Lin S.D. Water and Wastewater Calculations Manual

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Oxygen requirements generally depend on the design peak hourly

BOD

, MLSS, and degree of treatment. Aeration equipment must be able

to maintain a minimum of 2.0 mg/L of dissolved oxygen concentration

in the mixed liquor at all times and provide adequate mixing. The

normal air requirements for all activated-sludge systems, except

extended aeration, are 1.1 kg of oxygen (93.5 m

of air) per kg BOD

1.1 lb of oxygen (1500 ft

of air) per lb BOD

, for design peak aeration

tank loading. That is 94 m

of air per kg of BOD

(1500 ft

/lb BOD

) at

standard conditions of temperature, pressure, and humidity. For the

extended aeration process, normal air requirements are 128 m

/kg BOD

or 2050 ft

/lb BOD

(GLUMRB, 1996).

For F/M ratios greater than 0.3 d

⫺1

, the air requirements for con-

ventional activated-sludge systems amount to 31 to 56 m

/kg (500 to

900 ft

/lb) of BOD

removed for coarse bubble (nonporous) diffusers

and 25 to 37 m

/kg (400 to 600 ft

/lb) BOD

removal for fine bubble

(porous) diffusers. For lower F/M ratios, endogenous respiration, nitri-

fication, and prolonged aeration increase air use to 75 to 112 m

/kg

(1200 to 1800 ft

/lb) of BOD

removal. In practice, air requirements

range from 3.75 to 15.0 m

air/m

water (0.5 to 2 ft

/gal) with a typical

value of 7.5 m

or 1.0 ft

/gal (Metcalf and Eddy, Inc. 1991).

Example 1a: Design a complete-mix activated-sludge system.

Given:

Average design flow ⫽ 0.32 m

/s (6.30 Mgal/d)

Peak design flow ⫽ 0.80 m

/s (18.25 Mgal/d)

Raw wastewater BOD

⫽ 240 mg/L

Raw wastewater TSS ⫽ 280 mg/L

Effluent BOD

ⱕ 20 mg/L

Effluent TSS ⱕ 24 mg/L

Wastewater temperature ⫽ 20⬚C

Operational parameters and biological kinetic coefficients:

Design mean cell residence time u

⫽ 10 d

MLVSS ⫽ 2400 mg/L (can be 3600 mg/L)

VSS/TSS ⫽ 0.8

TSS concentration in RAS ⫽ 9300 mg/L

Y ⫽ 0.5 mg VSS/mgBOD

⫽ 0.06/d

BOD

/ultimate BOD

⫽ 0.67

Wastewater Engineering 635

Assume:

1. BOD (i.e. BOD

) and TSS removal in the primary clarifiers are 33% and

67%, respectively.

2. Specific gravity of the primary sludge is 1.05 and the sludge has 4.4% of

solids content.

3. Oxygen consumption is 1.42 mg per mg of cell oxidized.

solution:

Step 1. Calculate BOD and TSS loading to the plant

Design flow Q ⫽ 0.32 m

/s ⫻ 86,400 s/d

⫽ 27,648 m

Since 1 mg/L ⫽ 1 g/m

⫽ 0.001 kg/m

BOD loading ⫽ 0.24 kg/m

⫻ 27,648 m

⫽ 6636 kg/d

TSS loading ⫽ 0.28 kg/m

⫻ 27,648 m

⫽ 7741 kg/d

Step 2. Calculate characteristics of primary sludge

Step 3. Calculate flow, BOD, and TSS in primary effluent (secondary influent)

4446 kg/d 3 1000 g/kg

27,536 m

5 4446 kg/d

BOD 5 6636 kg/d 2 2190 kg/d

5 27,536 m

/d 5 Q for Step 6

Flow 5 design flow, 27,648 m

/d 2 112 m

5 112 m

Sludge flow rate 5

5186 kg/d

1.05 3 1000 kg/m

4 0.044 kg/kg

Solids concentration 5 4.4% 5 0.044 kg/kg

Specific gravity of sludge 5 1.05

TSS removed 5 7741 kg/d 3 0.67 5 5186 kg/d

BOD removed 5 6636 kg/d 3 0.33 5 2190 kg/d

636 Chapter 6

Step 4. Estimate the soluble BOD

escaping treatment, S, in the effluent

Use the following relationship

Effluent BOD ⫽ influent soluble BOD escaping treatment,

S ⫹ BOD of effluent suspended solids

(a) Determine the BOD

of the effluent SS (assuming 63% biodegradable)

Biodegradable effluent solids ⫽ 24 mg/L ⫻ 0.63 ⫽ 15.1 mg/L

Ultimate BOD

of the

biodegradable effluent solids ⫽ 15.1 mg/L ⫻ 1.42 mg O

/mg cell

⫽ 21.4 mg/L

BOD

⫽ 0.67 BOD

⫽ 0.67 ⫻ 21.4 mg/L

⫽ 14.3 mg/L

(b) Solve for influent soluble BOD

escaping treatment

20 mg/L ⫽ S ⫹ 14.3 mg/L

S ⫽ 5.7 mg/L

Step 5. Calculate the treatment efficiency E using Eq. (6.90)

