Question 1 |

A catchment may be idealized as a circle of radius 30 \mathrm{~km}. There are five rain gauges, one at the center of the catchment and four on the boundary (equi-spaced), as shown in the figure (not to scale).

The annual rainfall recorded at these gauges in a particular year are given below.

\begin{array}{|c|c|c|c|c|c|} \hline Gauge & \mathrm{G}_{1} & \mathrm{G}_{2} & \mathrm{G}_{3} & \mathrm{G}_{4} & \mathrm{G}_{5} \\ \hline Rainfall (mm) & 910 & 930 & 925 & 895 & 905 \\ \hline \end{array}

Using the Thiessen polygon method, what is the average rainfall (in \mathrm{mm}, rounded off to two decimal places) over the catchment in that year ? ___

The annual rainfall recorded at these gauges in a particular year are given below.

\begin{array}{|c|c|c|c|c|c|} \hline Gauge & \mathrm{G}_{1} & \mathrm{G}_{2} & \mathrm{G}_{3} & \mathrm{G}_{4} & \mathrm{G}_{5} \\ \hline Rainfall (mm) & 910 & 930 & 925 & 895 & 905 \\ \hline \end{array}

Using the Thiessen polygon method, what is the average rainfall (in \mathrm{mm}, rounded off to two decimal places) over the catchment in that year ? ___

912.56 | |

521.32 | |

5463.25 | |

472.36 |

Question 1 Explanation:

Radius of basin =30 \mathrm{~km}

We know that in Theissen polygon method, the polygons are obtained by joining the perpendicular bisectors of triangles formed, when various rain gauge stations are joined.

The Theissen polygon in the given basin can be drawn as:

Area under influence of station GI

=30 \times 30=900 \mathrm{~km}^{2}

Area under influence of stations

G2, G3, G4, G5

=\frac{\pi \times 30^{2}-900}{4}

=481.86 \mathrm{~km}^{2}

mean rainfall =\frac{\begin{array}{c}910 \times 900+481.86 \times 930+ \\ 481.86 \times 925+481.86 \times 895+ \\ 481.86 \times 905\end{array}}{900+481.86 \times 4} =912.56 \mathrm{~mm}

We know that in Theissen polygon method, the polygons are obtained by joining the perpendicular bisectors of triangles formed, when various rain gauge stations are joined.

The Theissen polygon in the given basin can be drawn as:

Area under influence of station GI

=30 \times 30=900 \mathrm{~km}^{2}

Area under influence of stations

G2, G3, G4, G5

=\frac{\pi \times 30^{2}-900}{4}

=481.86 \mathrm{~km}^{2}

mean rainfall =\frac{\begin{array}{c}910 \times 900+481.86 \times 930+ \\ 481.86 \times 925+481.86 \times 895+ \\ 481.86 \times 905\end{array}}{900+481.86 \times 4} =912.56 \mathrm{~mm}

Question 2 |

A tube-well of 20 cm diameter fully penetrates a horizontal, homogeneous and isotropic confined aquifer of infinite horizontal extent. The aquifer is of 30 m uniform thickness. A steady pumping at the rate of 40 litres/s from the well for a long time results in a steady drawdown of 4 m at the well face. The subsurface flow to the well due to pumping is steady, horizontal and Darcian and the radius of influence of the well is 245 m. The hydraulic conductivity of the aquifer (in m/day,round off to integer) is ______________

45 | |

82 | |

36 | |

12 |

Question 2 Explanation:

\begin{aligned} 40 \times 10^{-3} \mathrm{~m}^{3} / \mathrm{s} &=\frac{2 \pi k \times 30 \mathrm{~m} \times 4 \mathrm{~m}}{\ln \left(\frac{245}{0.1}\right)} \\ k &=4.14 \times 10^{-4} \mathrm{~m} / \mathrm{s} \\ \text{or, } \qquad \qquad \qquad \qquad k &=35.77 \mathrm{~m} / \mathrm{d} \\ &\approx 36 \mathrm{~m} / \mathrm{d} (\text{ round off to nearest integer}) \end{aligned}

Question 3 |

Two reservoirs are connected through a homogeneous and isotropic aquifer having hydraulic conductivity (K) of 25 m/day and effective porosity (\eta) of 0.3 as shown in the figure (not to scale). Ground water is flowing in the aquifer at the steady state.

If water in Reservoir 1 is contaminated then the time (in days, round off to one decimal place) taken by the contaminated water to reach to Reservoir 2 will be _____________

If water in Reservoir 1 is contaminated then the time (in days, round off to one decimal place) taken by the contaminated water to reach to Reservoir 2 will be _____________

2000 | |

2400 | |

2800 | |

3200 |

Question 3 Explanation:

\begin{aligned} k&=25 \mathrm{~m} / \mathrm{d} \\ n&=0.3 \\ i&=\frac{20 \mathrm{~m}}{2000 \mathrm{~m}}=0.01 \\ V&=k i \\ V_{s}&=\frac{V}{n} \\ t&=\frac{l}{V_{s}} \\ t&=2400 \mathrm{days} \end{aligned}

Question 4 |

A catchment may be idealised as a rectangle. There are three rain gauges located inside the catchment at arbitrary locations. The average precipitation over the catchment is estimated by two methods:

(i) Arithmetic mean (P_A), and

(ii) Thiessen polygon (P_T).

Which one of the followingstatements is correct?

(i) Arithmetic mean (P_A), and

(ii) Thiessen polygon (P_T).

Which one of the followingstatements is correct?

P_A is always smaller than P_T | |

P_A is always greater than P_T | |

P_A is always equal to P_T | |

There is no definite relationship between P_A \; and \; P_T |

Question 4 Explanation:

The result from Thiessen polygon method is more accurate than arithmetic mean method.
But there is no any close relationship between values obtained by Thiessen polygon method and Arithmetic mean method.

Therefore, there is no any relation between P_A and P_T.

Therefore, there is no any relation between P_A and P_T.

Question 5 |

Rainfall depth over a watershed is monitored through six number of well distributed rain gauges. Gauged data are given below

The Thiessen mean value (in mm, up to one decimal place) of the rainfall is ______

The Thiessen mean value (in mm, up to one decimal place) of the rainfall is ______

479.1 | |

258.8 | |

147.3 | |

369.4 |

Question 5 Explanation:

Thiessen mean value of rainfall;

P_{avg}=\frac{\sum_{i=1}^{6}P_{i}A_{i}}{\sum_{i=1}^{6}A_{i}}

P_{avg}=\frac{470\times 95+465\times 100+435\times 98+525\times 80+480\times 85+510\times 92}{95+100+98+80+85+92} =479.09 mm

P_{avg}=\frac{\sum_{i=1}^{6}P_{i}A_{i}}{\sum_{i=1}^{6}A_{i}}

P_{avg}=\frac{470\times 95+465\times 100+435\times 98+525\times 80+480\times 85+510\times 92}{95+100+98+80+85+92} =479.09 mm

There are 5 questions to complete.