Question 1 |

A wastewater sample contains two nitrogen species, namely ammonia and nitrate. Consider the atomic weight of N, H, and O as 14 g/mol, 1 g/mol, and 16 g/mol, respectively. In this wastewater, the concentration of ammonia is 34 mg NH_3/liter and that of nitrate is 6.2 mg NO_3^{-}/liter. The total nitrogenconcentration in this wastewater is _______ milligram nitrogen per liter. (round off to one decimal place)

29.4 | |

12.8 | |

36.2 | |

16.4 |

Question 1 Explanation:

Total nitrogen concentration (mg/liter)in this
wastewater =\left ( \frac{34}{17}+\frac{6.2}{62} \right ) \times 14=29.4 mg/l as Nitrogen.

Question 2 |

The total hardness in raw water is 500 milligram per liter as CaCO_3. The total
hardness of this raw water, expressed in milligram equivalent per liter, is

(Consider the atomic weights of Ca, C, and O as 40 g/mol, 12 g/mol, and 16 g/mol, respectively.)

(Consider the atomic weights of Ca, C, and O as 40 g/mol, 12 g/mol, and 16 g/mol, respectively.)

10 | |

100 | |

1 | |

5 |

Question 2 Explanation:

TH=500\frac{mg}{L} \text{ as }CaCO_3=\frac{500}{(100/2)}\frac{meq}{L}=10meq/L

Question 3 |

Ammonia nitrogen is present in a given wastewater sample as the ammonium ion \left(N H_{4}^{+}\right) and ammonia \left(N H_{3}\right). If pH is the only deciding factor for the proportion of these two constituents, which of the following is a correct statement?

At pH above 9.25, only \left(N H_{4}^{+}\right) will be present | |

At pH above 9.25, \left(N H_{3}\right) will be predominant | |

At pH 7.0, \left(N H_{4}^{+}\right) and \left(N H_{3}\right) will be found in equal measures | |

At pH 7.0, \left(N H_{4}^{+}\right) will be predominant |

Question 3 Explanation:

From the above curve, it is evident that at pH 7.0, \mathrm{NH}_{4}^{+} will be predominant.

Question 4 |

A 50 mL sample of industrial wastewater is taken into a silica crucible. The empty weight of the crucible is 54.352 g. The crucible with the sample is dried in a hot air oven at 104 ^{\circ}C till a constant weight of 55.129 g. Thereafter, the crucible with the dried sample is fired at 600 ^{\circ}C for 1 h in a muffle furnace, and the weight of the crucible along with residue is determined as 54.783 g. The concentration of total volatile solids is

15540 mg/L | |

8620 mg/L | |

6920 mg/L | |

1700 mg/L |

Question 4 Explanation:

\begin{aligned} V&= 50 \; ml\\ W_1&=54.352\; gm \\ W_2&=55.129 \; gm \\ W_3&=54.783 \; gm \\ &\text{Total volatile solids,} \\ &=\frac{W_2-W_3}{V} \\ &= \frac{(55.129-54.783)}{50} \times 10^6 \; mg/lit\\ &=6920\; mg/l \end{aligned}

Question 5 |

A completely mixed dilute suspension of sand particles having diameters 0.25, 0.35, 0.40, 0.45 and 0.50 mm are filled in a transparent glass column of diameter 10 cm and height 2.50 m. The suspension is allowed to settle without any disturbance. It is observed that all particles of diameter 0.35 mm settle to the bottom of the column in 30 s. For the same period of 30 s, the percentage removal (round off to integer value) of particles of diameters 0.45 and 0.50 mm from the suspension is ____

50 | |

75 | |

90 | |

100 |

Question 5 Explanation:

Since sand particle of size 0.35 mm settles to the bottom of the column in 30 sec
particles having size greater than 0.35 mm i.e. 0.45 and 0.50 mm will also settle in
suspension at the bottom of column by 100% in 30 sec, infact these bigger sized particle
will settle by 100% in less than 30 sec. So answer is 100%.

