Residence time distribution: E and F curves

Why it matters

Real reactors are neither perfect plug flow nor perfectly mixed: fluid channels, short-circuits and stagnates in dead corners. A simple tracer test gives the residence time distribution (RTD), which diagnoses these faults and, combined with kinetics, predicts the conversion a real vessel will achieve.

Key ideas

Exit-age distribution E(t). Fluid elements leaving a vessel have spent different times inside. E(t)·dt is the fraction of the exit stream that has been inside for between t and t + dt. E has units of time⁻¹ and its total area is one: ∫E dt = 1. It is a property of the flow pattern (assuming steady flow and a closed vessel — fluid enters and leaves once, by plug flow at the boundaries).

Pulse experiment → E. Inject a quick pulse of tracer (amount M) into the feed and record the exit concentration C(t). Then E(t) = C(t)/∫C dt. A mass balance checks the test: M = v·∫C dt.

Step experiment → F. Switch the feed abruptly to a stream containing tracer at C0 and record C(t). F(t) = C(t)/C0 is the fraction of exit fluid younger than t. F rises from 0 to 1, never decreases, and F(t) = ∫(0→t) E dt, so E = dF/dt.

Mean and variance.

• Mean residence time t̄ = ∫t·E dt. For a closed vessel with constant density, t̄ = V/v = τ. If the measured t̄ is shorter than V/v, part of the volume is dead (stagnant); if the curve shows an early sharp peak, some fluid bypasses.
• Variance σ² = ∫(t − t̄)²·E dt = ∫t²·E dt − t̄² measures spread.

Dimensionless form. θ = t/t̄, Eθ = t̄·E(t), F(θ) = F(t). Plotting in θ lets vessels of different sizes be compared; σθ² = σ²/t̄².

Ideal reactors.

• PFR: E is a spike (Dirac delta) at t = τ; F is a step from 0 to 1 at τ; σ² = 0.
• CSTR: E = (1/τ)·exp(−t/τ); F = 1 − exp(−t/τ); σθ² = 1. About 63.2 % of the fluid leaves before τ.
• Laminar flow in a pipe (no diffusion): E = τ²/(2t³) for t ≥ τ/2 and 0 before; the first fluid leaves at τ/2.

Diagnosing faults. Early peak plus long tail → channelling or bypassing; t̄ less than V/v → dead zones; multiple peaks → internal recirculation; a delayed response → plug-flow section in series.

From RTD to conversion. For first-order reactions the RTD alone predicts the conversion exactly (next topic: segregation model). For other orders the RTD is necessary but not sufficient — the degree of micromixing also matters.

Formulas

E(t) = C(t) / ∫(0→∞) C dt (pulse tracer)

• C: exit tracer concentration (mol/m³ or kg/m³); E in s⁻¹.

F(t) = C_step(t)/C0 = ∫(0→t) E dt; E = dF/dt

t̄ = ∫ t·E dt = Σ t_i·C_i·Δt_i / Σ C_i·Δt_i (discrete, equal or unequal intervals)

σ² = ∫ t²·E dt − t̄² = Σ t_i²·C_i·Δt_i / Σ C_i·Δt_i − t̄²

• σ² in s²; t̄ in s.

t̄ = V/v (closed vessel, constant density, no dead volume)

θ = t/t̄, Eθ = t̄·E, σθ² = σ²/t̄²

E_CSTR = (1/τ)·exp(−t/τ), F_CSTR = 1 − exp(−t/τ)

E_laminar = τ²/(2t³) for t ≥ τ/2; F_laminar = 1 − τ²/(4t²) for t ≥ τ/2

M = v·∫ C dt (tracer balance for a pulse)

• M: tracer injected (mol or kg); v: volumetric flow (m³/s).

Worked examples

Example 1 (standard, pulse data). A pulse test on a vessel gives (t in min, C in g/m³): 0, 0; 5, 3; 10, 5; 15, 5; 20, 4; 25, 2; 30, 1; 35, 0. Find E, t̄ and σ².

1. Equal intervals Δt = 5 min: ∫C dt ≈ ΣC·Δt = (3 + 5 + 5 + 4 + 2 + 1) × 5 = 100 g·min/m³.
2. E = C/100: 0.03, 0.05, 0.05, 0.04, 0.02, 0.01 min⁻¹ at t = 5 … 30 min.
3. t̄ = Σt·C/ΣC = (15 + 50 + 75 + 80 + 50 + 30)/20 = 300/20 = 15 min.
4. Σt²·C/ΣC = (75 + 500 + 1125 + 1600 + 1250 + 900)/20 = 5450/20 = 272.5 min².
5. σ² = 272.5 − 15² = 47.5 min²; σθ² = 47.5/225 = 0.211.

t̄ = 15 min; σ² = 47.5 min². If the vessel has V = 2 m³ and v = 0.1 m³/min, V/v = 20 min > 15 min, so about (20 − 15)/20 = 25 % of the volume is dead.

Example 2 (GATE level, ideal curves). For an ideal CSTR with τ = 10 min, find (a) the fraction of exit fluid that stayed less than 5 min, (b) the fraction that stayed more than 20 min. For a laminar-flow tube with τ = 10 min, find the fraction younger than 10 min.

1. (a) F(5) = 1 − exp(−5/10) = 1 − 0.607 = 0.393.
2. (b) 1 − F(20) = exp(−20/10) = 0.135.
3. Laminar: F(10) = 1 − τ²/(4t²) = 1 − 100/400 = 0.75.

CSTR: 39.3 % leaves within 5 min and 13.5 % stays longer than 20 min; laminar tube: 75 % leaves before t = τ.

Common mistakes

• Forgetting to normalise C(t) before calling it E, or normalising with ΣC instead of ΣC·Δt.
• Treating F as a probability density — F is cumulative and dimensionless; E has units of 1/time.
• Assuming t̄ always equals V/v: dead zones and open-vessel boundaries break this.
• Using unequal time intervals without weighting by Δt.
• Thinking a measured RTD fixes conversion for all kinetics — only for first order.

For GATE CH

Expect NAT questions on t̄ and σ² from tabulated pulse data, F or E values for an ideal CSTR or PFR (or combinations in series), the fraction of fluid in a given age band, dead-volume fraction from t̄ vs V/v, and interpreting curve shapes. Practise the discrete summations quickly and remember the exponential results for the CSTR.

Quick check

1. What is the area under any E curve?
2. For a CSTR with τ = 4 min, what is E at t = 0?
3. A vessel with V/v = 10 min shows t̄ = 8 min. What fraction is dead?
4. What is σθ² for an ideal CSTR and for a PFR?

Answers: 1. one; 2. 1/τ = 0.25 min⁻¹; 3. 0.2 (20 %); 4. 1 and 0.