Fluid Mechanics and Hydraulic Machinery Miscellaneous


Fluid Mechanics and Hydraulic Machinery Miscellaneous

Fluid Mechanics and Hydraulic Machinery

  1. A U-tube manometer with a small quantity of mercury is used to measure the static pressure difference between two locations A and B in a conical section through which an in compressible fluid flows. At a particular flow rate, the mercury column appears as shown in the figure. The density of mercury is 13600 kg/ m3 and g - 9.81 m/s2. Which of the following is correct?









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    PA – PB = rg. Dh
    = 13600 × 9.81 × .15
    = 20 kPa
    As pressure is decreasing from A to B, so flow direction is A to B.

    Correct Option: A


    PA – PB = rg. Dh
    = 13600 × 9.81 × .15
    = 20 kPa
    As pressure is decreasing from A to B, so flow direction is A to B.


  1. A venturimeter of 20 mm throat diameter is used to measure the velocity of water in a horizontal pipe of 40 mm diameter. If the pressure difference between the pipe and throat sections is found to be 30 kPa then, neglecting frictional losses, the flow velocity is









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    We know, A1 V1 = A2 V2

    ƥ V2 =
    D21
    V1 =
    16
    V1
    D224

    ∴ V2 = 4V1
    Applying Bernoulli's Equation
    p1
    +
    v21
    + z1 =
    p2
    +
    p22
    + z2
    ρg2gρg2g

    P1 - P2
    =
    V22 - V21
    eg2g

    ƥ
    15V21
    =
    30 × 103
    21000

    ƥ V21 = 4
    ƥ V1 = 2.0 m/s
    So velocity of flow is 2.0 m/sec

    Correct Option: D

    We know, A1 V1 = A2 V2

    ƥ V2 =
    D21
    V1 =
    16
    V1
    D224

    ∴ V2 = 4V1
    Applying Bernoulli's Equation
    p1
    +
    v21
    + z1 =
    p2
    +
    p22
    + z2
    ρg2gρg2g

    P1 - P2
    =
    V22 - V21
    eg2g

    ƥ
    15V21
    =
    30 × 103
    21000

    ƥ V21 = 4
    ƥ V1 = 2.0 m/s
    So velocity of flow is 2.0 m/sec


  1. Air flows through a venturi and into atmosphere. Air density is ρa; atmospheric pressure is ρa; throat diameter is Dt; exit diameter is D and exit velocity is U. The throat is connected to a cylinder containing a friction less piston attached to a spring. The spring constant is k. The bottom surface of the piston is exposed to atmosphere. Due to the flow, the piston moves by distance x. Assuming in compressible friction less flow, x is









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    Applying Bernoulli’s equation at points (1) and (2), we have

    p1
    +
    v21
    + z1 =
    p2
    +
    p22
    + z2
    ρg2gρg2g

    Since venturi is horizontal
    z1 = z2
    Now
    p1
    -
    p2
    =
    v22
    -
    p21
    ρgρg2g2g

    ⇒(p1 - p2) =
    ρg
    (v22 - v21) =
    ρ
    = (v22 - v21)
    2gg

    Since P2 = Pa = atmospheric pressure
    ∴ (P2 - Pa)
    ρ
    (v22 - v21) ........(i)
    2

    Applying continuity equation at points (i) and (ii), we have
    A1 v1 = A2 v2
    ⇒ v1 =
    A2
    v2 since V2 = U
    A1


    v1 =
    D
    2U
    Dt

    From equation (i),

    At point P
    Spring force = pressure force due air

    Correct Option: D


    Applying Bernoulli’s equation at points (1) and (2), we have

    p1
    +
    v21
    + z1 =
    p2
    +
    p22
    + z2
    ρg2gρg2g

    Since venturi is horizontal
    z1 = z2
    Now
    p1
    -
    p2
    =
    v22
    -
    p21
    ρgρg2g2g

    ⇒(p1 - p2) =
    ρg
    (v22 - v21) =
    ρ
    = (v22 - v21)
    2gg

    Since P2 = Pa = atmospheric pressure
    ∴ (P2 - Pa)
    ρ
    (v22 - v21) ........(i)
    2

    Applying continuity equation at points (i) and (ii), we have
    A1 v1 = A2 v2
    ⇒ v1 =
    A2
    v2 since V2 = U
    A1


    v1 =
    D
    2U
    Dt

    From equation (i),

    At point P
    Spring force = pressure force due air


  1. Water flows though a vertical contraction from a pipe of diameter d to another of diameter d/2 (see fig.) The flow velocity at the inlet to the contraction is 2 m/s and pressure 200 kN/m2. If the height of the contraction measures 2m,then pressure at the exit of the contraction will be very nearly









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    From continuity equation,
    A1 v1 = A2 v2

    or
    π
    × d2 × 2 =
    π
    d
    2× v2
    442

    or v2 = 8 m/s.
    Applying Bernoulli's theorem,
    p1
    +
    v21
    + z1 =
    p2
    +
    p22
    + z2
    w2gw2g

    or
    200 × 1000
    +
    (2)2
    + 0
    98102 × 9.81

    or
    p2
    +
    (8)2
    + 2
    w2 × 9.81

    or p2 =150.38 kN/m2

    Correct Option: C

    From continuity equation,
    A1 v1 = A2 v2

    or
    π
    × d2 × 2 =
    π
    d
    2× v2
    442

    or v2 = 8 m/s.
    Applying Bernoulli's theorem,
    p1
    +
    v21
    + z1 =
    p2
    +
    p22
    + z2
    w2gw2g

    or
    200 × 1000
    +
    (2)2
    + 0
    98102 × 9.81

    or
    p2
    +
    (8)2
    + 2
    w2 × 9.81

    or p2 =150.38 kN/m2


  1. In a venturimeter, the angle of the diverging section is more than that of converging section. State: (T/F)









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    False

    Correct Option: A

    False