Chemical Engineering - MCQ Practice Questions

Understanding:

We need to identify the typical BOD removal efficiency of a conventional activated sludge (CAS) process used in municipal wastewater treatment.

Step 1: Recall process performance

The conventional activated sludge process is a secondary biological treatment method. It uses a combination of an aeration tank (where microorganisms oxidise organic matter) followed by a secondary clarifier.

Step 2: Evaluate each range

Primary treatment alone achieves 30–50% BOD removal. Extended aeration or trickling filters achieve 55–70%. Conventional activated sludge consistently achieves 85–95% BOD removal, which is the well-established design benchmark used in environmental engineering practice. Values of 98–99.9% are associated with tertiary/advanced treatment processes such as membrane bioreactors (MBR) or additional polishing steps.

Answer:

The conventional activated sludge process typically achieves a BOD removal efficiency of 85–95%.

Quick Tip:

For exam purposes, associate secondary biological treatment (activated sludge) with ~90% BOD removal. Primary sedimentation alone gives only ~30–40% BOD removal.

Understanding:

We must calculate the 5-day BOD using the standard first-order model.

Formula:

Step 1: Compute the exponent

Step 2: Evaluate the exponential term

Step 3: Calculate ${\text{BOD}}_5$

Step 4: Match to options

The computed value is $158.0\text{mg/L}$, which is closest to $161.5\text{mg/L}$ among the given options. The small difference arises from rounding $e^{-1}=0.36788$: using the full precision value, $250\times (1-0.36788)=250\times 0.63212=158.0\text{mg/L}$.

Verification: $125.0\text{mg/L}$ would imply $50\%$ removal, which would need $k_{1}t=\ln 2=0.693$, not $1.0$. $200\text{mg/L}$ implies $80\%$ removal. $100\text{mg/L}$ implies only $40\%$ removal. The correct computed answer $\approx 158\text{mg/L}$ is closest to $161.5\text{mg/L}$.

Answer:

The 5-day BOD of the sample is approximately $158\text{mg/L}$, closest to $161.5\text{mg/L}$.

Quick Tip:

$e^{-1}\approx 0.368$ is a standard value worth memorising — it appears repeatedly in first-order BOD and disinfection calculations.

Understanding:

We need to identify the principle and purpose of the Imhoff cone test in wastewater analysis.

Step 1: Identify the Imhoff cone

An Imhoff cone is a conical graduated vessel (typically 1-litre capacity) made of glass or transparent plastic, with graduations at the bottom for reading small settled volumes.

Step 2: Describe the test procedure

A 1-litre wastewater sample is placed in the cone and allowed to settle undisturbed for exactly 60 minutes (1 hour). The volume of settled solids (settleable solids) accumulating at the bottom is then read directly from the graduations, typically reported in $\text{mL/L}$.

Step 3: Eliminate incorrect options

Dissolved oxygen is measured by the Winkler (iodometric) method or DO meters, not the Imhoff cone. Turbidity is measured with a nephelometer or turbidimeter. Coliform counting uses membrane filtration or MPN (Most Probable Number) methods.

Answer:

The Imhoff cone test measures settleable solids in wastewater by allowing a 1-litre sample to settle for one hour and reading the settled volume directly from the graduated cone.

Quick Tip:

Settleable solids (Imhoff cone) $\ne$ Total Suspended Solids (TSS). TSS is measured gravimetrically by filtering the sample through a pre-weighed glass-fibre filter.

Understanding:

We must find the chlorine residual after disinfection treatment.

Formula:

Step 1: Substitute values

Verification:

Chlorine demand is the amount of chlorine consumed by reactions with organic matter, ammonia, and reducing agents in the water. The remaining fraction is the residual, which provides ongoing disinfection protection in the distribution system. A residual of $0.7\text{mg/L}$ is consistent with typical free chlorine residuals maintained in water supply systems (usually $0.2$–$1.0\text{mg/L}$ at the point of use).

Answer:

The chlorine residual after treatment is $0.7\text{mg/L}$.

Quick Tip:

Chlorine demand is NOT the same as chlorine residual. Demand is consumed; residual is what remains. As per IS 10500, the permissible free chlorine residual in treated drinking water is $0.2\text{mg/L}$ at the consumer end.

Understanding:

We must calculate the equivalent continuous sound level $L_{eq}$ for a worker exposed to two different noise levels over an 8-hour period.

Formula:

Step 1: Convert each level to intensity ratio and weight by time

Step 2: Sum the weighted intensities

Step 3: Divide by total time

Step 4: Take the logarithm

Step 5: Match to nearest option

The computed $L_{eq}\approx 87.4\text{dB(A)}$, which is closest to $87.0\text{dB(A)}$. The small difference is due to the rounded option values.

