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Irrigation Engineering Test 2

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Irrigation Engineering Test 2
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  • Question 1
    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?
    Solution

    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
    1 / -0

    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

    Solution

    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
    1 / -0
    Which one of the following is not the major factor influencing seepage from a canal?
    Solution

    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 4
    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?
    Solution

    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 5
    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?
    Solution

    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 6
    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? 
    Solution

    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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