Theoretical Jet Engine

All jet engines produce thrust by increasing the momentum of a stream of fluid and ejecting it at the rear. They are grouped by the way the working fluid is compressed and by whether oxygen is taken from the atmosphere.

Air-breathing engines (take oxygen from the air)

1. Turbine-powered (continuous combustion with compressor):
   • Turbojet: all air passes through compressor, combustor, turbine and nozzle. Thrust comes from the hot jet.
   • Turbofan: a fan driven by the turbine pushes extra (bypass) air around the core. Good efficiency at subsonic speed.
   • Turboprop/turboshaft: most turbine power drives a propeller or shaft; jet thrust is small.
2. Ram-powered:
   • Ramjet: compression only by ram effect of the inlet diffuser; no compressor or turbine; works at $Ma > 1$ but cannot start at rest.
   • Scramjet: combustion in supersonic flow, for $Ma > 5$.
3. Non-continuous (intermittent) combustion: pulsejet with a valved inlet; fuel is burnt in cycles of intake, ignition, exhaust (Humphrey cycle).

Non-air-breathing engine

4. Rocket: carries both fuel and oxidiser; works in space; very high thrust, very high fuel consumption. Liquid, solid and hybrid propellant types.

Hybrid (combined) engines

5. Combine two principles in one unit, e.g. turbo-ramjet and air-turbo-rocket (turbojet at low speed, ramjet at high speed), or the hybrid rocket (solid fuel with a liquid oxidiser).

 Turbojet:   air->[C]->[Comb]->[T]->nozzle->jet
 Ramjet:     air->diffuser->[Comb]->nozzle->jet
 Rocket:     fuel+oxidiser->[Comb]->nozzle->jet

A ramjet is a jet engine with no compressor or turbine. The forward speed of the vehicle compresses the incoming air in an inlet diffuser (ram effect), fuel is burnt in the combustor, and the hot gas is expelled through the nozzle.

 flight -->  +-----------------------------+
 air ------->| diffuser | burner  | nozzle |------> jet
             |  (ram)   | + flame |        |
             +-----------------------------+
              0  1    2     3        4   5

Cycle (T-s)

1. 0-2: isentropic compression in the diffuser (ram compression), speed falls to near zero.
2. 2-3: heat addition at constant pressure in the combustor.
3. 3-5: isentropic expansion in the nozzle, converting enthalpy into jet velocity.
4. Heat rejection to atmosphere in the exhaust jet.

The ram pressure ratio is $\dfrac{p_{02}}{p_0} = \left(1 + \dfrac{\gamma-1}{2}Ma_0^2\right)^{\gamma/(\gamma-1)}$; at $Ma_0 = 2$ it is 7.8 and at $Ma_0 = 3$ it is 36.7.

Ideal thrust

$$F = \dot m_a\left[(1 + f)C_j - V_0\right] + (p_j - p_0)A_j$$

and the ideal cycle efficiency, same as Brayton: $\eta_{th} = 1 - \dfrac{1}{\text{(ram pressure ratio)}^{(\gamma-1)/\gamma}} = 1 - \dfrac{T_0}{T_{02}}$.

Advantages

• No moving parts: simple, light, cheap.
• Efficient at high supersonic speed; temperature not limited by turbine blades, so combustion temperatures can be high.

Limitations

• Zero thrust at rest: needs a booster or launch to reach the starting speed.
• Poor efficiency at low speed because of the low ram pressure ratio.
• Large fuel consumption below about $Ma$ 1.5; diffuser shock losses at high $Ma$.
• Intense heating of the structure at high speed; difficult flame stabilisation.

Applications

Missiles and target drones, as the second stage in rocket-boosted vehicles, and in combined-cycle propulsion (turbo-ramjet) for high-speed aircraft. At $Ma > 5$ the scramjet replaces it.

Data: $V = 250$ m/s, $\dot m_a = 40$ kg/s, $f = 0.015$, $C_j = 600$ m/s, $p_j = p_0$, $LHV = 43000$ kJ/kg.

