# Irrigation Engineering Questions and Answers – Reservoir Capacity Determination Using Hydrograph and Mass-curve – 2

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This set of Irrigation Engineering Questions and Answers for Experienced people focuses on “Reservoir Capacity Determination Using Hydrograph and Mass-curve – 2”.

1. The design annual rainfall for the catchment of a proposed reservoir has been computed to be 99 cm. The catchment area contributing to the proposed reservoir is 1000 sq.km having a mean annual temperature of 20°C. Calculate the annual design catchment yield for the reservoir using Khosla’s formula.
a) 89.4 M.m3
b) 8.94 M.m3
c) 894 m3
d) 894 M.m3

Explanation: Using Khosla’s formula –
Q = P – 0.48 Tm where, Rainfall (P) = 99 cm and Mean annual temperature (Tm) = 20°C
Q = 99 – 0.48 x 20 = 89.4 cm = 0.894 m
The total yield produced from the given catchment = 0.894 x 1000 x 106 = 894 M.m3.

2. The reservoir capacity cannot exceed the catchment yield.
a) True
b) False

Explanation: The reservoir capacity is fixed at a value which is lesser of the value of –
i. The accessed gross storage required to meet the demand
ii. The accessed dependable yield for the reservoir site.

3. With the reduction in reservoir capacity over the passage of time, the trap efficiency ___________
a) increases
b) decreases
c) remains unaffected
d) may increase or decrease depending upon the reservoir characteristics

Explanation: If the reservoir capacity reduces with constant inflow value, the trap efficiency reduces. Hence, for small reservoirs having small capacity on large rivers having large inflow rates, the trap efficiency is extremely low.
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4. Trap efficiency of a storage reservoir is defined as the ratio of ___________________
a) total annual sediment inflow to the reservoir capacity
b) total sediment deposited in a given period to the total sediment inflow in that period
c) total annual sediment deposited in the reservoir to the dead storage capacity of the reservoir
d) reservoir capacity to the total annual sediment

Explanation: Trap efficiency of a storage reservoir is defined as the ratio of the total sediment deposited in a given period to the total sediment inflow in that period. It can be defined as the percentage of sediment deposited in the reservoir in spite of taking precautionary control measures.

5. Capacity inflow ratio for a storage reservoir is defined as the ratio of _______________________
a) reservoir capacity to the average annual flood inflow
b) reservoir capacity to the average annual sediment inflow
c) the dead storage capacity of the reservoir to the average annual sediment deposited
d) total annual sediment inflow to the reservoir capacity

Explanation: Capacity-inflow ratio can be defined as the ratio of the reservoir capacity to the total inflow of water. The trap efficient is a function of the capacity-inflow ratio which is represented by a graph between them.

6. The capacity-inflow ratio for a reservoir _________________
a) is a constant factor overtime
b) increases with time
c) decreases with time
d) may increase or decrease with time

Explanation: The silting rate in the reservoir will be more in the beginning as its capacity reduces due to silting, the silting rate will also reduce. The trap efficiency is a function of capacity. The capacity reduces when trap efficiency reduces and lesser sediment is trapped.

7. A sequent peak algorithm is a plot between __________________
a) Accumulated flow v/s time
b) Discharge v/s time
c) (Cumulative Inflow – Cumulative Outflow) v/s time
d) Mass outflow v/s time

Explanation: It is a plot between time on X-axis and cumulative net flow on Y-axis. It is an excellent alternative to the mass-curve method of determining reservoir capacity. The positive value of cumulative net flow indicates a surplus of inflow and a negative value indicates a deficit of inflow.

8. A flow duration curve is a curve plotted between ________________
a) Accumulated flow v/s time
b) Discharge v/s time
c) (Cumulative Inflow – Cumulative Outflow) v/s time
d) Streamflow v/s Percent of time the flow is equaled or exceeded

Explanation: A flow-duration curve is a curve plotted between the streamflow and the percent of the time the flow is equaled or exceeded. It is also called a discharge-frequency curve and it represents the cumulative frequency distribution.

9. A steep slope of the flow duration curve indicates a stream with _________________
a) highly variable discharge
b) small variability of flow
c) considerable base flow
d) large flood plains

Explanation: A steep slope represents a stream with highly variable discharge and a flat slope represents a small variability of flow. The considerable base flow is indicated by a flat portion on the lower end of the curve and the upper end of the curve is of river basins having large flood plains.

10. The lowest portion of the capacity-elevation curve of a proposed irrigation reservoir draining 20 km2 of catchment is represented by the following data:
i. The rate of silting for the catchment = 300 m3 / km2 / year
ii. Life of the reservoir = 50 years
iv. The FSD of the canal at the head = 80 cm
v. The crop water requirement = 250 ha.m
vi. Dependable yield of the catchment = 0.29 m
Calculate the gross capacity of the reservoir.
a) 287.5 ha.m
b) 317.5 ha.m
c) 580 ha.m
d) 37.5 ha.m

Explanation: Net water demand = 250 ha.m and Reservoir losses = 15% x 250 = 37.5 ha.m
Live storage to meet the given demand = 250 + 37.5 = 287.5 ha.m
Gross storage required to meet the demand = live storage + dead storage = 287.5 + 30 = 317.5 ha.m
Dependable yield = 0.29 x 20 x 106 = 580 ha.m
The gross capacity is fixed at the lesser value of the gross storage and the dependable yield. Hence, the reservoir capacity = 317.5 ha.m.

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