Thrust equation and propulsive, thermal and overall efficiency

Why it matters

Every air-breathing engine is judged by two numbers: how much thrust it makes and how much fuel it burns to make it. The thrust equation tells you where thrust comes from (momentum change plus a pressure term), and the three efficiencies tell you where the fuel energy is lost — in the thermodynamic cycle or in the kinetic energy left behind in the jet. This split explains why airliners use big fans, why fighters use small hot jets, and how every later cycle-analysis topic is scored.

Key ideas

Control volume and thrust. Draw a control volume around the engine. Air enters at the flight speed V₀ with mass flow ṁₐ; fuel ṁ_f is added from inside the aircraft (it carries no forward momentum in the engine frame); gas leaves the nozzle exit at velocity Vₑ, pressure pₑ and area Aₑ. The net force on the engine, from Newton's second law for a steady flow, is the uninstalled thrust:

• momentum thrust — the rate of increase of momentum of the flow, (ṁₐ + ṁ_f)·Vₑ − ṁₐ·V₀;
• pressure thrust — (pₑ − p₀)·Aₑ, non-zero only when the nozzle is not fully expanded (pₑ ≠ p₀), for example a choked convergent nozzle. The term ṁₐ·V₀ is the ram drag: the momentum the engine must give back to the air it swallows. Installed thrust further subtracts inlet spillage drag and nacelle drag; those are airframe-integration effects and are not part of this equation.

Simplified form. When f = ṁ_f/ṁₐ is small (typically 0.01–0.03) and the nozzle is fully expanded, F ≈ ṁₐ·(Vₑ − V₀). This is the form used to explain trends, not for precise answers.

Energy flow. Fuel releases ṁ_f·Q_R of chemical power (Q_R = heating value). The engine turns part of it into an increase of the gas's kinetic energy; the rest leaves as heat in the hot exhaust. Of the kinetic energy added, only F·V₀ (the thrust power) does useful work on the aircraft; the rest is left behind as a moving wake.

• Thermal efficiency η_th — kinetic energy added to the gas ÷ fuel power. It is set by the cycle: pressure ratio, turbine inlet temperature and component losses (Brayton-cycle topics).
• Propulsive efficiency η_p — thrust power ÷ kinetic energy added. It depends only on the velocity ratio Vₑ/V₀. Thrust needs Vₑ > V₀, but every m/s of excess jet speed is kinetic energy wasted in the wake. Accelerating a lot of air by a little (a fan, a propeller) is efficient; accelerating a little air by a lot (a turbojet) is not.
• Overall efficiency η₀ — thrust power ÷ fuel power = η_th·η_p.

The thrust–efficiency trade-off. For fixed ṁₐ, thrust rises with (Vₑ − V₀) but η_p falls. At Vₑ = V₀, η_p = 1 but thrust is zero. This is why bypass ratio has grown: more mass flow at a lower jet velocity gives the same thrust with higher η_p.

Thrust specific fuel consumption (TSFC) = ṁ_f/F, in kg/(N·s) or more readably mg/(N·s) or kg/(N·h). Lower is better. For a given flight speed TSFC = V₀/(η₀·Q_R), so TSFC and η₀ carry the same information.

Specific thrust F/ṁₐ (N·s/kg = m/s) measures engine size for a given thrust: high specific thrust means a small, light engine but low η_p.

Formulas

F = (ṁₐ + ṁ_f)·Vₑ − ṁₐ·V₀ + (pₑ − p₀)·Aₑ F thrust (N); ṁₐ air mass flow (kg/s); ṁ_f fuel mass flow (kg/s); Vₑ nozzle exit velocity (m/s); V₀ flight speed (m/s); pₑ, p₀ exit and ambient static pressure (Pa); Aₑ nozzle exit area (m²). Steady flow, single stream; for a separate-flow turbofan add one such term per stream.

F ≈ ṁₐ·(Vₑ − V₀) — fuel mass neglected, nozzle fully expanded (pₑ = p₀).

η_p = F·V₀ / [½(ṁₐ + ṁ_f)·Vₑ² − ½ṁₐ·V₀²] — general definition.

η_p = 2V₀/(Vₑ + V₀) = 2/(1 + Vₑ/V₀) — Froude formula, valid for f ≈ 0 and pₑ = p₀.

η_th = [½(ṁₐ + ṁ_f)·Vₑ² − ½ṁₐ·V₀²] / (ṁ_f·Q_R) — Q_R heating value of fuel (J/kg), about 43 MJ/kg for kerosene (take from your data book).

η₀ = F·V₀ / (ṁ_f·Q_R) = η_th·η_p

TSFC = ṁ_f / F (kg/(N·s)); TSFC = V₀ / (η₀·Q_R)

f = ṁ_f / ṁₐ — fuel–air ratio (dimensionless).

