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by: Lesley Lynch V


Marketplace > Idaho State University > Environmental > ENVE 410 > INTRO TO ENVIRONMENTAL ENGR
Lesley Lynch V
GPA 3.67

C. Sato

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C. Sato
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This 79 page Class Notes was uploaded by Lesley Lynch V on Monday October 12, 2015. The Class Notes belongs to ENVE 410 at Idaho State University taught by C. Sato in Fall. Since its upload, it has received 35 views. For similar materials see /class/222173/enve-410-idaho-state-university in Environmental at Idaho State University.




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Date Created: 10/12/15
17Air Pollution7F09doc AIR POLLUTION 3rd DC 459 4th 548 INTRODUCTION a Human and most other animal can survive 1 considerable number of days wo food 2 few days wo water 3 few minutes wo air b An adult human must inhale about 400 ft3 11 m3 of air each day 400 f39t37439x 74 x 7439 1m33531 its Air Dry air contains by volume Oxygen 209 23 by wt Nitrogen 781 Carbon dioxide 003 Argon 003 Other gases Trace Ne He CH4 etc Standard Conditions Standard Temperature and Pressure STP 0 C 273 K 1 atm 101325 kPa Units Concentration ugm3 masstovolume ratio ppm volumetovolume ratio Particle size um 6 3 10 m 1 ppm 1 m3 Trivia What is the concentration in ppm of pure He gas in a 1 ma balloon Answer 10396 m3 He 1 ppm 1 m3 1 m3 He 106 ppm 1m3 17Air Pollution7F09doc Ideal gas 1 mole of an ideal gas occupies 22414 L at STP For Ideal gas PV n RT where R universal gas constant P absolute pressure T absolute temperature V volume of gas n mole number P2 V 2 n R T2 condition 2 P1 V1 n R T1 condition 1 T2 P1 V2 1 T1 P2 Example 61 3rd DC 463 Example 71 4th DC 550 A onecubicmeter sample of air was found to contain 80 ugm3 of 802 under standard condition The temperature and pressure were 25 C and 103193 kPa when the air sample was taken What was the 802 concentration in ppm Solution MW of 802 3206 2160 6406 gmol T1 0 C 273 K P1 101325 kPa standard condition T2 25 C 273 C 298 K P2 103193 kPa at T298 K Under the standard condition condition 1 1 mol of 802 gas occupies 22414 L At condition 2 1 mol of 802 gas occupies T2 P1 298 K 101325 kPa V2 V1 22414 L T1 P2 273 K 103193 kPa 2400 L per mole at 25 C Thus 17Air Pollution7F09doc 80 pg 80 pg g mol 2400 L m3 106 ppm 13quot 1661138163 quot33quot quot13965 7153quot 002997 ppm 003 ppm of 802 at 25 C or At STP condition 1 1 mol of 802 gas occupies 22414 L 80 pg 80 pg g mol 22414 L m3 106 ppm quoti 39 g quot516513 quotquot ESquot quot165139 002799 ppm at STP V1 At T2 298 K P2 103193 kPa T2 P1 298 K 101325 kPa V2 V1 002799 ppm T1 P2 273 K 103193 kPa 003 ppm of 802 Note at STP ppm 3 106 pg MW mole 103 L m3 m3 thus 1 x ug ppm 0022414 MW m3 or IJQ 446150MWx ppm m3 Since T2 P1 V2 V1 V1 and V2 are in ppm T1 P2 1 x pg T2 101325 kPa v2 0022414 1 MW m3 273 K p2 ppm at273 K at 101325 kPa or x pg MW 273 K pg 3 0 022414 T2 101325 kPa 17Air Pollution7F09doc Example Calculate the concentration of 802 in ppm if x 80 ug m3 of 802 MW of 802 64 gmole T2 25 C 273 25 lC 298 K p2 103193 kPa 1 x ug T2 101325 kPa V2 0022414 MW m3 273 K p2 1 80 pg 298 K 101325 kPa 0022414 64 m 273 K 103193 kPa v2 003 ppm Example King p 36 Sulfur dioxide has been measured at 35 ppm in a stack gas sample Determine the concentration of 802 in mgma Solution Given V2 35 ppm MW 64 T2 273 25 298 K p2 p1101325 kPa MW 273 K p2 0022414 T2 101325 kPa x pg 64 273 K 101325 kPa 0022414 298K 101325kPa 915535 pg m3 916 mgm3 Or T2 P1 v2 v1 Tl P2 1 T2 101325 kPa 0022414 ppm MW m 273 K P2 If x 80 ugma T2 298 K P2 103193 kPa 1 80 pg 128K 101325 kPa 0022414 ppm 64 m 273 K 103193 kPa 00300 ppm 17Air Pollution7F09doc d Type of air pollutants 1 Natural 2 Manmade Natural Pollutants 1 pollens spores expelled by flowers and fungi 2 natural dusts picked up by the wind These substances especially ragweed pollen can be extremely distressing to those who are allergic Manmade Pollutants results from man s activities 1 Industrial origin a combustion for industrial heatpower plant and various processes b incineration 2 Personal origin a individual home heating furnaces fireplaces b motor vehicles Air pollutants of concern Examples Particulate gases CO 002 RH Sulfur oxide SOx Nitrogen oxide NOx Ozonedepleting substances ODSs vapors odors Ozonedepleting substances ODSs Chlorofluorocarbons CFCs Carbon tetrachloride halons hydrochloro uorocarbons HCFCs methyl bromide methyl chloroform Chlorofluorocarbons CFCs CF2CI2 CFCla used as propellants in aerosol cans foamblowing agents solvents coolants for refrigerators air conditioners be phased out for nearly all uses in industrial countries by 1996 in developing countries by 2010 Brown et al 1997 halons fluorocarbons that contain bromine atoms used in some fire extinguishers stratospheric concentrations of chlorine nitrogen and bromine act as catalysts speeding up the O3 removal process CCIZ F2 hv gt Cl CCI F2 17Air Pollution7F09doc Primary Concerns a Health damages cancer cardiovascular emphysema b Property damages cracking deterioration of materials eg concrete stone metals c Agricultural vegetation damages agricultural crops yields deforestation d Ecological damages aquatic life reduced species diversity e Weather modification greenhouse effects Global warming flooding drought f Water quality acidity drywet deposition of heavy metals Hg Pb Cd Sources of air pollutants and Solutions a Sources of air pollutants combustion sources 1 stationary factories power plants houses 2 mobile autos b Solutions 1 switch to clean burning fuels 2 treatment 3 conservation modification of life style less driving car pool 17Air Pollution7F09doc Air Pollution Standards Clean Air Act CAA of 1970 required the US EPA to investigate and describe the environmental effects of any air pollutant emitted by stationary or mobile sources that may adversely affect human health or the environment The EPA established the National Ambient Air Quality Standards NAAQS The Primary standards was established to protect human health with an adequate margin of safetyquot The secondary standards were intended to prevent environmental and property damage Note The Ambient air the outdoor air that normally surrounds us Criteria pollutants were developed on healthbased criteria by the EPA CO Lead N02 Ozone Particulate matter PM10 PM25 802 listed in Table 71 4th DC 552 PM10 particulate matter standard applies to particles with an aerodynamic diameter 5 10 um PM25 particulate matter standard applies to particles with an aerodynamic diameter 5 25 pm Air Quality Control Regions AQRs States are divided into Air Quality Control Regions AQRs An attainment area An AQR that has air quality equal to or better than the primary standard A nonattainment area An AQR that does not meet the primary