Question
In a direct current arc welding process, the power source has an open circuit voltage of 100 V and short circuit current of 1000 A. Assume a linear relationship between voltage and current. The arc voltage (V) varies with the arc length (l) as V = 10 + 5l, where V is in volts and l is in mm. The maximum available arc power during the process is _________ kVA (in integer).
Answer :
Correct answer is : 25
Vo = 100 V, Is = 1000 A
V = 10 + 5l .... (1)
l = arc length
arc voltage is also V = Vo -
V = 100 - ...... (2)
for stable arc equate voltage equation 1 and 2 we get
10 + 5l = 100 -
= 900 - 50l
P = V × I = (10 + 5l )(900 - 50l)
for maximum power dP/dl = 0
dP/dl = 4500 - 500 - 500 l = 0 ⇒ l = 8 mm
Pmax = (10 + 5 × 8 )(900 - 50 × 8) = 50 × 500 = 25 kW
Copyright © 2025 Test Academy All Rights Reserved