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Engineering & Technology · Reaction Engineering, Separations & Transport

Batch, flow and mixed-reactor models

A stirred tank and a long tube can contain the same liquid volume and receive the same feed, yet produce different outlet compositions. The difference can come from how fluid elements encounter concentration over their time in the reactor.

Use Process Balances & Design for amount, flow, energy and boundary accounts. Review material structure and properties. Chemistry’s reaction rates and equilibrium explain the molecular and thermodynamic ideas further. Exponentials are evaluated explicitly; each model states its scope before calculation.

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Compare ideal batch, continuously mixed and plug-flow models without turning a conditional result into a universal reactor ranking.

01

Specify what enters, leaves and mixes

An ideal batch reactor contains a charge during reaction, with no material flow across its boundary in that interval. A continuously stirred tank reactor, CSTR, has feed and outlet flow. Its ideal mixing assumption makes outlet composition equal to the uniform composition inside. Steady state means the amount of each species inside is no longer changing, even though reaction continues.

An ideal plug-flow reactor, PFR, carries fluid along a path with no axial backmixing and with uniform conditions across each cross-section. Composition can change along the path. These are models of flow and reaction, not labels that real vessels automatically satisfy. Bypass, dead regions, dispersion and heat gradients can invalidate a simple idealisation.

02

Apply the same chemistry and feed

Use an original constant-density liquid model A → B with first-order consumption k C_A, k=0.20 min⁻¹. Take feed C_A0=1.00 mol/L, total active volume 100 L and volumetric flow 20 L/min. Space time τ=V/q is 5 min. In a general real flow system this nominal volume-to-flow ratio need not describe every fluid element’s residence time.

For an ideal steady CSTR, feed minus outlet minus consumption is zero: q C_A0−q C_A−k C_A V=0. Therefore C_A/C_A0=1/(1+kτ), and conversion X=1−C_A/C_A0 is 0.50. For an ideal isothermal PFR under the stated constant-density assumptions, C_A/C_A0=exp(−kτ), giving X≈0.6321. A batch at reaction time 5 min has that same fraction in this particular model.

Same total space time, different mixingHorizontal kτ is dimensionless. At kτ = 1: one CSTR has X = 0.5; two equal CSTRs have X = 0.555556; plug flow or batch has X = 0.632121. Curves span kτ from 0 to 3, all using equal total space time or batch reaction time under the stated model.Same total space time, different mixingConversion X / fraction012300.250.50.751Orange: PFR / batchViolet: two equal CSTRsBlue: one CSTRDimensionless kτOriginal supplied teaching model
Horizontal kτ is dimensionless. At kτ = 1: one CSTR has X = 0.5; two equal CSTRs have X = 0.555556; plug flow or batch has X = 0.632121. Curves span kτ from 0 to 3, all using equal total space time or batch reaction time under the stated model.
03

Interpret the contact history

The CSTR feed immediately joins fluid at the lower outlet concentration, so the entire model volume reacts at that lower concentration. The PFR begins at the higher inlet concentration and declines along its length. A positive first-order dependence makes that earlier high concentration contribute a higher rate. This reasoning explains the result rather than declaring that a tube is always better.

Different rate laws, reversibility, selectivity, temperature effects, controllability and service requirements can change the comparison. Batch production also includes charging, unloading, cleaning and other time outside the reaction interval. Comparing only reaction time with continuous throughput would hide those tasks. Identify the required product, flow history and evidence before selecting actual equipment.

Divide the same volume into two mixed stages

Put two ideal 50 L CSTRs in series at the same 20 L/min flow. Each has space time 2.5 min and kτ_stage=0.50. The first gives C_A1=1/1.5=0.6667 mol/L; the second gives C_A2=0.6667/1.5≈0.4444 mol/L. Overall conversion is 1−0.4444≈0.5556.

The two-stage value lies between 0.50 for one tank and 0.6321 for plug flow under these assumptions. Both tanks together still contain 100 L; using five minutes in each would silently double the total volume. The diagram plots conversion against the dimensionless product kτ so that the conditional comparison and the equal-total-space-time basis remain visible.

Try the reasoning

  1. Does steady state mean that reaction has stopped?

    Show the answer and reasoning

    No. Inflow, outflow and reaction can balance while the stored amount stays constant.

  2. Is every molecule in an ideal CSTR exactly five minutes old?

    Show the answer and reasoning

    No. V/q is a space time, while complete mixing produces a distribution of residence times.

  3. Why must the two tanks each use half the total space time here?

    Show the answer and reasoning

    Their volumes divide the same 100 L at the same flow; otherwise the comparison would use different total capacity.

What to carry forward

Compare reactors through their stated balances, mixing histories and equal service; keep the assumptions beside the ranking.

References

Teaching resources and research supporting this lesson’s explanations and examples.

  1. 12.4 Integrated Rate LawsOpenStax · Chemistry 2eTeaching resourceopenstax.orgRead source
  2. Chemical Reaction Engineering, Lecture 1 (2013 lecture set), slides18–31H. Scott Fogler · University of MichiganTeaching resource · PDFwebsites.umich.eduRead source