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Monday, July 25, 2016

HOW TO DESIGN A STP USING MBBR FBBR FMR TECHNOLOGY



An introduction to MBBR (moving bed biofilm reactor )/ FM Reactor/ FAB /FMR Reactor wastewater treatment

When communities of microorganisms grow on surfaces, they are called biofilms. Microorganisms in a biofilm wastewater treatment process are more resilient to process disturbances compared to other types of biological treatment processes.  Thus, biofilm wastewater treatment technologies can be considerably more robust especially when compared to conventional technologies like activated suldge process..
In the MBBR biofilm technology the biofilm grows protected within engineered plastic carriers, which are carefully designed with high internal surface area. These biofilm carriers are suspended and thoroughly mixed throughout the water phase. With this technology it is possible to handle extremely high loading conditions without any problems of clogging, and treat industrial and municipal wastewater on a relatively small footprint.

System description

The MBBR™ biofilm technology is based on specially designed plastic biofilm carriers or biocarriers that are suspended and in continuous movement within a tank or reactor of specified volume. The design of associated aerators, grids, sieves, spray nozzles and other integral parts to the reactor is also of great importance in making up the system as a whole .
The industrial and municipal wastewater is led to the MBBR™ treatment reactor where biofilm, growing within the internal structures of the biocarriers, degrade the pollutants.  These pollutants that need to be removed in order to treat the wastewater are food or substrate for growth of the biofilm.  The biocarrier design is critical due to requirements for good mass transfer of substrate and oxygen to the microorganisms  .  Excess biofilm sloughs off the biocarrier in a natural way .

An aeration grid located at the bottom of the reactor supplies oxygen to the biofilm along with the mixing energy required to keep the biocarriers suspended and completely mix within the reactor.

Treated water flows from reactor through a grid or a sieve, which retains the MBBR™ biocarriers in the reactor. Depending on the wastewater, the reactors are may be equipped with special spray nozzles that prevent excessive foam formation.

The MBBR is a biological aerobic degradation of organic pollutants. The process utilizes millions of tiny, polyethylene biofilm elements that provide a high surface area as a home for a vast, highly active bacteria culture. This fixed film process features a flexible reactor design, the ability to handle load increases without the need for extra tankage, and remains stable under large load variations, including temperature, strength or pH. Like the activated sludge process, the MBBR process utilizes the whole volume of an open tank. Unlike an activated sludge reactor, it does not require sludge return to operate effectively. In MBBR , addition of media quantity and Air Quantity is the Key Factor.

Total reactor volume of the MBBRs is designed for different hydraulic retention time for different types of waste water at average flows and than checked against peak flows. Essentially nutrient levels and DO levels are the only control points for the system.



Moving Bed Biofilm Bioreactor (MBBR) process uses the whole tank volume for biomass growth. It uses simple floating media, which are carriers for attached growth of biofilms. Biofilm carrier movement is caused by the agitation of air bubbles. This compact treatment system is effective in removal of BOD as well as nitrogen and phosphorus while facilitating effective solids separation.

Design and Construction Principles

Neutralised and settled wastewater passes through MBBR for reduction in BOD/COD. Most of the MBBR plants are provided with vertically or horizontally mounted rectangular mesh sieves or cylindrical bar sieves. Biofilm carriers are made up of high density (0.95 g/cm3) polyethelene. These are normally shaped as small cylinders with a cross inside and fins outside. The standard filling of carrier is  not more than 465 m2/m3. Generally, design load for COD-BOD removal is 20 g COD / m2d. Smaller carriers need smaller reactor volume at a given loading rate (as g/m2d) when the carrier filling is same. 

It is advisable to use MBBR in combination with a DEWATS  as a pre-treatment unit, depending on the local conditions and input characteristics. It is a very robust and compact alternative for secondary treatment of municipal wastewater, having removal efficiency for BOD 90 – 95% (low rate) and that of 75 – 80% for high rate. Average nitrogen removal is about 85%. There is no need for sludge recirculation. Phosphorus and faecal coliform reduction is feasible with additional passive (non-mechanical) or active (mechanical) system components.

A constantly operating MBBR does not require backwashing or return sludge flows. It has minimal head-loss. Coarse-bubble aeration in the aeration zone in the wastewater treatment tank provides ease of operation at low-cost. Agitation continuously moves the carrier elements over the surface of the screen thus preventing clogging. Maintenance of MBBR system includes screening, influent equalisation, clarifier system, sludge handling and integrated control system. There is no need to maintain f/M ratio as there is self-maintenance of an optimum level of productive biofilm. Skilled labour is required for routine monitoring and operations of pumps and blowers.


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

Developing MLSS during Startup; How?
We are commissioning 200 m3/hr capacity wastewater treatment plant. Type of industry: Pulp and paper mode of treatment: Primary clarifier, Aeration tank; secondary clarifier.  Please let us know how to develop MLSS in aeration tank during start-up of the plant.

Satyanarayna
satyanarayna.cheerla@gmail.com
Showing: 1-5 of 5  
Responses

06 Jan, 2009   |  Nazimuddin


Dear Satyanarayna,

Fill the aeration tank with 50% primary treated wastewater + 50 % water and add some bacterial seed material (say 500 litre mixed liquor from an operating ASP plant or 100 kg cattle dung) aerate for 12 hours (as a batch reactor) to acclimatize and develop more biomass. Allow to settle for an hour and replace 50% supernatant with primary treated wastewater. Aerate for 6 more hours (as a batch reactor) to develop more biomass.

Now put the aeration tank to operation in continuous mode by feeding continuously with primary treated wastewater with sludge re-circulation as per design re-circulation rate but without any sludge wasting till the biomass concentration reaches to design MLSS. During this phase collect and analyze MLSS samples at every 12 hour interval.

Once the design MLSS concentration is achieved start sludge wasting as per design solids retention time (SRT). For example for a design SRT of 10 days one-tenth of the biomass is to be wasted per day and for a design SRT of 5 days one-fifth of the biomass is        to be wasted per day. Volume of sludge wasting is calculated based on concentration of biomass in the sludge wasted considering whether sludge is wasted directly from aeration tank of from re-circulation line.

However, the total sludge wasting should be spread / split throughout a day, but at least in three parts if done in batches.
It is also worth noting that in a plant to treat industrial wastewater such as of paper mill it is very important that primary sedimentation should work very efficiently all the time else the share of inert solids in the aeration tank would be substantial and will reflect as part of MLSS/MLVSS, although not being active biomass.

Views are expressed in personal capacity.

Nazimuddin
Environmental Engineer
Pollution Control Implementation Division - III
Central Pollution Control Board
New Delhi



The F/M ratio calculation is just one of the process control parameters you should monitor in your activated sludge system. The following are suggested operating parameters for F/M Ratio. The optimum F/M ratio for your plant can only be determined by monitoring facility performance through regular process control testing.
Conventional Activated Sludge (F/M)
0.2  -  0.5
Contact Stabilization (F/M)
0.15  -  0.2
Extended Aeration (F/M)
0.01  - 0.07
Step Aeration (F/M)
0.2  - 0.5
Two-Stage Aeration (F/M)
0.07  -  0.15

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


Solution...


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


  1. 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
  2. The mass of inert SS added in 1st day = (200 - 150 g/m3)(10-3 kg/g)(3,800 m3/day) = 190 kg/day
  3. 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.


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.


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)


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.


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

SOURCE OF THE ARTICLE :http://web.deu.edu.tr/atiksu/toprak/ani414.html




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