# CSTR Reaction Engineering Problem: Step-by-Step Solution --- ## **Given Data** | Parameter | Symbol | Value | |------------------------------|------------------|---------------------------| | Feed concentration of A | \( C_{A} \) | 6 mol/L | | Feed temperature | \( T_ \) | 320 K | | Volumetric flow rate | \( v_ \) | 15 L/min | | Reactor volume | \( V \) | 250 L | | Rate constant at 320 K | \( k_{320} \) | .08 L/(mol·min) | | Activation energy | \( E \) | 90000 kJ/kmol = 90 kJ/mol | | Heat of reaction | \( \Delta H_{rxn} \) | -95000 kJ/kmol = -95 kJ/mol | | Density | \( \rho \) | 100 kg/m³ | | Specific heat | \( C_p \) | 4.2 kJ/(kg·K) | | Coolant temperature | \( T_c \) | 290 K | | UA (overall heat transfer) | \( UA \) | 30000 kJ/(min·K) | --- ## **1. Space Time and Mean Residence Time** - **Space Time (\( \tau \))**: \[ \tau = \frac{V}{v_} \] - \( V = 250 \) L - \( v_ = 15 \) L/min \[ \tau = \frac{250}{15} = 16.67 \ \text{min} \] - **Mean Residence Time**: For a CSTR, this is equal to space time: \[ t_{res} = \tau = 16.67 \ \text{min} \] --- ## **2. Isothermal Conversion at 320 K** ### **2.1. Steady-State Mole Balance for CSTR** \[ F_{A} - F_A + r_A V = \] \[ v_ C_{A} - v_ C_A - k C_A^2 V = \] \[ C_{A} - C_A - k \tau C_A^2 = \] Let \( X \) be the conversion: \[ X = \frac{C_{A} - C_A}{C_{A}} \implies C_A = C_{A}(1-X) \] Plug into previous equation: \[ C_{A} - C_{A}(1-X) - k \tau [C_{A}(1-X)]^2 = \] \[ C_{A} X = k \tau C_{A}^2 (1-X)^2 \] \[ X = k \tau C_{A} (1-X)^2 \] Let’s denote \( \alpha = k \tau C_{A} \): \[ X = \alpha (1 - X)^2 \] \[ X = .08 \times 16.67 \times 6 \times (1-X)^2 \] \[ .08 \times 16.67 = 1.3336 \] \[ 1.3336 \times 6 = 8.0016 \approx 8 \] \[ X = 8 (1-X)^2 \] ### **2.2. Solve Quadratic for \( X \)** \[ X = 8 (1 - 2X + X^2) \] \[ X = 8 - 16X + 8X^2 \] \[ 8X^2 - 17X + 8 = \] Quadratic equation: \( aX^2 + bX + c = \) Here, \( a = 8 \), \( b = -17 \), \( c = 8 \): \[ X = \frac{17 \pm \sqrt{(-17)^2 - 4 \times 8 \times 8}}{2 \times 8} \] \[ = \frac{17 \pm \sqrt{289 - 256}}{16} = \frac{17 \pm \sqrt{33}}{16} \] \[ \sqrt{33} \approx 5.7446 \] So, \[ X_1 = \frac{17 + 5.7446}{16} = \frac{22.7446}{16} = 1.4215 \ (\text{not physical, } X > 1) \] \[ X_2 = \frac{17 - 5.7446}{16} = \frac{11.2554}{16} = .7035 \] **Final Isothermal Conversion at 320 K:** \[ \boxed{X_{320} = .70} \] (rounded to two decimal places) ### **2.3. Outlet Concentration of A** \[ C_A = C_{A} (1-X) = 6 \times (1 - .7035) = 6 \times .2965 = 1.779 \ \text{mol/L} \] \[ \boxed{C_{A,\,out,\,320} = 1.78 \ \mathrm{mol/L}} \] --- ## **3. Rate of Heat Generation at 320 K** \[ \text{Heat generated} = -\Delta H_{rxn} \cdot r_A V \] At steady-state outlet: \[ r_A = -k C_A^2 = -.08 \times (1.78)^2 = -.08 \times 3.1684 = -.25347 \ \text{mol/L·min} \] Total rate (for whole reactor): \[ r_A V = -.25347 \times 250 = -63.37 \ \text{mol/min} \] \[ \text{Heat generation rate} = -(-95\,\text{kJ/mol}) \times (-63.37\,\text{mol/min}) = 95 \times 63.37 = 602 \ \text{kJ/min} \] \[ \boxed{q_{gen,320} = 6{,}020 \ \text{kJ/min}} \] --- ## **4. Adiabatic Temperature Rise for Complete Conversion** \[ \Delta T_{ad} = \frac{(-\Delta H_{rxn}) \cdot C_{A}}{\rho