Let's break down and solve the two main problems shown in the image: --- # 1. **Counter-Current Heat Exchanger** **Given:** - Hot oil enters at 200°C - Mass flow rate of oil, \( m = 2 \) kg/s - Specific heat of oil, \( c_p = 25 \) kJ/kg·K **Formula:** \[ Q = m \cdot c_p \cdot \Delta T \] ### **Step 1: Find the heat transferred (\( Q \))** - \( \Delta T \) is the temperature difference (assume outlet temperatures are provided or can be found) \[ Q = 2 \ \text{kg/s} \times 25\, \text{kJ/kg·K} \times \Delta T \] \[ Q = 50 \ \text{kJ/s} \times \Delta T \] - If \( \Delta T \) is known, plug in the value to get \( Q \). ### **Step 2: Log Mean Temperature Difference (LMTD)** - For counter-current exchangers: \[ Q = U \cdot A \cdot \Delta T_{LMTD} \] \[ \Delta T_{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln\left(\frac{\Delta T_1}{\Delta T_2}\right)} \] **From the image:** \[ Q = U \cdot A \cdot \frac{1}{2} \ln 21:2 \] (this is likely a shorthand; for real calculation, you need \( U \), \( A \), and inlet/outlet temperatures.) --- # 2. **Continuous Reactor with Recycle** **Given:** - First-order reaction \( A \rightarrow B \) - Rate constant \( k = .5 \) min\(^{-1}\) - Recycle ratio \( R = .8 \) ## **Step 1: Overall Mass Balance** \[ F = F_{A} \cdot X_A \quad \text{(with Recycle)} \] ## **Step 2: Design Equation for CSTR** \[ C_A = X_A \left[ C_{A} - (\text{something}) + kC_A^* V \right] \] (The formula in the image appears to have some transcription errors. Standard CSTR with recycle equation is:) \[ C_A = \frac{C_{A}}{1 + k \tau} \] Where \(\tau\) is the residence time. **With recycle:** The effective flow increases by \(1+R\): \[ \tau = \frac{V}{F_ (1+R)} \] ## **Step 3: Steady-State Conversion** \[ X_A = 1 - \frac{C_A}{C_{A}} \] ### **Calculate Ratio** Calculate the ratio of the steady-state conversion of A to the final concentration of B. --- ## **Final Answers (Symbolic Form):** ### **Heat Exchanger** \[ Q = 2 \cdot 25 \cdot \Delta T = 50 \Delta T \ \text{kJ/s} \] ### **CSTR with Recycle** - Effective flow rate: \( F = F_ (1+R) \) - Residence time: \( \tau = \frac{V}{F_ (1+R)} \) - Outlet concentration: \( C_A = \frac{C_{A}}{1 + k \tau} \) - Conversion: \( X_A = 1 - \frac{C_A}{C_{A}} \) --- **If you provide specific values (e.g., volume, inlet concentration), I can compute exact numerical answers.**