Diffusion+Advection+Reaction (solution) Pure diffusion in a 1D infinite pipe: 𝑎𝑎 – radius of the tube • A mass of tracer 𝑀𝑀 is injected at the point 𝑥𝑥 = 0 at time 𝑡𝑡 = 0, uniformly across the cross-section area 𝐴𝐴 = 𝜋𝜋𝑎𝑎2 • How does the tracer spreads in time, due to molecular diffusion alone? Boundary conditions: Initial condition: 𝛿𝛿(𝑥𝑥) = � 0, 𝑥𝑥 > 0 +∞, 𝑥𝑥 = 0 0, 𝑥𝑥 < 0 Conservation of mass: Dirac delta function � −∞ +∞ 𝛿𝛿 𝑥𝑥 𝑑𝑑𝑥𝑥 = 1 Recall dimensional analysis: • Buckingham π-theorem – Consider a process that can be described by 𝑚𝑚 dimensional variables (parameters) – This full set of variables contains 𝑛𝑛 different physical dimensions (e.g. 𝐿𝐿, 𝑇𝑇, 𝑀𝑀) – Then there are (𝑚𝑚− 𝑛𝑛) independent non-dimensional groups that can be formed from these variables… – …and the variables can be related according to 𝜋𝜋1 = 𝑓𝑓(𝜋𝜋2, … ,𝜋𝜋𝑚𝑚−𝑛𝑛) One-dimensional pure diffusion in an infinite pipe: Number of parameters: 𝑚𝑚 = 5 Number of dimensions: 𝑛𝑛 = 3 Number of (independent) dimensionless groups: 𝑚𝑚 − 𝑛𝑛 = 2 𝜋𝜋1 = 𝑓𝑓(𝜋𝜋2) 𝐶𝐶 ⁄𝑀𝑀 (𝐴𝐴 𝐷𝐷𝑡𝑡) = 𝑓𝑓( 𝑥𝑥 𝐷𝐷𝑡𝑡 ) 𝐶𝐶 = ⁄𝑀𝑀 (𝐴𝐴 𝐷𝐷𝑡𝑡)𝑓𝑓( 𝑥𝑥 𝐷𝐷𝑡𝑡 ) One-dimensional pure diffusion in an