Fluid Mechanics and Hydraulic Machinery Miscellaneous


Fluid Mechanics and Hydraulic Machinery Miscellaneous

Fluid Mechanics and Hydraulic Machinery

Direction: The gap between a moving circular plate and a stationary surface is being continuously reduced, as the circular plate comes down at a uniform speed V towards the stationary bottom surface, as shown in the figure. In the process, the fluid contained between the two plates flows out radially. The fluid is assumed to be incompressible and inviscid.

  1. The radial velocity Vr at any radius r, when the gap width is h, is









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    V × πr² = Vr × 2πrh

    Vr =
    Vr
    2h

    Correct Option: A

    V × πr² = Vr × 2πrh

    Vr =
    Vr
    2h


  1. In a steady flow through a nozzle, the flow velocity on the nozzle axis is given by v = u0 (1 + 3x/L)i, where x is the distance along the axis of the nozzle from its inlet plane and L is the length of the nozzle. The time required for a fluid particle on the axis to travel from the inlet to the exit plane of the nozzle is









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    v = u01 +
    3x
    L

    dx
    = u01 +
    3x
    dtL

    dx
    = dt
    u01 +
    3x
    L


    ⇒ T =
    L
    ln(4) =
    L
    ln(4)
    3u03u0

    Correct Option: B


    v = u01 +
    3x
    L

    dx
    = u01 +
    3x
    dtL

    dx
    = dt
    u01 +
    3x
    L


    ⇒ T =
    L
    ln(4) =
    L
    ln(4)
    3u03u0



  1. A leaf is caught in a whirlpool. At a given instant, the leaf is at a distance of 120 m from the centre of the whirlpool. The whirlpool can be described by the following velocity distribution:
    Vr = -
    60 × 103
    m/s and Vθ =
    300 × 103
    m/s
    2πr2πr

    where r (in meters) is the distance from the centre of the whirlpool. What will be the distance of the leaf from the centre when it has moved through half a revolution?









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    Radial distance = 120 m

    Vr = -
    60 × 103
    m /s
    2πr

    & Vθ =
    300 × 103
    m /s
    2πr

    & θ = π
    Vr
    =
    -1
    Vθ5

    we know that
    Vr =
    dr
    ...(i)
    dt

    & Vθ = rw = r
    dt

    -5Vr = r
    dt

    Vr =
    -r
    ×
    ...(ii)
    5dt

    By equating (i) & (ii), we get
    dr
    =
    -r
    .
    dt5dt

    Integrating both sides, we get

    ln
    r
    =
    1205

    By solving above, we get
    r = 64 m

    Correct Option: B

    Radial distance = 120 m

    Vr = -
    60 × 103
    m /s
    2πr

    & Vθ =
    300 × 103
    m /s
    2πr

    & θ = π
    Vr
    =
    -1
    Vθ5

    we know that
    Vr =
    dr
    ...(i)
    dt

    & Vθ = rw = r
    dt

    -5Vr = r
    dt

    Vr =
    -r
    ×
    ...(ii)
    5dt

    By equating (i) & (ii), we get
    dr
    =
    -r
    .
    dt5dt

    Integrating both sides, we get

    ln
    r
    =
    1205

    By solving above, we get
    r = 64 m


  1. The velocity components in the x and y directions of a two dimensional potential flow are u and v, respectively.
    Then ,
    ∂u
    is equal to
    ∂x










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    ∂²φ
    +
    ∂²φ
    = 0 ...(i) (Laplace Equation should be satisfied)
    ∂x²∂y²

    ∴ u = -
    ∂φ
      and   v = -
    ∂φ
    ∂x∂y

    ∂u
    = -
    ∂²φ
     and  
    ∂v
    = -
    ∂²φ
    ∂x∂x²∂y∂y²

    Putting values in equation (i), we get
    -∂u
    +
    -∂v
    = 0
    ∂x∂y

    ∂u
    = -
    ∂v
    ∂x∂y

    Correct Option: D

    ∂²φ
    +
    ∂²φ
    = 0 ...(i) (Laplace Equation should be satisfied)
    ∂x²∂y²

    ∴ u = -
    ∂φ
      and   v = -
    ∂φ
    ∂x∂y

    ∂u
    = -
    ∂²φ
     and  
    ∂v
    = -
    ∂²φ
    ∂x∂x²∂y∂y²

    Putting values in equation (i), we get
    -∂u
    +
    -∂v
    = 0
    ∂x∂y

    ∂u
    = -
    ∂v
    ∂x∂y



  1. The velocity components in the x and y directions are given by u = λxy3 – x2y, v = xy2 – ( 3 / 4 )y4. The value of λ for a possible flow field involving an incompressible fluid is









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    u = λxy3 – x2y ;
    ∂u
    = λy3 – 2xy
    ∂x

    v = xy2
    3
    y4 ;
    ∂v
    = 2xy - 3y3
    4∂y

    For 2D incompressible flow,
    ∂u
    +
    ∂v
    = 0
    ∂x∂y

    ⇒ xy3 – 2xy + 2xy - 3y3 = 0
    λ = 3

    Correct Option: D

    u = λxy3 – x2y ;
    ∂u
    = λy3 – 2xy
    ∂x

    v = xy2
    3
    y4 ;
    ∂v
    = 2xy - 3y3
    4∂y

    For 2D incompressible flow,
    ∂u
    +
    ∂v
    = 0
    ∂x∂y

    ⇒ xy3 – 2xy + 2xy - 3y3 = 0
    λ = 3