GATE CE · Environmental Engineering

Treatment Of Waste Water – GATE CE Environmental EngineeringPractice Questions & PYQs

Generate GATE-level questions on Treatment Of Waste Water in Environmental Engineering. Focus on core concepts, previous year patterns, and numerical problem-solving techniques.

43 questions · 20 PYQs · 0 AI practice · GATE CE 2027

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Q1GATE 2025MSQ

A compound has a general formula $\text{C}_a\text{H}_b\text{O}_c\text{N}_d$ and molecular weight 187. A 935 mg/l solution of the compound is prepared in distilled deionized water. The Total Organic Carbon (TOC) is measured as 360 mg/l (as C). The Chemical Oxygen Demand (COD) and the Total Kjeldahl Nitrogen (TKN) are determined as 600 mg/l (as $\text{O}_2$ ) and 140 mg/l (as N), respectively (as per the chemical equation given below). Which of the following options is/are CORRECT? $\text{C}_a\text{H}_b\text{O}_c\text{N}_d + \frac{4a + b - 2c - 3d}{4} , \text{O}_2 \rightarrow a , \text{CO}_2 + \frac{b - 3d}{2} , \text{H}_2\text{O} + d , \text{NH}_3$ Atomic weight: C(12), H(1), O(16), N(14)

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📖 Explanation

Given: molecular weight of $\mathrm{C}_{\mathrm{a}} \mathrm{H}_{\mathrm{b}} \mathrm{O}_{\mathrm{c}} \mathrm{N}_{\mathrm{d}}=187 \mathrm{gm}$ Solution of compound in distilled water $=935 \mathrm{mg} / \mathrm{t}$

\begin{aligned} \mathrm{TOC} & =360 \mathrm{mg} / \mathrm{lt}(\text { as } \mathrm{C}) \\ \mathrm{COD} & =600 \mathrm{mg} / \mathrm{tt}\left(\text { as } \mathrm{O}_2\right) \\ \mathrm{TKN} & =140 \mathrm{mg} / \mathrm{lt}(\mathrm{As} \mathrm{~N}) \end{aligned}

$\mathrm{C}_{\mathrm{a}} \mathrm{H}_{\mathrm{b}} \mathrm{O}_{\mathrm{c}} \mathrm{~N}_{\mathrm{d}}+\frac{(4 \mathrm{a}+\mathrm{b}-2 \mathrm{c}-3 \mathrm{~d})}{4} \mathrm{O}_2 \rightarrow \mathrm{aCO}_2+\frac{\mathrm{b}-3 \mathrm{~d}}{2} \mathrm{H}_2 \mathrm{O}+\mathrm{dNH}_3$ Now, $d=\frac{\left(\frac{140}{935} \times 187\right)}{\text { Atomic weight of } \mathrm{N}}=\frac{\frac{140}{935} \times 187}{14}=2$ $a=\frac{\frac{360}{935} \times 187}{12}=6$ Now, $\quad \frac{4 a+b-2 c-3 d}{4}=\frac{\left(\frac{660 \times 187}{935}\right)}{\text { Atomic weight of } \mathrm{O}_2}=\frac{\left(\frac{600 \times 187}{935}\right)}{32}$

\begin{aligned} \frac{4 \times 6+b-2 c-3 \times 2}{4} & =3.75 \\ b-2 c & =-3 \;\;\;\; (i) \end{aligned}

Now, Molecular water of compound $\left(\mathrm{C}_{\mathrm{a}} \mathrm{H}_{\mathrm{b}} \mathrm{O}_{\mathrm{c}} \mathrm{N}_{\mathrm{d}}\right)=187$

\begin{aligned} & 12 a+b \times 1+16 c+14 d=187 \\ & 12 \times 6+b+16 c+14 \times 2=187 \end{aligned}

$b+16 c=87 \;\;\;\; (ii)$ On solving equation (i) and (ii), we get

\begin{aligned} & \mathrm{b}=7 \\ & \mathrm{c}=5 \end{aligned}

Hence, $a=6, b=7, c=5, d=2$ Correct options are (A), (B) and (C).

