Aircraft and space propulsion principles and performance

Name the momentum flow and the energy source

A propeller accelerates surrounding air, using shaft power supplied by a piston engine, turbine or electric motor. A gas-turbine engine compresses incoming air, adds energy and expands the flow; different architectures distribute work between exhaust thrust and shaft-driven propellers or fans. The atmosphere supplies reaction mass and, for combustion, oxidiser.

A rocket carries its expelled propellant and can create thrust in vacuum; it does not push against ambient air. Chemical, electrical and other energy-conversion paths differ. An electric thruster uses electrical energy to accelerate expelled material. Power supply, thermal rejection, thrust magnitude, propellant, mass and time all constrain mission suitability.

Aircraft thrust: more mass flow or a larger velocity increment?

Take a control volume moving steadily with an ideal propulsor. Let V be the upstream air speed relative to the device and V_e the far-wake speed in the same direction. Assume one uniform streamtube, equal inlet/outlet mass flow ṁ, equal ambient static pressure at the far boundaries, negligible swirl and no stored momentum. The magnitude of thrust is F=ṁ(V_e−V). The air gains downstream momentum; the device receives the opposite reaction. Nearby blade pressures need not equal ambient just because the far control surfaces do.

The corresponding flow kinetic-power increase is P_flow=½ṁ(V_e²−V²). Useful propulsive power at flight speed is P_use=FV. Factoring the difference of squares gives P_flow=½F(V_e+V), so the ideal propulsive ratio is P_use/P_flow=2V/(V_e+V). A greater velocity increment leaves more kinetic energy in the wake for the same useful force and flight speed. At rest FV=0 despite nonzero thrust and flow power; zero propulsive ratio at that instant is not zero force.

For an original common requirement F=600 N at V=50 m/s, option A has V_e=70 m/s and needs ṁ=600/20=30 kg/s. It adds 36 kW to the flow while delivering 30 kW of useful power, ratio 0.8333. Option B uses V_e=90 m/s and ṁ=15 kg/s: flow power is 42 kW and the ratio 0.7143. Lower velocity increment looks better at these declared boundaries, but moving more air can require a larger disk/duct, installation space, mass and noise trade. This is not a universal propeller-versus-jet ranking.

Shaft, fuel and electrical inputs are separate boundaries. If a supplied aggregate conversion factor from bus input to this ideal flow-power increment is 0.75, the electrical inputs would be 48 and 56 kW. A combustion engine additionally needs a fuel-energy/shaft or jet-energy account; the momentum ratio alone supplies no fuel consumption. Actual inlet losses, pressure thrust, nonuniform flow and Mach effects require a richer model.

Thrust and propellant-use metrics answer different questions

For an idealised rocket control volume, thrust magnitude is ṁv_e+(p_e−p_a)A_e. Here ṁ is expelled mass per second, v_e is exhaust speed relative to the vehicle, p_e and p_a are exit and ambient pressure and A_e is exit area. Define effective exhaust velocity c=F/ṁ, which includes any pressure contribution consistently.

Specific impulse is I_sp=F/(ṁg_0)=c/g_0, expressed in seconds, using a conventional reference acceleration g_0. It measures impulse per propellant weight under the stated conditions; it is not a dimensionless energy efficiency, a thrust magnitude or a guaranteed mission time. A high value can coexist with low thrust or demanding electrical-power requirements.

Why changing mass creates a logarithm

For a short pressure-matched ideal ejection, let the vehicle have mass m and velocity v. It loses a positive small amount of propellant, so its own change dm is negative. The expelled mass is −dm and leaves with velocity approximately v−c in an inertial frame; here c is the constant relative exhaust speed. The remaining vehicle is (m+dm) at (v+dv). Conservation of the combined vehicle-plus-new-exhaust momentum, after cancelling terms and taking the differential limit, gives m dv=−c dm.

Thus dv=−c dm/m: losing the same small mass creates a larger velocity increment once less vehicle mass remains. Summing these increments from m_i to m_f gives Δv=−c∫(dm/m)=c ln(m_i/m_f). The natural logarithm is the integral of 1/m, not an arbitrary correction factor. The ratio is dimensionless and exceeds one for propellant loss. External gravity/drag and variable thrust direction require additional trajectory terms. Using effective c with pressure thrust needs its consistently defined conditions; this short derivation chose c equal to the relative speed to make the momentum boundary explicit.

A changing vehicle mass affects velocity change

With constant effective exhaust speed, no external force and all expelled propellant accounted, momentum conservation gives ideal Δv=c ln(m_i/m_f). Initial and final total mass include structure, payload and remaining contents. This velocity-change budget is not an altitude and not the speed reached in a real ascent with gravity, drag and steering.