standard 17Air Pollution7F09doc METEOROLOGY and CLIMATOLOGY Macrometerology earth atmosphere Stratosphere Troposphere Ozone layer provides a barrier to UV radiation 17Air Pollution7F09doc Continental and Oceanic Influences land wind Upwelling Land See Breezes Daytime Rising air Sea breeze Air Ocean Moon Night time in Warm air rises Rapid cooling of air Land breeze E Air Ocean 17Air Pollution7F09doc Frontal Movement 1 Cold front Cold front Cold air Q Col air Warm air more dense less dense Ground surface Cold air moving in Cold front moves more rapidly than warm front Possibility of precipitation at localized area 2 Warm Front Warm front Warm air Warm air Cool air Warm air moving in Air movement is slower because it moves by warming the cold air underneath Possibility of precipitation at wide area Cooling Wet air Warming Dry air 17Air Pollution7F09doc Valley Breeze Sun Daytime 3 Warm air rising i gt Cool air E a Moon Night time 9 Rapidly cooling Warm air rsisng Cool air Cool air l 17Air Pollution7F09doc Stability of Atmosphere the tendency of the atmosphere to resist or enhance vertical motion related to both wind speed and the change of air temperature with height Three 3 primary stability categories of atmosphere Atmosphere is said to be 1 Unstable if the thermal structure enhances mechanical turbulence wind speeds 2 Neutral if the thermal structure neither enhances nor resists mechanical turbulence 3 Stable if the thermal structure inhibits mechanical turbulence Neutral Atmosphere neutral stability Lapse rate the change of air temperature with height A parcel of dry air expands adiabaticay as it is raised through the atmosphere parcel of gt results in the rate of temperature decrease dry air The rate of temperature decrease is called the dry adiabatic lapse rate DALRI gamma I 100 C 100 m with no interchanging of heat with surrounding air Adiabatically without the addition or loss of heat 17Air Pollution7F09doc 3 Isothermal 5 Psudo 4 Inversion 1 DALR Negative 2 Supe Elevation m Stable Unstable Temperature deg C See the Elevation vs Temperature Diagram 1 Dry air adiabatic lapse rate 100 C100 m I atmospheric stability is neutral 2 Super adiabatic unstable and desirable in terms of air pollution summer condition 3 lsothernmal stable atmospheric condition because the lapse rate is less negative than I 4 Negative adiabatic Inversion very stable atmospheric condition undesirable situation in terms of air pollution 5 Pseudo adiabatic Subadiabatic stable atmospheric condition because the lapse rate is less negative than I when air is saturated with water Unstable Atmosphere Superadiabatic the temperature of the atmosphere falls at a rate greater than I39 good from pollution stand point Stable Atmosphere Subadiabatic the lapse rate is less negative than I39 bad from pollution stand point 17Air Pollution7F09doc Example 63 3rd DC 497 Example 73 4th DC 585 Given the following temperature and elevation data determine the stability of the atmosphere Elevation m Temperature C 20 1435 3240 1113 Solution Elevation m Temperature C Z1 20 T1 1435 22 3240 T2 1113 1 Determine the existing lapse rate AT T2 T1 1113 1435 322 quotC Dry air adiabatic lapse rate P Thus the atmospheric stability is neutral 17Air Pollution7F09doc Examples Ts smoke temperature Ta ambient air temperature 1 Isothermal Ts gt Ta and Isothermal Polluted gases could accumulate in this zone Elevation Temperature N Super adiabaticDry air adiabatic Ts lt Ta Super adiabacic dry adiabatic Good situation smoke keep rising Elevation Temperature 17Air Pollution7F09doc 3 Variations from Ideal Temperature Gradients a Surface inversion b Inversion aloft c Super adiabatic condition near surface a Surface inversion Advection inversion Warm air passes over a cold surface Surface invasion Surface inverstion Satable air polluted air lt could accumulate x in this zone Elevation Elevation Temperature Tenperature Drainage inversion Valley Surface inversion When air in contact with the ground along the slopes of a valley air cools at night and drains down the slopes into the valley to create a pool of cool air overlain by warm air aloft Elevation Temperature 17Air Pollution7F09doc a Inversion aloft deep layer of cold air passes over a warmer surface Inversion aloft eg Los Angeles smog Elevation Tenperature c Super adiabatic condition near surface Super adiabatic condition near surface Dry adiavatic Elevation Super adiavatic Tenperature 2 Plume Types It has been found that the smoke trail or plume from atall stack located on flat terrain exhibits a characteristic shape that is dependent on the stability ofthe atmosphere 17Air Pollution7F09doc Pollutants examples Nitrogen oxides NOx N20 NO N02 Sources 0 Natural source Bacterial action in the soil releases nitrous oxide N20 to the atmosphere Nitrous oxide reacts with atomic oxygen in the upper troposphere and stratosphere to form nitric oxide NO N20 O 9 2N0 Note atomic oxygen results from the dissociation of ozone The nitric oxide NO further reacts with ozone to form nitrogen dioxide N02 NO 03 9 N02 02 o Combustion processes 96 of the anthropogenic sources The NOx emissions in the US 40 45 from transportation 30 35 from power plants 20 o from industrial sources Ultimately N02 is converted to either N0239 or N0339 in particulate forms Dissolution of nitrate in a water droplet results in the formation of nitric acid HNO3 Acid rainquot Sulfur oxides SOx 02 03 304239 Sources Natural volcanoes oceans power plants other industries 0 Natural source Bacterial action releases hydrogen sulfide H28 to the atmosphere Reaction of hydrogen sulfide with ozone yields sulfur dioxide H28 03 9 H2O 802 0 The combustion of fossil fuels containing sulfur yields sulfur dioxide in direct proportion to the sulfur content of the fuel SO29SO2 The ultimate fate of most of the 802 in the atmosphere is conversion to sulfate salts which are removed by sedimentation or by washout with precipitation Catalytic oxidation 2802 02 9 2H2SO4 Catalyst 17Air Pollution7F09doc Most effective if Fe Mn or NHS is present in water droplets Photochemical oxidation 802 hv 9 802 802 02 9 803 O SOS H20 9 H2804 This reaction in large part account for acid rain Acid rain precipitation with a pH value less than 56 Normal precipitation has a pH of 56 due to the carbonate buffer system Emissions Example 62 339d DC 479 Example 72 4m DC 565 An Illinois coal is burned at a rate of 10 kgsec If the analysis of the coal reveals a sulfur content of 30 what is the annual rate of emission of 802 Assume that 5 of the sulfur in the fuel ends up in the ash Solution 8 Mass Balance on S Accum Inputs Outputs i Rxs Inputs Outputs Sin Sash Sin Sash Satm Satrn Sin 39 Sash 10 kg Coal 003 kg S 003 kg S Sin sec 1 kg Coal sec For 1 yr 003 kg 8 86400 sec 365 days 946 X 105 kg 8 Sin sec day yr yr 946x105kgS 473x104kgS Sash 005 W