(a) The efficiency of biological treatment based on soluble BOD is

5 96.5%

E 5

2 S

3 100 5

s161.5 2 5.7 mg/Ld 3 100%

161.5 mg/L

5 92.8 mg/L

5 92.8 g/m

2555 kg/d 3 1000 g/kg

27,536 m

5 2555 kg/d

TSS 5 7741 kg/d 2 5186 kg/d

5 161.5 mg/L 5 S

5 161.5 g/m

Wastewater Engineering 637

(b) The overall plant efficiency including primary treatment is

Step 6. Calculate the reactor volume using Eq. (6.79)

Step 7. Determine the dimensions of the aeration tank

Provide 4 rectangular tanks with common walls. Use width-to-length ratio

of 1:2 and water depth of 4.4 m with 0.6 m freeboard

w ⫻ 2w ⫻ (4.4 m) ⫻ 4 ⫽ 5586 m

w ⫽ 12.6 m

width ⫽ 12.6 m

length ⫽ 25.2 m

water depth ⫽ 4.4 m (total tank depth ⫽ 5.0 m)

Note: In the Ten States Standards, liquid depth should be 3 to 9 m (10 to

30 ft). The tank size would be smaller if a higher design value of MLVSS

were used.

Step 8. Calculate the sludge wasting flow rate from the aeration tank

5 1.48 Mgal

5 5586 m

V 5

s10 daysds27,536 m

/dds0.5ds161.5 2 5.7d mg/L

s2400 mg/Lds1 1 0.06 day

3 10 daysd

5 0.06 d

X 5 2400 mg/L

S 5 5.7 mg/L sfrom Step 4bd

5 161.5 mg/L sfrom Step 3d

Y 5 0.5 mg/mg

Q 5 27,536 m

/d sfrom Step 3d

5 10 d

V 5

QYsS

2 Sd

Xs1 1 k

5 91.7%

E 5

s240 2 20d mg/L 3 100%

240 mg/L

638 Chapter 6

Using Eq. (6.71), also V ⫽ 5586 m

and VSS ⫽ 0.8 SS

Step 9. Estimate the quantity of sludge to be wasted daily

(a) Calculate observed yield

Using Eq. (6.81)

(b) Calculate the increase in the mass of MLVSS for Eq. (6.91)

⫽ Y

obs

Q(S

– S) ⫻ (1 kg/1000 g)

⫽ 0.3125 ⫻ 27,536 m

/d ⫻ (161.5 – 5.7) g/m

⫻ 0.001 kg/g

⫽ 1341 kg/d

Note: A factor of 8.34 lb/Mgal is used if Q is in Mgal/d.

⫽ 1341 kg/d ⫼ 0.8

⫽ 1676 kg/d

(d) Calculate TSS lost in the effluent, p

⫽ (27,536 ⫺ 270) m

/d ⫻ 24 g/m

⫼1000 g/kg

⫽ 654 kg/d

Note: Flow is less sludge wasting rate from Step 8.

(e) Calculate the amount of sludge that must be wasted

Wastewater sludge ⫽ p

⫺ p

⫽ 1676 kg/d ⫺ 654 kg/d

⫽ 1022 kg/d

5 0.3125

obs

1 1 K

0.5

1 1 0.06 3 10

5 0.0715 Mgal/d

5 270 m

10 days 5

s5586 m

ds2400 mg/Ld

s3000 mg/Ld 1 s27,536 m

/dds24 mg/L 3 0.8d

X 1 Q

Wastewater Engineering 639

Step 10. Estimate return activated sludge rate

Using a mass balance of VSS, Q and Q

are the influent and RAS flow rates,

respectively.

VSS in aerator ⫽ 2400 mg/L

VSS in RAS ⫽ 9300 mg/L ⫻ 0.8 ⫽ 7440 mg/L

2400 (Q ⫹ Q

) ⫽ 7440 Q

/Q ⫽ 0.4762

⫽ 0.4762 ⫻ 27,536 m

⫽ 13,110 m

⫽ 0.152 m

Step 11. Check hydraulic retention time (HRT ⫽ u)

u ⫽ V/Q ⫽ 5586 m

/(27,536 m

/d)

⫽ 0.203 d ⫻ 24 h/d

⫽ 4.87 h

Note: The preferred range of HRT is 5–15 h.

Step 12. Check F/M ratio using U in Eq. (6.83)

Step 13. Check organic loading rate and mass of ultimate BOD

utilized

Step 14. Compute theoretical oxygen requirements

The theoretical oxygen required is calculated from Eq. (6.94a)

5 6403 kg/d

27,536 m

/d 3 s161.5 2 5.7d g/m

0.67 3 1000 g/kg

BOD

used 5 QsS

–Sd/0.67

BOD

5 0.67 BOD

sgivend

Loading 5

27,536 m

/d 3 161.5 g/m

5586 m

3 1000 g/kg

5 0.80 kg BOD

/sm

5 0.32 day

U 5

2 S

161.5 mg/L 2 5.7 mg/L

s0.203 dayds2400 mg/Ld

640 Chapter 6

Step 15. Compute the volume of air required

Assume that air weighs 1.202 kg/m

(0.075 lb/ft

) and contains 23.2% oxygen

by weight; the oxygen transfer efficiency for the aeration equipment is 8%;

and a safety factor of 2 is used to determine the actual volume for sizing the

blowers.

