Aerospace Propulsion: Cycles, Engines, Turbomachinery and Rockets
1. Thermodynamics for propulsion, and the Brayton cycle
Every engine component is an open system in steady flow, and the steady-flow energy equation per unit mass is q − w = Δ(h + V²/2 + gz). Grouping enthalpy and kinetic energy into the stagnation enthalpy h0 = h + V²/2 = c_pT0 makes each component a one-line statement: an intake or nozzle does no work and exchanges no heat, so T0 is constant through it; a compressor’s work is c_p(T02 − T01); a combustor adds heat at nearly constant pressure. For an ideal gas undergoing an isentropic process T2/T1 = (p2/p1)^((γ−1)/γ). The second law enters through component efficiencies: the compressor isentropic efficiency is η_c = (T02s − T01)/(T02 − T01), ideal work over actual, and the turbine’s is η_t = (T03 − T04)/(T03 − T04s), actual over ideal.
The gas-turbine engine runs on the Brayton (Joule) cycle: isentropic compression 1-2, constant-pressure heat addition 2-3, isentropic expansion 3-4 and constant-pressure heat rejection 4-1. Its ideal thermal efficiency depends only on the pressure ratio π: η = 1 − 1/π^((γ−1)/γ) = 1 − T1/T2, so η = 48.2% at π = 10 with γ = 1.4. The specific net work, however, depends on the maximum temperature too, and for given T1 and T3 it is greatest when π^((γ−1)/γ) = √(T3/T1), that is when T2 = T4 = √(T1T3). With real component efficiencies the thermal efficiency also peaks at a finite pressure ratio, and raising the turbine inlet temperature raises both the work and the optimum pressure ratio — which is why turbine inlet temperature is the single most important engine parameter.
2. Thrust, efficiencies and range
A control volume round an air-breathing engine gives the thrust F = (ṁ_a + ṁ_f)V_e − ṁ_aV0 + (p_e − p_a)A_e: the momentum thrust of the jet minus the ram drag of the captured air, plus a pressure thrust when the nozzle is not fully expanded. The efficiencies split the energy chain. Thermal efficiency η_th is the rate at which kinetic energy is added to the stream divided by the fuel energy ṁ_fQ_R. Propulsive efficiency η_p is the thrust power FV0 divided by that kinetic-energy rate; for a fully expanded nozzle with f ≪ 1, η_p = 2V0/(V_e + V0) = 2/(1 + V_e/V0). The overall efficiency is their product, η_o = η_th·η_p = FV0/(ṁ_fQ_R), and the thrust-specific fuel consumption is TSFC = ṁ_f/F = V0/(η_oQ_R).
Propulsive efficiency explains the shape of modern engines: the thrust is ṁ(V_e − V0), and for a given thrust a small jet-velocity excess over a large mass flow wastes far less kinetic energy in the wake than a fast jet from a small one. Range connects to the engine through η_o: substituting TSFC into the Breguet equation gives R = (η_oQ_R/g)(L/D) ln(W0/W1), so for a given airframe the range is set by the product of overall efficiency and lift-to-drag ratio.
3. Ramjet, turbojet, turbofan, turboprop and turboshaft engines, and afterburners
- Ramjet — no rotating machinery: the intake compresses the air by ram effect alone, fuel burns in the duct, and the nozzle expands the products. It produces no static thrust and must be boosted to speed; its ideal thermal efficiency, 1 − 1/(1 + ((γ − 1)/2)M0²), rises with flight Mach number, so it is a supersonic engine, with the scramjet (supersonic combustion) taking over at hypersonic speeds.
- Turbojet — intake, compressor, combustor, turbine, nozzle. The turbine drives the compressor, so c_p,g(T04 − T05) = c_p,a(T03 − T02)/η_m, and the gas left over expands through the nozzle to make a fast jet — good for high speed, poor in propulsive efficiency at subsonic speed.
