Combustion Chambers Questions and Answers

Answer: B

The statement that is not always true is Total number of molecules is conserved.

Combustion rearranges atoms into new molecules, and the number of molecules usually changes. For example, \(2H_{2} + O_{2} \rightarrow 2H_{2}O\) turns three molecules into two, while \(CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O\) happens to keep three.

The other three are always true in a closed system. Chemical reactions neither create nor destroy atoms, so the total number of atoms and the number of atoms of each element are conserved, and mass is conserved because no matter crosses the boundary (the mass equivalent of the heat released is negligible).

Answer: B

The fuel-to-air ratio is 0.018.

An energy balance on the combustor, per kg of air, with fuel-to-air ratio f: the heat released by the fuel raises the air from its inlet enthalpy to the exit enthalpy of \((1 + f)\) kg of hot gas.

\[ \begin{aligned} f\,Q &= (1 + f)c_{pg}T_{02} - c_{pa}T_{01} \\ 44000f &= 1376.4(1 + f) - 603 \\ 42623.6f &= 773.4 \\ f &= 0.0181 \end{aligned} \]

with \(c_{pg}T_{02} = 1.147 \times 1200 = 1376.4\) kJ/kg and \(c_{pa}T_{01} = 1.005 \times 600 = 603\) kJ/kg. The other options are the same digits shifted by factors of ten; a real gas turbine combustor runs at about 0.02, so 0.18 and 1.18 are far too rich and 0.0018 far too lean.

Answer: C

Combustion in a gas turbine is ideally an isobaric process.

The ideal gas turbine cycle is the Brayton cycle: isentropic compression, heat addition at constant pressure in the combustor, isentropic expansion and heat rejection at constant pressure. The combustor is an open duct with a steady flow, so the pressure stays roughly constant while the fuel burns.

Adiabatic and isentropic processes add no heat, which is the opposite of combustion. Constant-volume (isochoric) heat addition belongs to the Otto cycle of piston engines, not to a continuous-flow gas turbine.

Answer: A

About 61.9 moles of dry air are needed per mole of fuel.

Balance the complete combustion of C₁₀H₁₂: 10 carbon atoms need 10 O₂ to form 10 CO₂, and 12 hydrogen atoms form 6 H₂O, which needs 3 more O₂.

\[ \begin{aligned} &\mathrm{C_{10}H_{12}} + 13\,\mathrm{O_{2}} \\ &\rightarrow 10\,\mathrm{CO_{2}} + 6\,\mathrm{H_{2}O} \end{aligned} \]

Air is 21% oxygen by volume, so

\[ n_{air} = \frac{13}{0.21} = 61.9\ \text{mol} \]

13 is the oxygen alone, not the air. 30.95 is half the correct value, as if the 79/21 ratio were halved, and 10 counts only the carbon.

Answer: A

The required fuel-air ratio is about 0.0177.

An energy balance on the combustor, with complete combustion and \(f \ll 1\) so that the mass flow is essentially that of the air, gives

\[ \begin{aligned} f\,Q_{HV} &= c_{p}(T_{exit} - T_{inlet}) \\ f &= \frac{1100 \times (1300 - 600)}{43.4 \times 10^{6}} \\ &= \frac{7.7 \times 10^{5}}{43.4 \times 10^{6}} \approx 0.0177 \end{aligned} \]

The other values do not come from this balance: 0.0127 would correspond to a temperature rise of only about 500 K, and 0.0215 and 0.0277 to rises of about 850 K and 1090 K, so they overshoot the 700 K rise asked for.

Answer: A

The value of X is 7.5.

Balance each fuel separately for complete combustion:

\[ \begin{aligned} CH_{4} + 2O_{2} &\to CO_{2} + 2H_{2}O \\ 2NH_{3} + 1.5O_{2} &\to N_{2} + 3H_{2}O \end{aligned} \]

Three moles of methane need 3 × 2 = 6 moles of O₂, and two moles of ammonia need 1.5 moles, so X = 7.5. Check by atoms: 3 C, 18 H and 4 N give 3 CO₂, 9 H₂O and 2 N₂, containing 6 + 9 = 15 O atoms, that is 7.5 O₂. None of the other values balances the oxygen atoms.