Recycle and purge: the boundary never sees what the equipment carries

Fresh feed goes in at 120 kmol an hour and product and purge come out at 120, so the plant balance closes on the figure it opened with. The reactor inlet reads 444. Recycle never crosses the outer boundary, so no overall balance can show it, and the purge fraction that stops the inert piling up is the same figure that sets how hard the loop circulates. Type four numbers and the whole loop solves in one pass, with no separator to configure and nothing looked up.

Live result

The reactor inlet reads 444.0 kmol/h, a circulation multiplier of 3.70× on the fresh feed.

fresh feed in120.0 kmol/h
inert held in the loop6.67%

None of that circulation appears in the balance across the boundary, which still closes.

Try a scenario

Fresh feedkmol/h

Inert in the fresh feed%

The mole fraction of fresh feed that is inert.

Conversion per pass%

The share of inlet reactant consumed in one pass.

Single pass 20.0% across the reactor. Overall 71.43% across the process, because what the reactor misses comes back and gets another pass.

Purge fraction%

The share of loop stream that leaves the process.

flow · against fresh feed

fresh feed	120.0 kmol/h	1.00×
reactor inlet	444.0 kmol/h	3.70×
recycle	324.0 kmol/h	2.70×
purge	36.0 kmol/h	0.30×
product	84.0 kmol/h	0.70×
Inert in the loop	24.0 kmol/h	0.20×

Accounting assumption

The separation between the reactor and the splitter is an accounting assumption and not a piece of equipment. Everything the reaction made leaves as product, and everything else goes back to the splitter whole. That is the only separation this page can describe, because a real one is a consequence of properties: no volatility, no solubility, no equilibrium constant, no density, no molecular weight. There is no efficiency to set, no stage count and no split fraction, and the four figures you type are the only numbers here that did not come from arithmetic on those four figures.

The splitter is arithmetic as well. Purge and recycle leave one stream through two openings, so they carry the same composition as each other and as the stream they came from. Nothing on this page can make those two compositions differ, and that single fact is what ties the purge fraction to the inert concentration.

inside the loop

120.0 + 324.0 = 444.0

fresh feedrecycle

reactor inlet

across the boundary

120.0 = 84.0 + 36.0

fresh feed

productpurge

circulation 3.70×

loop inert 6.67%

steady

Horizontal: the three places a flow is counted: entering the process, leaving it, and entering the reactor
Vertical: molar flow, in kilomoles an hour
Units: Kilomoles an hour to one decimal place; compositions and ratios to two, which is the precision the four typed figures support
Assumptions: The separation is the accounting assumption stated above the controls; purge and recycle carry the same composition; the loop is at steady state

Fixed panels

purge 5.0%

purge 10.0%

purge 25.0%

purge 50.0%

What the loop costs to keep clean

The purge is the only door the inert has, and a splitter cannot change a composition. Make the door smaller and the loop has to hold more inert behind it to push the same amount through.

The reactant pays for it twice: a smaller purge sends less reactant out unconverted and raises the reactor inlet that every pump, exchanger, vessel and line inside the loop must carry. Neither movement is visible in the overall balance, which is why sizing anything inside a loop needs a second boundary around the reactor.

Here is the maths

Scale: The four fixed panels share one scale, 0.0 to 560.0 kmol/h, and it does not move. The live chart above rescales with what you type.
Basis: One reference reactant and one inert, over one hour of steady running, on the boundary named beside each column
Model version: 1.0

