GATE AE 2017 • Q45 MCQ • 2 Marks Propulsion

Propulsion

Inlets, Combustors & Nozzles

Subtopic: Combustor Design & Flame Stabilization

Question Statement:

In a combustor, gaseous Octane ($\text{C}_8\text{H}_{18}$) and air are to be burned in a stoichiometric ratio. Assuming air contains 21% $\text{O}_2$ and 79% $\text{N}_2$ by volume, the stoichiometric fuel-to-air mass ratio $(f/a)_{stoich}$ is approximately:

Options:

  • (A) 0.033
  • (B) 0.066
  • (C) 0.099
  • (D) 0.132

Official Answer Key:

B

Step-by-Step Solution & Explanation

### **Detailed Step-by-Step Solution (GATE AE 2017 Q.45)** **1. Intuitive Physical Setup & Given Parameters**: - Fuel: Octane, $\text{C}_8\text{H}_{18}$. - Molar mass of Carbon: $M_C = 12\text{ g/mol}$, Hydrogen: $M_H = 1\text{ g/mol}$. - Molecular weight of Octane: $M_{fuel} = 8(12) + 18(1) = 96 + 18 = 114\text{ g/mol}$. - Apparent molecular weight of dry air: $M_{air} \approx 28.97\text{ g/mol}$. **2. Step-by-Step Chemical Combustion Derivation**: 1. **Balanced Stoichiometric Combustion Equation**: \text{C}_8\text{H}_{18} + 12.5\left(\text{O}_2 + \frac{79}{21}\text{N}_2\right) \longrightarrow 8\text{CO}_2 + 9\text{H}_2\text{O} + 12.5\left(\frac{79}{21}\right)\text{N}_2 Moles of air per mole of fuel: n_{air} = 12.5 \times \left(1 + \frac{79}{21}\right) = 12.5 \times \frac{100}{21} \approx 59.52\text{ moles of air} 2. **Mass of Air per Mole of Fuel**: m_{air} = 59.52 \times 28.97 \approx 1724.3\text{ g} 3. **Stoichiometric Fuel-to-Air Mass Ratio**: (f/a)_{stoich} = \frac{M_{fuel}}{m_{air}} = \frac{114}{1724.3} \approx 0.0661 \text{Air-to-Fuel Ratio } (A/F)_{stoich} = \frac{1}{0.0661} \approx 15.12 Therefore, $(f/a)_{stoich} \approx 0.066$. **3. Comprehensive Options & Distractor Analysis**: - **(A)** 0.033: ❌ Incorrect. Half the correct ratio. - **(B)** 0.066: ✅ **CORRECT**. Exactly matches standard stoichiometric octane value $\approx 1/15.1 = 0.066$. - **(C)** 0.099: ❌ Incorrect. Corresponds to overly fuel-rich mixture. - **(D)** 0.132: ❌ Incorrect. Twice the stoichiometric ratio. **4. Official Master Key & Exam Takeaway**: - **Official Answer**: **B** - **Exam Rule**: For standard hydrocarbon aviation fuels, $(f/a)_{stoich} \approx 0.066$ to $0.068$ ($A/F \approx 14.7 - 15.1$).
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