- Afterburner (reheat) — extra fuel is burned downstream of the turbine, which is possible because the main combustor runs very lean and plenty of oxygen remains. It raises the jet temperature and velocity and hence the thrust substantially, at a much higher TSFC; it needs a variable-area nozzle, since the hotter gas needs a larger throat to pass the same mass flow without back-pressuring the turbine.
- Turbofan — a fan driven by the low-pressure turbine sends a bypass stream round the core. The bypass ratio is ṁ_bypass/ṁ_core. A high bypass ratio lowers the mean jet velocity, raising propulsive efficiency, lowering TSFC and noise — the engine of every modern airliner; low-bypass, often afterburning, turbofans power combat aircraft.
- Turboprop — a gas generator drives, through a power turbine and a reduction gearbox, a propeller that produces most of the thrust; the small residual jet thrust is counted through the equivalent shaft power. Very efficient below about Mach 0.6, limited above it by propeller-tip compressibility.
- Turboshaft — all the useful output is shaft power from a free power turbine, with negligible jet thrust: helicopters, auxiliary power units, ships and generating sets.
4. Intakes, nozzles and the combustor
The intake must deliver air to the compressor face at the right Mach number (about 0.4 to 0.6) with the least loss of stagnation pressure; its figure of merit is the pressure recovery p02/p0∞. A subsonic intake is a gently diverging diffuser: at take-off the engine draws air in from a stream tube larger than the lip, and at cruise the captured stream tube is smaller than the lip and the excess spills round it. Supersonic intakes cannot diffuse isentropically. A normal-shock (pitot) intake is simple and acceptable at low supersonic Mach numbers, where the normal-shock loss is small; faster, external- or mixed-compression intakes use ramps or a centre-body cone to decelerate the flow through one or more oblique shocks before a final weak normal shock, because several weak shocks lose far less stagnation pressure than one strong one. A mismatched supersonic intake can unstart or buzz.
The propelling nozzle converts the remaining enthalpy into jet velocity. A convergent nozzle chokes once the nozzle pressure ratio p0/p_a reaches 1.893; beyond that the exit stays sonic, the exit pressure exceeds ambient and the excess appears as pressure thrust, with further expansion outside. A convergent-divergent nozzle expands fully to ambient at higher pressure ratios and is needed on supersonic and afterburning engines, usually with variable geometry.
The combustor must burn fuel completely, stably and compactly with small pressure loss and an acceptable exit temperature profile. Its types are the can (tubular) combustor, separate flame tubes each in its own casing — easy to develop, but heavy and long; the can-annular (cannular), flame tubes inside a common annular casing; and the annular combustor, a single annular flame tube — shortest, lightest, lowest pressure loss and most uniform exit temperature, and the usual choice today. Inside, a swirler and the primary zone create a recirculation region where a near-stoichiometric mixture burns and the flame is stabilised; the secondary (intermediate) zone completes combustion and recombines dissociated products; the dilution zone mixes in the remaining air to bring the gas down to a temperature the turbine can accept. The overall fuel-air ratio is therefore far leaner than stoichiometric. For a hydrocarbon CₓHᵧ, the stoichiometric reaction needs (x + y/4) mol of O₂, each accompanied by 3.76 mol of N₂; for kerosene, roughly (CH₂)ₙ, the stoichiometric fuel-air ratio is about 0.068.
5. The axial compressor — work, reaction, characteristics and multi-staging
Torque on a rotor equals the rate of change of the fluid’s angular momentum, which gives the Euler turbomachinery equation: the work per unit mass is w = U2C_w2 − U1C_w1, and for an axial stage at constant radius w = U(C_w2 − C_w1) = UΔC_w. With constant axial velocity C_a, and α and β the absolute and relative flow angles from the axial direction, the stage temperature rise is ΔT0 = UC_a(tan β1 − tan β2)/c_p = UC_a(tan α2 − tan α1)/c_p; a work-done factor λ slightly below 1 is often applied for the non-uniform axial velocity near the annulus walls. The degree of reaction is the fraction of the stage’s static enthalpy (temperature) rise that occurs in the rotor, Λ = (C_a/(2U))(tan β1 + tan β2); with symmetric blading (α1 = β2, α2 = β1) it is 0.5, sharing the diffusion equally between rotor and stator. The stage pressure ratio is (1 + η_sΔT0/T01)^(γ/(γ−1)).
