Yeast Growth on Glucose — Yield Coefficients and Oxygen Requirement: 04-BS-13, December 2013, Q4

Question 4: Yeast Growth on Glucose — Yield Coefficients and Oxygen Requirement (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.

Quantity	Value
Balanced equation coefficients	1 glucose : 3 O₂ : 0.48 NH₃ → 0.48 biomass : 4.32 H₂O : 3.12 CO₂
MW glucose / O₂ / yeast / NH₃	180 / 32 / 144 / 17 g/mol
Reactor volume, $V$	100 000 L
Final yeast concentration, $X_f$	50 g/L
Exponential-phase growth rate, $r_X$	0.7 g biomass/(L·h)

Find. (a) $Y_{XS}$, $Y_{XO_2}$; (b) total O₂ required for the batch; (c) O₂ consumption rate during exponential growth.

Approach. The equation is already fully balanced, so both yield coefficients follow directly from mass ratios of the stoichiometric coefficients (moles × MW). Part (b) scales the total biomass to be produced by $1/Y_{XO_2}$; part (c) applies the same yield coefficient to convert the given biomass-formation rate into an O₂-consumption rate.

Part (a): yield coefficients directly from the balanced equation (basis 1 mol glucose). $$Y_{XS}=\frac{0.48(144)}{1(180)}=\frac{69.12}{180}=\boxed{0.384\ \text{g biomass/g glucose}},$$ $$Y_{XO_2}=\frac{0.48(144)}{3(32)}=\frac{69.12}{96}=\boxed{0.720\ \text{g biomass/g O}_2}.$$
Part (b): total biomass to be produced, then total O₂. $$m_{X,\text{total}}=X_f\cdot V = 50(100\,000)=5.0\times10^6\ \text{g}=5000\ \text{kg}.$$ $$m_{\text{O}_2,\text{total}}=\frac{m_{X,\text{total}}}{Y_{XO_2}}=\frac{5.0\times10^6}{0.720}=\boxed{6.944\times10^6\ \text{g} = 6944\ \text{kg O}_2}.$$
Part (c): O₂ consumption rate at $r_X=0.7$ g/(L·h). $$r_{\text{O}_2}=\frac{r_X}{Y_{XO_2}}=\frac{0.7}{0.720}=\boxed{0.972\ \text{g O}_2/\text{(L}\cdot\text{h)}}.$$

Quantity	Result
$Y_{XS}$	0.384 g biomass/g glucose
$Y_{XO_2}$	0.720 g biomass/g O₂
Total biomass produced	5000 kg
Total O₂ required	6944 kg (2.17×10⁵ mol)
O₂ consumption rate (exp. phase)	0.972 g O₂/(L·h)

Concept & Theory

Topic. Yield coefficients from a fully-specified fermentation stoichiometric equation.

Key relations. $Y_{XS}=\Delta X/\Delta S$, $Y_{XO_2}=\Delta X/\Delta O_2$, both read directly off the balanced equation's mole ratios once converted to mass via molecular weight; a rate relation $r_{O_2}=r_X/Y_{XO_2}$ follows immediately because yield coefficients are (to first approximation) constant ratios independent of absolute rate.

Why this works. Because the given equation already satisfies every atom balance, no further stoichiometric solving is required — the yield coefficients are pure unit conversions from the reaction's own mole ratios, and treating them as constant lets any instantaneous rate (or any total extent of reaction) be converted between biomass, substrate and oxygen bases interchangeably.

Common pitfall. Using the wrong basis when computing $Y_{XO_2}$ — the O₂ mass must use its own coefficient (3 mol) and MW (32), not be confused with the glucose-basis $Y_{XS}$ calculation. Also, part (b)'s "total O₂ required" should use the final total biomass concentration ($X_f\cdot V$), not the instantaneous rate from part (c) — the two parts intentionally use different bases (extent vs. rate).

Source. Shuler & Kargi, Bioprocess Engineering: Basic Concepts, Ch. 6 (Yield coefficients in fermentation stoichiometry).

Question 4 of 10: Yeast Growth on Glucose — Yield Coefficients and Oxygen Requirement

Question 4: Yeast Growth on Glucose — Yield Coefficients and Oxygen Requirement (20 marks)