Question 6 |

The ultimate BOD (L_{0}) of a wastewater sample is estimated as 87% of COD. The COD of this wastewater is 300 mg/L. Considering first order BOD reaction rate constant k (use natural log) = 0.23 per day and temperature coefficient \theta = 1.047, the BOD value (in mg/L,
up to one decimal place) after three days of incubation at 27^{\circ}C for this wastewater will be______

120.4 | |

160.2 | |

180.6 | |

220.2 |

Question 6 Explanation:

\begin{aligned} \text{Ultimate BOD} &=0.87\; \text{COD} \\ &=0.87\times 300=261\: mg/l \\ BOD_{3}&=L_{0}\left ( 1-e^{-k_{27}\times 3} \right ) \\ k_{27}&=k_{20}\left ( 1.047 \right )^{T-20} \end{aligned}

For municipal sewage, at standard temperature, value of k(base e)=0.23 per day. Thus, value 0.23 per day given is w.r.t. the standard temperature of 20^{\circ}C.

\begin{aligned} &=0.23\left ( 1.047 \right )^{27-20} \\ &=0.317\; day^{-1} \\ BOD_{3} &=261\left ( 1-e^{-0.317\times 3} \right ) \\ &= 160.226\; mg/l \end{aligned}

For municipal sewage, at standard temperature, value of k(base e)=0.23 per day. Thus, value 0.23 per day given is w.r.t. the standard temperature of 20^{\circ}C.

\begin{aligned} &=0.23\left ( 1.047 \right )^{27-20} \\ &=0.317\; day^{-1} \\ BOD_{3} &=261\left ( 1-e^{-0.317\times 3} \right ) \\ &= 160.226\; mg/l \end{aligned}

Question 7 |

For a given water sample, the ratio between BOD_{5-day,20^{\circ}C} and the ultimate BOD is 0.68. The value of the reaction rate constant k (on base e) (in day^{-1} up to two decimal places) is_____

0.23 | |

0.13 | |

0.34 | |

0.63 |

Question 7 Explanation:

\begin{aligned} BOD/20^{\circ}C &=C_{0}\left ( 1-e^{-k\: 20^{\circ}C\times 5} \right ) \\ \frac{BOD/20^{\circ}C}{C_{0}} &=0.68=\left ( 1-e^{-k\: 20^{\circ}C\times 5} \right ) \\ K,20^{\circ}C &=0.227\; day^{-1}\simeq 0.23\; day^{-1} \end{aligned}

Question 8 |

Two wastewater streams A and B, having an identical ultimate BOD are getting mixed to form the stream C. The temperature of the stream A is 20^{\circ}C and the temperature of the stream C is 10^{\circ}C. It is given that

The 5-day BOD of the stream A measured at 20^{\circ}C=50 mg/l

BOD rate constant (base 10) at 20^{\circ}C=0.115 per day

Temperature coefficient = 1.135

The 5-day BOD (in mg/l, up to one decimal place)of stream C, calculated at 10^{\circ}C, is ______

The 5-day BOD of the stream A measured at 20^{\circ}C=50 mg/l

BOD rate constant (base 10) at 20^{\circ}C=0.115 per day

Temperature coefficient = 1.135

The 5-day BOD (in mg/l, up to one decimal place)of stream C, calculated at 10^{\circ}C, is ______

25.2 | |

54.8 | |

21.2 | |

68.1 |

Question 8 Explanation:

\begin{aligned} \text{At A,}\; \; \; \left ( BOD_{5} \right )_{20^{\circ}C}&=L\left ( 1-10^{-k_{D}20^{\circ}C\times 5} \right ) \\ 50 &=L\left ( 1-10^{-0.115\times 5} \right ) \\ L&= 68.129\; mg/l\\ \text{Here,} \; \; k_{D_{10}}&=k_{D_{20}}\times \left [ 1.135 \right ]^{10-20} \\ &=0.115\times \left ( 1.135 \right )^{-10} \\ &=0.0324 \\ \text{For C,}\; \;\left ( BOD_{5} \right )_{10^{\circ}C} &=L\left [ 1-10^{-k_{D_{10\times 4}}} \right ] \\ &=68.13\left [ 1-10^{-0.0324\times 5} \right ] \\ &=21.21\; mg/l \end{aligned}

