DEPARTMENT OF CIVIL & ENVIRONMENTAL ENGINEERING; LAFAYETTE COLLEGE
Homework Assignment 5 SOLUTIONS
CE 321, Fall 2008
1. Describe a set of natural conditions (i.e. no wastewater discharge) for which would you expect
to find low DO in a stream?
High levels of natural organic matter (leaves, goose poop, deer poop) , dead/decomposing plant matter, warm
water (low solubility), high salinity or dissolved solids (reduces solubility of O2), deep water (long time to
reach equilibrium with atmosphere), slow-moving water (little turbulence for aeration), dark colored water
(no sunlight penetration for plants/algae, which produce O2)
2. To meet discharge permit requirements, wastewater containing a BOD5 of 300 mg/L must be
treated to have BOD5 less than 30 mg/L (i.e. 90% removal). A sample of treated waste has initial
DO of 8.0 mg/L and final DO of 4.4 mg/L, using a 25-mL sample and a 300-mL BOD bottle,
and 5-day test.
(a) Does this sample meet the discharge requirement?
BOD5 = (DOi - DOf)/DF = (8 mg/L - 4.4 mg/L)/(25 mL/300 mL) = 43.2 mg/L (doesn’t meet standard)
(b) If the ultimate BOD for the sample is 50 mg/L, and the test is continued for 5 more days,
what will be the final reading for DO and the BOD10?
Using the BOD equation, BODt = BODult (1 - e-k t), we can solve for k:
k = -ln(1 - BOD5/BODult)/5 days = -ln(1 - 43.2 mg/L/50 mg/L)/5 days = 0.4 day-1
Now we can find BOD10:
BOD10 = 50 mg/L (1 - e-0.4 day-1 10 day) = 49.1 mg/L
Now we can find DOf:
BOD10 = 49.1 mg/L = (DOi - DOf)/DF = (8 mg/L - x mg/L)/(25 mL/300 mL)
x = 3.9 mg/L
3. The following figure shows a plot of BOD remaining versus time for a sample of the effluent
taken from a wastewater treatment plant. Assume the data is asymptotic to 5 mg/L.
(a) What is the ultimate BOD (Lo)? 35 mg/L if you assume asymp to 5 mg/L, or 40 mg/L if not
(b) What is the five-day BOD? 40 – 15 = 25 mg/L
(c) What is Lt for 7 days? Read it right off the graph ~10 mg/L
