Question:easy
An arc welding process is being carried out with a power source of 6 kW to weld two similar metals. The total energy loss is 50%, the area of cross-section of the weld is \( 10 \ mm^2 \), and the specific energy needed to melt the metal is \( 15 \ J/mm^3 \).
The speed of welding is ______ mm/s.
An arc welding process is being carried out with a power source of 6 kW to weld two similar metals. The total energy loss is 50%, the area of cross-section of the weld is \( 10 \ mm^2 \), and the specific energy needed to melt the metal is \( 15 \ J/mm^3 \).
The speed of welding is ______ mm/s.
The speed of welding is ______ mm/s.
Show Hint
Net power = input power x efficiency. Melt rate = net power / specific energy. Speed = melt rate / weld area.
Updated On: Aug 5, 2026
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The Correct Option is B
Solution and Explanation
Step 1: Write one combined formula for speed.
We can skip the middle step and connect welding speed directly to power, loss, area and specific energy in a single equation.
$P_{in} \times \eta = v \times A \times U$
This says the useful power equals the rate of energy used to melt metal moving at speed $v$ over area $A$.
Step 2: Plug in the known numbers.
$P_{in} = 6000 \ W$, $\eta = 0.5$, $A = 10 \ mm^2$, $U = 15 \ J/mm^3$
$6000 \times 0.5 = v \times 10 \times 15$
Step 3: Solve for the speed.
$3000 = 150v$
$v = \frac{3000}{150} = 20$
So the welding speed comes out to 20 mm/s using this single-equation route, matching the earlier stepwise method.
Step 4: Sanity check with rough numbers.
Half of 6 kW is 3 kW, and melting 1 mm^3 needs 15 J, so about 200 mm^3 gets melted every second.
Spread over a 10 mm^2 cross-section, that gives a speed close to 20 mm/s, confirming the answer holds up under a quick rough check too.
Final Answer:
Solving the combined power balance equation gives a welding speed of 20 mm/s. \[ \boxed{v = 20 \ mm/s} \]
We can skip the middle step and connect welding speed directly to power, loss, area and specific energy in a single equation.
$P_{in} \times \eta = v \times A \times U$
This says the useful power equals the rate of energy used to melt metal moving at speed $v$ over area $A$.
Step 2: Plug in the known numbers.
$P_{in} = 6000 \ W$, $\eta = 0.5$, $A = 10 \ mm^2$, $U = 15 \ J/mm^3$
$6000 \times 0.5 = v \times 10 \times 15$
Step 3: Solve for the speed.
$3000 = 150v$
$v = \frac{3000}{150} = 20$
So the welding speed comes out to 20 mm/s using this single-equation route, matching the earlier stepwise method.
Step 4: Sanity check with rough numbers.
Half of 6 kW is 3 kW, and melting 1 mm^3 needs 15 J, so about 200 mm^3 gets melted every second.
Spread over a 10 mm^2 cross-section, that gives a speed close to 20 mm/s, confirming the answer holds up under a quick rough check too.
Final Answer:
Solving the combined power balance equation gives a welding speed of 20 mm/s. \[ \boxed{v = 20 \ mm/s} \]
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