Question 9 |

For a wastewater sample, the three-day biochemical oxygen demand at incubation temperature of 20\: ^{\circ}C \; (BOD_{3days,20{\circ}C}), is estimated as 200 mg/L. Taking the value of the first order BOD reaction rate constant as 0.22\: day^{-1}, the five-day BOD (expressed in mg/L) of the wastewater at incubation temperature of 20\: ^{\circ}C (BOD_{5days,20{\circ}C}) would be ___________

276.19 | |

265.24 | |

413.95 | |

224.25 |

Question 9 Explanation:

\begin{aligned} BOD_{5} &=L_{0}\left ( 1-e^{-0.22\times 5} \right ) \\ BOD_{3}&=L_{0}\left ( 1-e^{-0.22\times 3} \right ) \end{aligned}

From equation (i) and (ii), we get,

\begin{aligned} \frac{BOD_{5}}{BOD_{3}} &=\frac{\left ( 1-e^{-0.22\times 5} \right )}{\left ( 1-e^{-0.22\times 3} \right )} \\ BOD_{5} &=200\times \frac{0.667}{0.483} \\ &=200\times 1.38 \\ &=276.19\: mg/l \end{aligned}

From equation (i) and (ii), we get,

\begin{aligned} \frac{BOD_{5}}{BOD_{3}} &=\frac{\left ( 1-e^{-0.22\times 5} \right )}{\left ( 1-e^{-0.22\times 3} \right )} \\ BOD_{5} &=200\times \frac{0.667}{0.483} \\ &=200\times 1.38 \\ &=276.19\: mg/l \end{aligned}

Question 10 |

The 2-day and 4-day BOD values of a sewage sample are 100 mg/L and 155 mg/L, respectively. The value of BOD rate constant (expressed in per day) is ___________

0.1 | |

0.2 | |

0.3 | |

0.4 |

Question 10 Explanation:

\begin{matrix} Y_{2}=100\: mg/l & t_{2}=2\: day\\ Y_{4}=155\: mg/l & t_{4}=4\: day \end{matrix}

\begin{aligned} K_{D} &=? \\ Y_{2} &=Y_{0}\left ( 1-e^{-K_{D}.t_{2}} \right ) \\ Y_{4} &=Y_{0}\left ( 1-e^{-K_{D}.t_{4}} \right ) \\ \frac{Y_{2}}{Y_{4}} &=\frac{1-e^{-K_{D}\times 2}}{1-e^{-K_{D}\times 4}} \\ \frac{100}{155} &=\frac{1-e^{-2K_{D}}}{1-e^{-4K_{D}}} \\ 1.55 &=\frac{1-e^{-4K}}{1-e^{-2K}} \\ 1.55-1.55x &=1-x^{2} \\ x^{2}-1.55x-0.55 &=0 \\ x &=1\; \; \text{or}\; \; x=0.55 \\ e^{-2k} &=1\; \; \text{or}\; \; e^{-2k}=0.55 \\ -2k &=\ln 1\; \; \text{or}\; \; -2k=\ln 0.55 \\ k &=0\; \; \text{or}\; \; k=0.299 \\ \therefore \; \; k &=0.299 \end{aligned}

\begin{aligned} K_{D} &=? \\ Y_{2} &=Y_{0}\left ( 1-e^{-K_{D}.t_{2}} \right ) \\ Y_{4} &=Y_{0}\left ( 1-e^{-K_{D}.t_{4}} \right ) \\ \frac{Y_{2}}{Y_{4}} &=\frac{1-e^{-K_{D}\times 2}}{1-e^{-K_{D}\times 4}} \\ \frac{100}{155} &=\frac{1-e^{-2K_{D}}}{1-e^{-4K_{D}}} \\ 1.55 &=\frac{1-e^{-4K}}{1-e^{-2K}} \\ 1.55-1.55x &=1-x^{2} \\ x^{2}-1.55x-0.55 &=0 \\ x &=1\; \; \text{or}\; \; x=0.55 \\ e^{-2k} &=1\; \; \text{or}\; \; e^{-2k}=0.55 \\ -2k &=\ln 1\; \; \text{or}\; \; -2k=\ln 0.55 \\ k &=0\; \; \text{or}\; \; k=0.299 \\ \therefore \; \; k &=0.299 \end{aligned}

There are 10 questions to complete.