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Irrigation Engineering Test 2
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Irrigation Engineering Test 2
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  • Question 1/10
    1 / -0

    According to Kennedy’s theory which of the following factor does take into account the effect of different silt grade while calculating the critical velocity?
    Solutions

    Concept:

    Kennedy’s Silt Theory: Kennedy’s slit theory is used for designing of unlined canal especially for the canals on alluvial soils. According to this theory, the critical velocity is given as:

    \({{\rm{V}}_0} = 0.55{\rm{m}}{{\rm{y}}^{0.64}}\) where V0 is the critical velocity, y is the depth of the water in the channel and m is critical velocity ratio.

    Critical Velocity Ratio (m): The ratio of mean velocity V to the critical velocity V0 is known as the critical velocity ratio (CVR). It is denoted by m i.e. \({\rm{CVR\;}}\left( {\rm{m}} \right) = \frac{{\rm{V}}}{{{{\rm{V}}_0}}}\).

    CVR accounts for the effect of different types of silt grade on critical velocity. Generally, m ranges 1.0 to 1.2 for coarse sand and 0.7 to 1.0 for fine sand.

    Note:

    Silt Factor (f): Used in Lacey’s Theory to calculate mean velocity.

    Rugosity Coefficient (n): Used in Kutter’s formula and Manning’s formula to calculate mean velocity.

    Shape and surface factor (C): Used in Chezy’s formula to calculate mean velocity.

  • Question 2/10
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    Different components diversion headworks of a canal is listed in column 1 with their functions in column 2 in the following table.

    Components (Column 1)

    Function (Column 2)

    1. Head sluices
    1. Helps in providing a less turbulent pocket near head regulator
    1. Silt Ejector
    1. Helps in removing silts near the head regulator
    1. Under sluices
    1. Helps in controlling the entry of silts in the canal
    1. Divide Wall
    1. Helps in removing silt from the water after it enters the canal

    The correct code is

    Solutions

    Concept:

    Head Sluices (or Canal Head Regulator): A head sluices or canal head regulator (CHR) is provided at the head of the off-taking canal, and serves the following functions:

    • It controls the entry of silt in the canal.
    • It regulates the supply of water entering the canal.
    • It prevents the river floods from entering the canal.

    Silt Ejectors (or Silt extractors): These are the devices that extract the silt from the canal water after the silted water has traveled a certain distance in the off-take canal. These works are, therefore, constructed on the bed of the canal, and a little distance downstream from the head regulator.

    Under-Sluices (or Scouring Sluices): The under sluices are the openings that are located on the same side as the off-taking canal and are fully controlled by gates, provided in the weir wall with their crest at a low level. Their main functions are:

    • It helps in controlling the silt entry into the canal.
    • It scours the silt deposited on the river bed above the approach channel.
    • It passes the low floods without dropping the shutter of the main weir.
    • It preserves a clear and defined river channel approaching the regulator.

    Divide Walls: Divide wall is a masonry or concrete wall constructed at the right angle to the axis of the weir. Its main objective is to form a still and comparatively less turbulent water pocket in front of the canal head regulator so that the suspended silt can be settled down which then later be cleaned through the scouring sluices from time to time.

    ∴ The correct combination is 1-C, 2-D, 3-B, 4-A.

     

  • Question 3/10
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    When the bed level of canal is higher than the highest flood level (HFL) of discharge, then the cross discharge work is said to be
    Solutions

    Concept:

    A cross drainage work is a structure constructed for carrying a canal across a natural drain or river intercepting the canal to dispose drainage water without mixing with the continuous canal supplies.

    Types of Cross Drainage Works:

    Based on the relative bed levels, water levels of the canal and the drain and their relative discharge the Cross Drainage works are of the following types:-

    (1) Cross drainage works carrying the Canal over the Natural Drain:

    (i) Aqueduct

    • An aqueduct is a hydraulic structure which carries a canal (through a trough) across and above the drainage similar to a bridge in which instead of the road or a railway, a canal is carried over a natural drain.

    • In the case of an aqueduct, HFL (highest flood level) of the drainage should remain lower than the level of the underside of the canal trough.
    • The drain flows at atmospheric pressure under the work.
    • Generally, an inspection road is provided along with the trough

    (ii) Syphon aqueduct

    • A syphon aqueduct is a cross drainage structure similar to an aqueduct except that the streambed is depressed locally where it passes under the trough of the canal and the barrels discharges the streamflow under pressure.
    • A syphon aqueduct is constructed where the water surface level of the drain at high flood is higher than the canal bed.
    • The horizontal floor of the barrels is provided with slopes at its ends to join the drain bed on either side.
    • The drain water flows under pressure through the barrels which act as inverted syphons and hence this cross drainage work is known as syphon aqueduct.

