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Two pipes of inner diameter 100 mm and outer diameter 110 mm each joined by flash butt welding using 30 V power supply. At the interface, 1 mm of the material melts from each pipe which has a resistance of 42.4 Ω. If the unit melt energy is 64.4 MJm-3, then the time required for welding in seconds is


1. 1
2. 5
3. 10
4. 20

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Correct Answer - Option 3 : 10

Concept:

Heat generated during welding process is given by:

\(Q=VIt=I^2Rt=\frac{V^2}{R}t\)

Heat required to melt welding portion = Unit melting energy (u) × Volume 

Calculation:

Given:

di = 100 mm, do = 110 mm, Voltage = 30 V, l (thickness) = 1 mm, R = 42.4 Ω, and u = 64.4 MJ/m3

Volume = \(\frac{\pi}{4}(d_o^2\;-\;d_i^2)\;×\;2l\)  [∵ thickness melts in both pipes]

Volume = \(\frac{\pi}{4}(110^2\;-\;100^2)\;×\;2\;×\;1\)

Volume = \(3298.672\;mm^3=3298.672×10^{-9}\;m^3\)

Melting Energy = Unit melting energy (u) × Volume

Melting Energy = 64.4 × 10× 3298.672 × 10-9 ⇒ 212.43 J 

Heat Generated = \(\frac{{{V^2}}}{R}t\)

Since no heat goes in waste,

Heat generated = Melting energy required

\(∴ \frac{{{30^2}}}{42.4}t=212.43\;\)

∴ t = 10 sec

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