DESIGN EXAMPLE OF ACTIVATED SLUDGE PROCESS

Design Examples...
Design Example of Conventional Activated Sludge Process...
An activated sludge system is to be used for secondary treatment of 10,000 m³/day of municipal wastewater. After primary clarification, the BOD is 150 mg/L, and it is desired to have not more than 5 mg/L of soluble BOD in the effluent. A completely mixed reactor is to be used, and pilot - plant analysis has established the following kinetic values ;

• Y = 0.5 kg / kg
• k_d = 0.05 1 / day

Assuming an MLSS concentration of 3,000 mg/L and an underflow concentration of 10,000 mg/L from the secondary clarifier, determine ;

• The volume of the reactor
• The mass and volume of solids that must be wasted each day
• The recycle ratio

"Schematic Diagram of the System"...

Solution...

"Reactor Volume"... "Excess Solids"... "Mass Balance"...

Design Example of Extended - Aeration Activated Sludge Process (Package Plant)...
A prefabricated package plant is to be used to treat the wastewater from a resort area consisting of 550 individual family residences. The average occupancy has been estimated to be 2.9 persons per residence. Use a flow of 230 L/person.day and a daily peaking factor of 2.5 for flow, BOD₅ and SS. Use an hourly peaking factor of 4 for sizing the sedimentation facilities. Select the type of package plant and size the principal components of the plant. Effluent BOD₅ concentration of the process must be or lower than 30 mg/L.
Solution...

1. The total number of person : (550 home)(2.9 person/home) = 1,595 person
2. The corresponding average flowrate : (1,595 person)(0.230 m³/person.day) = 366.85 m³/day
3. The corresponding peak daily flowrate : (366.85 m³/day)(2.5) = 917.13 m³/day
4. The average BOD₅ load : (1,595 persons)(80 g BOD₅/person.day) = 127,600 g BOD₅/day
5. The corresponding BOD₅ concentration : (127,600 g/day) / (366.85 m³/day) = 348 g/m³ (mg/L)
6. The average SS load : (1,595 persons)(90 g SS/person.day) = 143,550 g SS/day
7. The corresponding SS concentration : (143,550 g/day) / (366.85 m³/day) = 391 g/m³ (mg/L)
8. The peak daily BOD₅ load : (127.60 kg BOD₅/day)(2.5) = 319.00 kg BOD₅/day
9. The peak daily SS load : (143.55 kg SS/day)(2.5) = 358.88 kg SS/day
10. An extended aeration activated sludge process package plant is selected
11. The aeration time : 1.0 day
12. The aeration tank volume : (366.85 m³/day)(1.0 day) = 366.85 m³
13. The oxygen transfer efficiency : 6 %
14. The specific weight of air : 1.26 kg/m³
15. The oxygen content : 23.2 %
16. The air requirement : (319.00 kg BOD₅/day) / (1.26 kg/m³)(0.232)(0.06) = 18,187.83 m³/day
17. The peak hour factor for the settling tank : 4
18. The overflow rate for settling tank : 24 m³/m².day
19. The surface area of settling tank : (366.85 m³/day)(4) / 24 m³/m².day = 61.14 m²
20. The hydraulic detention time for settling tank : 0.5 h
21. The volume of the settling tank : (366.85 m³/day)(4)(0.5 h) / 24 h/day = 30.57 m³

Design Example of SBR...
A sequencing batch reactor activated - sludge process is to be used to treat wastewater with the characteristics given below. Determine the mass of suspended solids in the reactor over a 7 - day operating period. The effluent is to have 20 mg/L of BOD₅ or less. Determine also the depth of clear liquid measured from the top of the settled sludge to the lowest liquid level reached during the decant cycle. Use the following design criteria and constraints.
Data...

1 - Influent flow-rate = 3,800 m3/day	10 - Concentration of settled sludge = 8,000 mg/L
2 - Influent suspended solids = 200 mg/L	11 - Settled sludge specific gravity = 1.02
3 - Influent VSS = 150 mg/L	12 - 60 % of reactor volume will be decanted each day
4 - Wastewater temperature = 20 °C	13 - Liquid depth of SBR = 6.60 m
5 - Hydraulic detention time = 24 h	14 - Sludge wasting is done once a week
6 - F/M = 0.1 kg BOD5/kg MLVSS.day	15 - 65 % of effluent is biodegradable
7 - MLVSS/MLSS = 0.80	16 - BOD5 = 0.68 BODL
8 - Y = 0.65 kg/kg	17 - BODL of one mole cells = 1.42 times of X
9 - kd = 0.05 1/day	18 - C : N : P is suitable

Day	Average BOD5 (mg/L)
1	250
2	400 *
3	400 *
4	400 *
5	400 *
6	250
7	250
*	Increase over 250 mg/L is soluble BOD

Solution...