The blades diffuse the flow against a rising pressure, and the boundary layers on them separate if the diffusion is too great, so an axial stage achieves only a modest pressure ratio and many stages are needed. A compressor’s performance is drawn as a characteristic, or map: pressure ratio against the corrected mass flow ṁ√T01/p01, with lines of constant corrected speed N/√T01 and contours of efficiency. At the low-flow end of each speed line lies the surge line, beyond which the flow becomes unstable — rotating stall and then surge, a violent flow reversal; at the high-flow end the passages choke. In a multi-stage compressor the overall isentropic efficiency is lower than the stage (polytropic) efficiency, because each stage’s losses heat the air that the next stage must compress. At low speed the front stages tend to stall and the rear stages to choke, which is handled by inlet guide vanes and variable stators, interstage bleed, and splitting the compressor into two or three spools.
6. The centrifugal compressor — inducer, impeller and diffuser
Air enters a centrifugal compressor axially through the inducer, the curved leading portion of the impeller vanes that turns the relative flow smoothly into the vane passages; the impeller then flings it radially outward, raising its angular momentum and static pressure; and the diffuser, vaneless or vaned, converts the large absolute kinetic energy leaving the impeller into pressure. With no inlet whirl, Euler’s equation gives w = C_w2U2, and because the fluid cannot follow the vanes exactly, C_w2 = σU2 with slip factor σ below 1; including friction and windage through a power input factor ψ, w = ψσU2² and ΔT0 = ψσU2²/c_p. With radial vanes about half of the static pressure rise occurs in the impeller and half in the diffuser. A single centrifugal stage reaches a much higher pressure ratio than an axial stage and is robust and short, but it has a large frontal area and is hard to multi-stage, so it suits small engines, helicopter turboshafts and the last stage of some compressors.
7. The axial turbine stage, blade cooling and compressor-turbine matching
An axial turbine stage is a row of nozzle guide vanes that accelerate and turn the hot gas, followed by a rotor that removes its swirl. By Euler’s equation the stage work is w = U(C_w2 + C_w3) when the exit swirl C_w3 is opposite to the inlet swirl, or UC_a(tan β2 + tan β3) in terms of relative angles; the blade-loading coefficient ψ = c_pΔT0/U² and flow coefficient φ = C_a/U describe the velocity triangles. In an impulse stage (Λ = 0) the whole enthalpy drop occurs in the nozzles and the rotor only turns the flow; in a 50% reaction stage the triangles are symmetric. Because the flow accelerates through turbine passages — a favourable pressure gradient — boundary layers stay attached and a turbine stage can take a far larger enthalpy drop than a compressor stage, so one or two turbine stages drive many compressor stages.
Turbine blade cooling lets the turbine inlet temperature exceed what the blade alloy could survive uncooled. Air bled from the compressor is passed through internal passages (convection cooling), directed as jets at the inside of the leading edge (impingement cooling), and ejected through rows of small holes to form a protective cool layer over the outer surface (film cooling); transpiration through a porous wall is the most effective in principle but hard to make durable. Thermal-barrier coatings add insulation. Cooling is not free: the bled air has absorbed compressor work and bypasses part of the cycle, so the gain from the higher temperature must exceed that penalty.
Compressor-turbine matching finds where on its map a single-spool gas generator actually runs. Three conditions tie the components together: they turn at the same speed; the turbine mass flow equals the compressor’s plus fuel minus bleed; and the turbine work equals the compressor work divided by the mechanical efficiency. The turbine nozzle guide vanes are usually choked, which fixes ṁ√T03/p03, and the propelling nozzle imposes its own flow condition. Solving these together for each speed gives one operating point per speed — the equilibrium running line on the compressor map, whose distance from the surge line is the surge margin. Changing the nozzle area moves the running line, which is one of the matching tools.
