Question 6 of 9: Cooling Requirement for Immobilized-Cell Glutamic Acid Production

Question 5: Cooling Requirement for Immobilized-Cell Glutamic Acid Production (20 marks)

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Given.

Find. (a) rate of heat removal required to hold the reactor at 25°C (kJ/h); (b) adiabatic temperature rise if cooling is not provided.

Approach. The liquid feed, product, reactor and off-gas are all at 25°C, so the liquid-stream sensible-heat terms cancel; the one exception is the sparged gas, which enters at 15°C and leaves at 25°C and therefore absorbs sensible heat that the cooling system does not have to remove. The dominant term is the heat of reaction itself — found via Hess's law from the given heats of combustion (CO₂ and H₂O, being fully oxidized, contribute $\Delta h_c^{\circ}=0$ and drop out of the products' combustion-heat sum). Multiplying by the molar rate of glucose consumption (from a glucose mass balance on the liquid stream) gives the heat released; subtracting the sensible heat taken up by the incoming 15°C gas gives the cooling duty for part (a); letting that surplus heat instead warm both outlet streams (liquid and gas) gives part (b).

1. Heat of reaction via Hess's law (heats of combustion). $\Delta H_{rxn}=\sum\Delta h_c^{\circ}(\text{reactants})-\sum\Delta h_c^{\circ}(\text{products})$, with CO₂ and H₂O contributing zero on the product side: $$\Delta H_{rxn}=\big[-2805+1.5(0)+(-382.6)\big]-\big[-2244.1+0+3(0)\big]=-3187.6-(-2244.1)=\boxed{-943.5\ \text{kJ/mol glucose reacted}}$$ (exothermic — 943.5 kJ released per mole glucose consumed).
2. Glucose consumption rate from a liquid-stream mass balance. Treating the liquid mass flow as essentially unchanged from feed to product (2000 kg/h; the dissolved-gas mass transferred is small on this scale): $$\text{glucose in}=2000(0.04)=80\ \text{kg/h},\qquad \text{glucose out}=2000(0.005)=10\ \text{kg/h},$$ $$\text{glucose consumed}=80-10=70\ \text{kg/h}=\frac{70\,000\ \text{g/h}}{180\ \text{g/mol}}=\boxed{389\ \text{mol/h}}.$$
3. Heat released by reaction. $$\dot Q_{rxn}=|\Delta H_{rxn}|\times(\text{mol glucose/h})=943.5(388.9)=3.669\times10^{5}\ \text{kJ/h}.$$
4. Sensible heat absorbed by the sparged gas (15°C → 25°C). Gas volumetric flow $=4\ \text{vvm}\times25\ \text{m}^3=100\ \text{m}^3/\text{min}$ at 1 atm and 15°C, so by the ideal-gas law $$\dot n_{gas}=\frac{PV}{RT}=\frac{101\,325(100)}{8.314(288.15)}=4230\ \text{mol/min}=2.538\times10^{5}\ \text{mol/h}$$ (12% NH₃ = 30 450 mol/h, 88% air = 223 300 mol/h). Taking molar heat capacities of air $\approx29.1$ and NH₃ $\approx35.1$ J/(mol·K) near room temperature (the given 4.18 kJ/(kg·K) applies to the liquid, not the gas), the gas heat-capacity rate is $\dot n c_p=223\,300(29.1)+30\,450(35.1)=7.57\times10^{3}$ kJ/(h·K), and $$\dot Q_{gas}=7.57\times10^{3}(25-15)=7.57\times10^{4}\ \text{kJ/h}.$$ The NH₃ (389 mol/h) and O₂ (583 mol/h) actually consumed are <1% of the gas flow, so the off-gas flow is taken equal to the inlet flow.
5. Part (a): cooling duty. Energy balance at steady state (all enthalpies referenced to 25°C): $$\dot Q_{cooling}=\dot Q_{rxn}-\dot Q_{gas}=3.669\times10^{5}-0.757\times10^{5}=\boxed{2.91\times10^{5}\ \text{kJ/h}\ (\approx81\ \text{kW})}.$$
6. Part (b): temperature rise with no cooling. Assuming the same reaction rate and conversion, the 2.91×10⁵ kJ/h no longer removed must instead raise the temperature of everything leaving the reactor (liquid product and off-gas) by $\Delta T$ above 25°C: $$\Delta T=\frac{\dot Q_{cooling}}{\dot m_{liq}C_p+\dot n_{gas}c_p}=\frac{2.912\times10^{5}}{2000(4.18)+7.57\times10^{3}}=\frac{2.912\times10^{5}}{1.593\times10^{4}}=\boxed{18.3^{\circ}\text{C}}$$ — the reactor would climb from 25°C to $\approx43^{\circ}$C, well above the optimum for this organism, which is why active cooling is essential. (Crediting the liquid alone, as if the gas did not warm, would overstate the rise at $2.912\times10^{5}/8360=34.8^{\circ}$C.)