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


This set of Irrigation Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Sediment Transport – Bed Load and its Measurement”.

1. On which factor does the movement of bedload depends?
a) Velocity of Flow
b) Type of Flow
c) Depth of Flow
d) Width of the River
View Answer

Answer: a
Explanation: Bedload is a sediment load, and it moves by the actions like rolling, sliding and hopping which in turn depend on the velocity of flow. The bed load remains in the bottom layers of the flow.

2. Commonly the bed load is 10% of total suspended load.
a) True
b) False
View Answer

Answer: a
Explanation: Sometimes bed load is estimated to be in between 3 to 25% of the total suspended load, depending upon the reflexes of the sediment to forces from the sectors like physical, chemical, and biological. In between this percentage we commonly take bed load is 10% of total suspended load.

3. What type of force is completely responsible for the bedload movement?
a) Forces of Turbulence
b) Drag Force
c) Capillary Force
d) Gravity Force
View Answer

Answer: b
Explanation: Generally we know that bedload is a type of sediment which moves adjacent to the bed of the channel. Shear stress (τo) is responsible for this movement of bedload along the bed if the channel, which is developed by the flowing water along the channel bed. This shear force is called drag force or tractive force.

4. Which type of force is needed for suspension of suspended load in flowing water?
a) Capillary Force
b) Drag Force
c) Forces of Turbulence
d) Gravity Force
View Answer

Answer: c
Explanation: Suspended load is a sediment load, but remains in suspension in the flowing water of the channel. The forces which cause this suspension are the forces of turbulence, which are generated by the flow of the channel itself.

5. A part of tractive force does not have any part in transporting the bed material, i. e bed load.
a) True
b) False
View Answer

Answer: a
Explanation: In order to transport the bed load the tractive force (τo) should exceed the critical tractive force (τc), therefore the rate of movement of bedload becomes a function of (τo – τc). But, as soon as the sediment starts moving it is opposed by the ripples generated by the channel bed, and to overcome these ripples a large part of tractive force is lost and cannot be used again. So, this part of the force does not have any part in transporting the bed load.

6. Given that the bed slope of a channel is 1 in 2000 and the discharge is 60 cumecs. The depth of the channel is fixed and is given as 2 m. The tractive force needed for the movement of bedload is? Take critical velocity ratio as 1.1.
a) 8.89 N/m3
b) 8.87 N/m3
c) 8.85 N/m3
d) 8.83 N/m3
View Answer

Answer: d
Explanation: Critical velocity (Vo) = 0.55my0.64
= 0.55 x 1.1 x (2)0.64
= 0.943 m/sec
Area = Q / Vo = 60 / 0.943 = 63.63 m2 (Given Q = 60 cumecs)
A = y (b + y (1/2)) (y = depth, b = base width)
63.63 = 2(b + 1) (for side slope as (1/2:1/2H:V))
b = 30.82 m
Perimeter (P) = b + √5y = 30.82 + 2 x √5
P = 35.3 m
R = A/P = 63.63 / 35.3 = 1.8 m (R = hydraulic mean depth)
Now, tractive force (τo) = γwRS = 9.81 x 103 x 1.8 x (1/2000) (γw = 9.81 x 103 N/m3)
= 8.83 N/m3.

7. What cause is prime responsible for the heavy movement of water from main canal to branch canal?
a) Existence of Favorable Gradient
b) Velocity of Flow
c) Type of Flow
d) Depth of the Canal
View Answer

Answer: a
Explanation: When a branch canal is connected to the main canal, the branch canal starts taking water discharge which reduces the discharge in main canal thereby reduces sediment load capacity. This leads to the deposition of sediment load which further diverts the flow towards branch canal. This process continues till all the water in the main canal gets discharged into branch canal. The prime cause of this phenomenon is due to the existence of favorable gradient in the branch canal.

8. Water in which condition or state carry a maximum amount of sediment?
a) Uniform State
b) Vapor State
c) Floods
d) Ice
View Answer

Answer: c
Explanation: According to Einstein’s equation (qb/q≼qS2/d3/2), the sediment carrying capacity of the channel depends upon the discharge per unit width. Therefore, this gives that floods can carry more amount of sediment than any other state of water because the discharge per unit width of channel is maximum in case of floods. Most of the annual sediment load is done by floods.

9. What minimum value of shear stress is needed to move the sediment?
a) Critical Velocity Ratio
b) Critical Shear Stress
c) Critical Velocity
d) Drag Force
View Answer

Answer: b
Explanation: Generally in nature, any two moving bodies oppose each other due to internal presence of friction between them. Here as the sediment load moves along the bed of the channel, then internal friction is developed between them depending upon the soil. Therefore to overcome this friction a minimum value of shear stress is needed. This minimum value is called critical shear stress (τc).

10. Design a channel carrying a 30 cumecs. The median grain diameter is taken as 0.5 mm. The bedload concentration is 60 p.p.m by weight. Use Lacey’s Regime perimeter and Meyer-peter’s formulas.
a) B = 26 m, S = 1/5600, y = 1.25 m
b) B = 22 m, S = 1/5800, y = 1.55 m
c) B = 24 m, S = 1/5500, y = 1.45 m
d) B = 25 m, S = 1/5700, y = 1.35 m
View Answer

Answer: d
Explanation: Quantity of bed load transported by weight = 40/106
Quantity of bed load transported per second = 40/106 (30 x 9.81 x 1000) = 11.8 N/sec
Lacey’s Regime perimeter = 4.75 x √Q = 26.03 m
Let us take channel bed width (B) as 25 m
Bed load per unit width = gb = 11.8 / 25 = 0.472 N/m/sec
Meyer Peter equation –
gb = 0.417 x [τo(n’/n) – τc]3/2
n’ = (1/24) x (0.5)1/6
= 0.011
n = 0.02
n’/n = 0.55
τc = 0.687 x da = 0.687 x 0.5 = 0.3435 N/m2
gb = 0.417 [9.81 x 1000 x RS x (0.55)3/2 – 0.3435]3/2
RS = 0.0002
Manning’s equation –
Q = 1/n x R2/3 x S1/2
R2/3 x S1/2 = 0.6
S = 0.0002/R
S = 1/5700
R = 1.15 m
Now P = 25+y√5, A = 25y + y2/2 for trapezoidal channel of 1/2:1 slopes
R = A/P = (25+√5y)/ (25+y2/2) = 1.15
From this y = 1.35 m
Therefore B = 25 m, y = 1.35 m, S = 1/5700.

Sanfoundry Global Education & Learning Series – Irrigation Engineering.

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