Combustion Chambers Questions and Answers

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.

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.

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.

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.

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.

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.

Answer: C

Correct answer: 8

A rise of 120 K is three intervals of 40 K, and each doubles the output, so \[ 2^3 = 8. \] The exponential dependence is why nitrogen oxide emission is so sensitive to peak flame temperature, and why every low-emission combustor design - lean premixed, staged, rich-quench-lean - works by holding that temperature down.

Answer: D

Correct answer: 5

\[ \frac{25}{5} = 5. \] The range matters because combustion behaviour changes with pressure: at the low pressure of high altitude idle the reaction rates fall, the fuel atomises less well and the weak extinction limit narrows, which is precisely why altitude relight is the demanding case rather than take-off.

Answer: A

Multiplying gives \( 8 \times 3 = 24 \) per cent of the total flow devoted to wall cooling. As turbine entry temperatures rise this share must fall, which is exactly why effusion cooling has displaced simple film slots.

Answer: B

Both quantities have risen by the same factor of four, so the ratio is unchanged at 50 to 1. Real engines do not hold it constant, and the way it varies across the range is what makes idle and take-off such different combustion problems.