This set of Engineering Hydrology Multiple Choice Questions & Answers (MCQs) focuses on “Floods – Rational Method – Set 2”.

1. In the Kirpich equation for estimating time of concentration, what the unit in which the time is obtained when length is entered in meters.

a) Seconds

b) Minutes

c) Hours

d) Days

View Answer

Explanation: The Kirpich equation is a popular empirical equation that relates the time of concentration to the length of travel and slope of the area. The length should be entered in meters, and the corresponding time of concentration estimated will be in minutes.

2. The elevation difference for the slope of the catchment to be used in Kirpich equation is taken between which two points of the catchment?

a) Inlet and outlet

b) Centroid and outlet

c) Most remote point and inlet

d) Most remote point and outlet

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Explanation: The difference in elevation of the most remote point of the catchment and its outlet divided by the length gives the slope of the catchment for estimating time of concentration. This is because the water from even the most remote location must reach the outlet of the catchment.

3. Find the time of concentration (in minutes) for length of maximum travel of 1.1 km in a catchment area of slope 0.0078. Use Kirpich equation.

a) 27.7

b) 28.5

c) 13.6

d) 32.3

View Answer

Explanation: The time of concentration by Kirpich equation is given as,

t

_{c}=0.01947 *L

^{0.77}*S

^{-0.385}=0.01947*(1100)

^{0.77}*(0.0078)

^{-0.385}=27.7 minutes

4. What is the general form of the rainfall intensity to be used in the rational formula for a particular area? T is the return period, D is the rainfall duration, K, a, b, x, n are coefficients for that area.

a) \(\frac{KT}{(D+a)^n}\)

b) \(\frac{T^x}{K(D+a)^n}\)

c) \(\frac{KT^x}{D^n}\)

d) \(\frac{KT^x}{(aD+b)^n}\)

View Answer

Explanation: The rainfall intensity is found based on the duration of rainfall (time of concentration) and the return period from rainfall-frequency-duration curves for the given area. The general relationship is,

Rainfall intensity=\(\frac{KT^x}{(D+a)^n}\).

5. In catchment area, the longest path of travel of water is 3200 m. The difference in elevation between the inlet-outlet, centroid-outlet and farthest point-outlet is 23.6 m, 17.8 m and 28.8 m, respectively. Find the time of concentration for the catchment.

a) 0.045 days

b) 1.19 hours

c) 63.32 minutes

d) 3582 seconds

View Answer

Explanation: The required elevation is between the farthest point and outlet, H = 28.8 m

Slope = H/L = 28.8/3200 = 0.009

So time of concentration by Kirpich equation is,

t

_{c}=0.01947*L

^{0.77}*S

^{-0.385}=0.01947*(3200)

^{0.77}*(0.009)

^{-0.385}

=59.7 minutes

=0.99 hours

=3582 seconds

=0.0415 days

6. The time of concentration of a catchment of area 600 hectares with a runoff coefficient of 0.32 is 81 minutes. The maximum depth of a particular 20-year rainfall in this area for time of concentrations above 1 hour is 9.6 cm. What is the peak discharge at the outlet for the given rainfall?

a) 37.9 m^{3}/s

b) 51.2 m^{3}/s

c) 3.8 m^{3}/s

d) 5.1 m^{3}/s

View Answer

Explanation: The rainfall intensity to be considered for duration of 81 minutes is,

R=\(\frac{9.6 cm}{81 min}*60 \frac{min}{hr}\)=7.111 cm/hr=71.11 mm/hr

By rational formula, the peak discharge is,

\(Q_p=\frac{AIR}{360}=\frac{600*0.32*71.11}{360}\)=37.9 m

^{3}/s

7. A 100 hectare area has a runoff coefficient of 0.4 and a time of 36 minutes. The maximum depth of a 50-year return period rainfall is given.

Duration (in min) | 10 | 20 | 30 | 40 | 50 |

Depth of rainfall (in mm) | 22 | 29 | 38 | 45 | 53 |

A spillway is to be designed for a return period of 50 years near the catchment. Estimate the peak flow to be considered for the design.

a) 78.14 m^{3}/s

b) 78.14 m^{3}/min

c) 468.87 m^{3}/s

d) 468.87 m^{3}/min

View Answer

Explanation: Since the time of concentration is 36 mins, finding the depth for 36 min by interpolation,

Required depth=38+\(\frac{(45-38)}{(40-30)}\)*(36-30)=42.2 mm

Rainfall intensity, R=\(\frac{42.2}{36}\)*60=70.33 mm/hr

So peak design discharge for the spillway by rational method is,

Q=\(\frac{AIR}{360}=\frac{100*0.4*70.33}{360}\)=7.814 m

^{3}/s=468.87 m

^{3}/min

8. The intensity duration frequency relation of a catchment (area 5.5 km^{2}, time of concentration 1.2 hours and runoff coefficient 0.48) is given as follows.

R =\(\frac{5.9T^{0.2}}{(D+0.6)^{0.8}}\)
R = rainfall intensity in cm/hr, T = return period in years, D = duration of rainfall in hours. Estimate the 30-year peak discharge (in m^{3}/s) from the catchment.

a) 27.7

b) 41.5

c) 53.4

d) 66.1

View Answer

Explanation: Using the given equation to compute the required rainfall intensity,

R =\(\frac{5.9T^{0.2}}{(D+0.6)^{0.8}} =\frac{5.9*(30)^{0.2}}{(1.2+0.6)^{0.8}} \)=7.28 cm/hr=72.8 mm/hr

Now, 30 year peak discharge as per rational formula is,

Q=\(\frac{AIR}{360}=\frac{550*0.48*72.8}{360}\)=53.4 m

^{3}/s

9. The rational formula is applicable to all catchment areas for any duration of rainfall of any return period.

a) True

b) False

View Answer

Explanation: The rational method, though popular, has limitations to its application. It is more suitable to smaller catchment area of size less than 50 km

^{2}. It can be applied for rainfall whose duration is at least equal to the time of concentration for the catchment.

10. A catchment of 7 km^{2} comprises of 72% cultivable land (runoff coefficient 0.4) and rest non-cultivable land (runoff coefficient 0.5). Its time of concentration is 100 minutes and intensity duration frequency relation is represented by the following equation.

\(R = \frac{65.44T^{0.1873}}{(0.017D+0.521)^{0.9154}}\)

R = rainfall intensity in mm/hr, T = return period in years, D = duration of rainfall in minutes. Estimate the 75-year peak discharge (in m^{3}/s) from the catchment.

a) 28.88 m^{3}/s

b) 38.88 m^{3}/s

c) 48.88 m^{3}/s

d) 58.88 m^{3}/s

View Answer

Explanation: Using the given equation to compute the required rainfall intensity,

\(R = \frac{65.44*(75)^{0.1873}}{(0.017*100+0.521)^{0.9154}}\) = 70.76 mm/hr

The equivalent runoff coefficient is, I=\(\frac{(72*0.4)+(28*0.5)}{100}\)=0.428

Now, 75 year peak discharge as per rational formula is,

Q=\(\frac{AIR}{360}=\frac{700*0.428*70.76}{360}\)=58.88 m

^{3}/s.

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