Certainly! Let's break down the problem and walk through the step-by-step solution as described in the text and figure: --- # **Step-by-Step Solution: FCC Regenerator Energy Balance** ## **Step 1: Define the Control Volume and State Variables** - **Control Volume:** Includes the fluidized bed and the freeboard region (see diagram). - **Inlets:** Spent catalyst (at \( T_{sp} \)), combustion air (at \( T_{air} \)). - **Outlets:** Regenerated catalyst (at \( T_{reg} \)), flue gas (at \( T_{fg} \)). --- ## **Step 2: Write the General Energy (Enthalpy) Balance** At steady-state, the total enthalpy entering the control volume = total enthalpy leaving + heat loss + heat generated by coke combustion. \[ \text{Inlet enthalpy} + Q_{\text{gen}} = \text{Outlet enthalpy} \] Where: - \( Q_{\text{gen}} \) is the heat generated by combustion of coke on the spent catalyst. --- ## **Step 3: Express Each Enthalpy Term** - **Spent Catalyst Inlet:** \( F_c C_{pc} (T_{sp} - T_{ref}) \) - **Combustion Air Inlet:** \( F_{air} C_{p,air} (T_{air} - T_{ref}) \) - **Regenerated Catalyst Outlet:** \( F_c C_{pc} (T_{reg} - T_{ref}) \) - **Flue Gas Outlet:** \( F_{fg} C_{p,fg} (T_{fg} - T_{ref}) \) Where: - \( F_c \): catalyst flow rate - \( C_{pc} \): specific heat of catalyst - \( T_{sp}, T_{reg} \): spent/regenerated catalyst temperatures - (Similar notation for air and flue gas) --- ### **Heat Generation by Coke Combustion** \[ Q_{\text{gen}} = F_c \cdot Y_{\text{coke}} \cdot (-\Delta H_{\text{comb}}) \] - \( Y_{\text{coke}} \): mass fraction of coke on spent catalyst - \( -\Delta H_{\text{comb}} \): heat of combustion of coke (positive, since combustion is exothermic) --- ## **Step 4: Assume Perfect Mixing and Complete Combustion** - All coke is converted to \( \text{CO}_2 \). - Temperatures in the bed are uniform. --- ## **Step 5: Write the Full Steady-State Balance** \[ F_c C_{pc} (T_{sp} - T_{ref}) + F_{air} C_{p,air} (T_{air} - T_{ref}) + F_c Y_{\text{coke}} (-\Delta H_{\text{comb}}) = F_c C_{pc} (T_{reg} - T_{ref}) + F_{fg} C_{p,fg} (T_{fg} - T_{ref}) \] - You can rearrange to solve for the unknown, typically \( T_{reg} \) (the bed temperature). --- ## **Step 6: Solve for Bed Temperature (\( T_{reg} \))** Assume all terms except \( T_{reg} \) are known or can be estimated. \[ T_{reg} = \frac{F_c C_{pc} T_{sp} + F_{air} C_{p,air} T_{air} + F_c Y_{\text{coke}} (-\Delta H_{\text{comb}}) - F_{fg} C_{p,fg} T_{fg}}{F_c C_{pc}} \] --- ## **Final Answer (Summary)** ### **The steady-state bed temperature (\( T_{reg} \)) of the FCC regenerator is given by:** \[ \boxed{ T_{reg} = \frac{F_c C_{pc} T_{sp} + F_{air} C_{p,air} T_{air} + F_c Y_{\text{coke}} (-\Delta H_{\text{comb}}) - F_{fg} C_{p,fg} T_{fg}}{F_c C_{pc}} } \] **This equation balances:** - The enthalpy of incoming catalyst and air, - The heat released by coke combustion, - The enthalpy carried away by the outgoing catalyst and flue gas. --- ### **Key Assumptions:** - Steady state - Complete coke combustion to \( \text{CO}_2 \) - Perfect mixing (uniform bed temperature) - Negligible heat losses to surroundings (unless specified) --- ### **What To Do Next:** - Plug in the given values for flow rates, specific heats, temperatures, coke content, and heat of combustion to calculate \( T_{reg} \) for an actual problem. --- **Let me know if you need a sample calculation or have specific values to work with!**