# Environmental Engineering

 Question 1
A sample of air analyzed at $25^{\circ}C$ and 1 atm pressure is reported to contain 0.04 ppm of $SO_2$. Atomic mass of S = 32, O = 16. The equivalent $SO_2$ concentration (in $\mu g/m^3$) will be__________. (round off to the nearest integer)
 A 105 B 118 C 138 D 162
GATE CE 2022 SET-2      Air and Noise Pollution
Question 1 Explanation:
Concentration of $SO_2$ in ppm = 0.04
Let's equivalent concentration in $=\mu g/m^3$ is $x$.
$x\mu g$ of $SO_2$ present in $1m^3$ of air.
We know, 1 mole of $SO_2$ (at $0^{\circ}C$, 1 atm) has volume of 22.4 L.
at $25^{\circ}C$ and 1 atm, volume of $SO_2$
$=\frac{22.4}{273+0}\times (273+25)=24.45 lit$
$\frac{x \times 10^{-6}}{64}$ mole of $SO_2$ has volume
$\frac{24.45 \times x \times 10^{-6}}{64} lit \; in \; 1m^3$ of air.
$0.382x \times 10^{-3}m^3$ of $SO_2$ present in $10^{6}m^3$ of air.
$0.382x \times 10^{-3}=0.04$
$x=104.7 \approx105$
0.04 ppm of $SO_2=105\mu g/m^3$ of $SO_2$
 Question 2
A sewage treatment plant receives sewage at a flow rate of 5000 $m^3/day$. The total suspended solids (TSS) concentration in the sewage at the inlet of primary clarifier is 200 mg/L. After the primary treatment, the TSS concentration in sewage is reduced by 60%. The sludge from the primary clarifier contains 2% solids concentration. Subsequently, the sludge is subjected to gravity thickening process to achieve a solids concentration of 6%. Assume that the density of sludge, before and after thickening, is 1000 $kg/m^3$.
The daily volume of the thickened sludge (in $m^3/day$) will be_________. (round off to the nearest integer)
 A 5 B 10 C 20 D 30
GATE CE 2022 SET-2      Disposal of Sewage Effluents
Question 2 Explanation:
Wt. of TSS at inlet of PST
$=5000 \times 10^3\frac{L}{d} \times 200 \times 10^{-6}Kg/L=1000Kg/d$
Wt. of solid in sludge from PST $=0.6 \times 1000=600 Kg/d$
Wt. of sludge before thikeninig $=\frac{600}{\frac{2}{100}}=30000 Kg/d$
Wt. of sludge before thikeninig $=\frac{600}{\frac{6}{100}}=10000 Kg/d$
Daily volume of thickened sludge $=frac{10000}{1000}=10m^3$
 Question 3
At a municipal waste handling facility, 30 metric ton mixture of food waste, yard waste, and paper waste was available. The moisture content of this mixture was found to be 10%. The ideal moisture content for composting this mixture is 50%. The amount of water to be added to this mixture to bring its moisture content to the ideal condition is _______metric ton. (in integer)
 A 12 B 18 C 24 D 30
GATE CE 2022 SET-2      Municipal Solid Waste Management
Question 3 Explanation:

Wt. of water initially present $=frac{10}{100} \times 30=3mT$
Wt. of solid $=30-3=27mT$
In ideal condition,
Wt. of solid $=27mT$
$M/C=50%$
Wt. of water $=27mT$
Amount of water needed to get ideal condition $= 27 - 3 = 24 mT$
 Question 4
The concentration $s(x,t)$ of pollutants in a one-dimensional reservoir at position $x$ and time $t$ satisfies the diffusion equation
$\frac{\partial s(x,t)}{\partial t}=D\frac{\partial^2 s(x,t)}{\partial x^2}$
on the domain $0 \leq x \leq L$, where $D$ is the diffusion coefficient of the pollutants. The initial condition $s(x, 0)$ is defined by the step-function shown in the figure.

