Question:
Which of the following components is/are NOT removed in the secondary treatment of sewage?
Which of the following components is/are NOT removed in the secondary treatment of sewage?
Show Hint
Secondary treatment is the biological stage; it targets dissolved/colloidal organic matter, not settleable solids, floatables, or pathogens.
Updated On: Jul 17, 2026
- Suspended settleable organic solids
- Colloids including dissolved organic matter
- Pathogens
- Fat and grease
Show Solution
Verified By Collegedunia
The Correct Option is A, C, D
Solution and Explanation
Step 1: Understanding the Question.
Sewage treatment is carried out in stages, primary, secondary, and tertiary/disinfection, each targeting a different class of pollutant. We need to identify which of the listed components are NOT the ones targeted (removed) specifically by the secondary treatment stage.
Step 2: Key Formula or Approach.
Primary treatment (screening, grit removal, primary sedimentation, and skimming) removes coarse and settleable solids and floating matter like fat, oil, and grease by physical means (settling and skimming). Secondary treatment is a biological process (activated sludge, trickling filter, or similar), where microorganisms consume the colloidal and dissolved organic matter left in the wastewater after primary treatment, converting it into biomass and stable end products; this is the stage that reduces BOD from dissolved/colloidal organics. Tertiary treatment and disinfection (chlorination, UV, ozonation) are the stages that specifically target pathogens (disease-causing microorganisms) and any remaining nutrients or fine solids.
Step 3: Detailed Explanation.
Suspended settleable organic solids: these are removed mainly in primary sedimentation, by gravity settling, before the flow even reaches the secondary stage; so secondary treatment is not the stage that removes them. Option (A) is a component NOT removed by secondary treatment.
Colloids including dissolved organic matter: these are too fine to settle out and are exactly what the biological secondary stage is designed for, the microorganisms break down and remove this dissolved and colloidal organic load. So option (B) IS removed by secondary treatment; it is correctly excluded from our answer.
Pathogens: secondary biological treatment reduces pathogen numbers somewhat as a side effect, but is not designed to reliably remove them, disinfection (chlorination/UV) is the dedicated stage for pathogen removal. So option (C) is NOT removed (reliably) by secondary treatment.
Fat and grease: these are floatable materials removed by skimming in primary treatment (and sometimes in preliminary grease traps), not by the biological secondary process. So option (D) is NOT removed by secondary treatment.
Step 4: Final Answer.
The components NOT removed by secondary treatment are suspended settleable organic solids, pathogens, and fat and grease.
\[ \boxed{\text{Options (A), (C), (D)}} \]
Sewage treatment is carried out in stages, primary, secondary, and tertiary/disinfection, each targeting a different class of pollutant. We need to identify which of the listed components are NOT the ones targeted (removed) specifically by the secondary treatment stage.
Step 2: Key Formula or Approach.
Primary treatment (screening, grit removal, primary sedimentation, and skimming) removes coarse and settleable solids and floating matter like fat, oil, and grease by physical means (settling and skimming). Secondary treatment is a biological process (activated sludge, trickling filter, or similar), where microorganisms consume the colloidal and dissolved organic matter left in the wastewater after primary treatment, converting it into biomass and stable end products; this is the stage that reduces BOD from dissolved/colloidal organics. Tertiary treatment and disinfection (chlorination, UV, ozonation) are the stages that specifically target pathogens (disease-causing microorganisms) and any remaining nutrients or fine solids.
Step 3: Detailed Explanation.
Suspended settleable organic solids: these are removed mainly in primary sedimentation, by gravity settling, before the flow even reaches the secondary stage; so secondary treatment is not the stage that removes them. Option (A) is a component NOT removed by secondary treatment.
Colloids including dissolved organic matter: these are too fine to settle out and are exactly what the biological secondary stage is designed for, the microorganisms break down and remove this dissolved and colloidal organic load. So option (B) IS removed by secondary treatment; it is correctly excluded from our answer.
Pathogens: secondary biological treatment reduces pathogen numbers somewhat as a side effect, but is not designed to reliably remove them, disinfection (chlorination/UV) is the dedicated stage for pathogen removal. So option (C) is NOT removed (reliably) by secondary treatment.
Fat and grease: these are floatable materials removed by skimming in primary treatment (and sometimes in preliminary grease traps), not by the biological secondary process. So option (D) is NOT removed by secondary treatment.
Step 4: Final Answer.
The components NOT removed by secondary treatment are suspended settleable organic solids, pathogens, and fat and grease.
\[ \boxed{\text{Options (A), (C), (D)}} \]
Was this answer helpful?
0
0
Top GATE CE Environmental Engineering Questions
- The number of degrees of freedom for a natural open channel flow with a mobile bed is
- The following table gives various components of Municipal Solid Waste (MSW) and a list of treatment/separation techniques.
Component of MSW Treatment/separation technique P - Ferrous metals i - Incineration Q - Aluminum and copper ii - Rapid composting R- Food waste iii - Eddy current separator S-Cardboard iv - Magnetic separator
The CORRECT match is - The primary air pollutant(s) is/are
- Activated carbon is used to remove a pollutant from wastewater in a mixed batch reactor, which follows first-order reaction kinetics.
At a reaction rate of 0.38 /day, the time (in days) required to remove the pollutant by 95% is ______ (rounded off to 1 decimal place). - A water treatment plant treats 25 MLD water with a natural alkalinity of 4.0 mg/L (as CaCO3). It is estimated that, during coagulation of this water, 450 kg/day of calcium bicarbonate (Ca(HCO3)2) is required based on the alum dosage.
Consider the atomic weights as: Ca-40, H-1, C-12, O-16.
The quantity of pure quick lime, CaO (in kg) required for this process per day is _____ (rounded off to 2 decimal places).
View More Questions
Top GATE CE Sewage Treatment Questions
- A circular settling tank is to be designed for primary treatment of sewage at a flow rate of 10 million liters/day. Assume a detention period of 2.0 hours and surface loading rate of 40000 liters/m2/day. The height (in meters) of the water column in the tank is __________ (rounded off to 2 decimal places).
The wastewater inflow to an activated sludge plant is $0.5\, \text{m}^3/\text{s}$, and the plant is to be operated with a food to microorganism ratio of $0.2\, \text{mg/mg-d}$. The concentration of influent biodegradable organic matter (after primary settling) is $150\, \text{mg/L}$, and the mixed liquor volatile suspended solids (MLVSS) to be maintained is $2000\, \text{mg/L}$. Assuming complete removal of biodegradable organics in the tank, the volume of aeration tank (in m$^3$, integer) required is.
- The grown algae are useful in the operation of
Top GATE CE Questions
- If \(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}....\frac{1}{29} =\frac{N}{29!}\) Find the remainder when N is divided by 19.
- What will be the relative density?
M = 0.5 percentage of air voids
γmin = 1.3 gm/cc
γmax = 1.8 gm/cc - Find RL of R?
RL of P [BM = 200.000Point Staff Readings
Back sideStaff Readings
Fore sideRemarks P -2.050 - 200.000 Q 1.050 0.95 Change Point R - -1.655 - - What will the angle of the failure plane from horizontal for sand be?
σ3=300
σ1=300+100=400 - Which of the following having negative skamptam Pore water coefficient?
View More Questions