Q2GATE 2025MSQ

A facultative pond system is employed for wastewater treatment. Which of the following statements is/are true:

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📖 Explanation

(A) The dissolved oxygen concentration will be high during daytime compared to night-time Reason: In a facultative pond, algae perform photosynthesis during the daytime, producing oxygen: $\text{CO}_2 + \text{H}_2\text{O} \xrightarrow{\text{light}} \text{O}_2 + \text{Carbohydrates}$ Thus, dissolved oxygen (DO) increases during daytime. At night, photosynthesis stops and only respiration occurs, consuming oxygen, which reduces DO levels. (B) The pH will be high during daytime compared to night-time Reason: During photosynthesis, algae consume $\text{CO}_2$ , which reduces carbonic acid in water and increases pH. At night, $\text{CO}_2$ is released during respiration, forming carbonic acid and lowering pH. (C) The dissolved oxygen concentration will be low during daytime compared to night-time This is incorrect. DO is higher during the day due to photosynthesis and lower at night due to respiration. (D) The pH will be low during daytime compared to night-time This is incorrect. pH is higher during the day due to $\text{CO}_2$ consumption and lower at night due to $\text{CO}_2$ production. Correct options: (A) and (B)

Q3GATE 2025MSQ

Aeration is employed as a treatment option for the removal of several pollutants from contaminated water. Identify the pollutant(s), where aeration is employed as a part of their removal:

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📖 Explanation

Aeration is a process where air is introduced into water to allow oxidation reactions to occur. It is commonly used to oxidize soluble iron and manganese into insoluble forms, which can then be removed by filtration. Let?s analyze the options: (A) Iron - Aeration oxidizes $\text{Fe(OH)}_3$ , which precipitates as $\text{Fe(OH)}_3$ . (B) Cadmium - Not removed by aeration; it requires chemical precipitation or adsorption. (C) Manganese - Aeration helps oxidize $\text{Mn}^{2+} \rightarrow \text{MnO}_2$ under suitable pH conditions. (D) Zinc - Not removed by aeration alone; other treatments are required. Correct answer: (A) and (C)

Q4GATE 2025NAT

A community generates 1 million litres/day (MLD) of wastewater. This wastewater is treated using an activated sludge process (ASP). The working volume of the aeration tank of the ASP is $250\ \text{m}^3$ , and the biomass concentration in the tank is $3000\ \text{mg/L}$ . Analysis results showed that a biomass concentration of $10\ \text{mg/L}$ is present in the treated effluent from the secondary sedimentation tank of the ASP. Sludge wastage from the system is at a rate of $5000\ \text{L/day}$ with a biomass concentration of $10000\ \text{mg/L}$ . The system is in steady-state condition. The biological sludge residence time (BSRT) of the system (in days) is ______ (round off to one decimal place).

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📖 Explanation

BSRT (also called Solids Retention Time, SRT) is given by: $\text{BSRT} = \frac{\text{Mass of biomass \in aeration tank}}{\text{Mass of biomass leaving the system per day}}$ Step 1: Biomass in aeration tank Volume = $250\ \text{m}^3 = 250 \times 1000 = 250000\ \text{L}$ Biomass concentration = $3000\ \text{mg/L}$ $\Rightarrow \text{Mass} = 250000 \times 3000 = 750 \times 10^6\ \text{mg} = 750\ \text{kg}$ Step 2: Biomass leaving the system per day Treated effluent flow = $1\ \text{MLD} = 10^6\ \text{L/day}$ , biomass = $10\ \text{mg/L}$ $10 \times 10^6 = 10^7\ \text{mg/day} = 10\ \text{kg/day}$ Sludge wastage = $5000\ \text{L/day}$ , biomass = $10000\ \text{mg/L}$ $5000 \times 10000 = 5 \times 10^7\ \text{mg/day} = 50\ \text{kg/day}$ Total biomass leaving per day = $10 + 50 = 60\ \text{kg/day}$ Step 3: BSRT calculation $\text{BSRT} = \frac{750}{60} = \boxed{12.5\ \text{days}}$