The formula, in words

A species with one door has one steady rate through it. The inert enters only in the fresh feed and leaves only in the purge, so at steady state those two rates are equal: $I_L = F \times \frac{z}{p}$ , where $I_L$ is inert in the loop stream, $F$ fresh feed, $z$ the inert fraction, and $p$ the purge fraction. That single division fixes the inert inventory of the whole loop before any internal flow is calculated.
A species with two doors splits between them. The reference reactant leaves either by reacting or in the purge, so $F_A = f \times A_I + p \times A_u$ , where $F_A$ is fresh feed reactant, $f$ conversion per pass, $A_I$ reactor inlet reactant, and $A_u$ unconverted reactant. This rearranges to $A_I = \frac{F_A}{f+p(1-f)}$ . The bracket is the only group on this page that appears more than once.
Both balances are written on the outer boundary, where the recycle does not exist. That is why this loop needs no iteration. Everything inside is then arithmetic: the loop stream is the unconverted reactant plus the inert, the recycle and the purge are two fractions of it, and the reactor inlet is the fresh feed plus the recycle.
The circulation multiplier is the reactor inlet divided by the fresh feed, and it is also one plus the recycle to feed ratio. In closed form it is $\frac{1-z}{f+p(1-f)}+\frac{z}{p}$ , which contains no flow at all. Doubling the plant does not move it.
Overall conversion is the single pass conversion divided by that same bracket. It contains neither the feed nor the inert fraction. Single pass conversion answers what the reactor did; overall conversion answers what the process did with the raw material; and a conversion figure reported without its boundary belongs to neither question.

I_L = F \times \frac{z}{p}

The denominator:

D = f + p \times (1 - f)

The reactor inlet reactant:

A_I = \frac{F_A}{D}

The circulation multiplier:

C = \frac{G}{F} = 1 + \frac{R}{F}

where

G

is reactor inlet and

R

recycle.

X_O = \frac{f}{D}

where

X_O

is overall conversion.

A worked example

A steady process runs on one reference reactant. Fresh feed is 120.0 kmol/h, of which 2.0% is an inert that takes no part in the reaction, so the feed contains 2.4 kmol/h inert and 117.6 kmol/h reactant. The reactor consumes 20.0% of the reactant reaching it on every pass.

Every figure below is kmol/h and the reaction is written one mole to one mole, so the product stream molar rate equals the reactant consumed.

Case A, purge 10.0%

z = 0.020\qquad f = 0.200\qquad p = 0.100\qquad D = 0.200 + 0.100 \times 0.800 = 0.280

\begin{gathered}z = 0.020 \\ f = 0.200 \\ p = 0.100 \\ D = 0.200 + 0.100 \times 0.800 = 0.280\end{gathered}

Step	Arithmetic	Result
reactant in fresh feed	$120.0 \times 0.980$	117.6
inert in fresh feed	$120.0 \times 0.020$	2.4
reactant at the reactor inlet	$117.6 \div 0.280$	420.0
inert in the loop stream	$2.4 \div 0.100$	24.0
reactor inlet, total	$420.0 + 24.0$	444.0
reactant handed to the splitter	$0.800 \times 420.0$	336.0
loop stream	$336.0 + 24.0$	360.0
recycle	$0.900 \times 360.0$	324.0
purge	$0.100 \times 360.0$	36.0
product	$0.200 \times 420.0$	84.0

reactor inlet by the mixing point $120.0 + 324.0 = 444.0$

reactor inlet by its components $420.0 + 24.0 = 444.0$

overall boundary $84.0 + 36.0 = 120.0$

purge reactant and inert $33.6 + 2.4 = 36.0$

circulation $444.0 \div 120.0 = 3.70\times$

circulation $1 + 324.0 \div 120.0 = 1 + 2.70 = 3.70\times$

circulation $0.980 \div 0.280 + 0.020 \div 0.100 = 3.500 + 0.200 = 3.70\times$

loop inert $24.0 \div 360.0 = 6.67\%$

reactor inlet inert $24.0 \div 444.0 = 5.41\%$

overall conversion $84.0 \div 117.6 = 71.43\%$ and $0.200 \div 0.280 = 71.43\%$

single pass conversion 20.00%

reactant lost unconverted 33.6 kmol/h and $33.6 \div 117.6 = 28.57\%$

Case B, purge 5.0%

D = 0.200 + 0.050 \times 0.800 = 0.240

Step	Arithmetic	Result
reactant at the reactor inlet	$117.6 \div 0.240$	490.0
inert in the loop stream	$2.4 \div 0.050$	48.0
reactor inlet, total	$490.0 + 48.0$	538.0
reactant handed to the splitter	$0.800 \times 490.0$	392.0
loop stream	$392.0 + 48.0$	440.0
recycle	$0.950 \times 440.0$	418.0
purge	$0.050 \times 440.0$	22.0
product	$0.200 \times 490.0$	98.0

