Question

During carburizing of a steel, the surface concentration is kept constant at 1.4 wt.% carbon. Diffusivity of carbon for the steel at \(950^{\circ}\text{C}\) is \(6.25 \times 10^{-11}\) m\(^2\)/s. At \(950^{\circ}\text{C}\), the time required to carburize the steel with an initial composition of 0.2 wt.% carbon to 0.8859 wt.% carbon at a depth of 0.2 mm is _______ seconds (approximate to the nearest integer).
Use the nearest value of the error function from the table given below for your calculation.
\[ \begin{array}{cc} z & \text{erf}(z) \\ 0.3 & 0.3268 \\ 0.4 & 0.4284 \\ 0.5 & 0.5205 \end{array} \]

Hint:

Use Fick's second law error-function solution \((C_s-C_x)/(C_s-C_0) = \text{erf}(x/(2\sqrt{Dt}))\) and match the left side against the given erf table to find the argument, then solve for t.

Answer 1

Correct Answer: 990

Step 1: Recall the diffusion equation for carburizing.
Carburizing is treated as diffusion of carbon into a semi-infinite solid whose surface composition is held fixed. Fick's second law for this case gives the standard error-function solution,
\[ \frac{C_s - C_x}{C_s - C_0} = \text{erf}\left(\frac{x}{2\sqrt{Dt}}\right) \]
where \(C_s\) is the fixed surface composition, \(C_0\) is the original bulk composition of the steel, \(C_x\) is the composition reached at depth \(x\) after time \(t\), and \(D\) is the diffusion coefficient of carbon at the carburizing temperature.

Step 2: Read off the given data.
\[ C_s = 1.4\%, \quad C_0 = 0.2\%, \quad C_x = 0.8859\%, \quad D = 6.25\times10^{-11} \text{ m}^2/\text{s}, \quad x = 0.2 \text{ mm} = 2\times10^{-4} \text{ m} \]

Step 3: Compute the left-hand side.
\[ \frac{C_s - C_x}{C_s - C_0} = \frac{1.4 - 0.8859}{1.4 - 0.2} = \frac{0.5141}{1.2} = 0.4284 \]

Step 4: Match this to the error function table.
The table gives \(\text{erf}(0.4) = 0.4284\), an exact match to the value found above. So
\[ \frac{x}{2\sqrt{Dt}} = 0.4 \]

Step 5: Solve for the time.
\[ 2\sqrt{Dt} = \frac{x}{0.4} = \frac{2\times10^{-4}}{0.4} = 5\times10^{-4} \text{ m} \]
\[ \sqrt{Dt} = 2.5\times10^{-4} \text{ m} \]
\[ Dt = \left(2.5\times10^{-4}\right)^2 = 6.25\times10^{-8} \text{ m}^2 \]
\[ t = \frac{6.25\times10^{-8}}{D} = \frac{6.25\times10^{-8}}{6.25\times10^{-11}} = 1000 \text{ s} \]

Step 6: Final Answer.
The time required is \(1000\) s, rounded to the nearest integer, which lies inside the accepted range of 990 to 1010 s.
\[ \boxed{t = 1000 \text{ s}} \]