Answer:
77362.56 J
163730.28571 J
Explanation:
A = Area = 25 mm²
l = Length = 300 mm
K = Constant = [tex]3.33\times 10^{-6}[/tex]
[tex]\eta[/tex] = Heat transfer factor = 0.75
[tex]f_m[/tex] = Melting factor = 0.63
T = Melting point of low carbon steel = 1760 K
Volume of the fillet would be
[tex]V=Al\\\Rightarrow V=25\times 300\\\Rightarrow V=7500\ mm^3=7500\times 10^{-9}\ m^3[/tex]
The unit energy for melting is given by
[tex]U_m=KT^2\\\Rightarrow U_m=3.33\times 10^{-6}\times 1760^2\\\Rightarrow U_m=10.315008\ J/mm^3[/tex]
Heat would be
[tex]Q=U_mV\\\Rightarrow Q=10.315008\times 7500\\\Rightarrow Q=77362.56\ J[/tex]
Heat required to weld is 77362.56 J
Amount of heat generation is given by
[tex]Q_g=\dfrac{Q}{\eta f_m}\\\Rightarrow Q_g=\dfrac{77362.56}{0.75\times 0.63}\\\Rightarrow Q_g=163730.28571\ J[/tex]
The heat generated at the welding source is 163730.28571 J