Question 1 |
A flood control structure having an expected life of n years is designed by
considering a flood of return period T years. When T=n , and n\rightarrow \infty , the
structure's hydrologic risk of failure in percentage is .
(round off to one decimal place)
25.2 | |
68.4 | |
78.2 | |
63.5 |
Question 1 Explanation:
Risk of failure =1-q^n=1-(1-p)^n=1-\left ( 1-\frac{1}{T} \right )^n
For T=n\rightarrow \infty
Risk of failure =1-\frac{1}{e}=0.632
% risk of failure = 0.632 x 100=63.2%
For T=n\rightarrow \infty
Risk of failure =1-\frac{1}{e}=0.632
% risk of failure = 0.632 x 100=63.2%
Question 2 |
In a certain month, the reference crop evapotranspiration at a location is
6 mm/day. If the crop coefficient and soil coefficient are 1.2 and 0.8, respectively,
the actual evapotranspiration in mm/day is
5.76 | |
7.2 | |
6.8 | |
8 |
Question 2 Explanation:
Actual evapotranspiration (ET_C)
=K_S \times K_C \times Reference evapotranspiration (ET_0)
=0.8 \times 1.2 \times 6=5.76mm
=K_S \times K_C \times Reference evapotranspiration (ET_0)
=0.8 \times 1.2 \times 6=5.76mm
Question 3 |
A two-hour duration storm event with uniform excess rainfall of 3 cm occurred
on a watershed. The ordinates of streamflow hydrograph resulting from this event
are given in the table.
\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \text{Time (hours)}& 0&1 &2 &3 &4 &5 &6 &7 \\ \hline \text{Streamflow}(m^3/s)& 10 &16 & 34 &40 & 31 & 25 &16&10 \\ \hline \end{array}
Considering a constant baseflow of 10 m^3/s , the peak flow ordinate (in m^3/s ) of one-hour unit hydrograph for the watershed is ________ . (in integer)
\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \text{Time (hours)}& 0&1 &2 &3 &4 &5 &6 &7 \\ \hline \text{Streamflow}(m^3/s)& 10 &16 & 34 &40 & 31 & 25 &16&10 \\ \hline \end{array}
Considering a constant baseflow of 10 m^3/s , the peak flow ordinate (in m^3/s ) of one-hour unit hydrograph for the watershed is ________ . (in integer)
12 | |
14 | |
22 | |
18 |
Question 3 Explanation:
C_1 = Time
C_2 = Ordinates of 2hr DRH
C_3 = Ordinates of 2hr UH=(Ordinates of 2hr DRH)/(Rianfall excess of 3cm)
C_4 = S-curve lag by 2hr
C_5 = S-curve ordinates S_2
C_6 = S_1 curve
C_7 = Ordinates of 1hr UH = \frac{S_2-S-1}{1/2}
\begin{array}{|c|c|c|c|c|c|c|} \hline C_1& C_2& C_3& C_4& C_5& C_6&C_7 \\ \hline 0&0 &0 &- &0 &- & 0\\ \hline 1&6 & 2& -&2 & 0& 4\\ \hline 2&24 & 8&0 &8 & 2&12 \\ \hline 3&30 &10 &2 &12 &8 &8 \\ \hline 4&21 & 7& 8& 15& 12& 6\\ \hline 5&15 &5 & 12& 17& 15& 4\\ \hline 6&6 & 2& 15& 17&17& 0\\ \hline 7 &0 & 0& 17& 17& 17&0\\ \hline & & & 17& 17& 17&0\\ \hline \end{array}
C_2 = Ordinates of 2hr DRH
C_3 = Ordinates of 2hr UH=(Ordinates of 2hr DRH)/(Rianfall excess of 3cm)
C_4 = S-curve lag by 2hr
C_5 = S-curve ordinates S_2
C_6 = S_1 curve
C_7 = Ordinates of 1hr UH = \frac{S_2-S-1}{1/2}
\begin{array}{|c|c|c|c|c|c|c|} \hline C_1& C_2& C_3& C_4& C_5& C_6&C_7 \\ \hline 0&0 &0 &- &0 &- & 0\\ \hline 1&6 & 2& -&2 & 0& 4\\ \hline 2&24 & 8&0 &8 & 2&12 \\ \hline 3&30 &10 &2 &12 &8 &8 \\ \hline 4&21 & 7& 8& 15& 12& 6\\ \hline 5&15 &5 & 12& 17& 15& 4\\ \hline 6&6 & 2& 15& 17&17& 0\\ \hline 7 &0 & 0& 17& 17& 17&0\\ \hline & & & 17& 17& 17&0\\ \hline \end{array}
Question 4 |
The hyetograph in the figure corresponds to a rainfall event of 3 cm.
