This set of Irrigation Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Sediment Transport – Mechanics”.
1. The soil is assumed to be coherent in the study of the mechanism of sediment transport.
Explanation: Coherent means that there is a cohesive force between particles such as in the case of clays. The soil is assumed to be incoherent so that each soil grain is studied individually. Most of our river beds are made up of gravels and sands in which there is no cohesion.
2. The basic mechanism behind the phenomenon of sediment transport is ______________
a) drag force opposite to the direction of the flow
b) drag force in the direction of the flow
c) free motion of the sediment particles
d) force exerted by water vertically
Explanation: The force exerted by water in the direction of flow is the drag force or tractive force. If this force on the bed particles exceeds the force opposing their movement then, the bed particles will start moving to lead to the sediment transport.
3. Which of the following statement is wrong?
a) Threshold condition is the one in which a few particles on the bed will just start moving
b) Knowledge of critical velocity helps in designing stable non-scouring channels
c) The critical tractive force approach helps in designing unstable channels in alluviums
d) The knowledge of threshold condition is required for the computation of sediment load
Explanation: Shield was the first person to analyze the experimental data on incipient motion condition using the critical tractive force approach. The assumption of the entry of clean and clear water in the channel is used to develop non-scouring highest possible flow velocity at the peak flow.
4. For the bed of the canal, the average shear stress is equal to the tractive force per unit area.
Explanation: The unit tractive force in channels is uniformly distributed along the wetted perimeter. For the bed of the canal, the average shear stress is equal to the tractive force per unit area. For side slopes of the canal, average shear stress is equal to 0.75 times the tractive force per unit area.
5. A wide unlined channel carrying silt-free water has a depth of 2.0 m. The maximum slope that can be given to a channel is 1 in 10,000. Calculate the maximum tractive stress permissible on the bed to prevent scouring.
a) 0.2 kg/m2
b) 1.962 kg/m2
c) 2 kg/m2
d) 1 kg/cm2
Explanation: The maximum tractive stress = Yw. R. S
= 9.81 x 1000 x 2.0 x 10-4
= 1.962 kg/m2
6. The water flows at a depth of 0.6 m in a wide stream having a bed slope of 1 in 2500. The critical tractive stress is 0.53 N/m2. Determine the motion of soil grains and the average shear stress.
a) 2.35 N/m2 and soil grains will be stationary
b) 2.35 N/m2 and soil grains is in motion
c) 0.235 N/m2 and soil grains will be stationary
d) 0.235 N/m2 and soil grains is in motion
Explanation: The average shear stress = Yw. R. S
= 9.81 x 1000 x 0.6 x 1/2500 = 2.35 N/m2.
This value is more than 0.53 N/m2, the soil grains will not be stationary and sediment transport and scouring will occur.
Sanfoundry Global Education & Learning Series – Irrigation Engineering.
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