Let's solve **Problem 2: Mass transfer coefficients** step by step. --- ## **Given** - **Tube diameter** \( = 1\, \text{} \) - **Tube length \( L = 10\, \text{cm} \) (unless otherwise stated- **T (temperature)** \( = 293\, \text{K} \) - **Total pressure** \( = 1\, \text{atm} = 1.013 \times 10^5\, \text{Pa} \) - **Vapor pressure of water at 293K** \( P_v = 17.54\, \text{mm Hg} = .0231\, \text{atm} \) - **Water evaporation rate** \( = 4 \times 10^{-4}\, \text{ml/s} \) - **Diffusivity of water vapor in air** \( D_{AB} \) (to be used; typically \( \sim 2.57 \times 10^{-5}\, \text{m}^2/\text{s} \) at room temp) - **R (ideal gas constant)** \( = .0821\, \text{atm·L}/(\text{mol·K}) \) --- ## **a. Determine the values of the mass transfer coefficients \( k_c, k_g, k_D \).** First, let's define these: - **\( k_c \)**: Mass transfer coefficient based on concentration gradient (mol/(L·s)) - **\( k_g \)**: Mass transfer coefficient based on partial pressure gradient (mol/(atm·s)) - **\( k_D \)**: Mass transfer coefficient based on mole fraction gradient (mol/(mol·s)) ### **Step 1: Calculate the mass transfer rate (\( N \))** Convert evaporation rate from ml/s to mol/s: - \( 4 \times 10^{-4} \) ml/s \( = 4 \times 10^{-7} \) L/s - Moles of water per second: \(\dfrac{4 \times 10^{-7}\ \text{L/s}}{18 \times 10^{-3}\ \text{L/mol}} = 2.22 \times 10^{-5}\ \text{mol/s}\) But let's use the more precise calculation: Water density: \( 1\, \text{g/ml} \) - \( 4\times 10^{-4}\, \text{ml} = 4\times 10^{-4}\, \text{g} \) - Moles: \( \frac{4 \times 10^{-4}}{18} \approx 2.22 \times 10^{-5}\, \text{mol/s} \) ### **Step 2: Surface Area** Tube diameter = 1 cm \( \Rightarrow \) radius = .5 cm = .005 m Area \( A = \pi r^2 = \pi (.005)^2 = 7.85 \times 10^{-5}\, \text{m}^2 \) --- ### **Step 3: Driving force** - \( C_{A,1} \): At the water surface, saturated vapor (\( P_v \)) - \( C_{A,2} \): At the tube opening, assumed to be (dry air) **Partial pressure gradient:** \( \Delta P = P_v - = .0231\, \text{atm} \) **Concentration at surface:** \[ C_{A,1} = \frac{P_v}{RT} = \frac{.0231}{.0821 \times 293} = 9.59 \times 10^{-4}\, \text{mol/L} = .959\, \text{mol/m}^3 \] --- ### **Step 4: Mass transfer coefficients** \[ N_A = k_c \cdot A \cdot (C_{A,1} - C_{A,2}) \\ N_A = k_g \cdot A \cdot (P_{A,1} - P_{A,2}) \\ N_A = k_D \cdot A \cdot (y_{A,1} - y_{A,2}) \] Since \( C_{A,2} \approx \), \( P_{A,2} = \), \( y_{A,2} = \), \[ k_c = \frac{N_A}{A \cdot C_{A,1}} \] \[ k_g = \frac{N_A}{A \cdot P_{A,1}} \] \[ k_D = \frac{N_A}{A \cdot y_{A,1}} \] Where \( y_{A,1} = P_{A,1}/P_{tot} = .0231 \) --- #### **Calculate:** - \( N_A = 2.22 \times 10^{-5} \) mol/s - \( A = 7.85 \times 10^{-5} \) m² - \( C_{A,1} = .959 \) mol/m³ - \( P_{A,1} = .0231 \) atm - \( y_{A,1} = .0231 \) \[ k_c = \frac{2.22 \times 10^{-5}}{7.85 \times 10^{-5} \times .959} = \frac{2.22 \times 10^{-5}}{7.53 \times 10^{-5}} = .295\, \text{m/s} \] \[ k_g = \frac{2.22 \times 10^{-5}}{7.85 \times 10^{-5} \times .0231} = \frac{2.22 \times 10^{-5}}{1.815 \times 10^{-6}} = 12.24\, \text{mol}/(\text{atm·m}^2\cdot \text{s}) \] \[ k_D = \frac{2.22 \times 10^{-5}}{7.85 \times 10^{-5} \times .0231} = 12.24\, \text{mol}/(\text{mol·m}^2\cdot \text{s}) \] **Final Answers:** - **\( k_c = .295\, \text{m/s} \)** - **\( k_g = 12.24\, \text{mol}/(\text{atm·m}^2\cdot \text{s}) \)** - **\( k_D = 12.24\, \text{m/s} \)** --- ## **b. If L is reduced to 2 cm, what are the new values?** The mass transfer coefficient for a tube is inversely proportional to \( L \): \[ k \propto \frac{1}{L} \] So, for \( L = 2\, \text{cm} \) (1/5 of the original): \[ k_{c,\text{new}} = .295 \times \frac{10}{2} = 1.475\, \text{m/s} \] \[ k_{g,\text{new}} = 12.24 \times \frac{10}{2} = 61.2\, \text{mol}/(\text{atm·m}^2\cdot \text{s}) \] \[ k_{D,\text{new}} = 12.24 \times \frac{10}{2} = 61.2\, \text{m/s} \] --- ## **c. If \( N_A = \), show \( k_ = \frac{D_{AB} P}{LRT} \)** **Derivation:** From Fick's Law for steady-state diffusion in a tube: \[ N_A = -D_{AB} \frac{dC_A}{dz} \] For a stagnant air column, integrate over length \( L \): \[ N_A = D_{AB} \frac{C_{A,1} - C_{A,2}}{L} \] But in terms of pressure, \[ N_A = \frac{D_{AB}}{LRT} (P_{A,1} - P_{A,2}) P \] So, \[ k_ = \frac{D_{AB} P}{LRT} \] --- ## **d. Sherwood Number, \( \text{Sh}_c = \frac{k_c L}{D_{AB}} \) when \( L = 2 \) cm** Assume \( D_{AB} = 2.57 \times 10^{-5} \, \text{m}^2/\text{s} \): \[ \text{Sh}_c = \frac{k_c L}{D_{AB}} \] \[ k_c = 1.475\, \text{m/s}, \quad L = .02\, \text{m} \] \[ \text{Sh}_c = \frac{1.475 \times .02}{2.57 \times 10^{-5}} = \frac{.0295}{2.57 \times 10^{-5}} \approx 1,148 \] --- ## **e. Water at the inlet, new Sherwood and mass transfer coefficients** If water is at the inlet, \( L = \), so: - **No diffusion path**; transfer is limited only by surface transfer (very fast), so \( k_c \to \infty \), \( \text{Sh}_c \to \infty \). --- ## **Summary Table** | Case | \( k_c \) (m/s) | \( k_g \) (mol/atm·m²·s) | \( k_D \) (m/s) | \( \text{Sh}_c \) | |-------------------|-----------------|--------------------------|-----------------|-------------------| | \( L = 10\, \text{cm} \) | .295 | 12.24 | 12.24 | 115 | | \( L = 2\, \text{cm} \) | 1.475 | 61.2 | 61.2 | 1,148 | | \( L \to \) | \( \infty \) | \( \infty \) | \( \infty \) | \( \infty \) | --- ### **Let me know if you need a detailed derivation for any step!**