Question 1 |
The concentration s(x,t) of pollutants in a one-dimensional reservoir at position x and time t satisfies the diffusion equation
\frac{\partial s(x,t)}{\partial t}=D\frac{\partial^2 s(x,t)}{\partial x^2}
on the domain 0 \leq x \leq L, where D is the diffusion coefficient of the pollutants. The initial condition s(x, 0) is defined by the step-function shown in the figure.
The boundary conditions of the problem are given by \frac{\partial s(x,t)}{\partial x}=0 at the boundary points x = 0 and x = L at all times. Consider D = 0. 1 m^2/s, s_0 = 5 \mu mol/m, and L = 10 m.
The steady-state concentration \tilde{S}\left ( \frac{L}{2} \right )=S\left ( \frac{L}{2},\infty \right ), at the center x=\frac{L}{2} of the reservoir (in \mumol/m) is ___________. (in integer)
\frac{\partial s(x,t)}{\partial t}=D\frac{\partial^2 s(x,t)}{\partial x^2}
on the domain 0 \leq x \leq L, where D is the diffusion coefficient of the pollutants. The initial condition s(x, 0) is defined by the step-function shown in the figure.
The boundary conditions of the problem are given by \frac{\partial s(x,t)}{\partial x}=0 at the boundary points x = 0 and x = L at all times. Consider D = 0. 1 m^2/s, s_0 = 5 \mu mol/m, and L = 10 m.
The steady-state concentration \tilde{S}\left ( \frac{L}{2} \right )=S\left ( \frac{L}{2},\infty \right ), at the center x=\frac{L}{2} of the reservoir (in \mumol/m) is ___________. (in integer)
1 | |
2 | |
3 | |
4 |
Question 1 Explanation:
From figure s(x,t)
at x=0\Rightarrow s(0,t)=s_0=5
at x=0.4L\Rightarrow s(0.4L,t)=s_0=5
at x=L\Rightarrow s(L,t)=0
From x = 0\; to \;x = 0.4 L
Concentration of pollutant 5\mu \; mol/m \times (0.4 \times 10m) =20 \mu \; mol
x = 0.4L\; to \;x = L
Concentration of pollutant = 0
Total concentration of pollutant in 10 m =20 \mu \; mol
In infinite time this concentration will be diluted so concentration of pollutant per m
=\frac{20}{10} \mu \; mol/m=2 \mu \; mol/m
Under steady state condition, concentration of pollutant will be uniformly distributed.
Steady state concentration at x=\frac{L}{2} =2\mu \; mol/m
at x=0\Rightarrow s(0,t)=s_0=5
at x=0.4L\Rightarrow s(0.4L,t)=s_0=5
at x=L\Rightarrow s(L,t)=0
From x = 0\; to \;x = 0.4 L
Concentration of pollutant 5\mu \; mol/m \times (0.4 \times 10m) =20 \mu \; mol
x = 0.4L\; to \;x = L
Concentration of pollutant = 0
Total concentration of pollutant in 10 m =20 \mu \; mol
In infinite time this concentration will be diluted so concentration of pollutant per m
=\frac{20}{10} \mu \; mol/m=2 \mu \; mol/m
Under steady state condition, concentration of pollutant will be uniformly distributed.
Steady state concentration at x=\frac{L}{2} =2\mu \; mol/m
Question 2 |
A lake has a maximum depth of 60 m. If the mean atmospheric pressure in the lake region is 91 kPa and the unit weight of the lake water is 9790 \mathrm{~N} / \mathrm{m}^{3}, the absolute pressure (in kPa,round off to two decimal places) at the maximum depth of the lake is ___________
678.4 | |
258.6 | |
458.2 | |
125.9 |
Question 2 Explanation:
Absolute pressure at maximum depth of the lake =P_{\text {atm }}+\rho g h
=91+\frac{9790(60)}{1000}=678.4 \mathrm{kPa}
=91+\frac{9790(60)}{1000}=678.4 \mathrm{kPa}
Question 3 |
Dupuit's assumptions are valid for
artesian aquifer | |
confined aquifer | |
leaky aquifer | |
unconfined aquifer |
Question 3 Explanation:
Dupuit's theory assumptions hold that groundwater flows horizontally in an unconfined aquifer and that ground water discharge is propotional to saturated aquifer thickness.
Question 4 |
A tracer takes 100 days to travel from Well-1 to Well-2 which are 100 m apart. The elevation of water surface in Well-2 is 3m below that in Well-1. Assuming porosity equal to 15%, the coefficient of permeability (expressed in m/day) is
0.3 | |
0.45 | |
1 | |
5 |
Question 4 Explanation:
Seepage velocity,
\begin{aligned} V_{s}&=\frac{v}{n}=\frac{\text { distance }}{\text { time }}\\ \text{as per Darcy},\\ v &=k i \\ \frac{k i}{n} &=\frac{100 \mathrm{m}}{100 \mathrm{days}} \\ i &=\frac{\text { head difference }}{\text { length }}=\frac{3}{100} \\ \frac{k \times 3}{0.15 \times 100} &=\frac{100}{100} \mathrm{m} / \mathrm{day} \\ k &=5 \mathrm{m} / \mathrm{day} \end{aligned}
\begin{aligned} V_{s}&=\frac{v}{n}=\frac{\text { distance }}{\text { time }}\\ \text{as per Darcy},\\ v &=k i \\ \frac{k i}{n} &=\frac{100 \mathrm{m}}{100 \mathrm{days}} \\ i &=\frac{\text { head difference }}{\text { length }}=\frac{3}{100} \\ \frac{k \times 3}{0.15 \times 100} &=\frac{100}{100} \mathrm{m} / \mathrm{day} \\ k &=5 \mathrm{m} / \mathrm{day} \end{aligned}
Question 5 |
Water table of an aquifer drops by 100 cm over an area of 1000 km^{2}. The porosity and specific retention of the aquifer material are 25% and 5%, respectively. The amount of water (expressed in km^{3} ) drained out from the area is ______ .
0.2 | |
0.3 | |
0.5 | |
0.7 |
Question 5 Explanation:
\begin{aligned} S_{\mathrm{r}}+S_{\mathrm{y}} &=n \\ \frac{5}{100}+\frac{V_{\mathrm{w}}}{10^{3} \times 10^{6} \times 1} &=\frac{25}{100} \\ V_{\mathrm{w}} &=0.2 \times 10^{9} \mathrm{m}^{3}=0.2 \mathrm{km}^{3} \end{aligned}
There are 5 questions to complete.