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23-Chem-B4 Biochemical Engineering · December 2016

Question 2 of 5: Geometric Bioreactor Scale-Up — Four Impeller-Speed Criteria

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exam 04-Chem-B4, Biochemical Engineering — December 2016. 3 hours, Closed-Book Exam (one approved Casio or Sharp calculator model permitted). Per the exam notes, FIVE (5) questions constitute a complete paper and all five must be answered; most require a short-essay-format answer.

Reference texts: Shuler & Kargi, Bioprocess Engineering: Basic Concepts, 2nd ed.; Bailey & Ollis, Biochemical Engineering Fundamentals, 2nd ed.; Madigan et al., Brock Biology of Microorganisms, 13th ed.

Question 2: Geometric Bioreactor Scale-Up — Four Impeller-Speed Criteria (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

QuantitySymbolValue
Lab-scale working volumeV150 L
Plant-scale working volumeV210,000 L
Lab-scale impeller speedN1300 RPM
Geometry (both scales)H=3D, D=3Digeometric similarity

Find. The plant-scale impeller speed N2 under each of the four scale-up criteria (a)–(d).

Approach. Under geometric similarity every linear dimension of the plant vessel is the same multiple R of the lab vessel's, and because H=3D and D=3Di at both scales, that includes the tank volume itself (V∝D³), so R=(V2/V1)1/3. Each scale-up rule holds a different turbulent-regime dimensionless (or physical) group constant across scales — impeller Reynolds number Re=NDi²/ν, power per volume P/V∝N³Di² (constant power number), pumping capacity per volume Q/V∝N (constant flow number, Q∝NDi³), and impeller tip speed vtip=πNDi — giving four different power-law relations N2=N1Rp.

D = D₁H = H₁=3D₁Di = Di₁=D₁/3Lab scale, V₁ = 50 LN₁ = 300 RPMD = D₂=R·D₁H = H₂=3D₂Di = Di₂=D₂/3Plant scale, V₂ = 10,000 LN₂ = ?Geometric similarity: H = 3D, D = 3Di ⇒ R = D₂/D₁ = (V₂/V₁)^(1/3)
Fig. 2 — lab- and plant-scale vessels under geometric similarity (H=3D, D=3Di at both scales); the plant vessel's every linear dimension is R times the lab vessel's.
  1. Scale ratio. Since V∝D³ under this fixed-proportion geometry, $$R=\frac{D_2}{D_1}=\left(\frac{V_2}{V_1}\right)^{1/3}=\left(\frac{10{,}000}{50}\right)^{1/3}=200^{1/3}=\boxed{5.848}$$
  2. (a) Equal impeller Reynolds number. Re=NDi²/ν constant (same fluid) ⇒ N1Di1²=N2Di2²: $$N_2=\frac{N_1}{R^2}=\frac{300}{5.848^2}=\boxed{8.77\ \text{RPM}}$$
  3. (b) Equal power per unit volume. P∝N³Di&sup5; (constant power number, turbulent regime) and V∝Di³, so P/V∝N³Di² constant: $$N_2=\frac{N_1}{R^{2/3}}=\frac{300}{5.848^{2/3}}=\boxed{92.4\ \text{RPM}}$$
  4. (c) Equal liquid pumping rate per unit volume. Pumping capacity Q∝NDi³ (constant flow number), and since V∝Di³ too, Q/V∝N directly — the Di³ cancels exactly, so this criterion is scale-independent: $$N_2=N_1\cdot R^{0}=\boxed{300\ \text{RPM}}$$ This is a genuinely counter-intuitive result worth stating explicitly: matching pumping rate per unit volume requires the SAME impeller speed at any scale under geometric similarity, not a reduced one.
  5. (d) Equal impeller tip speed. vtip=πNDi constant ⇒ N1Di1=N2Di2: $$N_2=\frac{N_1}{R}=\frac{300}{5.848}=\boxed{51.3\ \text{RPM}}$$

The four criteria span more than a 30-fold range (8.8–300 RPM) for the identical scale-up, which is the central lesson of this problem: there is no single "correct" scale-up rule — the choice must be driven by whichever physical phenomenon (shear damage, gas–liquid mass transfer, blend/circulation time) actually limits performance in the specific process being scaled.

CriterionN2 at plant scale
Scale ratio, R5.848
(a) Equal impeller Reynolds number8.77 RPM
(b) Equal power per unit volume92.4 RPM
(c) Equal pumping rate per unit volume300 RPM (=N1)
(d) Equal impeller tip speed51.3 RPM