23-Chem-B4 Biochemical Engineering · Undated paper
Question 4 of 5: Geometric Scale-Up of a Cell-Culture Bioreactor
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 16-Chem-B4, Biochemical Engineering — May 2019 (the header on page 1 reads "16-Chem-B4/May 2019"). 3 hours, Closed-Book Exam (approved Casio or Sharp calculator
permitted). Per the exam notes, FIVE (5) questions constitute a complete paper and all five must be answered;
each question is of equal value (20 marks) and short-essay-format answers are marked for clarity and
organization.
Interpretation notes: Two points where the printed question itself needs an interpretation are flagged where they are used: the meaning of p in the Question 1 solubility equation, and the definition of the observable Thiele modulus in Question 3.
Question 4: Geometric Scale-Up of a Cell-Culture Bioreactor (20 marks)
Find. The large-vessel tank diameter T2, impeller diameter D2, and
agitation speed N2 under geometric similarity (constant H/T and constant D/T) with equal impeller
tip speed.
Fig. 3 — geometric scale-up from a 0.01 m³ bench vessel to a 50 m³
production vessel (schematic, not to true relative scale); every linear dimension scales by the same factor,
agitation speed scales inversely to preserve tip speed.
Approach. Geometric similarity means every linear dimension of the vessel (tank diameter,
impeller diameter, liquid height) scales by the same factor S, found from the volume ratio; then equal impeller
tip speed fixes the large-vessel agitation speed.
Linear scale factor from the volume ratio. Since $V\propto T^3$ under geometric similarity
(constant H/T),
$$S=\left(\frac{V_2}{V_1}\right)^{1/3}=\left(\frac{50}{0.01}\right)^{1/3}=(5000)^{1/3}$$
$$\boxed{S=17.10}$$
Large-vessel tank and impeller diameters. Every linear dimension scales by S, so the
impeller-to-tank ratio D/T=0.1/0.3=1/3 is automatically preserved:
$$T_2=S\,T_1=17.10\times0.3=\boxed{5.130\ \text{m}}$$
$$D_2=S\,D_1=17.10\times0.1=\boxed{1.710\ \text{m}}$$
(Check: $T_2/D_2=5.130/1.710=3.00=T_1/D_1$ — geometric similarity confirmed.)
Agitation speed from equal impeller tip speed. Impeller tip speed is $u_{tip}=\pi ND$;
setting $u_{tip,1}=u_{tip,2}$,
$$\pi N_1D_1=\pi N_2D_2\quad\Rightarrow\quad N_2=N_1\frac{D_1}{D_2}=\frac{N_1}{S}$$
$$N_2=\frac{100}{17.10}=\boxed{5.848\ \text{rpm}}$$
The tip speed held constant in both vessels is $u_{tip}=\pi\times(100/60)\times0.1=0.524$ m/s.