This set of Engineering Hydrology Multiple Choice Questions & Answers (MCQs) focuses on “Partial Duration Series”.

1. How many values of floods are selected for a region per year as data points in the annual flood series?

a) 1

b) 2

c) 3

d) 5

View Answer

Explanation: For the annual series consisting of hydrologic flood data, only one value of peak flood is selected as a data point per year for a region. This value may or may not re[resent the actual maximum flood for that region in that year.

2. Which of the following is true regarding partial duration series for a region?

a) One maximum flood value is selected for each year

b) All floods lower than a specific value are selected for each year

c) There may be more than one flood value from each year

d) Not more than five floods per year are to be considered

View Answer

Explanation: In a year, there may be many number of large peak flood values that need to be considered for a flood series. For this reason, all floods larger than a specified value are to be considered for each year. The number of floods has no limit, but the base value should be selected properly.

3. Which of the following is true regarding the floods of a yearly partial duration series?

a) They should be independent

b) They should be dependent

c) They should be mutually exclusive

d) They should be exhaustive

View Answer

Explanation: Any peak flood considered in the partial duration series should be completely unaffected by the occurrence of any other flood in the series. This implies that the floods should be independent. Occurrence of one flood should not affect the occurrence of any other flood.

4. Partial duration series is preferred for rainfall analysis over flood analysis.

a) True

b) False

View Answer

Explanation: Partial duration series requires that all events considered are independent. In the case of rainfall, it is easier to establish the conditions of independency of the events in an area. However, the proof of independency of floods is difficult and hence the use of partial duration series for flood analysis is very rare.

5. Let T_{A} be the return period of a flood obtained by annual series, and T_{P} be the return period of the same flood as obtained by partial duration series. How are T_{A} and T_{P} related?

a) \(T_A=\frac{1}{ln T_P-ln(T_P-1)}\)

b) \(T_A=\frac{1}{ln(T_P+1)-ln T_P}\)

c) \(T_P=\frac{1}{ln T_A-ln(T_A-1)}\)

d) \(T_P=\frac{1}{ln(T_A+1)-ln T_A}\)

View Answer

Explanation: The return period of an event estimated by annual series (T

_{A}) and partial duration series (T

_{P}) can be related by the following equation,

\(T_P=\frac{1}{ln T_A-ln(T_A-1)}\).

6. For which of the following return periods (as estimated from annual series) is the percentage error with the return period as estimated form partial duration series, the highest?

a) 5 years

b) 10 years

c) 15 years

d) > 20 years

View Answer

Explanation: From the equation relating return periods of annual series and partial duration series, it can be observed that as the magnitude of annual series return period increases, the error between the two time periods is negligible. However, for return periods < 10 years, the difference is significant.

7. The return period of a flood as estimated using the annual series was found to be 220 years. What would be the return period of the same flood when obtained using partial duration series?

a) 2628 months

b) 2634 months

c) 2640 months

d) 2646 months

View Answer

Explanation: The relation between annual series and partial duration series return period is,

\(T_P=\frac{1}{ln T_A-ln(T_A-1)}=\frac{1}{ln 220-ln(220-1)}\)

=\(\frac{1}{ln 220-ln219}\)=219.49962 years=2633.995 months≅2634 months.

8. What is the percentage error in annual series return period of 4.5 years with respect to partial duration series return period of the same flood?

a) 11.6%

b) 13.1%

c) 22.2%

d) 28.6%

View Answer

Explanation: The relation between annual series and partial duration series return period is,

\(T_P=\frac{1}{ln T_A-ln(T_A-1)}=\frac{1}{ln 4.5-ln(4.5-1)}\)

=\(\frac{1}{ln 4.5-ln3.5}\) = 3.9791 years

∴Percentage error=\(\frac{T_A-T_P}{T_P}*100=\frac{4.5-3.9791}{3.9791}*100\)=13.09%≈13.1%

**Sanfoundry Global Education & Learning Series – Engineering Hydrology.**

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