W 17Air Pollution7F09doc Sam sin sash 946 x 105 kgyr 473 x 104 kg Syr 899 x105 kg 8 yr Amount of 802 formed 8 02 gt 802 MW 32 64 g 802 1 mole 64 gmole 2 g 802 2 kg 802 g S 1 mole 32 gmole 1 g S 1 kg S The annual rate of 802 emission 899x105 kgS 2kg 802 18x106kg SOgyl yr 1kgS Point Source Gaussian Dispersion Model 339d DC 5024m DC 590 2 2 CXayaOaH L exp 1 i ex 1 i 7139039y 039z u 2 0y 2 oZ where C x y 0 H or X x y 0 H downwind concentration at ground level gm3 Q or E emission rate of pollutant gs u wind speed ms H effective stack height m X y and z distances in X y and 2 directions respectively m oy c7Z or sy sZ plume standard deviation in y and 2 directions respectively m 2 2 Cltxy0Hgt L exp 1 y 2 112 7ZO39yO39z u 2 o y oz 17Air Pollution7F09doc Assumptions 1 Atmospheric stability is uniform Atmospheric stability is uniform throughout the layer into which the contaminated gas stream is discharged 2 Diffusion by the Gaussian or normal distribution Turbulent diffusion is a random activity the diffusion of the contaminated gas stream in both the horizontal and vertical direction can be described by the Gaussian or normal distribution equation 3 Diffusion 1u The degree of diffusion of the effluent plume is inversely proportional to the wind speed u 4 Mirror Image Reflection The pollutant material that reaches ground level is totally reflected back into the atmosphere like a beam of light striking a mirror at an angle 5 Effective Stack Height H h Ah The contaminated gas stream is released into the atmosphere at a distance above ground level that is equal to the physical stack height plus the plume rise Effective Stack Height H see Fig 42 H h Ah where h physical stack height m Ah plume rise m Holland s formula Ah Vs d 15 268x10 2pEd u T S 621 where v5 stack velocity ms d stack diameter m u wind speed ms p atmospheric pressure kPa TS stack temperature K Ta air temperature K Note 1 atm 14696 b inz 1 lb inz 68948 Pa 1 atm 14696 bin268948 Pa1 ib inz 101 326 Pa 101326 kPa 17Air Pollution7F09doc 6 The values of 0 and c7Z depend upon the turbulent structure or stability of the temperature Horizontal dispersion coefficient on i Determine from Fig 619 3rd DC 504 or Fig 722 4th DC 592 or ii 039y sy a x0894 622 3rd DC 507 722 4th DC 594 where X downwind distance km a constant obtained from Table 67 3rd DC 506 Table 79 4th DC 595 Vertical dispersion coefficient c7Z i Determined from Fig 620 3rd DC 505 or Fig 723 4th DC 593 or ii oz sZ c xd f 623 3rd DC 507 723 4th DC 594 where X downwind distance km c d and f constants obtained from Table 67 3rd DC 506 Table 79 4th DC 595 Figures 46 and 47 handout provide graphical relationships between the downwind distance X and TY and oz respectively Stability Categories See Table 66 3rd DC 506 Table 78 4th DC 594 or Table 41 handout A very unstable atmospheric conditions B unstable C slightly unstable to neutral D slightly stable E stable F very stable 17Air Pollution7F09doc Example 64 3rd DC 507 Example 74 4th DC 595 It has been estimated that the emission of 802 from a coalfired power plant is 16562 gs At 3 km downwind on an overcast summer afternoon what is the centerline concentration of 802 if the wind speed is 45 ms Note A centerline implies y 0 Stack parameters Height 1200 m Stack Diameter 120 m Stack velocity Exit velocity v5 100 ms Temperature 315 C Atmospheric Conditions Pressure 950 kPa 1 atmosphere 101325 kPa Temperature 25 C Solution 1 Effective stack height H Holland s formula Ah Vsd15268 x 1011 TST T MM 11 s where vS stack velocity 100 ms d stack diameter 12 m u wind speed 45 ms p atmospheric pressure 950 kPa TS stack temperature 273 315 C 588 K Ta air temperature 273 25 C 298 K Ahml5268 x10z950wj12 80 m 45 588 HhAh1200m80m128m 2 Atmospheric Stability Class Overcast condition gt select D 0894 3 0 sy a X 622 3rd DC 507 722 4 h DC 594 c7Z sZ c Xd f 623 3rd DC 507 723 4 h DC 594 17Air Pollution7F09doc From Table 67 3rd DC 506 Table 79 4th DC 595 under the given conditions ie X gt 1 km Stability class D a 68 c 445 d 0516 and f 130 o sy 68 300894 181 oz s2 445 300516 13 65 lt From Figs 619 and 620 3rd DC or Figs 722 and 723 4th DC 0 sy 190 m c7Z sZ 65 m 4 2 2 Cxy0HL explij 1 exp1 7139039y O39Z u 2 oy 2 oz CxyoH M 7139 190 m65 m45 ms mi rwiirwi rmi 136x 10393gm3 of 802 136x103 9 m3 of 802 Note 1 g 106 pg 5 Compare with National Ambient Air Quality Standards NAAQS in Table 71 4th DC 552 Primary standard for 802 annual arithmetric mean 80 pigm3 Secondary standard for 802 maximum 3hr concentration 1300 pigm3 17Air Pollution7F09doc Parameter Input data T rm 1 Term 2 Term 3 emigrate ugs 16562E09 vdistance Nistance m m C ugm3 C ugm3 700 0 949082 000 0 1 154 600 0 949082 001 0 14 500 0 9490 82 003 0 1 42 80 400 0 9490 82 011 0 14 148 87 300 0 9490 82 029 0 14 392 53 200 0 9490 82 057 0 14 784 56 100 0 9490 82 087 014 1188 73 50 0 9490 82 097 014 1318 85 20 0 9490 82 099 0 14 1357 78 0 0 9490 82 100 0 14 1365 32 20 0 9490 82 099 0 14 1357 78 50 0 9490 82 097 014 1318 85 100 0 9490 82 087 014 1188 73 200 0 9490 82 057 0 14 784 56 300 0 9490 82 029 0 14 392 53 400 0 9490 82 011 014 148 87 500 0 949082 003 0 1 42 80 600 0 949082 001 014 700 0 9490 82 000 0 14 1 54 9 with reflection 1600 1400 1200 1000 A m E 800 c 3 V o 600 400 200 0 1000 500 0 500 1000 Cross Wid y Distance m 7 r7rAttachGr0wthyF09 doc Biological Treatment Types of Secondary Treatment Systems 1 Suspended Growth Systems Reactors eg Activated Sludge processes Conventional Completely mixed 2 Attached Growth Systems Reactors eg Trickling Filters Rotating Biological Contactor RBC Submerged Rotating Biological Contactor SBC Attached Growth System Principles Schematic diagram showing principles of the biological process in a trickling filter Nutrient 7 Organic carbo 3 I c 0 E 9 I m 0 Air quot 1 O2 4 02 H quot W astewater MIL g 39 l 939 39 Endproducts 002 7 Elt gt Anaerobic lAerobic 01 02 mm Biofilm Biological process on medium surface in a Trickling Filter 7 7 AttachGrowthFO9doc Biofilm images x 1000 dead red and live green bacterial cells grown on an MFC electrode Left heavy growth with clustered live cells on the anode membrane Right sparse attachment to a control membrane an anodematerial membrane unattached to the electrical circuit Examples Trickling filters Rotating Biological Contactor RBC Submerged Rotating Biological Contactor SBC 7r7rAtta ththiw9 doc Trickling Fillers Major components Three 1 Filter media 2 Rotary distributor 3 Under drain system Rotary distributor Distributor arm 3 it z 3 Filler media Ellluenl channel Feed pipe Under drain system Filter Media provide surface for bacterial growth provide voids for pa sage of wastewater and air Comm n media crushed rock eld stone plastic Rotary distributor provides uniform hydraulic load and organic load Under drain system carries away the ef uent passage of air Filter Media WEFTEC exmbtt 7 