Materials Science and Manufacturing Engineering Miscellaneous


Materials Science and Manufacturing Engineering Miscellaneous

Materials Science and Manufacturing Engineering

  1. In a sand casting operation, the total liquid head is maintained constant such that it is equal to the mould height. The time taken to fill the mould with a top gate is tA. If the same mould is filled with a bottom gate, then the time taken is tB. Ignore the time required to fill the runner and frictional effects, Assume Atmospheric pressure at the top molten metal surfaces. The relation between tA and tB is









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    Top gate tf1 =
    A × H
    Ag √( 2ghm )

    hm = H
    tf1 =
    A × √hm
    .....(1)
    Ag √2g

    Bottom gate tf2 =
    2A
    ( √hm - √hm - H )
    Ag √2g

    tf2 =
    2A
    hm .....(2)
    Ag √2g

    tf2
    = 2
    tf1

    tf2 = 2tf1

    Correct Option: B

    Top gate tf1 =
    A × H
    Ag √( 2ghm )

    hm = H
    tf1 =
    A × √hm
    .....(1)
    Ag √2g

    Bottom gate tf2 =
    2A
    ( √hm - √hm - H )
    Ag √2g

    tf2 =
    2A
    hm .....(2)
    Ag √2g

    tf2
    = 2
    tf1

    tf2 = 2tf1


  1. A mould has downsprue whose length is 20 cm and the cross sectional area at the base of the downsprue is 1 cm2. The downsprue feeds a horizontal runner leading into the mould cavity of volume 1000 cm3. The time required to fill the mould cavity will be









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    Height of sprue, h = 20 cm = 0.2m
    Area of down sprue (Ingate) = 1m2 = 1 × 10–4 m2
    Volume of casting = 1000 cm3 (Mould cavity)
    Velocity of sprue base V = √2gh = √2 × 9.81 × 0.2 = 1.98 m/s
    10–3 = 10–4 × 1.98 × t

    t =
    10–3 × 10–4
    = 5.05 sec
    1.98

    Correct Option: B

    Height of sprue, h = 20 cm = 0.2m
    Area of down sprue (Ingate) = 1m2 = 1 × 10–4 m2
    Volume of casting = 1000 cm3 (Mould cavity)
    Velocity of sprue base V = √2gh = √2 × 9.81 × 0.2 = 1.98 m/s
    10–3 = 10–4 × 1.98 × t

    t =
    10–3 × 10–4
    = 5.05 sec
    1.98



  1. Gray cast iron blocks 200 x 100 x 10 mm are to be cast in sand moulds, Shrinkage allowance for pattern making is 1%. The ratio of the volume of pattern to that of the casting will be









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    1% shrinkage allowance
    ⇒ 200 × 0.01 = 2mm
    ⇒ 100 × 0.01 = 1 mm
    ⇒ 10 × 0.01 = 0.1 mm
    Gray cast iron expands after solidification.
    Vp = pattern volume = (200 – 2) × (100 – 1)
    Vc = Casting volume = (200 × 100 × 10)

    Vp
    =
    (200 – 2) × (100 – 1) × 999
    Vc
    200 × 100 × 10

    Vp
    = 0.97
    Vc

    Correct Option: A

    1% shrinkage allowance
    ⇒ 200 × 0.01 = 2mm
    ⇒ 100 × 0.01 = 1 mm
    ⇒ 10 × 0.01 = 0.1 mm
    Gray cast iron expands after solidification.
    Vp = pattern volume = (200 – 2) × (100 – 1)
    Vc = Casting volume = (200 × 100 × 10)

    Vp
    =
    (200 – 2) × (100 – 1) × 999
    Vc
    200 × 100 × 10

    Vp
    = 0.97
    Vc


  1. With a solidification factor of 0.97 × 106 s/m2, the solidification time (in seconds) for a spherical casting of 200 mm diameter is









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    t = k
    V
    2
    A

    D = 200 mm
    t = 0.97 × 106
    (4 / 3)π(0.1)3
    2
    4π(0.1)2

    t = 0.97 × 106
    0.1
    2
    3

    t = 1077.8 sec
    t ≈ 1078 sec

    Correct Option: B

    t = k
    V
    2
    A

    D = 200 mm
    t = 0.97 × 106
    (4 / 3)π(0.1)3
    2
    4π(0.1)2

    t = 0.97 × 106
    0.1
    2
    3

    t = 1077.8 sec
    t ≈ 1078 sec



  1. Hardness of green sand mould increases with









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    decrease in permeability

    Correct Option: C

    decrease in permeability