GATE Papers >> Mechanical >> 2017 >> Question No 23

Question No. 23 Mechanical | GATE 2017

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


Answer : (D) reduce by 50%


Solution of Question No 23 of GATE 2017 Mechanical Paper

Heat given to weld pool $ ={\mathrm H}_\mathrm s=\frac{\mathrm V\vert{\mathrm\eta}_\mathrm h}{\mathrm\nu\;\mathrm A} $

$ \mathrm V= $ arc voltage

$ \mathrm I= $ current

$ {\mathrm\eta}_\mathrm h= $ heat transfer efficiency

$ \mathrm v= $ Velocity of welding (speed)

$ \mathrm A= $ Cross sectional area of weld.

$ {\mathrm v}_1=\mathrm v\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\mathrm A}_1=\mathrm A $

$ {\mathrm v}_1=2\;\mathrm v $

Now $ \mathrm{vA}=\frac{{\mathrm H}_2}{\mathrm V\vert{\mathrm\eta}_\mathrm h}= $ constant (RHS unchanged)

$ {\mathrm v}_1{\mathrm A}_1\;=\;{\mathrm v}_2{\mathrm A}_2 $ (Material deposition rate is constant)

$ {\mathrm A}_2\;=\;\frac{{\mathrm A}_1}2 $

Area of cross section reduced by 50%

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