r7rAttachGrowthiF09 doc Characteristics 1 Facultative system Aerobic to anaerobic 2 Biofilm bacteria protozoans fungi rotifers algae sludge worms filterfly larvae 3 Thin aerobic film 0102 mm 4 Biological variation with depth of the filter Algae uptake at the upper surface Nitrification near the bottom 5 Sloughing breaking off of biofilm Advantages and Disadvantages of TF Advantages Low energy input Accepts qualitative and quantitative shock loads Accepts toxic load to some extent Good sludge settling at secondary clarifier Disadvantages Poor performance in winter Fair performance in summer Land requirement 7 rAttacthwthiF09 doc National Research Council NRC formula SI Unit 3rd DC 379 4 h DC 486 For a singlestage filter or the first stage of a twostage filter 1 05 1 412 V F E1 where E BOD5 removal efficiency forthe first stage filter at 20 C including recirculation and sedimentation fraction Q wastewater flow rate m3s Cin influent BOD5 mgL V volume of filter media m3 F recirculation factor 1R 101R2 F where R recirculation ratio QrQ Qr recirculation flow rate m3s Q wastewater flow rate m3s For the second stage filter 1 05 1 412 ch 1 E1 VF where E2 BOD5 removal efficiency for the second stage filter at 20 C fraction E1 BOD5 removed in first stage fraction Ce effluent BOD5 from first stage mgL Effect of Temperature on the efficiency ET 2 E206T 20 where E BOD removal efficiency at temperature T C E20 BOD removal efficiency at 20 C 9 1035 7 rAttacthwthiF09 doc QCe F 1 E241121212T 1 05 1 412 V F Rearranging for V 05 E1 1 VF E1 05 1 E1 VF 412E1 V QC 2 1 E1 412E1 E2 4121 C 05 1 er 1 E1 VF 7 rAttacthwthiF09 doc 412EZQC 051 E 7 7 2 2 QC 2 1E 1 E1 VF 412152 2 1E1 2 QCE VF 1 E2 412 QCZ F 1 EZ V 7 rAttacthwthiF09 doc Example 55 3rd DC 380 Example 612 4th DC 487 Using the NRC equations determine the BOD5 of the effluent from a singlestage lowerrate trickling filter that has a filter volume of 1443 m3 a hydraulic loading of 1900 m3d and a recirculation factor of 278 The influent BOD5 is 150 mgL Solution To use the NRC equation the hydraulic loading must first be converted to the correct units 1900 MI day J 0022 m3 day 86400 sec sec E1 1 05 1 05 08943 1 412 LCM 1 41 z O Ozms 0 VF 1443278 where Q 0022 m3s Cin 150 V 1443 m F 278 The concentration of BOD5 in the effluent is Ce 1 08943150 mgL 158 mgL Check QCi 0022150 1442 m3 1 E 2 1 0 8943 2 F 1 J 278 412E1 41208943 V 7 rAttacthwthiF09 doc National Research Council NRC formula English Unit For a singlestage filter or the first stage of a twostage filter E1 100 W 05 1 00561 VF where E BOD5 removal efficiency for the first stage filter at 20 C W BOD5 load applied lbd V volume of filter media 1000 ft3 F recirculation factor 1 R 1 01122 where R recirculation ratio QrQ Qr recirculation flow rate MGD Q wastewater flow rate MGD For the second stage filter 100 05 1 00561 W2 1 E1 W where E2 BOD5 removal efficiency for the second stage filter at 20 C E1 fraction of BOD5 removal removed in first stage fraction W2 BOD5 loading applied to the second stage filter lbday W V volumetric BOD loading lb1000 ft3day Effect of Temperature on the Efficiency ET 192090 20 where ET BOD removal efficiency at temperature T C E20 BOD removal efficiency at 20DC e 1035 7 rAttacthwthiF09 doc Exam pl e Given Design Criteria Design BOD loa 15lb1000 f39t3 day Hydraulic loading 2 4 Mgalacre day Filter Depth 5 7 ft Over flow rate for primary sedimentation tank 600 gpdf39t2 Overflow rate for final clarifier 800 gpdft2 Given Data Dave 180 mgL Or 0 MGD T 17 C 15 GD BO BOD removal efficiency In the primary sedimentation tank 35 Design the single stage TF and compute the overall WWTP efficiency for BOD removal Primary sedimentation Secondary sedimentation e Trickling filler gt O 15 MGD BOD 180 mgL Solution 1 BOD loading to the primary sedimentation tank 180 mgL 15 MGD 834 2252 lbd 2 BOD loading to TF 1 0352252bd1464lbd 1464 lbd 3 Volume of TF media required v 15 Ib1000 ft3 d 97600 r13 4 Surface area of TF with the TF depth h 5 ft A 5 ft AVh97000f39t35ft19400ft2 A 19400 ft2 1 acre43560 ftz 045 acre 15 MGD 34 Mgal acred 045 acre 5 Hydraulic loading QA 7 rAttacthwthiF09 doc 439 Surface area of TF with the TF depth h 6 ft A Vh 97000 ftaG ft 16200 it2 A 16200 ft2 1 acre43560 ftz 037 acre 539 Hydraulic loading QA 41 Mgal awed 037 acre 6 Depth of TF Let s build two 2 TF units with the diameter 00 ft v 139rr2h314 50 ft2h97600 fta248 800 fta Solve forthe depth h h 62 ft 7 BOD removal efficiency E1 100 100 822 05 05 1 00561 1 00561Ej VF 1 Note Design BOD load W V 15lb1000fta day 1R 10 2 72 1 01R 1 0 8 Effect of temperature on the efficiency ET E206T20 82210351720 74 9 Overall plant efficiency ET including primary and secondary treatment at 20 C E 100 100 1 Prim Sed eff 1 TF eff 100 100 1 035 1 082 883 7 AttachGrowthFO9doc Rotating Biological Contactors RBCs 4th DC 492 The RBC process consists of a series of closely spaced discs 3 35 m in diameter which are mounted on a horizontal shaft and rotated About onehalf of their surface area is immersed in wastewater Fig 633 4th DC The discs are typically constructed of light weight plastic The speed of rotation of the disk is adjustable The surface of the discs is covered with a 1 to 3mm layer of biological slime The excess growth of microbes is sheared from the discs as they move through the reservonr The purpose of the discs Provide media for the buildup of attached microbial growth Bring the growth into contact with the wastewater food Aerate the wastewater and suspended bacteria Rotating Biological Contactor American Falls Idaho 2003 12 llrllrSO ningiF09doc Water Softening Precipitation Softening 3rd DC 178 4 h DC 235 1 Introduction Hardness Multivalent metal ions which will form precipitates with soaps eg Caz soap lt gt Casoap2 s Complexation reaction a Caused by the ions of Ca2 and Mg2 Hardness in water is caused by the ions of Ca2 and M92 b Other hardness constituents Iron Fe manganese Mn strontium Sr aluminum Iron lZe manganese Mn strontium Sr aluminum Al also produce hardness 0 Sources Largely the result of geological formations of the water source Precipitation 4 W M Top organic soil microbial activity W W CHZO Oz 9 C02 H20 organics Subsoil C02 H20 9 H2C03 Limestone formation weathering CaCO3s H2C03 9 CaHCO32 MgC03s H2C03 9 MgHC032 Types of Hardness with respect to cations metallic ion with respect to anions nonmetallic ion 1 With respect to cations metallic ion Ca2 Mgz a Calcium Hardness CaH0032 CaSO4 CaClz b Magnesium Hardness MgH0032 MgSO4 MgClz Total Hardness TH Calcium Hardness Magnesium Hardness llrllrSo ningiF09doc 2 With respect to anions nonmetallic ion HC0339 804239 Cl39 a Carbonate Hardness CH b Noncarbonate hardness NCH Carbonate Hardness Temporary Hardness heating the water removes it Calcium bicarbonate CaHC032 Magnesium bicarbonate MgHC032 Carbonate hardness alkalinity when