Worked examples

Example 1 (standard). A turbojet flies at V₀ = 250 m/s. ṁₐ = 50 kg/s, fuel–air ratio f = 0.012, exhaust velocity Vₑ = 650 m/s, nozzle fully expanded, Q_R = 43 MJ/kg. Find thrust, the three efficiencies and TSFC.

1. Fuel flow: ṁ_f = f·ṁₐ = 0.012 × 50 = 0.6 kg/s.
2. Thrust: F = (ṁₐ + ṁ_f)·Vₑ − ṁₐ·V₀ = 50.6 × 650 − 50 × 250 = 32 890 − 12 500 = 20 390 N.
3. Thrust power: F·V₀ = 20 390 × 250 = 5.098 MW.
4. Kinetic energy added: ½ × 50.6 × 650² − ½ × 50 × 250² = 10 689 250 − 1 562 500 = 9.127 MW.
5. Fuel power: ṁ_f·Q_R = 0.6 × 43 × 10⁶ = 25.8 MW.
6. η_p = 5.098/9.127 = 0.559; η_th = 9.127/25.8 = 0.354; η₀ = 5.098/25.8 = 0.198 (check: 0.559 × 0.354 = 0.198).
7. TSFC = ṁ_f/F = 0.6/20 390 = 2.94 × 10⁻⁵ kg/(N·s) = 29.4 mg/(N·s) ≈ 0.106 kg/(N·h).

Answer: F = 20.4 kN, η_p = 0.559, η_th = 0.354, η₀ = 0.198, TSFC = 29.4 mg/(N·s). The Froude formula gives η_p = 2/(1 + 2.6) = 0.556, within 1% — the fuel mass makes little difference.

Example 2 (GATE level, pressure thrust). At an altitude where p₀ = 26.5 kPa, an engine flying at 240 m/s swallows 80 kg/s of air and burns 1.6 kg/s of fuel. Its convergent nozzle is choked: Vₑ = 560 m/s, pₑ = 60 kPa, Aₑ = 0.45 m². Find the thrust and TSFC.

1. Momentum thrust: (80 + 1.6) × 560 − 80 × 240 = 45 696 − 19 200 = 26 496 N.
2. Pressure thrust: (pₑ − p₀)·Aₑ = (60 000 − 26 500) × 0.45 = 15 075 N.
3. Total: F = 26 496 + 15 075 = 41 571 N.
4. TSFC = 1.6/41 571 = 3.85 × 10⁻⁵ kg/(N·s) = 38.5 mg/(N·s).

Answer: F ≈ 41.6 kN, TSFC ≈ 38.5 mg/(N·s). Dropping the pressure term would understate thrust by more than a third — always check whether the nozzle is choked.

Common mistakes

• Using the Froude formula η_p = 2/(1 + Vₑ/V₀) when the nozzle is under-expanded or when the question gives fuel flow explicitly; use the general definition then.
• Forgetting ram drag ṁₐ·V₀ — writing F = ṁ·Vₑ for a flying engine (that is only the static thrust, V₀ = 0).
• Writing η_p as a ratio of "exhaust to flight" speed the wrong way up; η_p must be ≤ 1 and falls as Vₑ/V₀ rises.
• Defining thermal efficiency with temperatures of one particular cycle; the propulsion definition is kinetic-energy gain ÷ fuel power.
• Mixing units for TSFC: 1 kg/(N·h) = 10⁶/3600 = 277.8 mg/(N·s); convert carefully.
• Using p₀ in gauge terms or kPa with m² and expecting newtons without converting to Pa.

For GATE AE

Expect direct numericals on thrust with and without the pressure term, propulsive efficiency from a velocity ratio, overall efficiency from fuel flow and heating value, and TSFC with unit conversion. Conceptual questions test why η_p falls as jet velocity rises, what happens at Vₑ = V₀, and why high-bypass engines have lower TSFC. Practise rearranging: given η_p find Vₑ/V₀; given thrust and TSFC find fuel flow.

Quick check

1. Write the full uninstalled thrust equation and name the ram-drag term.
2. An engine has Vₑ/V₀ = 1.5 and a fully expanded nozzle; what is η_p (neglect fuel mass)?
3. If η_th = 0.45 and η_p = 0.80, what is η₀?
4. Why does a turbofan have higher propulsive efficiency than a turbojet of the same thrust?
5. Static thrust test: ṁₐ = 20 kg/s, Vₑ = 500 m/s, pₑ = p₀, fuel neglected. What is F?

Answers: 1. F = (ṁₐ + ṁ_f)Vₑ − ṁₐV₀ + (pₑ − p₀)Aₑ; ram drag is ṁₐV₀. 2. 2/2.5 = 0.80. 3. 0.36. 4. It moves more air with a smaller velocity increase, so less kinetic energy is wasted in the jet. 5. 10 000 N.