Let's break down and solve the problem step by step: --- ## Problem Statement **If Pentane (C₅H₁₂) is burned with an air to fuel ratio by weight of 19, determine:** 1. **a) Volumetric analysis of the flue gas on a wet basis.** 2. **b) Orsat analysis of the combustion products.** 3. **c) Volumetric flow rate of flue gases formed at 110 kPa, 45°C if burning 250 kg/min of fuel.** 4. **d) Average molecular weight of flue gases.** --- ## Step 1: Write the Balanced Combustion Equation ### General Combustion Equation For **Pentane:** \[ \text{C}_5\text{H}_{12} + a(\text{O}_2 + 3.76\text{N}_2) \rightarrow b\text{CO}_2 + c\text{H}_2\text{O} + d\text{O}_2 + e\text{N}_2 \] ### Stoichiometric Combustion Balanced equation for complete combustion: \[ \text{C}_5\text{H}_{12} + 8\text{O}_2 \rightarrow 5\text{CO}_2 + 6\text{H}_2\text{O} \] Air required per mole pentane (stoichiometric): \[ \text{O}_2: 8~\text{mol},~\text{Air} = \frac{8}{.21} = 38.1~\text{mol} \] But let's use the **given air-to-fuel ratio by weight**. --- ## Step 2: Air-to-Fuel Ratio by Weight ### Molecular Weights - **Pentane (C₅H₁₂):** \( 5 \times 12 + 12 \times 1 = 72~\text{g/mol} \) - **O₂:** \( 32~\text{g/mol} \) - **Air:** \( \approx 28.96~\text{g/mol} \) ### Mass of Air per 1 kg Fuel \[ \text{Air-to-fuel ratio} = 19 \implies 19~\text{kg air}~\text{per}~1~\text{kg pentane} \] ### Moles of Pentane in 1 kg \[ n_{\text{C}_5\text{H}_{12}} = \frac{100}{72} = 13.89~\text{mol} \] ### Mass and Moles of Air \[ \text{Mass of air} = 19~\text{kg} \] \[ n_{\text{air}} = \frac{19000}{28.96} = 656.~\text{mol} \] ### Moles of O₂ in Air \[ \text{Air: } 21\%~\text{O}_2 \text{ by volume (and moles)} \] \[ n_{\text{O}_2} = .21 \times 656. = 137.76~\text{mol} \] ### Moles of N₂ in Air \[ n_{\text{N}_2} = .79 \times 656. = 517.24~\text{mol} \] --- ## Step 3: Combustion Reaction with Actual Air ### Combustion of 1 kmol Pentane Scale everything to **1 kmol C₅H₁₂** for simplicity. \[ \text{For 1 kmol C}_5\text{H}_{12}:\\ \] \[ \text{Moles of air supplied: } 19 \times 72 = 1368~\text{kg air} \] \[ n_{\text{air}} = \frac{1368}{28.96} = 47.25~\text{kmol air} \] \[ n_{\text{O}_2} = .21 \times 47.25 = 9.92~\text{kmol} \] \[ n_{\text{N}_2} = .79 \times 47.25 = 37.33~\text{kmol} \] **Stoichiometric O₂ needed:** \( 8~\text{kmol} \) **O₂ supplied:** \( 9.92~\text{kmol} \) **Excess O₂:** \( 9.92 - 8 = 1.92~\text{kmol} \) ### Balanced Equation (per 1 kmol C₅H₁₂): \[ \text{C}_5\text{H}_{12} + 9.92\text{O}_2 + (9.92 \times 3.76)\text{N}_2 \rightarrow 5\text{CO}_2 + 6\text{H}_2\text{O} + 1.92\text{O}_2 + 37.33\text{N}_2 \] --- ## Step 4: Calculate Flue Gas Composition ### Products per 1 kmol C₅H₁₂: - CO₂: 5 kmol - H₂O: 6 kmol - O₂: 1.92 kmol (excess) - N₂: 37.33 kmol - **Total:** \( 