This set of Irrigation Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Tank or Surplus Escape Weirs Design”.
1. The usual difference between F.T.L and M.T.L is rarely allowed to exceed _____________
a) 0.3 m
b) 0.6 m
c) 0.9 m
d) 0.5 m
Explanation: This difference is fixed on a compromise basis in order to obtain the overall economy and efficiency. It is kept from 0.3 to 0.6 m for small and medium-sized tanks and is rarely allowed to exceed 0.9 m.
2. Which of the following type of tank weirs are provided with a number of vertical steps instead of horizontal or sloping downstream apron?
a) Type A: Masonry weir with a vertical drop
b) Type B: Rock-fill weirs with a sloping apron
c) Type C: Masonry weirs with a glacis
d) Type D: Combination of Type A and Type C
Explanation: Weirs of Type-D are called as weirs with stepped aprons as they are provided with a number of vertical steps as in case of a stepped fall instead of providing a horizontal or sloping downstream apron. Type-A and Type-D weirs are the most widely adopted.
3. The width of the horizontal floors of type A and D weirs from the foot of the drop wall to the d/s edge of the floor should never be less than ________________
a) 5 (D + H)
b) 3 (D + H)
c) 2.5 (D + H)
d) 2 (D + H)
Explanation: The width of the horizontal floor of Type-A and Type-D weirs should not be less than 2(D + H) and in important works, the width can be increased to 3(D + H). where D is the height of drop wall and H is the maximum head of water over the wall. The rough stone apron forming a talus below the last curtain wall generally vary from 2.5(D+ H) to 5(D + H) depending upon the nature of the soil and the velocity and annual probable quantity and intensity of run-off.
4. Which among the following is a correct representation of Ryve’s formula?
a) Qp = C1. A2/3
b) Qp = C1. A1/3
c) Qp = C1. L A2/3
d) Qp = C1. A
Explanation: All tank weirs generally in South India are designed on the basis of Ryve’s formula which is given by:
Qp = C1.A2/3 where Qp is the peak flood discharge, C1 is the Ryve’s coefficient, A is the catchment area
The formula is directly applicable for free catchments in all isolated tanks.
5. The capacity of an irrigation tank is sometimes increased by installing a temporary stone wall over the top of the tank weir. This fixture is known as____________________
b) dam stone
c) breast wall
d) divide wall
Explanation: A common practice has been resorted to in South India is to fix dam stones at the crest wall of the weir. The dam stones cause obstruction to the discharge and if no dam stones or shutters are provided above F.T.L, the flood water will start spilling over the surplus work immediately after the tank water level exceeds F.T.L.
6. The discharge through the sluice of a small irrigation tank is usually controlled by _______________
a) dam stone
c) special balanced valves
Explanation: Flat shutters working in grooves and regulated by screw spears are used in the case of large tanks where the quantity of water to be released is great. Special balanced valves or shutters moving on rollers are generally installed for heads over 9 m or so. A Dam stone causes obstruction to flow and is used sometimes to increase the capacity.
7. What will the discharging capacity of a masonry waste weir of 55 m length and 1.5 m width provided in an earthen bund for passing a flood with 1.2 m water depth over its crest?
a) 120 cumecs
b) 254 cumecs
c) 59.4 cumecs
d) 65.8 cumecs
Explanation: The discharge over the weir without any velocity of approach is given by –
Q = C.L H3/2 where C = 1.66 when the width of rectangular crest > 0.9 m, L is the length of the weir and H is the head of water over the weir.
Q = 1.66 x 55 x 1.2
8. Pipe sluices are generally not adopted in tank bunds where the depth below F.T.L exceeds ____________
a) 1.5 m
b) 1 m
c) 2.5 m
d) 5 m
Explanation: Pipe sluices are earthenware or cement or cast iron pipes which may be used in place of masonry culverts in case of very small slices. The earthenware pipes may get fractured or leakage through their joints may take place resulting in a breach. The pipes can neither be examined nor repaired easily without cutting open the bund. They are not adopted when the depth below F.T.L exceeds 2.5 m or so.
9. Calculate peak discharge for a combined catchment of a tank forming a constituent unit of a group with the following given data –
Combined catchment = 26 sq. km Intercepted catchment = 20 sq. km Ryve’s coefficient for combined catchment = 9.0 Ryve’s coefficient intercepted = 1.8
a) 66 cumecs
b) 59 cumecs
c) 50 cumecs
d) 69 cumecs
Explanation: Using Ryve’s equation –
Peak flood discharge Qp = C1A2/3 – c1a2/3 where C1 = 9.0, c1 = 1.8, A = 26 sq.km and a = 20 sq.km
Qp = (9 x 262/3) – (1.8 x 202/3) = 65.9 cumecs.
10. What is the approximate value of Ryve’s coefficient for combined catchment having limited areas near hills?
d) Up to 40
Explanation: For areas within 80 km from the east coast, the value of constant is taken as 6.5 and for areas within 80 to 160 km from the east coast; the value is taken as 8.5. In the case of limited areas near hills, the value of the coefficient is 10.2 and the actual observed values are always up to 40.
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