300+ REAL TIME Irrigation Engineering Objective Questions & Answers 2023

250+ TOP MCQs on Energy Dissipation Below Overflow Spillway – 1 and Answers

Irrigation Engineering Multiple Choice Questions on “Energy Dissipation Below Overflow Spillway – 1”.

1. The energy dissipation at the toe of the spillway is affected basically by the use of hydraulic jump in _______________________
a) roller bucket
b) a ski-jump bucket
c) a sloping apron below the downstream river bed
d) both roller and ski-jump bucket
Answer: c
Clarification: Most of the kinetic energy is destroyed by creating a condition suitable for the formation of a hydraulic jump. Sometimes the depth of tail-water may be more than that necessary to create the hydraulic jump. The depth of water can be reduced to create a hydraulic jump by providing a sloping apron.

2. The most ideal condition for energy dissipation in the design of spillways is the one when the tail-water rating curve coincides with the jump rating curve at all discharge.
a) True
b) False
Answer: a
Clarification: The most ideal condition for jump formation is when TWC coincides with JHC at all discharge. To ensure protection in the region of a hydraulic jump, a simple concrete apron of apron length 5 (y₂ – y₁) is provided.

3. When the tail-water depths in the river downstream of a spillway are quite low such that the tail-water curve at all discharges lies below the post jump depth curve, then the energy dissipation can be affected best by ___________________
a) a roller bucket
b) a ski-jump bucket
c) either roller or ski-jump bucket
d) a sloping apron
Answer: b
Clarification: Energy dissipation bucket called ski-jump bucket is used when the tail-water depth is insufficient or low at all discharge. It requires sound and rocky river bed. Water may shoot up out of the bucket and fail harmlessly into the river at some distance downstream of the bucket.

4. The device which does not help in energy dissipation at the bottom of a hydraulic structure over which water spills is ________________
a) chute block
b) dentated sill
c) morning glory
d) baffle piers
Answer: c
Clarification: A flared inlet called morning glory is often used in large projects. The horizontal tunnel is either taken through the dam body or below the foundations. Chute blocks, dentated sills and baffle piers are all auxiliary devices which help in energy dissipation.

5. The formation of hydraulic jump at the foot of a spillway is one of the common methods of energy dissipation because ______________________
a) it destroys more than 90% of total energy by the turbulence produced in the jump
b) it reduces the kinetic energy by increasing the depth of flow
c) its action is not understood
d) it reduces the kinetic energy by decreasing the depth of flow
Answer: a
Clarification: Hydraulic jump is generally accompanied by large scale turbulence dissipating most of the kinetic energy of the super-critical flow. It is the most suitable method because the energy is lost in the impact of the water against water. Most of the kinetic energy is destroyed by creating a condition suitable for the hydraulic jump.

6. A ski-jump bucket is also known as _____________________
a) flip bucket
b) solid roller bucket
c) slotted roller bucket
d) flexible bucket
Answer: a
Clarification: A ski-jump bucket is also called flip bucket is used for energy dissipation when tail-water depth is insufficient or low at all discharge. A part of energy dissipation takes place by impact and some of the energy is dissipated in the air by diffusion or aeration.

7. The percentage of energy dissipation in a hydraulic jump increases with the increase in the Froude number.
a) True
b) False
Answer: a
Clarification: The energy dissipation in the jump depends upon the Froude number, if this Froude number is higher, the greater energy dissipation can take place.

S NO.	Froude number	% loss in energy
1.	2.5	17
2.	4.5	45
3.	9.0	70

8. Which of the following stilling basin help in stabilizing the flow and improve the jump performance?
a) dentated sills
b) chute blocks
c) baffle piers
d) friction blocks
Answer: b
Clarification: Chute blocks are a row of small projections like teeth of saw and are provided at the entrance of the silting basin. It produces a shorter length of jump and stabilizes the flow. Hence, they improve jump performance.

9. What is the expected solution for the case when the T.W.C is lying above the J.H.C curve at all discharges?
a) By providing a simple concrete apron of length 5(Y₁ – Y₂)
b) By providing a sloping apron above the river bed
c) By providing a sloping apron below the river bed
d) Provision of a ski-jump bucket
Answer: b
Clarification: When the TWC is lying above the JHC at all discharges, the problem can be solved by-
1) By constructing a sloping apron above the river bed
2) By providing a roller bucket type of energy dissipator.
In this case, the jump is formed at the toe will be drowned by the tail-water and little energy will be dissipated.

