Let's tackle **Q2** step by step, as requested. --- # Q2 Solution: Calculate the Heating Requirement (in kW) ## **Step 1: Material Balances** Given: - Feed: 132 mol/s, 34. mol% benzene, rest toluene, at 25°C - Output (liquid): 20.5 mol% benzene - Output (vapor): 69.5 mol% benzene - Both liquid and vapor leave at 80°C Let: - \( F \) = feed = 132 mol/s - \( x_F \) = mole fraction benzene in feed = .34 - \( L \), \( x_L \) = liquid flow, mole fraction benzene = ? , .205 - \( V \), \( y_V \) = vapor flow, mole fraction benzene = ? , .695 ### **Overall Mole Balance:** \[ F = L + V \implies 132 = L + V \] ### **Benzene Balance:** \[ F \cdot x_F = L \cdot x_L + V \cdot y_V \] \[ 132 \times .34 = L \times .205 + V \times .695 \] Substitute \( V = 132 - L \): \[ 132 \times .34 = L \times .205 + (132 - L) \times .695 \] \[ 448.8 = .205L + 132 \times .695 - .695L \] \[ 448.8 = .205L + 917.4 - .695L \] \[ 448.8 - 917.4 = (.205 - .695)L \] \[ -468.6 = -.49L \] \[ L = \frac{-468.6}{-.49} = 956.7 \text{ mol/s} \] \[ V = 132 - 956.7 = 363.3 \text{ mol/s} \] --- ## **Step 2: Energy Balance** ### **Assumptions:** - Reference temperature: 25°C - The process is at constant pressure - Neglect kinetic and potential energy changes - Use average/approximate values for enthalpies (from steam tables or chemical engineering handbooks). ### **We need:** \[ Q = \text{Total enthalpy out} - \text{Total enthalpy in} \] ### **Mole Flows:** - Feed: 132 mol/s (34% benzene, 66% toluene) at 25°C, liquid - Output (liquid): 956.7 mol/s, 20.5% benzene, 79.5% toluene, at 80°C, liquid - Output (vapor): 363.3 mol/s, 69.5% benzene, 30.5% toluene, at 80°C, vapor ### **Enthalpies:** Assume ideal mixing and use heat capacities and latent heats: | Component | Cp(l) (J/mol·K) | Cp(g) (J/mol·K) | Hvap (kJ/mol) | Tb (°C) | |---|---|---|---|---| | Benzene | 136 | 82 | 30.8 | 80.1 | | Toluene | 156 | 95 | 33.2 | 110.6 | #### **Feed Enthalpy (at 25°C, liquid):** Reference: #### **Liquid Out (at 80°C, liquid):** \[ \Delta H = n \cdot x_{\text{benz}} \cdot Cp_{\text{benz(l)}} \cdot \Delta T + n \cdot x_{\text{tol}} \cdot Cp_{\text{tol(l)}} \cdot \Delta T \] \[ \Delta T = 80 - 25 = 55\,^\circ\text{C} \] \[ x_{\text{benz}} = .205,\quad x_{\text{tol}} = .795 \] \[ \Delta H_{L} = 956.7 \left[.205 \times 136 \times 55 + .795 \times 156 \times 55\right] \text{ J} \] Calculate each term: - Benzene: \(.205 \times 136 \times 55 = 1532.8\) - Toluene: \(.795 \times 156 \times 55 = 6821.4\) - Total per mole: \(1532.8 + 6821.4 = 8354.2\) J/mol \[ \Delta H_{L} = 956.7 \times 8354.2 = 7,992,128 \text{ J/s} = 7992 \text{ kW} \] #### **Vapor Out (at 80°C, vapor):** For each mole, need to: 1. Heat from 25°C to boiling point (benzene: 80.1°C, toluene: 110.6°C) 2. Vaporize at boiling point (at 1 atm) 3. Heat the vapor from boiling point to 80°C (for benzene, negligible; for toluene, must condense first before vaporizing, but at 80°C, toluene will not vaporize at 1 atm — let's assume it's an equilibrium vapor mixture). But, since both vapor and liquid are at 80°C, and the vapor contains mostly benzene, let's approximate: - For benzene (boiling point at 80.1°C): heat from 25°C to 80°C (liquid), vaporize (Hvap at 80°C) - For toluene (boiling pt 110.6°C): at 80°C, it's not supposed to be in vapor, but it's in the vapor phase per the problem statement, so we proceed similarly. ##### **Benzene in vapor:** - Moles: \(363.3 \times .695 = 252.7\) mol/s \[ \Delta H_{\text{benz, vapor}} = [Cp_{\text{l}} \cdot (80-25) + H_{\text{vap}}] \text{ per mole} \] \[ = 136 \times 55 + 30,800 = 748 + 30,800 = 38,280 \text{ J/mol} \] \[ \text{Total: } 252.7 \times 38,280 = 9,674,556 \text{ J/s} \] ##### **Toluene in vapor:** - Moles: \(363.3 \times .305 = 110.6\) mol/s \[ \Delta H_{\text{tol, vapor}} = [Cp_{\text{l}} \cdot (80-25) + H_{\text{vap}}] \text{ per mole} \] \[ = 156 \times 55 + 33,200 = 858 + 33,200 = 41,780 \text{ J/mol} \] \[ \text{Total: } 110.6 \times 41,780 = 4,620,268 \text{ J/s} \] \[ \Delta H_{V} = 9,674,556 + 4,620,268 = 14,294,824 \text{ J/s} = 14,295 \text{ kW} \] --- ### **Feed Enthalpy (reference at ):** \[ \Delta H_F = \] --- ### **Total Heating Requirement:** \[ Q = (\Delta H_L + \Delta H_V) - \Delta H_F \] \[ Q = (7,992,128 + 14,294,824) \text{ J/s} \] \[ Q = 22,286,952 \text{ J/s} = 22,287 \text{ kW} \] --- ## **Final Answer** ### **The heating requirement for this process is:** \[ \boxed{22,300\ \text{ kW}} \] *(rounded to 3 significant figures)* --- **Note:** - The answer assumes the enthalpy values are calculated using constant heat capacities and latent heats, and the vapor composition is as given. For a more precise answer, use enthalpy tables or detailed phase equilibrium data. - If you need the solution to Q3, let me know!