Let's break down the question and options: ### **Question Analysis** - **Given:** - Hydrocarbon fuel: \( C_nH_{2n+2} \), where \( n = 8 \) (so, \( C_8H_{18} \), i.e., octane) - Burned with unknown air - 3 kmoles of free oxygen present in products - Find the **quantity of air** involved and **AFR (Air-Fuel Ratio)** ### **Combustion Reaction** For octane (\( C_8H_{18} \)): General combustion: \[ C_8H_{18} + \text{O}_2 + \text{N}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} + \text{O}_2 (\text{excess}) + \text{N}_2 \] **Balanced equation for complete combustion (theoretical air):** \[ C_8H_{18} + 12.5 O_2 \rightarrow 8CO_2 + 9H_2O \] ### **Theoretical Oxygen Requirement** - 1 kmol octane needs **12.5 kmol O₂** for complete combustion. ### **Given:** - 1 kmol octane was burned and **3 kmol O₂** is left as excess in products. ### **Actual O₂ supplied:** \[ \text{O}_2 \text{ supplied} = \text{O}_2 \text{ required} + \text{excess O}_2 = 12.5 + 3 = 15.5 \text{ kmol} \] ### **Theoretical Air Required** - Air contains 21% O₂ by volume, so per kmol O₂, air required = \( \frac{100}{21} \approx 4.76 \) kmol air. - For 12.5 kmol O₂: \[ \text{Theoretical air required} = 12.5 \times 4.76 = 59.5 \text{ kmol} \] ### **Actual Air Supplied** \[ \text{Actual air supplied} = 15.5 \times 4.76 = 73.78 \text{ kmol} \] ### **Excess Air** \[ \text{Excess air} = \frac{73.78 - 59.5}{59.5} \times 100 = \frac{14.28}{59.5} \times 100 \approx 24\% \] ### **Air-Fuel Ratio (AFR)** \[ \text{AFR (by kmol)} = \frac{\text{Actual air supplied}}{\text{Fuel supplied}} = \frac{73.78}{1} = 73.78 \text{ (by kmol)} \] But **AFR is usually reported by mass.** - 1 kmol \( C_8H_{18} \) = \( 8 \times 12 + 18 \times 1 = 114 \) kg - 1 kmol air = 28.96 kg - Mass of air supplied: \( 73.78 \times 28.96 = 2138 \) kg - Mass of fuel: 114 kg \[ \text{AFR (by mass)} = \frac{2138}{114} = 18.76 \] --- ## **Match with Options** - **24% of excess air and AFR = 18.76** (This matches our calculation.) ## **Correct Option:** > **24% of excess air and AFR = 18.76**