Looking at the NCEES Exam Specifications, I'm having a little difficulty at locating the subjects below within the CERM. Can anyone identify the specific sections in the CERM I should focus for the subjects below?
D. Wastewater Treatment
1. Collection systems (e.g., lift stations, sewer networks, infiltration, inflow)
E. Water Treatment
1. Hydraulic loading
2. Distribution systems
Much Appreciated.
Problems are not difficult. See some sample problems below
1. Two lagoons 700 ft by 500 ft, operated in parallel, are receiving an
organic load of 203.8 mg/l of BOD5 from a community of 1700 people.
What is the organic load, in lbs of BOD5 per acre per day?
Organic load/lbs/BOD5/Day/ Acre
Lbs/BOD5/day
Area/Acres
Lbs BOD5 = Flow /mgd X 8.34 X mg/l/BOD5
Flow = 1,700 pop X 100 gal/day/person =170,000 gpd
Convert 170,000 to mgd = 17 mgd
0.17 mgd X 8.34 X 203.8 mg/l = 288.9 lbs/day
Area in acres
Length ft X width ft X 2 lagoons/
43,560 ft2/ Acre
700 ft X 500 ft X
= 700,000 ft2/
43,560 ft2
= 16 Acres
288.9 lbs BOD5/day/
16 Acres
= 18.05 lbs BOD5 / Acre/day
2. Calculate the organic loading on an extended aeration treatment
plant with a flow of 0.2 MGD with a population of 1250 (assume 0.17
lb BOD5/person/day).
Organic loading formula:
Lbs of BOD5/ day/
Aeration tank volume in 1000 ft3
Lbs BOD5 formula:
Population x 0.17 lb/BOD5/day
=1250 people X 0.17 lb BOD5/capita =212.5 lbs
AT volume formula:
AT volume in gallons/
Gallons per ft3 (7.48)
Convert 0.2 mgd to 200,000 gpd
200,000 gpd/
7.48
= 26,738 ft3
Convert bottom of AT volume formula to 1000 ft3 by dividing by 1000
(the 1000 ft3 is a unit of measure in this calculation).
26,738 ft3/
1000 ft3
= 26.738, 1000 ft3
212.5/
26.738,1000 ft3
= 7.94 lbs BOD5,1000 ft3
