Two mild steel plates of similar thickness, in butt-joint configuration, are welded by gas tungsten arc welding process using the following welding parameters.

Welding valtage	20 V
Welding current	150 A
Welding speed	5 mm/s

A filler wire of the same mild steel material having 3 mm diameter is used in this welding process. The filler wire feed rate is selected such that the final weld bead is composed of 60% volume of filler and 40% volume of plate material. The heat required to melt the mild steel material is 10 J/mm³. The heat transfer factor is 0.7 and melting factor is 0.6. The feed rate of the filler wire is __________ mm/s (round off to one decimal place).

The correct answer is 10.7 (or 10.695 before rounding).

1. Identify the given parameters:
We are given the following welding parameters from the problem statement:
Welding Voltage, V = 20 V
Welding Current, I = 150 A
Welding Speed, v = 5 mm/s
Diameter of the filler wire, d = 3 mm
Heat required to melt the material per unit volume, H_m = 10 J/mm³
Heat transfer efficiency (heat transfer factor), η_h = 0.7
Melting efficiency (melting factor), η_m = 0.6
Fractional volume of filler wire in the weld bead, f = 60% = 0.6

2. Calculate the cross-sectional area of the weld bead (A):
The melting efficiency (η_m) is defined as the ratio of the energy required to melt the metal to the actual heat energy delivered to the weld pool:

$${\eta }_m=\frac{H_m\times A\times v}{V\times I\times {\eta }_h}$$
Where:
- A is the total cross-sectional area of the weld bead (mm²)
- v is the welding speed (mm/s)
- V is the welding voltage (V)
- I is the welding current (A)
- η_h is the heat transfer efficiency
- H_m is the energy required to melt a unit volume of metal (J/mm³)

Substituting the given values into the equation:

$$0.6=\frac{10\times A\times 5}{20\times 150\times 0.7}$$
Simplify the terms in the equation:

$$0.6=\frac{50A}{2100}$$
Solving for the cross-sectional area of the weld bead A:

$$A=\frac{0.6\times 2100}{50}=25.2{\text{mm}}^2$$

3. Use volume flow balance to determine the filler wire feed rate (f_e):
Let f_e be the feed rate of the filler wire (in mm/s).
The volume flow rate of the filler wire supplied must equal the volume flow rate of the filler material deposited in the weld bead.

The cross-sectional area of the filler wire (A_e) is:

$$A_e=\frac{\pi }{4}{d}^2=\frac{\pi }{4}{\left(3\right)}^2=2.25\pi \approx 7.0686{\text{mm}}^2$$
The volume flow rate of filler wire fed per second is:

$$Q_{\text{fed}}=A_e\times f_e$$
The total volume flow rate of the weld bead deposited per second is:

$$Q_{\text{bead}}=A\times v$$
Since the filler wire accounts for 60% (f = 0.6) of the total weld bead volume, the volume flow rate of the filler material in the weld bead is:

$$Q_{\text{filler-deposited}}=A\times v\times f$$
Equating the rate of filler wire fed to the rate of filler material deposited:

$$A_e\times f_e=A\times v\times f$$
Substituting the calculated and given values:

$$\frac{\pi \times 3^2}{4}\times f_e=25.2\times 5\times 0.6$$
$$7.0686\times f_e=75.6$$
$$f_e=\frac{75.6}{7.0686}\approx 10.695\text{mm/s}$$

Rounding to one decimal place as requested in the question, the feed rate of the filler wire is 10.7 mm/s.