Q5GATE 2024NAT

A circular settling tank is to be designed for primary treatment of sewage of a flow rate of 10 million liters/day. Assume a detention period of 2.0 hours and surface loading rate of 40000 liters $/ \mathrm{m}^{2} /$ day. The height (in meters) of the water column in the tank is _____ (rounded off to 2 decimal places)

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📖 Explanation

Given, Discharge, $\mathrm{Q}=10 \mathrm{MLD}$ Detention period, $\mathrm{t}_{\mathrm{d}}=2 \mathrm{hr}$ Over flow rate, $\mathrm{V}_{0}=40,000$ litres $/ \mathrm{m}^{2} /$ day $\mathrm{H}=$ ? As, Surface area of tank $=\frac{\text { Discharge }}{\text { Over flow rate }}=\frac{\mathrm{Q}}{\mathrm{V}_{0}}$ So,

\begin{aligned} & \text { S.A. }=\frac{10 \times 10^{6} \mathrm{~L} / \text { day }}{40,000 \mathrm{~L} / \mathrm{m}^{2} / \text { day }} \\ & \text { S.A }=250 \mathrm{~m}^{2} \end{aligned}

Volume of tank $(V)=Q \times t_{d}$

\begin{aligned} & =\frac{10 \times 10^{6} \times 10^{-3}}{24} \times 2 \mathrm{~m}^{3} \\ & \mathrm{~V}=833.33 \mathrm{~m}^{3} \end{aligned}

Hence, Height of water in tank or Height of setting zone

\begin{aligned} H & =\frac{833.33 \mathrm{~m}^{3}}{250 \mathrm{~m}^{2}} \\ H & =3.33 \mathrm{~m} \end{aligned}

Note: In reality, height of water in tank OR height of settling zone is estimated by above approach but if the bottom of sedimentation tank is assumed to be sloping or hoppered (i.e sludge zone is also considered) then height of water in tank is determined as follows: Using, Volume of tank $(V)=D^{2}\left(\frac{\pi}{4} H+0.011 D\right)$ As,

\begin{aligned} \text { S.A. } & =\frac{\pi}{4} D^{2}=250 \mathrm{~m}^{2} \\ D & =17.835 \mathrm{~m} \end{aligned}

So, $833.33=(17.835)^{2}\left(\frac{\pi}{4} H+0.011(17.835)\right)$ So, $\mathrm{H}=3.085 \mathrm{~m}$

Q6GATE 2024NAT

Activated carbon is used to remove a pollution 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).

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📖 Explanation

For first order kinetics, $N_{t}=N_{0} e^{-k t}$ For efficiency of $95 \%$ ,

\begin{aligned} N_{t} & =\left(1-\frac{95}{100}\right)^{L_{0}} \\ & =0.05 N_{0} \end{aligned}

$\therefore \quad 0.05=e^{-0.38 t}$ Taking In on both sides,

\begin{array}{ll} & \ln 0.05 =-0.38 t \times \operatorname{lne} \\ \Rightarrow &t =7.9 \text { days } \end{array}

Q7GATE 2023MSQ

Identify the waterborne diseases caused by viral pathogens:

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Q8GATE 2023MSQ

In the context of water and wastewater treatments, the correct statements are:

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Q9GATE 2023NAT

The wastewater inflow to an activated sludge plant is $0.5 \mathrm{~m}^{3} / \mathrm{s}$ , and the plant is to be operated with a food to microorganism ratio of $0.2 \mathrm{mg} / \mathrm{mg}-\mathrm{d}$ . The concentration of influent biodegradable organic matter of the wastewater to the plant (after primary settling) is $150 \mathrm{mg} / \mathrm{L}$ , and the mixed liquor volatile suspended solids concentration to be maintained in the plant is $2000 \mathrm{mg} / \mathrm{L}$ . Assuming that complete removal of biodegradable organic matter in the tank, the volume of aeration tank (in $\mathrm{~m}^{3}$ , in integer) required for the plant is _____.

