The stagnation temperature after the diffuser is found from

T02 = T0 (1 + 0.2 Ma0^2) = 244.4 K

T05 = T04 - (T03 - T02), because turbine work equals compressor work

The jet velocity of 1039 m/s is found from

C_j = sqrt(2 cp (T05 - T6)) after complete expansion of the gas to ambient pressure

The fuel-air ratio is computed as

f = cp (T04 - T03)/LHV = 907.5/43000 = 0.0211

Why is propulsive efficiency (0.370) much lower than thermal efficiency (0.564)?

Jet velocity 1039 m/s is much larger than flight speed 236 m/s, so much jet kinetic energy is wasted

Practice · 8 marks

Ideal turbojet at altitude: thrust and TSFC

Short answer

Flight speed is 236.0 m/s, jet velocity is 1039.0 m/s, and specific thrust is 803.0 N s/kg. Fuel-air ratio is 0.0211 with TSFC 94.6 kg/(kN h). Efficiencies are eta_p = 0.370, eta_th = 0.564, eta_o = 0.209, and thrust at 50 kg/s is 40.15 kN.

An ideal turbojet flies at Mach 0.8 at an altitude where the ambient temperature is 216.65 K and pressure is 22.63 kPa. The compressor pressure ratio is 12 and the maximum cycle temperature is 1400 K. Assume isentropic diffuser, compressor, turbine and nozzle, no pressure loss in the combustor, complete expansion in the nozzle, turbine work equal to compressor work, $c_p = 1.005$ kJ/kg K and $\gamma = 1.4$ throughout, and neglect the fuel mass in the thrust. The fuel has a lower calorific value of 43 000 kJ/kg. Find (a) the flight speed, (b) the stagnation temperature and pressure after the diffuser, (c) the temperature after the turbine, (d) the jet velocity, (e) the specific thrust, (f) the fuel-air ratio and TSFC, and (g) the propulsive, thermal and overall efficiencies. Also find the thrust for an air mass flow of 50 kg/s.

Data: $Ma_0 = 0.8$ , $T_0 = 216.65$ K, $p_0 = 22.63$ kPa, $r_c = 12$ , $T_{04} = 1400$ K.

(a) Flight speed

V = Ma_0\sqrt{\gamma R T_0} = 0.8\sqrt{1.4 \times 287 \times 216.65} = 236.0\ \text{m/s}

(b) Diffuser exit (station 2)

T_{02} = T_0\left(1 + 0.2 Ma_0^2\right) = 216.65 \times 1.128 = 244.4\ \text{K}, \qquad p_{02} = p_0\left(\frac{T_{02}}{T_0}\right)^{3.5} = 34.50\ \text{kPa}

(c) Compressor, turbine

T_{03} = T_{02} r_c^{0.2857} = 244.4 \times 12^{0.2857} = 497.1\ \text{K}, \qquad p_{03} = 12 \times 34.50 = 413.9\ \text{kPa}

Turbine work equals compressor work:

T_{04} - T_{05} = T_{03} - T_{02} \;\Rightarrow\; T_{05} = 1400 - (497.1 - 244.4) = 1147.3\ \text{K}

p_{05} = p_{04}\left(\frac{T_{05}}{T_{04}}\right)^{3.5} = 413.9 \times \left(\frac{1147.3}{1400}\right)^{3.5} = 206.25\ \text{kPa}

(d) Jet velocity

Complete expansion to $p_0$ :

T_6 = T_{05}\left(\frac{p_0}{p_{05}}\right)^{0.2857} = 610.2\ \text{K}

C_j = \sqrt{2c_p(T_{05} - T_6)} = \sqrt{2 \times 1005 \times 537.1} = 1039.0\ \text{m/s}

(e) Specific thrust

\frac{F}{\dot m_a} = C_j - V = 1039.0 - 236.0 = 803.0\ \text{N s/kg}

(f) Fuel-air ratio and TSFC

q_{in} = c_p (T_{04} - T_{03}) = 1.005 \times (1400 - 497.1) = 907.5\ \text{kJ/kg}, \qquad f = \frac{q_{in}}{LHV} = 0.0211

TSFC = \frac{f}{F/\dot m_a} \times 3600 \times 1000 = 94.6\ \text{kg/(kN h)}

(g) Efficiencies

\eta_p = \frac{2V}{C_j + V} = \frac{2 \times 236.0}{1039.0 + 236.0} = 0.370

\eta_{th} = \frac{(C_j^2 - V^2)/2}{q_{in}} = \frac{511.9}{907.5} = 0.564, \qquad \eta_o = \eta_p\eta_{th} = 0.209

Thrust

F = 50 \times 803.0 = 40.15\ \text{kN}

Answer: $V = 236.0$ m/s; $C_j = 1039.0$ m/s; specific thrust = 803.0 N s/kg; $f = 0.0211$ ; TSFC = 94.6 kg/(kN h); $\eta_p = 0.370$ , $\eta_{th} = 0.564$ , $\eta_o = 0.209$ ; $F = 40.15$ kN.

✓ Answer checked by Claude Opus 5.5 (high effort)

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