Question

The steady-state diffusion flux through a metal plate is 5.4 x 10-10 kg/m2 s at a temperature of 827 °C (1100 K) and when the concentration gradient is -350 kg/m4 . Calculate the diffusion flux at 1027 °C (1300 K) for the same concentration gradient and assuming an activation energy for diffusion of 150,000 J/mol.

Answer 1

J = -D d/dx

Given ,J = 5.4 x 10^-10 Kg/m²s

d/dx = -350 Kg/m⁴

Hence, the diffusion coefficient D is

1.542 x 10^-12 m²/s

now consider the temperature dependence equation of diffusion coefficient

D = D_o e^{(-Q_d / RT)}

so D_o = D e^-Q_d/RT

At Temp. 1100K D₀ = 1.542 x 10^-10 e^{-150000/ 8.31 x 1100}

                                        =                    m²/s

At Temp 1300K   D = D_o e^{(-Q_d/ RT)}

                                      =          e^{-150000/ 8.31 x 1300}

                                      =                      m²/s

thus find J at temp 1300K,    J = -D d/dx

                                             = -            (-350)

                                             =                        Kg/m² s