10. A sloping apron is provided partly above the river bed and partly below the river bed in case of ____________________________
a) when TWC coincides with the JHC at all discharges
b) when TWC lies above the JHC at all discharges
c) when TWC lies below the JHC at all discharges
d) when TWC lies above the JHC at low discharges and below the JHC at high discharges
Answer: d
Clarification: At low discharges, the jump will be drowned and at high discharges tail-water depth is insufficient. When TWC lies above the JHC at low discharges and below the JHC at high discharges, the solution is the provision of sloping apron partly above and partly below the river bed. The horizontal apron and end-sill are also provided.

11. When the TWC lies below the JHC at all discharges, the problem can be solved by which of the following provisions?
i. Constructing a sloping apron above the river bed
ii. Provision of roller bucket type of energy dissipator
iii. Provision of a ski-jump bucket
iv. A sloping apron below the river bed
v. Construction of a subsidiary dam
vi. A sloping apron partly above and partly below the river bed
a) i, iii and v
b) i, ii and vi
c) iii, iv and v
d) i, iii, iv and v
Answer: c
Clarification: When TWC lies below the JHC at all discharges, the expected solution is –
i. Provision of a ski-jump bucket
ii. A sloping apron below the river bed of length 5 (y₂ – y₁)
iii. Construction of a subsidiary dam below the main dam.

250+ TOP MCQs on Gravity Dams – Stability and Answers

Irrigation Engineering Multiple Choice Questions on “Gravity Dams – Stability”.

1. The factor of safety against overturning generally varies between ___________
a) 2 to 3
b) 1.5 to 2
c) 0.5 to 1.5
d) 1 to 2
Answer: a
Clarification: Factor of safety against overturning can be determined by the ratio of righting moments about the toe to the overturning moments about the toe. The value generally varies between 2 to 3.

2. What is the maximum permissible tensile stress for high concrete gravity dam under worst conditions?
a) 500 KN/m²
b) 500 kg/cm²
c) 5 kg/m²
d) 50 KN/m²
Answer: a
Clarification: The masonry and concrete gravity dams are usually designed in such a way that no tension is developed anywhere in the structure. The maximum permissible tensile stress for high gravity dams is taken as 500 KN/m² under worst conditions. If subjected to such tensile stresses crack develops near the heel.

3. Which failure occurs when the net horizontal force above any plane in the dam or at the base of the dam exceeds the frictional resistance developed at that level?
a) Overturning
b) Crushing
c) Sliding
d) BY development of tension
Answer: c
Clarification: Sliding should always be fully resisted. At any horizontal section of the dam, the factor of safety against sliding is –
FOS = u P_h / P_v where u = coefficient of friction, P_h = Sum of horizontal forces causing sliding and P_v = Algebraic sum of vertical forces.

4. Which failure occurs when the minimum stress exceeds the allowable compressive stress of the dam material?
a) Overturning
b) Crushing
c) Sliding
d) By development of tension
Answer: b
Clarification: The compressive stress produced if exceeds the allowable stresses then the dam material may get crushed, a dam may fail by the failure of its own material. The allowable compressive stress of concrete is generally taken as 3000 KN/m².

5. Tension cracks in the dam may sometimes lead to the failure of the structure by?
a) Sliding of the dam at the cracked section
b) Overturning about the toe
c) Crushing of concrete starting from the toe
d) Both overturning and crushing
Answer: c
Clarification: When tension prevails, cracks develop near the heel and uplift pressure increases, reducing the net salinizing force. This crack by itself does not fail the structure but it leads to failure of the structure by producing excessive compressive stresses.

6. The major principal stress at the toe of a gravity dam under full reservoir condition neglecting the tailwater effect is given by ____________________
a) P_v
b) P_v tanQ²
c) P_v secQ²
d) P_v sinQ²
Answer: c
Clarification: When there is no tailwater, the principal stress in such a case is P_v secQ² where P_v is the intensity of vertical pressure. This value of principal stress should not be allowed to exceed the maximum allowable compressive stress of dam material.