5 + 6 + 1.92 + 37.33 = 50.25~\text{kmol} \) --- ### **a) Volumetric Analysis of Flue Gas (Wet Basis)** \[ \begin{align*} \text{CO}_2: &~ \frac{5}{50.25} \times 100 = 9.95\% \\ \text{H}_2\text{O}: &~ \frac{6}{50.25} \times 100 = 11.94\% \\ \text{O}_2: &~ \frac{1.92}{50.25} \times 100 = 3.82\% \\ \text{N}_2: &~ \frac{37.33}{50.25} \times 100 = 74.29\% \\ \end{align*} \] --- ### **b) Orsat Analysis (Dry Basis: CO₂, O₂, N₂ only)** Remove water vapor: - CO₂: 5 kmol - O₂: 1.92 kmol - N₂: 37.33 kmol - **Total (dry):** \( 5 + 1.92 + 37.33 = 44.25~\text{kmol} \) \[ \begin{align*} \text{CO}_2: &~ \frac{5}{44.25} \times 100 = 11.3\% \\ \text{O}_2: &~ \frac{1.92}{44.25} \times 100 = 4.34\% \\ \text{N}_2: &~ \frac{37.33}{44.25} \times 100 = 84.36\% \\ \end{align*} \] --- ### **c) Volumetric Flow Rate of Flue Gases** Given: **250 kg/min fuel burned** #### 1. Find kmol C₅H₁₂ burned per min: \[ n = \frac{250}{72} = 3.472~\text{kmol/min} \] #### 2. Total kmol flue gas per min: \[ \text{Total kmol flue gas/min} = 3.472 \times 50.25 = 174.4~\text{kmol/min} \] #### 3. Use ideal gas law to get volume: \[ P = 110~\text{kPa},~T = 45^\circ\text{C} = 318.15~\text{K} \] \[ R = 8.314~\text{kPa}\cdot\text{L}/(\text{mol}\cdot\text{K}) \] \[ V = \frac{nRT}{P} \] \[ V = \frac{174,400~\text{mol/min} \times 8.314 \times 318.15}{110,000} \] Calculate numerator: \[ 174,400 \times 8.314 \times 318.15 = 4.610 \times 10^8 \] \[ V = \frac{4.610 \times 10^8}{110,000} = 4,190~\text{L/min} = 4.19~\text{m}^3/\text{min} \] But this seems low; let's check units: \[ V = \frac{nRT}{P} \] - \( n = 174.4~\text{kmol/min} = 174,400~\text{mol/min} \) - \( R = 8.314~\text{kPa}\cdot\text{L}/(\text{mol}\cdot\text{K}) \) - \( T = 318.15~\text{K} \) - \( P = 110~\text{kPa} \) \[ V = \frac{174,400 \times 8.314 \times 318.15}{110} \] \[ = \frac{461,006,330}{110} \] \[ = 4,191,876~\text{L/min} = 4,192~\text{m}^3/\text{min} \] --- ### **d) Average Molecular Weight of Flue Gases** \[ \text{Average MW} = \frac{\sum (\text{mol}_i \times \text{MW}_i)}{\text{total mol}} \] \[ \text{CO}_2: 5 \times 44 = 220 \\ \text{H}_2\text{O}: 6 \times 18 = 108 \\ \text{O}_2: 1.92 \times 32 = 61.44 \\ \text{N}_2: 37.33 \times 28 = 1,045.24 \] \[ \text{Sum} = 220 + 108 + 61.44 + 1,045.24 = 1,434.68 \] \[ \text{Average MW} = \frac{1,434.68}{50.25} = 28.56~\text{g/mol} \] --- ## **Final Answers Summary** ### a) **Volumetric (Wet) Analysis of Flue Gas** - CO₂: **9.95%** - H₂O: **11.94%** - O₂: **3.82%** - N₂: **74.29%** ### b) **Orsat (Dry) Analysis** - CO₂: **11.3%** - O₂: **4.34%** - N₂: **84.36%** ### c) **Volumetric Flow Rate at 110 kPa, 45°C** - **4,192 m³/min** ### d) **Average Molecular Weight of Flue Gases** - **28.56 g/mol** --- Feel free to ask for any step's clarification!