This set of Engineering Hydrology Multiple Choice Questions & Answers (MCQs) focuses on “Flood Empirical Formulas – Set 2”.

1. Estimate the peak flood flow (in m^{3}s) for a 32 km^{2} catchment area in the Gangetic plains region?

a) 80.7

b) 91.5

c) 112.1

d) 147.9

View Answer

Explanation: Since the area lies in the north Indian plains, Dickens formula should be used with a constant of 6. The peak flood is given as,

Q

_{p}=C

_{D}*\(A^{\frac{3}{4}}\)=6*32

^{0.75}=80.7 m

^{3}s

2. Estimate the peak flood flow (in m^{3}/s) for a 2500 hectare catchment area in the Chola Nadu region Tamil Nadu?

a) 51.3

b) 58.1

c) 72.7

d) 87.2

View Answer

Explanation: Since the area lies within 80 km of the east coast, Ryves formula should be used with a constant of 6.8. The peak flood is given as,

Q

_{p}=C

_{R}*\(A^{\frac{2}{3}}=6.8*(25)^{\frac{2}{3}}\)=58.1 m

^{3}/s

3. Find the peak flow of a flood with 25-year return period in a 5030 ha area of the western ghats?

a) 500 m^{3}/s

b) 600 m^{3}/s

c) 700 m^{3}/s

d) 800 m^{3}/s

View Answer

Explanation: Using Inglis formula,

\(Q=\frac{124A}{\sqrt{A+10.4}}=\frac{124*50.3}{\sqrt{50.3+10.4}}\)=800.56 m

^{3}/s≅800 m

^{3}/s

4. The peak flood in a 10 km^{2} catchment in Karnataka was mistakenly found using Dickens formula with a constant of 11. Find the percentage error in the calculated value. Take Ryves constant as 10.2.

a) 12%

b) 23%

c) 31%

d) 62%

View Answer

Explanation: Incorrect value using Dickens formula is,

Q

_{pD}=C

_{D}*\(A^{\frac{3}{4}}\)=11*10

^{0.75}=61.86 m

^{3}/s

Correct value using Ryves formula,

Q

_{pR}=C

_{R}*\(A^{\frac{2}{3}}=10.2*10^{\frac{2}{3}}\)=47.34 m

^{3}/s

∴Percentage error=\(\frac{Q_{pD}-Q_{pR}}{Q_{pR}} *100=\frac{61.86-47.34}{47.34}\)*100=30.67%≈31%

5. The peak flood of an area as estimated by Inglis formula was found to be 200 m3/s. Find the area of the region (in hectares)?

a) 340

b) 666

c) 533

d) 178

View Answer

Explanation: As per Inglis formula,

\(Q=\frac{124A}{\sqrt{A+10.4}}⇒200=\frac{124A}{\sqrt{A+10.4}}⇒200*\sqrt{A+10.4}\)=124A

⇒40000*(A+10.4)=15376*A

^{2}⇒15376A

^{2}-40000A-416000=0

Solving for A by using quadratic formula,

\(A=\frac{40000±\sqrt{40000^2+(4*15376*416000)}}{2*15376}=\frac{40000±164880.76}{30752}\) \(A=\frac{204880.76}{30752}\)or-\(\frac{124880.76}{30752}\)=6.66 or-4.06 (rejected)

Therefore, the area of the catchment is 6.66 km

^{2}or 666 hectares.

6. Estimate the maximum 24 hour flood (in m^{3}/s) with a return period of 50 years for an area of 27 km^{2}. Take Fuller’s constant as 0.8.

a) 26.36

b) 46.14

c) 59.31

d) 103.82

View Answer

Explanation: The maximum flood as per Fuller’s formula is,

Q

_{p}=C

_{f}A

^{0.8}(1+0.8 log T) = 0.8*27

^{0.8}*(1+0.8*log(50))

⇒Q

_{p}=26.36 m

^{3}/s

7. Envelope curve method of peak flood estimation involves plotting graph between which two data?

a) Peak discharge and rainfall intensity

b) Rainfall duration and catchment area

c) Rainfall intensity and catchment area

d) Peak discharge and catchment area

View Answer

Explanation: In envelope curve method, the available flood data from suitable catchments is collected and plotted on a log-log graph in the form of a graph between the peak flood discharges and the catchment area.

8. Estimate the maximum flood discharge (in m^{3}/s) with a return period of 100 years for an area of 4760 ha. Take Fuller’s constant as 1.4.

a) 80

b) 46

c) 144

d) 112

View Answer

Explanation: The maximum flood as per Fuller’s formula is,

Q

_{p}=C

_{f}A

^{0.8}(1+0.8 log T) = 1.4* 47.6

^{0.8}*(1+0.8*log(100))=80 m

^{3}/s

9. What is the peak flood discharge (in m^{3}/s) for a 45 km^{2} area as per maximum world flood experience?

a) 893.2

b) 1502.3

c) 1786.7

d) 2836.4

View Answer

Explanation: The peak discharge as per maximum world flood experience is given as,

\(Q_p=\frac{3025A}{(278+A)^{0.78}} = \frac{3025*45}{(278+45)^{0.78}}\) = 1502.3 m

^{3}/s

10. What is the area for which Ryves and Dickens formula will give same peak flood? Assume the constants as equal.

a) 1 ha

b) 10 ha

c) 100 ha

d) 1000 ha

View Answer

Explanation: Equating the Ryves and Dickens formulae,

\(C_D*A^{\frac{3}{4}}=C_R*A^{\frac{2}{3}} ⇒ A^{\frac{3}{4}}=A^{\frac{2}{3}}\) ⇒A=1 km

^{2}=100 ha

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