Question 5 of 8: Citric Acid Fermentation — Cooling Requirement
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2016 — 04-BS-13, Biology. Three-hour, closed-book exam (one double-sided aid sheet permitted, approved calculator allowed). Format: Part I offers 5 questions (any 3 constitute a complete answer, 20 marks each) and Part II offers 3 questions (any 2 constitute a complete answer, 20 marks each) — a full paper is 5 questions. All 8 numbered questions are solved below for completeness. Q2–Q5 and Q7 are calculation/derivation questions; Q1, Q6, and Q8 are essay questions.
Reference texts: Shuler & Kargi, Bioprocess Engineering: Basic Concepts (2nd ed., Prentice Hall) — elemental/electron balances, yield coefficients, fermenter mass and energy balances, growth kinetics; Madigan et al., Brock Biology of Microorganisms (15th ed., Pearson) — bacterial/eukaryotic cell structure, fungi, protozoa/algae, Gram-stain cell envelope; Toledo, Fundamentals of Food Process Engineering (3rd ed., Springer) — plant tissue structure and cereal grain morphology.
Given. Batch, 2 days (48 h); O2 consumed 860 kg; heat of reaction $-460$ kJ/mol O2; agitation input $8.64\times10^6$ kJ over the culture period; water evaporated 100 kg; enthalpy of evaporation 2430.7 kJ/kg.
Find. The cooling duty required to hold the fermenter at 30 °C.
Approach. Perform an energy balance on the fermenter at (quasi-)steady temperature: heat generated by the exothermic metabolic reaction plus heat added mechanically by agitation must be removed either by evaporative cooling (already accounted for by the given water loss) or by the cooling system — solve for the cooling duty as the residual.
Check
The problem states agitation power as "15 kW (8.64×106 kJ)"; note that 15 kW sustained over the full 48-hour batch would integrate to only $15\times172{,}800\,\mathrm s\approx2.59\times10^6$ kJ, not $8.64\times10^6$ kJ (that total instead corresponds to ≈50 kW average). Since the problem supplies the total energy figure explicitly in parentheses, it is used directly below as the given agitation-energy input rather than recomputed from the 15 kW figure — this discrepancy is a feature of the source exam, not an error introduced here. If a marker instead takes the 15 kW literally, $Q_{agitation}=2.592\times10^6$ kJ and the same balance gives $Q_{cooling}=1.236\times10^7+2.592\times10^6-2.431\times10^5=1.471\times10^7$ kJ (≈85 kW average); state whichever reading you adopt.
Heat removed by evaporation.
$$Q_{evap}=100\ \mathrm{kg}\times2430.7\ \mathrm{kJ/kg}=2.431\times10^5\ \mathrm{kJ}.$$
Energy balance to isolate the cooling duty. At the (assumed) quasi-steady 30 °C operating temperature, heat in (metabolic $+$ agitation) equals heat out (evaporation $+$ cooling water):
$$Q_{metabolic}+Q_{agitation}=Q_{evap}+Q_{cooling}$$
$$\boxed{Q_{cooling}=Q_{metabolic}+Q_{agitation}-Q_{evap}=1.236\times10^7+8.64\times10^6-2.431\times10^5=2.076\times10^7\ \mathrm{kJ}}$$
over the full 2-day (48 h) batch.
Express as an average cooling rate. Dividing by the batch duration ($48\,\mathrm h=172{,}800$ s):
$$\boxed{\dot Q_{cooling}=\frac{2.076\times10^7\ \mathrm{kJ}}{172{,}800\ \mathrm s}\approx120\ \mathrm{kW}}$$
— the cooling jacket/coil must remove heat at an average rate of roughly 120 kW throughout the batch.