7. Which of the following criteria has to be satisfied for no tension at any point on a gravity dam?
a) The resultant of all the forces must always pass through the mid-point of the base of the dam
b) The resultant force for all conditions of loading must pass through the middle third of the base
c) The resultant of all the forces must pass through the upstream extremity of the middle third of the base
d) The resultant of all the forces must pass through the downstream extremity of the middle third of the base
Answer: b
Clarification: The minimum vertical stress P_min is equal to zero in order to ensure that no tension is developed anywhere. If P_min = 0, e = B/6 i.e. the maximum value of eccentricity that can be permitted on either side of the center is equal to B/6. This concludes the fact that the resultant of all forces must lie within the middle third of the joint width.

8. The bottom portion of a concrete or a masonry gravity dam is usually stepped in order to _______
a) increase the overturning resistance of the dam
b) increase the shear strength
c) decrease the shear strength
d) increase the frictional resistance
Answer: b
Clarification: The foundation is stepped at the base to increase the shear strength at the base and at other joints and measures is taken to ensure a better bond between the dam and the rock foundation. By ensuring a better bond between the surfaces the shear strength of these joints should be made as good as possible.

9. The governing compressive stress in a concrete gravity dam which should not be allowed to exceed the permissible value of about 3000 KN/m² while analyzing full reservoir case is ____________________
a) the vertical maximum stress at the toe
b) the major principal stress at toe
c) the shear stress at the toe
d) the principal stress at the heel
Answer: b
Clarification: In reservoir full case, the resultant is nearer to the toe and hence, maximum compressive stress is produced at the toe. The vertical direct stress distribution at the base is the sum of the direct stress and the bending stress and is given by the equation –
P_max = V/B [1 + 6e/B] where V is the total vertical force, e is the eccentricity of the resultant force from the center of the base and B is the base width.

10. If the uplift increases and the net effective downward force reduces, the resultant will shift towards the toe.
a) True
b) False
Answer: a
Clarification: The resultant shifts towards the toe if the uplift increases and the net effective downward force reduces. This further increases the compressive stress at the toe and further lengthening the crack due to the development of tension. It finally leads to the failure of the toe by direct compression.

250+ TOP MCQs on Factor Controlling Occurrence of Ground Water and Answers

Irrigation Engineering Questions and Answers for Entrance exams on “Factor Controlling Occurrence of Ground Water”.

1. An irrigation project is classified as a major project when the CCA involved in the project is more than ___________
a) 2000 hectares
b) 5000 hectares
c) 10000 hectares
d) 2500 hectares
Answer: c
Clarification: Culturable Command Area is the basis for the design of watercourse and the basis for the design of an irrigation project. The irrigation schemes in India are classified into three parts viz. Minor, Medium and Major Irrigation schemes depending upon the areas involved. Major irrigation scheme is the one where CCA involved in the project is greater than 10,000 hectares.

2. A minor irrigation scheme serves up to ________________
a) 100 hectares
b) 500 hectares
c) 1000 hectares
d) 2000 hectares
Answer: d
Clarification: The irrigation schemes in India are classified into three parts depending upon the areas involving culturable command area.

Type of scheme	Areas involving CCA
1. Minor irrigation scheme	< 2000 hectares
2. Medium irrigation scheme	2000 to 10000 hectares
3. Major irrigation scheme	> 10000 hectares

3. Energy is required in the utilisation of _____________
a) groundwater
b) surface water
c) both groundwater and surface water
d) capillary water
Answer: a
Clarification: For irrigation purposes, groundwater is largely tapped in India through wells and tube wells. Manual, wind, diesel or electric power can be used for lifting water from open wells. The subsequent development in the technique of tapping groundwater is the use of tube wells which requires diesel or electric power.

4. Which of the following property of geological formation represents its water storage capacity?
a) Permeability
b) Porosity
c) Both permeability and porosity
d) Transmissibility
Answer: b
Clarification: Porosity is the quantitative measurement of the interstices of voids present in the rock. In other words, it is the maximum amount of water that can be stored in the rock.

5. The zone of aeration in a groundwater profile does not include ___________
a) capillary zone
b) soil water zone
c) intermediate zone
d) saturation zone
Answer: d
Clarification: Depending upon the number of interstices present, the aeration zone is divided into three classes. The capillary fringe is a continuation of the zone of saturation and does contain some interstitial water. Soil zone is the depth from the surface penetrated by the roots of vegetation and the remaining intermediate part is intermediate zone.

6. In which of the following zone the stresses are beyond the elastic limits?
a) Zone of rock fracture
b) Zone of rock flowage
c) Zone of saturation
d) Zone of aeration
Answer: b
Clarification: In the zone of rock flowage, interstices are absent because the stresses are beyond the elastic limits. The rocks remain in a state of plastic flow and water present in this zone is known as internal water.