8. Rockets — thrust, specific impulse, staging, solid and liquid propellants
A rocket carries its oxidiser, so there is no ram drag: F = ṁV_e + (p_e − p_a)A_e = ṁc, where c is the effective exhaust velocity. Specific impulse is thrust per unit weight flow of propellant, I_sp = F/(ṁg0) = c/g0, in seconds. Chamber performance and nozzle performance separate neatly: the characteristic velocity c* = p_cA_t/ṁ depends on the propellant combination and chamber conditions, the thrust coefficient C_F = F/(p_cA_t) on the nozzle expansion, and c = C_Fc*. Integrating Newton’s law with no gravity or drag gives the Tsiolkovsky rocket equation Δv = c ln(m0/m_f), where m0/m_f is the ratio of initial to burn-out mass; gravity and drag losses are subtracted in real flight.
Because Δv grows only logarithmically with mass ratio, a single stage cannot carry the dead weight of empty tanks and engines all the way to orbit. Multi-staging discards that structure as each stage burns out, so the later stages accelerate a much smaller mass; for stages of equal specific impulse and structural fraction, the lightest vehicle for a given Δv splits it equally among the stages.
| Feature | Solid propellant | Liquid propellant |
|---|---|---|
| Propellant | A cast grain: composite (ammonium perchlorate, aluminium, polymer binder) or double-base | Fuel and oxidiser in tanks: cryogenic (LOX/LH₂), semi-cryogenic (LOX/kerosene) or storable hypergolic |
| Thrust control | Set by grain geometry and burning rate r = a·p_cⁿ (n < 1 for stable burning); no throttling, usually no restart | Throttleable and restartable by valves |
| Feed system | None — the grain burns in the case | Pressure-fed, or pump-fed (gas-generator, staged-combustion or expander cycle) |
| Specific impulse | Lower | Higher, LOX/LH₂ the highest of the common chemical combinations |
| Strengths | Simple, storable, ready, high thrust and density — boosters | Performance and control — core and upper stages |
Key takeaways
- Brayton: η = 1 − 1/π^((γ−1)/γ) (48.2% at π = 10); maximum specific work when T2 = T4 = √(T1T3). η_c = ideal/actual work, η_t = actual/ideal.
- F = ṁ(V_e − V0) + (p_e − p_a)A_e; η_p = 2/(1 + V_e/V0); η_o = FV0/(ṁ_fQ_R) = η_thη_p; range ∝ η_o(L/D).
- Ramjet: no static thrust. Afterburner: more thrust, much higher TSFC, variable nozzle. High bypass: lower jet velocity, higher η_p, lower TSFC. Turboprop and turboshaft deliver shaft power.
- Combustor zones: primary (stabilise, near-stoichiometric), secondary (complete), dilution (cool for the turbine); annular is lightest and most uniform; kerosene stoichiometric f ≈ 0.068.
- Axial stage: w = UΔC_w, Λ = (C_a/2U)(tan β1 + tan β2), symmetric blading Λ = 0.5; centrifugal: w = ψσU2²; multi-stage compressor overall efficiency < polytropic; matching gives the running line.
- Rocket: F = ṁc, I_sp = c/g0, Δv = c ln(m0/m_f); staging drops dead mass; solids are simple and fixed, liquids throttle and restart with higher I_sp.
Practice questions (23)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
What is the thermal efficiency of an ideal Brayton cycle with a pressure ratio of 10 and γ = 1.4, in per cent (to one decimal place)?
