Backyard Fermenter — Cooling-Water Energy Balance: 04-BS-13, December 2018, Q3

Question 3: Backyard Fermenter — Cooling-Water Energy Balance (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.

Quantity	Value
Ethanol yield	0.45 g ethanol/g glucose
Ethanol production rate	0.4 kg/h = 400 g/h
Cooling water	2.5 L/h at 10°C
Biomass formula (8% ash)	CH_1.75O_0.58N_0.18
$\Delta h_c^{\circ}$: glucose, NH₃, biomass, ethanol	−2805, −382.6, −21.2, −1366 kJ/gmol
Water $C_p$	75.4 J/(gmol·°C)

Find. Final temperature of the jacket water.

Approach. Fix the ethanol coefficient $f$ from the given mass yield, close the remaining C/H/O/N elemental balances for $b,c,d,e$, apply Hess's law with the given heats of combustion ($\Delta H_{rxn}=\sum\nu_i\Delta h_{c,reactants}-\sum\nu_j\Delta h_{c,products}$, since CO₂/H₂O have $\Delta h_c=0$ by definition) to get the heat released per mole of glucose consumed, scale by the actual glucose consumption rate, and finally close a simple sensible-heat balance on the cooling-water stream.

Control-volume energy balance on the fermenter jacket: glucose + NH₃ in, biomass + CO₂ + H₂O + ethanol out, with the heat of fermentation transferred into the once-through cooling-water stream.

Fix the ethanol coefficient $f$ and close the elemental balances. From the mass yield, $f=0.45(180.2)/46.1=1.759$ mol ethanol/mol glucose. With $b=0.18c$ (N-balance) substituted into the C, H, O balances, the resulting $4\times4$ linear system gives $$b=0.1285,\qquad c=0.7141,\qquad d=1.768,\qquad e=0.291\ \ (\text{mol per mol glucose}).$$
Heat of reaction (Hess's law). $\text{CO}_2$ and $\text{H}_2\text{O}$ have $\Delta h_c=0$ by definition, so only glucose, NH₃, biomass, and ethanol contribute: $$\Delta H_{rxn}=\big[1(-2805)+b(-382.6)\big]-\big[c(-21.2)+f(-1366)\big]$$ $$=\big[-2805-49.2\big]-\big[-15.14-2402.8\big]=-2854.2+2417.9=\boxed{-436.2\ \text{kJ/mol glucose}}$$ (exothermic, as expected for any fermentation).
Scale to the actual glucose consumption rate. $r_{glucose}=400/0.45=888.9$ g/h $=888.9/180.2=4.933$ mol/h, so the heat release rate is $$\dot Q=|\Delta H_{rxn}|\,r_{glucose}=436.2(4.933)=\boxed{2152\ \text{kJ/h}}\ (\approx 598\ \text{W}).$$
Cooling-water sensible-heat balance. Water flow $=2.5$ L/h $\times1$ kg/L $=2500$ g/h $=2500/18=138.9$ mol/h. With all the fermentation heat absorbed by this once-through stream ($\dot Q=n_{water}C_p\Delta T$): $$\Delta T=\frac{2152\times10^3\ \text{J/h}}{138.9\ \text{mol/h}\times75.4\ \text{J/(mol}\cdot^\circ\text{C)}}=\boxed{205.5^\circ\text{C}}$$ $$T_{final}=10+205.5=\boxed{215.5^{\circ}\text{C}}.$$

Quantity	Result
Ethanol coefficient $f$	1.759 mol/mol glucose
$b,c,d,e$ (per mol glucose)	0.1285, 0.7141, 1.768, 0.291
$\Delta H_{rxn}$	−436.2 kJ/mol glucose
Heat release rate $\dot Q$	2152 kJ/h (≈598 W)
Water temperature rise $\Delta T$	205.5°C
Final water temperature	215.5°C (as computed from the given ideal balance)

Check: taking the given numbers at face value (negligible heat loss, no phase change considered, entire heat of fermentation absorbed by only 2.5 L/h of water) yields a final temperature of 215.5°C, which is above the normal boiling point of water (100°C at 1 atm). Physically, the jacket water would boil (or the fermenter would need to run at elevated pressure, or heat losses/a larger water flow would in practice cap the real outlet temperature at or below 100°C) well before this value is reached. The computed 215.5°C is reported as the direct answer to the stated idealization — the physical infeasibility is flagged here rather than silently adjusted.

Topic: Hess's-law energy balance on a fermentation reaction, coupled to a sensible-heat balance on a cooling stream.

Key relations: $\Delta H_{rxn}=\sum\nu_i\Delta h_{c,reactants}-\sum\nu_j\Delta h_{c,products}$ (CO₂/H₂O contribute zero); $\dot Q=n_{water}C_p\Delta T$ for a once-through sensible-heat exchanger with no phase change.

Why this works: heat of combustion is a state function referenced to the same fully-oxidized products (CO₂, H₂O, N₂) for every species, so subtracting the products' combustion heats from the reactants' isolates exactly the heat released by the partial oxidation that is fermentation — the same Hess's-law device used throughout this subject's degree-of-reduction balances.

Common pitfall: treating the computed $\Delta T$ as automatically physically valid without checking it against water's boiling point — a numerically "correct" application of $\dot Q=nC_p\Delta T$ can still describe a scenario (boiling) the simple sensible-heat model no longer applies to.

Source: Shuler & Kargi, Bioprocess Engineering: Basic Concepts, Ch. 5 (energy balances on fermentation processes via heats of combustion).

Question 3 of 8: Backyard Fermenter — Cooling-Water Energy Balance

Question 3: Backyard Fermenter — Cooling-Water Energy Balance (20 marks)