Question 1 |
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 1 Explanation:
From the above curve, it is evident that at pH 7.0, \mathrm{NH}_{4}^{+} will be predominant.
Question 2 |
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 2 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 3 |
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 3 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 4 |
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 4 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 5 |
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 5 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 6 |
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 6 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 7 |
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 7 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 8 |
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 8 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}
Question 9 |
Ultimate BOD of a river water sample is 20 mg/L. BOD rate constant (natural log) is 0.15 day^{-1}. The respective values of BOD (in %) exerted and remaining after 7 days are:
45 and 55 | |
55 and 45 | |
65 and 35 | |
75 and 25 |
Question 9 Explanation:
BOD remaining FTER 7 DAYS
\begin{aligned} &=BOD_{u}e^{-kt} \\ &=20\times e^{-0.15\times 7} \\ &=20\times 0.35=7 \\ \%: remaining &=\frac{7\times 100}{20}=35\% \end{aligned}
BOD exerted in 7 days
=\left ( 100-35 \right )\%=65\%
\begin{aligned} &=BOD_{u}e^{-kt} \\ &=20\times e^{-0.15\times 7} \\ &=20\times 0.35=7 \\ \%: remaining &=\frac{7\times 100}{20}=35\% \end{aligned}
BOD exerted in 7 days
=\left ( 100-35 \right )\%=65\%
Question 10 |
A student began experiment for determination of 5-day, 20^{\circ}C BOD on Monday. Since the 5^{th} day
fell on Saturday, the final DO readings were taken on next Monday. On calculation, BOD (i.e. 7
day, 20^{\circ}C) was found to be 150 mg/L. What would be the 5-day, 20^{\circ}C BOD (in mg/L)?
Assume value of BOD rate constant (k) at standard temperature of 20^{\circ}C as 0.23/day (base e). __________
Assume value of BOD rate constant (k) at standard temperature of 20^{\circ}C as 0.23/day (base e). __________
96.97 | |
158.68 | |
146.45 | |
128.13 |
Question 10 Explanation:
\begin{aligned} BOD_{7} &=L\left [ 1-e^{-k\times t} \right ] \\ \therefore \; \; 150 &=L\left [ 1-e^{-0.23\times 7} \right ] \\ \Rightarrow \; \; L &=187.54\: mg/L \\ \therefore \; \; BOD_{5} &=L\left [ 1-e^{-k\times t} \right ] \\ &=187.539\times \left [ 1-e^{-0.23\times 5} \right ] \\ &= 128.13\: mg/L \end{aligned}
There are 10 questions to complete.