alkalinity lt TH Carbonate hardness TH when alkalinity 3 TH where TH total hardness Alkalinity measured as the amount of acid required to titrate to PH 43 eg H0 003239 HCOJ39 Noncarbonate hardness Permanent hardness not removed when water is heated will not precipitate when the water is boiled Calcium sulfates CaSO4 Magnesium sulfates MgSO4 Calcium chlorides CaClz Magnesium chlorides MgClz Total Hardness TH Carbonate Hardness CH Noncarbonate hardness NCH d Expressed in mgL as CaCO3 The sum of calcium and magnesium concentrations expressed in mgL as CaCOa eq wt of CaCO3 Hardness mgL as CaCO3 mgL of M2 eq wt of M2 Hardness mgL as CaCog meqL Caz meqL Mgz ew wt of CaCO3 eq wt of CaCO3 50 mgmeq Note Caco3 9 Ca cos cos 2H e HZCO3 22 llrllrSO ningiF09doc Example Given Ca2 70 mgL and Mgz 97 mgL Determine calcium hardness magnesium hardness and total hardness as CaCO3 Solutions EW equivalent weight EW of Ca2 20 mgmeq EW of Mg2122 mgmeq EW of CaCO3 50 mgmeq 50 mgmeq CaCO3 Calcium hardness 70 mgL Ca2 20 mgmeq Caz 175 mgL hardness as CaCO3 50 mgmeq CaCO3 Magnesium hardness 97 mgL Mgz 122 mgmeq Mgz 40 mgL hardness as CaCO3 Total hardness 175 40 215 mgL as CaCO3 Example Given Caz 35 meqL and Mg2 0795 meqL Determine total hardness as CaCO3 Solution Hardness 35 meqL 0795 meqL 50 mgmeq CaCO3 215 mgL as CaCO3 Hard Water Classification Table 313 DC 179 Table 414 4th DC 236 Hardness Range Description mg L as CaCO3 0 75 Soft 75 100 Moderately hard 100 300 Hard gt300 Very Hard 1171 lrSo ningiF09doc Goal of water treatment is 75 120 mgL as CaCO3 339d DC 179 o Hardness gt300 mgL as Ca003 is considered excessive for public water supply results in a high soap consumption b scale in heating vessels and pipes o Mg2 in excess of 40 mgL as Ca003 forms scale on heat exchange elements in hot water heaters Other sources a High excessive hardness gt300 mgL as CaCO b Hard 150 300 mgL as CaCO 100 300 mgL as CaCOJ c Moderate hardness 60120 mgL as CaCO 75 150 mgL as CaCO is considered moderately hard d Soft 0 75 mgL as CaCOa e Acceptable 80100 mgL as CaCO acceptable for a public water suppl but magnesium content should not exceed 40 mgL as CaCO 2 Chemistry of Water Softening Limesoda ash processes a The limesoda ash watersoftening process uses Lime CaOH 2 or CaO and Soda ash Nazcog to precipitate hardness as Calcium carbonate Ca003s Magnesium hydroxide MgOH2s b Chemical reactions a 2 is not hardness but it consumes lime and must therefore be considered in calculating the amount required 002 CaOH2 lt gt CaCOssl H20 1 llrllrSO ningiF09doc b Carbonate hardness is precipitated by lime CaHCOs2 Ca OH2 H 20a00351 2HZO 2 MgHCO2 CaOH2 H CaCOHsl MgCO3 2H20 3 M9003 CaOH2 H Mg OHMSI Ca003 s 4 Note 1 mole of lime is needed for each mole of calcium bicarbonate Rxn 2 2 moles of lime are required for each mole of magnesium bicarbonate Rxns 3 and 4 c Noncarbonate hardness requires the addition of soda ash for precipitation MgSO4CaOH2 H MgOH2s CaSO4 5 CaSO4 Na2003 lt gt 08005 NazSO4 MgC2CaOH2 H MgOHms CaCIz 7 CaClz Na2003 H Ca0035 2NaC 8 Note 1 mole of lime CaOH2 and 1 mole of soda ash Na2003 are needed to each mole of M9804 or MgCl 2 1 mole of soda ash Na2003 is needed to each mole of CaSO4 or CaCl 2 Solubility of CaCO 3 s and MgOH 2 s Precipitation softening cannot produce water completely free of hardness because of a Solubility of Ca003s and MgOH2s 06 meqL of CaCog02 meqL of MgOH2 30 mgL CaCog 10 mgL of MgOH 2 as CaCog Total limiting hardness 40 mgL 1171 lrSoftningiF09doc The minimum practical limits of precipitation softening 30 mgL of Ca003 and10 mgL of MgOH2 expressed as Ca003 Goal is 75 120 mgL hardness as 05100 b Limited completion of the chemical reactions by physical considerations eg adequate mixing limited detention time in settling basins Advantages of LimeSoda ash Softening a Hardness is taken out of solution b Lime added is also removed when soda ash is applied Na remains in the finished water noncarbonate hardness requiring soda ash is generally a small portion of the total hardness c TDS is reduced Lime also precipitates the soluble Fe and Mn TDS total dissolved solids may be significantly reduced d Disinfection Excess lime treatment provides disinfection e Aids in coagulation Excess lime treatment provides aids in coagulation for removal of turbidity Schemes of limesoda ash softening three different basic schemes may be used to provide a finished water with the desired hardness a Excess lime treatment b Selective calcium removal c Split treatment 1171 lrSo ningiF09doc Excess Lime Treatment 1 Carbonate hardness associated with Ca2 can be effectively removed to the practical limit of Ca003 solubility 30 mgL by stoichiometric additions of lime CaHCOs2 CaOH2 a 20acogs 2HZO 2 Precipitation of Mgz calls for a surplus of approximately 125 meqL 30 mgL of CaO above stoichiometric requirements 3 The practice of excesslime treatment reduces the total hardness to about 40 mgL as CaCO3 Le 30 mgL of Ca003 as CaCO3 10 mgL of MgOH2 as CaCO3 4 After excesslime treatment the water is scale forming and must be neutralized to remove caustic alkalinity OH39 Recarbonation and soda ash are regularly used to stabilize the water 5 COZ neutralizes excess lime as follows CaOH2 002 9 CaCOss H20 excess lime this reaction precipitates calcium hardness and reduces the pH from near 11 to about 102 6 Further recarbonation of the clarified water converts a portion say 12 of the remaining carbonate ions to bicarbonate by the reaction CaCO 35 CO 2 H20 9 CaHCOs 2 the final pH is in the range 95 to 85 depending on the desired carbonate to bicarbonate ratio Bar Diagram Bar Graph purpose is to visualization of the chemical composition data can be expressed in meqL milliequivalents per liter a Top row of the bar graph consists of major cations arranged in the order of Ca Mg Na K b Bottom row of the aligned in the sequence of OH39 003 239 o3 39 804239 CI 39 NOg39 b The sum of the positive meqL must equal the sum of the negative meqL for a water in equilibrium 1171 lrSo ningiF09doc lon Balance or ZCations 2 Anions l Charge Balance X 100 ZCations ZAnions lt 5 OK Example WATER SOFTENING Excess Lime Treatment The water defined by the analysis given below is to be softened by excess lime treatment in a twostage system Given chemical Analysis Data 002 88 mgL as 002 Ca 2 700 mgL Mg2 97 mgL Na 69 mgL HCOg 1150 mgL as CaCog 804239 960 mgL CI39 106 mgL 1 Sketch a bar graph for a the raw water b softened water after chemical addition and settling but before recarbonation and filtration c softened water after 1St stage recarbonation d softened water after 2nd stage recarbonation and filtration assuming that onehalf of the alkalinity is in the bicarbonate form N List the hypothetical combinations of chemical compounds in the raw water 9 Calculate the quantity of softening chemicals required in lbMG of water P Calculate