M.M. Ghangrekar, IIT Kharagpur

Transcription

1 BOD Model I is generally assumed ha he rae a which he oxygen is consumed is direcly proporional o he concenraion of degradable organic maer remaining a any ime. The kineics of BOD reacion can be formulaed in accordance wih firs order reacion kineics as: d L / d = - K L Where, L = amoun of firs order BOD remaining in wasewaer a ime K = BOD reacion rae consan, ime -1 Inegraing dl = KL. d i.e., [ log ] = K 0 0 L. L / Lo = e -K. or 10 -K. Where Lo or BOD u a ime = 0, i.e., ulimae firs sage BOD iniially presen in he sample. The relaion beween K(base e) and K (base 10) is K(base 10) = K(base e) / The amoun of BOD remaining a ime equals (Figure 11.1) L = Lo (e -k. ) The amoun of BOD ha has been exered (amoun of oxygen consumed) a any ime is given by BOD = Lo L = Lo (1 e -k. ) And he five day BOD is equal o BOD 5 = Lo L 5 = Lo (1 e -5k ) For pollued waer and wasewaer, a ypical value of K (base e, 20 o C) is 0.23 per day and K (base 10, 20 o C) is 0.10 per day. These values vary widely for he wasewaer in he range from 0.05 o 0.3 per day for base 10 and 0.23 o 0.7 for base e. The ulimae BOD (Lo) is defined as he maximum BOD exered by he wasewaer. I is difficul o assign exac ime o achieve ulimae BOD, and heoreically i akes infinie ime. From he pracical poin of view, i can be said ha when he BOD curve is approximaely horizonal (Figure 11.1) he ulimae BOD has been achieved. The ime required o achieve he ulimae BOD depends upon he characerisics of he wasewaer, i.e., chemical composiion of he organic maer presen in he wasewaer and is biodegradable properies and emperaure of incubaion. A higher emperaure for same concenraion and naure of organic maer ulimae BOD will be achieved in shorer ime as compared o lower

2 2 emperaures, where i will require more ime. The ulimae BOD bes expresses he concenraion of degradable organic maer based on he oal oxygen required o oxidize i. However, i does no indicae how rapidly oxygen will be depleed in receiving waer. Oxygen depleion is relaed o boh he ulimae BOD and he BOD rae consan (K). The ulimae BOD will increase in direc proporion o he concenraion of biodegradable organic maer. The BOD reacion rae consan is dependen on he following: 1. The naure of he wase 2. The abiliy of he organisms in he sysem o uilize he wase 3. The emperaure Naure of he wase: Thousands of organic maer exiss wih differen chemical composiion in naure. All organic maer will no have same degradaion rae. Simple sugar and sarches are rapidly degraded and will herefore have a very large BOD rae consan. Cellulose degrades much more slowly and hairs are almos undegradable during BOD es or during biological reamen of wasewaer. Oher compounds are inermediae degradable beween hese exremes. For complex wase, like sewage, he BOD rae consan depends upon he relaive proporions of he various componens. The BOD rae consan is high for he raw sewage (K (base e) = per day) and low for he reaed sewage (K (base e) = per day), owing o he fac ha, during wasewaer reamen he easily degradable organic maer will ge more compleely removed han he less degradable organics. Hence, in he reaed wasewaer, relaive proporion of he less biodegradable organic maer will be higher, giving lower BOD rae consan. Abiliy of organisms o uilize wase: Every microorganism is limied in is abiliy o uilize organic compounds. Many organic maers can only be uilized by paricular group of microorganisms. In naural environmen, where he waer course is receiving paricular organic compound, he microorganisms which have capabiliy o degrade ha organic maer will grow in predominan. However, he culure used during BOD es may have very small fracion of he organisms which can degrade ha paricular organic compounds in he wase. As a resul he BOD value, for limied incubaion duraion, and he rae consan would be lower in he laboraory es han in he naural waer where he wase is regularly discharged. Therefore, he BOD es should be conduced wih organisms which have been acclimaed o

