Question

A fillet weld has a cross-sectional area of 25.0 mm2and is 300 mm long. (a) What quantity of heat (in joules) is required to accomplish the weld, if the metal to be welded is low carbon steel? (b) How much heat must be generated at the welding source, if the heat transfer factor is 0.75 and the melting factor=0.63?(Ans: ?, 163,700)

Answer 1

77362.56 J

163730.28571 J

Step-by-step explanation:

A = Area = 25 mm²

l = Length = 300 mm

K = Constant =
`3.33* 10^(-6)`

`\eta` = Heat transfer factor = 0.75

`f_m` = Melting factor = 0.63

T = Melting point of low carbon steel = 1760 K

Volume of the fillet would be

`V=Al\\\Rightarrow V=25* 300\\\Rightarrow V=7500\ mm^3=7500* 10^(-9)\ m^3`

The unit energy for melting is given by

`U_m=KT^2\\\Rightarrow U_m=3.33* 10^(-6)* 1760^2\\\Rightarrow U_m=10.315008\ J/mm^3`

Heat would be

`Q=U_mV\\\Rightarrow Q=10.315008* 7500\\\Rightarrow Q=77362.56\ J`

Heat required to weld is 77362.56 J

Amount of heat generation is given by

`Q_g=(Q)/(\eta f_m)\\\Rightarrow Q_g=(77362.56)/(0.75* 0.63)\\\Rightarrow Q_g=163730.28571\ J`

The heat generated at the welding source is 163730.28571 J