Question

a) A groove weld has a cross-sectional area = 0.05 in- and is 12 inches long. (1What quantity of heat (in Btu) is required to accomplish the weld, if the metal to be welded is medium carbon steel? (11) How much heat must be generated at the welding source, if the heat transfer factor = 0.9 and the melting factor = 0.7? b) Solve part a), except that the metal to be welded is aluminum, and the corresponding melting factor is half the value for steel.

Answer 1

Given,

A= 0.05 in2 , L= 12 in , so Volume of weld,V= 0.05*12 = 0.6 in3

(a) For Medium carbon steel:

Heat transfer factor, f1=0.9, Melting factor,f2=0.7

Melting temperature Tm= 3060R ( Rankine Temperature Scale)

Unit Heat required, Um = KTm2 ( K = 1.467 * 10-5) (Rankine Temperature scale constant)

So, Um= 1.467 * 10-5 * (3060)2 = 137.36 Btu/in3

Heat required for welding, Hw = Um * V = 137.36 * 0.6 = 82.42 Btu at weld.

Heat required at source, H = Hw/(f1*f2) = 82.42/(0.9*0.7) = 130.825 Btu at source.

(b) For Aluminium:

Heat transfer factor, f1= 0.9, Melting factor, f2=0.35

Melting temperature Tm = 1680R (Rankine Temperature scale)

Unit Heat required, Um= 1.467 * 10-5 * (1680)2 = 41.40 Btu/in3

Heat required for welding, Hw= Um* V = 41.40 * 0.6 = 24.84 Btu at Weld.

Heat required at source, H = Hw/(f1*f2) = 24.84/(0.9*0.35) = ​​​​​​​78.86 Btu at source.