7. The rate of flow of water through ground strata can be estimated by _____________
a) Manning’s formula
b) Strickler’s formula
c) Dupuit’s equation
d) Darcy’s formula
Answer: d
Clarification: A French Scientist Mr. H Darcy on the basis of experimental evidence gave a law governing the discharge through soils. According to the law, the discharge is directly proportional to the head loss and the area of cross-section of the soil and inversely proportional to the length of the soil sample.

8. The relation between Transmissibility (T) and Permeability (K) for an aquifer of depth d is _______
a) K = T.d
b) T = K.d
c) T = K.log d
d) T = ln (Kd)
Answer: b
Clarification: Transmissibility is measured by the coefficient of transmissibility (T) and was introduced by Theiss. It can be defined as the rate of flow of water through an aquifer of unit width and full-depth under a hydraulic gradient and at a temperature of 20°C. The relation between K and T is given as T=K.d.

9. Darcy’s law is valid when the flow is ___________
a) laminar and steady
b) non-uniform
c) turbulent
d) both laminar and turbulent
Answer: a
Clarification: The Darcy’s law has been demonstrated to be valid only for laminar flow conditions because the flow in sands, silts, and clays is invariably laminar.
Mathematically, v = K.i where v = discharge velocity, K = coefficient of permeability and I = hydraulic gradient.

10. The coefficient of permeability indicates the ease with which water can flow through a soil mass. The soil type which has less permeability is __________
a) gravelly soil
b) clayey soil
c) sandy soil
d) both sandy and gravel soil
Answer: b
Clarification: The approximate average value of the coefficient of permeability in clayey soil is 0.04 x 10^-5 cm/sec. The permeability coefficient of gravel soil is of the order of 4cm/sec and for sandy soil is 0.04 cm/sec.

11. Which of the following is the most important zone for a groundwater hydraulic engineer?
a) Zone of rock fracture
b) Zone of rock flowage
c) Zone of saturation
d) Zone of aeration
Answer: c
Clarification: This is the most important zone for a hydraulic engineer as this water has to be tapped out and the water in this zone is under hydrostatic pressure. In this zone, water exists within the interstice which is nothing but groundwater.

12. Which zone contains water that is under molecular attraction?
a) Zone of rock fracture
b) Zone of rock flowage
c) Zone of saturation
d) Zone of aeration
Answer: d
Clarification: Zone of aeration is the space above the water table and below the surface and the water exists in this zone by molecular attraction. The water in this zone is not at hydrostatic pressure and the gravity water moves through this zone.

Irrigation Engineering for Entrance exams,

250+ TOP MCQs on Canal Falls – Location and Answers

Irrigation Engineering Multiple Choice Questions on “Canal Falls – Location”.

1. Canal drops are provided to _______________
a) dissipate the excess energy
b) increase the driving head of flow of water
c) dissipate excess land slope
d) dissipate inadequate land slope
Answer: c
Clarification: when a canal fall or drop is provided at a suitable section, it brings down the canal bed line. In this process, water comes down with a great force and all the potential energy is converted into kinetic energy. The excess energy of the flow is destroyed with some suitable energy dissipation method.

2. For a canal that irrigates the area directly the fall should be provided at a location FSL outstrips the G.L before the bed of the canal comes into the filling.
a) True
b) False
Answer: a
Clarification: This is done in case of branch canals and distributary channels which irrigate the area directly, the FSL is fixed and marked at the head of the off-taking channel and outlets. In the case of the main canal which does not irrigate the area directly, the site of the fall is determined by considerations of economy.

3. In case of branch canals and distributary channels, the falls are located with consideration to _________________
a) command area
b) topography
c) cost economy
d) availability of earth material
Answer: a
Clarification: In the case of branch canals and distributary canals, the procedure is to mark the FSL required on the L-section of the canal. It is done so as to cover all the commanded points that decide the suitable locations for fall in canal FSL and in canal beds.

4. By providing a natural drop in one drop _______________________
a) the quantity of unbalanced earthwork decreases
b) the quantity of unbalanced earthwork increases
c) the quantity of balanced earthwork increases
d) number of fall increases
Answer: b
Clarification: The number of fall decreases and the quantity of unbalanced earthwork increases by providing a larger drop in one step. All these factors have to be worked out before deciding the locations and extent of falls.