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Answer: 48.2
η = 1 − π^(−(γ−1)/γ) = 1 − 10^(−0.2857) = 1 − 0.5179 = 0.482, i.e. 48.2%. Equivalently the compressor temperature ratio is 10^0.2857 = 1.931 and η = 1 − 1/1.931. Using the exponent γ/(γ − 1) = 3.5 in place of (γ − 1)/γ gives an efficiency of practically 100%, impossible.For a simple gas-turbine (Brayton) cycle, which statements are correct?
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Answer: A — The ideal-cycle thermal efficiency depends only on the pressure ratio and γ; B — For fixed minimum and maximum temperatures, the ideal specific work is greatest when the compressor exit and turbine exit temperatures are equal; C — Compressor isentropic efficiency is the ideal work divided by the actual work
η_ideal = 1 − 1/π^((γ−1)/γ) contains nothing else; the specific work c_p[(T3 − T4) − (T2 − T1)] is maximised at T2 = T4 = √(T1T3); and a compressor needs more work than ideal, so η_c = ideal/actual. A higher T3 increases the work — it is the main route to a smaller, more powerful engine.A turbojet takes in 50 kg/s of air at a flight speed of 200 m/s and exhausts it at 600 m/s through a fully expanded nozzle. Neglecting the fuel mass flow, what is the thrust, in kN?
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Answer: 20
F = ṁ(V_e − V0) = 50 × (600 − 200) = 20 000 N = 20 kN; there is no pressure thrust because p_e = p_a. Forgetting the ram drag ṁV0 gives 30 kN.For the same engine (flight speed 200 m/s, jet speed 600 m/s), what is the propulsive efficiency?
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Answer: 0.5
η_p = 2V0/(V_e + V0) = 400/800 = 0.5. Check from its definition: thrust power FV0 = 20 000 × 200 = 4 MW, kinetic-energy rate ½ṁ(V_e² − V0²) = 25 × 320 000 = 8 MW, ratio 0.5. Writing V0/V_e = 0.333 is not the propulsive efficiency.An engine produces 20 kN of thrust at a flight speed of 250 m/s while burning 0.5 kg/s of fuel with a heating value of 43 MJ/kg. What is its overall efficiency (to three decimal places)?
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Answer: 0.233
η_o = FV0/(ṁ_fQ_R) = 20 000 × 250/(0.5 × 43 × 10⁶) = 5 × 10⁶/2.15 × 10⁷ = 0.233. Its TSFC is 0.5/20 000 = 2.5 × 10⁻⁵ kg/(N·s), and V0/(TSFC·Q_R) gives the same 0.233. Leaving Q_R in MJ gives a meaningless efficiency above 1.Which engine cannot produce thrust when the aircraft is stationary on the ground?
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Answer: C — Ramjet
A ramjet has no compressor: its compression comes entirely from decelerating the incoming air, and at zero flight speed there is none, so no cycle and no thrust. The others all have rotating compressors (or a propeller) that work at zero speed.Lighting the afterburner of a turbojet at a fixed flight condition:
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Answer: A — Increases the thrust; B — Increases the thrust-specific fuel consumption; C — Requires the nozzle throat area to be increased
Reheat raises the jet temperature and velocity, so thrust rises; the heat is added at low pressure, so the fuel buys thrust inefficiently and TSFC rises sharply; and the hotter, less dense gas needs a larger throat to keep the turbine’s operating point. A faster jet LOWERS the propulsive efficiency, 2/(1 + V_e/V0).A turbofan takes in 300 kg/s of air in total, of which 50 kg/s passes through the core. What is its bypass ratio?
