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1 Department of Civil Engineering-I.I.T. Delhi CEL 212: Environmental Engineering Example (Disinfection) Q1. The Chick’s Law: For disinfection, assume N (0) is initial number of pathogens and N (t) is remaining number of pathogens at time and given by: N (t) =N (0) × exp (-K×t); where K is disinfection rate (unit=1/unit of time) and depends on disinfectant-pathogen interaction and solution characteristics. Here R- log removal: R=-log 10 [N (t)/N (0)]. If disinfectant concentration (C disinfectant ) and contact time (t c ) are related to each other by following equation (the Watson’s Law): (C disinfectant ) n × (t c ) = constant (standard unit: C in mg/L and t c in minute) Calculate R-log removal value for 99.9% removal? What is the remaining pathogen concentration at this removal after 1 minutes of contact time? (Assume K=0.046/min) [3+3=6 points] Solution: R-log removal = -log 10 [1-N t /N 0 ] Given removal = 99.9%, so N t /N 0 = (1-99.9/100) =0.001 R-log removal = -log 10 [0.001] =3 (answer) For calculating remaining fraction of pathogens after 1 minute of disinfection with K=0.046/min, use the Chick’s Law: N t /N 0 = exp (-k ×t) = exp (-0.046/min×1 min) =0.9550 (i.e., 95.50%) (answer) Q2. An experiment shows that a concentration of 0.1g/m 3 of free available chlorine yield a 99% kill of bacteria in 8 minutes. What contact time is required to achieve a 99.9% kill at a free available chlorine concentration of 0.05 g/m 3 ? Assume that Chick’s Law and Watson’s Law hold with n=1. [2+2=4 points] Solution: Given: For 99% kill: C= 0.1 g/m 3 and time (t) =8 minutes Chick’s Law: N t =N 0 ×exp (-k ×t) Calculation of disinfection rate constant: N t /N 0 = (1-99/100) =0.01 in 8 minutes From Chick’s Law: 0.01 = exp (-k ×8) => k = - (1/8) ln (0.01) = 0.5756/min (answer) Using calculated k value, calculate time for getting 99.9% kill: N t /N 0 = (1-99.9/100)=0.001 Using Chick’s Law: 0.001 = exp (-0.5756×t) =>t = - (1/0.5756) ln (0.001) =12 min (answer) Note: Watson’s Law: C n ×t=constant = > C×t=constant (as n=1) For 99.9% kill: C= 0.1 g/m 3 and time (t) =8 minutes. So, C×t value =(0.1 *1000mg/1000L)*(8 minutes) = 0.8 (mg/L)(min.) For 99.9% removal: Ct value= (0.05 ×1000mg/1000L)*(12 minutes) = 0.6 (mg/L) (min.) Q3. Look at the following relationship between concentration of free residual chlorine and contact time required for 99% kill (Watson’s Law: C 0.86 t p = λ (constant) for different pathogens). Pathogen type Adenovirus 3 E.coli Coxsackievirus A2 λ (constant) 0.098 0.24 6.3 For given chlorine dose, how long would you like to disinfect to achieve maximum removal of all pathogens? Explain the result. [6+4=10 points] Solution: For given chlorine dose: high contact time is required for high λ value. So we need longest contact time for Coxsackievirus A2 than that for other pathogens. Thus to achieve maximum removal of all pathogens we need contact time equal to that of Coxsackievirus A2.

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Page 1: Department of Civil Engineering-I.I.T. Delhi CEL 212 ...web.iitd.ac.in/~arunku/files/CEL795_Y13/Ex_Disinfection.pdf · Q7. Comment on dose requirements for inactivation of viruses

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Department of Civil Engineering-I.I.T. Delhi

CEL 212: Environmental Engineering

Example (Disinfection)

Q1. The Chick’s Law: For disinfection, assume N (0) is initial number of pathogens and N (t) is remaining

number of pathogens at time and given by: N (t) =N (0) × exp (-K×t); where K is disinfection rate

(unit=1/unit of time) and depends on disinfectant-pathogen interaction and solution characteristics. Here R-

log removal: R=-log10 [N (t)/N (0)].