1. Biodegradable portion of effluent biological solids = (0.65)(20 mg/L) = 13.0 mg/L
2. Ultimate BOD of the biodegradable effluent solids = (13.0 mg/L)(1.42 mg/mg) = 18.5 mg/L
3. BOD₅ of effluent suspended solids = (18.5 mg/L)(0.68) = 12.6 mg/L
4. Influent soluble BOD₅ escaping treatment = 20.0 mg/L - 12.6 mg/L = 7.4 mg/L
5. Tank volume = (3,800 m³/day)(1.0 day) / 0.60 = 6,333.33 m³
6. MLVSS = (3,800 m³/day)(250 g/m³)(10^{- 3} kg/g) / (6,333.33 m³)(0.1 kg/kg.day) = 1,497 g/m³
7. Total SS in the reactor = (200 - 150 mg/L) + (1,497 mg/L) / 0.80 = 1,921 mg/L
8. The mass of VSS in the reactor = (6,333.33 m³)(1,497 g/m³)(10^-3 kg/g) = 9,481 kg
9. The total mass of SS in the reactor = (6,333.33 m³)(1,921 g/m³)(10^-3 kg/g) = 12,166 kg
   
   "Mass of SS in the Reactor"...
   "Solid Production"...
   
10. The net mass of VSS in the system at the beginning of 1^st day = (0.65)(250 - 7.4 g/m³)(10^-3 kg/g)(3,800 m³/day) - (0.05 1/day)(9,481 kg) = 125 kg/day
11. The mass of inert SS added in 1^st day = (200 - 150 g/m³)(10^-3 kg/g)(3,800 m³/day) = 190 kg/day
12. The mass of SS at the end of 1^st day = 12,166 kg/day+125 kg/day / 0.80+190 kg/day = 12,512 kg

Results...

Day	BOD (mg/L)	Px (kg/day)	SSi (kg/day)	VSST (kg/day)	SST (kg/day)
1	250	125	190	9,590	12,512
2	400	487	190	10,077	13,289
3	400	462	190	10,539	14,055
4	400	439	190	10,978	14,794
5	400	417	190	11,395	15,505
6	250	27	190	11,422	15,728
7	250	26	190	11,448	15,950

Calculation of Oxygen Requirement...
Calculate oxygen requirement of a complete - mix activated sludge process treating domestic wastewater having flowrate of 0.25 m³/sec. BOD₅ concentration of settled wastewater is 250 mg/L. The effluent soluble BOD₅ is 6.2 mg/L. Increase in the mass of MLVSS is 1,646 kg/day. Assume that the temperature is 20 °C and the conversion factor, BOD₅ / BOD_L is 0.68. "Oxygen Requirement"...

Design Example of Final Settling Tank...
A column analysis was run to determine the settling characteristics of an activated sludge suspension. The results of the analysis are shown below. "Results of the Column Analysis"...

The influent concentration of MLSS is 3,000 mg/L, and the flow rate is 8,000 m³/day. Determine the size of the clarifier that will thicken the solids to 10,000 mg/L.
Solution...
1. Calculate the solids flux from the above data : G = MLSS (kg/m³) . Velocity (m/h) "Results of the Solid Flux"...

2. Plot solids flux vs. MLSS concentration as shown below. Draw a line from the desired underflow concentration, 10,000 mg/L, tangent to the curve and intersecting the ordinate. The value of G at the intersection, 2.4 kg/m².h, is the limiting flux rate and governs the thickening function. "Solid Flux vs MLSS Concentration"...

3. Determine total solids loading to the clarifier.
(8,000 m³/day)(1/24 day/h)(3,0 kg/m³) = 1,000 kg/h
4. Determine the surface area of the clarifier.
(1,000 kg/h) / (2.4 kg/m².h) = 416.7 m²
5. Calculate the diameter of the clarifier.
SQRT[(4/3.14)(416.7 m²)] = 23 m
6. Check clarification function.
(8,000 m³/day)(1/24 day/h) = 333 m³/h
(333 m³/h) / (1.21 m/h) = 275 m²
Result...
Because 275 m² < 416.7 m², the thickening function governs the design.
Design Results of an Extended Aeration ASP...

Parameter	Result
	Value	Unit
Influent parameters
Population equivalent	45,000	pe
Average daily flow	11,500	m3 / day
Average daily flow	133	L / s
Peaking factor	1.80	-
Design flow	863	m3 / hr
Design flow	240	L / s
PWWF	25,000	m3 / day
PWWF	289	L / s
PWWF	1,042	m3 / hr
BOD design load	3,150	kg / day
Aeration tank
F / M	0.05	kg BOD / kg MLSS . day
MLSS concentration	4,000	mg / L
Aeration volume required	15,750	m3
HRT @ avg flow	32.9	hr
HRT @ design flow	18.3	hr
Approximate tank size
Depth	4.0	m
Width	35.0	m
Length	112.5	m
Secondary clarifier
SVI	150	g / mL
Sludge volume load	0.06	m3 / m2 . day
Surface area required	8,625	m2
Overflow rate	0.10	m3 / m2 . hr
Diameter	104.8	m
Average depth	3.5	m
Volume	30,188	m3
HRT @ average flow	6.3	hr
HRT @ design flow	3.5	hr
Sludge handling
Sludge production rate	0.95	kg TS / kg BOD
Sludge production	2,993	kg TS / day
Days storage in aeration tank	2.6	by incr. MLSS by 0.5 kg / m3 . day
Days storage in aeration tank	5.3	by incr. MLSS by 1.0 kg / m3 . day
Volume @ 0.8 %	374	m3 / day
HRT thickener	18.0	day
Volume thickener	6,733	m3
Height	4.0	m
Diameter	46.3	m
HRT @ 2 %	45	hr
Volume @ 3 %	99.8	m3 / day
HRT	67	day
Dewatering capacity	15.0	m3 / hr
Dewatering duration	46.6	hr / week
Dewatering volume @ 18 %	16.6	m3

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