     

    2) Cross drainage works carrying the natural drain over the canal-

    (i) Super passage

    A super passage is also similar to a bridge in which the natural drain is carried over the canal.

    • A super passage is reverse of an aqueduct.
    • A super passage is constructed where the bed of the drain is well above the canal F.S.L
    • The canal flows at atmospheric pressure under the work.

    • In this case it is not possible to provide an inspection road along the canal.

    (ii) Syphon

    • A syphon is similar to a syphon aqueduct with the difference that in the case of a syphon the canal water is carried through the barrels under the drain.
    • A syphon is constructed where the full supply level of the canal is higher than the bed of the drain.
    • The barrels in this case also act as inverted syphons through which the canal water flows under pressure.

    3) Cross drainage works admitting the drain water into the canal-

    In this type of cross drainage works, the canal water and the drain water are allowed to intermingle with each other. This may be achieved by the following two types of cross drainage works:-

    (i) Level crossing

    (ii) Inlet and outlet

  • Question 4/10
    1 / -0

    Which one of the following is not the major factor influencing seepage from a canal?
    Solutions

    Concept:

    The process by which a liquid leaks through a porous substance is known as the process of seeping.

    According to darcy’s law,

    Q = kiA where K = permeability of soil

    i = hydraulic gradient

    A = Area of percolation

    So in seepage, the discharge depends upon the wetted area.

    Location of the canal also affects the hydraulic gradient available for seepage. So location of canal also influences the seepage.

    The time period for which the canal is having flow, during that time seepage loss will also occur. When there is no flow or the canal is dry, there will be no seepage loss in the canal. So frequency of canal usage is also a major factor in seepage loss.

    Characteristics of the soil traversed by the canal system does not influence the seepage in canal.

  • Question 5/10
    1 / -0

    A canal has a flowrate of 4.5 m3/sec and has medium size silts with an average grain size of 0.016 cm. What will be the velocity through the canal as per Lacey’s theory?
    Solutions

    Concept:

    Lacey’s Theory: Lacey’s theory is based on the concept of regime condition of the channel. The regime condition will be satisfied if, a) the channel flows uniformly in unlimited incoherent alluvium of the same character which is transported by the channel, b) the silt grade and silt charge remains constant and c) the discharge remains constant. According to this theory, the velocity is given by,

    \({\rm{V}} = {\rm{\;}}{\left( {\frac{{{\rm{Q}}{{\rm{f}}^2}}}{{140}}} \right)^{\frac{1}{6}}}\) where V is the velocity in m/ sec, f is the Lacey’s silt factor and Q is the discharge in m3/sec.

    The silt factor is related with average particle size and given by, \({\rm{f}} = 1.76\sqrt {{{\rm{d}}_{{\rm{mm}}}}}\) where dmm is the average particle size in mm.

    Calculation:

    Given, average particle size, dmm = 0.016 cm = 0.16 mm and discharge, Q = 4.5 m3/sec.

    ∴ Silt factor, \({\rm{f}} = 1.76 \times \sqrt {0.16} = 0.704\).

    ∴ Velocity, \({\rm{V}} = {\rm{\;}}{\left( {\frac{{4.5\; \times \;{{0.704}^2}}}{{140}}} \right)^{\frac{1}{6}}} = 0.502{\rm{\;m}}/{\rm{sec}}\)

    Mistake Point:

    Lacey’s formulas are empirical in nature. So, all the data must be substituted in the exact units as mentioned above, otherwise the constants given in the formulas are invalid. So, if in the formula the average particle size is substituted in cm, the answer will be wrong.
  • Question 6/10
    1 / -0

    Consider the following statements regarding cross drainage structures:

    1) The crossing arrangement, preferably made at the junction of a huge canal and a river stream carrying short lived high flood discharge at almost equal bed level is called level crossing.

    2) Crest Wall is provided in a level crossing across the drainage just at the upstream side of the crossing point.

    3) Drainage regulator and canal regulator are provided on the downstream point of crossing in a level crossing.

    Which of the following statements are correct?

    Solutions

    Level Crossing

    The level crossing is an arrangement provided to regulate the flow of water through the drainage and the canal when they cross each other approximately at the same bed level. The level crossing consists of the following components:

    Crest Wall: It is provided across the drainage just at the upstream side of the crossing point. The top level of the crest wall is kept at the full supply level of the canal.

    Drainage Regulator: It is provided across the drainage just at the downstream side of the crossing point. The regulator consists of adjustable shutters at different tiers.

    Canal Regulator: It is provided across the canal just at the downstream side of the crossing point. This regulator also consists of adjustable shutters at different tiers.

  • Question 7/10
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    The dimensions of trapezoidal concrete lined channel having discharge Q = 350 m3/s, bed slope S= 1/5000, side slope 1.5H:1V, n = 0.014.

