Q21GATE 2017NAT

The ordinates of a 2-hour unit hydrograph for a catchment are given as The ordinate (in $m^{3}/s$ ) of a 4-hour unit hydrograph for this catchment at the time of 3 h would be ______

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📖 Explanation

$\therefore$ Ordinate of 4 hr U.H. at 3 hr duration $=15.0 m^{3}/sec.$

Q22GATE 2017NAT

During a storm event in a certain period, the rainfall intensity is 3.5 cm/hour and the $\Phi$ -index is 1.5 cm/hour. The intensity of effective rainfall (in cm/hour, up to one decimal place) for this period is

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📖 Explanation

Let the rainfall duration is 't' hours $\therefore$ Total rainfall =3.5 cm/hr $\times t$ hr =3.5 t Total infitration = $\phi$ -index $\times t$ =1.5 t $\therefore$ Intensity of rainfall excess $=\frac{\text{Total rainfall excess}}{\text{Time period}}$ $=\frac{3.5t-1.5t}{t}$ cm/hr =2.0 cm/hr

Q23GATE 2017NAT

An effective rainfall of 2-hour duration produced a flood hydrograph peak of 200 $m^{3}/s$ . The flood hydrograph has a base flow of 20 $m^{3}/s$ . If the spatial average rainfall in the watershed for the duration of storm is 2 cm and the average loss rate is 0.4 cm/hour, the peak of 2-hour unit hydrograph (in $m^{3}/s-cm$ , up to one decimal place) is _____

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📖 Explanation

Peak of 2 hr direct runoff hydrograph = Peak of 2 hr flood hydrograph-Base flow =200-20 =180 $m^{3}/sec.$ The rainfall excess depth for a given duration of 2 hr effective storm/rainfall is given by, Rainfall excess depth $=2 cm-0.4 \frac{cm}{hr}\times 2 hr$ $=1.2 cm$ $\therefore$ Peak of 2 hr UH $=\frac{180}{1.2}=150 m^{3}/s-cm$

Q24GATE 2017NAT

The infiltration capacity of a soil follows the Horton's exponential model, $f=c_{1}+c_{2} e^{-kt}$ . During and experiment, the initial infiltration capacity was observed to be 200 mm/h. After a long time, the infiltration capacity was reduced to 25 mm/h. If the infiltration capacity after 1 hour was 90 mm/h, the value of the decay rate constant, k (in $h^{-1}$ , up to two decimal places) is____

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📖 Explanation

Horton's expression for infiltration capacity of a soil is given as, $f=C_{1}+C_{2}.e^{-kt}$ Here, $C_{1}=$ Constant or steady stage infiltration capacity $C_{2}=$ Difference between initial and steady stage infiltration capacity $=\left ( f_{0}-f_{c} \right )$ $f=$ Infiltration capacity of any time 't' $\therefore \;\; f_{p}=f_{c}+\left ( f_{0}-f_{c} \right )e^{-k_{h}t}$ $\Rightarrow \;\; 90=25+\left ( 200-25 \right )e^{-k_{h}1}$ $\Rightarrow \;\; k_{h}=0.9904\, hr^{-1}$

Q25GATE 2016NAT

The direct runoff hydrograph in response to 5 cm rainfall excess in a catchment is shown in the figure. The area of the catchment (expressed in hectares) is ___________.

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📖 Explanation

\begin{aligned} \frac{1}{2}\times 1\times 6\times 3600&=A\times \frac{5}{100} \\ \Rightarrow \;\; 2160\times 100&=A \\ \Rightarrow \;\; A&=216000 m^{2} \\ &=21.6 ha \end{aligned}

Q26GATE 2016NAT

The ordinates of a one-hour unit hydrograph at sixty minute interval are 0, 3, 12, 8, 6, 3 and 0 $m^{3}/s$ . A two-hour storm of 4 cm excess rainfall occurred in the basin from 10 AM. Considering constant base flow of 20 $m^{3}/s$ , the flow of the river (expressed in $m^{3}/s$ ) at 1 PM is_____________

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📖 Explanation

$\therefore$ Ordinate of 01:00 pm $=60 m^{3}/sec$

Q27GATE 2015NAT

The 4-hr unit hydrograph for a catchment is given in the table below. What would be the maximum ordinate of the S-curve (in $m^{3}/s$ ) derive from this hydrograph?