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📖 Explanation

Given,

\begin{aligned} Q_1&=6.5 m^3/s \\ &=0.5 \times 24 \times 60 \times 60 \\ &=43200 m^3/d\\ \frac{F}{M} & =0.2 (mg/mg-d)\\ \mathrm{S}_0 & =150 \mathrm{mg} / \mathrm{l} \\ \mathrm{X} & =2000 \mathrm{mg} / \mathrm{l} \\ \because \quad \frac{F}{M} & =\frac{\mathrm{Q}_0 \mathrm{~S}_0}{\mathrm{VX}} \\ 0.2 & =\frac{43200 \times 150}{\mathrm{~V} \times 2000} \\ V & =16200 \mathrm{~m}^3\end{aligned}

Q10GATE 2022NAT

A 2% sewage sample (in distilled water) was incubated for 3 days at $27^{\circ}C$ temperature. After incubation, a dissolved oxygen depletion of 10 mg/L was recorded. The biochemical oxygen demand (BOD) rate constant at $27^{\circ}C$ was found to be $0.23 \; day^{-1}$ (at base e). The ultimate BOD (in mg/L) of the sewage will be_____________. (round off to the nearest integer)

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📖 Explanation

Depletion of dissolved oxygen after 3 days incubation = 10 mg/L $\text{Dilution factor }=\frac{\text{Vol. of dilutedsample}}{\text{Vol.of undiluted sewage sample}}=\frac{100}{2}50$ $BOD_3= \text{DO consumed} \times \text{dilution factor}=10 \times 50=500 mg/L$

\begin{aligned} BOD_3&= BOD_u(1-e^{-kt}) \\ 500 &=BOD_u(1-e^{-0.23 \times 3}) \\ BOD_u &= 1003.162 mg/L \end{aligned}

Q11GATE 2021NAT

A grit chamber of rectangular cross-section is to be designed to remove particles with diameter of 0.25 mm and specific gravity of 2.70. The terminal settling velocity of the particles is estimated as 2.5 cm/s. The chamber is having a width of 0.50 m and has to carry a peak wastewater flow of $9720 \mathrm{~m}^{3} / \mathrm{d}$ giving the depth of flow as 0.75 m. If a flow-through velocity of 0.3 m/s has to be maintained using a proportional weir at the outlet end of the chamber, the minimum length of the chamber (in m,in integer) to remove 0.25 mm particles completely is _________

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📖 Explanation

Minimum length of chamber

\begin{aligned} L_{\min } &=V_{\text {flow }} \times t_{d} \\ V_{\text {fow }} &=0.30 \mathrm{~m} / \mathrm{s} \\ t_{d} &=\frac{H}{V_{\text {settling }}}=\frac{0.75}{2.5 \times 10^{-2}}=30 \mathrm{sec} \\ L_{\text {min }} &=0.3 \mathrm{~m} / \mathrm{sec} \times 30 \mathrm{sec} \\ &=9 \mathrm{~m} \end{aligned}

Q12GATE 2021NAT

A secondary clarifier handles a total flow of $9600 \mathrm{~m}^{3} / \mathrm{d}$ from the aeration tank of a conventional activated-sludge treatment system. The concentration of solids in the flow from the aeration tank is 3000 mg/L. The clarifier is required to thicken the solids to 12000 mg/L, and hence it is to be designed for a solid flux of $3.2 \frac{\mathrm{kg}}{\mathrm{m}^{2} \cdot \mathrm{h}}$ . The surface area of the designed clarifier for thickening (in $\mathrm{~m}^{2}$ , in integer) is _________________

