Worked problem

Recycle and Purge: Preventing Inert Buildup

Calculate purge, recycle and reactor feed flows for a process with 30% single-pass conversion and a 20% inert limit, and close the inert balance.

Chemical & Process · Last reviewed

Given

Fresh feed is 100 kmol/h: 99 kmol/h of reactant A and 1 kmol/h of an inert gas I. In the reactor A → B (one mole of B per mole of A) with a single-pass conversion of 30%. The separator removes all of the product B as a liquid, and every other component leaves as one gas stream. A purge takes part of that gas away. The rest is recycled to the reactor inlet. To keep the loop workable, the inert may not exceed 20 mole % of the gas stream.

Find

The purge flow, the gas flow leaving the separator, the recycle flow, the reactor feed and the product rate.

Solution

  1. Inert balance, overall: the inert enters at 1 kmol/h and can only leave in the purge. At 20% inert in the purge, the purge flow P = 1 ÷ 0.20 = 5.0 kmol/h.
  2. The purge also carries A: 5.0 × 0.80 = 4.0 kmol/h of A is lost in the purge.
  3. Overall A balance: 99 = B produced + 4.0, so the product B = 95.0 kmol/h.
  4. Let G be the gas flow from the separator, with 20% inert and 80% unreacted A. The A leaving the reactor is a(1 − 0.30) = 0.80 G, so the reactor inlet A is a = 0.80 G ÷ 0.70 = 1.1429 G.
  5. A balance at the mixing point: a = 99 + (recycle fraction) × 0.80 G, where the recycle is G − 5.0. So 1.1429 G = 99 + 0.80 G − 4.0, which gives G = 95 ÷ 0.3429 = 277.1 kmol/h.
  6. Recycle = G − P = 277.1 − 5.0 = 272.1 kmol/h. Reactor feed = 100 + 272.1 = 372.1 kmol/h, made up of 316.7 of A and 55.4 of inert.

Answer

purge 5.0 kmol/h | gas 277.1 | recycle 272.1 | reactor feed 372.1 | product 95.0 kmol/h

What it shows

  • Without a purge the inert has no way out, so it accumulates until it dilutes the reactant and stops the reaction.
  • The purge is small (1.8% of the gas stream) but it removes all of the inert and 4% of the reactant.
  • A smaller allowed inert fraction needs a larger purge and loses more reactant. At 10% inert the purge doubles to 10 kmol/h.
  • The recycle flow is almost three times the fresh feed. Its size is set by the low per-pass conversion.

Common mistakes

  • Writing a balance on the reactor alone and forgetting that the inert recirculates.
  • Setting the purge from the product flow instead of the inert balance.
  • Choosing the inert limit without a reason. In practice it comes from reaction rate, equilibrium, pressure and equipment limits.
  • Forgetting that the purge composition equals the recycle composition (both come from the same gas stream).

Related tools and labs

Calculators, labs and simulators are part of the All-Access Pass.

More worked problems

Questions

Why not use a perfect separator for the inert too?

It can be done, for example with a membrane or an absorber, but it adds cost. A purge is the simplest way to stop the inert from building up.

Does the conversion per pass change the purge?

The purge in this simplified case depends only on the inert limit and the inert in the fresh feed. The conversion changes the recycle and the reactor size.

Can the purge be recovered?

Often yes. Plants may burn it for heat, treat it or recover the reactant from it.

Educational material. Examples use simplified, hypothetical inputs and are not a design, a construction document or professional engineering advice. Confirm requirements with the codes and standards adopted where the work is built and with a licensed engineer.

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