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04-BS-13 · December 2013

Question 3 of 10: Baker's Yeast Fermenter — Rate of Heat Removal

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

Notes on this paper

National Exams — December 2013 — 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. Most questions require an essay-format answer; Q1–Q4 and Q7 are calculation questions.

Reference texts: Shuler & Kargi, Bioprocess Engineering: Basic Concepts (2nd ed., Prentice Hall) — elemental/electron balances, yield coefficients, fermenter energy balances, growth kinetics; Madigan et al., Brock Biology of Microorganisms (15th ed., Pearson) — bacterial/viral morphology, physiology and growth control; Toledo, Fundamentals of Food Process Engineering (3rd ed., Springer) — plant/animal tissue structure.

Question 3: Baker's Yeast Fermenter — Rate of Heat Removal (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.

Check: assumes the standard bioprocess-engineering rule of thumb of ≈460 kJ of metabolic heat released per mole of O₂ consumed during aerobic growth (Cooney's correlation) — the exam does not supply a heat-of-reaction value directly, so this textbook constant (Shuler & Kargi, Ch. 6) is taken as given.
QuantityValue
Fermenter volume, $V$50 000 L
Biomass yield, $Y_{XS}$0.5 g/g (dry, incl. 5% ash)
Specific growth rate, $\mu$0.45 h⁻¹
Yeast concentration, $X$10 g/L
MW sucrose / biomass (ash-free)342 / 25.01 g/(c)mol
Heat evolved per mole O₂ consumed≈460 kJ/mol (assumption, see callout)

Find. The rate of heat removal (kW, or kJ/h) needed to hold the fermenter at constant temperature.

Approach. First close the stoichiometric equation (C, H, N, O atom balances, with $Y_{XS}$ and the ash correction fixing $c$) to get the O2 demand per mole sucrose consumed. Then convert the given growth rate and cell concentration to a volumetric substrate-consumption rate via $Y_{XS}$, scale by the O2 stoichiometric coefficient and the fermenter volume to get a total O2 uptake rate, and finally apply the heat-per-mole-O2 rule (since at steady temperature, all metabolic heat generated must be removed).

  1. Biomass coefficient $c$ (ash-corrected, per mole sucrose). Dry biomass per mole sucrose $=Y_{XS}(342)=171$ g; ash-free $=171(0.95)=162.45$ g: $$c=\frac{162.45}{25.01}=\boxed{6.495\ \text{cmol biomass/mol sucrose}}.$$
  2. N, C, H balances → $a$, $d$, $e$. $$a=0.17c=1.104,\qquad d=12-c=5.505,$$ $$e=\frac{22+3a-1.83c}{2}=\frac{22+3.312-11.887}{2}=6.713.$$
  3. Oxygen balance → $b$. $$2b = 0.55c+2d+e-11 = 3.572+11.009+6.713-11=10.295\ \Rightarrow\ b=\boxed{5.147\ \text{mol O}_2/\text{mol sucrose}}.$$
  4. Volumetric growth and substrate-consumption rates. $$r_X=\mu X = 0.45(10)=4.5\ \text{g biomass/(L}\cdot\text{h)},\qquad r_S=\frac{r_X}{Y_{XS}}=\frac{4.5}{0.5}=9.0\ \text{g sucrose/(L}\cdot\text{h)}=\frac{9.0}{342}=0.02632\ \text{mol/(L}\cdot\text{h)}.$$
  5. Volumetric and total O₂ consumption rate. $$r_{\text{O}_2}=b\cdot r_S=5.147(0.02632)=0.1355\ \text{mol O}_2/\text{(L}\cdot\text{h)},$$ $$\dot n_{\text{O}_2}=r_{\text{O}_2}\cdot V = 0.1355(50\,000)=\boxed{6773\ \text{mol O}_2/\text{h}}.$$
  6. Heat generation (= heat removal) rate. $$\dot Q = \dot n_{\text{O}_2}\times 460\ \frac{\text{kJ}}{\text{mol O}_2} = 6773(460)=3.116\times10^6\ \text{kJ/h} = \boxed{865\ \text{kW}}.$$
QuantityResult
Biomass coefficient, $c$6.495 cmol/mol sucrose
O₂ coefficient, $b$5.147 mol/mol sucrose
Substrate consumption rate9.0 g/(L·h)
Total O₂ consumption rate6773 mol O₂/h
Rate of heat removal required3.12×10⁶ kJ/h ≈ 865 kW