3 3 he wase so ha he rae consan deermined in he laboraory can be compared o ha in he river. Temperaure: The biochemical reacions are emperaure dependen and he aciviy of he microorganism increases wih he increase in emperaure up o cerain value, and drop wih decrease in emperaure. Since, he oxygen uilizaion in BOD es is caused by microbial meabolism, he rae of uilizaion is similarly affeced by he emperaure. The sandard emperaure a which BOD is deermined is usually 20 o C. However, he waer emperaure may vary from place o place for he same river; hence, he BOD rae consan is adjused o he emperaure of receiving waer using following relaionship: K T = K 20 θ (T-20) Where T = emperaure of ineres, o C K T = BOD rae consan a he emperaure of ineres, day -1 K 20 = BOD rae consan deermined a 20 o C, day -1 θ = emperaure coefficien. This has a value of in general and for higher emperaure greaer han 20 o C. This is because increase in reacion rae is higher when emperaure increases from 10 o 20 o C as compared o when emperaure is increased from 20 o 30 o C. Example: 3 The wasewaer is being discharged ino a river ha has a emperaure of 15 o C. The BOD rae consan deermined in he laboraory for his mixed waer is 0.12 per day. Wha fracion of maximum oxygen consumpion will occur in firs four days? Soluion Deermine he BOD rae consan a he river waer emperaure: K 15 = K 20 (1.056) (T-20) = 0.12 (1.056) (15-20) = per day Using his value of K o find he fracion of maximum oxygen consumpion in four days: BOD 4 = Lo (1 e x4 ) Therefore, BOD 4 / Lo = Example: 4

4 4 The dissolved oxygen in an unseeded sample of dilued wasewaer having an iniial DO of 9.0 mg/l is measured o be 3.0 mg/l afer 5 days. The diluion fracion is 0.03 and reacion rae consan k = 0.22 day -1. Calculae a) 5 day BOD of he wase, b) ulimae carbonaceous BOD, and c) Wha would be remaining oxygen demand afer 5 days? Soluion a) Oxygen demand for firs 5 days BOD 5 = (DO i Do f ) / p = ( ) / 0.03 = 200 mg/l b) Ulimae BOD BOD u = Lo = BOD / (1 e -k ) = 200 / (1 e x 5 ) = 300 mg/l c) Afer 5 days, 200 mg/l of oxygen demand ou of oal 300 mg/l would be saisfied. Hence, he remaining oxygen demand would be = 100 mg/l Example: 5 Deermine ulimae BOD for a wasewaer having 5 day BOD a 20 o C as 160 mg/l. Assume reacion rae consan as 0.2 per day (base 10). Soluion BOD 5 = Lo ( k. ) 160 = Lo ( x 0.2 ) Therefore, Lo = mg/l ~ 232 mg/l Example: 6 The BOD of a sewage incubaed for one day a 30 o C has been found o be 100 mg/l. Wha will be he five day 20 o C BOD? Assume K = 0.12 (base 10) a 20 o C, and θ = Soluion: BOD a 30 o C = 100 mg/l K 20 = 0.12 Now K 30 = K 20 θ (T-20) K 30 = 0.12 (1.056) = per day BOD = Lo (1 10 -k ) 100 = Lo ( x 1 ) Lo = mg/l This is ulimae BOD, he value of which is independen of incubaion emperaure.

5 5 Now BOD 5 a 20 o C can be calculaed as: BOD 5 a 20 o C = Lo (1 10 -k ) = ( x 5 ) = mg/l Example: 7 Deermine he 1 day BOD and ulimae firs sage BOD for a wasewaer whose 5 day 20 o C BOD is 200 mg/l. The reacion rae consan k (base e) = 0.23 per day. Soluion Ulimae BOD BOD u = Lo = BOD / (1 e -k ) = 200 / (1 e x 5 ) = 293 mg/l Therefore, one day BOD Y 1 = Lo L1 = Lo (1 e -k ) = 293 (1 e x 1 ) = 60 mg/l Inerpreaion of he BOD es Resul Following facors mus be considered in he inerpreaion of he BOD for indusrial wasewaers: 1. The seed is acclimaed o he wasewaer and lag period required for acclimaion is eliminaed. 2. The rae consan should be esablished based on long erm BOD ess on boh wasewaer and reaed effluens. The rae consan for unreaed and reaed wasewaer is no same for many wasewaers. The rae consan value is higher for unreaed wasewaer and lower for reaed wasewaer. For example, for raw sewage rae consan is abou 0.15 o 0.3 and ha for reaed sewage i is around 0.05 o 0.15 (base 10). Hence, direc comparison of BOD may no be valid. The value of K (base e) for raw sewage varies in he range 0.35 o 0.7 and ha for reaed sewage i will be 0.12 o In case of acidic wase, all samples mus be neuralized before incubaion. 4. When organic maer is presen in suspended form, inerpreaion of he es resul is difficul due o lag ime involved in hydrolysis of organic maer before acual oxidaion sars during BOD es Nirificaion in BOD Tes

