Wastewater Calculations - Mathematics and Statistics for the Environment - page 745

■ EXAMPLE 24.139
Problem: A BOD 5 test has been completed. Bottle 1 of the test had dissolved oxygen (DO) of 7.1
mg/L at the start of the test. After 5 days, bottle 1 had a DO of 2.9 mg/L. Bottle 1 contained 120
mg/L of sample. Determine the unseeded BOD 5 .
Solution:

( ) ×

7.1 mg/L − 2. mg/L 300 mL

BOD unseeded) = = 10.5 mg/L

5

120 mL

24.15.3.2 BOD 5 (Seeded)
If the BOD 5 sample has been exposed to conditions that could reduce the number of healthy, active
organisms, the sample must be seeded with organisms. Seeding requires the use of a correction fac-
tor to remove the BOD 5 contribution of the seed material:

Seed material BOD eed in × dilution (mL)

5

(24.173)

Seed correction =

300 mL

( )

DO (mg/L) − DO (mg/L) − S eed correction × 300 mL

start final

BOD seeded) = (24.174)

5

Sample volume( mL)

■ EXAMPLE 24.140
Problem: Using the data provided below, determine the BOD 5 :
BOD 5 of seed material = 90 mg/L
Seed material = 3 mL
Sample = 100 mL
Start DO = 7.6 mg/L
Final DO = 2.7 mg/L

Solution:

90 mg/L 3mL
300 mL

×

Seed correction = = 0.90 mg/ L

( ) ×

76 . − − 30 0

mg/L 2.7 mg/L 0.90

BOD seeded) = = 12 mg/L

5

Sample volume(mL)

24.15.3.3 BOD 7-Day Moving Average
Because the BOD characteristic of wastewater varies from day to day, even hour to hour, operational
control of the treatment system is most often accomplished based on trends in data rather than indi-
vidual data points. The BOD 7-day moving average is a calculation of the BOD trend.

Key Point: The 7-day moving average is called a moving average because a new average is calcu-
lated each day by adding the new day's value to the six previous days' values.

BOD + OD + BODB + OD

day1 day2 day3 day4

+ BOD + OD + BOD

day5 day6 day7

7-dayaverage BOD = (24.175)

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