Heat Transfer

Question. Which of the following operational modifications will increase the inside convective heat transfer coefficient ($h$) in a stirred tank bioreactor? (a) Increasing the impeller agitation speed (b) Increasing the broth temperature (which lowers viscosity) (c) Accumulation of a dense mycelial wall cake (d) Decreasing the broth flow velocity past the heat exchange surface.

Solution.

1. (a) Increasing agitation speed — raises turbulence and flow velocity at the wall, thinning the boundary film. Increases $h$.
2. (b) Increasing broth temperature (lower viscosity) — less resistance to flow near the wall, effectively raising the Reynolds number. Increases $h$.
3. (c) Mycelial wall cake — a solid, low-conductivity layer stuck to the wall acts as fouling: it adds resistance and dampens turbulence at the surface. Decreases $h$.
4. (d) Decreasing flow velocity — less turbulence, thicker boundary film. Decreases $h$.

Answer: (a) and (b) increase $h$; (c) and (d) decrease it.

Question. A composite bioreactor wall has an inner stainless steel sheet ($x = 5\ \text{mm}$, $k = 15\ \text{W}/(\text{m}\cdot\text{K})$) and outer insulation ($x = 25\ \text{mm}$, $k = 0.05\ \text{W}/(\text{m}\cdot \text{K})$). Convective resistances are negligible. Find the total conductive resistance per unit area.

Solution.

1. Steel: $R_1 = 0.005/15 = 0.000333\ (\text{m}^2\cdot\text{K})/\text{W}$.
2. Insulation: $R_2 = 0.025/0.05 = 0.500\ (\text{m}^2\cdot\text{K})/\text{W}$.
3. $R_{total} = R_1 + R_2 = 0.500\ (\text{m}^2\cdot\text{K})/\text{W}$ (to 3 d.p.) — the insulation dominates since its conductivity is $300\times$ lower than the steel's.

Answer: $R_{total} \approx 0.500\ (\text{m}^2\cdot\text{K})/\text{W}$.

Question. A hot fluid enters a heat exchanger at $120\,^\circ\text{C}$ and leaves at $80\,^\circ\text{C}$. A cold fluid enters at $20\,^\circ\text{C}$ and leaves at $50\,^\circ\text{C}$. Find the LMTD for (a) parallel flow and (b) counter-current flow.

Solution.

1. Parallel flow (co-current) — both fluids enter at the same end, so the inlet ends pair together and the outlet ends pair together:

   

   • $\Delta T_1 = 120 - 20 = 100\,^\circ\text{C}$ (both inlets)
   • $\Delta T_2 = 80 - 50 = 30\,^\circ\text{C}$ (both outlets)
   • $\Delta T_{LMTD} = \dfrac{\Delta T_2 - \Delta T_1}{\ln(\Delta T_2 / \Delta T_1)} = \dfrac{30 - 100}{\ln(30/100)} = \dfrac{-70}{-1.204} \approx 58.14\,^\circ\text{C}$

2. Counter-current flow — the hot inlet pairs with the cold outlet, and the hot outlet pairs with the cold inlet:

   

   • $\Delta T_1 = 120 - 50 = 70\,^\circ\text{C}$ (hot inlet, cold outlet)
   • $\Delta T_2 = 80 - 20 = 60\,^\circ\text{C}$ (hot outlet, cold inlet)
   • $\Delta T_{LMTD} = \dfrac{60 - 70}{\ln(60/70)} = \dfrac{-10}{-0.1542} \approx 64.87\,^\circ\text{C}$

Answer: LMTD $\approx 58.14\,^\circ\text{C}$ for parallel flow, and $\approx 64.87\,^\circ\text{C}$ for counter-current flow — the counter-current arrangement gives the larger LMTD for the same four terminal temperatures.

Question. A counter-current heat exchanger cools hot oil from $120\,^\circ\text{C}$ to $80\,^\circ\text{C}$ using cooling water that heats from $30\,^\circ\text{C}$ to $70\,^\circ\text{C}$. Find the LMTD.

Solution.

1. Counter-current flow: the hot inlet pairs with the cold outlet, and the hot outlet pairs with the cold inlet.
   • $\Delta T_1 = 120 - 70 = 50\,^\circ\text{C}$ (hot inlet, cold outlet)
   • $\Delta T_2 = 80 - 30 = 50\,^\circ\text{C}$ (hot outlet, cold inlet)
2. $\Delta T_1 = \Delta T_2$, so $\dfrac{\Delta T_2 - \Delta T_1}{\ln(\Delta T_2/\Delta T_1)}$ is $0/0$. By the equal-terminal-differences limit, $\Delta T_{LMTD} = \Delta T_1 = \Delta T_2$.

Answer: LMTD $= 50.0\,^\circ\text{C}$.

Question. Select the correct statements regarding heat generation and removal in aerobic bioprocesses: (a) Aerobic microbial growth is accompanied by heat release proportional to the rate of oxygen consumption. (b) The heat of fermentation can be estimated using Cooney's correlation: approximately $460\,\text{kJ}$ per mole of $O_2$ consumed (or $14\,\text{kJ}$ per g of $O_2$). (c) Scale-up of stirred fermenters creates heat transfer challenges because reactor volume scales with $D^3$, while surface area for cooling scales with $D^2$. (d) Internal cooling coils are preferred over external jackets for highly shear-sensitive animal cell cultures to minimize volume.

Solution.

1. (a) Metabolic heat comes from the same respiratory reactions that consume $O_2$, so heat release tracks the oxygen uptake rate. Correct.
2. (b) This is exactly Cooney's correlation above: $460\,\text{kJ/mol}\ O_2 \approx 14\,\text{kJ/g}\ O_2$. Correct.
3. (c) $V \propto D^3$ and $A \propto D^2$, so specific cooling area falls as $1/D$ — the standard scale-up cooling bottleneck. Correct.
4. (d) Backwards on both counts: external jackets, not internal coils, are preferred for shear-sensitive cultures, precisely because coils add internal turbulence/shear and reduce usable working volume rather than minimizing it. Incorrect.

Answer: (a), (b), and (c) are correct; (d) is incorrect.