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Answer: 5
Bypass ratio = bypass flow/core flow = (300 − 50)/50 = 250/50 = 5. Dividing the total by the core flow gives 6, and bypass over total gives 0.83 — neither is the bypass ratio as defined.The main reason a high-bypass turbofan has a lower TSFC than a turbojet at subsonic cruise is that it:
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Answer: B — Accelerates a much larger mass of air to a smaller velocity, raising propulsive efficiency
For a given thrust ṁ(V_e − V0), a large ṁ with small V_e − V0 wastes less kinetic energy in the jet, so η_p = 2/(1 + V_e/V0) is higher and less fuel is needed per unit thrust. Core temperature matters too, but is not what distinguishes the two; the fan’s captured air still carries ram drag.In a gas-turbine combustor, the dilution zone mainly serves to:
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Answer: C — Reduce the gas temperature and shape the temperature profile entering the turbine
The primary zone burns a near-stoichiometric mixture far too hot for the turbine; the dilution zone mixes in the remaining compressor air to bring the mean temperature down to the turbine limit with an acceptable radial profile. Flame stabilisation is the primary zone’s job, and dilution makes the mixture leaner, not richer.Methane (CH₄) burns stoichiometrically in air. Taking molar masses C = 12, H = 1, O₂ = 32, N₂ = 28 kg/kmol and air as 3.76 mol of N₂ per mol of O₂, what is the stoichiometric air-to-fuel ratio by mass (to two decimal places)?
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Answer: 17.16
CH₄ + 2(O₂ + 3.76N₂) → CO₂ + 2H₂O + 7.52N₂. Air per kmol of fuel = 2 × (32 + 3.76 × 28) = 2 × 137.28 = 274.56 kg against 16 kg of fuel, so A/F = 274.56/16 = 17.16 (fuel-air ratio 0.0583). Counting only the oxygen (64/16 = 4) ignores the nitrogen that comes with it.An axial compressor stage has a mean blade speed of 300 m/s and raises the whirl velocity by 100 m/s, with a work-done factor of 1. Taking c_p = 1005 J/(kg·K), what is the stagnation temperature rise across the stage, in K (to two decimal places)?
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Answer: 29.85
Euler: w = UΔC_w = 300 × 100 = 30 000 J/kg, so ΔT0 = w/c_p = 30 000/1005 = 29.85 K. Using U² (90 000 J/kg) treats the stage as if the whirl rose by the whole blade speed.An axial compressor stage with an inlet stagnation temperature of 300 K has a stagnation temperature rise of 30 K and an isentropic stage efficiency of 0.9. Taking γ = 1.4, what is the stage stagnation pressure ratio (to three decimal places)?
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Answer: 1.352
π = (1 + η_sΔT0/T01)^(γ/(γ−1)) = (1 + 0.9 × 30/300)^3.5 = 1.09^3.5 = exp(3.5 × 0.08618) = 1.352. Leaving out the efficiency gives 1.1^3.5 = 1.396, the ideal-stage value.An axial compressor stage has symmetric velocity triangles (α1 = β2 and α2 = β1). Its degree of reaction is:
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Answer: B — 0.5
Λ = (C_a/(2U))(tan β1 + tan β2), and with symmetric triangles tan β1 + tan β2 = tan α2 + tan β2 = U/C_a, so Λ = 0.5: the static temperature rise is shared equally between rotor and stator. Zero reaction is the impulse case, all diffusion in the stator.For a multi-stage axial compressor in which every stage has the same small-stage (polytropic) efficiency, the overall isentropic efficiency of the compressor is:
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Answer: C — Lower than the polytropic efficiency, and falling as the overall pressure ratio rises
The losses of each stage appear as heat in the air passed to the next, which must then compress warmer air and so does more work for the same pressure ratio; the penalty accumulates with pressure ratio. For a turbine the effect reverses — reheat is partly recovered downstream — and the overall isentropic efficiency exceeds the polytropic one.A centrifugal compressor with radial vanes has an impeller tip speed of 400 m/s, a slip factor of 0.9, a power input factor of 1 and no inlet whirl. What is the specific work input, in kJ/kg?
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Answer: 144
w = ψσU2² = 1 × 0.9 × 400² = 0.9 × 160 000 = 144 000 J/kg = 144 kJ/kg. Without slip the work would be U2² = 160 kJ/kg; slip reduces the whirl leaving the impeller to σU2.In the steady matching of the compressor and turbine of a single-spool turbojet, which conditions must hold?
