Reactors in series and parallel

Why it matters

Real plants rarely use one reactor. Cascades of stirred tanks cut the volume penalty of mixed flow, parallel trains add capacity and redundancy, and combinations of PFRs and CSTRs appear wherever heat removal or staging matters. Knowing how to combine them — and in what order — avoids oversizing and lost conversion.

Key ideas

Plug flow reactors in series. N PFRs in series with total volume V behave exactly like a single PFR of volume V: the fluid simply continues along a longer tube. Space times add.

PFRs (or any identical reactor type) in parallel. For the best overall conversion, split the feed so that every branch has the same space time, V_i/v_i = constant. Then each branch reaches the same conversion and the system behaves like one reactor of the total volume. Any unequal split gives a lower mixed exit conversion. The same rule holds for CSTRs in parallel.

CSTRs in series. Each tank operates at its own exit concentration, which falls stage by stage. For positive-order kinetics the total volume of N tanks is smaller than one CSTR for the same conversion, and as N → ∞ the cascade approaches a PFR. For first order with equal tanks, the overall result has a closed form. Two to four tanks capture most of the saving.

Defining conversion in a series. The overall conversion is based on the original feed: XN = 1 − CN/C0. If each stage i converts a fraction xi of what enters it, then 1 − XN = (1 − x1)(1 − x2)…(1 − xN). Do not add stage conversions.

Unequal-size CSTRs in series. For first order, the arrangement order does not matter. For n > 1, put the smaller tank first; for n < 1, put the larger tank first. For the best total volume of two tanks for a given conversion, the rectangles on a Levenspiel plot are chosen to maximise the area saved; graphical methods handle arbitrary kinetics (plot −rA vs CA and step from tank to tank with lines of slope −1/τi).

Mixed arrangements (PFR + CSTR). For first order, the order of arrangement does not affect the final conversion. For n > 1, keep concentrations high where the reaction is fastest: PFR first, then the CSTR (more generally: PFR, small CSTR, large CSTR). For n < 1 do the reverse. The reasoning: a CSTR "wastes" less when the concentration is already low if the rate is strongly concentration-dependent.

Practical reasons. Series arrangements allow inter-stage heating, cooling or feed addition; parallel trains allow maintenance on one train while others run and let a plant expand in steps.

Formulas

τ_total = Σ τ_i (PFRs in series; behaves as one PFR of total volume)

V_i/v_i = V/v0 for every branch (optimum parallel split)

• V_i: branch volume (m³); v_i: branch feed rate (m³/s).

C_N/C_0 = 1/(1 + k·τ_i)^N (N equal CSTRs, first order, constant density)

• τ_i = V_i/v0: space time per tank (s); k in s⁻¹.

N·k·τ_i = N·[(C_0/C_N)^(1/N) − 1] (total kτ for N equal tanks)

lim(N→∞) C_N/C_0 = exp(−k·τ_total) (cascade → PFR)

1 − X_N = Π(1 − x_i) (overall vs per-stage conversion)

k·τ_i·C_i² = C_(i−1) − C_i (stage i, second order −rA = k·CA²)

Worked examples

Example 1 (standard). A first-order liquid reaction (k = 0.1 min⁻¹, v0 = 1 m³/min) needs XA = 0.9. Compare one CSTR, two equal CSTRs in series and a PFR.

1. One CSTR: kτ = X/(1 − X) = 9 → τ = 90 min → V = 90 m³.
2. Two equal CSTRs: C2/C0 = 0.1 = 1/(1 + kτi)² → 1 + kτi = √10 = 3.162 → kτi = 2.162 → τi = 21.62 min.
3. Total V = 2 × 21.62 × 1 = 43.2 m³.
4. PFR: τ = ln 10/k = 23.0 min → V = 23.0 m³.

One CSTR 90 m³; two CSTRs 43.2 m³ total; PFR 23.0 m³.

Example 2 (GATE level, order of arrangement). A second-order liquid reaction (−rA = k·CA²) is processed in a PFR and a CSTR, each with k·CA0·τ = 1. Find the exit conversion for (a) PFR then CSTR, (b) CSTR then PFR. Work with c = CA/CA0.

(a) PFR: 1/c1 − 1 = 1 → c1 = 0.5. CSTR: c2² = c1 − c2 (since k·CA0·τ = 1) → c2² + c2 − 0.5 = 0 → c2 = (−1 + √3)/2 = 0.366. X = 0.634.

(b) CSTR: c1² + c1 − 1 = 0 → c1 = (−1 + √5)/2 = 0.618. PFR: 1/c2 = 1/c1 + 1 = 2.618 → c2 = 0.382. X = 0.618.

PFR first gives X ≈ 0.634; CSTR first gives X ≈ 0.618 — for n > 1, keep the plug-flow section where concentration is high.

Example 3 (stage conversions). Three reactors in series convert 50 %, 60 % and 80 % of what enters each. Overall: 1 − X = 0.5 × 0.4 × 0.2 = 0.04 → X = 0.96.

Common mistakes

• Adding stage conversions (0.5 + 0.6 + 0.8) or averaging them.
• Using the overall feed CA0 instead of the stage inlet concentration in a stage balance.
• Splitting parallel feed equally between unequal reactors instead of in proportion to volume.
• Assuming the arrangement order never matters — it matters for every order except first.
• Taking the wrong (negative or > 1) root of the stage quadratic.

For GATE CH

Expect NAT problems on N equal CSTRs for first order (conversion or total volume), comparison with a PFR, overall conversion from stage conversions, PFR–CSTR arrangement order for second-order kinetics, and the optimum split of feed between parallel reactors. Practise the quadratic for second-order stages quickly.

Quick check

1. Two PFRs of 2 m³ and 3 m³ in series behave like what?
2. Two parallel PFRs of 1 m³ and 3 m³: what fraction of the feed should go to the larger one?
3. Two equal CSTRs, first order, kτi = 1 each: overall conversion?
4. For n = 2 and two unequal CSTRs, which goes first?

Answers: 1. one PFR of 5 m³; 2. 3/4; 3. 1 − 1/4 = 0.75; 4. the smaller one.