# Combustion of Ethylene: Step-by-Step Solution ## **Given Data** - Ethylene flow rate: \( \dot{m}_{C_2H_4} = .01\, \mathrm{kg/s} \) - Ethylene inlet: \( T = 25^\circ\mathrm{C} = 298\, \mathrm{K} \), \( P = 1\, \mathrm{atm} \) - Air supplied: 150% of stoichiometric, \( T = 500\, \mathrm{K} \), \( P = 1\, \mathrm{atm} \) - Combustion products exit: \( T = 800\, \mathrm{K} \), \( P = 1\, \mathrm{atm} \) - Burner wall temperature: \( T_b = 600\, \mathrm{K} \) --- ## **a. Air Mass Flow Rate (\( \dot{m}_\text{air} \))** ### 1. **Balanced Combustion Reaction** Ethylene combustion with oxygen: \[ \mathrm{C_2H_4} + 3\,\mathrm{O_2} \rightarrow 2\,\mathrm{CO_2} + 2\,\mathrm{H_2O} \] - **Stoichiometric O₂ per 1 mol C₂H₄:** 3 mol O₂ ### 2. **Stoichiometric Air Required** - **Air composition:** 21% O₂ by mole (or 1 mol O₂ in 4.76 mol air) \[ \text{Air per 1 mol C}_2\mathrm{H}_4 = 3 \text{ mol O}_2 \times \frac{4.76 \text{ mol air}}{1 \text{ mol O}_2} = 14.28 \text{ mol air} \] ### 3. **Mole Calculation for Given Mass Flow** - **Molar mass C₂H₄:** \( 2 \times 12.01 + 4 \times 1.008 = 28.05\,\mathrm{g/mol} \) - **Moles per second:** \[ n_{C_2H_4} = \frac{.01\,\mathrm{kg/s}}{.02805\,\mathrm{kg/mol}} \approx .3566\,\mathrm{mol/s} \] ### 4. **Air Required (150% Stoichiometric)** - **Stoichiometric air required:** \[ n_\text{air,stoich} = .3566\,\mathrm{mol/s} \times 14.28 = 5.095\,\mathrm{mol/s} \] - **Actual air supplied (150%):** \[ n_\text{air,actual} = 1.5 \times 5.095 = 7.643\,\mathrm{mol/s} \] ### 5. **Convert to Mass Flow** - **Molar mass of air:** \( .21 \times 32 + .79 \times 28 = 28.96\,\mathrm{g/mol} \) - **Mass flow rate:** \[ \dot{m}_\text{air} = n_\text{air,actual} \times .02896\,\mathrm{kg/mol} = 7.643 \times .02896 = \boxed{.221\,\mathrm{kg/s}} \] --- ## **b. Rate of Heat Transfer (\( \dot{Q} \), in kW)** ### 1. **Energy Balance (Steady Flow)** \[ \dot{Q} = \sum n_\text{out} h_\text{out} - \sum n_\text{in} h_\text{in} \] - **Neglect kinetic and potential energy changes.** - **Reference state:** \( 298\,\mathrm{K} \), \( 1\,\mathrm{atm} \) ### 2. **Find Moles of Each Product (per s)** #### Combusted C₂H₄: - Consumed: \( .3566\,\mathrm{mol/s} \) - O₂ consumed: \( .3566 \times 3 = 1.0698\,\mathrm{mol/s} \) - CO₂ formed: \( .3566 \times 2 = .7132\,\mathrm{mol/s} \) - H₂O formed: \( .3566 \times 2 = .7132\,\mathrm{mol/s} \) #### Air Supplied: - Air supplied: \( n_\text{air,actual} = 7.643\,\mathrm{mol/s} \) - O₂ supplied: \( 7.643 \times .21 = 1.605\,\mathrm{mol/s} \) - O₂ excess (not consumed): \( 1.605 - 1.0698 = .535\,\mathrm{mol/s} \) - N₂ supplied: \( 7.643 \times .79 = 6.038\,\mathrm{mol/s} \) #### Product Mixture (per second): - CO₂: \( .7132 \) - H₂O: \( .7132 \) - O₂ (excess): \( .535 \) - N₂: \( 6.038 \) - **Total:** \( .7132 + .7132 + .535 + 6.038 = 8.00 \) mol/s ### 3. **Enthalpy Calculations** #### Reference: \( 298\,\mathrm{K} \) ##### a) **Inlet Streams** - **Ethylene:** at 298 K (reference), \( h = \) - **Air:** at 500 K - Enthalpy of air at 500 K relative to 298 K (using \( C_p \) for air ≈ 29 J/mol·K): \[ \Delta h_\text{air,in} = C_p \Delta T = 29 \times (500-298) = 29 \times 202 = 5858\,\mathrm{J/mol} \] - Total for air: \( 7.643 \times 5858 = 44,791\,\mathrm{J/s} \) ##### b) **Outlet Streams** (Products at 800 K) - For each gas, use average \( C_p \) values: - CO₂: \( C_p \approx 44\,\mathrm{J/mol\cdot K} \) - H₂O (g): \( C_p \approx 36\,\mathrm{J/mol\cdot K} \) - O₂, N₂: \( C_p \approx 29\,\mathrm{J/mol\cdot K} \) - \( \Delta T = 800 - 298 = 502\,\mathrm{K} \) - Enthalpy change for each product: - CO₂: \( .7132 \times 44 \times 502 = 15,736\,\mathrm{J/s} \) - H₂O: \( .7132 \times 36 \times 502 = 12,889\,\mathrm{J/s} \) - O₂: \( .535 \times 29 \times 502 = 7,784\,\mathrm{J/s} \) - N₂: \( 6.038 \times 29 \times 502 = 87,925\,\mathrm{J/s} \) - **Sum:** \( 15,736 + 12,889 + 7,784 + 87,925 = 124,334\,\mathrm{J/s} \) ##### c) **Standard Enthalpy of Reaction (\( \Delta H_{rxn}^ \))** - For complete combustion at 298 K: \[ \mathrm{C_2H_4} + 3\,\mathrm{O_2} \rightarrow 2\,\mathrm{CO_2} + 2\,\mathrm{H_2O} \] - \( \Delta H_f^ \) (kJ/mol): - C₂H₄: 52.3 - O₂: - CO₂: -393.5 - H₂O (g): -241.8 \[ \Delta H_{rxn}^ = [2 \times (-393.5) + 2 \times (-241.8)] - [52.3 + ] = [-787 - 483.6] - 52.3 = -127.6 - 52.3 = -1322.9\,\mathrm{kJ/mol} \] (But proper calculation: products minus reactants) \[ = [2 \times (-393.5) + 2 \times (-241.8)] - [52.3] = (-787. - 483.6) - 52.3 = -127.6 - 52.3 = -1322.9\,\mathrm{kJ/mol} \] - For .3566 mol/s: \( -1322.9 \times .3566 = -471.7\,\mathrm{kJ/s} \) ##### d) **Sum Up for \(\dot{Q}\)** \[ \dot{Q} = \text{Total } h_\text{products} - \text{Total } h_\text{reactants} \] - \( h_\text{products} = \) Sensible enthalpy rise (from 298 to 800 K) + formation enthalpy - \( h_\text{reactants} = \) Sensible enthalpy of air (from 298 to 500 K) + formation enthalpy of ethylene \[ \dot{Q} = [\text{sensible } h_\text{products}] + [n_{C_2H_4} \Delta H_{rxn}^] - [\text{sensible } h_\text{air}] \] \[ = 124,334\,\mathrm{J/s} + (-471,700\,\mathrm{J/s}) - 44,791\,\mathrm{J/s} \] \[ = (-471,700 + 124,334 - 44,791)\,\mathrm{J/s} \] \[ = (-471,700 + 79,543)\,\mathrm{J/s} \] \[ = -392,157\,\mathrm{J/s} \] \[ \boxed{\dot{Q} = -392\,\mathrm{kW}} \] (The negative sign means heat is **lost** from the system, i.e., transferred **out**.) --- ## **c. Rate of Entropy Production (\( \dot{S}_{gen} \), in kW/K)** ### 1. **Entropy Balance (Steady State)** \[ \dot{S}_{gen} = \sum n_\text{out} s_\text{out} - \sum n_\text{in} s_\text{in} - \frac{\dot{Q}}{T_b} \] ### 2. **Estimate Entropy Terms** - Specific entropy change for each species (ideal gas): \[ \Delta s = \int_{T_}^{T} \frac{C_p}{T} dT \approx C_p \ln \frac{T}{T_} \] #### a) **Inlet Air (from 298 K to 500 K):** - \( \Delta s_\text{air,in} = 29 \ln \frac{500}{298} = 29 \times .517 = 15.\,\mathrm{J/mol\cdot K} \) - Total: \( 7.643 \times 15. = 114.6\,\mathrm{J/s\cdot K} \) #### b) **Products (from 298 K to 800 K):** - For each product: - \( \ln \frac{800}{298} = .975 \) - CO₂: \( 44 \times .975 = 42.9 \) - H₂O: \( 36 \times .975 = 35.1 \) - O₂: \( 29 \times .975 = 28.3 \) - N₂: \( 29 \times .975 = 28.3 \) - Multiply by moles/s: - CO₂: \( .7132 \times 42.9 = 30.6 \) - H₂O: \( .7132 \times 35.1 = 25. \) - O₂: \( .535 \times 28.3 = 15.1 \) - N₂: \( 6.038 \times 28.3 = 171. \) - **Total out:** \( 30.6 + 25. + 15.1 + 171. = 241.7\,\mathrm{J/s\cdot K} \) #### c) **Net Entropy Flow** \[ \Delta \dot{S}_\text{flow} = 241.7 - 114.6 = 127.1\,\mathrm{J/s\cdot K} \] #### d) **Heat Transfer Entropy Change** \[ \frac{\dot{Q}}{T_b} = \frac{-392,000}{600} = -653.3\,\mathrm{J/s\cdot K