Question

A Gas Metal Arc Welding process is used to join two mild steel plates with the following welding process parameters: Current = 220 A; Voltage = 24 V; Welding speed: 19 cm/min; Wire diameter 1.2 mm; wire feed rate: 4 m/min; Thermal efficiency of the process 63%. Then the heat input per unit length of the weld (in kJ/cm) and the area of or cross section of weld bead (in $m{m}^2$) are respectively [Assume duty cycle is 1]

• A. 7.3, 18.91
• B. 1.1, 20.01
• C. 4.3, 28.35
• D. 10.5, 23.80

Answer 1

The correct option is D 10.5, 23.80

$\text{Current (I) = 220 A}$

$\text{Voltage (V) = 24 V}$

$\text{Welding speed (v) = 19 cm/min}$

$\text{=}\frac{19}{60}\text{0.3166 cm/sec = 19}\times \text{10mm/min})$

$\text{Wire diameter}\left(d_{wire}\right)\text{1.2 mm}$

$\text{Wire feed rate (f) = 4 m/min}$

$\text{= 4}\times \text{1000 mm/min}$

$\text{Thermal efficiency}{\eta }_{th}\text{= 63\% = 0.63}$

We know,

(i) Heat require per unit length

$\frac{H}{l}=\frac{\eta DVI}{v}=\frac{0.63\times 1\times 24\times 220}{0.3166}=10.506kJ/cm$

(ii)
$\frac{\text{volume of metal deposited}}{t}=\frac{\text{volume of electrode wire consumed}}{t}\frac{A\times l}{t}=\frac{\pi }{4}{d}^2\times fAv=\frac{\pi }{4}{d}^2\times fA=\frac{(\frac{\pi }{4}\times ({1.2}^2)\times 4\times 1000)}{19\times 10}A=\frac{\pi }{4}{d}^2\times f\times v$

$\text{The area of the cross section of weldbead is = 23.80 mm}{}^2$