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Environment Problem - Morning Exam

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I have not seen a single environment problem in NCEES practice exam for morning portion.... Please correct me if I am wrong..

I spent whole day studying environment and now want to kick myself......

What kind of problems can come in morning exam?

Thanks...

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 passpe

I can tell you that last October's PE exam had an environmental question, and I believe that Problem 140 on the practice exam is an environmental question (water supply and distribution).

Edited by jharris

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you are correct. But do we get eviron problems.. like calculation oxygen requirements, pH value, BOD or fire water demand in exam?

Although, I have studied but I am not sure wt to expect in morning....

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Just look at the NCEES exam specs.

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How is fire water demand an envl problem? :blink:

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The NCEES specs (for 2012), currently state the two sections on environmental are:

D. Wastewater treatment;

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

E. Water Treatment

1. Hydraulic loading

2. Distribution systems

It seems to me that the 'environmental' portion of the breadth exam is, for the most part, an extension of hydraulics (closed conduit and open channel). i.e. Water distribution systems will be closed conduit hydraulics, sewer networks is mostly open channel flow, infiltration and inflow is a bit of hydrology, and a lift station is just a type of pump. The question one I see that is truly unique here is the hydraulic loading (which is the quantity of water - or wastewater- a plant takes on). Of course, there will be industry specific jargon that may help feel more comfortable with these problems, and some of the questions may be 'definition-type' questions.

It would be worth while to do some reading/ skimming on a few environmental chapters to get a good feel for the concepts and know where things are for these two sub-sections. It would make sense to me to work the problems in these chapters that are primarily associated with the "hydraulics" aspect of water and sewer networks, as well as the hydraulic loading component. I would also review concepts such as water storage and demand, and fire flow as these are quite important concepts that are tied to the idea of a distribution system.

I do not think there will be a whole lot of actual treatment questions, but there may be a simple one or two on the morning exam. The NCEES specs does make a disclaimer that 'these specifications are not exhaustive or exclusive..." (a clause we all just love). The point is that you will need to expect a couple "curveballs" on the exam, and a few questions will appear that seemingly did not meet the exam specs. Do not spend too much time preparing for these outlier questions, but be ready to know where to look to solve them (if you have the time....)

Do you agree with this approach to preparing for these two Sub-Sections of the Civil PE Exam??

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1. Calculate the BOD loading (lbs/day) on a pond if the influent flow is 390,000 gpd with a BOD of 245 mg/L.

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CONCENTRATIONS

AND SOLUTIONS

1 mg/l = 1 ppm

1 mg/l = 8.34 Lbs/ Million Gal

Lbs, Chemical = Desired ppm X Flow, MGD X 8.34/Purity of Chemical

1 Mole = 1 Gram Molecular Weight per Liter of Solution

1 Normal = 1 Gram Equivalent Weight per Liter of Solution

ppm =Lbs Chemical Fed /MGD X 8.34

1 % Solution = 10,000 mg/l

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1. Calculate the BOD loading (lbs/day) on a pond if the influent flow is 390,000 gpd with a BOD of 245 mg/L.

No fancy formulas needed, just basic unit conversion until the units cancel. Work mg into lbs and equate gallons with liters.

390,000 gal/d * 3.785 L/gal = 1,476,150 L/day

1,476,150 L/day * 245 mg/L = 3,616,567 mg/day

3,616,567 mg/day = 3.617 kg/day

3.617 kg/day * 2.2 lb/kg = 7.96 lb/day --> Say 8 lb/day

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1. Calculate the BOD loading (lbs/day) on a pond if the influent flow is 390,000 gpd with a BOD of 245 mg/L.

No fancy formulas needed, just basic unit conversion until the units cancel. Work mg into lbs and equate gallons with liters.

390,000 gal/d * 3.785 L/gal = 1,476,150 L/day

1,476,150 L/day * 245 mg/L = 3,616,567 mg/day This should be 361.6 x 106 (off by a factor of 100)

3,616,567 mg/day = 3.617 kg/day This should be 361.6 kg/day (off by a factor of 100)

3.617 kg/day * 2.2 lb/kg = 7.96 lb/day --> Say 8 lb/day This should be 800 lb/day (off by a factor of 100)

Also, since this particular combination occurs frequently, it i worthwhile remembering:

Concentration (mg/L) x Flow rate (MGD) x 8.3454 = Load (lb/day)

245 x 0.39 x 8.3454 = 797 lb/day (instead of going step by step)

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That's what I get for using a crappy 4 function calculator, it just showed the numbers it could on the screen and chopped off the rest.

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A wastewater treatment plant has an average flow of 12 M-gal/day and a peak to average

flow ratio of 2.5. Determine the total volume of an aerated grit chamber if detention time

is 3 min.

a. 8316 cu.ft.

b. 4500 cu.ft.

c. 6127 cu.ft.

d. 5512 cu.ft.

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This is a very simple problem. If you are stumbling, I suggest you brush up in basic water/wastewater concepts.

Vol = Qpeak * t = 8,300 ft^3

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This is a very simple problem. If you are stumbling, I suggest you brush up in basic water/wastewater concepts.

Vol = Qpeak * t = 8,300 ft^3

An important question is whether the flow goes through any kind of flow equalization before getting to the grit chamber (this is typically the case). In that case, using the average flow rate is more appropriate than the peak. The peak flow rate is more appropriate for sizing components that are further upstream in the system, such as sewer lines

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Au contraire, my friend. Headworks processes, such as grit chambers and screens, should be sized to handle peak flows that can get to the plant. This is standard practice.

Not many plants have flow equalization facilities either, as that is not the typical case at all. In those cases where there are equalization basins, they are usually downstream of the headworks, to avoid having rags, grit, etc., in the basin.

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That didn't come out right at all!

I meant to say that, in my experience, most wastewater treatment plants lack equalization basins and when they do have them, they are usually downstream of the plant's headworks, to remove the grit and screenings from the influent to the equalization basin.

If the plant does not have flow equalization, the headworks facilities (grit chambers and screens) must be sized to handle all the flow that can get to the plant, which means using peak flow rates.

There, that's better.

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