the theoretical quantity of 002 needed to provide finished water with 12 of the alkalinity converted to bicarbonate ion 1171 lrSo ningiF09doc SOLUTIONS 1 Express the concentrations in meqL b Check ion balance lon Balance or ZCations 2 Anions l Charge Balance X 100 ZCations ZAnions l4595 46l x 100 005 lt 5 4595 46 lon Balance is OK b Calculate the softening chemicals required List the combination and concentration meqL of chemical compounds from the bar graph Raw water 117117S0 ning7F09doc Lime Required coz co2 CaOH2 a Cacongzo 1 04 04 CaH0032 CaH0032 CaOH2 a 2Cacom2H20 2 23 23 M9804 M9804 CaOH2 a MgOH2SCaSO4 5 08 08 08 meqL 35 35 Soda Ash Required CaSO4 CaSO4 Na2003 9 CaCOqs NaZSO4 6 Raw water 12 12 Produced wLime 08 08 meqL 20 20 Calculate eqwt of lime and soda ash MW z eqwt mgmeq Quick Lime CaO 561 2 280 Soda Ash Na2003 106 2 530 Lime as CaO required stoichiometric requirement excess lime 35 meqL28 mgmeq 125 meqL28 mgmeq 133 mgL CaO 133 mgL834 lbMG per mgL 1100 lbMG Soda ash required 20 meqL53 mgmeq 106 mgL Na2003 106 mgL834 lbMG per mgL 884 lbMG c Sketch an meqL bar graph for the water after lime and soda ash additions and settling but before recarbonation 1 Calculate solubilities in meqL 1171 lrSoftningiF09doc mgL eqwt mgmeq meqL Cacog as Ca 003 30 50 06 MgOH2 as Ca 003 10 50 02 After the addition of softening chemicals raw excess raw excess Recarbonation 1 converts the excess OH39 to 003239 CaOH2 002 9 CaCOss H20 Excess OH39 OH39 from excess lime OH39 from MgOH2 125 meqL 02 meqL 145 meqL 22 mg 145 meqL 002 319 mgL of 002 meq Draw a bar graph for the softened water after recarbonation and filtration assuming that onehalf of the alkalinity is in the bicarbonate form 2 After secondstage processing final recarbonation convert 12 of remaining 003239 to 00339 Ca003 002 H20 9 CaHCOs2 06 06 1171 lrSO ning7F09doc M9003 COZ H20 9 MgHCog2 02 02 22 mg 208 meqL 002 88 mgL of 002 meq Total 002 Reacted 319 88 407 mgL 407 mgL 834 IbMG per mgL 340 lb COZ MG d Draw a bar graph for the softened water after recarbonation and filtration raw excess I PW Ionic 11 11 SoftningF09doc 35 43 46 Ca Mg Na39 CO 3 5 08 05 HCO3 5042 C 25 20 03 L04 I 23 meq 2395 I2 meq I 08 megquot3 1 I CO I CaHCO31 CaSO I MgSO4 03 a N39aCI I Mm oz M r Z QM 539 I IS 639quot A0 39 07 mfarm natdf 3 i 0 06 08 31 r I Caz Caz Mgz Na39 25 06 22 23 I I I OH oH 3032 39 042 Cl L 15 Io2I ac 20 as I 125 mcq I 08 meq I 20 meq I I excess lime I hardiness I soda ash addition I b 3 AH j 39s4u5g reCarIaoMZim 0 06 08 3 l I Cal Mgp Na 01 Io 23 3032 39 so1 2 CI39 68 20 03 I 08 mcq I 20 meq quotI 03 I hardness I soda ash addition I XML5r hm recurLond lion anlb39n4f A39Hr jmn 0 06 08 I I Cal Mg NaO Q6 oi239 23 co HCO3 5041 CI 04 me I 20 03 0 04 39 08 28 C 88Anaerobic ProcessesiFOQdoc Anaerobic Processes 4 h DC 515521 Anaerobic complete absence of molecular oxygen 02 anaerobic more new bacteria anaerobic bacteria Organic gt CH4 002 other end products sludge Example Anaerobic Digester to treat primary and secondary sludge Purposes of Anaerobic Digester 1 Reduction of sludge volume 2 Stabilization of organic solids 3 Disinfection Applications a Strong organic wastes extremely high BOD very high 02 demand b Domestic wastewater treatment primary sludge secondary sludge c Industrial wastes meet packing feed lots d Hazardous wastes Research level chlorinated compounds need more research 88Anaerobic ProcessesiFOQdoc Advantages and Disadvantages Advantages a CH4 as usable products b No 02 aeration costs c Less nutrients N P are required d Greater degree of waste stabilization e Smaller amount of sludge synthesis aerobic W 50 kg of bacteria W Yobs 005 5 kg of bacteria 95 kg of CH4 amp 002 100 kg BOD Disadvantages a Slow growth slow reaction b Higher temperature energy Methophilic 3035 C Thermophilic 5055 C c System is sensitive to change eg pH Temp Loading difficult to operate d CH4 gas is explosive safety 88Anaerobic ProcessesiFO9doc Types of Anaerobic Digestion Processes 1 StandardRate Anaerobic Digester 4th DC 517 Single tank No sludge mixing Intermittent sludge feeding and withdrawal Generally heated td 30 60 days for heated digester Organic loading rate 048 16 kg TVS m3 of digester volume day Gas relief Influent Effluent solids 1 Solid removal 2 HighRate Anaerobic Digester 4th DC 518 Continuous sludge feeding and withdrawal Organic loading rate 16 80 kg TVSm3 of digester volume day td 10 15 days Two tanks in series 1St Tank and 2nd Tank 1St Tank 2nd Tank Fermentation gas Solid Liquid separation Thoroughly mixed No mixing Heated Not heated 88Anacrobic ProcesscsF09doc Gas outlet Gas outlet ower TFloating cover Gas storage Gas storage Sludge inlet QB m Scum layer gt Supernatant Supernatant outlet lnfluent heater lgt Digested sludge First stage Completely mixed Characteristics Sludge outlet Second stage Stratified Acid formers Produce acids and alcohols SRT gt 3 days Less PH sensitive Methane formers Produce CH4 SRT gt10 days 95 F gt 20 days 65 F Very pH sensitive PH 70 71 88Anaerobic ProcessesiFOQdoc Stage 1 Stage 2 Stage 3 Hydrolysis Acid Methane Fermentation fermentation Liquifaction Acid formation Gasification Heterogeneous Acid formers Methane formers roups of facultative and anaerobic bacteria Complex Smaller Simple CH4 organics gt Soluble gt organic gt C02 sludge VSS organics acids alcohols Fats Proteins Triglycerides Propionic Polysaccharides Fatty aci s aci Amino acids Acetic Sugars acid Stage 1 Stage 2 Stage 3 Acid Methane Fermentation fermentation Hydrolysis Acetogenesis Methanogenesis Fermentation Acid enesis Liquifaction Acid formation Gasification Heterogeneous Acid formers Methane formers groups of facultative and anaerobic bacteria Complex Smaller Simple CH4 organics gt Soluble gt organic gt C02 sludge VSS organics acids alcohols Fats Proteins Triglycerides Propionic Polysaccharides Fatty acids acid Amino acids Acetic Sugars acid 88Anaerobic ProcessesiFOQdoc Conversion of organics BODL to methane gas Use glucose as representative organic compound CGH1206 9 3002 30H4 MW 180 344 316 30H4 602 9 6 H2O MW 316 632 344 Combine the two reactions CeH1206 9 3002 3 CH4 30H4 602 9 6 H2O CGH1206 602 9 6002 6 H2O g Oz 6 mol 32 gmol 192 g BODL 1067 g BODL g glucose 1 mol 180 gmol 180 g glucose g glucose g CH4 3 mol 16 gmol 48 g CH4 025 g CH4 produced g BODL 6 mol 32 gmol 192 g BODL g BODL destroyed 025 g CH4 1 mol CH4 224 L m3 103 g BOD 035 m3 CH4 produced g BODL 16 g CH4 1 mol 103 L kg BODL kg BODL destroyed 025 lb CH4 454 g 1 mol 224 L ft3 562 ft3 CH4 produced lb BODL lb 16 g 1 mol 2832 L lb BODL destroyed 88Anaerobic ProcessesiFOQdoc Under the standard conditions 20 C 1 atm 035 m3 CH4 produced quot1 EBBL39JJSIEJQAE quotquotquot quot 562 ft3 CH4 produced lb BODL destroyed Useful Values SI unit Y 093 m3gas produced gas kg VSS destroyed 034 m3 CH4 produced kg BOD destroyed English Unit 8 12 SCF gas