Question

Consider the gas carburizing of a gear of 1018 steel (0.18 wt %) at 927°C (1700°F). Calculate the time necessary to increase the carbon content to 0.35 wt % at 0.40 mm below the surface of the gear. Assume the carbon content at the surface to be 1.15 wt % and that the nominal carbon content of the steel gear before carburizing is 0.18 wt %. D (C in  iron) at 927°C = 1.28  10-11 m2 /s.

Answer 1

Answer:

t = 56.6 min

Explanation:

Fick's second law is used to calculate time required for diffusion

$\frac{C_s - C_x}{C_s - C_o} = erf( \frac{x}{2\sqrt{Dt}})$

where

$C_s$= 1.15%

$C_o$ = 0.18%

$C_x$= 0.35%

x = 0.40 mm = 0.0004 n

$D_{927^O\ C } = 1.28\times 10^{11} m^2/s$

therefore we ahave

$\frac{1.15-0.35}{1.15- 0.18} = erf[\frac{4\times 10^{-4}}{2\sqrt{1.28\times 10^{-11} t}}]$

$0.8247 = erf [\frac{55.90}{\sqrt{t}}] = erf z$

from error function table we hvae following result

for erf z                          z

     0.8209                      0.95

      0.8247                   x

      0.8427                    1

therefore

$\frac{0.8247 - 0.8209}{0.8427 - 0.8209} = \frac{x - 0.95}{1 - 0.95}$

x = 0.959

thus

$z = \frac{55.90}{\sqrt{t}}$

$0.959 = \frac{55.90}{\sqrt{t}}$

t = 56.6 min