5. A canal fall is provided when the available natural slope is flatter than the designed bed slope of the channel.
a) True
b) False
Answer: b
Clarification: When the ground has a steep slope heavy earth filling is required to construct the canal with a flatter bed slope. The difference between the natural slope and the designed bed slope is adjusted by constructing vertical falls or drops in the canal at a suitable interval.

6. Consider the following statements.
I. Possibility of combining some other structure with a fall eg. regulator, road bridge etc.
II. Cutting and filling are required below and above the fall should not be necessarily equal.
III. The command should not be reduced due to the lowering of F.S.L and the fall may be located below the outlets.
Which of the following statement is correct for site selection of a fall?
a) I only
b) III only
c) I and II
d) I and III
Answer: d
Clarification: The unbalanced earthwork is quite expensive so the excavation and filling on two sides of the fall should be balanced. The considerations of the economy in ‘cost of cutting and filling’ versus ‘cost of fall’ form a basis in deciding the site of a fall in case of the main canal.

250+ TOP MCQs on Classification of Rivers and Answers

Irrigation Engineering Multiple Choice Questions on “Classification of Rivers”.

1. Depending on the topography of the river basin, into how many classifications the river reaches are divided?
a) 3
b) 4
c) 5
d) 1
Answer: a
Clarification: The classifications given on the basis of topography of river reaches are river in hills (upper reaches), rivers in plains (lower reaches), and tidal rivers.

2. The rivers in hills are further classified into how many groups?
a) 4
b) 2
c) 3
d) 6
Answer: b
Clarification: Before entering the plains the upper reaches of the rivers flow through the hills, hence the name river in hills. These are further subdivided into a rocky river stage, and boulder river stage.

3. Which stage of the river is formed by erosion?
a) Aggrading
b) Tidal Rivers
c) Boulder River Stage
d) Rocky River Stage
Answer: d
Clarification: In this type of stage of the river the flow is formed by rapid erosion. These river reaches have high steep with swift flow, and form rapids along their courses. The bed load carried by these reaches cannot be determined by usual bed load transportation formulas.

4. What type of river stage widens the bed?
a) Degrading
b) Incised River Stage
c) Boulder River Stage
d) Stable Type
Answer: c
Clarification: The river bed in these reaches is created by itself, consists a mixture of boulders, gravels, shingles, and alluvial sand deposits. In latter stage, the river flows through deep well defined beds and wider floodplains. Here the river flows in a straight course. Due to the floods, the boulders and material travels downward and gets deposited there after floods. Because of the opposition to the flow due to these obstructions the river takes new course and thus widens the river bed.

5. In what reaches the river flows in a zigzag manner?
a) Boulder River Stage
b) Lower Reaches
c) Degrading
d) Tidal Rivers
Answer: b
Clarification: The chief character of the river in these reaches is that it flows in a zigzag manner called meandering. This meandering is due to the difference in carrying the sediment which is similar to the bed load from one bank to the other bank of the river.

6. Into how many groups the rivers in floodplains can further be divided?
a) 5
b) 6
c) 4
d) 3
Answer: a
Clarification: River in the floodplains is further divided into aggrading, degrading, stable, braided, and deltaic types.

7. In the meandering process on which side of the bank erosion takes place?
a) Convex Side
b) Bottom of the River
c) Concave Side
d) Due to Sediment Load
Answer: c
Clarification: In these reaches there is a constant erosion of the river bed on the concave side and the deposition of the eroded soil on the convex side of the successive bends or between two successive bends.

8. In which type of river building up of slope is present?
a) Deltaic
b) Stable type
c) Degrading Type
d) Aggrading Type
Answer: d
Clarification: This type of river is formed due to the silting action which increases the bed slope. This silting action is due to various factors like heavy sediment load, construction of obstructions like dam or a weir, and sudden intrusion from a tributary.

9. In which type of river there is a reduction of the available slope?
a) Braided
b) Stable Type
c) Aggrading Type
d) Degrading Type
Answer: d
Clarification: This type of river is formed due to the constant erosion to reduce and finally dissipate available excess land scape. It is found below a dam or weir or barrage or above a cut off.

10. What type of river does the diagram represent?

a) Braided River
b) Aggrading Type River
c) Degrading type River
d) Stable Type River
Answer: a
Clarification: This is braided river as the flow is around alluvial islands. These islands are developed due deposition of coarser material which cannot be transported under the existing conditions of the flow, consisting of coarse and fine material.

