A combustor passes \( 60 \ \mathrm{kg/s} \) of air and burns \( 1.0 \ \mathrm{kg/s} \) of fuel. The overall air-fuel ratio is
Answer: A
Dividing gives 60 kilograms of air per kilogram of fuel, roughly four times leaner than stoichiometric for kerosene. Only the primary zone burns near 15 to 1, and the rest of the air cools the walls and dilutes the products.
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Kerosene has a stoichiometric air-fuel ratio near 15 to 1. An overall ratio of 60 to 1 gives an equivalence ratio of
Answer: B
Equivalence ratio is the stoichiometric ratio divided by the actual one, \( 15/60 = 0.25 \). Values well below one describe a lean mixture, and the whole architecture of the chamber exists to burn a globally lean flow in a locally stoichiometric zone.
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A combustor loses 4 per cent of a total pressure of \( 2400 \ \mathrm{kPa} \). The delivery pressure to the turbine is
Answer: C
The loss is \( 0.04 \times 2400 = 96 \), leaving 2304 kilopascals. That lost pressure is never recovered, and because thrust responds more than proportionally it costs rather more than four per cent of the engine's output.
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Fuel of calorific value \( 43 \ \mathrm{MJ/kg} \) burns at \( 0.8 \ \mathrm{kg/s} \) with 99 per cent combustion efficiency. The heat released is approximately
Answer: D
The ideal release is \( 43 \times 0.8 = 34.4 \) megawatts and 99 per cent of that is about 34.1. At cruise the efficiency is so close to unity that it barely matters, which is why the parameter is quoted mainly for idle and relight conditions.
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Air enters a combustor at 700 K and leaves at 1500 K, with \( c_p = 1150 \ \mathrm{J/(kg\,K)} \) for the products. The heat added per kilogram is approximately
Answer: A
Correct answer: 920 kJ/kg
\[ q = c_p \Delta T = 1150 \times (1500 - 700) \] \[ = 1150 \times 800 = 920\,000\ \mathrm{J/kg} = 920\ \mathrm{kJ/kg.} \] Note that the temperature difference is what matters, so the same answer follows whether the temperatures are in kelvin or Celsius. Using the exit temperature alone gives 1725 kJ/kg, which is the distractor.
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A combustor exit has a mean temperature of 1500 K, a peak of 1650 K and an inlet of 700 K. The pattern factor \( (T_{max}-T_{mean})/(T_{mean}-T_{inlet}) \) is
Answer: B
Substituting gives \( (1650-1500)/(1500-700) = 150/800 \). Values around 0.15 to 0.25 are typical, and reducing the figure lengthens turbine blade life directly because the hottest blade sets the life of the whole row.
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Air entering a combustor is split 20 per cent to the primary zone, 20 per cent intermediate and the rest to dilution and cooling. With \( 60 \ \mathrm{kg/s} \) total, the dilution and cooling flow is
Answer: C
The remaining fraction is 60 per cent, giving \( 0.6 \times 60 = 36 \) kilograms per second. The large share devoted to cooling and dilution is exactly why the overall mixture is so much leaner than the primary zone's.
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A primary zone receives \( 12 \ \mathrm{kg/s} \) of air with a fuel flow of \( 0.8 \ \mathrm{kg/s} \). Its local air-fuel ratio is
Answer: D
Dividing gives 15 to 1, essentially stoichiometric for kerosene. That is precisely the point of dividing the air: the zone where combustion must be self sustaining burns at the ratio that gives the fastest and hottest flame.
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A chamber has a residence time of 4 milliseconds at cruise. At idle the mass flow falls to a quarter with the same volume, giving a residence time of approximately
Answer: A
Residence time varies inversely with mass flow for a fixed volume, so quartering the flow quadruples it. The longer time helps completeness at idle but the low pressure and temperature there work against it, which is why carbon monoxide peaks at idle regardless.
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A combustor delivers 1500 K where the turbine limit is 1450 K. The required reduction in exit temperature is
Answer: B
Correct answer: 50 K, about 3.3 per cent
The difference is \( 1500 - 1450 = 50 \) K, and as a fraction of the delivered temperature \[ \frac{50}{1500} = 0.033 = 3.3\ \text{per cent.} \] Taking it as a fraction of 1450 gives 3.4 per cent, which is the other option offered - the base must be stated, and here it is the temperature to be reduced.
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