Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 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 4 |
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 4 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 5 |
The probability that a 50 year flood may NOT occur at all during 25 years life of a project
(round off to two decimal places), is _______.
0.6 | |
0.25 | |
0.78 | |
0.44 |
Question 5 Explanation:
\begin{aligned} P&=\frac{1}{T}=\frac{1}{50}=0.02 \\ q&=1-P=0.98 \end{aligned}
\therefore \; Probability of non-occurance of an event is given by
\begin{aligned} \text{Assurance} &= q^n \\ &= (0.98)^{25}\\ &= 0.603 \end{aligned}
\therefore \; Probability of non-occurance of an event is given by
\begin{aligned} \text{Assurance} &= q^n \\ &= (0.98)^{25}\\ &= 0.603 \end{aligned}
Question 6 |
The ordinates, u, of a 2-hour unit hydrograph (i.e., for 1 cm of effective rain), for a catchment are shown in the table.
A 6-hour storm occurs over the catchment such that the effective rainfall intensity is 1 cm/hour for the first two hours, zero for the next two hours, and 0.5 cm/hour for the last two hours. If the base flow is constant at 5 m^3/s, the peak flow due to this storm (in m^3/s,round off to 1 decimal place) will be _____
A 6-hour storm occurs over the catchment such that the effective rainfall intensity is 1 cm/hour for the first two hours, zero for the next two hours, and 0.5 cm/hour for the last two hours. If the base flow is constant at 5 m^3/s, the peak flow due to this storm (in m^3/s,round off to 1 decimal place) will be _____45.6 | |
74.6 | |
64.3 | |
97.0 |
Question 6 Explanation:
Rainfall excess in 1st two hours,
R_1=1cm/hr \times 2 hr=2cm
Rainfall excess in 2nd two hours
R_2=0
Rainfall excess in 3rd two hours,
R_3=0.5cm/hr \times 2 hr=1cm
R_1=1cm/hr \times 2 hr=2cm
Rainfall excess in 2nd two hours
R_2=0
Rainfall excess in 3rd two hours,
R_3=0.5cm/hr \times 2 hr=1cm
Question 7 |
An inflow hydrograph is routed through a reservoir to produce an outflow hydrograph. The peak flow of the inflow hydrograph is P_I and the time of occurrence of the peak is t_I. The peak flow of the outflow hydrograph is P_O and the time of occurrence of the peak is t_O. Which one of the following statements is correct?
P_I \lt P_O \; and \; t_I \lt t_O | |
P_I \lt P_O \; and \; t_I \gt t_O | |
P_I \gt P_O \; and \; t_I \lt t_O | |
P_I \gt P_O \; and \; t_I \gt t_O |
Question 7 Explanation:
The outflow from the reservoir is uncontrolled therefore peak of outflow hydrograph will occur at the junction of inflow and outflow hydrograph.
P_I \gt P_0
t_I \lt t_0
Question 8 |
The hyetograph of a storm event of duration 140 minutesis shown in the figure.
The infiltration capacity at the start of this event (t=0) is 17 mm/hour, which linearly decreases to 10 mm/hour after 40 minutes duration. As the event progresses, the infiltration rate further drops down linearly to attain a value of 4 mm/hour at t=100 minutes and remains constant thereafter till the end of the storm event. The value of the infiltration index, \phi (in mm/hour, round off to 2 decimal places), is _______
The infiltration capacity at the start of this event (t=0) is 17 mm/hour, which linearly decreases to 10 mm/hour after 40 minutes duration. As the event progresses, the infiltration rate further drops down linearly to attain a value of 4 mm/hour at t=100 minutes and remains constant thereafter till the end of the storm event. The value of the infiltration index, \phi (in mm/hour, round off to 2 decimal places), is _______
7.24 | |
6.44 | |
9.26 | |
10.64 |
Question 8 Explanation:
\begin{aligned} P&=(4+8+15+10+8+3+1) \times \frac{20}{60} \\ &=16.33\; minute \\ Q&=(15-10) \times \frac{20}{60} +(10-8) \times \frac{20}{60} \\ &+(8-6) \times \frac{20}{60} +\frac{1}{2}\times 2 \times \frac{20}{60}+\frac{1}{2} \\ &+\frac{1}{2}\times 2 \times \frac{20}{60}+frac{1}{2}\times 2 \times \frac{20}{60}\\ &=4mm\\ W-index&=\frac{P-Q}{t}=\frac{16.33-4}{\frac{140}{60}}=5.28 mm/hr\\ \text{Since,} \; \phi & \geq W\\ \text{Assume, }\; \phi &=5.28 mm/hr\\ \Rightarrow \; \text{Corrected, }\phi &=\frac{16.33-4-4 \times \frac{20}{60}-3 \times \frac{20}{60}-1 \times \frac{20}{60}}{\left ( \frac{140-30-20-20}{60} \right )} \\ &=7.2475 mm/hr \end{aligned}
Question 9 |
The total rainfall in a catchment of area 1000\: km^{2}, during a 6 h storm, is 19 cm. The surface runoff due to this storm computed from triangular direct runoff hydrograph is 1 \times 10^8 m^{3}. The \phi _{index} for this storm (in cm/h, up to one decimal place) is ______
1 | |
0.5 | |
1.5 | |
2 |
Question 9 Explanation:
Surface runoff =\frac{1\times 10^{8}m^{3}}{1000\times 10^{6}m^{2}} =0.1 m=10 cm
Total rainfall = 19cm
Rainfall intensity =\frac{19}{6}=3.167 cm/hr
w-Index = \frac{P-Q}{t} =\frac{\text{Total Infiltration}}{\text{Total duration of storm}}
\therefore \;\; w-Index=\frac{19-10}{6}=1.5 cm/hr
As intensity of rainfall \gt w-Index
And rainfall intensity is uniform therefore \phi-index = w-Index= 1.5 cm/hr.
Total rainfall = 19cm
Rainfall intensity =\frac{19}{6}=3.167 cm/hr
w-Index = \frac{P-Q}{t} =\frac{\text{Total Infiltration}}{\text{Total duration of storm}}
\therefore \;\; w-Index=\frac{19-10}{6}=1.5 cm/hr
As intensity of rainfall \gt w-Index
And rainfall intensity is uniform therefore \phi-index = w-Index= 1.5 cm/hr.
Question 10 |
The infiltration rate f in a basin under ponding condition is given by f=30+10e^{-2t}, where, f is in mm/h and t is time in hour. Total depth of infiltration (in mm, up to one decimal place) during the last 20 minutes of a storm of 30 minutes duration is ______
7.36 | |
11.74 | |
25.74 | |
34.85 |
Question 10 Explanation:
Infiltration rate f(t)= 30+10e^{-2t}
Total infiltration depth in time 10 min. to 30 min.
i.e., 0.166 hour to 0.5 hour
=\int_{0.166}^{0.5}\left ( 30+10e^{-2t} \right )dt =11.74 mm
Total infiltration depth in time 10 min. to 30 min.
i.e., 0.166 hour to 0.5 hour
=\int_{0.166}^{0.5}\left ( 30+10e^{-2t} \right )dt =11.74 mm
There are 10 questions to complete.