Materials Science and Manufacturing Engineering Miscellaneous

Materials Science and Manufacturing Engineering Miscellaneous

Easy Questions
Moderate Questions
Difficult Questions

Materials Science and Manufacturing Engineering

Materials Science and Manufacturing Engineering Miscellaneous
  1. A gas t ungst en ar c wel di ng oper at i on i s performed using a current of 250 A and an arc voltage of 20 V at a welding speed of 5 mm/s. Assuming that the arc efficiency is 70%, the net heat input per unit length of the weld will be ____ kJ/mm (round off to one decimal place).








  1. View Hint View Answer Discuss in Forum

    Net heat input per unit length,

    Ht =
    V.I
    		× η
    v

    V = voltage, I = current, A
    v = welding speed
    Ht =
    20 × 250
    		× 0.70
    5

    Ht = 0.7kJ/ mm

    Correct Option: C

    Net heat input per unit length,

    Ht =
    V.I
    		× η
    v

    V = voltage, I = current, A
    v = welding speed
    Ht =
    20 × 250
    		× 0.70
    5

    Ht = 0.7kJ/ mm

  1. A welding operation is being performed with voltage = 30 V and current = 100 A. The crosssectional area of the weld bead is 20 mm2. The work-piece and filler are of titanium for which of the specific energy of melting is 14 J/mm³.
    Assuming a thermal efficiency of the welding process 70%, the welding speed (in mm/s) is (correct to two decimal places).








  1. View Hint View Answer Discuss in Forum

    Voltage = 30 V,
    I = 100 A Area = 30 mm2
    Specific Energy = 14 J/mm3
    ηth = 70%
    welding speed V =?
    Power = V × I = 30 × 100 = 300 J/s

    Specific Energy =
    2100
    A × V

    Transferred Power = 0.7 × 3000 = 2100 J/S
    V =
    2100
    14 × 20

    V = 7.5mm/s

    Correct Option: D

    Voltage = 30 V,
    I = 100 A Area = 30 mm2
    Specific Energy = 14 J/mm3
    ηth = 70%
    welding speed V =?
    Power = V × I = 30 × 100 = 300 J/s

    Specific Energy =
    2100
    A × V

    Transferred Power = 0.7 × 3000 = 2100 J/S
    V =
    2100
    14 × 20

    V = 7.5mm/s

  1. In an arc welding process, welding speed is doubled. Assuming all other process parameters to be constant, the cross sectional area of the weld bead will








  1. View Hint View Answer Discuss in Forum

    If welding speed is doubled then original cross sectional area would become half if volume remains constant. Therefore, area reduces by 50%.

    Correct Option: D

    If welding speed is doubled then original cross sectional area would become half if volume remains constant. Therefore, area reduces by 50%.

  1. The voltage-length characteristic of a direct arc in an arc welding process is V = (100 + 40l), where l is the length of the arc in mm and V is arc voltage in volts. During a welding operation, the arc length varies between 1 and 2 mm and the welding current is in the range 200-250 A. Assuming a linear power source, the short circuit current is ____ A.








  1. View Hint View Answer Discuss in Forum

    Given, V = 100 + 40 l
    l1 = 1mm ⇒  V1 = 140V
    l1 = 250A
    I2 = 2mm ⇒  V2 = 180V
    I2 = 200A

    V
    		 +
    I
    		 = 1
    OCVSSC

    140
    		 +
    250
    		 = 1
    OCVSSC

    180
    		 +
    200
    		 = 1
    OCVSSC

    On solving we get
    SSC = 424.6 A

    Correct Option: B

    Given, V = 100 + 40 l
    l1 = 1mm ⇒  V1 = 140V
    l1 = 250A
    I2 = 2mm ⇒  V2 = 180V
    I2 = 200A

    V
    		 +
    I
    		 = 1
    OCVSSC

    140
    		 +
    250
    		 = 1
    OCVSSC

    180
    		 +
    200
    		 = 1
    OCVSSC

    On solving we get
    SSC = 424.6 A

  1. The welding process which uses a blanket of fusible granular flux is








  1. View Hint View Answer Discuss in Forum

    In submerged arc welding, the arc is completely submerged inside the granular flux powder and thus forming as blanket.

    Correct Option: B

    In submerged arc welding, the arc is completely submerged inside the granular flux powder and thus forming as blanket.