Recycle, bypass and purge stream calculations

Why it matters

Few reactors convert all their feed in one pass, and few products are made at exactly the right concentration. Recycle returns unreacted feed to the reactor, bypass blends part of a stream around a unit to hit a target composition or temperature, and purge bleeds off inerts that would otherwise build up. Ammonia and methanol loops, juice concentration and heat-exchanger temperature control all depend on these balances.

Key ideas

Recycle. Part of a downstream stream is returned and mixed with the fresh feed. The fresh feed and the net product are unchanged by recycle at steady state; what grows is the internal flow through the reactor and separator.

Two conversions.

• Single-pass conversion = (reactant in reactor feed − reactant in reactor outlet) / reactant in reactor feed.
• Overall conversion = (reactant in fresh feed − reactant in net process outlet) / reactant in fresh feed. With perfect separation and total recycle of unreacted reactant, the overall conversion is 100 % whatever the single-pass conversion; the price is a larger reactor feed and recycle compressor or pump.

Recycle ratio. Usually recycle flow / fresh feed flow, but some books use recycle / net product. State the definition you use.

Bypass. Part of the feed is split off before a unit and rejoins its outlet. The bypassed fraction keeps the feed composition; mixing it back with the processed stream gives an intermediate composition or temperature that can be adjusted by changing the split.

Purge. If the fresh feed contains an inert or impurity that the separator cannot remove, it recirculates and accumulates without limit unless part of the recycle is withdrawn. At steady state the purge must remove exactly the inert that enters: P·y_I = F·x_I,F. The purge also removes valuable reactant (it has the recycle composition), so a small purge means a high inert level in the loop and a large purge means a high reactant loss – a design trade-off.

Solution strategy.

1. Overall balance first: recycle and bypass streams are internal and cancel, often giving the net product and the purge directly.
2. Then the mixing point, reactor or separator, and splitter (all splitter outlets have the same composition).
3. Recycle problems often need simultaneous equations; avoid trial and error by writing the reactor balance in terms of the unknown recycle.

Formulas

• X_single-pass = (n_A,reactor in − n_A,reactor out) / n_A,reactor in
• X_overall = (n_A,fresh − n_A,process out) / n_A,fresh
• Recycle ratio = R / F (recycle / fresh feed; state the definition)
• Reactor feed = F + R – mixing point.
• Purge: P·y_I = F·x_I,F – steady state for an inert I.
• Total recycle, perfect separation: n_A,reactor in = n_A,fresh / X_single-pass
• Bypass: x_product = (B·x_F + C·x_C) / (B + C) – B bypass flow, C processed stream leaving the unit.

Worked examples

Example 1 (standard, bypass). 1000 kg/h of fresh juice with 14 % solids is to give a product with 42 % solids. Part of the juice bypasses an evaporator that concentrates the rest to 58 % solids, and the two streams are blended. Find the bypass flow and the water evaporated.

1. Overall solids balance: 0.14 × 1000 = 0.42·P, so P = 333.3 kg/h; water evaporated W = 1000 − 333.3 = 666.7 kg/h.
2. Evaporator: feed E, concentrate C = E − 666.7, solids 0.14·E = 0.58·C.
3. 0.14·E = 0.58·(E − 666.7), so 0.44·E = 386.7 and E = 878.8 kg/h.
4. Bypass = 1000 − 878.8 = 121.2 kg/h; C = 212.1 kg/h.
5. Check the blend: (0.14 × 121.2 + 0.58 × 212.1) / 333.3 = (17.0 + 123.0) / 333.3 = 0.42 ✓.

Answer: Bypass ≈ 121 kg/h; water evaporated ≈ 667 kg/h

Example 2 (GATE level, recycle with purge). A fresh feed of 100 mol/h contains 99 % A and 1 % inert I. The reaction A → B has a single-pass conversion of 25 %. All B is removed in the separator; the remaining A and I are recycled, with part purged so that the recycle contains 10 mol % I. Find the purge rate, product rate, recycle rate, overall conversion and inert fraction in the reactor feed.

1. Overall I balance: 1 = 0.10·P, so P = 10 mol/h (9 A + 1 I).
2. Overall A balance: 99 = 9 + (A reacted), so B produced = 90 mol/h; X_overall = 90 / 99 = 90.9 %.
3. Reactor A balance: A into reactor = 99 + 0.90·R; A leaving = 0.75 × (99 + 0.90·R), which must equal A in purge + recycle = 0.90 × (10 + R).
4. 74.25 + 0.675·R = 9 + 0.90·R, so 0.225·R = 65.25 and R = 290 mol/h.
5. Reactor feed = 100 + 290 = 390 mol/h; inert = 1 + 0.10 × 290 = 30 mol/h, i.e. 7.7 %.
6. Check single pass: A in = 99 + 261 = 360, reacted 0.25 × 360 = 90 ✓.

Answer: P = 10 mol/h, B = 90 mol/h, R = 290 mol/h, X_overall = 90.9 %, 7.7 % inert in reactor feed

Common mistakes

• Treating recycle as an extra input in the overall balance – it is internal.
• Confusing single-pass and overall conversion.
• Subtracting the purge from the reactor feed: purge leaves after the splitter, not before the mixer.
• Forgetting that the purge has the same composition as the recycle.
• Using a recycle-ratio definition different from the one in the question.

For GATE CH

Expect NAT questions on recycle flow for a given single-pass and overall conversion, purge rate for a given inert level, bypass fraction for a target composition, and ammonia or methanol loops. Practise the overall-balance-first approach and writing the reactor balance with the recycle as the unknown.

Quick check

1. With perfect separation and total recycle, what is the overall conversion?
2. Fresh feed 100 mol/h pure A, single-pass conversion 40 %, total recycle of A. What is the recycle flow?
3. Why must a loop with inert in its feed have a purge?
4. What is the composition of a purge stream compared with the recycle?
5. Does recycle change the net product rate at steady state?

Answers: 1. 100 %; 2. 150 mol/h (reactor feed 250 mol/h); 3. otherwise the inert accumulates without limit; 4. identical; 5. no – it changes internal flows only.