The CT values for inactivation of E. coli is given in the following table: Log removal CT (mg/L.min) 0.5 log 0.75 1.0 log 1.40 2.0 log 3.0 log 4.0 log 2.10 2.95 3.65 What is the required time for disinfection to achieve 99.9% removal if the concentration of disinfectant is 2 mg/L7
The CT values for inactivation of E. coli is given in the following table: Log removal CT (mg/L.min) 0.5 log 0.75 1.0 log 1.40 2.0 log 3.0 log 4.0 log 2.10 2.95 3.65 What is the required time for disinfection to achieve 99.9% removal if the concentration of disinfectant is 2 mg/L7
Chapter2: Loads On Structures
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![### CT Values for Inactivation of *E. coli*
The following table provides the CT values (mg/L·min) required for different log removals of *E. coli*:
| Log Removal | CT (mg/L·min) |
|-------------|---------------|
| 0.5 log | 0.75 |
| 1.0 log | 1.40 |
| 2.0 log | 2.10 |
| 3.0 log | 2.95 |
| 4.0 log | 3.65 |
#### Explanation
- **Log Removal**: This column represents the log reduction (a logarithmic scale used to show the relative number of live microbes eliminated). For example, 1.0 log represents a 90% reduction, 2.0 log represents a 99% reduction, and so forth.
- **CT (mg/L·min)**: This value represents the concentration of the disinfectant (mg/L) multiplied by the contact time (minutes) needed to achieve the corresponding log removal efficiency.
### Problem Statement
To determine the required time for disinfection to achieve 99.9% removal of *E. coli* if the concentration of disinfectant is 2 mg/L.
#### Calculation:
1. **99.9% removal** corresponds to a 3.0 log reduction.
2. From the table, the CT value for 3.0 log removal is 2.95 mg/L·min.
3. Given that the concentration \( C \) of the disinfectant is 2 mg/L, we need to find the required contact time \( T \).
\[ CT = C \times T \]
\[ 2.95 \, \text{mg/L·min} = 2 \, \text{mg/L} \times T \]
\[ T = \frac{2.95}{2} \]
\[ T = 1.475 \, \text{minutes} \]
So, the required time for disinfection to achieve 99.9% removal of *E. coli* with a disinfectant concentration of 2 mg/L is approximately **1.475 minutes**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2b7dd8f1-dbab-4607-8b33-bde9974f18c9%2F2cdc6b59-54e3-4e6e-aab6-956a1b1fd87c%2Fmnj9qgs_processed.png&w=3840&q=75)
Transcribed Image Text:### CT Values for Inactivation of *E. coli*
The following table provides the CT values (mg/L·min) required for different log removals of *E. coli*:
| Log Removal | CT (mg/L·min) |
|-------------|---------------|
| 0.5 log | 0.75 |
| 1.0 log | 1.40 |
| 2.0 log | 2.10 |
| 3.0 log | 2.95 |
| 4.0 log | 3.65 |
#### Explanation
- **Log Removal**: This column represents the log reduction (a logarithmic scale used to show the relative number of live microbes eliminated). For example, 1.0 log represents a 90% reduction, 2.0 log represents a 99% reduction, and so forth.
- **CT (mg/L·min)**: This value represents the concentration of the disinfectant (mg/L) multiplied by the contact time (minutes) needed to achieve the corresponding log removal efficiency.
### Problem Statement
To determine the required time for disinfection to achieve 99.9% removal of *E. coli* if the concentration of disinfectant is 2 mg/L.
#### Calculation:
1. **99.9% removal** corresponds to a 3.0 log reduction.
2. From the table, the CT value for 3.0 log removal is 2.95 mg/L·min.
3. Given that the concentration \( C \) of the disinfectant is 2 mg/L, we need to find the required contact time \( T \).
\[ CT = C \times T \]
\[ 2.95 \, \text{mg/L·min} = 2 \, \text{mg/L} \times T \]
\[ T = \frac{2.95}{2} \]
\[ T = 1.475 \, \text{minutes} \]
So, the required time for disinfection to achieve 99.9% removal of *E. coli* with a disinfectant concentration of 2 mg/L is approximately **1.475 minutes**.
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