🧪 Plug-flow Reactor T-profile

A long pipe with chemicals flowing through it. As they travel, A slowly turns into B (the product you want), but if heat lingers, B keeps reacting and becomes C (waste). The pipe is split into ten sections, each with its own thermostat between 300 K and 480 K. Pick the ten temperatures that give you the most B coming out the far end.

The twist: heat speeds up both reactions, but it speeds the over-reaction (B → C) up more than the good reaction (A → B). So the textbook answer is a decreasing profile — start hot to make B quickly, then cool the rest of the pipe so it doesn't keep cooking into C. Classic chem-eng problem; 10-D continuous.

What's happening

Two first-order reactions in series with Arrhenius temperature dependence:

    A → B,  k₁(T) = exp[ B₁·(1 − T_ref/T) ],  B₁ =  6
    B → C,  k₂(T) = exp[ B₂·(1 − T_ref/T) ],  B₂ = 22

The two activation-energy parameters give k₂ much sharper temperature dependence than k₁ — a 100 K increase at T_ref = 400 K multiplies k₁ by ~5 but k₂ by ~80. Inside each axial zone the temperature is constant, so the linear ODE system has a closed-form solution and the integration is numerically stable at any rate.

Score is the mole fraction of B at the reactor outlet, expressed as a percentage. Cold all the way leaves most of the A unreacted; hot all the way burns everything through to C; the best isothermal temperature (~400 K) reaches B ≈ 37 %. A decreasing T-profile — hot to convert A early, cooled at the tail to lock in B — pushes past 42 %.

Real reactor design adds heat transfer, axial dispersion, catalyst deactivation, multi-reaction networks, and constraints on jacket temperature step changes. The structure — find a T(z) profile that maximises a downstream yield — is the same.

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