An arc-welding DC source has linear characteristic with open-circuit voltage 60 V and short-circuit current 600 A. Melting energy is 10 J/mm³; heat-transfer and melting efficiencies are 80% and 25%. At maximum source power, a weld cross-section of 20 mm² is produced. Find speed in mm/s to one decimal.
Voltage is V = 60−0.1I, so power $P=60I-0.1I^2$ is maximized at I = 300 A and V = 30 V. Useful melting power = 9000×0.8×0.25 = 1800 W. Volume rate = 1800/10 = 180 mm³/s. Divide by area 20 mm² to obtain **9.0 mm/s**.
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