Question:hard

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.
\[ \begin{array}{cc} z & \text{erf}(z) \\ 0.3 & 0.3268 \\ 0.4 & 0.4284 \\ 0.5 & 0.5205 \end{array} \]

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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.
Updated On: Jul 28, 2026
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Correct Answer: 990

Solution and Explanation

Step 1: Write the fixed-surface diffusion formula in a slightly different form.
Instead of isolating $\sqrt{Dt}$ step by step, square both sides of the standard relation right away and solve for $t$ directly. Starting from
\[ \frac{C_s-C_x}{C_s-C_0}=\text{erf}(z), \qquad z=\frac{x}{2\sqrt{Dt}} \]
we can rearrange once, symbolically, to
\[ t=\frac{x^2}{4Dz^2} \]

Step 2: Get the fraction reacted.
\[ \frac{C_s-C_x}{C_s-C_0}=\frac{1.4-0.8859}{1.4-0.2}=\frac{0.5141}{1.2}=0.4284 \]

Step 3: Read $z$ off the table.
The table lists $\text{erf}(0.4)=0.4284$, which lines up exactly with the value from Step 2, so no interpolation between rows is needed: $z=0.4$.

Step 4: Plug directly into the rearranged formula.
With $x=2\times10^{-4}$ m and $D=6.25\times10^{-11}$ m$^2$/s,
\[ t=\frac{x^2}{4Dz^2}=\frac{(2\times10^{-4})^2}{4(6.25\times10^{-11})(0.4)^2}=\frac{4\times10^{-8}}{4\times10^{-11}}=1000\text{ s} \]

Step 5: Verify by plugging back in.
At $t=1000$ s, $\sqrt{Dt}=\sqrt{6.25\times10^{-11}\times1000}=\sqrt{6.25\times10^{-8}}=2.5\times10^{-4}$ m, so $z=x/(2\sqrt{Dt})=2\times10^{-4}/(5\times10^{-4})=0.4$, which matches Step 3 and confirms the arithmetic.

Step 6: Conclude.
\[ \boxed{t=1000\text{ s}} \]
This sits comfortably inside the accepted range of 990 to 1010 s.
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