Question 6 of 10: Unsteady-State Mass Balances on a Continuous Stirred-Tank Reactor
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2015 — 04-BS-13, Biology. Three-hour, closed-book exam (one double-sided aid sheet permitted, approved Casio/Sharp calculator allowed). Format: Part I offers 6 questions (any 3 constitute a complete answer, 20 marks each) and Part II offers 4 questions (any 2 constitute a complete answer, 20 marks each) — a full paper is 5 questions. All 10 are solved below for completeness. Q1–Q4, Q7, and Q8 are calculation questions; Q5, Q9, and Q10 are essay questions; Q6 is a derivation.
Reference texts: Shuler & Kargi, Bioprocess Engineering: Basic Concepts (2nd ed., Prentice Hall) — elemental/electron balances, yield coefficients, fermenter mass balances, growth kinetics; Madigan et al., Brock Biology of Microorganisms (15th ed., Pearson) — bacterial classification, fungal reproduction, plasmid biology; Toledo, Fundamentals of Food Process Engineering (3rd ed., Springer) — plant/animal tissue structure and mechanical properties.
Question 6: Unsteady-State Mass Balances on a Continuous Stirred-Tank Reactor (20 marks)
Fig. 6.1 — Control volume for the CSTR: feed stream $(F_i,\rho_i,C_{Ai})$ in, product stream $(F_o,\rho_o,C_{Ao})$ out, well-mixed tank contents $(V,\rho,C_A)$, first-order consumption of $A$ inside the boundary.
Find. The unsteady-state differential (unsteady-state) mass balance equations for (a) total mass in the tank and (b) mass of species $A$ in the tank.
Approach. Apply the general conservation statement Accumulation = In − Out ± Generation/Consumption to the control volume shown, once for total mass (no generation/consumption term, since total mass is conserved even though $A$ reacts to $B$) and once for the mass of species $A$ specifically (which does have a consumption term, since $A$ is destroyed by reaction inside the tank).
Part (a): total mass balance. The mass of liquid held in the tank at any instant is $V\rho$ (volume × density — density is not assumed constant here, since $\rho_i\ne\rho_o$ is allowed). Total mass is neither created nor destroyed by the $A\rightarrow B$ reaction (it only converts one species into another of possibly different molecular weight, but the total mass entering/leaving the reaction terms cancels), so accumulation equals in minus out:
$$\boxed{\frac{d(V\rho)}{dt}=F_i\rho_i-F_o\rho_o.}$$
If the liquid is treated as incompressible with $\rho_i=\rho_o=\rho=$ constant, this reduces to the familiar $\dfrac{dV}{dt}=F_i-F_o$.
Part (b): mass balance on species $A$. The mass of $A$ held in the tank at any instant is $VC_A$. Unlike total mass, $A$ has a sink term: it is consumed by the first-order reaction at volumetric rate $r_C=k_1C_A$ (mass or moles of $A$ consumed per unit volume per unit time), so the total consumption rate throughout the well-mixed tank is $V\,k_1C_A$. Accumulation = in − out − consumption:
$$\boxed{\frac{d(VC_A)}{dt}=F_iC_{Ai}-F_oC_{Ao}-Vk_1C_A.}$$
If $V$ is constant (steady liquid holdup, part (a) at steady state with $F_i=F_o=F$), this may be expanded as $V\dfrac{dC_A}{dt}=F(C_{Ai}-C_{Ao})-Vk_1C_A$, the standard unsteady-state CSTR design equation for a first-order reaction.
