# Step-by-Step Solution: Semi-Batch Reactor Project ## **1. Reaction Modeling** ### **a. Reaction and Mole Balance Equations** #### **Reaction:** \[ A + B \rightarrow C + D \] #### **Given:** - Reactor volume: 5 dm³ (constant) - Initial concentration of A: \(C_{A} = .05\) mol/dm³ - Initial concentration of B: - B is fed at \(C_{B,feed} = .025\) mol/dm³ - Volumetric flow rate of feed: \(F = .05\) dm³/s - Rate constant: \(k = 2.2\) dm³/(mol·s) #### **Assumptions:** - Perfect mixing - Constant temperature - Constant volume (since feed is much less than volume over time interval) #### **Mole Balance:** Let \(N_A, N_B, N_C, N_D\) be moles of A, B, C, D. For **semi-batch** (B is fed, A is present initially): - **A:** Only consumed in reaction - **B:** Fed and consumed - **C, D:** Produced \[ \text{Let } V = 5 \text{ dm}^3 \text{ (constant)} \] \[ C_A = \frac{N_A}{V},\; C_B = \frac{N_B}{V},\; C_C = \frac{N_C}{V} \] Let the **rate of reaction** (disappearance of A, B): \[ -r_A = -r_B = r_C = r_D = k C_A C_B \] ##### **Material Balances:** - **A:** \[ \frac{dN_A}{dt} = -r_A V = -k C_A C_B V \] \[ \frac{dC_A}{dt} = -k C_A C_B \] - **B:** \[ \frac{dN_B}{dt} = F C_{B,feed} - k C_A C_B V \] \[ \frac{dC_B}{dt} = \frac{F}{V} C_{B,feed} - k C_A C_B \] - **C:** \[ \frac{dN_C}{dt} = r_C V = k C_A C_B V \] \[ \frac{dC_C}{dt} = k C_A C_B \] - **D:** \[ \frac{dC_D}{dt} = k C_A C_B \] ### **Initial Conditions:** At \(t = \): - \(C_A = .05\) mol/dm³ - \(C_B = \) - \(C_C = \) - \(C_D = \) --- ## **2. Numerical Solution** ### **Discretization (Euler or RK4 Example)** Let’s define time steps, \( \Delta t \). For each time step: \[ C_A^{n+1} = C_A^{n} + \left(-k C_A^n C_B^n\right) \Delta t \] \[ C_B^{n+1} = C_B^{n} + \left(\frac{F}{V} C_{B,feed} - k C_A^n C_B^n\right) \Delta t \] \[ C_C^{n+1} = C_C^{n} + \left(k C_A^n C_B^n\right) \Delta t \] \[ C_D^{n+1} = C_D^{n} + \left(k C_A^n C_B^n\right) \Delta t \] Where: - \(k = 2.2\) dm³/(mol·s) - \(F = .05\) dm³/s - \(V = 5\) dm³ - \(C_{B,feed} = .025\) mol/dm³ --- ## **3. Graphical Representation** **Plot:** \(C_A(t)\), \(C_B(t)\), \(C_C(t)\), and \(C_D(t)\) for \(t = \) to \(500\) s. - x-axis: Time (s) - y-axis: Concentration (mol/dm³) --- ## **4. Interpretation of Results** - **A** decreases monotonically (only consumed). - **B** increases at first (from feed), then reaches a plateau or decreases as reaction consumes it. - **C** and **D** increase as products. **Physical Explanation:** - **B**: Initially increases due to feed, but as reaction accelerates, more B is consumed; thus, B concentration may plateau or slightly decrease. - **A**: Only decreases, as not fed. - **C/D**: Increase as reaction proceeds. --- # **Final Answers** ## **a. Differential Equations:** \[ \boxed{ \begin{align*} \frac{dC_A}{dt} &= -k C_A C_B \\ \frac{dC_B}{dt} &= \frac{F}{V} C_{B,feed} - k C_A C_B \\ \frac{dC_C}{dt} &= k C_A C_B \\ \frac{dC_D}{dt} &= k C_A C_B \\ \end{align*} } \] With \(k = 2.2\) dm³/(mol·s), \(F = .05\) dm³/s, \(V = 5\) dm³, \(C_{B,feed} = .025\) mol/dm³, initial conditions above. --- ## **b. Concentration Profiles** - **Use Euler or RK4 to integrate from \(t = \) to \(500\) s.** - **Plot \(C_A, C_B, C_C, C_D\) vs. time.** --- If you need a sample code (Python/Excel) for numerical solution, let me know! If you need example plots or further explanation, just ask.