Gas Turbine Cycles of Aircraft Propulsion

Cycle

Station numbers: 0 ambient, 1 inlet, 2 compressor inlet, 3 compressor exit, 4 turbine inlet, 5 turbine exit, 6 nozzle exit.

 flight->  [diffuser][C][ Comb ][T][nozzle]--> jet
            0-2      2-3  3-4   4-5   5-6

 T-s: 0-2 ram rise (isentropic, up), 2-3 compression (up),
      3-4 heat addition (right and up along p3 line),
      4-5 turbine expansion (down; drop = 2-3 rise),
      5-6 nozzle expansion (down to p0), 6-0 exhaust cooling.

 T          4
 |         /|
 |      3 / |
 |      |/  5
 |      /   |
 |     2    6
 |     |   /
 |     0--/
 +------------ s

Thrust

Momentum equation on the engine, with fuel-air ratio $f$ and air mass flow $\dot m_a$:

$$F = \dot m_a\left[(1 + f)C_j - V\right] + (p_j - p_0)A_j$$

For a fully expanded nozzle ($p_j = p_0$) and small $f$, $F \approx \dot m_a (C_j - V)$. The specific thrust is $F/\dot m_a = C_j - V$.

Propulsive efficiency

Useful thrust power $= FV = \dot m_a V (C_j - V)$. Rate of increase of kinetic energy of the stream $= \tfrac12\dot m_a (C_j^2 - V^2)$.

$$\eta_p = \frac{\dot m_a V(C_j - V)}{\tfrac12\dot m_a (C_j^2 - V^2)} = \frac{2V}{C_j + V} = \frac{2}{1 + C_j/V}$$

It approaches 1 as $C_j \to V$, but then the thrust goes to zero. Hence the large mass flow with a lower jet velocity in the turbofan.

Thermal and overall efficiency

$$\eta_{th} = \frac{\tfrac12\dot m_a (C_j^2 - V^2)}{\dot m_f \, LHV}, \qquad \eta_o = \eta_p \eta_{th} = \frac{F V}{\dot m_f \, LHV}$$

For the ideal Brayton-type engine, $\eta_{th} = 1 - T_0/T_{03}$, i.e. it depends on the overall pressure ratio (ram plus compressor). The thrust specific fuel consumption is $TSFC = \dot m_f/F = f/(C_j - V)$.

Higher turbine inlet temperature increases specific thrust (more $C_j$); higher pressure ratio improves $\eta_{th}$ and so lowers TSFC.

Engines

• Turbojet: all the air passes through the core; thrust comes only from the high-velocity jet.
• Turbofan: a large fan, driven by the turbine, accelerates extra air that bypasses the core. Bypass ratio $\beta = \dot m_{bypass}/\dot m_{core}$ (0.3 to 1 for military, 5 to 12 for modern airliners).
• Turboprop: the turbine drives a propeller through a reduction gear; most thrust (about 90%) is from the propeller, only a small amount from the exhaust jet.

 Turbofan:   fan->[ bypass duct ]----------->
                 ->[C][Comb][T]----------->
 Turboprop:  prop<-gear<-[C][Comb][T]------> small jet

Comparison

Variation with flight speed

$$\eta_p = \frac{2V}{C_j + V}$$

• A turboprop accelerates a large mass of air by a small amount; $\eta_p$ is about 80 to 85% at low speed but drops above $Ma \approx 0.6$ because of propeller tip shock losses.
• A turbofan has intermediate jet velocity, giving good $\eta_p$ at $Ma$ 0.8.
• A turbojet has very high jet velocity compared with the flight speed, so $\eta_p$ is low at subsonic speed but improves as the flight speed approaches the jet velocity at supersonic flight.

 eta_p
 1 |  prop.
   |    \  fan
   |      \   /\    jet
   |        \/   \  /
   |              \/
   +------------------- flight Mach no.

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.

The simple Brayton cycle has a low work ratio and, at moderate pressure ratio, a high exhaust temperature. These modifications address that.

1. Regeneration (heat exchanger)

Hot turbine exhaust pre-heats compressor delivery air before the combustor, which reduces the fuel needed.

• Work output is unchanged; heat supplied falls, so efficiency rises.
• Useful only when the turbine exhaust is hotter than the compressor exit, i.e. at low pressure ratio. Effectiveness $\epsilon = (T_x - T_2)/(T_4 - T_2)$ is 0.6 to 0.85.
• For aircraft it adds too much weight; used in stationary and marine plants.

2. Intercooling

Compression is done in two stages with cooling in between to near the inlet temperature.

• Compressor work falls (work per kg $\propto$ absolute inlet temperature), so net work and work ratio rise.
• Efficiency alone may fall because more fuel is needed to heat the cooler air, unless a regenerator is also used.

3. Reheating

The gas is expanded in two turbine stages with an additional combustor between them.

• Turbine work rises, so net work increases (and the exhaust temperature too).
• Efficiency without regeneration usually falls slightly; with regeneration it improves.

 Arrangement with all three modifications:

 air->[LPC]->[Intercooler]->[HPC]->[Regenerator]
                                      |
        [Comb 1]->[HPT]->[Comb 2 (reheat)]->[LPT]
                                      |
                       exhaust <------+ (heat to regenerator)

4. Afterburning (thrust augmentation)

Extra fuel is burnt in the turbine exhaust, which still contains much unburnt oxygen, before the propelling nozzle.

• Jet velocity rises; thrust increases by 40 to 70%.
• Fuel consumption rises sharply (TSFC two to three times higher), so used only for take-off, climb and combat.
• A variable-area nozzle is needed.

Combined arrangement

Intercooling + reheating + regeneration approach the Ericsson cycle (isothermal compression and expansion) and so give the highest efficiency of all, at the price of weight, cost and complexity.