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📖 Explanation

\begin{aligned} Q &=9600 \mathrm{~m}^{3} / \mathrm{d} \\ X &=3000 \mathrm{mg} / l \\ X_{u} &=12000 \mathrm{mg} / l \\ \text {Solid flux }& =3.2 \mathrm{~kg} / \mathrm{m}^{2}-\mathrm{h}\\ \text{Surface area,} \qquad \mathrm{A}&=\frac{\text { Total quantity of solids entering }}{\text { Solid flux }}\\ A &=\frac{9600 \mathrm{~m}^{3} / \mathrm{d} \times 3000 \mathrm{mg} / l}{3.2 \times 10^{6} \mathrm{mg} / \mathrm{m}^{2}-\mathrm{h}} \\ &=\frac{9600 \times 10^{3} \mathrm{lld} \times 3000 \mathrm{mg} / \mathrm{l}}{3.2 \times 10^{6} \mathrm{mg} / \mathrm{m}^{2}-\mathrm{h} \times 24 \mathrm{~h} / \mathrm{d}} \\ &=375 \mathrm{~m}^{2} \end{aligned}

Q13GATE 2021NAT

An activated sludge process (ASP) is designed for secondary treatment of $7500 m^{3} / day$ of municipal wastewater. After primary clarifier, the ultimate BOD of the influent, which enters into ASP reactor is 200 mg/L. Treated effluent after secondary clarifier is required to have an ultimate BOD of 20 mg/L. Mix liquor volatile suspended solids (MLVSS) concentration in the reactor and the underflow is maintained as 3000 mg/L and 12000 mg/L, respectively. The hydraulic retention time and mean cell residence time are 0.2 day and 10 days, respectively. A representative flow diagram of the ASP is shown below. The underflow volume (in $m^{3} / day$ , round off to one decimal place) of sludge wastage is _________

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📖 Explanation

\begin{aligned} \text{Given,} \qquad X &=3000 \mathrm{mg} / l, \quad X_{\mathrm{u}}=12000 \mathrm{mg} / l \\ \text{Since} \qquad \mathrm{HRT} &=\frac{V}{Q_{0}}\\ \Rightarrow &\text{Volume of reactor,}\\ V &=Q_{0} \times H R T \\ &=7500 \times 0.2 \mathrm{~m}^{3} \\ &=1500 \mathrm{~m}^{3} \\ &\text{Sludge age,} \\ \theta_{\mathrm{C}} &=\frac{V X}{\left(Q_{0}-Q_{w}\right) X_{e}+Q_{w} X_{u}} \qquad \left(X_{e} \simeq 0\right)\\ 10&=\frac{1500 \times 3000}{Q_{W} \times 12000}\\ Q_{w}&=37.5 \mathrm{~m}^{3} / day \end{aligned}

Explanation Figure 2

Q14GATE 2019NAT

A wastewater is to be disinfected with 35 mg/Lof chlorine to obtain 99% kill of micro-organisms. The number of micro-organisms remaining alive ( $N_t$ ) at time t, is modelled by $N_t=N_oe^{-kt}$ , where $N_o$ is number of micro-organisms at t= 0, and k is the rate of kill. The wastewater flow rate is 36 $m^3/hr$ , and k= 0.23 $min^{-1}$ . If the depth and width of the chlorination tank are 1.5 m and 1.0 m, respectively, the length of the tank (in m, round off to 2 decimal places) is ________

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📖 Explanation

\begin{aligned} N_t&=N_0e^{-kt}\\ \therefore \; \eta &=\$[1-e^{-kt}]\$100\\ 99&=\$[1-e^{-0.23\times t}]\$100\\ \therefore \;t&=20.22min\\ Q&=36m^3/hr\\ V&=Q \times t\\ &=\frac{36}{60}\times 20.22=12.0132\\ \text{Length}&=\frac{V}{B \times H}=\frac{12.0132}{1 \times 1.5}\\ &=8.008\cong 8m \end{aligned}

Q15GATE 2018NAT

At a small water treatment plant which has 4 filters, the rates of filtration and backwashing are 200 $m^{3}/d/m^{2}$ and 1000 $m^{3}/d/m^{2}$ , respectively. Backwashing is done for 15 min per day. The maturation, which occurs initially as the filter is put back into service after cleaning, takes 30 min. It is proposed to recover the water being wasted during backwashing and maturation. The percentage increase in the filtered water produced (up to two decimal places) would be ______