    Assume limiting velocity = 2m/sec

    Solutions

    Area and perimeter for trapezoidal is given by:

    A = y2 (θ + cot θ) + By

    P = 2y (θ + cot θ) + B

    Cot θ = 1.5

    Q = 0.59 radian

    A = y2 (θ + cotθ) + By = y2 (0.59 + 1.5) + By

    ⇒ 2.09y2 + By                                                     ----------------(1)

    P = 2y (θ + cotθ) + B = 2y (0.59 + 1.5) + B

    ⇒ 4.18y + B                                                         ----------------(2)

    \(\begin{array}{l}V = \frac{1}{n}{R^{\frac{2}{3}}}\;{S^{\frac{1}{2}}} = 2\\ = \left( {\frac{1}{{014}}} \right){R^{\frac{2}{3}}}{\left( {\frac{1}{{5000}}} \right)^{\frac{1}{2}}} = 2\end{array}\)

    R = 2.79

    \(\begin{array}{l}A = \frac{Q}{V} = \frac{{350}}{2} = 175\;{m^2}\\R = \frac{A}{P} = \frac{{175}}{{4.18y + B}} = 2.79\end{array}\)

    Putting value of B from equation (1)

    \(R = \frac{{175}}{{4.18y + \left( {\frac{{175 - 2.09{y^2}}}{y}} \right)}} = 2.79\)

    Solving above equation,

    y = 3.2 m

    B = 48 m
  • Question 8/10
    1 / -0

    Consider the following weir floor on a permeable soil with three sheet piles. What will be the ratio (corrected up to three decimal points) of the creep length according to Lane’s creep theory to the creep length according to Bligh’s creep theory?

    Solutions

    Concept:

    A weir with three sheet piles is shown below. The seeping path of the water along the bottom contour of the structure is also shown with arrows.

    Here, d1, d2 and d3 are the depth of upstream, intermediate and downstream sheet piles respectively. b is the total length of the weir floor.

    Bligh's Creep Theory for Seepage Flow:

    • According to Bligh's Theory, seeping water creeps along the bottom contour of the structure. The length of the path thus traversed by water is called the length of the creep.
    • According to Bligh’s Theory from the diagram, the creep length is given by,

     \({{\rm{L}}_{{\rm{bligh}}}} = \left( {{{\rm{d}}_1} + {{\rm{d}}_1}} \right) + {{\rm{L}}_1} + \left( {{{\rm{d}}_2} + {{\rm{d}}_2}} \right) + {{\rm{L}}_2} + \left( {{{\rm{d}}_3} + {{\rm{d}}_3}} \right) = {\rm{b}} + 2\left( {{{\rm{d}}_1} + {{\rm{d}}_2} + {{\rm{d}}_3}} \right)\)

    Lane's Weighted Creep Theory:

    • Bligh, in his theory, had calculated the length of the creep, by simply adding the horizontal creep length and the vertical creep length, thereby making no distinction between the two creeps.
    • However, Lane suggested a weighting factor of 1/3 for the horizontal creep, as against 1.0 for the vertical creep.
    • According to Lane’s Theory from the diagram, the creep length is given by,

    \({{\rm{L}}_{{\rm{lane}}}} = \left( {{{\rm{d}}_1} + {{\rm{d}}_1}} \right) + \frac{1}{3}{{\rm{L}}_1} + \left( {{{\rm{d}}_2} + {{\rm{d}}_2}} \right) + \frac{1}{3}{{\rm{L}}_2} + \left( {{{\rm{d}}_3} + {{\rm{d}}_3}} \right) = \frac{1}{3}{\rm{b}} + 2\left( {{{\rm{d}}_1} + {{\rm{d}}_2} + {{\rm{d}}_3}} \right)\)

    Calculation:

    Given, upstream sheet pile depth, d1 = 10 m; intermediate sheet pile depth, d2 = 8 m; downstream sheet pile depth, d3 = 12 m;

    Total width, b = 8+12 = 20 m.

    ∴ Bligh’s creep length, \({{\rm{L}}_{{\rm{bligh}}}} = 20 + 2\left( {10 + 8 + 12} \right) = 80m\) 

    ∴ Lane’s creep length, \({{\rm{L}}_{{\rm{lane}}}} = \frac{1}{3} \times 20 + 2\left( {10 + 8 + 12} \right) = 66.6667m\) 

    ∴ The ratio of creep length = \(\frac{{{{\rm{L}}_{{\rm{lane}}}}}}{{{{\rm{L}}_{{\rm{bligh}}}}}} = \frac{{66.667}}{{80}} = 0.833\)

  • Question 9/10
    1 / -0

    Flow depth and bed slope of a wide rectangular alluvial canal are 3.8 m and 1.5 × 10-4 respectively. The velocity is found out to be 1.9 m/s. The mean size of the sediment particles is 1.6 mm. What is the ratio (corrected up to three decimal point) of rugosity coefficient given by Strickler’s formula to the actual rugosity coefficient?
    Solutions

    Concept:

    Manning’s Formula: This formula is widely used for finding out mean stream/ canal velocity which is dependent upon hydraulic mean radius of flow and bed slope of the stream/canal. According to Manning’s formula,

    \({\rm{V}} = \frac{1}{{\rm{n}}}{{\rm{R}}^{\frac{2}{3}}}{{\rm{S}}^{\frac{1}{2}}}\) where V is the mean velocity of the canal, R is the hydraulic mean radius of flow, S is the bed slope of the canal and n is the actual rugosity coefficient.

