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The operator of an extended-aeration activated sludge treatment facility running at a 10-day solids retention time (SRT) completes a monthly microscopic examination of a sample of mixed liquor suspended solids (MLSS) from the effluent end of the last aeration tank before the clarifiers. He looks at three slides and documents the data in the table below. Based on all the information provided, what is the best assumption about the current plant operating condition?

Organism Slide 1 Slide 2 Slide 3 Total Average
Amoeba, blobs 15 18 27 60 20
Amoeba, shelled 1 0 1 2 0.7
Flagellates 21 18 23 62 21
Free-swimming ciliates 15 14 10 39 13
Stalked ciliates 3 4 1 8 2.7
Rotifers 0 2 0 2 0.7
Worms 0 0 0 0 0
Other 0 0 0 0 0

A. The facility is running optimally and no changes need to be made.

B. The MLSS and SRT are too high and the waste rate should be increased.

C. The food/microorganism (F/M) ratio is too low and the waste rate should be increased.

D. The MLSS and SRT are too low and the waste rate should be decreased.

Answer: D. Remember that the question asks you to base your answer on all the information described, including the stem of the question, which states that the extended aeration plant is operating at a 10-day SRT. This is on the low side of the normal operating range for extended aeration, which is 20 to 30 days with an F/M ratio from 0.05 to 0.15 pounds of BOD per pound of mixed liquor volatile suspended solids (MLVSS) per day. Typical MLSS values range from 2,000 to 6,000 mg/L.

A microscopic examination of typical extended aeration MLSS will normally reveal a predominance of rotifers, nematodes (roundworms) and stalked ciliates, with some free-swimming ciliates, but few (if any) flagellates and blob (proteus) amoeba. There may be many shelled (testate) amoeba observed, and they resemble brown doughnuts. The sample results shown are more representative of a young sludge age commonly found in conventional or contact-stabilization activated sludge facilities. The operator will need to decrease the waste rate to build  MLSS up, which raises the SRT and decreases the F/M ratio. On a side note, a microscopic exam of MLSS is recommended more often than once per month. A few times per week is more acceptable.


An 18-inch water transmission main is 1 mile long. If the velocity in the main is 4 feet per second, what is the flow rate in gallons per minute (gpm)? Select the closest answer.

A. 52.9 gpm
B. 708.0 gpm
C. 3,177.6 gpm
D. 5,296.4 gpm

Answer: C. Before reviewing the steps used to arrive at the answer, it is important to understand a few things about solving math questions. There is usually more than one method to arrive at a correct answer. In solving this word problem, we used the standard rounding method: for anything over 5, round the preceding digit to next highest value, and round out to the hundredth digit (two values to the right of the decimal place — x.xx). It is also valuable to identify how the distractors (“fake answers”) are generated in many exam situations. As you work the math, you might recognize some of the numbers used to make the incorrect choices, but also recognize that the units shown with those choices are not correct. For example, answer B, 708.0 gpm, is actually the flow rate as cubic feet per second with the decimal in the wrong space. Watch out for traps along the way to solving the question that might draw you into picking an incorrect answer.

Calculation steps:

Use the formula shown, as adapted from State of Florida Drinking Water Operator and Water Distribution System Operator Math Formula Sheets:

Flow Rate, gpm = (Area, sq. ft.)(Velocity, ft/sec)(7.48 gal/cu ft)(60 sec/min) or Q = V x A x 7.48 x 60

  • Convert diameter (D) inches to feet: 18-inch pipe opening ÷ 12 inches/ft = 1.5 ft.
  • Calculate the cross-section area, in square feet, of the pipe: A = 0.785 x D x D.  A = 0.785 x 1.5 x 1.5.  Area = 1.77 sq. ft.

Next, calculate the cubic foot volume of just the 4-foot length given as velocity, ft/sec. The total pipe length of 1 mile is not needed. This question is about the how fast the 4-foot section of water in the pipe is moving, not the total gallons.

Volume, V = A x length. V = 1.77 sq. ft. x 4 . Volume = 7.08 cu ft. Note that this is actually 7.08 cubic feet per second (cfs) because the 4-foot number used was the velocity as 4 feet per second. The second units remain with the value as 7.08 cfs.

Now, calculate the flow rate in gallons per second (gps) using the conversion value of 7.48 gal/cu.ft. Gallons/second = 7.08 cfs x 7.48 gal/cu.ft. Flow rate = 52.96 gps.

Lastly, convert to gpm by multiplying the gps by 60 seconds per minute: 52.96 gps x 60 sec/min = 3,177.6 gpm.

That’s a lot of steps for one answer that may be only worth one point on an exam — but it may be the one extra point that rewards you with a 70 percent passing score. If you struggle with math, the best advice is to practice, practice, and practice some more. Use the steps above, but change the initial numbers a bit to make it a new question (pipe diameter, velocity, etc.) and then work it out again. State exams usually have math formula sheets, so practice with them to help you gain confidence in your math skills. You can do it!

About the author

Ron Trygar, a certified environmental trainer, is the senior training specialist for water and wastewater programs at the University of Florida’s TREEO Center. He has worked in the wastewater industry for more than 30 years in a variety of locations and positions. He holds a Florida Class A wastewater treatment operator license and a Florida Class B drinking water operator license.


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