Let's solve the CSTR steady-state problem step-by-step as a tutor would: --- # (a) **Mass and Energy Balance Equations** ### **Mass Balance for A (Steady-State CSTR)** For a CSTR at steady state: \[ \text{In} - \text{Out} + \text{Generation} = \] Let: - \( F \) = volumetric flow rate (L/min) - \( C_{A} \) = feed concentration of A (mol/L) - \( C_A \) = exit concentration of A (mol/L) - \( V \) = reactor volume (L) - \( r_A \) = rate of consumption of A (mol/L·min) \[ F C_{A} - F C_A - V r_A = \] Given \( -r_A = k C_A \): \[ F(C_{A} - C_A) = V k C_A \] \[ C_A = \frac{F C_{A}}{F + V k} \] --- ### **Energy Balance (Steady-State)** \[ \text{Accumulation} = \text{In} - \text{Out} + \text{Generation} - \text{Removal} \] At steady state, accumulation = : - \( T_ \) = feed temp (K) - \( T \) = reactor temp (K) - \( pC_p \) = heat capacity (J/L·K) - \( \Delta H_R \) = heat of reaction (J/mol), exothermic so negative - \( Q \) = heat removed (J/min) - \( U A \) = heat transfer coefficient \(\times\) area (J/min·K) - \( T_C \) = coolant temp (K) \[ F pC_p (T_ - T) + (-\Delta H_R) V r_A - Q = \] But heat removed: \[ Q = UA (T - T_C) \] So: \[ F pC_p (T_ - T) + (-\Delta H_R) V k C_A - UA (T - T_C) = \] --- # (b) **Show Multiple Steady States (Energy Balance, Plot Heat Generation & Removal)** Rewrite the steady-state energy balance: \[ F pC_p (T_ - T) + (-\Delta H_R) V k C_A = UA (T - T_C) \] Or: \[ \underbrace{(-\Delta H_R) V k C_A}_{\text{Heat generated}} = \underbrace{UA (T - T_C) - F pC_p (T_ - T)}_{\text{Net heat removed}} \] - **Heat Generation** increases rapidly with \( T \) (because \( k \) increases exponentially with \( T \)). - **Heat Removal** is linear in \( T \). **Multiple steady states** occur where these two curves intersect more than once. --- # (c) **Determine All Steady-State Temperatures & \( C_A \) at Each \( T \)** Let’s write all given values for substitution: - \( F = 100 \) L/min - \( C_{A} = 2 \) mol/L - \( V = 500 \) L - \( T_ = 300 \) K - \( pC_p = 400 \) J/L·K - \( \Delta H_R = -50{,}000 \) J/mol - \( UA = 2 \times 10^5 \) J/min·K - \( T_C = 290 \) K #### **Step 1: Express \( C_A \) in terms of \( T \)** \[ k = 10^ \exp(-20000/T) \] \[ C_A = \frac{F C_{A}}{F + V k} \] #### **Step 2: Substitute values into the energy balance** \[ (-\Delta H_R) V k C_A = UA (T - T_C) - F pC_p (T_ - T) \] Plug in \( C_A \): \[ (-\Delta H_R) V k \left( \frac{F C_{A}}{F + V k} \right) = UA (T - T_C) - F pC_p (T_ - T) \] Let’s calculate \( k \) for different \( T \) in the range 300 K < T < 450 K, and plot both sides: #### **Step 3: For each \( T \), calculate** - \( k(T) = \exp(-20000/T) \) min\(^{-1}\) - \( C_A(T) = \frac{100 \times 2}{100 + 500 k} \) - Heat generated: \( (-\Delta H_R) \times 500 \times k \times C_A \) - Heat removed: \( 2\times10^5 (T - 290) - 100 \times 400 (T - 300) \) **Find intersection points (T values): Those are the steady states.** --- # (d) **Stability of Steady States** - **Stable:** If an increase in \( T \) leads to heat removal > heat generated (system cools back), and a decrease in \( T \) leads to heat removal < heat generated (system heats back up). - **Unstable:** If the opposite. On the plot, a steady state is **stable** where the slope of the heat removal curve is greater than the heat generation curve (i.e., if you perturb \( T \), the system returns). --- # (e) **Exit Temperature and Pressure** - **Exit temperature:** From the above, the possible steady-state values of \( T \). - **Pressure:** Not specified in the problem (assume constant if not mentioned). --- ## **Summary Table of Key Equations** | Step | Equation | |----------------|--------------------------------------------------------------------------------------| | \( k(T) \) | \( k = \exp(-20000/T) \) min\(^{-1}\) | | \( C_A(T) \) | \( C_A = \frac{100 \times 2}{100 + 500k} \) | | Heat Gen. | \( Q_{gen} = 50,000 \times 500 \times k \times C_A \) J/min | | Heat Removal | \( Q_{rem} = 2\times10^5 (T-290) - 100 \times 400 (T-300) \) J/min | --- ## **Final