1. 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.
2. 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%.
3.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.
4. 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.
5. 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.
6. 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.
7. Design of alluvial channels in India is based on Kennedy and Lacey theories.
a) True
b) False
Explanation: The direct accurate mathematical solution based on the resistance equations given by Chezy’s formula and Manning’s formula is very complicated in Indian based alluvial channels. So, therefore hypothetical theories given by Kennedy and Lacey are used, as these theories are based on experiments and experience from existing channels over the years.
8. On what condition does the resistance equations of Chezy’s formula and manning’s formula are applicable?
a) τo
b) τc
c) τo < τc
d) τo > τc
Explanation: The average shear stress (τo) which is acting on the boundary of an alluvial channel should be less than the critical shear stress (τc), then the channel shape remains unchanged, and hence the channel is considered having rigid boundary. So based on this condition Chezy’s formula and Manning’s formula can be applied.
9. What is the problem in India for artificial channels?
a) Formation of Depressions
b) Formation of Alluvial Soil
c) Untimely Rains
d) Improper Usage of Channels
Explanation: In prehistoric periods, the area starting from the Himalayas to Vindhya mountains is used to in the form of depressions with water flowing over it. But over the ages these depressions got filled with fine silt particles, thereby forming into alluvial soil. So the rivers present in this area of north India flow in alluvial soil carrying silt. So, therefore the artificial channels carrying water from such rivers thus have to carry silt sediment.
10. The velocity of flow in a channel should not more or less, but instead it should be adequate.
a) True
b) False
Explanation: The given velocity of flow of a channel and certain depth can carry in suspension, some amount of silt. But if the velocity is more and depth is not fully charged with silt it will erode the bed and sides of the channel. And if the velocity is less, the silt cannot be carried in suspension by the flow, hence it is dropped.