If disinfectant concentration (Cdisinfectant) and contact time (tc) are related to each other by following equation

(the Watson’s Law):

(Cdisinfectant)n × (tc) = constant (standard unit: C in mg/L and tc in minute)

Calculate R-log removal value for 99.9% removal? What is the remaining pathogen concentration at this

removal after 1 minutes of contact time? (Assume K=0.046/min) [3+3=6 points]

Solution: R-log removal = -log10 [1-Nt/N0]

Given removal = 99.9%, so Nt/N0= (1-99.9/100) =0.001

R-log removal = -log10 [0.001] =3 (answer)

For calculating remaining fraction of pathogens after 1 minute of disinfection with K=0.046/min, use the Chick’s

Law: Nt/N0 = exp (-k ×t) = exp (-0.046/min×1 min) =0.9550 (i.e., 95.50%) (answer)

Q2. An experiment shows that a concentration of 0.1g/m3 of free available chlorine yield a 99% kill of

bacteria in 8 minutes. What contact time is required to achieve a 99.9% kill at a free available chlorine

concentration of 0.05 g/m3? Assume that Chick’s Law and Watson’s Law hold with n=1. [2+2=4 points]

Solution: Given: For 99% kill: C= 0.1 g/m3 and time (t) =8 minutes

Chick’s Law: Nt=N0×exp (-k ×t)

Calculation of disinfection rate constant:

Nt/N0= (1-99/100) =0.01 in 8 minutes

From Chick’s Law: 0.01 = exp (-k ×8)

=> k = - (1/8) ln (0.01) = 0.5756/min (answer)

Using calculated k value, calculate time for getting 99.9% kill:

Nt/N0= (1-99.9/100)=0.001

Using Chick’s Law: 0.001 = exp (-0.5756×t) =>t = - (1/0.5756) ln (0.001) =12 min (answer)

Note: Watson’s Law: Cn×t=constant = > C×t=constant (as n=1)

For 99.9% kill: C= 0.1 g/m3 and time (t) =8 minutes. So, C×t value

=(0.1 *1000mg/1000L)*(8 minutes) = 0.8 (mg/L)(min.)

For 99.9% removal: Ct value= (0.05 ×1000mg/1000L)*(12 minutes) = 0.6 (mg/L) (min.)

Q3. Look at the following relationship between concentration of free residual chlorine and contact time

required for 99% kill (Watson’s Law: C0.86

tp= λ (constant) for different pathogens).

Pathogen type Adenovirus 3 E.coli Coxsackievirus A2 λ (constant) 0.098 0.24 6.3

For given chlorine dose, how long would you like to disinfect to achieve maximum removal of all pathogens?

Explain the result. [6+4=10 points]

Solution: For given chlorine dose: high contact time is required for high λ value. So we need longest contact time

for Coxsackievirus A2 than that for other pathogens. Thus to achieve maximum removal of all pathogens we need

contact time equal to that of Coxsackievirus A2.

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Q4. Comment on decay of adenovirus (survival = Nt/N0) using low pressure (LP) and medium pressure

(MP) UV rays. Note UV dose is given in milli joules/cm2.

Solution:

Example: Log (survival)= (-4) (i.e., 4-log removal)

Log (Nt/N0)=-4; Nt=N0 (10-4) or Removal = (1-0.0001)100= 99.99%

LP UV rays produce high decay of adenovirus compared to MP UV rays. In initial regions, LP UV rays give high

disinfection rate than MP UV rays, but for high dose, both produces similar extent of virus decay.

Q5. Comment on effect of light and dark on concentration of five fecal indicators. Concentration values are

shown in box plots where middle line show median value, below 25th

percentile value (i.e., 25% of values

lower than this) and above 75th

percentile.