The boundary conditions of the problem are given by $\frac{\partial s(x,t)}{\partial x}=0$ at the boundary points $x = 0$ and $x = L$ at all times. Consider $D = 0. 1 m^2/s, s_0 = 5 \mu mol/m$, and $L = 10 m$.
The steady-state concentration $\tilde{S}\left ( \frac{L}{2} \right )=S\left ( \frac{L}{2},\infty \right )$, at the center $x=\frac{L}{2}$ of the reservoir (in $\mu$mol/m) is ___________. (in integer)
 A 1 B 2 C 3 D 4
GATE CE 2022 SET-2      Source of Water Supply, Distribution System and Well Hydraulics
Question 4 Explanation:
From figure s(x,t)
at $x=0\Rightarrow s(0,t)=s_0=5$
at $x=0.4L\Rightarrow s(0.4L,t)=s_0=5$
at $x=L\Rightarrow s(L,t)=0$
From $x = 0\; to \;x = 0.4 L$
Concentration of pollutant $5\mu \; mol/m \times (0.4 \times 10m) =20 \mu \; mol$
$x = 0.4L\; to \;x = L$
Concentration of pollutant = 0
Total concentration of pollutant in 10 m $=20 \mu \; mol$
In infinite time this concentration will be diluted so concentration of pollutant per m
$=\frac{20}{10} \mu \; mol/m=2 \mu \; mol/m$
Under steady state condition, concentration of pollutant will be uniformly distributed.
Steady state concentration at $x=\frac{L}{2} =2\mu \; mol/m$
 Question 5
In a city, the chemical formula of biodegradable fraction of municipal solid waste (MSW) is $C_{100}H_{250}O_{80}N$. The waste has to be treated by forced-aeration composting process for which air requirement has to be estimated.
Assume oxygen in air (by weight) = 23 %, and density of air = 1.3 $kg/m^3$. Atomic mass: C = 12, H = 1, O = 16, N = 14.
C and H are oxidized completely whereas N is converted only into $NH_3$ during oxidation.
For oxidative degradation of 1 tonne of the waste, the required theoretical volume of air (in $m^3/tonne$) will be (round off to the nearest integer)
 A 4749 B 8025 C 1418 D 1092
GATE CE 2022 SET-2      Municipal Solid Waste Management
Question 5 Explanation:
$C_{100}H_{250}O_{80}N+xO_2\rightarrow 100CO_2+\frac{247}{2}H_2O+NH_3$
\begin{aligned} 80+x \times 2&=200+\frac{247}{2}\\ x&=121.75 \end{aligned}
1 mole of MSW required 121.75 mole of oxygen No. of moles of MSW
$=\frac{1 \; ton}{(100 \times 12+250+80 \times 16+14)g}$
No. of moles of O2 required $=121.75 \times \frac{1 \; tonne}{2744g}$
\begin{aligned} \text{wt. of }O_2&= \frac{121.75}{2744}\times 32 \; tonne\\ \text{Mass of air }&= \frac{1.42}{0.23}\times 10^3 Kg\\ &=6.174 \times 10^3kg\\ \text{Density of air }&= 1.3 Kg/m^3\\ \text{Volume of air }&= \frac{6.174}{1.3}\times 10^3 m^3\\ =4749.2 m^3 &\text{per tonne of MSW} \end{aligned}
 Question 6
A 100 mg of $HNO_3$ (strong acid) is added to water, bringing the final volume to 1.0 liter. Consider the atomic weights of H, N, and O, as 1 g/mol, 14 g/mol, and 16 g/mol, respectively. The final pH of this water is (Ignore the dissociation of water.)
 A 2.8 B 6.5 C 3.8 D 8.5
GATE CE 2022 SET-2      Quality Characteristics of Water
Question 6 Explanation:
$HNO_3\rightarrow H^++NO_3^{-}$
\begin{aligned} \text{1 mole of }HNO_3&=\text{1 mole of }H^+\\ \text{No. of mole of }HNO_3&=\frac{100}{(1+14+16 \times 3)}\\ &=1.587 m\; mol\\ [H^+] \text{ in 1 lit. of water }&=1.587 m\;mol/Lit\\ &=1.587 \times 10^{-3}\;mol/Lit\\ pH&=-\log[H^+]\\ &=\log[1.587 \times 10^{-3}]\\ &=2.8 \end{aligned}
 Question 7
To design an optimum municipal solid waste collection route, which of the following is/are NOT desired:
 A Collection vehicle should not travel twice down the same street in a day. B Waste collection on congested roads should not occur during rush hours in morning or evening. C Collection should occur in the uphill direction. D The last collection point on a route should be as close as possible to the waste disposal facility.
GATE CE 2022 SET-2      Municipal Solid Waste Management
Question 7 Explanation:
Some heuristic guidelines that should be taken into consideration when laying out routes are as follows :
1. Existing policies and regulations related to such items as the point of collection and frequency of collection must be identified.
2. Existing system characteristics such as crew size and vehicle types must be coordinated.
3. Wherever possible, routes should be laid out so that they begin and end near arterial streets, using topographical and physical barriers as route boundaries.
4. In hilly area, routes should start at the top of the grade and proceed downhill as the vehicle becomes loaded.
5. Routes should be laid out so that the last container to be collected on the route is located nearest to the disposal site.
6. Waste generated at traffic-congested locations should be collected as early in the day as possible.
7. Sources at which extremely large quantities of waste are generated should be serviced during the first part of the days.
8. Scattered pickup points (where small quantities of solid waste are generated) that receive the same collection frequency should, if possible, be serviced during one trip or on the same day.
 Question 8
In a solid waste handling facility, the moisture contents ($MC$) of food waste, paper waste, and glass waste were found to be $MC_f$, $MC_p$, and $MC_g$, respectively. Similarly, the energy contents ($EC$) of plastic waste, food waste, and glass waste were found to be $EC_{pp}$, $EC_f$, and $EC_g$, respectively. Which of the following statement(s) is/are correct?
 A $MC_f \gt MC_p \gt MC_g$ B $EC_{pp} \gt EC_f \gt EC_g$ C $MC_f \lt MC_p \lt MC_g$ D $EC_{pp} \lt EC_f \lt EC_g$
GATE CE 2022 SET-2      Municipal Solid Waste Management
Question 8 Explanation:
Typical speicifiv weight and moisture content data for residential, commercial, industraial, and agricultural wastes.

