Irrigation Engineering Questions and Answers – Sediment Transport – Suspended Load and its Measurement

This set of Irrigation Engineering Questions & Answers for Exams focuses on “Sediment Transport – Suspended Load and its Measurement”.

1. Whose equation was the first on the rate of bed load transport?
a) DU-Bois empirical formula
b) Shield’s formula
c) Meyer-Peter’s formula
d) Einstein’s formula
View Answer

Answer: a
Explanation: The first equation was proposed by DU-Bois on the rate of bedload transport. He assumed that the bed load transportation rate was proportional to the excess of existing tractive force over the critical value required to initiate the movement.

2. Which of the following statement is not correct about Einstein’s theory?
a) He put forward a mathematical approach to the problem of bed load transport
b) He further postulated probability of the grain being dislodged is directly proportional to the lift force which the flow can exert on the grain
c) He introduced a factor called Einstein’s bedload function
d) He gave a curve graph which can be used to compute the bedload transport rate of a given channel
View Answer

Answer: a
Explanation: He put forward a semi-theoretical approach to the problem of bed load transport. It was assumed that every particle travels a certain minimum distance before coming to rest after it is dislodged from the bed.

3. If the width of a river increases, the discharge per unit width will decrease and therefore, sediment carrying capacity will increase.
a) True
b) False
View Answer

Answer: b
Explanation: The sediment carrying capacity will reduce if the width of the river increases as the discharge per unit width will decrease. The deposition of the sediment will start which will increase the bed slope.

4. Which of the following statement is not correct about sediment load phenomenon and its measurement?
a) The material is kept in suspension by the turbulence or by the generation of eddies
b) In laminar flow, the shear stress is caused due to the difference of the velocities at the top and the bottom
c) In turbulent flow, momentum transfer is not very significant
d) Due to the formation of eddies, the sediment transfer from high concentration regions to the low concentration regions takes place
View Answer

Answer: c
Explanation: In turbulent flow, momentum transfer or mass exchange takes place due to the jumping of particles from higher velocity region to lower velocity region. Due to this transfer of momentum between two adjacent fluid layers, effective shear stress is caused at the interface between the layers.

5. The sediment confined along and above the bed up to a depth ‘2d’ (d being grain size) is treated as bed load.
a) True
b) False
View Answer

Answer: a
Explanation: The sediment moves only on the top layers of the bed. Hence, the lower limit can be considered as equal to the thickness of the bed layer which is approximately equal to 2d and the upper limit as the total water depth. The sediment concentration at this depth can be considered equal to the concentration of bed load.
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6. In a wide stream, a suspended load sample taken at a height of 0.30 m from the bed. The stream is 5.0 m deep and has a bed slope of 1/4000. The bed material can be of uniform size. Estimate the shear friction velocity.
a) 0.111 m/s
b) 0.211 m/s
c) 0.311 m/s
d) 0.711 m/s
View Answer

Answer: a
Explanation: V*= Shear friction velocity = (S0/Dw)1/2 where Dw is the density of water and So (Yw.R.S) is shear stress at the bottom.
V* = [(Dw. g.R.S)/Dw]1/2 = (g.R.S)1/2
V* = (9.81 x 5 x 1/4000)1/2 = 0.111 m/s.

7. For the usual turbulent flow, calculate the critical shear stress if the mean diameter of the grain particle of bed material is 0.3 mm.
a) 0.300 N/m2
b) 0.106 N/m2
c) 0.206 N/m2
d) 0.116 N/m2
View Answer

Answer: c
Explanation: For usual turbulent flow, the value of critical shear stress is given by 0.687d where d is the arithmetic average diameter of sediment in mm.
S0 = 0.687 d
S0 = 0.687 x 0.3 = 0.206 N/m2.

8. Calculate the corresponding hydraulic mean depth that would exist in the channel if the bed was unrippled. The rugosity coefficient in an unrippled channel is 0.015 and the rugosity coefficient actually observed by experiments on the rippled bed of channel is 0.020. Consider the value of hydraulic mean depth of the channel as 1.5 m.
a) 1.5 m
b) 2.5 m
c) 0.97 m
d) 0.77 m
View Answer

Answer: c
Explanation: The formula that is used to take into account the effect of bed ripples is-
R’ = R (n’/n) 3/2 where, R = 1.5 m, n’ = 0.015, and n = 0.020
R’ = 1.5 x (0.015/0.020)3/2 = 0.97 m.

9. Calculate the quantity of bed load transport by using Meyer and Peter formula if the effective tractive force that causes bed load transportation is 2.5 N/m2.
a) 1.04 N/m/second
b) 1.5 N/m/second
c) 2.5 N/m/second
d) 0.476 N/m/second
View Answer

Answer: a
Explanation: Meyer and Peter has suggested the following formula for calculating the quantity of bed load transport –
Gb = 0.417 x Teff where Gb is the rate of bed transport by weight in N/m/second and Teff is an effective unit tractive force.
Gb = 0.417 x 2.5 = 1.04 N/m/second.

10. Calculate the effective unit tractive force that causes bed load transportation if the unit tractive force produced by the flowing water is 3.20 N/m2. The Manning’s coefficient on an unrippled bed is 0.0108 and the actual observed value of Manning’s coefficient on ripped channels is 0.0222. Take the value of critical shear stress that is required to move the grain particle as 0.5 N/m2.
a) 1.58 N/m2
b) 0.222 N/m2
c) 1.08 N/m2
d) 0.58 N/m2
View Answer

Answer: d
Explanation: The unit tractive force causing bed load to move is reduced by ripples in the ratio of
S0‘= S0 (n’/n) 3/2 = 3.2 (0.0108/0.0222)3/2 = 1.08 N/m2.
The effective unit tractive force = S0’ – Sc = 1.08 – 0.5 = 0.58 N/m2.

Sanfoundry Global Education & Learning Series – Irrigation Engineering.

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