Fluid Mechanics and Hydraulic Machinery Miscellaneous


Fluid Mechanics and Hydraulic Machinery Miscellaneous

Fluid Mechanics and Hydraulic Machinery

  1. Two pipes of uniform section but different diameters carry water at the same flow rate. Water properties are the same in the two pipes. The Reynolds number, based on the pipe diameter









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    Reynolds number, Re =
    Vd
    υ

    Discharge, Q = AV =
    π
    d2V
    4

    From above, =
    4Q
    πd2

    ∴  Reynolds No. =
    4Q
    ×
    d
    πd2 2

    Reynolds No. =
    4Q
    ....(1)
    πd

    From (1), Re
    1
    2

    Reynolds number is large in the Narrow pipe.

    Correct Option: B

    Reynolds number, Re =
    Vd
    υ

    Discharge, Q = AV =
    π
    d2V
    4

    From above, =
    4Q
    πd2

    ∴  Reynolds No. =
    4Q
    ×
    d
    πd2 2

    Reynolds No. =
    4Q
    ....(1)
    πd

    From (1), Re
    1
    2

    Reynolds number is large in the Narrow pipe.


Direction: Consider a linear programming problem with two variable and two constraints. The objective function is maximize x1 + x2. The corner points of the feasible region are (0, 0), (0,2) (2, 0) and (4/3, 4/3)

  1. Oil is being pumped through a straight pipe, the pipe length, diameter and volumetric flow rate are all doubled in a new arrangement. The pipe friction factor, however, remains constant. The ratio of pipe frictional losses in the new arrangement to that in the original configuration would be









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    For original configuration
    l1 = l
    d1 = d
    Q1 = Q
    For new configuration
    l2 = 21
    d2 = 2d
    Q2 = 2Q
    Head loss due to friction,

    hf =
    flV²
    2gb

    where f = friction factor
    l = Length of pipe
    v = Average velocity
    d = Diameter pipe
    Discharge: Q = AV =
    π
    a²V
    4

    or   V =
    4Q
    πa²

    hf =
    fl
    ×
    4Q
    ² =
    8flQ²
    2gdπa²πga5

    For new Pipe,
    hf2 =
    8fl1Q2²
    =
    8f × 2f × 4Q²
    =
    2fQ²
    πgda5π × 25 × a5πga5

    h1
    =
    2flQ²
    ×
    πga5
    =
    1
    h2²πga52flQ²4

    Correct Option: A

    For original configuration
    l1 = l
    d1 = d
    Q1 = Q
    For new configuration
    l2 = 21
    d2 = 2d
    Q2 = 2Q
    Head loss due to friction,

    hf =
    flV²
    2gb

    where f = friction factor
    l = Length of pipe
    v = Average velocity
    d = Diameter pipe
    Discharge: Q = AV =
    π
    a²V
    4

    or   V =
    4Q
    πa²

    hf =
    fl
    ×
    4Q
    ² =
    8flQ²
    2gdπa²πga5

    For new Pipe,
    hf2 =
    8fl1Q2²
    =
    8f × 2f × 4Q²
    =
    2fQ²
    πgda5π × 25 × a5πga5

    h1
    =
    2flQ²
    ×
    πga5
    =
    1
    h2²πga52flQ²4



  1. The ratio  
    PA − PB
    .
    (1/2)ρu0²

    (where pA and pB are the pressures at section A and B, respectively, and ρ is the density of the fluid), is









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    Applying Bernoulli’s equation at sections A and B, we get

    PA
    +
    u0²
    + ZA =
    PA
    +
    Vm²
    + ZB
    ρg2gρg2g

    PA
    +
    u0²
    =
    PB
    +
    Vm²
    ρg2gρg2g
    or  
    PA
    +
    PB
    =
    Vm²
    +
    u0²
    Pgρg2g2g

    Now, the ratio
    PA − PB/(1/2)ρ0² =
    ρVm²
      −
    ρu0²
    22
    1
    ρu0²
    2

    =
    Vm² − u0²
    =
    Vm
    ² − 1
    u0²u0

    Substituting the Value of
    Vm
    =
    1
    u01 − (δ/H)

    we get
    PA − PB
    =
    1
    − 1
    (1/2)ρ0² [1 − (δ/H)]²

    Correct Option: B

    Applying Bernoulli’s equation at sections A and B, we get

    PA
    +
    u0²
    + ZA =
    PA
    +
    Vm²
    + ZB
    ρg2gρg2g

    PA
    +
    u0²
    =
    PB
    +
    Vm²
    ρg2gρg2g
    or  
    PA
    +
    PB
    =
    Vm²
    +
    u0²
    Pgρg2g2g

    Now, the ratio
    PA − PB/(1/2)ρ0² =
    ρVm²
      −
    ρu0²
    22
    1
    ρu0²
    2

    =
    Vm² − u0²
    =
    Vm
    ² − 1
    u0²u0

    Substituting the Value of
    Vm
    =
    1
    u01 − (δ/H)

    we get
    PA − PB
    =
    1
    − 1
    (1/2)ρ0² [1 − (δ/H)]²


Direction: Consider a steady incompressible flow through a channel as shown below:

The velocity profile is uniform with a value of u0 at the inlet section A. The velocity profile at section B downstream is

  1. The ratio Vm/u0 is









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    Applying continuity equation at sections A and B, mass flow rate at inlet section A = mass rate at section B.

    ∴  u =
    VmY
    δ

    u0H = Vm(H − 2δ) + 2
    Vm
    δ
    δ20

    u0H = Vm(H − 2δ) + 2
    Vm
    ×
    δ²
    δ2

    u0H = Vm(H − 2δ) + Vmδ
    u0H = VmH − 2Vmδ + Vmδ
    u0H = VmH − Vmδ
    u0H = Vm(H − δ)
    or  
    Vm
    =
    H
    =
    1
    u0H − δ1 − (δ/H)

    Correct Option: C

    Applying continuity equation at sections A and B, mass flow rate at inlet section A = mass rate at section B.

    ∴  u =
    VmY
    δ

    u0H = Vm(H − 2δ) + 2
    Vm
    δ
    δ20

    u0H = Vm(H − 2δ) + 2
    Vm
    ×
    δ²
    δ2

    u0H = Vm(H − 2δ) + Vmδ
    u0H = VmH − 2Vmδ + Vmδ
    u0H = VmH − Vmδ
    u0H = Vm(H − δ)
    or  
    Vm
    =
    H
    =
    1
    u0H − δ1 − (δ/H)



  1. Consider steady laminar incompressible axisymmetric fully developed viscous flow through a straight circular pipe of constant crosssectional area at a Reynolds number of 5. The ratio of inertia force to viscous force on a fluid particle is









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    Reynold’s number =
    Inertia force
    = 5
    Viscous force

    Correct Option: A

    Reynold’s number =
    Inertia force
    = 5
    Viscous force