Wastewater Operator Biological Treatment Processes Questions and Answers — Questions and Answers
Question 1: An operator at an activated sludge plant notices the sludge in the secondary clarifier is settling poorly, has a high Sludge Volume Index (SVI), and appears light and fluffy. Microscopic examination reveals an excessive growth of long, thread-like bacteria. Which of the following is the MOST likely cause of this condition, known as filamentous bulking?
- Low Food-to-Microorganism (F/M) ratio and low dissolved oxygen. (Correct answer)
- High Mean Cell Residence Time (MCRT) and high pH.
- High wasting rates and excessive aeration.
- Low influent Biochemical Oxygen Demand (BOD) and high alkalinity.
Correct answer: Low Food-to-Microorganism (F/M) ratio and low dissolved oxygen.
Filamentous bulking is often caused by conditions that favor the growth of filamentous bacteria over floc-forming bacteria. Key causes include low dissolved oxygen (less than 2.0 mg/L), a low Food-to-Microorganism (F/M) ratio, nutrient deficiency, or low pH. These conditions give filamentous organisms a competitive advantage, leading to poor sludge settling and compaction.
Question 2: Which of the following conditions is essential for the denitrification process to occur effectively in an anoxic zone?
- High dissolved oxygen (DO) and a readily available carbon source.
- Low pH, high temperature, and the presence of ammonia.
- Presence of nitrate, an absence of free oxygen, and a sufficient carbon source. (Correct answer)
- High alkalinity, presence of nitrites, and vigorous mixing.
Correct answer: Presence of nitrate, an absence of free oxygen, and a sufficient carbon source.
Denitrification is an anoxic process where facultative bacteria convert nitrate (NO3) to nitrogen gas (N2). This requires three key conditions: the presence of nitrate as an electron acceptor, the absence of free molecular oxygen (anoxic conditions), and a source of readily biodegradable organic carbon (food for the bacteria).
Question 3: A Sequencing Batch Reactor (SBR) is a unique form of the activated sludge process. Which phase of the SBR cycle involves adding air to promote the biological breakdown of organic matter and the conversion of ammonia to nitrate?
- Settle
- React (Correct answer)
- Decant
- Fill
Correct answer: React
The 'React' phase is the primary aeration stage in an SBR cycle. During this phase, oxygen is supplied to the mixed liquor, allowing aerobic bacteria to consume BOD and convert ammonia to nitrate (nitrification). This is the main biological treatment step.
Question 4: During the nitrification process, ammonia is converted to nitrate. This biological process has a significant impact on the wastewater chemistry. Which of the following is a key requirement and consequence of nitrification?
- It generates a large amount of alkalinity.
- It consumes a significant amount of alkalinity. (Correct answer)
- It requires anaerobic conditions to proceed.
- It significantly increases the wastewater's pH.
Correct answer: It consumes a significant amount of alkalinity.
The nitrification process consumes approximately 7.14 parts of alkalinity (as CaCO3) for every one part of ammonia-nitrogen oxidized. This consumption of alkalinity can lead to a drop in pH if not buffered adequately. Therefore, maintaining sufficient alkalinity is crucial for a stable nitrification process.
Question 5: An operator of an anaerobic digester is monitoring process stability. Which parameter provides the EARLIEST indication of a potential process upset, often referred to as the digester 'going sour'?
- A rapid drop in pH.
- An increase in methane gas production.
- A decrease in sludge temperature.
- An increasing volatile acid to alkalinity ratio. (Correct answer)
Correct answer: An increasing volatile acid to alkalinity ratio.
The volatile acid to alkalinity ratio is the most sensitive indicator of an impending upset in an anaerobic digester. An increase in this ratio shows that volatile acids are being produced faster than they are being converted to methane, consuming the system's buffering capacity (alkalinity). This change occurs well before a significant drop in pH, providing an early warning for operators to take corrective action.
Question 6: A wastewater treatment plant is operating a conventional activated sludge process. If the Food-to-Microorganism (F/M) ratio is consistently too high, what is the most likely operational consequence?
- The microorganisms will be starved, leading to poor floc formation.
- Sludge bulking due to the growth of old, slow-growing filaments.
- Incomplete BOD removal and a turbid, poorly settling effluent. (Correct answer)
- A very stable process with high-quality effluent and low sludge production.
Correct answer: Incomplete BOD removal and a turbid, poorly settling effluent.
A high F/M ratio means there is an excess of food (BOD) for the available population of microorganisms. The microorganisms cannot consume all the available food in the given detention time. This leads to under-treatment, resulting in incomplete BOD removal and a high concentration of suspended solids in the effluent, which appears turbid and does not settle well.
Wastewater Operator Biological Treatment Processes Questions and Answers
An operator at an activated sludge plant notices the sludge in the secondary clarifier is settling poorly, has a high Sludge Volume Index (SVI), and appears light and fluffy.
Microscopic examination reveals an excessive growth of long, thread-like bacteria.
Which of the following is the MOST likely cause of this condition, known as filamentous bulking?