EXAMPLE: A 6.0-ml sample of wastewater is diluted to 300 mL with distilled water in a standard BOD bottle. The initial DO in the bottle is determined to be 8.5 mg/L, and the DO after 5d at 20oC is found to be 5.0 mg/L.
Determine the BOD5 of the wastewater and compute its BODL. Assume that k = 0.1/d
BOD5 = ((8.5-5.0) X 300)/ 6.0 = 3.5 X 300/6.0 = 180 mg/L
Now applying BODt = BODL X (1 – 10-kt)
180 = BODL X (1 – 10-0.1X5) and
BODL = 180/0.684 = 260mg/L