infinite pipe: 𝐶𝐶 𝑥𝑥, 𝑡𝑡 = 𝑀𝑀 𝐴𝐴 𝐷𝐷𝑡𝑡 𝑓𝑓( 𝑥𝑥 𝐷𝐷𝑡𝑡 ) 𝜂𝜂 = 𝑥𝑥 𝐷𝐷𝑡𝑡 New variable: 𝜕𝜕𝜂𝜂 𝜕𝜕𝑥𝑥 = 1 𝐷𝐷𝑡𝑡 𝜕𝜕𝜂𝜂 𝜕𝜕𝑡𝑡 = 𝑥𝑥 − 1 2 𝐷𝐷𝑡𝑡 − 1 2−1𝐷𝐷 = − 𝜂𝜂 2𝑡𝑡 𝐶𝐶 𝑥𝑥, 𝑡𝑡 = 𝑀𝑀 𝐴𝐴 𝐷𝐷𝑡𝑡 𝑓𝑓(𝜂𝜂) 𝜕𝜕𝐶𝐶 𝜕𝜕𝑡𝑡 = 𝜕𝜕 𝜕𝜕𝑡𝑡 𝑀𝑀 𝐴𝐴 𝐷𝐷𝑡𝑡 𝑓𝑓(𝜂𝜂) = 𝜕𝜕 𝜕𝜕𝑡𝑡 𝑀𝑀 𝐴𝐴 𝐷𝐷𝑡𝑡 𝑓𝑓 𝜂𝜂 + 𝑀𝑀 𝐴𝐴 𝐷𝐷𝑡𝑡 𝜕𝜕𝑓𝑓 𝜕𝜕𝜂𝜂 𝜕𝜕𝜂𝜂 𝜕𝜕𝑡𝑡 = − 𝑀𝑀 2𝐴𝐴𝑡𝑡 𝐷𝐷𝑡𝑡 𝑓𝑓 + 𝜂𝜂 𝜕𝜕𝑓𝑓 𝜕𝜕𝜂𝜂 𝜕𝜕2𝐶𝐶 𝜕𝜕𝑥𝑥2 = 𝑀𝑀 𝐴𝐴𝐷𝐷𝑡𝑡 𝐷𝐷𝑡𝑡 𝜕𝜕2𝑓𝑓 𝜕𝜕𝜂𝜂2 (diffusion equation in 1D)𝐷𝐷 𝜕𝜕2𝐶𝐶 𝜕𝜕𝑥𝑥2 = 𝜕𝜕𝐶𝐶 𝜕𝜕𝑡𝑡 One-dimensional pure diffusion in an infinite pipe: 𝐷𝐷 𝑀𝑀 𝐴𝐴𝐷𝐷𝑡𝑡 𝐷𝐷𝑡𝑡 𝜕𝜕2𝑓𝑓 𝜕𝜕𝜂𝜂2 = − 𝑀𝑀 2𝐴𝐴𝑡𝑡 𝐷𝐷𝑡𝑡 𝑓𝑓 + 𝜂𝜂 𝜕𝜕𝑓𝑓 𝜕𝜕𝜂𝜂 𝜕𝜕2𝑓𝑓 𝜕𝜕𝜂𝜂2 + 1 2 𝑓𝑓 + 𝜂𝜂 𝜕𝜕𝑓𝑓 𝜕𝜕𝜂𝜂 = 0 𝑑𝑑2𝑓𝑓 𝑑𝑑𝜂𝜂2 + 1 2 𝑓𝑓 + 𝜂𝜂 𝑑𝑑𝑓𝑓 𝑑𝑑𝜂𝜂 = 𝑑𝑑 𝑑𝑑𝜂𝜂 𝑑𝑑𝑓𝑓 𝑑𝑑𝜂𝜂 + 1 2 𝑓𝑓𝜂𝜂 = 0 �𝑑𝑑 𝑑𝑑𝑓𝑓 𝑑𝑑𝜂𝜂 + 1 2 𝑓𝑓𝜂𝜂 = �𝑑𝑑𝜂𝜂 𝑑𝑑𝑓𝑓 𝑑𝑑𝜂𝜂 + 1 2 𝑓𝑓𝜂𝜂 = 𝐶𝐶0 = 0 1 𝑓𝑓 𝑑𝑑𝑓𝑓 = − 1 2 𝜂𝜂𝑑𝑑𝜂𝜂 𝑓𝑓 = 𝐶𝐶1𝑒𝑒𝑥𝑥𝑒𝑒 − 𝜂𝜂2 4 (it can be shown that 𝐶𝐶0 = 0 satisfies initial and boundary conditions) One-dimensional pure diffusion in an infinite pipe: 𝐶𝐶 𝑥𝑥, 𝑡𝑡 = 𝑀𝑀 𝐴𝐴 𝐷𝐷𝑡𝑡 𝑓𝑓(𝜂𝜂) 𝑀𝑀 = � 𝑉𝑉 𝐶𝐶 𝑥𝑥, 𝑡𝑡 𝑑𝑑𝑑𝑑 = � −∞ ∞ � 0 𝑎𝑎 𝑀𝑀 𝐴𝐴 𝐷𝐷𝑡𝑡 𝑓𝑓(𝜂𝜂) 2𝜋𝜋𝜋𝜋𝑑𝑑𝜋𝜋𝑑𝑑𝑥𝑥 𝑑𝑑𝑥𝑥 = 𝑑𝑑𝜂𝜂 𝐷𝐷𝑡𝑡 (balance of mass) 1 = � −∞ ∞ 𝑓𝑓 𝜂𝜂 𝑑𝑑𝜂𝜂 𝑓𝑓(𝜂𝜂) = 𝐶𝐶1𝑒𝑒𝑥𝑥𝑒𝑒 − 𝜂𝜂2 4 � −∞ ∞ 𝐶𝐶1𝑒𝑒𝑥𝑥𝑒𝑒 − 𝜂𝜂2 4 𝑑𝑑𝜂𝜂 = 1 𝐶𝐶1 = 1 ∫−∞ ∞ 𝑒𝑒𝑥𝑥𝑒𝑒 −𝜂𝜂 2 4 𝑑𝑑𝜂𝜂 = 1 2 𝜋𝜋