The figure shows diminishing additional ideal Δv for equal increments of mass ratio. Structure and payload constrain the achievable ratio, while finite thrust sets how quickly a manoeuvre occurs. In an aircraft, useful thrust power is FV at the selected flight speed, whereas shaft or fuel input is measured at another boundary. Compare metrics only after naming those endpoints.

High exhaust speed exchanges propellant for power and time

For a constructed pressure-matched, single-speed exhaust with no unaccounted propellant, jet kinetic power is P_jet=½ṁc². With F=ṁc this becomes P_jet=Fc/2. Declare jet-power efficiency η_j=P_jet/P_in rather than calling it useful spacecraft efficiency: exhaust kinetic power at this relative boundary is different from FV at a chosen inertial speed. At fixed input power and η_j, F=2η_jP_in/c; doubling c halves thrust and quarters mass flow.

Supply P_in=1000 W and η_j=0.60 continuously available, so P_jet=600 W. At c=10000 m/s, F=0.12 N and ṁ=0.000012 kg/s. At c=20000 m/s, F=0.06 N and ṁ=0.000003 kg/s. The remaining 400 W belongs to other energy channels and losses; locating those losses is necessary before allocating hardware cooling. No beam voltage, propellant preparation or operating procedure is specified.

For the same unforced 100 kg initial vehicle and required ideal Δv=10 m/s, rearrange the log relation: propellant used is m_i[1−exp(−Δv/c)]. The two options use about 0.09995 and 0.04999 kg. Constant mass flow gives durations about 8329 s (2.31 h) and 16663 s (4.63 h). Under an illustrative 3 h continuous-power window, the lower-c option fits time while the higher-c option does not, despite using about half the propellant. Thrust direction, external acceleration and real duty limits are excluded, so this is a comparison of services within the ideal model.

Separate force, specific impulse and ideal velocity change

For an invented already-characterised thruster, supply F=20 N, ṁ=.010 kg/s and zero unaccounted flow. Effective exhaust speed is 2000 m/s and, using g_0=9.80665 m/s², I_sp≈203.943 s. These are supplied aggregate performance values, not a method for constructing or operating an engine.

Separately, an ideal unforced vehicle with m_i=100 kg, m_f=80 kg and c=2000 m/s has Δv=2000 ln(1.25)≈446.287 m/s. This calculation needs no assumption about a particular propellant chemistry. It cannot specify orbit insertion, burn duration or required thermal hardware. Holding mass ratio fixed while changing c changes ideal Δv; it does not by itself determine the cost or feasibility of the propulsion system.

Try the reasoning

1. Does a rocket need air behind it to provide thrust?

   Show the answer and reasoning

   No. Expelled momentum and the associated pressure account provide the reaction force.

2. Is specific impulse the same as energy efficiency?

   Show the answer and reasoning

   No. It is impulse per propellant weight, with units of seconds.

3. Does 446 m/s ideal delta-v mean 446 m of altitude?

   Show the answer and reasoning

   No. Velocity change and distance are different quantities; trajectory forces and time matter.

4. At V=40 m/s, two ideal aircraft streams both supply 400 N. One has V_e=60 m/s, the other 80 m/s. Compare mass flow, flow power and useful power.

   Show the answer and reasoning

   Mass flows are 400/20=20 and 400/40=10 kg/s. Flow powers are F(V_e+V)/2=20 and 24 kW. Useful power is FV=16 kW for either, giving ratios 0.80 and 0.667. The larger stream has a smaller ideal wake-power penalty, but installation and the input-energy boundary remain separate.

5. An analyst inserts final mass 80 kg and initial mass 100 kg into ln(80/100) yet reports a positive 446.287 m/s for c=2000 m/s. Reconcile the sign with momentum.

   Show the answer and reasoning

   With c a positive relative exhaust-speed magnitude and propellant expelled backward, m dv=−c dm is positive because dm<0. The integrated expression is c ln(m_i/m_f)=2000 ln(100/80)=+446.287 m/s along the chosen thrust direction. ln(80/100) is negative; dropping its sign without restoring the correct mass ratio hides the boundary/sign error.

6. Keep the electric example’s 1000 W input, but reduce η_j from 0.60 to 0.45. Which original option can still supply the 10 m/s ideal service within 3 h?

   Show the answer and reasoning

   Jet power falls to 450 W. At c=10000, F=0.09 N and ṁ=0.000009 kg/s; the same 0.09995 kg propellant now needs about 11106 s=3.08 h. At c=20000 the same 0.04999 kg needs about 22217 s=6.17 h. Neither fits 3 h. The propellant requirement at fixed c and Δv is unchanged, while reduced power conversion lengthens the service.

What to carry forward

Compare momentum change, power, propellant and mission time at explicitly defined boundaries.