Solution: Dark conditions give higher CFU values, i.e., lesser removal of pathogens than that due to light conditions.

Highest removal was observed for Enterococci than other microorganisms as lowest remaining concentration.

Lowest removal was observed for F-RNA phages than other microorganisms.

Higher removal for bacterial indicator (i.e., for fecal coliforms) was observed than viral indicators (i.e., for somatic

coliphage and F-RNA phages).

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Q6. Comment on effect of UV dose on virus survival for two sets of data. Note that two linear models are fit

to data (this is first order data).

Solution: Adeno 2 virus is disinfected at slower rate than other pathogens with increasing UV dose. For high log-removal,

high UV dose is required for adeno 2 virus than other pathogens.

Both linear models appear to fit the observed decay data, however, it is difficult to comment on model fitting due

to difficulty in determining goodness-of-fit of fit models to data.

Q7. Comment on dose requirements for inactivation of viruses at different levels by UV light for each of

viruses studied.

Solution:

With high UV dose, high log-removal can be achieved.

For 99.99% removal (i.e., for 4-log removal), highest UV dose is required for adenovirus type 2 than other

pathogens with lowest dose required is for echovirus 2.

These data also indicate that adenovirus type 2 is most resistant to UV dose (i.e., hard to kill). Further, adenovirus

2 requires relatively higher UV dose than other pathogens (at least 4-5 times higher).

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+++++++++

Q1. Calculate the quantity of chlorine consumed/day for 3-log removal of rotaviruses from secondary

effluent (pH7), where free chlorine residual is 2 mg/L, the first-order decay rate of chlorine is 0.2/h and

flow rate is 1000m3/day. Given that Ct value is 243 for pH7.0. [15 points]

Solution: free residual chlorine desired at desired end = 2mg/L; chlorine decay rate =0.2/h (first order)

supplied free residual chlorine (Cresidual supplied)?

Cresidual desired =2 mg/L= Cresidual supplied *exp(-kdecay chlorine*time)

Ct=243 for 3-log removal (Cresidual chlorine=2 mg/L)

t=243/2 min=121.5 min (this is time for disinfection; this is also the time when chlorine decays in first

order)

Cresidual desired =2 mg/L= Cresidual supplied *exp(-0.2*121.5/60)

Cresidual supplied= 2.99 mg/L

Q2. Name three disinfection kinetics models generally used to model disinfection process. Can CT

concept be applied to all three kinetic models? Why or why not? [10 points]

Solution: see lecture notes

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Q3. For wastewater consists of ammonia, organic matter and microorganisms), use following

information:

Chlorine dosage (mg/L) 0.1 0.5 1.0 1.5 2.0 2.5 3.0

Chlorine residual (mg/L) 0.0 0.4 0.8 0.4 0.4 0.9 1.4

(i) Discuss the significance of different regions. [8 points]

(ii) Calculate chlorine dose to achieve 0.75 mg/L free available chlorine? [4 points]

Solution

Page 7: Department of Civil Engineering-I.I.T. Delhi CEL 212 ...web.iitd.ac.in/~arunku/files/CEL795_Y13/Ex_Disinfection.pdf · Q7. Comment on dose requirements for inactivation of viruses

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Q4. Look at the following relationship between concentration of free residual chlorine and contact time

required for 99% kill (Watson’s Law: C0.86

tp= λ (constant) for different pathogens).

Pathogen

type

Adenovirus 3 E.coli Coxsackievirus A2

λ (constant) 0.098 0.24 6.3

For given chlorine dose, how long would you like to disinfect to achieve maximum removal of all

pathogens? Explain the result. [6+4=10 points]

Solution: For given chlorine dose: high contact time is required for high λ value. So we need longest

contact time for Coxsackievirus A2 than that for other pathogens. Thus to achieve maximum removal of

all pathogens we need contact time equal to that of Coxsackievirus A2.