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Standard X

Physics

Ohm's Law

Two 8 mm thic...

Question

Two $8 m m$ thick steel plates are placed $5 m m$ apart and welded by a butt joint. Welding is carried out at $20 V$ and speed of $5 m m / s$ . Heat transfer efficiency is $0.80$ . If the heat required to melt steel is $10 J / m m^{3}$ and melting efficiency is $0.625$ . The weld current (in amperes) will be

100

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300

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400

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200

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Solution

The correct option is D $200$
$Vol of weld puddle = 8 \times 5 \times x = 40 x m m^{3}$

$x = a gap$

$Heat required = 10 \times 40 \times x = 400 x J ... (i)$

$Heat generated = η_{1} η_{2} \times \frac{V I x}{v}$

$= \frac{0.625 \times 20 \times I \times x \times 0.8}{5}$

$400 x = \frac{0.625 \times 20 \times I \times x \times 0.8}{5}$

$[Given η_{1} = 0.8, η_{2} = 0.62]$

$I = 200 A$

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Two 8mm thick steel plates are placed 5mm apart and welded by a butt joint. Welding is carried out at 20V and speed of 5mm/s. Heat transfer efficiency is 0.80. If the heat required to melt steel is 10J/mm3 and melting efficiency is 0.625. The weld current (in amperes) will be

The correct option is D 200Vol of weld puddle =8×5×x=40xmm3 x = a gap Heat required =10×40×x=400xJ ... (i) Heat generated =η1η2×VIxv =0.625×20×I×x×0.85 400x=0.625×20×I×x×0.85 [Given η1=0.8,η2=0.62] I=200A