11. A river before joining the sea or ocean gets divided into branches.
a) False
b) True
Answer: b
Clarification: As a river approaches the sea or ocean, the velocity of the flow gets reduced, and in turn silting action takes place resulting in rising of water levels and the formation of new channels. These channels or branches multiply in numbers as the river approaches the sea or ocean.

12. A river which does not change its alignment, slope is called a stable river.
a) True
b) False
Answer: a
Clarification: In this type of river changes like silting action, scouring etc in alignment or slope occurs, but these changes are negligible may fail to produce any changes in the regime of the channel. In this type most of the sediment is brought to the sea or ocean.

250+ TOP MCQs on Stable Channels Design in India and Answers

Irrigation Engineering Multiple Choice Questions on “Stable Channels Design in India”.

1. Design of alluvial channels in India is based on Kennedy and Lacey theories.
a) True
b) False
Answer: a
Clarification: 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.

2. On what condition does the resistance equations of Chezy’s formula and manning’s formula are applicable?
a) τ_o
b) τ_c
c) τ_o < τ_c
d) τ_o > τ_c
Answer: c
Clarification: The average shear stress (τ_o) which is acting on the boundary of an alluvial channel should be less than the critical shear stress (τ_c), 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.

3. 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
Answer: b
Clarification: 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.

4. The velocity of flow in a channel should not more or less, but instead it should be adequate.
a) True
b) False
Answer: a
Clarification: 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.

5. What is the effect of scouring in channels?
a) The Channel Section gets reduced
b) Reduces Discharge Capacity
c) Improper Working of Channel
d) Breaching of Canal Banks
Answer: d
Clarification: Scouring is not a rare phenomenon in channels. It is very understandable and can be controlled or avoided by proper designing of channels. Scouring causes loss of command in channels and lowers full supply level of channel. This causes breaching of canal banks and in turn causes failure of foundations of irrigation structures.

6. What is the effect of silting in channels?
a) Reduced Discharge Capacity of Channel
b) Causes Loss of Command
c) Breaching of Canal Banks
d) Failure of Irrigation Structures
Answer: a
Clarification: The silting problem is very common in any kind of artificial channels, or we can say there is no natural or artificial channel with the silting problem. Silting interferes with the proper working of a channel, as it causes reduction in the channel section due to siltation, which thereby reduces the discharge capacity of the channel.

7. Which type of state is achievable in artificial channels but not in rivers?
a) Turbulent State
b) Laminar State
c) Regime State
d) Uniform State
Answer: c
Clarification: The channel is said to be in regime state, when the flow of the channel is such that silting and scouring effects need no special attention. This state is not easy to achieve in rivers but can be achieved in artificial channels, by proper designing of the channels.

8. On what basis for designing the regime state is obtained?
a) Silt and Velocity of Flow
b) Silt
c) Velocity
d) Depth of Channel
Answer: a
Clarification: The basis for designing a channel to be in regime state, whatever the amount of silt has entered the channel has to be kept in suspension, so that it does not settle down and get deposited in the channel anywhere. And the velocity of the flow in the channel should be such that it does not produce silt by eroding the banks and beds of the channel.

9. What vital factor is not considered in regime theories during the design of regime channels?
a) Velocity of Flow in the Channel
b) Type of Flow in the Channel
c) Quantity of Sediment Entering a channel
d) Area in which the Channel flows
Answer: c
Clarification: The quantity of sediment load entering the channel from the head works plays an important role in designing the regime channels, as it controls the cross-section and shape of the regime channel. Moreover the design of regime channel is not complete until provisions are made for the effects produced by the actual quantity of sediment load present in the channel.

10. According to Kennedy the silting action in channels is due to?
a) Generation of Eddies
b) Slope of the Bed
c) Area of Channel
d) Type of Flow in the Channel
Answer: a
Clarification: From the observations of his research, he concluded that the silting action in channels is due to the generation of eddies, rising to the surface. These eddies are in turn generated by the friction of flowing water with the channel surface.

11. Based on his research what factor is given by Kennedy for free silting and scouring actions in a channel?
a) Critical Velocity (V_o)
b) Bed Slope of Channel
c) Hydraulic Mean Depth
d) Rugosity Coefficient
Answer: a
Clarification: Eddies affect the formation of silting action in a channel brief investigation shows that the vertical component of these eddies try to move sediment up and the weight of sediment tries to bring it down, thus keeping the sediment in suspension. So, silting is avoided if sufficient velocity is generated to form eddies and keep the sediment in suspension. This velocity is called critical velocity.

