An iron-carbon alloy initially containing 0.275 wt% C is exposed to an oxygen-rich and virtually carbon-free atmosphere at 1110°C. Under the

Question

An iron-carbon alloy initially containing 0.275 wt% C is exposed to an oxygen-rich and virtually carbon-free atmosphere at 1110°C. Under these circumstances the carbon diffuses from the alloy and reacts at the surface with the oxygen in the atmosphere; that is, the carbon concentration at the surface position is maintained essentially at 0.0 wt% C. At what position will the carbon concentration be 0.206 wt% after a 5 h treatment? The value of D at 1110°C is 5.6 × 10-10 m2/s.

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Mộc Miên 5 years 2021-08-14T20:39:09+00:00 1 Answers 122 views 1

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  1. diemkieu
    1
    2021-08-14T20:40:16+00:00
    Reply

    Answer:

    5.12 mm

    Explanation:

    Let’s convert the time taken from hours to seconds

    t = 5 h

    t =  5h * \frac{3600s}{1h}

    t = 18000 s

    The relation expressing the concentration, position and time together is given by the formula:

    \frac{C_x-C_0}{C_s-C_0} = 1 - erf (\frac{x}{2\sqrt{Dt} } )

    where:

    C_x = concentration at depth  = 0.206  wt%

    C_0 = initial concentration = 0.275 wt%

    C_s = concentration at the surface position = 0.0 wt%

    D = diffusion coefficient = 5.6*10 ^{-10}m^2/s

    t = time = 18000 s

    Replacing the value into the previous formula; we have:

    \frac{0.206-0.275}{0.0-0.275} = 1 - erf (\frac{x}{2\sqrt{5.6*10^{-10}*18000}})

    0.2509 = 1 – erf (157.85 x)

    erf (157.85 x) = 1 – 0.2509

    erf (157.85 x) = 0.7491

    So, Let’s assume the value of z to be 157.485x ; we have:

    z = 157.485 x       —————— Equation (1)

    We obtain the value of Z corresponding to erf (Z) = 0.7491 from the Table 5.1 , ‘Table of  Error Function Values’

    \frac{0.75-x}{0.75-0.70} = \frac{0.7112-0.7491}{0.7112-0.6778}

    \frac{0.75-x}{0.05} = \frac{-0.0379}{0.0334}

    – 0.001895 = (0.75 – z ) 0.0334

    – 0.001895 = 0.02505 – 0.0334 z

    0.0334 z = 0.02505 + 0.001895

    0.0334 z = 0.026945

    z = \frac{0.026945}{0.0334}

    z = 0.806737

    Substituting 0.806737 for z in equation (1)

    0.806737 = 157.485 x

    x = \frac{0.806737}{157.485}

    x = 0.00512 m     to mm; we have

    x = 0.00512 m * \frac{1000mm}{1m}

    x = 5.12 mm

    Thus, the position at which the carbon concentration is 0.206 wt% after a 5 h treatment = 5.12 mm

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