C_p} \] Convert to consistent units: - \( \Delta H_{rxn} = -95 \) kJ/mol - \( C_{A} = 6 \) mol/L \( = 600 \) mol/m³ - \( \rho = 100 \) kg/m³ - \( C_p = 4.2 \) kJ/(kg·K) \[ \Delta T_{ad} = \frac{95 \times 600}{100 \times 4.2} = \frac{570,000}{4,200} \approx 135.7 \ \text{K} \] \[ \boxed{\Delta T_{ad,\,max} = 136\ \text{K} \ \text{(rounded)}} \] --- ## **5. Steady-State Outlet Temperature (Adiabatic, No Cooling)** Adiabatic energy balance: \[ v_ \rho C_p (T - T_) = -\Delta H_{rxn} \cdot r_A V \] But for full conversion (adiabatic, batch): \[ T_{ad} = T_ + \Delta T_{ad} = 320 + 136 = 456 \ \text{K} \] For partial conversion, use \( X_{ad} \): \[ T_{out} = T_ + X_{ad} \Delta T_{ad} \] If all A is converted, maximum temperature is 456 K. For 70% conversion: \[ \Delta T_{ad,\,actual} = .7035 \times 136 = 95.7 \ \text{K} \] \[ T_{out} = 320 + 95.7 = 415.7 \ \text{K} \] \[ \boxed{T_{out,\,ad,\,320} = 416 \ \text{K}} \] --- ## **6. Cooling Duty to Maintain 320 K** \[ Q_{removal} = -\Delta H_{rxn} \cdot r_A V = 6,020\ \text{kJ/min} \] **Is the cooling system sufficient?** \[ Q_{max} = UA (T_{R} - T_c) = 30,000 \times (320 - 290) = 30,000 \times 30 = 900,000 \ \text{kJ/min} \] But this is unreasonably large compared to the heat generated. In reality, you use the actual temperature difference: \[ Q_{removal,\,needed} = 6,020 \ \text{kJ/min} \] \[ Q_{removal,\,max} = 30,000 \times (320 - 290) = 900,000 \ \text{kJ/min} \] So, \[ \boxed{UA \text{ is sufficient to remove the generated heat at 320 K.}} \] --- ## **7. New Rate Constant at 340 K** Arrhenius equation: \[ k = k_ \exp\left(-\frac{E}{RT}\right) \] \[ \frac{k_{340}}{k_{320}} = \exp\left(\frac{E}{R}\left(\frac{1}{320} - \frac{1}{340}\right)\right) \] - \( E = 90,000 \) kJ/kmol \( = 90,000 \) J/mol - \( R = 8.314 \) J/(mol·K) Calculate: \[ \Delta \left(\frac{1}{T}\right) = \frac{1}{320} - \frac{1}{340} = \frac{340 - 320}{320 \times 340} = \frac{20}{108,800} = .0001839 \] \[ \frac{E}{R} \Delta \left(\frac{1}{T}\right) = \frac{90,000}{8.314} \times .0001839 = 10,825 \times .0001839 = 1.992 \] \[ k_{340} = k_{320} \times \exp(1.992) = .08 \times 7.33 = .586 \ \text{L/(mol·min)} \] \[ \boxed{k_{340} = .59 \ \text{L/(mol·min)}} \] --- ## **8. Conversion at 340 K (Isothermal)** \[ \alpha_{340} = k_{340} \tau C_{A} = .586 \times 16.67 \times 6 = 58.6 \times 1.667 = 97.7 \] \[ \alpha_{340} = 58.6 \times 1.667 = 97.7 \] So, \[ X = \alpha_{340} (1-X)^2 \] \[ X = 58.6 (1-X)^2 \] Let’s define \( y = 1 - X \): \[ X = 58.6 y^2 \] \[ X + 58.6 y^2 = 1 \] But \( X = 1 - y \): \[ 1 - y + 58.6 y^2 = 1 \] \[ - y + 58.6 y^2 = \] \[ y (58.6 y - 1) = \implies y = , \ y = \frac{1}{58.6} = .0171 \] Only physical solution is \( y = .0171 \), so \[ X_{340} = 1 - .0171 = .9829 \] \[ \boxed{X_{340} = .98} \] --- ## **9. Rate of Heat Generation at 340 K** \[ C_A = C_{A} (1 - X) = 6 \times (1 - .9829) = 6 \times .0171 = .1024 \ \text{mol/L} \] \[ r_A = -k C_A^2 = -.586 \times (.1024)^2 = -.586 \times .01049 = -.00614 \ \text{mol/L·min} \] \[ r_A V = -.00614 \times 250 = -1.535 \ \text{mol/min} \] \[ \text{Heat generation rate} = -(-95) \times (-1.535) = 145.8 \ \text{kJ/min} \] \[ \boxed{q_{gen,\,340} = 