FE Chemical: Chemical Engineering study guide

The Chemical Engineering domain of the FE Chemical exam tests integrated use of balances, thermodynamics, transport, reaction engineering, control, safety, and economics in short self-contained problems. Marks usually turn on a clear basis, consistent units, and the correct limiting assumption.

Balances and units

Select a basis (1 h, 100 mol feed, or 1 kg) and state steady state. Write total and component balances; inert components pass through unless purged.
Track mole versus mass units. For gases use partial pressure or molar flow; at standard conditions 1 lbmol ideal gas occupies 379 ft³ and 1 kmol occupies 22.414 m³.
Define conversion X, selectivity, and yield against the correct limiting reactant. Recycle raises overall conversion only when a purge or separation prevents inert buildup.

Thermodynamics and transport

For open steady systems, Q − W_s = Σ out n H − Σ in n H; use enthalpy paths and latent heats for phase change. Vapor-liquid equilibrium for ideal mixtures uses Raoult's law, y_i P = x_i P_i^sat.
Mechanical energy balance with friction: P1/ρ + v1²/2 + gz1 = P2/ρ + v2²/2 + gz2 + h_f. Reynolds number Re = ρvD/μ; Darcy-Weisbach h_f = f(L/D)(v²/2).
Heat exchanger duty Q = U A ΔT_lm F; add fouling resistances. Mass transfer uses Fick's law and two-film resistances; absorption columns often use NTU/HTU.

Reaction, control, and safety

Arrhenius k = A exp(−Ea/RT). For liquid constant-density first order: CSTR V = F_A0 X/[k C_A0(1−X)]; PFR V = (v0/k) ln[1/(1−X)].
PID controllers: proportional gain, integral eliminates offset, derivative anticipates; dead time destabilizes.
Check flammability limits, relief scenarios, and HAZOP deviations before accepting a design.

Worked example

A liquid feed of 100 L/min with C_A0 = 2 mol/L enters a CSTR where A → B is first order with k = 0.10 min⁻¹. Find V for 80% conversion. At the outlet, C_A = 2(1 − 0.80) = 0.40 mol/L, so −r_A = kC_A = 0.040 mol/(L·min). F_A0 = 100 × 2 = 200 mol/min. The CSTR design equation gives V = F_A0 X/(−r_A) = 200 × 0.80/0.040 = 4000 L. A PFR would need only (100/0.10) ln(1/0.20) ≈ 1609 L because its rate is not evaluated at the outlet.

Traps

Mixing gauge and absolute pressure, or mass and molar flow.
Using the CSTR rate at feed conditions instead of outlet conditions.
Assuming ΔT_lm applies without the correction factor F in multipass exchangers.
Forgetting inert accumulation in recycle loops and purge design.