Fluid Mechanics and Hydraulic Machinery Miscellaneous

Fluid Mechanics and Hydraulic Machinery Miscellaneous

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Fluid Mechanics and Hydraulic Machinery

Fluid Mechanics and Hydraulic Machinery Miscellaneous

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. Assuming ideal flow, the force F in newtons required on the plunger to push out the water is








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    ρwater =1000 kg/m3
    Velocity at points 1 = velocity of plunger = 10 mm/ s = 0.014 m/s

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

    P1
    		 +
    v1²
    		 + z1 =
    P2
    		 +
    v2²
    		 + z2
    ρg2gρg2g

    Since z1 = z2 and P2 = 0
    P1
    		 =
    v2²
    		 -
    v1²
    ρq2g2g

    P1 =
    ρ
    		(v2² - v1²) ...(i)
    2

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

    v2 =
    [ (π / 4) × (0.01)2 ]
    		v1
    [ (π / 4) × (0.001)2 ]

    = 100 v1 = 100 × 0.01 = 1 m/s
    Now from equation (i),
    P1 =
    1000
    		[ (1)² - (0.01)² ] = 499.95 N / m²
    2

    Force required on plunger = P1 × v1
    = 499.95 ×
    11
    		 × (0.01)2 = 0.04 N
    4

    Correct Option: B

    ρwater =1000 kg/m3
    Velocity at points 1 = velocity of plunger = 10 mm/ s = 0.014 m/s

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

    P1
    		 +
    v1²
    		 + z1 =
    P2
    		 +
    v2²
    		 + z2
    ρg2gρg2g

    Since z1 = z2 and P2 = 0
    P1
    		 =
    v2²
    		 -
    v1²
    ρq2g2g

    P1 =
    ρ
    		(v2² - v1²) ...(i)
    2

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

    v2 =
    [ (π / 4) × (0.01)2 ]
    		v1
    [ (π / 4) × (0.001)2 ]

    = 100 v1 = 100 × 0.01 = 1 m/s
    Now from equation (i),
    P1 =
    1000
    		[ (1)² - (0.01)² ] = 499.95 N / m²
    2

    Force required on plunger = P1 × v1
    = 499.95 ×
    11
    		 × (0.01)2 = 0.04 N
    4

  1. The discharge velocity at the pipe exit in figure is









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    By applying Bernoulli’s equation at (1) and (2), we get

    P1
    		 +
    V1²
    		 + Z1 =
    P2
    		 +
    V2²
    		 + Z2
    ρg2gρg2g

    P1 = P2 = Patm
    V1 = 0
    V2 = ?
    Z2 = H – h
    ∴ H =
    V2²
    		 + H - h
    2g

    V2 = √2gh

    Correct Option: B


    By applying Bernoulli’s equation at (1) and (2), we get

    P1
    		 +
    V1²
    		 + Z1 =
    P2
    		 +
    V2²
    		 + Z2
    ρg2gρg2g

    P1 = P2 = Patm
    V1 = 0
    V2 = ?
    Z2 = H – h
    ∴ H =
    V2²
    		 + H - h
    2g

    V2 = √2gh

  1. For laminar flow through along pipe, the pressure drop per unit length increases








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    P1 - P2 =
    32μUL
    D²

    Correct Option: C

    P1 - P2 =
    32μUL
    D²

  1. In fully developed laminar flow in the circular pipe, the head loss due to friction is directly proportional to _____ (mean velocity/square of the mean velocity)








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    f ∝ Mean velocity

    Correct Option: A

    f ∝ Mean velocity

  1. Fluid is flowing with an average velocity of V through a pipe of diameter d. Over a length of L,
    the head loss is given by
    fLV²
    		 .The friction factor, f for laminar flow in terms of Reynolds
    2gD

    number (Re) is _______ (fill in the blanks)








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    For Laminar flow,

    f =
    64
    Re

    Correct Option: B

    For Laminar flow,

    f =
    64
    Re