Fluid mechanics and hydraulics miscellaneous


Fluid mechanics and hydraulics miscellaneous

Fluid Mechanics and Hydraulics

  1. Critical depth at a section of a rectangular channel is 1.5 m. The specific energy at that section is









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    yc = 1.5 m

    Ec =
    3
    yc
    2

    =
    3
    × 1.5 = 2.25 m
    2

    Correct Option: D

    yc = 1.5 m

    Ec =
    3
    yc
    2

    =
    3
    × 1.5 = 2.25 m
    2


  1. A partially open sluice gate discharges water into a rectangular channel. The tail water depth
    in the channel is 3 m and Froude number is
    1
    . If a free hydraulic jump is to be
    2 √2
    formed at a downstream of the sluice gate after the vena contracta of the jet coming out from the sluice gae, the sluice gate opening should be (coefficient of contraction Cc = 0.9)









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    y1 =
    y2
    [-1 + √1 + 8Fr22]
    2


    = 0.62 m
    Coeff. of contraction, Cc
    Cc =
    area of vena contract
    area of opening

    ⇒ 0.9 =
    0.62 × b
    y × b

    ∴ y =
    0.62
    = 0.69 m
    0.9

    Correct Option: C

    y1 =
    y2
    [-1 + √1 + 8Fr22]
    2


    = 0.62 m
    Coeff. of contraction, Cc
    Cc =
    area of vena contract
    area of opening

    ⇒ 0.9 =
    0.62 × b
    y × b

    ∴ y =
    0.62
    = 0.69 m
    0.9



  1. A stream function is given by Ψ = 2x2y + (x + 1)y2
    The flow rate across a line joining points A(3, 0) and B(0,2) is









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    Ψ = 2x2y + (x + 1)y2
    At A (3, 0)
    Ψ(3, 0) = 2(3)2 - (0) + (3 + 1)(0)2
    = 0
    At B(2, 0)
    Ψ(2, 0) = 2(0)2(2) + (0 + 1)(2)2
    = 0 + 4 = 4
    Flow rate across line joining A and B = |ΨA - ΨB|
    = |0 - 4| units

    Correct Option: C

    Ψ = 2x2y + (x + 1)y2
    At A (3, 0)
    Ψ(3, 0) = 2(3)2 - (0) + (3 + 1)(0)2
    = 0
    At B(2, 0)
    Ψ(2, 0) = 2(0)2(2) + (0 + 1)(2)2
    = 0 + 4 = 4
    Flow rate across line joining A and B = |ΨA - ΨB|
    = |0 - 4| units


  1. Cross-section of an object (having same section normal to the paper) submerged into a fluid consists of a square of sides 2 m and triangle as shown in the figure. The object is hinged at point P that is one meter below the fluid free surface. If the object is to be kept in the position as shown in the figure, the value ‘x’ should be










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    Take moment of vertical component of hydrastatic force on two section

    V10 +
    2
    = V2 ×
    x
    23

    ρ gv1 × 1 = ρgv2 ×
    x
    3

    ∴ v1 = = v2
    x
    3

    (2 × z × w) =
    1
    x × 2(w) ×
    x
    23

    ⇒ x = √12 = 2√3

    Correct Option: A

    Take moment of vertical component of hydrastatic force on two section

    V10 +
    2
    = V2 ×
    x
    23

    ρ gv1 × 1 = ρgv2 ×
    x
    3

    ∴ v1 = = v2
    x
    3

    (2 × z × w) =
    1
    x × 2(w) ×
    x
    23

    ⇒ x = √12 = 2√3



  1. The circulation ‘Γ’ around a circle of radius 2 units for the velocity field u = 2x + 3y and v = 2y is









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    V = ui + vj
    u = 2x + 3y , v = -2y

    ∂u
    = 3 ,
    ∂v
    = 0
    ∂y∂x

    circulation =
    ∂v
    -
    ∂u
    × Area
    ∂x∂y

    = (0 - 3) × πR2
    = (-3)π × 22
    = – 12π units

    Correct Option: B

    V = ui + vj
    u = 2x + 3y , v = -2y

    ∂u
    = 3 ,
    ∂v
    = 0
    ∂y∂x

    circulation =
    ∂v
    -
    ∂u
    × Area
    ∂x∂y

    = (0 - 3) × πR2
    = (-3)π × 22
    = – 12π units