Let's solve the problem step by step. --- ## **Given Composition (by volume)** - **CO₂:** 6.5% - **CH₄:** 49.5% - **C₂H₆:** 16.% - **H₂:** 25.% - **N₂:** 2.5% - **O₂:** .5% - **Excess air:** 25% Assume 100 moles of fuel gas for simplicity. --- ### **Step 1: Write Combustion Reactions** #### **Combustion Equations (per mole):** 1. **CH₄ + 2O₂ → CO₂ + 2H₂O** 2. **C₂H₆ + 3.5O₂ → 2CO₂ + 3H₂O** 3. **H₂ + .5O₂ → H₂O** 4. **CO₂, N₂, O₂:** Inert (except O₂, which adds to the requirement) 5. **N₂:** Passes through --- ### **Step 2: Calculate Theoretical Oxygen Requirement** #### **Moles in 100 moles of fuel:** - CH₄: 49.5 moles - C₂H₆: 16. moles - H₂: 25. moles - CO₂: 6.5 moles - N₂: 2.5 moles - O₂: .5 moles #### **O₂ required:** - For CH₄: \(49.5 \times 2 = 99.\) moles O₂ - For C₂H₆: \(16. \times 3.5 = 56.\) moles O₂ - For H₂: \(25. \times .5 = 12.5\) moles O₂ **Total O₂ required:** \(99. + 56. + 12.5 = 167.5\) moles O₂ #### **Subtract O₂ in fuel:** O₂ in fuel = .5 moles **Net O₂ required from air:** \(167.5 - .5 = 167.\) moles O₂ #### **Theoretical Air Required:** Air contains 21% O₂ by volume (mole). \[ \text{Theoretical air (moles)} = \frac{167.}{.21} = 795.24 \text{ moles} \] #### **Actual Air (with 25% excess):** \[ \text{Actual air} = 795.24 \times 1.25 = 994.05 \text{ moles} \] --- ### **Step 3: Air to Fuel Ratio by Volume (a)** \[ \text{Air to fuel ratio (by volume)} = \frac{994.05}{100} = 9.94 : 1 \] --- ### **Step 4: Volume of Flue Gas for 10 moles of Fuel (b)** Let's use 10 moles of fuel. Scale all mole numbers by .1 (since calculations so far were done for 100 moles). #### **Products from Combustion:** - **CH₄:** 4.95 moles → 4.95 CO₂ + 9.90 H₂O - **C₂H₆:** 1.6 moles → 3.2 CO₂ + 4.8 H₂O - **H₂:** 2.5 moles → 2.5 H₂O - **CO₂ (in fuel):** .65 moles (unchanged) - **N₂ in fuel:** .25 moles (unchanged) - **N₂ from air:** Air supplied for 10 moles of fuel: \(994.05 \times .1 = 99.405\) moles air N₂ in air: 79% N₂ from air: \(99.405 \times .79 = 78.53\) moles **Total N₂ in exhaust:** \(.25 + 78.53 = 78.78\) moles #### **O₂ supplied:** From air: \(99.405 \times .21 = 20.88\) moles O₂ in fuel: .05 moles Total O₂ supplied: 20.93 moles O₂ consumed: \(9.90 + 5.6 + 1.25 = 16.75\) moles O₂ remaining: \(20.93 - 16.75 = 4.18\) moles #### **Sum of Products:** - CO₂: \(4.95 + 3.2 + .65 = 8.8\) - H₂O: \(9.90 + 4.8 + 2.5 = 17.2\) - N₂: \(78.78\) - O₂: \(4.18\) **Total moles of flue gas:** \(8.8 + 17.2 + 78.78 + 4.18 = 108.96\) moles --- ### **Step 5: Volume of Air at 21°C, 101.325 kPa for 10 moles of Fuel (c)** **Air supplied for 10 moles fuel:** \(994.05 \times .1 = 99.405\) moles Use ideal gas law: \[ V = \frac{nRT}{P} \] - \(n = 99.405\) - \(R = 8.314 \, \text{J/mol·K}\) - \(T = 273 + 21 = 294\, \text{K}\) - \(P = 101325\, \text{Pa}\) \[ V = \frac{99.405 \times 8.314 \times 294}{101325} = \frac{242,496}{101325} = 2.40 \, \text{m}^3 \] --- ### **Step 6: Partial Pressure of H₂O at 200°C, 101.325 kPa (d)** - H₂O moles: 17.2 - Total exhaust moles: 108.96 \[ \text{Partial pressure of H}_2\text{O} = 101.325 \times \frac{17.2}{108.96} = 16.\, \text{kPa} \] --- ### **Step 7: Volume of Exhaust Gas at 200°C, 101.325 kPa per m³ of Fuel (e)** First, find how many moles in 1 m³ of fuel at standard conditions (assume 1 mole ≈ 22.414 L at STP, so 1 m³ has ~44.6 moles): - For 44.6 moles of fuel, flue gas moles: \(44.6 \times \frac{108.96}{10} = 485.9\) moles At 200°C (\(T = 473\) K), 101325 Pa: \[ V = \frac{485.9 \times 8.314 \times 473}{101325} = \frac{1,907,363}{101325} = 18.83\, \text{m}^3 \] --- ## **Final Answers** ### **a) Air to fuel ratio by volume:** \[ \boxed{9.94 : 1} \] ### **b) Volume of flue gas from 10 moles of fuel:** \[ \boxed{108.96 \text{ moles}} \] ### **c) Volume of air at 21°C, 101.325 kPa for 10 moles of fuel:** \[ \boxed{2.40 \text{ m}^3} \] ### **d) Partial pressure of H₂O in exhaust at 200°C, 101.325 kPa:** \[ \boxed{16. \text{ kPa}} \] ### **e) Volume of exhaust gas at 200°C, 101.325 kPa per m³ of fuel at 20°C, 101.325 kPa:** \[ \boxed{18.83 \text{ m}^3} \] --- Let me know if you need any clarification or a deeper explanation of any step!