Let's tackle each part step by step, with equations and explanations: --- ### **Given Data** - **Tank diameter (\(D_T\))**: 2 m - **Liquid volume (\(V_L\))**: 0.9 m³ - **Stirrer speed (\(N\))**: 5 rpm = 5/60 = 0.0833 s⁻¹ - **Baffles**: Four, each 10% of tank diameter - **Air flow rate (\(Q_{air}\))**: 0.5 vvm - **Density (\(\rho\))**: 1000 kg/m³ - **Viscosity (\(\mu\))**: 1.2 cP = 1.2 × 10⁻³ Pa·s - **Impeller diameter (\(D\))**: 0.8 m - **Impeller type**: Rushton turbine --- ## **a) Is the air dispersed effectively in this system?** **Effective air dispersion** occurs if the **power input from the impeller is sufficient to disperse the air bubbles** (i.e., not in the "flooded" regime). We use the **flooding-loading transition** criterion, usually described by the **critical superficial gas velocity (\(u_{gc}\))** or the ratio \(Q_{air}/Q_{critical}\). ### **Step 1: Calculate air flow rate in m³/s** \[ Q_{air} = 0.5 \times V_L \text{ per minute} = 0.5 \times 0.9 = 0.45 \text{ m}^3/\text{min} \] \[ Q_{air} = 0.45/60 = 0.0075 \text{ m}^3/\text{s} \] --- ### **Step 2: Calculate superficial gas velocity (\(u_g\))** \[ u_g = \frac{Q_{air}}{A} \] Where \(A\) is the cross-sectional area of the tank: \[ A = \frac{\pi D_T^2}{4} = \frac{\pi \times 2^2}{4} = \pi \text{ m}^2 \approx 3.142 \text{ m}^2 \] \[ u_g = \frac{0.0075}{3.142} = 0.00239 \text{ m/s} \] --- ### **Step 3: Calculate impeller power input (ungassed)** For a **Rushton turbine**, the **power number (\(N_P\))** ≈ 5 (for turbulent regime). \[ P = N_P \cdot \rho N^3 D^5 \] \[ N = 0.0833 \text{ s}^{-1},\ D = 0.8 \text{ m} \] \[ P = 5 \times 1000 \times (0.0833)^3 \times (0.8)^5 \] Calculate step-by-step: - \(N^3 = (0.0833)^3 = 0.000578\) - \(D^5 = (0.8)^5 = 0.32768\) - \(P = 5 \times 1000 \times 0.000578 \times 0.32768\) - \(5 \times 1000 = 5000\) - \(0.000578 \times 0.32768 = 0.0001897\) - \(5000 \times 0.0001897 = 0.9485\ \text{W}\) \[ P \approx 0.95\ \text{W} \] **This is a low value**—in reality, this is because the stirrer speed is very low for a tank of this size. But let's proceed. --- ### **Step 4: Compare to air flow (flooding criterion)** The **flooding velocity** for a Rushton turbine is (van’t Riet, 1979): \[ Q_{flood} = K \cdot N D^3 \] Where \(K\) ≈ 0.2 for Rushton turbine. \[ Q_{flood} = 0.2 \times 0.0833 \times (0.8)^3 = 0.2 \times 0.0833 \times 0.512 = 0.2 \times 0.0427 = 0.00854\ \text{m}^3/\text{s} \] Our actual \(Q_{air} = 0.0075\ \text{m}^3/\text{s}\) < \(Q_{flood} = 0.00854\ \text{m}^3/\text{s}\) **Therefore, the air flow is just below the flooding limit.** - **Air is (just) being dispersed effectively, but very close to flooding.** --- ## **b) Is the flow turbulent?** Calculate the **Reynolds number (\(Re\))** for the impeller. \[ Re = \frac{\rho N D^2}{\mu} \] Given: - \(\rho = 1000\ \text{kg/m}^3\) - \(N = 0.0833\ \text{s}^{-1}\) - \(D = 0.8\ \text{m}\) - \(\mu = 1.2 \times 10^{-3}\ \text{Pa·s}\) \[ Re = \frac{1000 \times 0.0833 \times 0.8^2}{1.2 \times 10^{-3}} \] \[ 0.8^2 = 0.64 \] \[ 1000 \times 0.0833 \times 0.64 = 53.312 \] \[ Re = \frac{53.312}{1.2 \times 10^{-3}} = 44,427 \] **Since \(Re \gg 10,000\), the flow is highly turbulent.** --- ## **c) What proportion of the power input is from sparging?** **Power input from sparging (gas):** \[ P_{gas} = Q_{air} \cdot \rho \cdot g \cdot h \] Where: - \(Q_{air} = 0.0075\ \text{m}^3/\text{s}\) - \(\rho = 1.2\ \text{kg/m}^3\) (for air!), - \(g = 9.81\ \text{m/s}^2\) - \(h\) = liquid height (\(V_L/A = 0.9/3.142 = 0.286\ \text{m}\)) \[ P_{gas} = 0.0075 \times 1.2 \times 9.81 \times 0.286 \] \[ 1.2 \times 9.81 = 11.77 \] \[ 11.77 \times 0.286 = 3.369 \] \[ 0.0075 \times 3.369 = 0.0252\ \text{W} \] **Impeller power = 0.95 W** (as above). **Total input = 0.95 + 0.025 = 0.975 W** \[ \text{Fraction from sparging} = \frac{0.025}{0.975} \approx 2.6\% \] --- ## **d) What should be the air flow rate for mixing at the flooding-loading transition?** Use the **flooding flow rate** calculated above: \[ Q_{flood} = 0.00854\ \text{m}^3/\text{s} \] Convert to vvm: \[ Q_{flood,\ min} = 0.00854\ \text{m}^3/\text{s} \times 60 = 0.512\ \text{m}^3/\text{min} \] \[ \text{vvm} = \frac{0.512}{0.9} = 0.569\ \text{vvm} \] --- ## **Summary Table** | Part | Answer | |---|---| | a) | **Yes, air is just being dispersed effectively (Q_air is slightly below flooding limit).** | | b) | **Flow is highly turbulent (Re ≈ 44,400).** | | c) | **Proportion of power from sparging is ≈ 2.6%.** | | d) | **Flooding-loading transition occurs at Q_air = 0.00854 m³/s (0.569 vvm).** | --- ### **Final Answers** **a)** Yes, air is being effectively dispersed, since the air flow is just below the flooding limit. **b)** The flow is highly turbulent (Re = 44,400). **c)** About 2.6% of the total power input is from sparging. **d)** The air flow rate for operation at the flooding-loading transition should be 0.00854 m³/s, which is 0.569 vvm.