4 16 2015 An experiment reveals that a concentration of 0 1 g m3 of free available chlorine yields a 99 kill of bacteria in 8 minutes Find the inactivation rate constant k Disinfection kinetics Chick s law dN kC n N k N dt N number of pathogens per unit volume C concentration of disinfectant n coefficient of dilution 1 k is coefficient of lethality depending on pH and T Because C is usually high enough that it does not change A B C D E kC n k inactivation rate constant dN kC n N k N dt N t N 0 exp k t N t exp k t N0 0 1min 1 1 min 1 0 58 min 1 5 8 min 1 None of the above N t is design criteria and is usually known N0 1 An experiment reveals that a concentration of 0 1 g m3 of free available chlorine yields a 99 kill of bacteria in 8 minutes What contact time is required to achieve 99 9 kill at the same free available chlorine concentration dN N k N exp k t dt N0 N ln k t N0 N 0 01 N0 k 2 A B C D E ln 0 01 0 58 min 1 8 1 min 12 min 6 min 20 min None of the above 3 N dN N k N exp k t ln k t dt N0 N0 N 1 0 001 k 0 58 min N0 N ln N 0 12 min t 0 58 min 1 g g min Ct 0 1 3 12 min 1 2 achieves 3 log inactivation 3 m m 5 4 An experiment reveals that a concentration of 0 1 g m3 of free available chlorine yields a 99 kill of bacteria in 8 minutes What contact time is required to achieve 99 9 kill at available chlorine concentration of 0 05 g m3 A B C D E 10min 12 min 60 min 24 min None of the above 6 1 4 16 2015 Solar Disinfection Example 2 well mixed reactors are put in series with V 5 L each and use solar radiation for disinfection Water is pumped continuously through the chambers at a flow rate Q If the disinfection rate constant k 7 8 s 1 and pathogen counts must be removed by 3 log how much water will be treated after 1 hour g min achieves 3 log inactivation m3 3 C 0 05g m g min 1 2 m3 24 min t g 0 05 3 m Ct 1 2 7 A B C D E 1700 L 4585 L 280 L 6020 L None of the above Match the contaminant to remove with the appropriate unit process and put them in order Mass balance on pathogens tank 1 QN 0 QN1 Vk N1 and N1 QN 0 Q Vk Contaminants remove tank 2 QN1 QN 2 Vk N 2 N2 N0 1 V Q k 1 8 A Dense organic matter in influent Low concentrations of microorganisms Large solids debris 1 Aerobic bioreactor 2 Secondary clarifier 3 Grit chamber D Dissolved and suspended organic C E Dense inert particles in influent F High concentrations of microorganisms G Water 4 Disinfection 5 Bar screens 6 Primary clarifier 7 Belt Press 2 B N2 N 2 0 001 N 0 1 V Q k 2 N 0 C 1 V Q k 1000 Unit process 2 V k 1000 1 30 6 Q 1 Vk 7 8s 5 L 1 27 L s 30 6 30 6 Qt 1 27 L s 3600 s 4585 L Q Contaminants to remove 9 Unit process C Large solids debris 5 Bar screens E Dense inert particles in influent 3 Grit chamber A Dense organic matter in influent 6 Primary clarifier D Dissolved and suspended organic C 1 Aerobic bioreactor F High concentrations of microorganisms 2 Secondary clarifier B Low concentrations of microorganisms 4 Disinfection Let s review BOD DO and D First BOD DO D The amount of oxygen required for bacteria to oxidize organic wastes aerobically with oxygen is the biochemical oxygen demand BOD expressed in terms of mg L The concentration of oxygen in water is the dissolved oxygen DO saturation DOs is the max conc of O2 that will dissolve in water The deficit D is the DOs DOactual Last 2 4 16 2015 Let s review BOD DO and D BOD degradation follows 1st order kinetics BOD ultimate oxygen carbonaceous BOD amount of oxygen already used up in degradation of waste by microorganisms in t days DO D Let s review BOD DO and D BOD degradation follows 1st order kinetics Lt Loe kt Lo BODt Lt BODt Loe kt BODt Lo 1 e kt or Lo BODt 1 e kt BODt DO0 DOt 16 A waste has an ultimate BOD of 1 000 mg L and a rate coefficient k of 0 1 day What is its 5 day BOD BOD5 A BOD5 393 4 mg L B BOD5 606 5 mg L C BOD5 2 4 mg L D BOD5 4332 4 mg L E None of the above Lt Loe kt Lo BODt Lt BODt Loe kt BODt Lo 1 e kt or Lo BODt 1 e kt BODt DO0 DOt What is the ultimate BOD of a waste that has a measured BOD5 of 20 mg L assuming a BOD rate coefficient of 0 15 day A 42 3 mg L B 4332 4 mg L C 37 9 mg L D 4 2 mg L E None of the above 17 Lt Loe kt Lo BODt Lt BODt Loe kt BODt Lo 1 e kt or Lo BODt 1 e kt BODt DO0 DOt 18 3 4 16 2015 A wastewater having an ultimate BOD L0 of 1000mg L is discharged to a river at a rate of 2 m3 s The river has an ultimate BOD of 10 mg L and is flowing at a rate of 8 m 3 s Assuming a reaction rate constant k 0 1 day 1 calculate the ultimate BOD of the waste 20 km downstream The river is flowing at a velocity of 10 km day Ignore reaeration Find BOD5 for the sample at 0 km Lt Loe kt Lo BODt Lt BODt Loe kt BODt Lo 1 e kt or Lo BODt 1 e kt BOD The amount of oxygen required for bacteria to oxidize organic wastes aerobically with oxygen expressed in terms of mg L A L0 x 17 0mg L 8 2mg L B L0 x 170mg L 82 mg L C L0 x 1 7mg L 0 82 mg L D L0 x 2080mg L 80 2 mg L E None of the above 19 20 Mass balance calculation to determine L5 in the river after BOD5 The amount of oxygen consumed by bacteria in 5 days by oxidizing organic wastes aerobically with oxygen expressed in terms of mg L mixing with the waste x 0 km L0 x 0Qdownstream L0 upstreamQupstream L0 inQin L0 x 0 L0 upstreamQupstream L0 inQin Qdownstream Qdownstream Qupstream Qin 2 …
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