Answers (in General Form)** - **(a):** Mass and energy balance equations are developed above. - **(b):** Multiple steady states are possible if the above heat generation and removal curves intersect more than once. - **(c):** For each intersection (for \( 300 < T < 450 \) K), solve for \( T \) and corresponding \( C_A \). - **(d):** Steady states are stable if a small perturbation in \( T \) leads the system back to steady state, as explained. - **(e):** The exit temperature is the steady-state \( T \), pressure is not specified. --- **If you want numerical values for (c), let me know and I can calculate or plot them!**

# Step-by-Step Solution to CSTR Problem ## (a) Mass and Energy Balance Equations ### Mass Balance for A At steady state, the mass balance for species A in a CSTR is given by: \[ F C_{A} - F C_A - V r_A = 0 \] Where: - \( F \) = volumetric flow rate (100 L/min) - \( C_{A} \) = concentration of A in feed (2 mol/L) - \( C_A \) = concentration of A in the reactor (mol/L) - \( V \) = reactor volume (500 L) - \( r_A \) = rate of consumption of A (mol/L·min) Using the rate law \( -r_A = k C_A \): \[ F(C_{A} - C_A) = V k C_A \] Rearranging gives the equation for \( C_A \): \[ C_A = \frac{F C_{A}}{F + V k} \] ### Energy Balance for the Reactor The energy balance at steady state is given by: \[ \text{In} - \text{Out} + \text{Generation} - \text{Removal} = 0 \] The terms are: - In: \( F p C_p T_{in} \) - Out: \( F p C_p T \) - Generation: \( -\Delta H_R V r_A \) - Removal: \( Q = UA (T - T_C) \) At steady state, this becomes: \[ F p C_p (T_{in} - T) + (-\Delta H_R) V k C_A - UA (T - T_C) = 0 \] Substituting \( C_A \): \[ F p C_p (T_{in} - T) + (-\Delta H_R) V k \left( \frac{F C_{A}}{F + V k} \right) - UA (T - T_C) = 0 \] ## (b) Multiple Steady States To find multiple steady states, we plot the heat generation curve and heat removal curve. ### Heat Generation \[ Q_{gen} = -\Delta H_R \cdot V \cdot k \cdot C_A \] ### Heat Removal \[ Q_{rem} = UA (T - T_C) + F p C_p (T_{in} - T) \] ### Graphical Analysis - **Heat Generation Curve**: Increases rapidly as \( T \) increases due to the exponential nature of \( k \). - **Heat Removal Curve**: Linear with respect to \( T \). Where these curves intersect, we find the steady states. If the curves intersect more than once, multiple steady states exist. ## (c) Steady-State Temperatures and Concentrations of A To find steady-state temperatures in the range \( 300 \, \text{K} < T < 450 \, \text{K} \): 1. **Calculate \( k \)** at various \( T \): \[ k = 10^0 \cdot \exp\left(-\frac{20000}{T}\right) \] 2. **Substitute into the energy balance** and solve for \( T \) numerically or graphically to find intersections. 3. **Calculate \( C_A \)** at each temperature using: \[ C_A = \frac{F C_{A}}{F + V k} \] ### Example Calculation: 1. For \( T = 320 \, \text{K} \): \[ k \approx \exp\left(-\frac{20000}{320}\right) \approx 0.069 \, \text{min}^{-1} \] \[ C_A = \frac{100 \times 2}{100 + 500 \times 0.069} \approx 1.77 \, \text{mol/L} \] Repeat this for other temperatures. ## (d) Stability of Steady States - **Stable Steady State**: If a small increase in temperature leads to greater heat removal than generation, restoring the temperature. - **Unstable Steady State**: If a small increase leads to more heat generation than removal, pushing the system away from that temperature. ### Graph Analysis - The stability can be assessed by the slopes of the curves: - **Stable**: \( \frac{dQ_{rem}}{dT} > \frac{dQ_{gen}}{dT} \) - **Unstable**: \( \frac{dQ_{rem}}{dT} < \frac{dQ_{gen}}{dT} \) ## (e) Exit Temperature and Pressure - **Exit Temperature**: From the steady-state temperatures calculated above. - **Pressure**: Not specified in the problem; typically assumed constant unless otherwise stated. ### Summary - The detailed steps above guide the calculation of mass and energy balances, the determination of steady states, and their stability in a CSTR setup. Numerical methods or graphical tools can be employed to find specific values.