Q41GATE 2004MCQ

The rainfall during three successive 2 hour periods are 0.5, 2.8 and 1.6 cm. The surface runoff resulting from this storm is 3.2 cm. The $\phi$ index value of this storm is

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📖 Explanation

Total rainfall = 0.5+2.8+1.6&=4.9 cm Excess rainfall =4.9+3.2&=1.7 cm Taking excess rainfall duration, $t_{e}=2\times 3=6$ hrs $\phi$ -index = $\frac{1.7}{6}=0.28$ cm/hr This value being more than $\frac{0.5}{2}=0.25$ cm/hr, the excess rainfall duration will reduce by 2-hrs. $\therefore \;\; t_{e}=$ 4 hrs Excess rainfall = 2.8 +1.6-3.2 =1.2 cm $\phi$ -index $=\frac{1.2}{4}=0.3$ cm/hr

Q42GATE 2003MCQ

While applying the Rational formula for computing the design discharge, the rainfall duration is stipulated as the time of concentration because

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Q43GATE 2003MCQ

An average rainfall of 16 cm occurs over a catchment during a period of 12 hours with a uniform intensity. The unit hydrograph (unit depth = 1cm, duration = 6 hours) of the catchment rises linearly from 0 to 30 cumecs in six hours and then falls linearly from 30 to 0 cumecs in the next 12 hours. $\phi$ index of the catchment is known to be 0.5 cm/hr. Base flow in the river is known to be 5 cumecs. Area of the catchment in hectares is

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📖 Explanation

Catchment area $\times$ rainfall excess =Area under UH

\begin{aligned} A\times \frac{1}{100}&=\frac{30\times 18\times 60\times 60}{2} \\ \Rightarrow \;\;A&=9720\times 10^{4}m^{2} \\ \Rightarrow \;\; A&=9720 ha\end{aligned}

Q44GATE 2003MCQ

The vertical hydraulic conductivity of the top soil at certain stage is 0.2 cm/hr. A storm of intensity 0.5 cm/hr occurs over the soil for an indefinite period. Assuming the surfaces drainage to be adequate, the infiltration rate after the storm has lasted for a very long time, shall be

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📖 Explanation

Permeability = Hydraulic conductivity $\left ( \frac{\mu }{\rho g} \right )$ A storm of i=0.5 cm/hr occurs over the soil for an indefinite period. What is the meaning of "indefinite period"? In order to achieve steady state condition, continuous application of water is necessary so that the water level is maintained constant. 'Surface drainage adequate' means at steady state condition some water flows as surface runoff. 'Storm has lasted for a very long time' means the hydraulic conductivity approaches saturated hydraulic conductivity (so that Darcy's law can be obeyed) and at steady state condition infiltration rate is also constant. The constant infiltration rate which is achieved after very long time, shall be more than 0.2 cm/hr (as hydraulic conductivity has increased) but should be less than storm intensity 0.5 cm/hr due to adequate drainage.

Q45GATE 2002MCQ

The rainfall on five successive days in a catchment were measured as 3, 8, 12, 6, and 2 cms. If the total runoff at the outlet from the catchment was 15 cm, the value of the $\phi$ -index (in mm/hour) is

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📖 Explanation

W-index $=\frac{P-Q}{t}=\frac{\left ( 3+8+12+6+2 \right )-15}{5} =3.2 cm/day$ Now, as $\phi$ -index $\geq$ w-Index $\therefore$ The rainfall of magnitude 3 cm and 2 cm will not participate in rainfall excess depth (i.e., these will completely infiltrate) $\therefore$

\begin{aligned} \phi-index&= \frac{\text{Infiltration during rainfall excess depth}}{\text{Time period of rainfall excess}} \\ &=\frac{\left ( 8+12+6 \right )-15}{3}=3.67cm/day\\ \text{i.e., } &=\frac{3.67\times 10}{24} mm/hr=1.525 mm/hr \end{aligned}

Q46GATE 2002MCQ

During a 6-hour storm, the rainfall intensity was 0.8 cm/hour on a catchment of area 8.6 k $m^{2}$ . The measured runoff volume during this period was 256000 $m^{3}$ . The total rainfall was lost due to infiltration, evaporation, and transpiration (in cm/hour) is

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📖 Explanation

Infiltration (per 6 hr)=Rainfall- Runoff $=0.8\times 6-\frac{25600}{8.6\times 10^{6}}\times 100$ $$ =1.823 cm $\therefore$ Infiltration (per hr) $=\frac{1.823}{6}=0.304 cm/hr$

Q47GATE 2002MCQ

A 6-hour Unit Hydrograph (UH) of a catchment is triangular in shape with a total time base of 36 hours and a peak discharge of 18 $m^3/s$ . The area of the catchment (in sq. km) is

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📖 Explanation

$\frac{1}{2}\times 36\times 3600\times 18=\frac{A\times 1}{100}$ $3.6\times 3.6\times 9\times 10^{6}=A$ $A=116.64 km^{2}\simeq 117 km^{2}$

Q48GATE 2001MCQ

The direct runoff hydrograph of a storm obtained from a catchment is triangular in shape and has a base period of 80 hours. The peak flow rate is 30 $m^3/sec$ and catchment area is 86.4 $km^2$ . The rainfall excess that has resulted the above hydrograph is

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📖 Explanation

$A=86.4 km^{2}$ Rainfall excess $=\frac{1}{2}\frac{30\times 80\times 60\times 60m^{3}}{86.4\times 10^{4}m^{2}}\times 100=5$ cm