Water Demand

Question 1
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
GATE CE 2020 SET-2   Environmental Engineering
Question 1 Explanation: 
\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
Question 2
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 ________.
GATE CE 2020 SET-1   Environmental Engineering
Question 2 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 3
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
GATE CE 2004   Environmental Engineering
Question 3 Explanation: 
Assuming arithmetic rate of growth of population.
Increase in population per year
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 3 questions to complete.

Leave a Comment