(a) The theoretical air required is

(b) The actual air required at an 8% oxygen transfer efficiency

Air ⫽ 16,200 m

/d ⫼ 0.08

⫽ 202,000 m

⫽ 140 m

/min

⫽ 4950 ft

/min

Air ⫽ 140 m

/min ⫻ 2

⫽ 280 m

/min

⫽ 9900 ft

/min

⫽ 165 ft

/s or (cfs)

Step 16. Check the volume of air required per unit mass BOD

removed, and

per unit volume of wastewater and aeration tank, using the actual value

obtained in Step 15b

(a) Air supplied per kg of BOD

removed

5 754 ft

/lb

5 47.1 m

of air/kg BOD

Air 5

202,000 m

/d 3 1000 g/kg

27,536 m

/d 3 s161.5 2 5.7d g/m

5 16,200 m

Air 5

4499 kg/d

1.202 kg/m

3 0.232 g O

/g air

5 4499 kg/d

5 6403 kg/d sfrom Step 13d 2 1.42 3 1341 kg/d sfrom Step 9bd

QsS

2 Sd

s1000 g/kgd f

2 1.42p

Wastewater Engineering 641

(b) Air supplied per m

of wastewater treated

of aeration tank

Example 1b: Design secondary clarifiers using the data in Example la and

the MLSS settling test results. The MLSS setting data is derived from a pilot

plant study and is shown below:

MLSS, mg/L 1200 1800 2400 3300 4000 5500 6800 8100

Initial settling 4.1 3.1 2.1 1.2 0.77 0.26 0.13 0.06

velocity, m/h

solution:

Step 1. Plot the MLSS settling curve (Fig. 6.16) on log-log paper from the

observed data

Step 2. Construct the gravity solid-flux curve from Fig. 6.17

Data in columns 1 and 2 of the following table is adopted from Fig. 6.16.

Values in col. 3 are determined by col. 1 ⫻ col. 2. Plot the solids–flux curve

using MLSS concentration and calculate solids flux as shown in Fig. 6.17.

(1) (2) (3)

MLSS concentration X, Initial settling velocity Solids flux XV

mg/L or g/m

, m/h kg/(m

⭈ h)

1000 4.2 4.20

1500 3.7 5.55

2000 2.8 5.60

2500 2.0 5.00

3000 1.5 4.50

4000 0.76 3.04

5000 0.41 2.05

6000 0.22 1.32

7000 0.105 0.74

8000 0.062 0.50

9000 0.033 0.30

5 36.2 ft

/sft

5 36.2 m

/sm

Air 5

202,000 m

5586 m

5 0.98 ft

air/gal wastewater

5 7.34 m

air/m

wastewater

Air 5

202,000 m

27,536 m

642 Chapter 6

Figure 6.16 Solids settling rate: experimental results for design of secondary clarifier.

Step 3. Determine the limiting solids flux value

From Fig. 6.17, determine the limiting solids flux (SF) for an underflow con-

centration of 9300 mg/L. This is achieved by drawing a tangent to the solids

flux curve from 9300 mg/L (the desired underflow) solids concentration. The

limiting solids flux value is

1.3 kg/(m

⭈ h) ; 31.2 kg/(m

⭈ d) ; 6.4 lb/(ft

⭈ d)

Step 4. Calculate design flow to the secondary clarifiers, Q

From Steps 8 and 10 of Example 1a

Q ⫽ average design flow ⫹ return sludge flow ⫺ MLSS wasted

⫽ (27,563 ⫹ 12,942 ⫺ 270) m

⫽ 40,235 m

⫽ 0.466 m

Use two clarifiers, each one with flow of 20,200 m

/d.

Wastewater Engineering 643

Step 5. Compute the area A and diameter d of the clarifier

(6.98)

where A ⫽ area of the secondary clarifier, m

or ft

Q ⫽ influent flow of the clarifier, m

/h or gal/h

X ⫽ MLSS concentration, kg/m

or lb/ft

SF ⫽ limiting solids flux, kg/(m

⭈ h) or lb/(ft

⭈ h)

From Eq. (6.98), for each clarifier

Q ⫽ 20,200 m

/d ⫽ 841.7 m

MLSS ⫽ (2400/0.8) mg/L ⫽ 3.0 kg/m

SF ⫽ 1.3 kg/(m

⭈ h) (from Step 3)

Therefore

A 5 psd/2d

5 1942 m

A 5

841.7 m

/h 3 3.0 kg/m

1.3 kg/sm

A 5

644 Chapter 6

Figure 6.17 Solids flux curve.