Baker's Yeast Growth — Batch Reactor Media Requirements: 04-BS-13, December 2018, Q2

Question 2: Baker's Yeast Growth — Batch Reactor Media Requirements (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.

Given. $V=10^5$ L, $X_{final}=50$ g/L, stoichiometry above (per mol glucose: 3 mol O₂, 0.48 mol NH₃ → 0.48 mol biomass, 3.12 mol CO₂, 4.32 mol H₂O), $r_X=0.7$ g/(L·h).

Find. (a) glucose and NH₃ concentrations/amounts; (b) $Y_{X/S}$, $Y_{X/O_2}$; (c) total O₂ mass; (d) $r_{O_2}$.

Approach. Work on a 1-mol-glucose basis to get the fixed molar ratios of every species to biomass directly from the balanced equation, scale up to the total biomass required in the $10^5$-L reactor, then read the yield coefficients straight off the same stoichiometric ratios (they don't depend on reactor size).

(a) Glucose and NH₃ requirements. Total biomass needed $=50(10^5)=5\times10^6$ g $=5\times10^6/144=3.472\times10^4$ mol. Since 1 mol glucose yields 0.48 mol biomass: $$n_{glucose}=\frac{3.472\times10^4}{0.48}=7.234\times10^4\ \text{mol}\ \Rightarrow\ m_{glucose}=7.234\times10^4(180)=\boxed{1.302\times10^7\ \text{g}}\ (13{,}021\ \text{kg}).$$ $$C_{glucose}=\frac{1.302\times10^7}{10^5}=\boxed{130.2\ \text{g/L}}.$$ NH₃ is consumed in the same 0.48:1 ratio to glucose, so $n_{NH_3}=n_{glucose}(0.48)=3.472\times10^4$ mol: $$m_{NH_3}=3.472\times10^4(17)=\boxed{5.903\times10^5\ \text{g}},\qquad C_{NH_3}=5.903\times10^5/10^5=\boxed{5.90\ \text{g/L}}.$$
(b) Yield coefficients. Directly from the mass ratio of stoichiometric coefficients: $$Y_{X/S}=\frac{0.48(144)}{1(180)}=\boxed{0.384\ \text{g biomass/g glucose}},\qquad Y_{X/O_2}=\frac{0.48(144)}{3(32)}=\boxed{0.720\ \text{g biomass/g O}_2}.$$ (Both check against the totals in part (a): $5\times10^6/1.302\times10^7=0.384$.)
(c) Total oxygen required. $n_{O_2}=3\,n_{glucose}=3(7.234\times10^4)=2.170\times10^5$ mol: $$m_{O_2}=2.170\times10^5(32)=\boxed{6.944\times10^6\ \text{g}}\ (6{,}944\ \text{kg}).$$
(d) Oxygen consumption rate. The instantaneous $Y_{X/O_2}$ ratio applies at any instant, including the exponential phase, so $$r_{O_2}=\frac{r_X}{Y_{X/O_2}}=\frac{0.7}{0.720}=\boxed{0.972\ \text{g O}_2/(\text{L}\cdot\text{h})}.$$

Quantity	Result
Glucose required	130.2 g/L (1.302×10⁷ g total)
NH₃ required	5.90 g/L (5.903×10⁵ g total)
$Y_{X/S}$	0.384 g/g
$Y_{X/O_2}$	0.720 g/g
Total O₂ required	6.944×10⁶ g (6944 kg)
$r_{O_2}$ at $r_X=0.7$ g/(L·h)	0.972 g O₂/(L·h)

Topic: Scaling a balanced fermentation stoichiometry from a 1-mole basis to reactor-scale mass requirements and yield coefficients.

Key relations: $Y_{X/S}=\Delta X/\Delta S$, $Y_{X/O_2}=\Delta X/\Delta O_2$, both fixed by the stoichiometric mass ratios and therefore scale-independent; $r_{O_2}=r_X/Y_{X/O_2}$ at any instant, not just at the endpoint.

Why this works: a balanced chemical equation fixes the molar ratios between every species regardless of reactor size or extent of reaction, so yield coefficients computed from the coefficients directly equal the yields computed from the total scaled-up masses — a useful self-consistency check.

Common pitfall: using the total oxygen consumed over the whole batch, instead of the instantaneous yield ratio, to answer a question about a specific-rate ($r_X$) supplied only for one point in time.

Source: Shuler & Kargi, Bioprocess Engineering: Basic Concepts, Ch. 5 (yield coefficients from fermentation stoichiometry).

Question 2 of 8: Baker's Yeast Growth — Batch Reactor Media Requirements

Question 2: Baker's Yeast Growth — Batch Reactor Media Requirements (20 marks)