Question 4 of 6: Aircraft Gas Turbine Engine

Question 4: Aircraft Gas Turbine Engine (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. The temperature and pressure at compressor inlet, compressor outlet and turbine exhaust, the nozzle outlet temperature, the engine mass flow, the thrust, and the thermal and propulsive efficiencies, with the whole cycle placed on a T–s diagram.

Approach. Number the stations 1 (ambient) → 2 (compressor inlet) → 3 (compressor outlet) → 4 (turbine inlet, maximum temperature) → 5 (turbine exhaust) → 6 (nozzle outlet), then march through them: the steady-flow energy equation for the diffuser and nozzle, the isentropic relation for every pressure change, and the shaft work balance to close the turbine.

1. Part (a) — ram compression in the diffuser. The diffuser is adiabatic and does no work, so the steady-flow energy equation converts the loss of kinetic energy into a rise in enthalpy: \[ c_p T_1 + \frac{V_1^2}{2} = c_p T_2 + \frac{V_2^2}{2} \quad\Rightarrow\quad T_2 = T_1 + \frac{V_1^2-V_2^2}{2c_p} \] Substituting, \[ T_2 = 250.15 + \frac{277.8^2 - 100^2}{2(1005)} = 250.15 + 33.41 = 283.6\ \text{K} \]
2. Get the compressor inlet pressure from the isentropic relation. Since the diffusion is isentropic, \[ p_2 = p_1\left(\frac{T_2}{T_1}\right)^{k/(k-1)} = 30\,(1.1336)^{3.5} \] \[ \boxed{T_2 = 283.6\ \text{K} = 10.4\ {}^{\circ}\text{C}, \quad p_2 = 46.5\ \text{kPa}} \] The ram effect alone has raised the pressure by 55% before the compressor does any work at all — the reason a turbojet gains efficiency with flight speed.
3. Part (b) — isentropic compression through the pressure ratio of 20. \[ p_3 = r_p\,p_2 = 20(46.53) = 930.5\ \text{kPa}, \qquad T_3 = T_2\,r_p^{(k-1)/k} = 283.56\,(20)^{0.2857} \] \[ \boxed{T_3 = 667.4\ \text{K} = 394.2\ {}^{\circ}\text{C}, \quad p_3 = 930.5\ \text{kPa}} \] The overall pressure ratio from ambient is \(930.5/30 = 31.0\), the ram contributing the difference from 20.
4. Part (c) — close the turbine on the shaft work balance. With no mechanical friction loss the turbine delivers exactly the compressor's work, and with a common \(c_p\) the temperature drops match: \[ c_p(T_4-T_5) = c_p(T_3-T_2) \quad\Rightarrow\quad T_5 = T_4 - (T_3-T_2) \] \[ T_5 = 1450.15 - (667.38-283.56) = 1450.15 - 383.82 = 1066.3\ \text{K} \] The specific work exchanged on the shaft is \(c_p(T_3-T_2) = 1.005(383.8) = 385.7\) kJ/kg. Since the combustion chamber has no pressure drop, \(p_4 = p_3 = 930.5\) kPa, and the expansion is isentropic: \[ p_5 = p_4\left(\frac{T_5}{T_4}\right)^{k/(k-1)} = 930.5\,(0.73523)^{3.5} \] \[ \boxed{T_5 = 1066.3\ \text{K} = 793.2\ {}^{\circ}\text{C}, \quad p_5 = 317.3\ \text{kPa}} \] The turbine has used only part of the available pressure ratio; the 317 kPa left over is what the nozzle turns into jet velocity, and this split between turbine and nozzle is what distinguishes a jet engine from a shaft-power gas turbine.
5. Part (d) — expand isentropically in the nozzle to ambient pressure. \[ T_6 = T_5\left(\frac{p_1}{p_5}\right)^{(k-1)/k} = 1066.33\left(\frac{30}{317.25}\right)^{0.2857} \] \[ \boxed{T_6 = 543.6\ \text{K} = 270.4\ {}^{\circ}\text{C}} \]
6. Part (e) — mass flow from the captured stream tube. The engine ingests whatever passes the diffuser inlet area at flight conditions, so the ambient density governs: \[ \rho_1 = \frac{p_1}{RT_1} = \frac{30}{0.287(250.15)} = 0.4179\ \text{kg/m}^3 \] \[ \dot m = \rho_1 A_1 V_1 = 0.4179(1.0)(277.78) \] \[ \boxed{\dot m = 116.1\ \text{kg/s}} \]
7. Compute the jet velocity, which the thrust needs. The nozzle is adiabatic and does no work, so all the enthalpy drop becomes kinetic energy. Neglecting the 100 m/s still held at station 5, \[ V_6 = \sqrt{2c_p(T_5-T_6)} = \sqrt{2(1005)(1066.33-543.55)} = 1025\ \text{m/s} \] The jet leaves at about Mach 2.2 relative to the aircraft, some 3.7 times the flight speed.
8. Part (f) — thrust from the momentum equation. Because the nozzle expands fully to ambient pressure there is no pressure-thrust term, and neglecting the fuel mass added, \[ F = \dot m\,(V_6 - V_1) = 116.07\,(1025.08 - 277.78) \] \[ \boxed{F = 86.7\ \text{kN per engine}} \]
9. Part (g) — thermal efficiency of the engine as a heat engine. A turbojet produces no shaft work, so its useful output is the rate at which it raises the kinetic energy of the air stream: \[ \dot Q_{in} = \dot m\,c_p(T_4-T_3) = 116.07(1.005)(1450.15-667.38) = 91\,314\ \text{kW} \] \[ \Delta\dot{KE} = \dot m\,\frac{V_6^2-V_1^2}{2} = 116.07\left(\frac{1025.08^2-277.78^2}{2}\right) = 56\,507\ \text{kW} \] \[ \boxed{\eta_{th} = \frac{56\,507}{91\,314} = 0.619 \;\;\text{(61.9\%)}} \] The cold-air-standard Brayton efficiency on the overall pressure ratio of 31.0 is \(1 - 31.0^{-0.2857} = 62.5\%\), so the two agree to within 0.6 percentage points; the small gap is the 5.0 kJ/kg of kinetic energy still held at the compressor inlet, which the accounting above does not credit as output.
10. Part (h) — propulsive efficiency. Only part of that kinetic energy gain propels the aircraft; the rest is left behind in the wake as the jet's own kinetic energy. The propulsive (Froude) efficiency is thrust power over kinetic-energy gain: \[ \eta_p = \frac{F V_1}{\Delta\dot{KE}} = \frac{2V_1}{V_1+V_6} = \frac{2(277.78)}{277.78+1025.08} \] \[ \boxed{\eta_p = 0.426 \;\;\text{(42.6\%)}} \] The thrust power is \(86\,743 \times 277.78 = 24\,095\) kW, so the overall efficiency is \[ \eta_{overall} = \eta_{th}\,\eta_p = 0.619 \times 0.426 = 0.264 \;\;\text{(26.4\%)} \] which also equals \(24\,095/91\,314\) directly. The low propulsive efficiency is the defining weakness of the pure turbojet: because \(\eta_p = 2V_1/(V_1+V_6)\) rises as the jet velocity approaches the flight speed, the same thrust delivered by accelerating far more air far less — a high-bypass turbofan — roughly doubles this term, which is why turbojets have disappeared from subsonic civil aviation.