The temperature profile (T in Kelvin) of an arc weld across its width is given asπ = 2000 exp(β0.3π₯2) where π₯ (in mm) is the distance from the weld centre. The melting point of the base material is 1500 K. The width of the fusion zone is Β _________(in mm tO two decimal places).
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Correct answer is 1.96 Given, π = 2000 exp(β0.3π₯2) Temperature in fusion zone β melting point of base metal πm = 2000 exp(β0.3π₯2) 1500=2000 exp(β0.3π₯2) x= 0.98 Total width of fusion zone = 2x= 2(0.98) =1.96
Correct answer is 1.96
Given, π = 2000 exp(β0.3π₯2)
Temperature in fusion zone β melting point of base metal
πm = 2000 exp(β0.3π₯2)
1500=2000 exp(β0.3π₯2)
x= 0.98
Total width of fusion zone = 2x= 2(0.98) =1.96
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