waste water treatment plant design

Sunday, May 2, 2010

1-Introduction

Oil industry represents an important part of the food industry.A large number of disposes of their wastes into the environment in enormous quantities,constituting a significant loss of resources and causing serious pollution problems.If any saving in energy waste can be coupled with a cheap form of waste utilization or treatment,it would further improve the benefits to the food industry.

In the activated sludge plant, the water to be treated is subject to the simultaneous contact with microorganisms and oxygen (air). These microorganisms convert the organic compounds in the raw water for the greater part into carbon dioxide,water ammonia and new activated sludge.
the carbon dioxide generated is released into the atmosphere by aeration.The organic compounds are partly converted into micro-organisms,whilst the over-crowd is disposed off as excess sludge.
Because of long standing aeration,the ammonia formed by the activated sludge will be converted into nitrite, and after that into nitrate(this process is called nitrification).
Total nitrogen-reduction is requird by denitrification process, and it takes place under anoxic circumstance and uses the carbon in the wastewater to convert nitrite and nitrate into nitrogen gas, and it is released into the atmosphere.

2-System Information
Here is the information needed for waste water treatment plant design.
2.1 design parameters
The design of the system quoted here is based information obtained from client and experience.The client should inform us, if he feel that the design parameters, listed below, do not reflect actual situation:

*Type of wastewater
Wastewater from oil industry
* Quantity of wastewater
Plant design capacity,
Max=30 m3/day

*physical-Chemical constituents of wastewater

*For Sedimentation/Flotation System
-Water temperature:30 c
-PH:8.4
-COD:1400 mg/l
-BOD: 720mg/l
-SS:714 mg/l
-Nitrogen Total: 21.6 mg/l
-Oil& grease:-- mg/l

Above mentioned figures for COD,BOD,TKN,and SS are estimated related to our experience with similar processing plants.

2.2.System Design

Calculations of elalameah for oils


Background...
Abbreviations...
Q (m3 / h)
Flow - rate
Qo (m3 / h)
Influent flow - rate
Qe (m3 / h)
Effluent flow - rate
Qr (m3 / h)
Recycled sludge flow - rate
Qw (m3 / h)
Wasted sludge flow - rate
BOD (mg / L)
Biochemical oxygen demand
BODo (mg / L)
Influent biochemical oxygen demand
Va (m3)
Aeration tank volume
SS (mg / L)
Suspended solids (SS)
SSe (mg / L)
Effluent SS
SSr , w (mg / L)
Recycled and wasted sludge SS
A (m3 / h)
Air flow - rate
MLSS (mg / L)
Mixed liquor suspended solids
t (h)
Hydraulic retention time
OL (kg BOD / m3 . day)
Organic loading
F / M (kg BOD / kg MLSS . day)
Food to microorganism ratio
r
Recycle ratio
SA (day)
Sludge age
ASR (m3 / kg BOD)
Air supply rate
E (%)
BOD removal efficiency
Equations...
(1) Normally ;Qw << Qo and Qe = Qo = Q(2) Aeration time or hydralic retention time = Volume / Flow - ratet = Va / QTypical aeration times : Conventional activated sludge process = 4 - 8 hrs ; Contact - Stabilization = 0.5 - 1.0 hrs (contact tank) ; Extended aeration = 24 hrs (aeration times are used to determine volume of aeration tanks)(3) BOD or organic loading = Mass of BOD applied per day / Tank volumeOL = ( Qo )( BODo ) / VaTypical OL values : Conventional activated sludge process = 0.3 - 0.6 kg BOD / m3 . day ; Contact - Stabilization = 1.0 - 1.2 kg BOD / m3 . day ; Extended aeration = 0.16 - 0.4 kg BOD / m3 . day (this is an alternative method for determining volume)(4) Food to microorganism ratio = Mass of BOD applied per day / Mass of suspended solids in aeration tankF / M = ( Qo )( BODo ) / ( Va ) ( MLSS )Typical F / M values : Conventional activated sludge process = 0.2 - 0.4 kg BOD / kg MLSS . day ; Contact - Stabilization = 0.2 - 0.6 kg BOD / kg MLSS . day ; Extended aeration = 0.05 - 0.15 kg BOD / kg MLSS . day (used to determine the needed concentration of MLSS)(5) Recycle ratio = Recycle flow - rate / Influent flow - rater = Qr / QoTypical recycle ratio values : Conventional activated sludge process = 0.25 - 0.50 ; Contact - Stabilization = 0.25 - 1.0 ; Extended aeration = 0.75 - 1.5 (used to control concentration of MLSS)(6) Mean cell residence time or sludge age = Mass of suspended solids in aeration tank / Mass of solids leaving the systemSA = [ ( Va )( MLSS ) ] / [ ( Qe ) ( SSe ) + ( Qw ) ( SSw ) ]Typical sludge age values : Conventional activated sludge process = 4 - 15 day ; Extended aeration = 20 - 30 day (an estimate of the length of time that average suspended solids or biomass stays in the process, used to select volume and MLSS and solids wasting rate)(7) Air supply rate = Volume of air per unit time / Influent flowASR = A / QoorASR = A / [ ( Qo ) ( BODo ) ](8) BOD removal efficiency ;E ( % ) = [ ( BODo - BODe ) / BODo ] 100 (BOD removal efficiency is a function of MLSS, sludge age, or F/M ; there is usually an optimum range of MLSS, sludge age, and F/M)
Basic Calculations...

