Fluid Mechanics and Hydraulic Machinery Miscellaneous


Fluid Mechanics and Hydraulic Machinery Miscellaneous

Fluid Mechanics and Hydraulic Machinery

  1. Consider steady, incompressible and irrotational flow through a reducer in a horizontal pipe where the diameter is reduced from 20 cm to 10 cm. The pressure in the 20 cm pipe just upstream of the reducer is 150 kPa. The fluid has a vapour pressure of 50 kPa and a specific weight of 5 kN / m3 . Neglecting frictional effects, the maximum discharge (in m3 / s) that can pass through the reducer without causing cavitation is









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    Considering potential head difference = 0,
    i.e z1 = z2
    Apply Bernoulli’s theorem

    p
    +
    v2
    = C
    ρg2g


    p1
    +
    v1²
    =
    p2
    +
    v2²
    w12gw22g

    But w1 = w2 = w = 5 (incompressible flow)
    150
    +
    v1²
    =
    50
    +
    v2²
    52g52g

    or
    v2² - v1²
    =
    150 - 50
    2g5

    or
    v2² - v1²
    = 20 m ....(1)
    2g

    Also, discharge, Q =
    π
    d1²v1 =
    π
    d2²v2
    44

    or
    v1
    =
    d2
    ² =
    10
    ²
    v2d120

    or v1 =
    v2
    ......(2)
    4

    From equations (1) and (2)
    v2² -
    v2²
    16 = 20
    2g

    or
    15v2²
    = 20
    32g

    or v2 = √(20 × 32g) / 15 = 20.45 m /s
    ∴ Discharge , Q =
    π
    d22v2
    4

    =
    π
    (0.1)2 × 20.45 = 0.16 m3 /sec
    4

    Correct Option: B

    Considering potential head difference = 0,
    i.e z1 = z2
    Apply Bernoulli’s theorem

    p
    +
    v2
    = C
    ρg2g


    p1
    +
    v1²
    =
    p2
    +
    v2²
    w12gw22g

    But w1 = w2 = w = 5 (incompressible flow)
    150
    +
    v1²
    =
    50
    +
    v2²
    52g52g

    or
    v2² - v1²
    =
    150 - 50
    2g5

    or
    v2² - v1²
    = 20 m ....(1)
    2g

    Also, discharge, Q =
    π
    d1²v1 =
    π
    d2²v2
    44

    or
    v1
    =
    d2
    ² =
    10
    ²
    v2d120

    or v1 =
    v2
    ......(2)
    4

    From equations (1) and (2)
    v2² -
    v2²
    16 = 20
    2g

    or
    15v2²
    = 20
    32g

    or v2 = √(20 × 32g) / 15 = 20.45 m /s
    ∴ Discharge , Q =
    π
    d22v2
    4

    =
    π
    (0.1)2 × 20.45 = 0.16 m3 /sec
    4


  1. The following data about the flow of liquid was observed in a continuous Chemical process plant:

    Mean flow rate of the liquid is









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    ∴ Mean flow rate =
    ∑fx
    =
    652.8
    = 8.16
    ∑f80

    Correct Option: C


    ∴ Mean flow rate =
    ∑fx
    =
    652.8
    = 8.16
    ∑f80


  1. Navier Stoke's equation represents the conservation of









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    Momentum

    Correct Option: D

    Momentum


  1. Bernoulli's equation can be applied between any two points on a stream line for a rotational flow field. State:









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    TRUE

    Correct Option: A

    TRUE


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 component of the fluid acceleration at r = R is









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    Radial acceleration

    ar = Vr ×
    ∂Vr
    +
    ∂Vr
    ∂r∂t

    ar =
    V.r
    ×
    V.r
    +
    V.r
    2h∂r2h∂t2h

    -∂h
    = V
    ∂t

    ∴ ar =
    V.r
    ×
    V.r
    +
    V.r
    ×
    -1
    ∂h
    2h2h2h2∂t

    ∴ ar =
    V2r
    +
    2V2r
    4h24h2

    ar =
    3V2r
    4h2

    Correct Option: A

    Radial acceleration

    ar = Vr ×
    ∂Vr
    +
    ∂Vr
    ∂r∂t

    ar =
    V.r
    ×
    V.r
    +
    V.r
    2h∂r2h∂t2h

    -∂h
    = V
    ∂t

    ∴ ar =
    V.r
    ×
    V.r
    +
    V.r
    ×
    -1
    ∂h
    2h2h2h2∂t

    ∴ ar =
    V2r
    +
    2V2r
    4h24h2

    ar =
    3V2r
    4h2