500 to 3000 mg/L. The aeration period ranges between 4 and

8 h with return activated-sludge flow ratios of 0.27 to 0.75. The cell res-

idence time (u

) is 5 to 15 days and the F/M ratio is 0.2 to 0.4 per day.

The process generally removed 85% to 95% of BOD

and produces highly

nitrified effluent (WEF and ASCE, 1991a).

Example: The operational records at a conventional activated-sludge plant

are shown as average values as below.

Calculate the following operational parameters: (1) volumetric BOD loading

rate; (2) F/M ratio; (3) hydraulic retention time u; (4) cell residence time u

; (5)

return activated-sludge ratio; and (6) removal efficiencies for BOD, TSS, and

total solids.

Wastewater flow Q 7570 m

/d (2.0 Mgal/d)

Wastewater temperature 20⬚C

Volume of aeration tanks 2260 m

(79,800 ft

)

Influent BOD

(say BOD) 143 mg/L

Influent total suspended solids 125 mg/L

Influent total solids 513 mg/L

Effluent total solids 418 mg/L

Effluent TSS 24 mg/L

Effluent BOD 20 mg/L

Return sludge flow Q

3180 m

/d (0.84 Mgal/d)

MLSS 2600 mg/L

TSS in waste sludge 8900 mg/L

Volume of waste sludge 200 m

/d (0.053 Mgal/d)