Fuel flow $\dot m_f = f\dot m_a = 0.015 \times 40 = 0.60$ kg/s; jet mass flow $\dot m_j = 40 + 0.60 = 40.60$ kg/s.

(a) Net thrust

$$F = \dot m_j C_j - \dot m_a V = 40.60 \times 600 - 40 \times 250 = 14360\ \text{N} = 14.36\ \text{kN}$$

(Specific thrust $F/\dot m_a = 359.0$ N s/kg.)

(b) Thrust (propulsive) power

$$P_T = F V = 14360 \times 250 = 3590\ \text{kW}$$

(c) Propulsive efficiency

Rate of increase of kinetic energy of the working fluid:

$$\Delta KE = \tfrac12 \dot m_j C_j^2 - \tfrac12 \dot m_a V^2 = \tfrac12 (40.60)(600^2) - \tfrac12 (40)(250^2) = 6058\ \text{kW}$$ $$\eta_p = \frac{P_T}{\Delta KE} = \frac{3590}{6058} = 0.593$$

(The simple formula $2V/(C_j + V) = 0.588$ ignores the fuel mass; the value above is the exact one.)

(d) Thermal and overall efficiency

Energy input $= \dot m_f \times LHV = 0.60 \times 43000 = 25800$ kW.

$$\eta_{th} = \frac{\Delta KE}{\dot m_f LHV} = \frac{6058}{25800} = 0.235, \qquad \eta_o = \eta_p\eta_{th} = \frac{P_T}{\dot m_f LHV} = 0.139$$

(e) Thrust specific fuel consumption

$$TSFC = \frac{\dot m_f}{F} = \frac{0.60 \times 3600}{14.36} = 150.4\ \text{kg/(kN h)}$$

Answer: $F = 14.36$ kN; $P_T = 3590$ kW; $\eta_p = 0.593$; $\eta_{th} = 0.235$; $\eta_o = 0.139$; TSFC = 150.4 kg/(kN h).

Thrust equation

A rocket expels propellant at a mass flow rate $\dot m_p$ with exhaust velocity $C_e$ (relative to the rocket) through a nozzle exit of area $A_e$ at pressure $p_e$, into an ambient pressure $p_a$. Applying the momentum equation to the control volume around the engine:

$$F = \dot m_p C_e + (p_e - p_a)A_e$$

The first term is the momentum thrust; the second is the pressure thrust, zero when the nozzle is perfectly expanded ($p_e = p_a$). The thrust is greatest in vacuum and increases as the vehicle climbs.

Effective exhaust velocity

$$C_{eff} = \frac{F}{\dot m_p} = C_e + \frac{(p_e - p_a)A_e}{\dot m_p}$$

Specific impulse

Thrust per unit weight flow of propellant:

$$I_{sp} = \frac{F}{\dot m_p g_0} = \frac{C_{eff}}{g_0}\ \text{(seconds)}$$

It is the main figure of merit of a propellant: about 200 to 250 s for solid propellants, 300 to 450 s for liquid propellants (hydrogen/oxygen near 450 s in vacuum).

Propulsive efficiency

For vehicle speed $v$, the useful power is $Fv$; the jet leaves with absolute velocity $(C_{eff} - v)$, so the kinetic energy lost in the jet is $\tfrac12 \dot m_p (C_{eff} - v)^2$:

$$\eta_p = \frac{F v}{F v + \tfrac12 \dot m_p (C_{eff} - v)^2} = \frac{2(v/C_{eff})}{1 + (v/C_{eff})^2}$$

It is maximum (1.0) when $v = C_{eff}$.

Rocket equation

Take a rocket of instantaneous mass $M$ with velocity $v$, in vacuum and no gravity. Momentum balance over a short time $dt$:

$$M\frac{dv}{dt} = C_{eff}\left(-\frac{dM}{dt}\right) \;\Rightarrow\; dv = -C_{eff}\frac{dM}{M}$$

Integrating from initial mass $M_0$ to final (burnout) mass $M_f$:

$$\boxed{\Delta v = C_{eff}\ln\frac{M_0}{M_f}}$$

With gravity and drag the velocity gain is reduced by $g\,t_b$ (gravity loss) and drag loss. A larger mass ratio $M_0/M_f$ is obtained by staging, which is why launch vehicles have several stages.

