Let's solve each part step by step, providing explanations and final answers. --- ## 1. **Leak Rate Calculation Using Ideal Gas Law** Given data: | Time (min) | | 10 | 20 | 30 | 40 | 50 | |------------|-----|-----|-----|-----|-----|-----| | Pressure (P, kPa) | 500 | 485 | 468 | 450 | 432 | 415 | Assume constant volume and temperature (isothermal expansion): The ideal gas law: \( PV = nRT \) For constant V and T: \( \frac{P}{n} = \text{constant} \Rightarrow n \propto P \) Let the initial number of moles be \( n_ \) at \( P_ = 500\, \text{kPa} \). At time \( t \), \( n = n_ \frac{P}{P_} \). **Mass flow rate (leak rate) calculation:** Change in moles over time interval \( \Delta t \): \[ \Delta n = n_ \left(\frac{P_2 - P_1}{P_}\right) \] But more generally: \[ \frac{dn}{dt} = \frac{V}{RT} \frac{dP}{dt} \] The negative sign indicates a decrease in pressure due to the leak: \[ \text{Leak rate} = -\frac{V}{RT} \frac{dP}{dt} \] You can estimate \(\frac{dP}{dt}\) from the slope between two points, e.g., from \(t=\) to \(t=50\) min: \[ \frac{dP}{dt} \approx \frac{415 - 500}{50 - } = \frac{-85}{50} = -1.7\ \text{kPa/min} \] If V and T are known, plug values into the above formula for leak rate. --- ## 2. **Relationship Between Cooling Water Temperature and Runaway Temperature** Given: \[ \rho V C_p \frac{dT}{dt} = -\Delta H_r V k_r e^{-E/RT} C_A - UA(T - T_c) \] At critical runaway, set \(\frac{dT}{dt}=\) and solve for \(T\): \[ = -\Delta H_r V k_r e^{-E/RT_{cr}} C_A - UA(T_{cr} - T_c) \] \[ UA(T_{cr} - T_c) = -\Delta H_r V k_r e^{-E/RT_{cr}} C_A \] \[ T_{cr} = T_c + \frac{-\Delta H_r V k_r e^{-E/RT_{cr}} C_A}{UA} \] This provides the relationship. --- ## 3. **Dynamic Cooling Water Temperature Threshold** The critical cooling water temperature threshold is \(T_{cr}\). Below this, the system cannot remove heat fast enough, so temperature will rise uncontrollably (runaway). --- ## 4. **Batch Distillation Light Component Hold-Up** Given: \[ \frac{W}{W_} = \alpha^{\left[\ln \left(\frac{x_D}{x_B} \right)/(\alpha - 1)\right]} \] To remove 90% of light component from a 100 kg charge, \(W_ = 100\) kg, \(W = 10\) kg. Set up equation and solve for time (assuming \(\alpha\), \(x_D\), \(x_B\) are known or provided). --- ## 5. **Solute Dispersion Through a Stagnant Gas Film** Fick's Law for semi-infinite medium: \[ C_A(z,t) = C_{A,s} \cdot \text{erfc}\left(\frac{z}{2\sqrt{Dt}}\right) \] Given \(C_A = .5 C_{A,s}\) at \(z\) after \(t = 1\) hour. Set up: \[ .5 = \text{erfc}\left(\frac{z}{2\sqrt{Dt}}\right) \] Use erfc value for .5: \[ \text{erfc}(x) = .5 \implies x \approx .477 \] So, \[ \frac{z}{2\sqrt{D \cdot 1}} = .477 \implies z = 2 \times .477 \times \sqrt{D} \] --- ## 6. **Temperature Profile in Spherical Catalyst Pellet** Governing equation: \[ \frac{\partial T}{\partial t} = \alpha \left( \frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{\partial T}{\partial r}\right) \right) \] Initial condition: \(T(r,) = T_\) Boundary conditions: Surface temperature step change at \(r = R\). --- ## 7. **Biot Number for Lumped Capacitance** The Biot number (\(Bi\)): \[ Bi = \frac{hL_c}{k} \] For lumped capacitance to be valid, \(Bi < .1\). --- ## **Final Answers Summary** 1. Leak rate: \(-\frac{V}{RT} \frac{dP}{dt}\); slope is \(-1.7\) kPa/min. 2. Critical runaway temperature: \(T_{cr} = T_c + \frac{-\Delta H_r V k_r e^{-E/RT_{cr}} C_A}{UA}\). 3. Cooling water temperature must be below \(T_{cr}\) to avoid runaway. 4. For 90% removal: \(\frac{W}{W_} = .1\), use equation to solve for time. 5. For 50% concentration at 1 hour: \(z = 2 \times .477 \times \sqrt{D}\). 6. Spherical pellet temperature: solve heat equation with step change at surface. 7. For lumped capacitance, Biot number must be less than .1. Let me know if you want a detailed solution for a specific part!