# Irrigation Engineering Questions and Answers – Irrigation Channels Design – Method for Design of Non-Scouring

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This set of Irrigation Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Irrigation Channels Design – Method for Design of Non-Scouring”.

1. 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.

2. 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).

3. 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.
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4. 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.

5. 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.

6. What is the limitation of the Shield’s expression?
a) It can be used when the diameter of a particle (d) 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.

7. Calculate the Manning’s rugosity coefficient in a coarse alluvium gravel with D-75 size of 5 cm.
a) 0.025 m
b) 0.035 m
c) 0.055 m
d) 0.1 m

Explanation: The strickler’s formula is n = d1/6/24 where d = 0.05 m.
n = 0.051/6/24 = 0.025 m.

8. The manning’s coefficient for a lined trapezoidal channel with a bed slope of 1 in 4000 is 0.014 and it will be 0.028 if the channel is unlined. The area in the case of the lined section is 19.04 m2 and for the unlined section is 29.09 m2. What percentage of earthwork is saved in a lined section relative to an unlined section, when a hydraulically efficient section is used in both the cases?
a) 24.44 %
b) 34.55%
c) 50%
d) 37.66%

Explanation: The percentage of earthwork saving due to the lining is (A2 – A1)/A2 x 100.
= (29.09 – 19.04) / 29.09 x 100
= 34.55%.

9. Calculate the critical tractive stress if the median diameter of the sand bed is 1.0 mm.
a) 0.53 N/m2
b) 0.61 N/m2
c) 0.73 N/m2
d) 1.61 N/m2

Explanation: The given size of the particle is less than 6 mm so; shield’s equation cannot be used. The general relation is given by Mittal and Swamee which gives results within +5% of the values of Shield’s curve for all particle sizes.
Tc = 0.155 + [0.409 d2/(1 + 0.177 d2)0.5]
= 0.155 + [0.409 x 12/(1 + 0.177 x 12)0.5] = 0.53 N/m2.

10. Calculate the ratio of the tractive critical stress to the average shear stress if the water flows at a depth of 0.8 m in a wide stream having a bed slope of 1 in 3000. The median diameter of the sand bed is 2 mm.
a) 0.53
b) 1.86
c) 0.86
d) 1.53

Explanation: The critical tractive stress is given by (d < 6 mm) –
Tc = 0.155 + [0.409 d2/(1 + 0.177 d2)0.5]
= 0.155 + [0.409 x 22/(1 + 0.177 x 22)0.5] = 1.40 N/m2.
The average shear stress = Yw. RS = 9.81 x 0.8 x 1/3000 = 2.616 N/m2
Ratio = 1.40/2.616 = 0.53.

11. Determine the shear stress required to move the single grain on the side slopes, if the critical shear stress required moving the similar grain on the horizontal bed is 2.91 N/m2. Consider the angle of the side slope with the horizontal as 30° and the angle of repose of soil as 37°.
a) 1.61 N/m2
b) 2.61 N/m2
c) 2.91 N/m2
d) 1.00 N/m2

Explanation: The equation required is Tc’ = (1 – Sin2Q/Sin2R)1/2.Tc where Q = 30°, R = 37° and Tc = 2.91 N/m2.
Tc’ = (1 – Sin230°/Sin237°)1/2x 2.91 = 1.61 N/m2.

12. The shear stress required to move grain on the side slopes is less than the shear stress required to move the grain on the canal bed.
a) True
b) False

Explanation: The actual shear stress on the channel bed is Yw.RS while on slopes this value is given by 0.75 Yw. R.S. The following equation shows that Tc’ < Tc.
Tc’ = {(1 – Sin2Q/Sin2R)1/2.Tc } where Tc’ = shear stress on the side slopes, Tc = shear stress on the horizontal bed, Q = Angle of side slope with the horizontal, and R = angle of repose of the soil.

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