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)

105 | |

118 | |

138 | |

162 |

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

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)

The daily volume of the thickened sludge (in m^3/day) will be_________. (round off to the nearest integer)

5 | |

10 | |

20 | |

30 |

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

=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)

12 | |

18 | |

24 | |

30 |

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 \mumol/m) is ___________. (in integer)

\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 \mumol/m) is ___________. (in integer)

1 | |

2 | |

3 | |

4 |

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

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)

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)

4749 | |

8025 | |

1418 | |

1092 |

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}

\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.)

2.8 | |

6.5 | |

3.8 | |

8.5 |

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}

\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:

Collection vehicle should not travel twice down the same street in a day. | |

Waste collection on congested roads should not occur during rush hours in morning or evening. | |

Collection should occur in the uphill direction. | |

The last collection point on a route should be as close as possible to the waste disposal facility. |

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.

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?

MC_f \gt MC_p \gt MC_g | |

EC_{pp} \gt EC_f \gt EC_g | |

MC_f \lt MC_p \lt MC_g | |

EC_{pp} \lt EC_f \lt EC_g |

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

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

0.25 | |

4.75 | |

2.5 | |

0.48 |

Question 9 Explanation:

VOC emission = 5 Kg/h

Capturing Efficiency =95%

Fugitive emission = (5/100)*5 Kg/h=0.25 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 )

(Consider the particle density as 2650 kg/m^3 and liquid density as 991 kg/m^3 )

0.4 | |

0.2 | |

0.1 | |

0.8 |

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.

Question 49 of environmental engineering is provided with flawed data

Thank You Kishan Agarwal,

We have updated the figure in question.