10.3 Batch and CSTR Reactors

10.3 Batch and CSTR Reactors

Quick Answer: Batch (const. V): integrate time vs conversion from the rate law. CSTR: well mixed; (V = F_{A0} X / (-r_A)_{\mathrm{exit}}). High conversion in one CSTR needs large V because the whole vessel sits at the low exit rate. Prefer batch for flexible/small campaigns; CSTR for steady continuous flow.

Rate laws (10.1) and Arrhenius k(T) (10.2) become equipment sizes through design equations. This section covers batch and continuous stirred-tank (CSTR) reactors—the two workhorses of conceptual CRE on licensing exams and common in specialty chemicals, treating, and many liquid-phase processes.

Constant-Volume Batch Design Equation

In an ideal batch reactor, charge reactants, run for time t, then discharge. No inflow/outflow during reaction. For species A:

[ N_A = N_{A0}(1-X_A) ]

The mole balance is accumulation = generation:

[ \frac{dN_A}{dt} = r_A V = -(-r_A)V ]

With conversion:

[ N_{A0}\frac{dX_A}{dt} = (-r_A)V \quad \Rightarrow \quad t = N_{A0}\int_{0}^{X_A}\frac{dX}{(-r_A)V} ]

Constant volume (good for most liquid-phase batches; also gas-phase if V fixed and you track concentration carefully):

[ t = C_{A0}\int_{0}^{X}\frac{dX}{(-r_A)} \quad \text{or equivalently} \quad t = \int_{C_{A0}}^{C_A}\frac{dC_A}{r_A} ]

with C_A = C_{A0}(1−X) when volume and density are constant.

Exam intuition: first-order time to high conversion grows as −ln(1−X)—the last few percent take longer, but not as brutally as a CSTR. Second-order stretches even more at high X because rate collapses with C_A².

Worked batch example (first order)

Liquid batch, constant density, first-order k = 0.5 h⁻¹, target X = 0.80:

[ t = \frac{1}{0.5}\ln\frac{1}{1-0.80} = 2\ln 5 \approx 2\times 1.609 = 3.22,\mathrm{h} ]

For X = 0.95:

[ t = 2\ln(1/0.05) = 2\ln 20 \approx 6.0,\mathrm{h} ]

Roughly double the time to go from 80% to 95% conversion versus 0 → 80%—the integral nature of high conversion.

Industrial notes: batch cycle time also includes fill, heat-up, cool-down, and cleanout—exam design equations usually mean reaction time only unless the stem expands the definition. Multiproduct fine-chemical plants favor batch flexibility.

Ideal CSTR: Well-Mixed Implication

A continuous stirred-tank reactor (CSTR) has continuous feed and effluent. Ideal CSTR assumptions:

1. Perfect mixing—uniform composition and temperature in the vessel.
2. Exit stream composition = reactor interior composition.
3. Steady state: accumulation = 0.

Therefore the rate (−r_A) used in the design equation is evaluated at exit (and interior) conditions—often the lowest reactant concentration and thus the slowest rate in the vessel for positive-order kinetics.

Steady mole balance on A:

[ F_{A0} - F_A + r_A V = 0 ]

With F_A = F_{A0}(1−X) and r_A = −(−r_A):

[ F_{A0} X = (-r_A)_{\mathrm{exit}} V ]

[ \boxed{V = \frac{F_{A0} X}{(-r_A)_{\mathrm{exit}}}} ]

Space time τ = V/v₀ (v₀ = inlet volumetric flow) is the usual sizing companion. For constant density, C_A = C_{A0}(1−X) and τ = C_{A0} X / (−r_A exit).

Worked CSTR example (first order, const. density)

(−r_A) = k C_A = k C_{A0}(1−X). Then:

[ \tau = \frac{V}{v_0} = \frac{C_{A0} X}{k C_{A0}(1-X)} = \frac{X}{k(1-X)} ]

With k = 0.5 h⁻¹ and X = 0.80:

[ \tau = \frac{0.80}{0.5\times 0.20} = \frac{0.80}{0.10} = 8.0,\mathrm{h} ]

Compare to the batch reaction time 3.22 h for the same k and X (ignoring downtime). The single CSTR needs much more holding time because it operates entirely at C_A = 0.20 C_{A0}, where the rate is only 20% of the feed rate value.

For X = 0.95:

[ \tau = \frac{0.95}{0.5\times 0.05} = 38,\mathrm{h} ]

Volume (or τ) explodes as X → 1 for a single CSTR with positive-order kinetics—a major exam takeaway.

When Batch vs CSTR Is Preferred

| Prefer batch when… | Prefer CSTR when… | |---|---|---| | Small production rates / specialty chemicals | Large continuous throughput | | Many products on the same equipment | Single steady product slate | | Recipe flexibility, quality holds | Tight steady control, automation | | Solids handling, multiphase lab-to-plant mimic | Easy heat management with jackets/coils at uniform T | | Very slow reactions where continuous V would be huge and campaigns fit a schedule | Mixing quality and uniform T are critical |

CSTR advantages: simple construction, good temperature uniformity (mixing), easy to instrument, continuous product. CSTR disadvantages for kinetics: large volume for high conversion with n > 0; poor for series reactions when intermediate is desired (low intermediate concentration environment—selectivity often better in PFR/batch; intro in 10.4).

Batch advantages: high conversion without giant volume for a given rate law (time replaces volume), flexibility. Batch disadvantages: downtime between charges, labor, batch-to-batch variability, scale-up heat-transfer limits.

High Conversion in a CSTR Means Large Volume

Graphical Levenspiel view (conceptual): plot 1/(−r_A) vs X.

• CSTR volume ∝ rectangle height = F_{A0} × [1/(−r_A) at exit X] × X
• PFR/batch “volume·time” ∝ area under the 1/(−r_A) curve

For kinetics where (−r_A) falls as X rises (usual positive order), 1/(−r_A) rises with X. The CSTR rectangle is tall at high X, so V is large. That is why plants use PFR, batch, or CSTR-in-series instead of one huge tank when high conversion is required (Section 10.4).

Zero-order exception (recognition): if (−r_A) = k until A vanishes, rate does not fall with concentration (while the form holds), so the CSTR penalty is milder—still know the general positive-order story for exams.

UPDA Exam Checklist for Section 10.3

1. Batch const. V: t from integrating dX/(−r_A) (or C–t integrals).
2. First-order batch: t = (1/k) ln[1/(1−X)].
3. CSTR: V = F_{A0} X / (−r_A exit); evaluate rate at exit.
4. Well mixed ⇒ no concentration gradient inside the ideal tank.
5. Single CSTR at high X is volume-hungry for n > 0.
6. Match reactor type to campaign size, continuity, and conversion needs.

Section 10.4 develops the PFR integral and systematic reactor selection (PFR vs CSTR, series tanks, selectivity intro).