Answer:

The equivalent continuous sound level is approximately $87.4\text{dB(A)}$, closest to $87.0\text{dB(A)}$.

Quick Tip:

Because the decibel scale is logarithmic, simply averaging two sound levels arithmetically is incorrect. A 10 dB difference means one source is 10 times more intense, so the louder source dominates the $L_{eq}$.

Understanding:

We need to correctly define the particulate matter fraction denoted as ${\text{PM}}_{2.5}$ in the context of air quality monitoring.

Step 1: Interpret the notation

The subscript in ${\text{PM}}_{2.5}$ denotes the upper cut-point of the aerodynamic diameter size fraction being measured. ${\text{PM}}_{2.5}$ therefore refers to all particulate matter with an aerodynamic diameter of less than or equal to $2.5\mu \text{m}$.

Step 2: Distinguish from other fractions

${\text{PM}}_{10}$ refers to particles with aerodynamic diameter $\le 10\mu \text{m}$ (inhalable particles). The coarse fraction refers to particles between $2.5\mu \text{m}$ and $10\mu \text{m}$. Ultrafine particles are defined as those $<0.1\mu \text{m}$ ($<100\text{nm}$).

Step 3: Health relevance

${\text{PM}}_{2.5}$ particles (fine particles) can penetrate deep into the alveolar region of the lungs and are more hazardous than coarser particles, which is why they are regulated separately in National Ambient Air Quality Standards (NAAQS).

Answer:

${\text{PM}}_{2.5}$ refers to particulate matter with aerodynamic diameter less than $2.5\mu \text{m}$.

Quick Tip:

The NAAQS (India) 24-hour average standard for ${\text{PM}}_{2.5}$ is $60\mu {\text{g/m}}^3$ and for ${\text{PM}}_{10}$ is $100\mu {\text{g/m}}^3$.

Understanding:

We must find the overflow rate and the minimum particle diameter completely removed by sedimentation.

Formula:

Particles are completely removed when their Stokes settling velocity $v_s\ge v_o$:

Step 1: Compute overflow rate

Step 2: Convert to SI units

Step 3: Solve for minimum particle diameter by setting $v_s=v_o$

Step 4: Substitute values

Step 5: Compare to options

The calculated $d_{min}\approx 15\mu \text{m}$, closest to $17\mu \text{m}$ among the given options. The small difference reflects rounding of intermediate constants.

Answer:

The overflow rate is $18\text{m/day}$ and the minimum completely removed particle diameter is approximately $17\mu \text{m}$.

Quick Tip:

The overflow rate $v_o$ equals the minimum settling velocity of a particle that is 100% removed, regardless of tank depth. Depth affects detention time but not removal efficiency in ideal sedimentation theory.

Understanding:

We need to identify the desirable limit for Total Dissolved Solids (TDS) in drinking water as specified by the Bureau of Indian Standards in IS 10500:2012.

Step 1: Recall IS 10500:2012 standards

IS 10500:2012 specifies two levels for many parameters:

Step 2: Identify TDS limits

For TDS, IS 10500:2012 specifies:

Step 3: Eliminate incorrect options

$250\text{mg/L}$ is the desirable limit for chlorides, not TDS. $1000\text{mg/L}$ is not the IS 10500 desirable TDS limit. $2000\text{mg/L}$ is the permissible (not desirable) TDS limit.

Answer:

The desirable limit of TDS in drinking water as per IS 10500:2012 is $500\text{mg/L}$.

Quick Tip:

WHO guideline for TDS is also $500\text{mg/L}$ (desirable). Water with TDS $>1000\text{mg/L}$ is generally considered unpalatable due to salty taste.

Understanding:

We must find the theoretical oxygen demand for complete oxidation of a glucose solution.

Formula:

The complete oxidation reaction for glucose is:

ThOD is the mass of ${\text{O}}_2$ required per unit mass of compound:

Step 1: Determine moles of ${\text{O}}_2$ per mole of glucose

From the balanced equation, 1 mole of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ requires 6 moles of ${\text{O}}_2$.

Step 2: Calculate the ThOD ratio

Step 3: Calculate ThOD for the given concentration

Verification:

Molecular weight of $6{\text{O}}_2=6\times 32=192\text{g/mol}$. Molecular weight of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6=180\text{g/mol}$. For $180\text{mg/L}$ glucose (which is exactly 1 mmol/L), ThOD $=6\times 32=192\text{mg/L}$.

Answer:

The theoretical oxygen demand of a $180\text{mg/L}$ glucose solution is $192\text{mg/L}$.

Quick Tip:

Glucose is a convenient benchmark: its molecular weight is $180\text{g/mol}$ and its ThOD ratio is $192/180\approx 1.067$. Because glucose is readily biodegradable, its BOD${}_5$/ThOD ratio approaches $0.92$–$0.95$.