Fluid Mechanics and Hydraulic Machinery Miscellaneous


Fluid Mechanics and Hydraulic Machinery Miscellaneous

Fluid Mechanics and Hydraulic Machinery

  1. Match the following pairs:
    EquationPhysical Interpretation
    P. ∇ × v̄ = 0I. Incompressible continuity equation
    Q. ∇ . v̄ = 0II. Steady flow
    R. ∆v̄/∆t = 0III. Irrotational flow
    S. δv̄/δt = 0IV. Zero acceleration of fluid particle









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    P-III, Q-I, R-IV, S-II

    Correct Option: C

    P-III, Q-I, R-IV, S-II


  1. For an in-compressible flow field, V, which one of the following conditions must be satisfied?









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    In compressible flow condition
    ∇.v̄ = 0

    Correct Option: A

    In compressible flow condition
    ∇.v̄ = 0



  1. For the continuity equation given by∆ . V̄ = 0 to be valid, where V̄ is the velocity vector, which one of the following is a necessary condition?









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    The basic equation of continuity for fluid flow is given by

    ρu
    ρv
    ρw
    ρ
    uyzt

    Now if ρ remains constant, then only we can write
    ∇.v̄ = 0
    i.e.
    δu
    +
    δv
    +
    δw
    = 0
    δxδyδz

    hence incompressible flow

    Correct Option: D

    The basic equation of continuity for fluid flow is given by

    ρu
    ρv
    ρw
    ρ
    uyzt

    Now if ρ remains constant, then only we can write
    ∇.v̄ = 0
    i.e.
    δu
    +
    δv
    +
    δw
    = 0
    δxδyδz

    hence incompressible flow


  1. Which combination of the following statements about steady incompressible forced vortex flow is correct?
    P: Shear stress is zero at all points in the flow.
    Q: Vorticity is zero at all points in the flow.
    R: Velocity is directly proportional to the radius from the centre of the vortex.
    S: Total mechanical energy per unit mass is constant in the entire flow field.
    Select the correct answer using the codes given below:









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    Clearly zero shear stress and vortex.

    Correct Option: B

    Clearly zero shear stress and vortex.



  1. The 2-D flow with, velocity
    V̄ = (x + 2y + 2) î + (4 – y) ĵ is









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    v̄ = (x + 2y +2)î + (4 - y)ĵ
    u =x + 2y + 2, v = 4 – y

    δv
    = 1,
    δv
    = - 1
    δxδy

    δv
    + 1,
    δv
    = 0
    δxδy

    hence in compressible.
    Again, ω =
    1
    δv
    -
    δv
    2δxδy

    =
    1
    (0 - 2) = - 1.
    2

    hence not irrotational.

    Correct Option: D

    v̄ = (x + 2y +2)î + (4 - y)ĵ
    u =x + 2y + 2, v = 4 – y

    δv
    = 1,
    δv
    = - 1
    δxδy

    δv
    + 1,
    δv
    = 0
    δxδy

    hence in compressible.
    Again, ω =
    1
    δv
    -
    δv
    2δxδy

    =
    1
    (0 - 2) = - 1.
    2

    hence not irrotational.