Data: $\dot m_p = 120$ kg/s, $C_e = 2800$ m/s, $A_e = 0.5$ m², $p_e = 80$ kPa.

(a) Sea-level thrust

$$F = \dot m_p C_e + (p_e - p_a)A_e = 120 \times 2800 + (80000 - 101300) \times 0.5$$ $$F = 336000 - 10650 = 325350\ \text{N} = 325.4\ \text{kN}$$

The nozzle is over-expanded ($p_e < p_a$), so the pressure term is negative.

$$C_{eff} = \frac{F}{\dot m_p} = \frac{325350}{120} = 2711.2\ \text{m/s}$$

(b) Specific impulse

$$I_{sp,SL} = \frac{C_{eff}}{g_0} = \frac{2711.2}{9.81} = 276.4\ \text{s}$$

In vacuum $p_a = 0$:

$$F_{vac} = 336000 + 80000 \times 0.5 = 376000\ \text{N}, \qquad C_{eff,vac} = 3133.3\ \text{m/s}, \qquad I_{sp,vac} = \frac{3133.3}{9.81} = 319.4\ \text{s}$$

(c) Propulsive efficiency and thrust power

$$\frac{v}{C_{eff}} = \frac{1500}{2711.2} = 0.5533$$ $$\eta_p = \frac{2(v/C_{eff})}{1 + (v/C_{eff})^2} = \frac{2 \times 0.5533}{1 + 0.3061} = 0.847$$ $$P_T = F v = 325350 \times 1500 = 488025\ \text{kW}$$

Answer: $F = 325.4$ kN, $C_{eff} = 2711.2$ m/s; $I_{sp} = 276.4$ s (sea level), 319.4 s (vacuum); $\eta_p = 0.847$; $P_T = 488025$ kW.

(a) Pulsejet engine

A pulsejet burns fuel intermittently in a tube. Air enters through spring-loaded flapper valves at the front.

 valves     combustor        tail pipe
 |>|=====+===========+=================>  exhaust

Cycle: (1) air and fuel enter when pressure inside is low; (2) a spark ignites the mixture and the pressure rise closes the valves; (3) hot gas leaves through the tail pipe, producing thrust, and inertia of the gas column lowers the pressure; (4) the valves reopen and the cycle repeats at 40 to 250 cycles per second. Later cycles self-ignite from the hot walls. Its cycle is close to the Humphrey (constant-volume) cycle.

• Advantages: simple, cheap, thrust at rest. Disadvantages: loud, strong vibration, low efficiency, poor valve life, speed limited to low subsonic. The German V-1 flying bomb used one.

(b) Hybrid rocket engine

It combines a solid fuel grain (e.g. rubber/HTPB or paraffin) in the combustion chamber with a liquid or gaseous oxidiser (e.g. liquid oxygen, nitrous oxide) injected into the port.

• Can be throttled, shut down and restarted by controlling the oxidiser valve.
• Safer to handle than solid motors (fuel is inert); simpler than liquid engines (one liquid only).
• Disadvantages: lower regression rate (lower thrust density), mixture ratio shift during burn, lower combustion efficiency.

(c) Combined-cycle (turbo-ramjet) engines

At low speed a ramjet gives no thrust, while a turbojet is limited at high Mach number by compressor temperature. A turbo-ramjet has a turbojet core inside a ramjet duct: the turbojet works from take-off to about $Ma$ 2.5, then it is shut off, bypassed, and the air passes to an afterburner/ramjet combustor up to $Ma$ 3 to 4. Related concepts are the air-turbo-rocket and the rocket-based combined cycle. They give a single engine over a wide speed range, at the cost of complexity and weight.