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📖 Explanation

Hence, maximum ordinate is 22 $m^{3}/s.$

Q28GATE 2014MCQ

An isolated 3-h rainfall event on a small catchment produces a hydrograph peak and point of inflection on the falling limb of the hydrograph at 7 hours and 8.5 hours respectively, after the start of the rainfall. Assuming, no losses and no base flow contribution, the time of concentration (in hours) for this catchment is approximately

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Q29GATE 2013MCQ

At a station, Storm I of 5 hour duration with intensity 2 cm/h resulted in a runoff of 4 cm and Storm II of 8 hour duration resulted in a runoff of 8.4 cm. Assume that the $\phi$ -index is the same for both the storms. The intensity of storm II (in cm/h) is:

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📖 Explanation

Let intensity of storm-II be P cm/hr, $\therefore \;\; 1.2=\frac{P\times 8-8.4}{8}$ $\Rightarrow \;\; 1.2\times 8+8.4=P\times 8$ $P=2.25 cm/hr$

Q30GATE 2012MCQ

Group I contains parameters and Group II lists methods/instruments. The CORRECT match of Group I with Group II is

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Q31GATE 2012MCQ

The drainage area of a watershed is 50 $km^2$ . The $\phi$ index is 0.5 cm/hour and the base flow at the outlet is 10 $m^3$ /s. One hour unit hydrograph (unit depth = 1 cm) of the watershed is triangular in shape with a time base of 15 hours. The peak ordinate occurs at 5 hours For a storm of depth of 5.5 cm and duration of 1 hour, the peak ordinate (in $m^3$ /s) of the hydrograph is

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📖 Explanation

Total infiltration $=0.5\times 1=0.5cm$ Total runoff =5.5-0.5=5.0 cm $\therefore$ Peak of the direct runoff hydrograph due to 5 cm of rainfall excess depth $=18.52\times 5=92.60cm$ Peak of the flood hydrograph =Peak of D.R.H. + Base flow =92.60+10=102.60 $m^{3}/s$

Q32GATE 2011MCQ

A watershed got transformed from rural to urban over a period of time. The effect of urbanization on storm runoff hydrograph from the watershed is to

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Q33GATE 2011MCQ

The ordinates of a 2-h unit hydrograph at 1 hour intervals starting from time t = 0, are 0, 3, 8, 6, 3, 2 and 0 $m^3$ /s. Use trapezoidal rule for numerical integration, if required. A storm of 6.6 cm occurs uniformly over the catchment in 3 hours. If $\phi$ -index is equal to 2mm/h and base flow is 5 $m^{3}/s$ , what is the peak flow due to the storm?

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📖 Explanation

Peak of 3-h UH $= 6 m^{3}$ /s Runoff due to storm $=6.6\times \frac{2}{10}\times 3=6$ cm Peak flow due to storm $=6\times 6+5=41\, m^{3}/s$

Q34GATE 2009MCQ

One hour triangular unit hydrograph of a watershed has the peak discharge of 60 $m^3/sec . cm$ at 10 hours and time base of 30 hours. The $\phi$ index is 0.4 cm per hour and base flow is 15 $m^3/sec$ . If there is rainfall of 5.4 cm in 1 hour, the ordinate of the flood hydrograph at $15^{th}$ hour is

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📖 Explanation

Total rainfall =5.4 cm Infiltration in 1 hour $=\phi -index \times 1 =0.4\times 1=0.4 cm$ Runoff depth $=5.4-0.4 =5 cm.$ The discharge in the unit hydrograph decreases from $60 m^{3}/sec$ to 0 in a period of 20 hours. $\therefore$ Discharge at the 15th hour $=60-\frac{60}{20}\times 5=45 m^{3} /sec$ $\therefore$ Ordinate of DRH of 1h $=45\times 5=225m^{3}/s$ Ordinate of flood hydrograph, Q=225+ Base flow = 225+15 = 240 $m^3/sec$

Q35GATE 2007MCQ

An isolated 4-hour storm occurred over a catchment as follows : The $\phi$ index for the catchment is 10 mm/h. The estimated runoff depth from the catchment due to the above storm is

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📖 Explanation

\begin{aligned} \phi-index &=\frac{\text { Total Infiltration }}{\text { Time of rainfall excess }} \\ \Rightarrow \quad 10&=\frac{(28+12)-\text { Runoff }}{\text { Time of rainfall excess }} \end{aligned}

