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Showing posts with label Aeration. Show all posts
Showing posts with label Aeration. Show all posts

Wednesday, January 07, 2009

Calculations!:Aeration Tank Volume, HRT, MLSS Values.

Calculations!:Aeration Tank Volume, HRT, MLSS Values.
I am working in a STP & ETP plant. I need to know how to calculate:

1. Volume of Aeration tank
2. HRT
3. F/M Ratio
4. MLSS value.

The relevant details of the ETP are as below:

Influent flow = 10 m3 / day
In. flow BOD = 1200 mg / L
In.flow COD = 2200 mg / L
TSS  = 850 mg / L

  Jeyaroopa
  jeyaroopa79@gmail.com
Dear Jayroopa

You may please refer Water & Waste Water Engineering by Metcalf & Eddy.
BOD of effluent is 1200 mg/l. It will be difficult to bring down the BOD by Activated Sludge process to desire permissible limits.

Simplified method of calculation is as below.

Process Activated sludge
Flo3, CUM 10
Type Extended aeration
Food to Microorganism ratio (F/M) 0.15
Total BOD load, Kg BOD*Flow/1000.= 12
Total mixed liquor suspended solids(MLSS) Total BOD/(F/M) = 80 kg
MLSS in Tank say mg/l 3000
Volume of tank, CUM Total MLSS/(MLSS/1000) = 26.6
Retention time, hr 24*Volume of Tank/Flow = 63.84

You can select F/M & MLSS values and optiimise the volume of a tank.
The waste water has high suspended solids therefore sedimentation is must.

Maximum efficiency of Activated sludge process is 95% therefore if you need lower value of BOD in treated effluent then you have to opt two stage process.
For further information please contact

Prof. R. V. Saraf
Director
Viraj Envirozing India Pvt. Ltd.
21 Radhakrishna Near SBI, Paud Road, Pune 38
watersgs@vsnl.net


13 Aug, 2008   |  Taral Kumar
Dear Jeyaroopa,

Though the calculation is not that simple, I will give you a thumb rule. Multiply BOD with Quantity of effluent and divide it by 500 to get aeration tank volume. For example, for 10 cum/day with 1200 BOD, aeration tank volume shall be 10 x 1200 / 500 = 24 cum. That is 2.5 days storage nearly. But that is because the BOD is excessively high.

With best regards,

Taral Kumar
Executive Director
Akar Impex Pvt. Ltd.
Noida, Uttar Pradesh
http://www.indiawaterportal.org/Network/askq/kb/?View=entry&EntryID=343
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Thursday, January 01, 1998

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
  • kd = 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 1st 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 1st 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 1st 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 / BODL 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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