Carbon is allowed to diffuse through a steel plate 1.5 cm thick at 700°C. The co

Question

Carbon is allowed to diffuse through a steel plate 1.5 cm thick at 700°C. The concentrations of carbon at the two faces are 0.85 and 0.40 kg/m", respectively, which are maintained constant. If the pre-exponential and activation energy for the diffusion of carbon in steel are 62 × 107 m/s and 80 kJ/mol, respectively, compute the diffusion area required in order for the passage of 1.25 × 10' g of carbon in 2 hours. (Hint: Use Fick's first law, i.e. the diffusion flux is proportional to the composition gradient. D-D, expl !, and R-8.31 J/mol-K.) composition gradient. D·D,esp( )andR,831

Explanation / Answer

Calculate the diffusion coefficient

D = D0 exp(-Q/RT)

= 6.2*10^-7 m2/s * exp (-80000 J/mol /8.314J/mol·K *973K)

= 3.145 * 10^-11 m2/s

From the Ficks law of diffusion, mass flux

J = - D (CB - CA) / (XB - XA)

= - 3.145 * 10^-11 m2/s * (0.40 - 0.85) kg/m3 / (0.015)m

J = 9.435 * 10^-10 kg/m2-s

J = M/A*t

9.435 * 10^-10 kg/m2-s = (1.25*10^-3 g x 1kg/1000g)/(A*7200s)

A = 0.184 m2

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---