But precipitation in the $1^{\text {st }}$ and $4^{\text {th }}$ hours is less than the $\phi$ -index i.e. the rainfall during these two hours is ineffective which means time of rainfall excess will be 2 hrs.

\begin{aligned} \therefore \quad 10&=\frac{40-\text { Runoff }}{2}\\ \Rightarrow \quad \text{ Runoff }&=20 \mathrm{mm} \end{aligned}

Q36GATE 2007MCQ

Ordinates of a 1-hour unity hydrograph at 1 hour intervals, starting from time t = 0 are 0, 2, 6, 4, 2, 1 and 0 $m^{3}/s$ . Ordinate of a 3-hour unit hydrograph for the catchment at t = 3 hours is

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📖 Explanation

Thus at time t = 3h, the ordinate of 3h - UH is $4 m^{3}/s$

Q37GATE 2006MCQ

During a 3 hour storm event, it was observed that all abstractions other than infiltration are negligible. The rainfall was idealized as 3 one hour storms of intensity 10 mm/hr, 20 mm/hr and 10 mm/hr respectively and the infiltration was idealized as a Horton curve, $f = 6.8 + 8.7 exp(- t)$ (f in mm/hr and t in hr). What is the effective rainfall ?

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📖 Explanation

From superposition of hyetograph and infiltration curve it is clear that first hour does not contribute to effective rainfall (i.e. rainfall excess). For t=1 to t=3, the effective rainfall is,

\begin{aligned} r_e&=20+10-\int_{1}^{3} (6.8+8.7e^{-t})dt\\ &=30-\left[ (6.8t)_1^{3+}8.7 \int_{1}^{3}e^{-t} dt\right]\$ \\ &= 30-\$[6.8 \times 2 -8.7(e^{-t})_1^3]\$\\ &=30-13.6+8.7\left[ \frac{1}{e^3}-\frac{1}{e^1} \right]\$ \\ &= 16.4-2.77\\ &= 13.63mm \end{aligned}

Q38GATE 2006MCQ

For a catchment, the S-curve (or S-hydrograph) due to a rainfall of intensity 1 cm/hr is given by $Q=1-(1+t)exp(-t)$ (t in hr and Q in $m^3/s$ ) What will be the ordinate of a 2-hour unit hydrograph for this catchment at t=3 hours ?

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📖 Explanation

The duration of S_curve =1 hr The ordinate of 2-hr UH is obtained by the following procedure: Step 1: The ordinate of S-curve at t=3 hr $S_3=1-(1-3)e^{-3}=0.8m^3/s$ Step 2: The ordinate of S-curve at $t=3-2=1 \;hr$ $S_1=1-(1+1)e^{-1}=0.26m^3/s$ Step 3: $S_3-S_1=0.8-0.26=0.54m^3/s$ Step 4: The ordinate of UH at t=3 hr is $\frac{S_3-S_2}{2}=0.27 m^3/s$

Q39GATE 2005MCQ

A four hour unit hydrograph of a catchment is triangular in shape with base of 80hours. The area of the catchment is 720 $km^2$ . The base flow and f-index are 30 $m^3/s$ and 1mm/h, respectively. A storm of 4cm occurs uniformly in 4 hours over the catchment. The peak flood discharge due to the storm is

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📖 Explanation

Depth of precipitation =4 cm Infiltration in 4 hrs $=4\times 1 =4 mm =0.4 cm$ Run off depth $=4-0.4=3.6 cm$ Peak of DRH of 4 hrs $=50\times 3.6 =180m^{3}/sec$ Peak of flood discharge, $Q=180+30 =210m^{3}/sec$

Q40GATE 2004MCQ

The average rainfall for a 3 hour duration storm is 2.7 cm and the loss rate is 0.3 cm/hr. The flood hydrograph has a base flow of 20 $m^{3}$ /s and produces a peak flow of 210 $m^{3}$ /s. The peak of a 3-h unity hydrograph is

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📖 Explanation

Rainfall excess $=2.7-0.3 \times 3 =1.8$ cm Peak discharge of DRH $= 210-20=190 m^{3}/s$ Peak of 3-h UH $=\frac{190}{1.8}=105.55 m^{3}/s$