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Advanced Waterworks Mathematics, 2019a

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Location and Type of<br />

Disinfection<br />

Treatment Plant<br />

Pipeline<br />

Actual CT Required CT CT Ratio<br />

3.0 mg/L · min<br />

712 mg/L · min<br />

Treatment Plant Actual CT:<br />

1.2 mg/L 35 min 42 mg/L min<br />

Pipeline Actual CT:<br />

0.4 mg/L 12 min 4.8 mg/L min<br />

Now you can finish populating the table and calculating the CT Ratios.<br />

Actual CT 42 mg/L min<br />

= 14<br />

Required CT 3.0 mg/L min <br />

Actual CT 4.8 mg/L min<br />

= 0.00674 = 0.0<br />

Required CT 712 mg/L min <br />

Location and Type of<br />

Disinfection<br />

Actual CT Required CT CT Ratio<br />

Treatment Plant 42 mg/L · min 3.0 mg/L · min 14<br />

Pipeline 4.8 mg/L · min 712 mg/L · min 0.0<br />

Total: 14.0<br />

Yes! This plant does meets compliance for CT inactivation of viruses.<br />

8. Does a water utility meet CT for viruses by disinfection if only the free chlorine<br />

concentration is 0.5 ppm through 200 ft of 24” diameter pipe at a flow rate of 730 gpm?<br />

Assume the water is 15°C and has a pH of 8.0.<br />

Pipeline:<br />

From Table 7.2 Treatment Credits and Log Inactivation Requirements you can see that:<br />

4 Log (Required) Pipeline Only – No credits<br />

Now look at Table C-7 Inactivation of Viruses by Free Chlorine. The solution is where the<br />

15°C column intersects with the 4 Log Inactivation row.<br />

4.0 mg/L min CT is required.<br />

333 | A dvanced <strong>Waterworks</strong> <strong>Mathematics</strong>

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