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Answer: A — The compressor and turbine rotate at the same speed; B — The turbine mass flow equals the compressor mass flow plus fuel minus any bleed; C — The turbine work equals the compressor work divided by the mechanical efficiency
Same shaft, same speed; continuity through the gas generator; and the turbine must supply the compressor’s work plus bearing losses. Together with the choked turbine and nozzle these fix the running line. The turbine’s pressure ratio is smaller than the compressor’s — the remainder, less the combustor loss, is left for the nozzle to make thrust.In film cooling of a turbine blade, the coolant air is:
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Answer: B — Ejected through rows of holes to form a cool layer over the blade’s outer surface
Compressor bleed air passes through internal passages (convection and impingement) and is then ejected through small holes so that it spreads as an insulating film between the hot gas and the metal. It must come from the compressor, at a pressure above that of the gas outside the blade — turbine-exit gas is too hot and at too low a pressure.A rocket engine has an effective exhaust velocity of 2500 m/s. Taking g0 = 9.81 m/s², what is its specific impulse, in s (to one decimal place)?
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Answer: 254.8
I_sp = c/g0 = 2500/9.81 = 254.8 s. Multiplying by g0 instead gives 24 525, and quoting 2500 gives the effective exhaust velocity itself, in m/s rather than seconds.A single-stage rocket with a specific impulse of 300 s has an initial-to-burn-out mass ratio of 4. Neglecting gravity and drag and taking g0 = 9.81 m/s², what is its velocity increment, in km/s (to two decimal places)?
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Answer: 4.08
c = I_sp·g0 = 300 × 9.81 = 2943 m/s, and Δv = c ln(m0/m_f) = 2943 × ln 4 = 2943 × 1.3863 = 4080 m/s = 4.08 km/s. Using log10 4 = 0.602 gives 1.77 km/s; the rocket equation uses the natural logarithm.Which of the following statements about chemical rockets are correct?
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Answer: A — A solid rocket motor is generally not throttleable once ignited; B — A burning-rate pressure exponent n below 1 is needed for stable solid-propellant combustion; C — Liquid oxygen with liquid hydrogen gives a higher specific impulse than common solid propellants
Once lit, a solid grain burns until it is consumed, at a rate r = a·p_cⁿ; with n < 1 a rise in chamber pressure increases the nozzle outflow faster than the gas generation, restoring equilibrium, while with n ≥ 1 the pressure runs away. LOX/LH₂ has the highest I_sp of the common chemical combinations. Staging raises the mission Δv by discarding dead mass; it does not change any engine’s I_sp.A convergent propelling nozzle expanding air (γ = 1.4) becomes choked when the ratio of the nozzle inlet stagnation pressure to ambient pressure reaches about:
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Answer: C — 1.893
Sonic flow at the exit needs p0/p* = ((γ + 1)/2)^(γ/(γ−1)) = 1.2^3.5 = 1.893, the inverse of p/p0 = 0.528. Above that the exit stays sonic, the exit pressure exceeds ambient, and a pressure-thrust term appears. 1.528 is a mangled version of 0.528; 1.2 is the temperature ratio T0/T.Which of the following statements about aircraft engine intakes are correct?
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Answer: A — At high flight Mach numbers, compressing through several oblique shocks followed by a weak normal shock recovers more stagnation pressure than a single normal shock; B — A subsonic intake at cruise typically captures a stream tube smaller than its lip area, with the excess spilling round the lip; C — The stagnation temperature of the air is essentially unchanged through the intake
Stagnation-pressure loss rises steeply with the normal Mach number of a shock, so several weak oblique shocks and a weak terminal normal shock lose much less than one strong normal shock. At cruise the engine needs less flow than the lip could capture, so air spills. The intake does no work and is adiabatic, so T0 is constant. At Mach 3 a single normal shock would keep only about a third of the stagnation pressure, which is why a pitot intake is not used there.