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

t = 56.6 min

Step-by-step explanation:

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

`(C_s - C_x)/(C_s - C_o) =  erf( (x)/(2√(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* 10^(11) m^2/s`

therefore we ahave

`(1.15-0.35)/(1.15- 0.18) =  erf[\frac{4* 10^(-4)}{2\sqrt{1.28* 10^(-11) t}}]`

`0.8247 = erf [(55.90)/(√(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

`(0.8247 - 0.8209)/(0.8427 - 0.8209) = (x - 0.95)/(1 - 0.95)`

x = 0.959

thus

`z = (55.90)/(√(t))`

`0.959 = (55.90)/(√(t))`

t = 56.6 min