12. Given to design the irrigation channel to carry 80 cumecs of discharge. The channel slope is 1 in 3000. The critical velocity ratio is 1.3. Take rugosity coefficient as 0.021.
a) Depth = 3 m, Base width = 15 m
b) Depth = 3.5 m, Base width = 15.5 m
c) Depth = 3.3 m, Base width = 15.2 m
d) Depth = 3.2 m, Base width = 15.1 m
Answer: d
Clarification: Given Q = 80 cumecs, S = 1/3000, m = 1.3, n = 0.021
Critical velocity (V_o) = 0.55my^0.64 (y = depth of the channel)
= 0.55 x 1.3 x (2)^0.64
= 1.114 m/sec
Area (A) = Q/V_o = 80 / 1.114 = 71.8 m²
Now assume side slope as (1/2): 1(1/2H: V)
Now, A = y (b + y x 1/2)
71.8 = 2(b + 2 x 1/2)
71.8 = 2(b + 1)
b = 34.9 m
Perimeter (P) = b + 2 x √5/2y
= 34.9 + 4.47
= 39.37 m
Now R = A/P = 71.8 / 39.37 = 1.82 m
V = ((1/n + (23 + 0.00155/s)) / (1 + (23 + 0.00155/s)n/(sqrt{RR})))(sqrt{RSRS})
V = 0.78 m/sec
Now calculating for critical velocity,
V_o = 0.55 x 1.3 x y^0.64
= 0.715 x y^0.64
Now assume y = 2.5 m
V_o = 0.715 x (2.5)^0.64 = 1.28 m/sec
A = 80 / 1.28 = 62.5 m²
62.5 = 2.5 (b + 1/2(2.5))
b = 23.75 m
P = 23.75 + 2 x √5/2(2.5) = 29.35 m
R = A/P = 2.13
√R = 1.46
V = 75.27/1.4 x (1.46/54.8) = 1.432 m/sec
V > V₀
Now assume y = 3.2 m
V_o = 0.715 x (3.2)^0.64 = 1.51 m/sec
A = 80 / 1.5 = 53.33 m²
53.33 = 3.2 (b + 1.6)
b= 15.1 m
P = 15.1 + √5(3.2)
P = 22.2 m
R = A/P = 53.33/22.2 = 2.4 m
√R = 1.56
V = (75.27/1.4) x (1.56/54.8)
= 1.53 m/sec
Therefore V = V_o
Hence use the depth equal to 3.2 m and base width 15.1 m with slope 1/2:1.

13. Design a regime channel for a discharge of 90 cumecs and silt factor is 1.3 use Lacey’s Theory.
a) Depth = 2.2 m, Base width = 40.5 m, Slope = 1/4670
b) Depth = 2.12 m, Base width = 40.42 m, Slope = 1/4675
c) Depth = 2.135 m, Base width = 40.3 m, Slope = 1/4676
d) Depth = 2.123 m, Base width = 40.1 m, Slope = 1/4672
Answer: c
Clarification: Given Q = 90 cumecs, f = 1.3
V = [Qf²/140]^1/6 = 1.014 m/sec
A = Q/V = 90/1.014 = 88.76 m²
R = 5/2 x V²/f = 2 m
P = 4.75√Q = 45.06 m
For a trapezoidal channel with 1/2H: 1V slopes
P = b +√5y and A = (b + y/2) y
45.06 = b + √5y, 88.76 = (b + y/2) y
b = 45.06 – 2.24y, substitute this value in area equation
88.76 = (45.06 – 2.24y + y/2) y
Y = 2.135 m
b = 45.06 – 2.24 x 2.135 = 40.3 m
S = (f^5/3 / 3340Q^1/6) = 1/4676.

14. According to Lacey what factor is needed to have a true regime in an artificial channel?
a) Flow is Uniform
b) Critical Velocity = Actual Velocity
c) No Silting or Scouring Action
d) Lining of the Channel Bed
Answer: a
Clarification: For a channel to be in regime condition it should not have silting and scouring action. To obtain this the silt entering the channel must be carried away through the channel, by the channel section. The artificial channel to be in a regime it must have a fixed slope and fixed section. So, therefore to achieve this flow in the channel should uniform flow.