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}

Question 6 |

The relationship between porosity (\eta), specific yield (S_{y}) and specific retention (S_{r}) of an unconfined aquifer is:

S_{y}+S_{r}=\eta | |

S_{y}+\eta=S_{r} | |

S_{r}+\eta=S_{y} | |

S_{r}+S_{y}+\eta=1 |

Question 7 |

In an aquifer extending over 150 hectare, the water table was 20m below ground
level. Over a period of time the water table dropped to 23m below the ground
level. If the porosity of aquifer is 0.40 and the specific retention is 0.15, what is
the change in ground water storage of the aquifer?

67.5 ha-m | |

112.5 ha-m | |

180.0 ha-m | |

450.0 ha-m |

Question 7 Explanation:

We know that,

\begin{aligned} n&=S_{y}+S_{r} \\ \Rightarrow\quad S_{y}&=S_{t}-n=0.15-0.40=-0.25 \end{aligned}

The negative sign signifies the decrease in storage. Change in volume of aquifer

=150 \times(23-20)=450 \mathrm{ha}-\mathrm{m}

Specific yield is nothing but the actual volume of water that can be extracted by force of gravity from a unit volume of aquifer.

\therefore Change in water storage

=450 \times 0.25=112.5 \mathrm{ha}-\mathrm{m}

\begin{aligned} n&=S_{y}+S_{r} \\ \Rightarrow\quad S_{y}&=S_{t}-n=0.15-0.40=-0.25 \end{aligned}

The negative sign signifies the decrease in storage. Change in volume of aquifer

=150 \times(23-20)=450 \mathrm{ha}-\mathrm{m}

Specific yield is nothing but the actual volume of water that can be extracted by force of gravity from a unit volume of aquifer.

\therefore Change in water storage

=450 \times 0.25=112.5 \mathrm{ha}-\mathrm{m}

Question 8 |

A well of diameter 20 cm fully penetrates a confined aquifer. After a long period
of pumping at rate of 2720 litres per minute, the observations of drawdown taken
at 10 m and 100 m distances from the centre of the well are found to be 3m and
0.5m respectively. The transmissivity of the aquifer is

676m^{2}/day | |

576m^{2}/day | |

526m^{2}/day | |

249m^{2}/day |

Question 8 Explanation:

\begin{aligned} Q &=\frac{2 \pi T\left(s_{1}-s_{2}\right)}{\ln \left(\frac{r_{2}}{r_{1}}\right)} \\ T &=\frac{2720 \times 10^{-3} \times 24 \times 60 \times \ln \left(\frac{100}{10}\right)}{2 \times \pi \times(3-0.5)} \\ &=574.26 \mathrm{m}^{2} / \mathrm{day} \simeq 576 \mathrm{m}^{2} / \mathrm{day} \end{aligned}

Question 9 |

The relationship among specific yield (S_{y}), specific retention (S_{r}) and porosity (\eta)
of an aquifer is

S_{y}=S_{r}+\eta | |

S_{y}=S_{r}-\eta | |

S_{y}=\eta-S_{r} | |

S_{y}=S_{r}+2\eta |

Question 10 |

A volume of 3.0 \times 10^{6} m^{3} of groundwater was pumped form an unconfined aquifer uniformly from an area of 5 km^{2}. The pumping lowered the water table form initial level of 102 m to 99 m. The specific yield of the aquifer is

0.2 | |

0.3 | |

0.4 | |

0.5 |

Question 10 Explanation:

Specific yield

\begin{aligned} &=\frac{\begin{array}{c} \text{Volume of water pumped }\\\text { out of the aquifer }\end{array}}{\text { Total volume of aquifer }} \\ &=\frac{3.0 \times 10^{6}}{5 \times 10^{6} \times(102-99)}=0.20 \end{aligned}

\begin{aligned} &=\frac{\begin{array}{c} \text{Volume of water pumped }\\\text { out of the aquifer }\end{array}}{\text { Total volume of aquifer }} \\ &=\frac{3.0 \times 10^{6}}{5 \times 10^{6} \times(102-99)}=0.20 \end{aligned}

There are 10 questions to complete.