mixing point $120.0 + 418.0 = 538.0$

components $490.0 + 48.0 = 538.0$

overall boundary $98.0 + 22.0 = 120.0$

purge reactant and inert $19.6 + 2.4 = 22.0$

circulation $538.0 \div 120.0 = 4.48\times$

loop inert $48.0 \div 440.0 = 10.91\%$

overall conversion $98.0 \div 117.6 = 83.33\%$

Halving the purge bought 14.0 kmol/h more product and cost 94.0 kmol/h more circulation. The inert fraction went from 6.67% to 10.91% because the same 2.4 kmol/h now has to leave through a door half the size. The page prints both movements and picks neither.

Case C, purge 25.0%

D = 0.200 + 0.250 \times 0.800 = 0.400

Step	Arithmetic	Result
reactant at the reactor inlet	$117.6 \div 0.400$	294.0
inert in the loop stream	$2.4 \div 0.250$	9.6
reactor inlet, total	$294.0 + 9.6$	303.6
reactant handed to the splitter	$0.800 \times 294.0$	235.2
loop stream	$235.2 + 9.6$	244.8
recycle	$0.750 \times 244.8$	183.6
purge	$0.250 \times 244.8$	61.2
product	$0.200 \times 294.0$	58.8

mixing point $120.0 + 183.6 = 303.6$

components $294.0 + 9.6 = 303.6$

overall boundary $58.8 + 61.2 = 120.0$

purge reactant and inert $58.8 + 2.4 = 61.2$

product and purged reactant $58.8 = 58.8$

circulation $303.6 \div 120.0 = 2.53\times$

loop inert $9.6 \div 244.8 = 3.92\%$

overall conversion $58.8 \div 117.6 = 50.00\%$

At a quarter purge, exactly half the fresh reactant becomes product and exactly half leaves unconverted: 58.8 and 58.8. The same 2.4 kmol/h of inert is still leaving through the purge, because that is the only thing about the inert a steady state can require.

Purge 0.0%

The inert enters at 2.4 kmol/h and has no exit. There is no steady inventory, so there is no steady concentration, no steady loop stream, no steady recycle and no multiplier, and the page prints none of them.

Check it yourself

With no inert there is no species without an exit, and the loop solves at total recycle with an overall conversion of 100.00% and a multiplier of $\frac{1}{f}$ . At the shipped 20.0% per pass that is $5.00\times$ , and the reactor inlet is 600.0 kmol/h against a fresh feed of 120.0.

What this does not model

Energy, in any form: this is a material balance and a loop that closes on moles can be running away on temperature. Every property a real separation depends on, none of which this page holds. Any real separator, its efficiency, its stages, its own recycle and the reactant it would recover from the purge. Pressure and compression duty. Reaction stoichiometry, selectivity, side products, catalyst and catalyst decay. The dynamics: this page solves the settled state and says nothing about how long settling takes or what the loop does on the way. Measurement, which is the usual reason a real loop's figures do not reconcile and the one thing arithmetic cannot tell apart from a stream nobody mentioned.

Recycle, purge and the distinction between single pass and overall conversion are standard chemical engineering, set out in Felder, Rousseau and Bullard, Elementary Principles of Chemical Processes, fourth edition; this page describes that material in its own words, reproduces no text, problem, figure or table from it, and its worked example is original. A loop that solves is an arithmetic result and not a statement that the purge fraction, the conversion or the feed behind those figures was the right one to choose.