Question 6 of 7: Load factor, turn radius, by-pass and afterburning engines, and jet performance parameters

Question 6: Load factor, turn radius, by-pass and afterburning engines, and jet performance parameters (20 marks — 5 marks each)

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.

(a) Load factor. The load factor $n$ is the ratio of the net aerodynamic force perpendicular to the flight path — in practice the lift — to the weight of the aircraft, $n=L/W$. It is dimensionless but is universally quoted in multiples of $g$, because an occupant of an aircraft pulling $n$ experiences an apparent weight $n$ times their real weight. In steady level flight $n=1$; in a steady banked turn at bank angle $\phi$ the vertical component of lift must still carry the weight, so $n=1/\cos\phi$, giving $2$ at a $60^{\circ}$ bank; in a symmetric pull-up of radius $r$ at speed $V$ the load factor is $n=1+V^{2}/(gr)$; and in a bunt or an inverted push-over it is negative. Load factor matters because it is the single number that scales every structural load on the airframe: a wing designed for a limit load factor of $2.5$ must carry two and a half times the aeroplane's weight without permanent deformation, and $1.5$ times that again — the ultimate load factor — without failing. It also scales the stalling speed, since the lift that must be produced has risen: the accelerated stall speed is $V_{s}\sqrt{n}$. The permissible combinations of load factor and speed are collected in the V-n or manoeuvre envelope, whose corner point — where the aerodynamic limit $C_{L,\max}$ and the structural limit meet — is the manoeuvre speed, the fastest speed at which full control deflection cannot overstress the aircraft. Gusts as well as manoeuvres produce load factor, which is why a gust envelope is superimposed on the manoeuvre envelope.

(b) Factors determining the minimum turn radius. In a steady, level, co-ordinated turn the lift is banked so that its vertical component carries the weight and its horizontal component supplies the centripetal force. Eliminating the bank angle between $L\cos\phi=W$ and $L\sin\phi=mV^{2}/r$ gives the two governing relations $$\begin{aligned} n &= \frac{1}{\cos\phi}, \\ r &= \frac{V^{2}}{g\sqrt{n^{2}-1}}, \\ \omega &= \frac{g\sqrt{n^{2}-1}}{V} \end{aligned}$$ so the radius falls if the aircraft turns more slowly or pulls more load factor. Neither can be pushed indefinitely, and at a given altitude four separate ceilings compete.

The first is structural: the load factor may not exceed the limit value for which the airframe was certified, and above the corner speed this is the binding constraint, giving a radius that grows as $V^{2}$. The second is aerodynamic: the wing cannot generate more lift than $\tfrac{1}{2}\rho V^{2}SC_{L,\max}$, so $n_{\max}=\rho V^{2}C_{L,\max}/(2W/S)$, which falls rapidly as speed is reduced and is the binding constraint below the corner speed. Combining the two, the tightest possible instantaneous turn occurs exactly at the corner speed, where the wing reaches $C_{L,\max}$ and the limit load factor simultaneously. The third is propulsive: to hold height and speed through the turn the thrust must equal the drag, and the induced drag has grown by $n^{2}$; the largest load factor that thrust can sustain gives the sustained turn, which is always gentler than the instantaneous one. The fourth is atmospheric: at a higher altitude the lower density reduces the lift available at a given true airspeed and reduces the thrust, so both the aerodynamic and the propulsive ceilings fall and the minimum radius grows; wing loading $W/S$ enters the aerodynamic limit directly, so a lightly loaded wing turns more tightly than a heavily loaded one at the same speed. In practice the pilot of a transport is bound by a fifth, non-physical limit: passenger comfort and certification rules that restrict normal manoeuvring to about $25^{\circ}$ to $30^{\circ}$ of bank, that is $n$ between $1.10$ and $1.15$.

(c) By-pass engines and afterburning. Both are answers to the same question — how should a given amount of momentum change be produced — but they answer it for opposite missions. Thrust is the product of mass flow and velocity increment, $F=\dot{m}(V_{e}-V)$, while the kinetic energy wasted in the jet is $\tfrac{1}{2}\dot{m}(V_{e}-V)^{2}$. For a required thrust, therefore, it is always more efficient to accelerate a large mass of air by a small increment than a small mass by a large one. A by-pass or turbofan engine does exactly that: only part of the air taken in passes through the core, while the rest is accelerated modestly by the fan and by-passes the combustion system. The propulsive efficiency rises sharply, the specific fuel consumption falls by a third or more relative to a pure turbojet at subsonic speeds, and as a valuable by-product the jet noise, which depends on a very high power of the exhaust velocity, falls dramatically and the cool by-pass stream shrouds the hot core. The penalties are frontal area, nacelle drag and weight, and a fan that is inefficient once the flight speed approaches the jet velocity — which is why by-pass ratios above about ten belong to subsonic transports and low ratios to supersonic aircraft. Afterburning, or reheat, addresses the opposite need. Because the flow leaving a turbine still contains a large proportion of unburnt oxygen, additional fuel can be injected and burnt in the jet pipe downstream of the turbine, where there are no blades to be temperature-limited. The exhaust temperature and hence the jet velocity rise, and thrust increases by roughly $50\%$ for a modest increase in engine size and weight. The cost is a specific fuel consumption two to three times the dry value, so afterburning is used only in short bursts: take-off from a short runway, acceleration through the transonic drag rise, and combat manoeuvring.

(d) Specific fuel consumption and propulsion efficiency. The thrust specific fuel consumption of a jet engine is the rate at which fuel is burnt per unit of thrust produced, $c=\dot{m}_{f}/F$, with units of $\text{kg}\,\text{N}^{-1}\,\text{s}^{-1}$ or, in the form usually quoted, $\text{kg}\,\text{N}^{-1}\,\text{h}^{-1}$; a modern high-by-pass turbofan achieves about $0.055$ in those units in cruise, an afterburning military engine perhaps four times that in reheat. It is the figure of merit that enters the Breguet range equation directly, and it is the number that a designer trades against weight and frontal area. Propulsion efficiency is a different and more fundamental quantity: it is the fraction of the mechanical power delivered to the airstream that appears as useful propulsive power. The useful power is thrust times flight speed, $FV=\dot{m}(V_{e}-V)V$, while the power added to the air is the rate of increase of its kinetic energy, $\tfrac{1}{2}\dot{m}(V_{e}^{2}-V^{2})$, so $$\begin{aligned} \eta_{p} &= \frac{FV}{\tfrac{1}{2}\dot{m}(V_{e}^{2}-V^{2})} \\ &= \frac{2V}{V+V_{e}} \\ &= \frac{2}{1+V_{e}/V} \end{aligned}$$ This is the Froude efficiency. It approaches unity only as the jet velocity approaches the flight speed — which is precisely the by-pass argument of part (c) — and it is zero on a static test bed, where the engine produces thrust but no useful power at all. The overall efficiency is the product of this and the thermal efficiency of the gas-turbine cycle, and the specific fuel consumption is inversely proportional to that product multiplied by the fuel's calorific value.