146 \ \text{kJ/min}} \] --- ## **10. Percentage Increase in Conversion (320 K to 340 K)** \[ \% \text{increase}

# CSTR Reaction Engineering Problem: Step-by-Step Solution --- ## **Given Data** | Parameter | Symbol | Value | |------------------------------|------------------|---------------------------| | Feed concentration of A | \( C_{A} \) | 6 mol/L | | Feed temperature | \( T_ \) | 320 K | | Volumetric flow rate | \( v_ \) | 15 L/min | | Reactor volume | \( V \) | 250 L | | Rate constant at 320 K | \( k_{320} \) | 0.08 L/(mol·min) | | Activation energy | \( E \) | 90,000 kJ/kmol = 90 kJ/mol | | Heat of reaction | \( \Delta H_{rxn} \) | -95,000 kJ/kmol = -95 kJ/mol | | Density | \( \rho \) | 1000 kg/m³ | | Specific heat | \( C_p \) | 4.2 kJ/(kg·K) | | Coolant temperature | \( T_c \) | 290 K | | UA (overall heat transfer) | \( UA \) | 30,000 kJ/(min·K) | --- ## **1. Space Time and Mean Residence Time** - **Space Time (\( \tau \))**: \[ \tau = \frac{V}{v_} = \frac{250 \ \text{L}}{15 \ \text{L/min}} = 16.67 \ \text{min} \] - **Mean Residence Time**: \[ t_{res} = \tau = 16.67 \ \text{min} \] --- ## **2. Isothermal Conversion at 320 K** ### **2.1. Steady-State Mole Balance for CSTR** \[ F_{A} - F_A + r_A V = 0 \] Where: \[ F_{A} = vC_A, \quad r_A = -kC_A^2 \] Substituting gives: \[ vC_{A} - vC_A - kC_A^2 \cdot V = 0 \] Rearranging: \[ C_A = C_{A}(1-X) \] Substituting in: \[ C_{A} X = k \tau C_{A}^2 (1-X)^2 \] Let \( \alpha = k \tau C_{A} \): \[ X = \alpha (1 - X)^2 \] Calculating \( \alpha \): \[ \alpha = 0.08 \times 16.67 \times 6 = 8 \] ### **2.2. Solve Quadratic for \( X \)** Setting up the equation: \[ X = 8 (1 - 2X + X^2) \] \[ 8X^2 - 17X + 8 = 0 \] Using the quadratic formula: \[ X = \frac{17 \pm \sqrt{(-17)^2 - 4 \cdot 8 \cdot 8}}{2 \cdot 8} \] Calculating the discriminant: \[ 17^2 - 4 \cdot 8 \cdot 8 = 289 - 256 = 33 \] \[ X = \frac{17 \pm \sqrt{33}}{16} \] Calculating the roots: \[ \sqrt{33} \approx 5.7446 \] \[ X_1 = \frac{22.7446}{16} \quad (\text{not physical, } X > 1) \] \[ X_2 = \frac{11.2554}{16} \approx 0.7035 \] **Final Isothermal Conversion at 320 K:** \[ \boxed{X_{320} = 0.70} \] ### **2.3. Outlet Concentration of A** \[ C_A = C_{A} (1-X) = 6 \times (1 - 0.7035) = 1.78 \ \text{mol/L} \] \[ \boxed{C_{A,\,out,\,320} = 1.78 \ \text{mol/L}} \] --- ## **3. Rate of Heat Generation at 320 K** Calculate the rate: \[ r_A = -k C_A^2 = -0.08 \times (1.78)^2 = -0.25347 \ \text{mol/L·min} \] For the whole reactor: \[ r_A V = -0.25347 \times 250 = -63.37 \ \text{mol/min} \] Heat generation rate: \[ \text{Heat generation rate} = -\Delta H_{rxn} \cdot r_A V = 95 \times 63.37 = 6020 \ \text{kJ/min} \] \[ \boxed{q_{gen,\,320} = 6020 \ \text{kJ/min}} \] --- ## **4. Adiabatic Temperature Rise for Complete Conversion** Using: \[ \Delta T_{ad} = \frac{-\Delta H_{rxn} \cdot C_{A}}{\rho C_p} \] Calculating: \[ \Delta T_{ad} = \frac{95 \times 600}{100 \times 4.2} = \frac{570000}{420} \approx 135.71 \ \text{K} \] \[ \boxed{\Delta T_{ad,\,max} = 136\ \text{K}} \] --- ## **5. Steady-State Outlet Temperature (Adiabatic, No Cooling)** Assuming complete conversion: \[ T_{ad} = T_ + \Delta T_{ad} = 320 + 136 = 456 \ \text{K} \] For partial conversion: \[ \Delta T_{ad,\,actual} = 0.7035 \times 136 = 95.7 \ \text{K} \] \[ T_{out} = 320 + 95.7 = 415.7 \ \text{K} \] \[ \boxed{T_{out,\,ad,\,320} = 416 \ \text{K}} \] --- ## **6. Cooling Duty to Maintain 320 K** Calculate the required cooling: \[ Q_{removal} = 6020 \ \text{kJ/min} \] Maximum possible cooling: \[ Q_{max} = UA (T_{R} - T_c) = 30,000 \times (320 - 290) = 900,000 \ \text{kJ/min} \] The system can effectively remove the required heat: \[ \boxed{UA \text{ is sufficient to remove the generated heat at 320 K.}} \] --- ## **7. New Rate Constant at 340 K** Using the Arrhenius equation: \[ k = k_{320} \exp\left(-\frac{E}{R}\left(\frac{1}{T_{320}} - \frac{1}{T_{340}}\right)\right) \] Where: - \( R = 8.314 \ \text{J/(mol·K)} = 0.008314 \ \text{kJ/(mol·K)} \) Calculating: \[ \frac{1}{320} - \frac{1}{340} = \frac{20}{108800} = 0.0001839 \] \[ k_{340} = 0.08 \times \exp\left(\frac{90000}{8.314} \times 0.0001839\right) \] Calculating the exponent: \[ \frac{90000}{8.314} \times 0.0001839 \approx 1.992 \] Thus: \[ k_{340} \approx 0.08 \times e^{1.992} \approx 0.08 \times 7.33 = 0.586 \ \text{L/(mol·min)} \] \[ \boxed{k_{340} = 0.59 \ \text{L/(mol·min)}} \] --- ## **8. Conversion at 340 K (Isothermal)** Calculating: \[ \alpha_{340} = k_{340} \tau C_{A} = 0.586 \times 16.67 \times 6 \approx 58.6 \] Solving for \( X \): \[ X = \alpha_{340} (1 - X)^2 \] Let \( y = 1 - X \): \[ X = 58.6y^2 \] Setting up: \[ X + 58.6y^2 = 1 \Rightarrow y(58.6y - 1) = 0 \] \[ y = 0 \quad \text{or} \quad y = \frac{1}{58.6} \] Calculating \( y \): \[ y \approx 0.0171 \quad \Rightarrow \quad X = 1 - 0.0171 \approx 0.9829 \] \[ \boxed{X_{340} = 0.98} \] --- ## **9. Rate of Heat Generation at 340 K** For outlet concentration: \[ C_A = C_{A} (1 - X_{340}) = 6 \times (1 - 0.9829) \approx 0.1024 \ \text{mol/L} \] Calculating \( r_A \): \[ r_A = -k_{340} C_A^2 = -0.586 \times (0.1024)^2 \approx -0.00614 \ \text{mol/L·min} \] Total rate: \[ r_A V = -0.00614 \times 250 \approx -1.535 \ \text{mol/min} \] Heat generation: \[ q_{gen} = -(-95) \times -1.535 = 145.8 \ \text{kJ/min} \] \[ \boxed{q_{gen,\,340} = 146 \ \text{kJ/min}} \] --- ## **10. Percentage Increase in Conversion (320 K to 340 K)** Calculating percentage increase: \[ \text{Percentage Increase} = \frac{X_{340} - X_{320}}{X_{320}} \times 100\% = \frac{0.98 - 0.70}{0.70} \times 100\% \approx 40\% \] \[ \boxed{\text{Percentage Increase in Conversion} \approx 40\%} \] --- ## **11. Discussion on Reactor Stability** **Susceptibility to Multiple Steady States or Thermal Runaway:** - The high activation energy (90 kJ/mol) indicates a relatively stable reaction with respect to temperature changes. - However, the significant heat release (6020 kJ/min) at full conversion raises concerns for thermal runaway if uncontrolled. - Given the exothermic nature of the reaction and the potential for high temperatures, proper cooling and monitoring are essential to prevent multiple steady states or thermal runaway conditions. ### Conclusion: - The reactor must be carefully controlled to maintain thermal stability, especially under varying operational conditions.