Question 1 |

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)

The ultimate BOD (in mg/L) of the sewage will be_____________. (round off to the nearest integer)

825 | |

1254 | |

1003 | |

1502 |

Question 1 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}

\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}

Question 2 |

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 _________

5 | |

7 | |

9 | |

10 |

Question 2 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}

\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}

Question 3 |

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 _________

The underflow volume (in m^{3} / day, round off to one decimal place) of sludge wastage is _________

25.4 | |

37.5 | |

65.2 | |

87.6 |

Question 3 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}

Question 4 |

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 _________________

125 | |

258 | |

525 | |

375 |

Question 4 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}

Question 5 |

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 ________

2.65 | |

8.65 | |

8.00 | |

6.74 |

Question 5 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}

Question 6 |

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 ______

The mean cell residence time (in days, up to one decimal place) is ______

2.5 | |

5.5 | |

7.5 | |

10.2 |

Question 6 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}\\

Question 7 |

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 ______

7.52 | |

2.25 | |

8.22 | |

9.85 |

Question 7 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 \%

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 \%

Question 8 |

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 ______

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 ______

1.26 | |

1.94 | |

2.65 | |

1.64 |

Question 8 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}

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}

Question 9 |

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

Autotrophs | |

Heterotrophs | |

Prototrophs | |

Photo-autotrophs |

Question 9 Explanation:

Activated sludge process is designed primarily for satisfaction of carbonaceous BOD which is done by heterotrophs.

Question 10 |

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 ______

17 | |

21 | |

20 | |

18 |

Question 10 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}

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}

There are 10 questions to complete.