Carbon at \(1273 \mathrm{~K}\) is contained in a \(7-\mathrm{cm}\)-innerdiameter cylinder made of iron whose thickness is \(1.2 \mathrm{~mm}\). The concentration of carbon in the iron at the inner surface is \(0.5 \mathrm{~kg} / \mathrm{m}^{3}\) and the concentration of carbon in the iron at the outer surface is negligible. The diffusion coefficient of carbon through iron is \(3 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\). The mass flow rate of carbon by diffusion through the cylinder shell per unit length of the cylinder is (a) \(2.8 \times 10^{-9} \mathrm{~kg} / \mathrm{s}\) (b) \(5.4 \times 10^{-9} \mathrm{~kg} / \mathrm{s}\) (c) \(8.8 \times 10^{-9} \mathrm{~kg} / \mathrm{s}\) (d) \(1.6 \times 10^{-8} \mathrm{~kg} / \mathrm{s}\) (e) \(5.2 \times 10^{-8} \mathrm{~kg} / \mathrm{s}\) 14-185 The surface of an iron component is to be hardened by carbon. The diffusion coefficient of carbon in iron at \(1000^{\circ} \mathrm{C}\) is given to be \(3 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\). If the penetration depth of carbon in iron is desired to be \(1.0 \mathrm{~mm}\), the hardening process must take at least (a) \(1.10 \mathrm{~h}\) (b) \(1.47 \mathrm{~h}\) (c) \(1.86 \mathrm{~h}\) (d) \(2.50 \mathrm{~h}\) (e) \(2.95 \mathrm{~h}\)

Step by step solution

01

1. Write down Fick's first law of diffusion

Fick's first law of diffusion states that: \(J = -D \frac{\Delta C}{\Delta x}\), where \(J\) is the mass flow rate per unit cross-sectional area, \(D\) is the diffusion coefficient, \(\Delta C\) is the difference in concentration, and \(\Delta x\) is the thickness of the material.

02

2. Determine the difference in concentration and the thickness

Given that the concentration of carbon in the iron at the inner surface is \(\frac{0.5 kg}{m^3}\) and the concentration at the outer surface is negligible, the difference in concentration is \(\Delta C = 0.5 \frac{kg}{m^3}\). The thickness of the iron cylinder is 1.2 mm, which is equal to \(\Delta x = 1.2 \times 10^{-3}m\).

03

3. Calculate the mass flow rate per unit cross-sectional area

Using Fick's first law of diffusion, we can determine the mass flow rate per unit cross-sectional area \(J\): \(J = -D \frac{\Delta C}{\Delta x} = - (3 \times 10^{-11} \frac{m^2}{s}) \frac{0.5 \frac{kg}{m^3}}{1.2 \times 10^{-3} m} = - 1.25 \times 10^{-5} \frac{kg}{m^2 \cdot s}\) The negative sign indicates that the mass flow rate is in the opposite direction of the concentration gradient.

04

4. Calculate the mass flow rate per unit length of the cylinder

Now, we need to find the mass flow rate per unit length of the cylinder by multiplying \(J\) by the cross-sectional area of the iron shell: Area of the iron shell = \(\pi (r_2^2 - r_1^2)\), where \(r_1 = 3.5 \times 10^{-2} m\) and \(r_2 = (3.5 + 1.2) \times 10^{-2} m\) are the inner and outer radii of the cylinder. Area of the iron shell = \(\pi ((3.7 \times 10^{-2} m)^2 - (3.5 \times 10^{-2} m)^2) = 4.494 \times 10^{-3} m^2\) Now, we can determine the mass flow rate per unit length (\(M\)) by multiplying \(J\) by the Area of the iron shell: \(M = J \times \mathrm{Area~of~the~iron~shell} = (- 1.25 \times 10^{-5} \frac{kg}{m^2 \cdot s}) (4.494 \times 10^{-3} m^2) = -5.62 \times 10^{-8} \frac{kg}{s}\) Since we are asked for the magnitude of the mass flow rate, the answer is \(5.62 \times 10^{-8} \frac{kg}{s}\). The closest answer in the given options is (e) \(5.2 \times 10^{-8} \mathrm{~kg} /\mathrm{s}\).

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