Moisture content
Mcf > Mcp > Mcg
Typical values for short residue and energy conent of residentail MSW.

Energy content
ECpp > ECf > ECg
 Question 9
A process equipment emits 5 kg/h of volatile organic compounds (VOCs). If a hood placed over the process equipment captures 95% of the VOCs, then the fugitive emission in kg/h is
 A 0.25 B 4.75 C 2.5 D 0.48
GATE CE 2022 SET-2      Air and Noise Pollution
Question 9 Explanation:
VOC emission = 5 Kg/h
Capturing Efficiency =95%
Fugitive emission = (5/100)*5 Kg/h=0.25 Kg/h
 Question 10
A water treatment plant has a sedimentation basin of depth 3 m, width 5 m, and length 40 m. The water inflow rate is 500 $m^3/h$. The removal fraction of particles having a settling velocity of 1.0 m/h is_______. (round off to one decimal place)
(Consider the particle density as 2650 $kg/m^3$ and liquid density as 991 $kg/m^3$)
 A 0.4 B 0.2 C 0.1 D 0.8
GATE CE 2022 SET-1      Water Treatment
Question 10 Explanation:
\begin{aligned} V_o&=\frac{500}{5 \times 40} =2.5m/h \\ \eta &=\frac{V_s}{V_o}\times 100=\frac{1}{2.5}=0.4 \; or \; 40% \end{aligned}

There are 10 questions to complete.

### 2 thoughts on “Environmental Engineering”

1. Question 49 of environmental engineering is provided with flawed data