This set of Engineering Hydrology Multiple Choice Questions & Answers (MCQs) focuses on “Flood Empirical Formulas – Set 2”.
1. Estimate the peak flood flow (in m3s) for a 32 km2 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,
Qp=CD*\(A^{\frac{3}{4}}\)=6*320.75=80.7 m3s
2. Estimate the peak flood flow (in m3/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,
Qp=CR*\(A^{\frac{2}{3}}=6.8*(25)^{\frac{2}{3}}\)=58.1 m3/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 m3/s
b) 600 m3/s
c) 700 m3/s
d) 800 m3/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 m3/s≅800 m3/s
4. The peak flood in a 10 km2 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,
QpD=CD*\(A^{\frac{3}{4}}\)=11*100.75=61.86 m3/s
Correct value using Ryves formula,
QpR=CR*\(A^{\frac{2}{3}}=10.2*10^{\frac{2}{3}}\)=47.34 m3/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*A2 ⇒15376A2-40000A-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 km2 or 666 hectares.
6. Estimate the maximum 24 hour flood (in m3/s) with a return period of 50 years for an area of 27 km2. 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,
Qp=Cf A0.8 (1+0.8 log T) = 0.8*270.8*(1+0.8*log(50))
⇒Qp=26.36 m3/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 m3/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,
Qp=Cf A0.8 (1+0.8 log T) = 1.4* 47.60.8*(1+0.8*log(100))=80 m3/s
9. What is the peak flood discharge (in m3/s) for a 45 km2 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 m3/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 km2=100 ha
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