Interactive Lecture
Mass Transport at the Atomic Level
by Dr Arnaldo Delli Carri
Diffusion is the net movement of atoms from regions of high concentration to regions of low concentration through random atomic motion. It is the fundamental mechanism by which materials mix, react, and change composition at elevated temperatures.
Diffusion is driven by:
For steady-state conditions where concentration profile doesn't change with time:
\(J = -D \frac{dC}{dx}\)
Where:
For time-dependent concentration changes:
\(\frac{\partial C}{\partial t} = D \frac{\partial^2 C}{\partial x^2}\)
This is the diffusion equation, showing how concentration changes with time.
For constant surface concentration and semi-infinite solid:
\(\frac{C_x - C_0}{C_s - C_0} = 1 - \text{erf}\left(\frac{x}{2\sqrt{Dt}}\right)\)
Where:
The diffusion coefficient varies exponentially with temperature:
\(D = D_0 \exp\left(-\frac{Q_d}{RT}\right)\)
Where:
Taking the natural logarithm:
\(\ln(D) = \ln(D_0) - \frac{Q_d}{RT}\)
This gives a linear relationship when plotting ln(D) vs 1/T
Watch atoms diffuse from high concentration (left) to low concentration (right). The histogram below shows the real-time concentration profile emerging from random atomic motion. Take snapshots to compare profiles at different times!
Simulate the carburising process where carbon diffuses into steel. The Arrhenius plot on the right shows how the diffusion coefficient changes with temperature. Use Play to animate the profile evolving over time, and take snapshots to compare different conditions!
Diffusion distance scales with √(Dt) — doubling the penetration depth requires 4× the time. That square-root dependence is what drives process planning in carburising and semiconductor fabrication.
| Concept | Equation | Description |
|---|---|---|
| Fick's First Law | \(J = -D \frac{dC}{dx}\) | Steady-state diffusion flux |
| Fick's Second Law | \(\frac{\partial C}{\partial t} = D \frac{\partial^2 C}{\partial x^2}\) | Non-steady-state diffusion |
| Arrhenius Equation | \(D = D_0 \exp\left(-\frac{Q_d}{RT}\right)\) | Temperature dependence of D |
| Diffusion Solution | \(\frac{C_x - C_0}{C_s - C_0} = 1 - \text{erf}\left(\frac{x}{2\sqrt{Dt}}\right)\) | Concentration profile (constant surface) |
| Diffusion Distance | \(x \propto \sqrt{Dt}\) | Characteristic penetration depth |
A steel component is carburised at 927°C (1200 K) for 10 hours. Using the Arrhenius parameters for carbon in austenite (\(D_0 = 2.3 \times 10^{-5}\) m²/s, \(Q_d = 148\) kJ/mol), the diffusion coefficient at this temperature is D = 8.3 × 10⁻¹² m²/s. Calculate the depth at which the carbon concentration is the average of the surface and initial concentrations (this occurs when erf(z) = 0.5, where z ≈ 0.48).
Show Solution
Given: \(D = 8.3 \times 10^{-12}\) m²/s, \(t = 10\) hours = 36,000 s
We need to find x where \(\text{erf}\left(\frac{x}{2\sqrt{Dt}}\right) = 0.5\)
This occurs when \(\frac{x}{2\sqrt{Dt}} = 0.48\)
Calculate \(\sqrt{Dt}\):
\(\sqrt{Dt} = \sqrt{8.3 \times 10^{-12} \times 36000}\)
\(\sqrt{Dt} = \sqrt{2.99 \times 10^{-7}} = 5.47 \times 10^{-4}\) m
Calculate depth x:
\(x = 2 \times 0.48 \times 5.47 \times 10^{-4}\)
\(x = 5.25 \times 10^{-4}\) m = 0.525 mm
The carbon concentration is half-way between surface and initial values at a depth of approximately 0.5 mm. This midpoint depth is close to the simulation's √(Dt) readout at 927°C, 10 h (here x = 0.96√(Dt)); it is not the same as the sim's "Case Depth" readout, which marks a fixed 0.4 wt% carbon rather than the surface–initial midpoint and so reads deeper.
The diffusion coefficient of copper in aluminium is D = 6.5 × 10⁻¹³ m²/s at 500°C. The activation energy for this diffusion is \(Q_d = 136\) kJ/mol. Calculate the diffusion coefficient at 600°C. How many times faster is diffusion at 600°C compared to 500°C?
Show Solution
Given: \(D_1 = 6.5 \times 10^{-13}\) m²/s at \(T_1 = 773\) K (500°C)
\(Q_d = 136,000\) J/mol, \(R = 8.314\) J/(mol·K), \(T_2 = 873\) K (600°C)
Using the Arrhenius equation ratio:
\(\frac{D_2}{D_1} = \exp\left[\frac{Q_d}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\right]\)
\(\frac{D_2}{D_1} = \exp\left[\frac{136000}{8.314}\left(\frac{1}{773} - \frac{1}{873}\right)\right]\)
\(\frac{D_2}{D_1} = \exp\left[16354 \times (0.001294 - 0.001146)\right]\)
\(\frac{D_2}{D_1} = \exp[16354 \times 0.000148] = \exp[2.42] = 11.2\)
Calculate D₂:
\(D_2 = 11.2 \times 6.5 \times 10^{-13} = 7.28 \times 10^{-12}\) m²/s
Diffusion at 600°C is approximately 11 times faster than at 500°C.
A carburising process achieves a penetration depth of 1.0 mm in 5 hours at a certain temperature. How long would it take to achieve a penetration depth of 2.5 mm at the same temperature?
Show Solution
The characteristic diffusion distance scales as \(x \propto \sqrt{Dt}\)
At constant temperature and D:
\(\frac{x_2}{x_1} = \sqrt{\frac{t_2}{t_1}}\)
Given: \(x_1 = 1.0\) mm, \(t_1 = 5\) hours, \(x_2 = 2.5\) mm
Solve for t₂:
\(\frac{2.5}{1.0} = \sqrt{\frac{t_2}{5}}\)
\(2.5 = \sqrt{\frac{t_2}{5}}\)
Squaring both sides:
\(6.25 = \frac{t_2}{5}\)
\(t_2 = 6.25 \times 5 = 31.25\) hours
To increase penetration depth by 2.5×, the time must increase by (2.5)² = 6.25×