In many situations, in practice, we are interested in determining the minimum heat input rate ‘Q’ in watts required to from a weld of a given width ‘w’ in a ‘V grove as shown in the Fig. 6.1. It can be calculated* for two dimensional heat source or a three dimensional heat source using equations (6.1) and (6.2) respectively.

Fig. 6.1 Plate geometry for calculating the heat input rate

w = weld width in (m) h = plate thickness in (m) v = welding speed (in m/s) 0m = MP of steel = 1530°C 00 = room temperature = 30°C (assumed) pc = 0.0044

The following symbols are used in these equations. a = thermal diffusivity of the work in (m2/s),

K = thermal conductivity of work material (W/m-°C)


1.2 x 10-5 m2/s = pc PC

43.6 W/m —°C a



P = density and C = specific heat 0m = M. P. of metal For two dimensional heat source


1 + vw 5 4a

Q = 8 K



and for three dimensional heat source

It can be observed from these equations that “vra/a” is the most important parameter Theoretical results fail to accomodate many practical difficulties e. g.

1. Inhomogeneous conducting medium (liquid pool + solid)

2. Absorption and rejection of the latent heat at the forward and rear edges, respec­tively, of the weld-pool.

Still the above two equations provide a good estimate.

In arc welding with short circuit transfer, the heat input is given by

Q _ CVI...(3)

where V = arc voltage, I = arc current and

C = fraction of total time for which the arc is on.

If the (actual) Heat input rate given by equation (3) is less than Q (Q = (CVI) < Qgiven by equations (1) or (2) a lack of side fusion occurs.

In a butt welding process using arc-welding, the arc-power was found to be 2.5 KVA. The process is used to weld 2 plates of steel 3 mm thick, with 60° V-edge preparation angle.

Determine the maximum possible welding speed. The metal transfer is short circuit type and the arc is on for 85% of the total time given.

Solution. The rate of heat input is given as


= 0.85 x 2.5 x 103 w = 2.12 x 103 w The minimum weld width to be maintained

w = AB _ 2y[3 mm. = 243 x 10-3 m.

0m = (1530 - 30) = 1500°C h _ 3 x 10-3 m As in the welding of thin plates, the source of heat can be approximated as a line source. Thus, using equation (1)

F1 vw

q_8 x k e„, h H 5+Ia

F1 vwl

2.12 x 103 = 8 x 43.6 x 1500 x 3 It + ^H x 10-3

5 4a

0. 2+4a K _ ,35

1.15 x 4a

v =


ain = 243 x 10-3 m,

_ 1.15 x 4 x 1.2 x 10-5 243 x 10-3 _ 0.0158 _ 0.016 m/sec. _ 0.95 m/min.

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