Treatment Efficiency
90%
Design of the Activated Sludge Process
Influent wastewater flow = 30m^3/day,
Volume of aeration tanks = 72 m3,
Influent total solids (TS) = mg / L
Influent suspended solids (SS) = 714 mg / L,
Influent BOD = 720 mg / L,
Effluent TS = mg / L,
Effluent SS = mg / L,
Effluent BOD =1oo mg / L,
MLSS = 2,500 mg/L,
Recirculated sludge flow = mgd,
Waste sludge flow = mgd,
SS in waste sludge = mg/L.
Calculation:

(a) Aeration period = Volume / Flow without recirculation =
26 m3 / 30 m3 / day = .8 day = 20 hr
(b) BOD organic loading
Ol=Q*BOD/V
=30 m3/day*720)/ 26
=8,3 mgBOD/m3
=.08 Kg BOD /m3
(c) F to M = Quantity of influent BOD / Mass of MLSS in Aeration BasinF to M = [(720 mg / L) (8.34 mg / m3 / mg / L) (30 m3/day)] / [(2,500 mg / L) (8.34 mg / m3 / mg /L) (30m3/day)] = 0.28 mg BOD / mg MLSS
(f) BOD removal efficiency = (720 - 100) / 720 = 87 %
(g) Sludge age = [(MLSS) (Volume)] / [(SSe) (Qe) + (SSw) (Qw)]Sludge age = [(2,500 mg / L) (30 m3/day) (8.34 mg / m3/ mg / L)] / [(22 mg / L) (30 m3 / day) (8.34 mg / m3 / mg / L) + (9,800 mg / L) (0.054 mil.gal / day) (8.34 lb / mil.gal / mg / L)] = 8.0 day
(h) calculating aeration tank volume
F/M=Q*BOD5/v*MLVSS
0.28=30*720/V*3000
V=26 m3
With surface area= 3*4
Depth= 2.5 m and the excess is not effective area

(j) calculating retention time at aeration tank
HRT=V/Q*24
=26/30*24
=20 hr
=,8 day

(h) calculating quantity of oxygen needed
=1.5*mg BOD5
=1.5*720
=1080 mg O/l


Calculations of sedimentation tank

Considering detention time =4 hrs
Volume=12 m3
Surface area=12/2=6 m2
= pi D2

Diameter^2=area/3,14
=6/3.14
=1.9 m
So let the diameter of settling tank be 2 meter

2.3 plant performance
Based on the given design data and our experience with similar wastewater we expected to meet your effluent requirements.
PH:6.5-7.5
Bod: 100mg/l
COD: mg/l
SS: mg/l
TN:<50mg/l
oil&grease:--mg/l
sulfite:<1mg/l
phosphorus:<5mg/l

the treated water compliance with environmental laws for discharged to sewer.
By ahmed elsayed elshwadfy
Environment and safety consulatant

2 comments:

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