gas lb VSS applied 15 18 SCF gas gas lb VSS destroyed 025 lb CH4 produced Z lb BOD destroyed 56 57 ft3 CH4 produced lb BOD destroyed SCF standard cubic foot 88Anaerobic ProcessesiFOQdoc Example BOD loading to a digester 682 kg BODday VSS loading to a digester 455 kg VSSday CH4 gas production 160 m3 CH4day Determine 1 the BOD removal kg BODd and efficiency 2 the VSS removal kg VSSd and efficiency Note Use CH4 23 gas Y 093 m3gas produced gas kg VSS destroyed 034 m3 CH4 produced kg BOD destroyed Solution 1 BOD removed 160 m3 CH4day 034 m3 CH4 kg BOD destroyed 470 kg BOD destroyedday 470 kg BODday BOD removal efficiency 0689 69 682 kg BODday BOD remaining 682 470 212 kg BODday m3 CH4 produced 093 m3 gas 2 CH4 kg VSS destroyed Kg VSS destroyed 062 m3 CH4 39tgv g39aggiai 160 m3 CH4day VSS removed 062 m3 CH4 kg vss destroyed 258 kg VSS destroyedday 88Anaerobic ProcessesiFOQdoc 258 kg VSS destroyedday VSS removal efficiency 455 kg VSSday 0567 57 V88 remaining 455 kg VSSd 258 kg VSSd 197 kg VSSd Example BOD loading to a digester 1500 lb BODday VSS loading to a digester 1000 lb VSSday CH4 gas production 5600 ft3 CH4day Determine 1 the BOD removal lb BODd and efficiency 2 the VSS removal lb VSSd and efficiency Note Use CH4 23 gas 56 3 CH4 produced Z lb BOD destroyed 15 SCF gas produced gas lb VSS destroyed Solution 1 5600 ft3 CH4day BOD removed 56 ft3 CH4 lb BOD destroyed 1000 lb BOD destroyedday 1000 lb BOD BOD removal efficiency 067 67 1500 lb BODday BOD remaining 1500 1000 500 lb BODday 88Anaer0bic ProcessesiFOQdoc 2 CH4 113 produced 15 SCF Gas 2 CH4 lb VSS destroyed lb VSS destroyed 10 ft3 CH4 quot1iquotv39 39at339y39e39aquot 5600 113 CH4day VSS removed 10 ft3 CH4 lb VSS destroyed 560 lb VSS destroyedday 560 lb VSS destroyedday VSS removal efficiency 1000 lb VSSday 056 56 V88 remaining 1000 lb VSSd 560 lb VSSd 440 lb VSSd Note 1 kg 22 lb 1 m3 35314 ft3 1 cfs 002832 m3s erastewateriF09 doc WASTEWATER TREATMENT Introduction Wastewater Dirty water or Liquid wastes need to be collected and conveyed in a system for treatment and ultimate disposal Types 1 Domestic Wastewater Sewage from housing relatively consistent in quality and quantity 2 Industrial wastewater Inconsistent in quality and quantity eg Metal plating vs food processing Domestic Wastewater 1 Composition of Domestic Wastewater Raw Sewage a 999 water b 01 is made up of trash grit grease solids dissolved matter microorganisms VH p457 2 Suspended Solids 024 lbcapitaday BOD5 02 lbcapitaday Phosphorus 3 lbcapitayr Wastewater flow 100120 galcapitaday Example Pocatello Wastewater Treatment Plant Qave 120 galcapday55000 people 66 x106 galday GPD 66 million gallonsday MGD BOD5 02 lbcapday55000 people 11000 lb d SS 024 lbcapday55000 people 13200 lbd P 3 lbcapYr55000 people 165000 lbyr Note 1 kg 2205 lb 1 m3 264172 gal mgL MGD834 Ibd Example VH 463 Q 01 MGD BOD 450 mgL What is the BOD equivalent population 450 mgL01 MGD834 375 lbd 02 lbcapd 02 lbcapd BOD equivalent population 1900 Wastewater Treatment 3rd DC 361 4th DC 419 Types of Wastewater Treatment 1 Seepage Systems Cesspool Septic Tank There are a total of about 3000 septic tanks in the Lower Portneuf River aquifer region estimated in 2006 2 Large Systems POTWs Publicly Owned Treatment Works WPCP Water Pollution Control Plant WWTP Wastewater Treatment Plant Wastewater Treatment Large System Preliminary Treatment Primary Treatment Secondary Treatment Advanced Wastewater Treatment Tertiary treatment PWNT 5 Disinfection 6 SludgeTreatment 6767Activated Sludge ModelsiF09 doc Bacterial Growth Kinetics KS 3 DC 350 4th DC 459 Fig 618 Monod growth rate constant as a function of limiting food concentration The Monod Model m5 K S 2 where p specific growth rate d391 m maximum specific growth rate d391 rate limiting substrate food concentration mgL eg soluble BOD halfsaturation coefficient mgL Monod Constant halfvelocity constant Ks Bacterialgrowth rate rg 8 dt dl 8 KS S where X bacteria concentration in VSS mgL ms 1 KSS 6767Activated Sludge ModelsiF09 doc dS Substrate UtIIIzatIon Rate r 7 t It le is dt 3 dt M le Hides dt 3 dt M where Y bacterial cell yield yield coefficient mg VSSmg BOD Bacterial Death Rate rd t rd dix kdX dt d d where kd death rate constant endogenous decay coefficient d 391 Endogenous Phase near starvation phase Net Growth Rate Growth Rate Death Rate ism e in 3 dt 3 dt 8 dt d where uX kdX dt 3 dt d dX quot k 73 7t 3 dX mX lnwhich uS YkS m k k m KSS d KSS d where k pmY 6767Activated Sludge ModelsiF09 doc maximum substrate utilization rate constant mt r LX x 8 dt KSS Substrate Utilization Rate rsu 5sz ZLLX 1 Aims X dtll Ydtg YKSS kSX Ks s Specific Substrate Utilization Rate U LS dt M S0 S S0S0 S X 0X 0XS0 Q 80 8 1 E VX So M where E 80 S 80 substrate removal efficiency 9 VQ hydraulic retention time aeration time hr FM Food to microorganisms ratio U 6767Activated Sludge ModelsiF09 doc Mass balance on X around the secondary treatment Accum Inputs Outputs i Rxns dX Vii QXU Q QWX6 QWXr rg V where r YkS kd X 8 KSS r9 net growth rate mgL d Y yield coefficient mg X mg 8 KS halfsaturation constant mgL kd endogenous decay rate constant d391 k maximum substrate utilization rate d391 k pm Y lJm k pm maximum specific growth rate at steady state dXdt 0 KS1kdac KS1kdac 00Yk kd 1 acwm kd 1 528 3rd DC 390 621 4th DC 467 S soluble BOD in effluent Y yield coefficient mg X mg 8 KS halfsaturation constant mgL kd endogenous decay rate constant d391 k maximum substrate utilization rate d391 k pm Y or pm Yk Mass balance on 8 around the aeration tank Accum Inputs Outputs i Rxns dS Vii Q50 QrS QQrS rm V kSX where r KSS SM rsu substrate utilization rate mgLd 6767Activated Sludge ModelsiF09 doc At steady state dSdt 0 acmso S 61 deC X 530 3rd DC 390 622 4th DC 467 Substituting 9 VQ solve for V V2 66 YQSo S X1kd 66 Example Q 10000 m3d 80 120 mgL 8 7 mgL X MLVSS 2000 mgL 90 6 days Y 06 mg VSSmg BOD kd 006 d391 What is the aeration tank volume Solution 2 60 Y Q Sn S X1kdec 3 06 mgVSS 10000m 120 7E BOD mgBOD d L 2000mg L1 60d 60 d V1495 m3 Excess Biological Solids Px PX Yobs So S or PX YobSSO SQ Y 7 Obs Helde 6767Activated Sludge ModelsiF09 doc Example Determine the excess sludge in mgL VSS for the given conditions 0 10000 m3d 80 120 mgL 8 7 mgL kd 006d ec 10 days Y 06 mg VSSmg BOD Solution 06mgVSS Y Y mg BOD 0375 mgVSS abs 7 16de 110d 036 mgBOD p 424 mgVSS L mgBOD L 3 Mn kg ld424 mgVSS10000m 11ng 12g J L d m 10 mg 420 kg VSS Id Example 57 3rd DC 391 Example 66 4th DC 468 The town of Gatesville has been directed to upgrade its primary WWTP to a secondary plant that can meet an effluent standard of 30 mgL BOD5 and 30 mgL suspended solids 88 They have selected a completely mixed activated sludge system Assuming that the BOD5 of the 88 may be estimated as equal to 63 percent of the 88 concentration estimate the required volume of the aeration tank The following