6 6 Non-carbonaceous maer, such as ammonia is produced during he hydrolysis of proeins. In addiion, when he living hings die, excrea wase, and nirogen organic compounds, he nirogen ied o organic molecule is convered o ammonia by bacerial and fungal acion. Under aerobic condiions, his ammonia will be convered o nirae, called as nirificaion as per he reacions given below: Nirosomonas - 2NH 3 + 3O 2 2 NO H + + 2H 2 O Nirobacer - - 2NO 2 + O 2 2 NO 3 Hence, he organic maer conaining nirogen will have oxygen requiremen for nirificaion. The oxygen demand associaed wih he oxidaion of ammonia o nirae is called he nirogenous BOD. Due o low growh rae of nirifying baceria, his BOD demand normally occurs from 6 o 10 days (Figure 11.2). This is one of he reasons o use incubaion period of 5 days for BOD deerminaion o eliminae oxygen demand for nirificaion and o find ou only carbonaceous oxygen demand. Incidenally, he five day period was chosen for he es because he Themes River requires five days from is origin o join sea, and if oxygen demand for hese five days is deermined and saisfied he river waer qualiy can be proeced. Ulimae carbonaceous BOD Oxygen consumed BOD 5 BOD Carbonaceous BOD 5 Time, days Figure 11.2 Nirificaion during BOD es Oher Measures of Oxygen Demand Chemical Oxygen Demand (COD) During COD deerminaion oal organic conen of he wase is oxidized by dichromae in acid soluion.

7 7 In his es o deermine he oxygen requiremen of he wasewaer, srong oxidizing agen poassium dichromae is used. Acidic environmen is provided o accelerae he reacions by addiion of sulphuric acid. The reflux flasks (closed reflux vials), used for he es, are heaed o 150 o C for wo hours wih silver sulphae as caalys. When silver sulphae caalys is used, he recovery of mos organic compounds is greaer han 92 percen. COD es measures virually all oxidizable organic compounds wheher biodegradable or no, excep some aromaic compounds which resiss dichromae oxidaion. The COD is proporional o BOD only for readily assimilable organic maer in dissolved form e.g. sugars. No correlaion beween BOD and COD exiss when: o Organic maer is presen in suspended form, under such siuaion filered samples should be used. o Complex wasewaer conaining refracory subsances. For readily biodegradable wase, such as dairy COD = BODu/0.92 The correlaion beween BOD and COD for sewage is presened in he Figure BODu BODu/COD b = 0.87 BOD/COD b raio BOD 5 BOD BOD 5 /BODu = 0.68 BOD 5 /COD b = Time, days Figure 11.3 Correlaion beween BOD and COD for sewage a 20 o C incubaion and k = 0.23 per day. The COD is faser deerminaion bu does no give idea abou he naure of organic maer wheher biodegradable or biorefracory organic maer. Hence, deerminaion of BOD is necessary for he wasewaer o know biodegradable organic maer fracion. The BOD is no very useful es for rouine plan conrol due o long incubaion period required, hence

8 8 requiring long ime (5 days) o obain resuls. Thus, i is imporan o develop correlaion beween BOD and COD (or TOC), so ha COD (or TOC) can be used as a parameer for rouine analysis and conrol of he reamen plan. Once COD values are known, he BOD can be esimaed using correlaion. The es resuls are more reproducible for COD. Theoreical Oxygen Demand (ThOD) Theoreical oxygen demand for he wasewaer is calculaed as oxygen required for oxidizing he organic maer o end producs. For example, for glucose, he heoreical oxygen demand can be worked ou as below: C 6 H 12 O 6 + 6O 2 6CO 2 + 6H 2 O ThOD = (6 M o2 ) / (M C6H12O6) = (6 x 16 x 2) / (12 x x x 16) = 1.07 For mos of he organic compounds (excep aromaics) COD is equal o ThOD. TOC is relaed o COD hrough carbon-oxygen balance. COD/TOC = (6 M o2 ) / (M C ) = 2.66 Depending on he organic maer in quesion COD/TOC raio may vary from zero (for organic maer resisan o dichromae oxidaion) o 5.33 for mehane (mos reduced organic compound). Since, organic conen undergoes changes during biological oxidaion, COD/TOC and BOD/TOC, BOD/COD values will change during reamen. Quesions 1. Explain objecives of conducing BOD es. 2. Explain BOD reacion rae consan and parameers on which i is dependen. 3. Draw a curve for BOD exered and remaining wih respec o ime for organic wasewaer and derive mahemaical expression for boh. 4. Why only abou 60% BOD is saisfied during BOD es deerminaion, whereas during acual wasewaer reamen in aerobic process more han 90% of BOD can be removed during 5 o 6 hours of reenion ime in biological reacor? 5. BOD of a sewage incubaed for 3 days a 27 o C was measured 110 mg/l. Calculae BOD5 a 20 o C. Consider k = 0.23 per day (base e) and emperaure coefficien = Describe nirificaion during BOD es.

9 9 7. Explain correlaion beween BOD, BODu and COD for sewage.