Question

53)Idem Eieercicio 51, empleando precalentamiento T2=300c EL. 51 dice: Calcular el tiempo de enfriamineto entre 800 y 500 C (1875) en un proceso de soldadura realizado con SMAW en un acero con los siguientes parámetros de soldadura: V=30 volts I=180 ampv-3mm 7seg To-20C p=7,85 gr/cm3 C=0,56 J/grº Junta en V dulcm 2 = 0,35/seg Cem

Same as exercise 51, however add To =300 C
Exercise 51 says: calculate the cooling time between 800°C and 500 °C in a welding process carried out with a smaw on a steel with the following welding parameters : voltage= 30 volts I= 180 Amp v( velocity) = 3mm/sec T0= 20°C p= 7.85 gr/ cm3 C= 0.56 J/ gr°C V- joint d= 1cm
Lambda( Greek letter) = 0.35 J/ sec°C cm

Answer 1

welding с SMAW on steel with the Ti = 800°c T2 =500c Caurrent, I= 180 and 500°c in a Given! Cooling time between 800c process carried out with following welding posameters: Voltage, V-30V, Amp Velocity, U = 3 mm/s, To 20c density, f= 7850 kgim specific heat, c= diameter d = 1cm x = 0.355 / see l cm 0.56 kriggoc from energy balance heat carried by heat lose by Welding materica envirarmeur mcp (dt) -2x (T-To) ot Skąd?x cp lat) = -Ax (5-To) at sx7d3cp s - Sot T2 dT T-To 3* * xdxGp in Tirana =-xt 7.85 x IxOX0:56 x.in 1500-20 = -0.35% € 800-20 ta 4.79 sec time required to cool body from 800°C to 500°C