# Water Morning Civil



## playboyman007 (Oct 9, 2010)

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.


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## pmblair (Oct 9, 2010)

Me too.

They barely touch on them. I am hoping its just going to be a theroy question(s), as environmental now has its own test.


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## PE_2009 (Oct 9, 2010)

playboyman007 said:


> 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
> 
> ...


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


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## pmblair (Oct 11, 2010)

Nice, thank you.

These examples look good.


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## Ambrug20 (Oct 14, 2010)

pmblair said:


> Nice, thank you.
> These examples look good.


And they are not :tardbang: "difficult" at all. Unfortunaly to me, it like a chinees' alphabet. :screwloose:


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## EnvEngineer (Oct 15, 2010)

D. Wastewater Treatment

1. Collection systems (e.g., lift stations, sewer networks, infiltration, inflow)

E. Water Treatment

1. Hydraulic loading

2. Distribution systems

My experience is they are mostly referring to the piping of these systems, sewer lines, flow type, slope and other design features. many of these topics are covered better in the water resources section of CERM rather than the environmental.


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