Question 1 |
Henry's law constant for transferring O_2 from air into water, at room temperature, is 1.3 \frac{mmol}{liter-atm}. Given that the partial pressure of O_2 in the
atmosphere is 0.21 atm, the concentration of dissolved oxygen (mg/liter) in
water in equilibrium with the atmosphere at room temperature is
(Consider the molecular weight of O_2 as 32 g/mol)
(Consider the molecular weight of O_2 as 32 g/mol)
8.7 | |
0.8 | |
198.1 | |
0.2 |
Question 1 Explanation:
Henry's Law C_g=KP
C_g= Concentration of dissolved gas
K= Henry?s law constant
P= Partial pressure of gas
C_g= 1.3 \;m\; mol/(lit-atm) \times 0.21 atm =0.273 \;m\; mol/lit=0.273 \times 32 \; mg/lit=8.736 \; mg/lit
C_g= Concentration of dissolved gas
K= Henry?s law constant
P= Partial pressure of gas
C_g= 1.3 \;m\; mol/(lit-atm) \times 0.21 atm =0.273 \;m\; mol/lit=0.273 \times 32 \; mg/lit=8.736 \; mg/lit
Question 2 |
For wastewater coming from a wood pulping industry, Chemical Oxygen
Demand (COD) and 5-day Biochemical Oxygen Demand (BOD_5) were
determined. For this wastewater, which of the following statement(s) is/are
correct?
COD \gt BOD_5 | |
COD \neq BOD_5 | |
COD \lt BOD_5 | |
COD = BOD_5 |
Question 2 Explanation:
We know COD is the oxygen demand for both biodegradable and non-biodegradable organic matter. Hence COD \gt BOD_5 also COD \neq BOD_5
Question 3 |
The relationship between oxygen consumption and equivalent biodegradable organic
removal (i.e. BOD) in a closed container with respect to time is shown in the figure.
Assume that the rate of oxygen consumption is directly proportional to the amount of degradable organic matter and is expressed as \frac{dL_t}{dt}=-kL_t, where, L_t (in mg/litre) is the oxygen equivalent of the organics remaining at time t and k (in d^{-1}) is the degradation rate constant. L_0 is the oxygen of organic matter at time, t = 0.
In the above context, the correct expression is
Assume that the rate of oxygen consumption is directly proportional to the amount of degradable organic matter and is expressed as \frac{dL_t}{dt}=-kL_t, where, L_t (in mg/litre) is the oxygen equivalent of the organics remaining at time t and k (in d^{-1}) is the degradation rate constant. L_0 is the oxygen of organic matter at time, t = 0.
In the above context, the correct expression is
BOD_5=L_5 | |
BOD_t=L_0-L_t | |
L_0=L_te^{-kt} | |
L_t=L_0(1-e^{-kt}) |
Question 3 Explanation:
Given,
\begin{aligned} \frac{dL_t}{dt}&=-kL_t \\ \Rightarrow \; \frac{dL_t}{L_t} &=-kdt \\ \Rightarrow \; \int_{L_o}^{L_t} \frac{dL_t}{L_t} &=-k\int_{0}^{t}dt \\ \Rightarrow \; \ln \left ( \frac{L_t}{L_o} \right ) &=-kt \\ \Rightarrow \; L_t&=L_o e^{-kt} \end{aligned}
The term L_o in the above equation represents the total oxygen equivalent of the organics at time t=0, while L_t represents the amount remaining at time t, and decays exponentially with time.
The amount of oxygen used in the consumption of the organics, the {BOD}_t can be found from the L_t value.
The difference between the value L_o and L_t is the oxygen equivalent consumed, or the BOD exerted. Mathematically
{BOD}_t=(L_o-L_t)=L_o-L_oe^{-kt}=L_o(1-e^{-kt})
\begin{aligned} \frac{dL_t}{dt}&=-kL_t \\ \Rightarrow \; \frac{dL_t}{L_t} &=-kdt \\ \Rightarrow \; \int_{L_o}^{L_t} \frac{dL_t}{L_t} &=-k\int_{0}^{t}dt \\ \Rightarrow \; \ln \left ( \frac{L_t}{L_o} \right ) &=-kt \\ \Rightarrow \; L_t&=L_o e^{-kt} \end{aligned}
The term L_o in the above equation represents the total oxygen equivalent of the organics at time t=0, while L_t represents the amount remaining at time t, and decays exponentially with time.
The amount of oxygen used in the consumption of the organics, the {BOD}_t can be found from the L_t value.
The difference between the value L_o and L_t is the oxygen equivalent consumed, or the BOD exerted. Mathematically
{BOD}_t=(L_o-L_t)=L_o-L_oe^{-kt}=L_o(1-e^{-kt})
Question 4 |
A stream with a flow rate of 5 m^3/s is having an ultimate BOD of 30 mg/litre. A wastewater
discharge of 0.20 m^3/s having BOD5 of 500 mg/litre joins the stream at a location and
instantaneously gets mixed up completely. The cross-sectional area of the stream is
40 m^2 which remains constant. BOD exertion rate constant is 0.3 per day (logarithm base
to e). The BOD (in mg/litre round off to two decimal places) remaining at 3 km downstream
from the mixing location, is ________.
53.62 | |
49.57 | |
23.45 | |
78.56 |
Question 4 Explanation:
\begin{aligned}
t &=\frac{d}{v} \\
\text{where,} \;\; v&=\frac{Q_s+Q_R}{A} \\
&= \frac{0.2+5}{40}=0.13 m/sec\\
t&=\frac{3 \times 10^3}{0.13 \times 86400}=0.26 \;day \\
BOD_5&=BOD_u(1-e^{-k\times 5}) \\
BOD_u&=\frac{500}{(1-e^{-0.5 \times 5})}=643.66\; mg/l \\
DO_{mix}&=\frac{Q_RBOD_u+Q_s BOD_u}{Q_s+Q_R} \\
&=\frac{5 \times 30+0.2\times 643.66}{5+0.2} \\
&= 53.6\; mg/l\\
L_t&=L_0 e^{-k \times t} \\
&=53.6 e^{-0.3 \times 0.26}\\
&=49.57\; mg/l
\end{aligned}
Question 5 |
The present population of a community is 28000 with an average water consumption of 4200 m^{3}/d. The existing water treatment plant has a design capacity of 6000 m^{3}/d. It is expected that the population will increase to 44000 during the next 20 years. The number of years from now when the plant will reach its design capacity, assuming an arithmetic rate of population growth, will be
5.5 years | |
8.6 years | |
15 years | |
16.5 years |
Question 5 Explanation:
Assuming arithmetic rate of growth of population.
Increase in population per year
=\frac{44000-28000}{20}=800
Discharge required to reach design capacity
=6000-4200=1800 m^{3} / d
Present average water consumption per head per day
=\frac{4200}{28000}=0.15 \mathrm{m}^{3}
Increase in water consumption each year
=0.15 \times 800=120 \mathrm{m}^{3} / \mathrm{day}
\therefore No. of years required to reach design capacity
=\frac{1800}{120}=15 \text { years }
Increase in population per year
=\frac{44000-28000}{20}=800
Discharge required to reach design capacity
=6000-4200=1800 m^{3} / d
Present average water consumption per head per day
=\frac{4200}{28000}=0.15 \mathrm{m}^{3}
Increase in water consumption each year
=0.15 \times 800=120 \mathrm{m}^{3} / \mathrm{day}
\therefore No. of years required to reach design capacity
=\frac{1800}{120}=15 \text { years }
There are 5 questions to complete.