If the rainfall event has produced a direct runoff of 1.6 cm, the \phi-index of the event (in mm/hour,round off to one decimal place) would be
If the rainfall event has produced a direct runoff of 1.6 cm, the \phi-index of the event (in mm/hour,round off to one decimal place) would be
1.2 | |
4.2 | |
3.4 | |
2.8 |
Question 4 Explanation:
\begin{aligned} \text { Total rainfall } &=3 \mathrm{~cm} \\ \text { Total runoff } &=1.6 \mathrm{~cm} \\ \therefore \qquad \qquad \qquad\text { Total infiltration } &=3-1.6=1.4 \mathrm{~cm}\\ \therefore \qquad \qquad \qquad \qquad\qquad\text { W-index }&=\frac{\text { Total infiltration }}{\text { Total duration of storm }}\\ &=\frac{1.4}{(210 / 60)} \mathrm{cm} / \mathrm{hr} \\ &=0.4 \mathrm{~cm} / \mathrm{hr}=4 \mathrm{~mm} / \mathrm{hr}\\ \text { As } \phi \text { -index }>\text { W-index }\qquad \qquad \end{aligned}
Hence storm of intensities 4 mm/hr and 3 mm/hr will not produce rainfall exam.
\begin{aligned} \phi \text { -index } &=\frac{\text { Total infiltration in which rainfall excess occur }}{\text { Time period in which rainfall excess occur }} \\ &=\frac{\text { Total infiltration }-\text { Infiltration in which no rainfall excess occur }}{T_{\text {excess }}} \\ &=\frac{14 \mathrm{~mm}-\left(4 \times \frac{30}{60}+3 \times \frac{30}{60}\right) \mathrm{mm}}{\left(\frac{150}{60}\right) \mathrm{hr}} \\ &=4.2 \mathrm{~mm} / \mathrm{hr} \end{aligned}
Question 5 |
A 12-hour unit hydrograph (of 1 cm excess rainfall) of a catchment is of a triangular shape with a base width of 144 hour and a peak discharge of 23 \mathrm{~m}^{3} / \mathrm{s}. The area of the catchment (in \mathrm{~km}^{2},round off to the nearest integer) is _______
412 | |
632 | |
596 | |
128 |
Question 5 Explanation:
Area of hydrograph = Total direct runoff volume
\begin{aligned} \Rightarrow \frac{1}{2} \times 23 \mathrm{~m}^{3} / \mathrm{sec} \times 144 \times &3600 \mathrm{sec}=\text { Area of catchment } \times \text { Runoff depth }\\ \Rightarrow \frac{1}{2} \times 23 \times 144 \times 3600 \mathrm{~m}^{3} &=A \times \frac{1}{100} \mathrm{~m} \\ A &=596.16 \times 10^{6} \mathrm{~m}^{2} \\ \therefore \quad \text{ Area of catchment }&=596.16 \mathrm{~km}^{2} \end{aligned}
Question 6 |
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 6 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 7 |
The value of abscissa (x) and ordinate (y) of a curve are as follows:
\begin{array}{|c|c|} \hline x & y \\ \hline 2.0 & 5.00 \\ \hline 2.5 & 7.25 \\ \hline 3.0 & 10.00 \\ \hline 3.5 & 13.25 \\ \hline 4.0 & 17.00 \\ \hline \end{array}
By Simpson's 1 / 3^{\mathrm{rd}} rule, the area under the curve (round off to two decimal places) is ______________
\begin{array}{|c|c|} \hline x & y \\ \hline 2.0 & 5.00 \\ \hline 2.5 & 7.25 \\ \hline 3.0 & 10.00 \\ \hline 3.5 & 13.25 \\ \hline 4.0 & 17.00 \\ \hline \end{array}
By Simpson's 1 / 3^{\mathrm{rd}} rule, the area under the curve (round off to two decimal places) is ______________
20.67 | |
54.62 | |
38.45 | |
66.22 |
Question 7 Explanation:
d = 0.5 unit
\begin{aligned} A &=\frac{d}{3}\left[\left(y_{1}+y_{5}\right)+4\left(y_{2}+y_{4}\right)+2 y_{3}\right] \\ &=\frac{0.5}{3}[(5+17)+4(7.25+13.25)+2 \times 10] \\ &=20.67 \text { unit }^{2} \end{aligned}
\begin{aligned} A &=\frac{d}{3}\left[\left(y_{1}+y_{5}\right)+4\left(y_{2}+y_{4}\right)+2 y_{3}\right] \\ &=\frac{0.5}{3}[(5+17)+4(7.25+13.25)+2 \times 10] \\ &=20.67 \text { unit }^{2} \end{aligned}
Question 8 |
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 8 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 9 |
A triangular direct runoff hydrograph due to a storm has a time base of 90 hours. The
peak flow of 60 m^3/s occurs at 20 hours from the start of the storm. The area of catchment
is 300 km^2. The rainfall excess of the storm (in cm), is
2 | |
3.24 | |
5.4 | |
6.48 |
Question 9 Explanation:
\left [ \frac{\frac{1}{2} \times 60 m^3/s \times 90 \times 3600s}{300 \times 10^6m^2} \times 100 \right ]cm= Rainfall excess
Rainfall excess = 3.24 cm
Question 10 |
Superpassage is a canal cross-drainage structure in which
natural stream water flows with free surface below a canal | |
natural stream water flows under pressure below a canal | |
canal water flows with free surface below a natural stream | |
canal water flows under pressure below a natural stream |
Question 10 Explanation:
Cross-section of a super passage
There are 10 questions to complete.