Fluid Mechanics and Hydraulic Machinery Miscellaneous


Fluid Mechanics and Hydraulic Machinery Miscellaneous

Fluid Mechanics and Hydraulic Machinery

  1. Water flows through a 10 mm diameter and 250 m long smooth pipe at an average velocity of 0.1 m/s. The density and the viscosity of water are 997 kg/m3 and 855 × 10–6 Ns/m2, respectively. Assuming fully- developed flow, the pressure drop (in Pa) in the pipe is ____?









  1. View Hint View Answer Discuss in Forum

    Re =
    ρVD
    μ

    Re =
    997 × 0.1 × 10 × 10-3
    = 1166.081 ( laminar flow )
    855 × 10-6

    f =
    64
    =
    64
    = 0.054
    Re1166.081

    ∆p =
    f l ρV2
    2D

    =
    0.054 × 250 × 997 × (0.1)2
    = 6840 pascal
    2 × 10 × 10-3

    Correct Option: A

    Re =
    ρVD
    μ

    Re =
    997 × 0.1 × 10 × 10-3
    = 1166.081 ( laminar flow )
    855 × 10-6

    f =
    64
    =
    64
    = 0.054
    Re1166.081

    ∆p =
    f l ρV2
    2D

    =
    0.054 × 250 × 997 × (0.1)2
    = 6840 pascal
    2 × 10 × 10-3


  1. Oil flows through a 200 mm diameter horizontal cast iron pipe (friction factor, f = 0.0225) of length 500 m, The volumetric flow rate is 0.2 m3/s, The head loss (in m) due to friction is (assume g = 9.81 m/s2)









  1. View Hint View Answer Discuss in Forum

    H =
    flv2
    2gD

    = 0.225 × 500 ×
    0.2
    ²
    { ∏ × (0.2² /4) }
    2 × 9.81 × 0.4

    = 116.18 m

    Correct Option: A

    H =
    flv2
    2gD

    = 0.225 × 500 ×
    0.2
    ²
    { ∏ × (0.2² /4) }
    2 × 9.81 × 0.4

    = 116.18 m


  1. A smooth pipe of diameter 200 mm carries water. The pressure in the pipe at section S1 (elevation: 10 m) ie 50 kPa, At section S2 (elevation: 12 m) the pressure is 20 kPa and velocity is 2 m/s. Density of water is 1000 kg/m3 and acceleration due to gravity is 9.8 ms–2, Which of the following is TRUE









  1. View Hint View Answer Discuss in Forum

    At section S1
    P1 = 50 kPa
    z1 = 10 m
    D = 200 mm,
    l = 1000 kg/m3
    V = 2 m/s
    g = 9.8 m/s2
    At section S2
    P2 = 20 kPa
    z2 = 12 m
    Applying Bernoulli’s equation at two sections, we have

    P1
    + z1 =
    P2
    + z2 + h2
    ρgρg

    where h2 = Loss of head from s1 to s2
    50 × 103
    + 10 =
    20 × 103
    + 12 + h2
    103 × 9.81103 × 9.81

    ⇒ h2 = 1.06

    Correct Option: C

    At section S1
    P1 = 50 kPa
    z1 = 10 m
    D = 200 mm,
    l = 1000 kg/m3
    V = 2 m/s
    g = 9.8 m/s2
    At section S2
    P2 = 20 kPa
    z2 = 12 m
    Applying Bernoulli’s equation at two sections, we have

    P1
    + z1 =
    P2
    + z2 + h2
    ρgρg

    where h2 = Loss of head from s1 to s2
    50 × 103
    + 10 =
    20 × 103
    + 12 + h2
    103 × 9.81103 × 9.81

    ⇒ h2 = 1.06


  1. A cubic block of side L and mass M is dragged over an oil film across table by a string connects to a hanging block of mass m as shown in figure. The Newtonian oil film of thickness h has dynamic viscosity μ and the flow condition is laminar. The acceleration due to gravity is g. The steady state velocity V of block is










  1. View Hint View Answer Discuss in Forum

    τ = μ
    du
    dy

    du = V – O – V
    dy = h
    τ = u
    V
    ...(1)
    h

    Also , τ =
    F
    A

    where F = mg , A = L2
    ∴ τ =
    mg
    ......(2)
    L2

    By (1) (2), we get
    V =
    mgh
    uL2

    Correct Option: C

    τ = μ
    du
    dy

    du = V – O – V
    dy = h
    τ = u
    V
    ...(1)
    h

    Also , τ =
    F
    A

    where F = mg , A = L2
    ∴ τ =
    mg
    ......(2)
    L2

    By (1) (2), we get
    V =
    mgh
    uL2


Direction: A syringe with a frictionless plunger contains water and has at its end a 100 mm long needle of 1 mm diameter. The internal diameter of the syringe is 10 mm. Water density is 1000 kg/m3. The plunger is pushed in at 10 mm/s and the water comes out as a jet.

  1. Neglect losses in the cylinder and assume fully developed laminar viscous flow throughout the needle: the Darcy friction factor is 64/Re. where Re is the Reynolds number. Given that the viscosity of water is 1.0 × 10-3 kg/ms, the force F in newtons required on the plunger is









  1. View Hint View Answer Discuss in Forum

    Given, v = velocity of water = 10 × 10–3 kg/sm

    Now , Re =
    ρv2d
    v

    ... since = v2 = 1
    Re = 1000
    Now, Darcy’s friction factor,
    4f =
    64
    =
    64
    = 0.064
    Re1000

    Head loss in needle = ht =
    flv2
    2gD

    =
    0.064 × 0.1 × (1)2
    = 0.3265
    2 × 9.8 × 0.001

    Applying Bernoulli’s equation at points 1 and 2, we have
    P1
    +
    v1²
    + z1 =
    P2
    +
    v2²
    + z2 + h1
    ρg2gρg2g

    Since z1 = z2 and P2 = 0
    P1
    =
    v2² - v1²
    + hl
    ρg2g

    P1 =
    ρ
    (v2² - v1²) + ρghl
    2

    P1 =
    1000
    [ (1)² - (0.01)²] + 1000 × 9.8 × 0.3265
    2

    = 499.95 + 3199.7 = 3699.65
    Now force required on plunger = P1 × A1
    = 3699.65 ×
    π
    × (0.01)2 = 0.3 N
    4

    Correct Option: C

    Given, v = velocity of water = 10 × 10–3 kg/sm

    Now , Re =
    ρv2d
    v

    ... since = v2 = 1
    Re = 1000
    Now, Darcy’s friction factor,
    4f =
    64
    =
    64
    = 0.064
    Re1000

    Head loss in needle = ht =
    flv2
    2gD

    =
    0.064 × 0.1 × (1)2
    = 0.3265
    2 × 9.8 × 0.001

    Applying Bernoulli’s equation at points 1 and 2, we have
    P1
    +
    v1²
    + z1 =
    P2
    +
    v2²
    + z2 + h1
    ρg2gρg2g

    Since z1 = z2 and P2 = 0
    P1
    =
    v2² - v1²
    + hl
    ρg2g

    P1 =
    ρ
    (v2² - v1²) + ρghl
    2

    P1 =
    1000
    [ (1)² - (0.01)²] + 1000 × 9.8 × 0.3265
    2

    = 499.95 + 3199.7 = 3699.65
    Now force required on plunger = P1 × A1
    = 3699.65 ×
    π
    × (0.01)2 = 0.3 N
    4