1.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.
2. For the bed of the canal, the average shear stress is equal to the tractive force per unit area.
a) True
b) False
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.
3. 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
4. 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.
5. Whose theory was the first to provide semi-theoretical analysis of the problem of incipient condition of bed motion?
a) Lacey’s theory
b) Kennedy’s theory
c) Shield’s theory
d) Strickler’s equation
Explanation: Shield was the first person who designed non-scouring channels by providing semi-theoretical analysis. He defined that the bed particles need a drag force greater than or equal to the resistance offered by the particle for its movement
6. Which one is the correct expression for Shield’s entrainment function?
a) Tc / Yw. d. (Gs – 1)
b) Tc.Yw / d. (Gs – 1)
c) Tc / Yw. d
d) Tc .Yw. d / (Gs – 1)
Explanation: Shield introduced a dimensionless number which is called entrainment function (Fs) which is a function of Reynold’s number at a critical stage of bed movement in alluviums. It is based on the experimental work done by the Shield.
Mathematically, Fs = Tc / Yw. d. (Gs – 1).
7. For the design of non-scouring channels in coarse alluviums, the shield’s entrainment function should be ____________
a) Fs > 0.056
b) Fs < 0.056
c) Fs = 0.056
d) Fs = 0
Explanation: According to Shield for coarse alluvium, Fs = 0.056 (d > 6 mm). When the particle Reynold’s number is more than 400, the application of the curve plotted by Shield between Reynold’s number and entrainment function becomes simpler. Also, the value of the entrainment function becomes constant and equal to 0.056.
8. Strickler’s formula is only applicable to the flexible boundary channels.
a) True
b) False
Explanation: Strickler’s formula is not used for moveable boundary channels as it does not account for the roughness due to undulations in the bed channel. It is used to calculate Manning’s rugosity coefficient and is valid for rivers with the bed of coarse material i.e. rigid boundary channels
9. What is the minimum size of the bed material that will remain at rest in a channel?
a) d > or = 11 R.S
b) d < or = 11 R.S
c) d = 11 R.S
d) d > 11 R.S
Explanation: For no movement of the sediment particles, Tc > or = T0.
0.056. Yw. d. (Gs – 1) >or = Yw. R. S where, for sand of gravel Gs = 2.65
d >or = 10.82 R.S
d >or = 11 R.S.
10. What is the limitation of the Shield’s expression?
a) It can be used when the diameter of a particle is < 6mm and Reynold’s number > 400
b) It can be used only when Reynold’s number > 400
c) It can be used only when the diameter of the particle is < 6 mm
d) It can be used when the diameter of a particle is > 6mm and Reynold’s number > 400
Explanation: The curve plotted by Shield between Reynold’s number and Entrainment function forms a suitable basis for the design of channels where it is required to prevent bed movement. When the particle size exceeds 6 mm such as for coarse alluvium soils, the particle Reynold’s number has found to be more than 400 representing roughness.