Question 3 of 10: Energy Balance on a Continuous Fermenter

Question 3: Energy Balance on a Continuous Fermenter (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. Metabolic heat-generation rate $q=27.6\ \text{kJ}/(\text{kg}\!\cdot\!\text{h})$; cell mass held up in fermenter $M=2$ kg; feed temperature $T_{\text{feed}}=25\,{}^{\circ}\text{C}$; mass flow rate $\dot m=1.025$ kg/h; specific heat $c_p=4\ \text{kJ}/(\text{kg}\!\cdot\!{}^{\circ}\text{C})$; growth ceiling $T_{\max}=42\,{}^{\circ}\text{C}$; fermenter well-insulated (no external heat loss); steady state (exit temperature = fermenter temperature, constant with time).

Find. The steady-state fermenter (exit) temperature, and whether the cells survive.

Approach. At steady state with an insulated vessel, all metabolic heat generated by the cell mass must leave as sensible heat carried by the flow between feed and exit temperature — a single algebraic energy balance gives the exit temperature directly.

1. Total heat generation rate. $$\dot Q_{\text{gen}}=q\cdot M = 27.6\left(\frac{\text{kJ}}{\text{kg}\!\cdot\!\text{h}}\right)(2\ \text{kg}) = 55.2\ \text{kJ/h}.$$
2. Steady-state energy balance. Well-insulated $\Rightarrow$ no heat loss to the surroundings, so all generated heat raises the stream temperature from feed to exit: $$\dot Q_{\text{gen}} = \dot m\,c_p\,(T_{\text{exit}}-T_{\text{feed}}).$$
3. Solve for the temperature rise. $$\Delta T = \frac{\dot Q_{\text{gen}}}{\dot m\,c_p} = \frac{55.2}{(1.025)(4)} = \frac{55.2}{4.1} = 13.46\,{}^{\circ}\text{C}.$$
4. Exit (fermenter) temperature. $$T_{\text{exit}} = 25 + 13.46 = \boxed{38.46\,{}^{\circ}\text{C}}.$$ Since $38.46\,{}^{\circ}\text{C} < 42\,{}^{\circ}\text{C}$, the fermenter operates below the growth ceiling and the cells will survive.