    Strickler’s Formula: This formula gives theoretical rugosity coefficient depending upon the grain size of the sediment particles and is given by,

    \(n' = \frac{1}{{24}}d_{mean}^{\frac{1}{6}}\) where n’ is the rugosity coefficient pertaining to Strickler’s formula and dmean is the mean size of the bed sediments in meter.

    Calculation:

    Given, Flow depth, D = 3.8 m; bed slope, S = 1.5 × 10-4 and velocity, V = 1.9 m/sec.

    Now, for a wide rectangular channel, it is known that hydraulic mean radius (R) is approximately equal to the depth of flow (D).

    ∴ Hydraulic mean radius, \({\rm{R}} \approx {\rm{D}} = 3.8{\rm{\;m}}\) .

    So, from Manning’s formula, \(1.9 = \frac{1}{{\rm{n}}} \times {3.8^{\frac{2}{3}}} \times {\left( {1.5 \times {{10}^{ - 4}}} \right)^{\frac{1}{2}}}\;\;\;\;\therefore n = 0.01569\)  

    Also, it is given, mean sediment particle size, dmean = 1.6 mm = 1.6 × 10-3 m

    So, from Strickler’s formula, \(n' = \frac{1}{{24}} \times {\left( {1.6 \times {{10}^{ - 3}}} \right)^{\frac{1}{6}}} = 0.01425\)

    ∴ The ratio of rugosity coefficient given by Strickler’s formula to the actual rugosity coefficient = \(\frac{{n'}}{n}\)

    ∴ \(\frac{{n'}}{n} = \frac{{0.01425}}{{0.01569}} = 0.908\)

    Mistake Point:

    Strickler’s formula is empirical in nature. So, the value for mean particle size must be substituted in m, otherwise, the constant given in the formula is invalid. So, if in the formula the mean particle size is substituted in mm, the answer will be wrong.

  • Question 10/10
    1 / -0

    A permeable weir floor has a length of 14 m with vertical downstream sheet pile depth of 8m. As per Khosla’s theory to maintain a safe exit gradient of 1 in 7 at downstream what should be the maximum water depth (in m corrected up to two decimal places) at upstream? 
    Solutions

    Concept:

    Exit Gradient: The exit gradient is the hydraulic gradient at the downstream end of the flow line where percolating water leaves the soil mass and emerges into the free water at the downstream.

    Khosla’s Exit Gradient: As per Khosla’s theory the exit gradient GE is dependent upon floor length and downstream vertical cut-off length and is given by,

    \({{\rm{G}}_{\rm{E}}} = \frac{{\rm{H}}}{{\rm{d}}}\frac{1}{{{\rm{\pi }}\sqrt {\rm{\lambda }} }}{\rm{\;\;\;where\;\lambda }} = \frac{{1 + \sqrt {1 + {{\rm{\alpha }}^2}} }}{2}{\rm{\;and\;\alpha }} = \frac{{\rm{b}}}{{\rm{d}}}{\rm{\;}}\)

    Where, GE is the Khosla’s Exit Gradient; H is the upstream water depth; d is the depth of vertical cut-off (or depth of downstream sheet pile) at downstream; b is the weir floor-length between upstream and downstream. 

    Calculation:

    Given, safe exit gradient, GE = 1 in 7 = 0.1428.

    downstream sheet pile depth = 8 m ⇒ d = 8m.

    Total weir floor-length, b = 14 m.

    \(\therefore {\rm{\;\alpha \;}} = \frac{{\rm{b}}}{{\rm{d}}}{\rm{\;}} = \frac{{14}}{8}{\rm{\;}} = 1.75{\rm{\;\;\;\;}}\therefore {\rm{\;\lambda }} = {\rm{\;}}\frac{{1 + \sqrt {1 + {{1.75}^2}} }}{2} = 1.5077\)

    \(\therefore 0.1428 = \frac{{\rm{H}}}{8} \times \frac{1}{{{\rm{\pi }}\; \times \;\sqrt {1.5077} }}{\rm{\;}} ⇒ {\rm{H}} = 4.41{\rm{\;m}}\)

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