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Example A city of 200,000 people discharges 37.0 cfs of treated sewage having an ultimate BOD of 28.0 mg/L and 1.8 mg/L DO into a river with a flow of.

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Presentation on theme: "Example A city of 200,000 people discharges 37.0 cfs of treated sewage having an ultimate BOD of 28.0 mg/L and 1.8 mg/L DO into a river with a flow of."— Presentation transcript:

1 Example A city of 200,000 people discharges 37.0 cfs of treated sewage having an ultimate BOD of 28.0 mg/L and 1.8 mg/L DO into a river with a flow of 250 cfs and velocity of 1.2 ft/sec. Upstream of the discharge point, the river has a BOD of 3.6 mg/L and a DO of 7.6 mg/L. The saturation DO is 8.5 mg/L, kd = 0.61 day-1, and kr = 0.76 day-1. Determine a) the critical DO and critical distance, and b) the DO at 10 miles downstream.

2 Determine Initial Conditions
Initial dissolved oxygen concentration Initial dissolved oxygen deficit

3 1. Determine Initial Conditions
Initial ultimate BOD concentration

4 Step 1. Variations Flow given in cfs, not m3/s – does not matter for mixing calculations Saturation DO given – no need to look up Ultimate BOD given – no need to calculate from BOD5

5 2. Determine Reaeration Rate
kr = 0.76 day-1 given no need to calculate from stream geometry assume given value is at the stream temperature (since not otherwise specified), so no need to correct

6 3. Determine the Deoxygenation Rate
kd = 0.61 day-1 given no need to calculate corrections from stream geometry assume given value is at the stream temperature (since not otherwise specified), so no need to correct

7 5. Calculate Critical time and DO

8 5. Calculate Critical time and DO
Problem asked for critical distance

9 5. Calculate Critical time and DO
Minimum DO

10 5. Calculate Critical time and DO

11 4. DO as function of time (at 10 miles)

12 4. DO as function of time (at 10 miles)


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