data are available from the existing primary plant Existing plant effluent characteristics Flow 0 0150 m3s 3005 850 mgL Assume the following values for the growth constants Ks 100 mgL BOD5 pm 25 d391kd 005 dquot Y 050 mg VSS mg BOD5 X 2000 mgL MLVSS additional assumption Approach Use 9 VQ 525 Determine 9c using 6767Activated Sludge ModelsiF09 doc Ks1kd6c 66Yk kd 1 Determine 9 using actso S 91kd96 Secondary Sedimentation Tank Aeration Tank Secondary Clarifier Q Q Qr Q Qr Q Qw So v x s x s 39 s TeL39 X MLSS X MLVSS Qr Qw Qw Qr Xr s s Xr 39 Note that So and S are soluble BOD in influent and effluent respectively Solution Note that S and So in the above eqns are soluble BOD5 BOD5 allowed soluble BOD5 S suspended solid BOD5 30 mgL S 063 30 mgL SS 1 Calculate soluble BOD5 in effluent S S BOD5 allowed suspended solid BOD5 30 mgL 06330 mgL 111 mgL 6767Activated Sludge ModelsiF09 doc 2 Obtain ec gmCd ac 100mgL1005d16ic cYkkd 1 6C25d1 005d1 1 111mgL solve for ac ec 50 d Note since k pm Y Y k pm 25 d391 3 Obtain 9 with X 2000 mgL VSS 03 YSo S 01kd00 50d05 mg VSS lmg B0D584mgL 111mgL o 1005d 150d 2000mgL Solve for e e 0073 d 18 hrs 4 V 9 Q 18 hrs015 m3s3600 sechr 972 m3 Note 6 YS 7S 6 6 P c 0 ic obsS07S c I 61kd66 6 6 where PX YobsS0 S Y Y 7 0 19ckd 9 9W Px PxQ a biomass in Aeration Tank biomass in Aeration Tank 7 biomass synthesized time 7 biomass wasted time XV XV es QwXQQwXe PXQ PXQ QWXQQwXe Activated Sludge TermsiF09doc Primary Treatment Purpose to remove settlable organic solids Efficiency BOD removal 30 30 40 SS removal 60 50 70 Secondary Treatment Purpose to remove soluble organics Efficiency BOD removal 90 85 95 SS removal 90 85 95 Biological Treatment Basic Reaction of Aerobic System X more new microbes microbes U Organics 02 nutrients gt 002 H20 Soluble air N P BOD5 8 Note all naturally present in domestic sewage except 02 Types of Secondary Treatment Systems 1 Suspended Growth Systems Reactors eg Activated Sludge processes Conventional Completely mixed 2 Attached Growth Systems Reactors eg Trickling Filters Rotating Biological Contactor RBC Submerged Rotating Biological Contactor SBC Activated Sludge TetmsiFO9tdoc Conventional Activated Sludge Process Dillused Aerator Waste sludge 4 4 D Log phase Endogenous phase Exponential growth Waste sludge Time or distance Completely Mixed Activated Sludge Process gm E CD lt3 C C C W a ste sludege 4 gt Endogenous phase BOD5 Ogdemand Distance Activated Sludge TermsiF09doc Completely Mixed Activated Sludge Process 3rd DC 387 4 h DC 462 Secondary Sedimentation Tank Aeration Tank Secondary Clarifier Q QQr QQr Qle So V X S X S V S Xe X MLSS X MLVSS Qr QWY Qw Qr Xr S S Xr 39 Terms and Definitions Hydraulic retention time HRT Hydraulic detention time Aeration period Liquid detention time HRTtdeVQ where V volume of the aeration tank Q flow rate MLSS MLVSS MLSS Mixed liquor suspended solids 88 in the aeration tank MLVSS Mixed liquor volatile suspended solids VSS in the aeration tank Note MLSS and MLVSS are indicative of biomass MLVSS is more indicative of true biomass than MLSS MLSS includes inorganic constituents MLVSS MLSS 07 08 Activated Sludge TermsiF09doc SVI Sludge Volume Index a The SVI is the volume in milliliters occupied by 1 g of suspended solids 88 after 30 min of settling O It is computed by Sludge volume after settling mLL 1000 mg x MLSS mgL g SVI 50 150 mLg indicates a good settling sludge lo6 Xr in mgL SVI Unit of Xr Organic BOD Loading BOD Loading Q So Volumetric BOD Loading FV Q So Volumetric BOD Loading V F lb BOD applied day V 1000 ft3 of aeration tank capacity This relationship has no information on microbial population density Activated Sludge TennsiF09doc FM Ratio FoodtoMicroorganism Ratio F Mass lb of BOD5 applied to the system day M Mass lb of microbes MLSS MLVSS in the system Aeration tank F Q 80 F Q So or M MLVSSV M MLSSV Q So So X v e X Note 1 Q 0 or Q 0 Range of FM value 005 10 lblbday 03 07 lblbday commonly 05 lblb day typically Unit of FM Ratio lb BOD5 d lb mg BOD5 d mg b MLVSS b d mg MLVSS mg d Example 59 3rd DC 394 Example 4th DC 468 Compute the FM ratio for the new activated sludge plant at Gaterville Given Q 015 m3s v 970 m3 80 84 mgL X MLVSS 2000 mgL Activated Sludge TermsiF09doc Solution F Mass lb of BOD5 applied to the system day M Mass lb of microbes MLSS MLVSS in the system Aeration tank Q 80 015 m3s 84 mgL 86400 sd X v 2000 mgL970 m3 056 mg BOD5d lb MLVSS Example The wastewater flow rate and BOD5 to the secondary treatment system is 1 MGD and 300 mgL respectively The hydraulic retention time of the aeration tank is 6 hrs and MLSS is 2500 mgL Calculate a BOD loading b volume of the aeration tank c volumetric BOD loading and d FM ratio Solution a BOD loading 80 o 300 mgL1 MGD 834 2500 lb BODd Note 1 mgL 834 lbMG 1 x 106 gal b v O 9 6 hrs 250000 gal 025 MG 24 hrs 250000 gal 1 ft3 748 gal 33500 ft3 80 Q Volumetric BOD Loading V O v 300 mgL1 MGD 834 75 lb 3005 d 33500 ft3 1000 ft3 Activated Sludge TermsiF09doc F 80 Q 300 mgL 1 MGD834 M x v 2500 mgL025 MG834 2500 lb BODd 048 lb 5200 lb MLSS lb d Mean cell residence time MCRT Solid retention time SRT Sludge age 90 a Definition Average length of time in which a solid particles ie biomass is retained in an aeration tank biomass MLVSS or MLSS in the aeration tank ec day sum of the biomass in the waste sludge and effluent day mass of solids in aeration tank MLSS or MLVSS V mass rate of sludge wasting waste sludgeday X V ec Qw Xr Q Qw Xe Rate of Rate of Intentionally Accidentally wasted wasted At steady state the rate of loss the rate of growth Activated Sludge TermsiF09doc Typical values ec 3 60 days commonly 10 15 days If 0c is high If 0c is low Low FM High FM Low sludge wasting High sludge wasting Process Efficiency E 80 S 100 So Example 527 3rd DC 452 Two activated sludge aeration tanks at Turkey Run Indiana are operated Each tank has the following dimensions 70 m wide by 300 m long by 43 m effective liquid depth The plant operating parameters are as follows Flow 00796 m3s Soluble BOD5 after primary settling 130 mgL MLVSS 1500 mgL MLSS 140 MLVSS Settled sludge volume after 30 min 230 mLL Aeration tank liquid temperature 15 C Determine the following 1 Aeration period 2 FM ratio 3 SW and 4 Solids concentration in return sludge Xr solution Q 00796 m3s 80 130 MLVSS 1500 mgL MLSS 14 MLVSS 14 1500 mgL 2100 mgL Settled sludge volume after 30 min 230 mLL T 15 C v each 70 m300 m43 m 903 m3 v Total 2903 m3 1806 m3 Activated Sludge TennsiF09doc 1 Aeration Period v 1806 m3 G 63 hrs Q 00796 m3s3600 shr 2 F Q 80 00796 m3s130 mgL86400 sd M X v 1500 mgL1806 m3 033 mg 033 mgd BOD mgd mg biomass Settled sludge volume after 30 min1000 MLSS 230 mLL 1000 mg 1095 2100 mgL g Note SVI 50150 mLg for good settling 4 1 1 1000 mgg Xr 91324 mgL SVI 1095 mLg L1000 mL Note 106 Xr in mgL SVI


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