Let's break down the solution step by step: --- ## **Step 1: Given Data** - **Air flow rate** (\(Q\)): 771 SCFM (Standard Cubic Feet per Minute) - **Benzene content**: 13 mol% in air - **Required benzene removal**: 90% - **Pressure**: 1 atm - **Heat of condensation of benzene** (\(\Delta H_{cond}\)): 13,236 Btu/lb-mole at 1 atm, 176°F - **Specific heats**: - Air: \(C_{p,air} = 6.96\) Btu/(lb-mole·°F) - Benzene: \(C_{p,benzene} = 19.65\) Btu/(lb-mole·°F) - Transfer medium: \(C_{p,med} = .65\) Btu/(lb-mole·°F) - **No heat loss to atmosphere** - **Overall heat transfer coefficient**: \(U = 40\) Btu/(hr·ft²·°F) - **Basis**: 1 hour of operation - **Critical temperature of benzene**: \(T_c = 1012^\circ R\) - **Condenser outlet pressure**: 1 atm --- ## **Step 2: Calculate Moles of Air and Benzene (per hour)** ### 1. **Standard Air Density (°C, 1 atm):** 1 SCFM = 1.325 kg/hr Molecular weight of air ≈ 29 g/mol = .029 kg/mol So, 1 SCFM ≈ 1.325/.029 = 45.7 mol/hr (approx) For 771 SCFM: \[ \text{Moles of air per hour} = 771 \times 45.7 \approx 35,238 \text{ mol/hr} \] ### 2. **Total Moles in Stream:** The stream contains both air and benzene (13 mol% benzene): Let \( n_{tot} \) = total moles/hr \( n_{benzene} = .13 n_{tot} \) \( n_{air} = .87 n_{tot} \) From air flow: \[ n_{air} = 35,238 = .87 n_{tot} \implies n_{tot} = \frac{35,238}{.87} \approx 40,510 \text{ mol/hr} \] \[ n_{benzene} = .13 \times 40,510 \approx 5,266 \text{ mol/hr} \] --- ## **Step 3: Amount of Benzene to Remove** \[ \text{Benzene to remove (90% of feed)} = .90 \times 5,266 \approx 4,739 \text{ mol/hr} \] --- ## **Step 4: Energy Balance** We assume the outlet temperature is the dew point of benzene in air (at 1 atm, after 90% removal). ### **Energy to Remove (Benzene condensation):** \[ Q_{cond} = \text{moles condensed} \times \Delta H_{cond} \] \[ = 4,739 \text{ mol/hr} \times 13,236 \text{ Btu/mol} \] \[ = 62,739,804 \text{ Btu/hr} \] But we also need to **cool the entire gas stream** from inlet temperature to the condenser outlet temperature. ### **Assume Inlet Temperature = 77°F** #### **Sensible Cooling (Gas Phase):** \[ Q_{cool} = [n_{air} C_{p,air} + n_{benzene,exit} C_{p,benzene}] \times (T_{in} - T_{out}) \] - \(n_{air}\) = 35,238 mol/hr - \(n_{benzene,exit}\) = .10 × 5,266 ≈ 527 mol/hr (remaining benzene in vapor) - \(C_{p,air}\) = 6.96 Btu/(mol·°F) - \(C_{p,benzene}\) = 19.65 Btu/(mol·°F) \[ Q_{cool} = [35,238 \times 6.96 + 527 \times 19.65] \times (77 - T_{out}) \] \[ = [245,457 + 10,351] \times (77 - T_{out}) \] \[ = 255,808 \times (77 - T_{out}) \text{ Btu/hr} \] --- ## **Step 5: Dew Point Calculation** ### **What is the Outlet Temperature (\(T_{out}\))?** After removing 90% of benzene, the vapor phase has 10% of the original benzene: \[ n_{benzene,exit} = .10 \times 5,266 = 527 \text{ mol/hr} \] \[ y_{benzene,out} = \frac{527}{35,238 + 527} \approx \frac{527}{35,765} \approx .0147 \] So, **benzene mole fraction in exit gas**: **.0147** At 1 atm total pressure, the partial pressure of benzene in the exit gas is: \[ P_{benzene} = y_{benzene,out} \times 1\text{ atm} = .0147\text{ atm} \] #### **Find Temperature Where \(P_{benzene,sat} = .0147\) atm** Use Antoine equation for benzene: \[ \log_{10} P^* = A - \frac{B}{C+T} \] Where \(P^*\) is in mmHg, \(T\) in °C. For benzene: - \(A = 6.87987\) - \(B = 1196.76\) - \(C = 219.161\) Convert atm to mmHg: \[ .0147 \text{ atm} \times 760 = 11.17 \text{ mmHg} \] Set \(P^* = 11.17\) mmHg, solve for \(T\): \[ \log_{10} 11.17 = 6.87987 - \frac{1196.76}{219.161 + T} \] \[ 1.048 = 6.87987 - \frac{1196.76}{219.161 + T} \] \[ 6.87987 - 1.048 = \frac{1196.76}{219.161 + T} \] \[ 5.83187 = \frac{1196.76}{219.161 + T} \] \[ 219.161 + T = \frac{1196.76}{5.83187} \] \[ 219.161 + T = 205.26 \] \[ T = 205.26 - 219.161 = -13.90^\circ \text{C} \] Convert to °F: \[ T_{out} = (-13.90 \times 9/5) + 32 = -24.82 + 32 = 7.18^\circ \text{F} \] --- ## **Step 6: Final Answer** ### **Summary Table** | Parameter | Value | |----------------------------------|-------------| | Air flow (mol/hr) | 35,238 | | Benzene in (mol/hr) | 5,266 | | Benzene condensed (mol/hr) | 4,739 | | Benzene out (mol/hr) | 527 | | Outlet benzene mole fraction | .0147 | | Outlet temperature (\(T_{out}\)) | **7.2°F** | --- ### **Final Answer** \[ \boxed{\text{The necessary temperature to achieve 90% removal of benzene is approximately 7.2°F.}} \] --- ### **Explanation** - We calculated the dew point of benzene for the final mole fraction in the exit air, which corresponds to the target removal. - The energy balance (cooling and condensation) can be calculated using the above values for condenser sizing, but the key answer is the condenser temperature. --- **Let me know if you want the heat duty or condenser area calculated as well!**