Propulsion Cycles and Engine Performance

Air-breathing engine performance

Air-breathing engines draw oxygen from the atmosphere, so their performance is described per unit air flow as well as for the whole engine. Specific thrust is thrust divided by the air mass flow rate, F/\dot m_a, and it measures the momentum change each kilogram of air receives. At a given flight speed, the ambient temperature and pressure fall with altitude. This raises the ratio of turbine-entry stagnation temperature to ambient temperature and the nozzle pressure ratio, so the jet velocity relative to the flight velocity grows and the specific thrust increases over the altitude range considered.

The Mach number is the ratio of flight speed to the local speed of sound, a=\sqrt{\gamma R T}, so a flight speed is obtained as V_0 = M a_0 once M and the local temperature are known.

Propulsive efficiency is the fraction of the kinetic energy supplied to the jet that becomes useful thrust power. For a simple jet, \eta_p = \dfrac{2}{1 + V_j/V_0}, so the efficiency rises as the jet velocity approaches the flight velocity. A velocity ratio of 3 gives \eta_p = 0.5, for example. Since \eta_p depends on the velocity ratio, it need not vary monotonically with a design variable. In a turbofan, changing the compressor pressure ratio moves the core and fan exit velocities and the bypass split in opposite directions, so \eta_p can first fall and then rise.

Overall efficiency is \eta_o = TV_0/(\dot m_f Q), where Q is the heating value of the fuel. It equals the product of thermal efficiency and propulsive efficiency, so \eta_o = \eta_{th}\eta_p and \eta_p = \eta_o/\eta_{th}. For a fixed thrust, a turbofan has higher propulsive efficiency than a turbojet because a larger bypass flow is accelerated through a smaller velocity change, which reduces the kinetic energy lost in the exhaust.

In a ramjet the air is compressed without any rotating machinery, using only the ram effect. The supersonic inlet compresses the air through an oblique shock system, a normal shock brings the flow to subsonic speed, and subsonic diffusion raises the static pressure before the combustor.

For an ideal ramjet with optimal expansion, the exit pressure equals ambient pressure, so the thrust is F = \dot m(V_e - V_0) when the fuel flow is negligible.

Specific impulse is I_{sp} = T/(\dot m_f g_0), and it links directly to overall efficiency through \eta_o = I_{sp} g_0 V_0/Q.

Static thrust is measured with zero flight speed. If each exit is at ambient pressure, the thrust is the sum of mass flow times exit velocity over all streams. For a turbofan, T/\dot m_{a,c} = V_{core} + BPR\,V_{fan}. The cold stream carries the fraction BPR/(1+BPR) of the total air flow, so its thrust is \dot m_c V_j with \dot m_c = \dot m\,BPR/(1+BPR). The ratio of fan thrust to core thrust depends on the bypass ratio, the velocity ratio, and the fuel-air ratio in the hot stream. The exhaust velocity itself follows from a steady-flow energy balance that includes fuel heat release and heat lost to the surroundings.

Brayton cycle ideal processes

The ideal Brayton cycle consists of isentropic compression, isobaric heat addition in the combustor, isentropic expansion, and isobaric heat rejection. The combustor of an ideal gas turbine, including a turboshaft engine, is therefore modelled as a constant-pressure heat-addition process.

For isentropic compression, T_2 = T_1 (p_2/p_1)^{(\gamma-1)/\gamma}. Heat added per kilogram of air passing through the combustor is q = c_p(T_3 - T_2), so the combustor exit temperature follows from T_3 = T_2 + q/c_p. When only a fraction of the compressed air enters the combustor, the heat quoted per kilogram must be interpreted on the air actually burned.

The ideal Brayton efficiency depends only on the pressure ratio, \eta = 1 - r_p^{-(\gamma-1)/\gamma}. Two cycles with the same pressure ratio therefore have equal efficiency even when their peak temperatures differ. The net work equals \eta\,q_{in}, so the cycle with the higher peak temperature, and hence greater heat input, produces more specific work.

Combustion process in closed system

In a closed combustion process, the chemical reaction rearranges atoms but does not create or destroy them. The total number of atoms, the number of atoms of each element, and the total mass are therefore conserved. The number of molecules is not necessarily conserved, because the number of moles of products can differ from the number of moles of reactants. For example, in 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} three molecules react to form two.

For a closed system the first law reads \delta Q = dU + \delta W. An adiabatic process has \delta Q = 0. In a closed combustor at constant pressure the volume changes as the mixture burns, so the boundary work \delta W = P\,dV is non-zero. The adiabatic constant-pressure case therefore gives dU = -\delta W, and both dU and \delta W are non-zero.

The standard heat of reaction is the sum of the standard enthalpies of formation of the products minus the sum for the reactants, with the enthalpy of formation of an element in its standard state taken as zero. Each species must be weighted by its stoichiometric coefficient in the balanced equation.

Combustor heat addition and burner efficiency

Burner efficiency \eta_b is the fraction of the fuel heating value that is actually released into the combustor flow. For a fuel-air ratio f and heating value Q, the energy balance per unit air flow is \eta_b f Q + c_{pa} T_{in} = (1+f)\,c_{pg} T_{out}, where the fuel mass joins the gas flow at the combustor exit. When the fuel flow cannot be neglected, the fuel flow rate follows from solving this equation for f and multiplying by the air flow.

When the fuel-air ratio is small and kinetic energy is negligible with 100% burner efficiency, the balance simplifies to f\,Q = c_p(T_3 - T_2). Thus f = c_p\Delta T/Q. A temperature rise of 500 K with c_p = 1000 J/kg K and Q = 40 MJ/kg, for instance, needs f \approx 0.0125.

Turbofan engine with afterburner

An afterburner adds fuel downstream of the turbine and burns it in the exhaust stream. The added heat raises the exhaust temperature and the jet velocity, giving a large thrust increase. The fuel flow rises much more than the thrust in proportion, so the specific fuel consumption increases when the afterburner is on. The nozzle exit area must be variable to keep the engine at a suitable operating point when the afterburner is lit. Heat addition in the afterburner also causes a loss of stagnation pressure through friction and Rayleigh-type effects, so the stagnation pressure does not increase across the engine.