Let's solve the problem step by step: --- ## **Given Data** - **Velocity (\( u \))**: .8 m/s - **Tube diameter (\( d \))**: 10 mm = .01 m - **Tube length (\( L \))**: 5 m - **Wall length coated (\( L_{\text{wall}} \))**: 1.8 m - **Solubility of acetic acid (\( C_{A,sat} \))**: 2.5 kg mol/m³ - **Viscosity of water (\( \mu \))**: \( 8.71 \times 10^{-4} \) Pa·s - **Density of water (\( \rho \))**: 997 kg/m³ - **Diffusion coefficient (\( D_{AB} \))**: \( 1.24 \times 10^{-9} \) m²/s --- ## **a) Determine the flow type** Use the **Reynolds number**: \[ \text{Re} = \frac{\rho u d}{\mu} \] Plug in the values: \[ \text{Re} = \frac{997 \times .8 \times .01}{8.71 \times 10^{-4}} \] \[ \text{Re} = \frac{7.976}{8.71 \times 10^{-4}} \] \[ \text{Re} \approx 9155 \] **Interpretation:** - Laminar: Re < 210 - Transitional: 210 < Re < 400 - Turbulent: Re > 400 **Answer:** The flow is **turbulent**. --- ## **b) Calculate the mass-transfer coefficient (\( k_L' \))** For turbulent flow inside a tube, use the **Dittus-Boelter correlation** (for analogy between heat and mass transfer): \[ \frac{Sh}{.023} = \text{Re}^{.8} \text{Sc}^{.33} \] \[ Sh = \frac{k_L' d}{D_{AB}} \] \[ Sc = \frac{\mu}{\rho D_{AB}} \] Let's calculate each: ### 1. **Schmidt number (Sc):** \[ Sc = \frac{\mu}{\rho D_{AB}} \] \[ Sc = \frac{8.71 \times 10^{-4}}{997 \times 1.24 \times 10^{-9}} \] \[ Sc = \frac{8.71 \times 10^{-4}}{1.236 \times 10^{-6}} \] \[ Sc \approx 705 \] ### 2. **Sherwood number (Sh):** \[ Sh = .023 \cdot Re^{.83} \cdot Sc^{.44} \] But for Dittus-Boelter: \( Sh = .023 Re^{.8} Sc^{.33} \) \[ Sh = .023 \times 9155^{.8} \times 705^{.33} \] Calculate each term: - \( 9155^{.8} \approx 1461 \) - \( 705^{.33} \approx 8.54 \) \[ Sh = .023 \times 1461 \times 8.54 \approx 287 \] ### 3. **Find \( k_L' \):** \[ Sh = \frac{k_L' d}{D_{AB}} \implies k_L' = \frac{Sh \cdot D_{AB}}{d} \] \[ k_L' = \frac{287 \times 1.24 \times 10^{-9}}{.01} \] \[ k_L' = \frac{355.88 \times 10^{-9}}{.01} \] \[ k_L' = 35.6 \times 10^{-6}~\text{m/s} \] \[ k_L' = 3.56 \times 10^{-5}~\text{m/s} \] --- ## **c) Calculate the flux of \( N_A \)** Assume dilute solution, so \( C_{A,\text{bulk}} \approx \): \[ N_A = k_L' (C_{A,sat} - C_{A,bulk}) \] \[ N_A = 3.56 \times 10^{-5} \times (2.5 - ) \] \[ N_A = 8.9 \times 10^{-5}~\text{kg mol}/(\text{m}^2\cdot\text{s}) \] --- ## **Summary of Answers** ### a) **Flow type** - **Turbulent** (Re = 9155) ### b) **Mass-transfer coefficient (\( k_L' \))** - \( \boxed{3.56 \times 10^{-5}~\text{m/s}} \) ### c) **Flux of \( N_A \)** - \( \boxed{8.9 \times 10^{-5}~\text{kg mol}/(\text{m}^2\cdot\text{s})} \)