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📖 Explanation

Let total area of filters be $1 \mathrm{m}^{2}$ Water used for backwashing $=1000 \times \frac{15}{24 \times 60}=10.4166 \mathrm{m}^{3}$ Water used for maturation $=200 \times \frac{30}{24 \times 60}=4.166 \mathrm{m}^{3}$ Total water wasted for backwashing and maturation $=10.4166+4.166=14.58 \mathrm{m}^{3}$ Water to be treated by filtered $=200 \times \frac{23.25}{24}=193.75 \mathrm{m}^{3} / \mathrm{day}$ % increase in filtered water produced $=\frac{14.58}{193.75} \times 100=7.525 \%$

Q16GATE 2018NAT

A waste activated sludge (WAS) is to be blended with green waste (GW). The carbon (C) and nitrogen (N) contents, per kg of WAS and GW, on dry basis are given in the table. The ratio of WAS to GW required (up to two decimal places) to achieve a blended C:N ratio of 20:1 on dry basis is ______

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📖 Explanation

Let 20 kg of C and 1 kg of N is required Let $x$ kg of WAS is taken

\begin{aligned} \therefore \quad \text{Carbon \in x } \mathrm{kg}&=0.054 x \mathrm{kg} \\ \text{ Nitrogen \in x } \mathrm{kg}&=0.010 x \mathrm{kg} \end{aligned}

Let y kg of GW is taken

\begin{aligned} \therefore \quad \text { Carbon \in } y \mathrm{kg} &=0.360 \mathrm{y} \mathrm{kg} \\ \text { Nitrogen \in } y \mathrm{kg} &=0.006 \mathrm{y} \mathrm{kg} \end{aligned}

Total Carbon, $0.054 x+0.36 y=20 \mathrm{kg}\quad \quad\ldots(i)$ Total Nitrogen,

\begin{aligned} 0.01 x+0.006 y &=1 \mathrm{kg} &\ldots(ii)\\ x &=73.26 \mathrm{kg} \\ y &=44.566 \mathrm{kg} \\ \frac{x}{y} &=\frac{\mathrm{WAS}}{\mathrm{GW}}\\ \therefore &=1.6438 \simeq 1.64 \end{aligned}

Q17GATE 2018NAT

A schematic flow diagram of a completely mixed biological reactor with provision for recycling of solids is shown in the figure. The mean cell residence time (in days, up to one decimal place) is ______

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📖 Explanation

\begin{aligned} \theta_{C} &=\frac{V X}{\left(Q_{0}-Q_{w}\right) X e+Q_{w} X_{u}} \\ &=\frac{Q_{0} H R T X}{\left(Q_{0}-Q_{w}\right) X_{e}+Q_{w} X_{u}}\left(X_{e}=0\right)\\ &=\frac{15000 \times \frac{2}{24} \times 3000}{50 \times 10000}=7.5 \text{ days} \end{aligned}\\

Q18GATE 2017MCQ

The most important type of species involved in the degradation of organic matter in the case of activated sludge process is

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Q19GATE 2017NAT

The wastewater having an organic concentration of 54 mg/l is flowing at a steady rate of 0.8 $m^{3}/day$ through a detention tank of dimensions $2m\times 4m\times 2m$ . If the contents of the tankare well mixed and the decay constant is 0.1 per day, the outlet concentration (in mg/l, up to one decimal place) is ______

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📖 Explanation

$Q=0.8 \mathrm{m}^{3} / \mathrm{d}$ Detention time,

\begin{aligned} D_{t}&=\frac{V}{Q}=\frac{4 \times 2 \times 2 \mathrm{m}^{3}}{0.8 \mathrm{m}^{3} / \mathrm{day}}=20 \mathrm{days} \\ L_{t}&=\frac{L_{0}}{1+k_{\mathrm{e}} t}=\frac{54}{1+0.1 \times 20}=18 \mathrm{mg} / l \end{aligned}

Q20GATE 2016MCQ

Match the items in Group-I with those in Group-II and choose the right combination.

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