Range and endurance of propeller aircraft: Breguet equations

Why it matters

How far an aircraft can fly and how long it can stay up on a tank of fuel are the two numbers customers care about most: they decide routes, payload–range trade-offs and patrol times. For propeller aircraft (piston and turboprop) the Breguet equations give both in closed form and show exactly which design choices — L/D, propeller efficiency, engine fuel consumption, fuel fraction — matter, and at which speed to fly.

Key ideas

Fuel consumption of a propeller engine. A piston or turboprop engine is rated by its specific fuel consumption c: fuel used per unit shaft power per unit time. Engine data usually give it on a mass basis in kg/(kW·h) (brake specific fuel consumption, BSFC; typically 0.2–0.35 kg/(kW·h)). In the Breguet equations it must be on a weight basis in SI: newtons of fuel per watt per second, which has units of 1/m: c [1/m] = BSFC [kg/(kW·h)] × g / 3.6 × 10⁶. For example, 0.25 kg/(kW·h) = 6.81 × 10⁻⁷ m⁻¹.

Weight loss rate. Shaft power needed is P_s = P_R/η_p = D·V/η_p, so the aircraft's weight falls at dW/dt = −c·P_s = −c·D·V/η_p.

Range. Distance flown is dR = V·dt. Using the weight-loss rate and D = W/(L/D): dR = −(η_p/c)·(L/D)·dW/W. With η_p, c and L/D held constant and integrating from initial weight W₀ to final weight W₁: R = (η_p/c)·(L/D)·ln(W₀/W₁). Key consequences:

• Maximum range at (L/D)_max, i.e. the minimum-drag lift coefficient C_L = √(C_D0/K).
• Speed does not appear explicitly: a propeller aircraft gets the same theoretical range at any altitude as long as it flies at (L/D)_max (it just flies faster at altitude). In practice η_p, c and engine power vary with altitude.
• Holding C_L constant as fuel burns means flying slower, or climbing slowly (a cruise climb), to keep W/(ρV²) constant.

Endurance. Time aloft is dt = −dW/(c·P_s) = −η_p·dW/(c·P_R). In level flight P_R = √(2W³C_D²/(ρSC_L³)). Integrating at constant C_L, ρ, η_p and c: E = (η_p/c)·(C_L^(3/2)/C_D)·√(2ρS)·(W₁^(−1/2) − W₀^(−1/2)). Consequences:

• Maximum endurance at maximum C_L^(3/2)/C_D, i.e. the minimum-power condition, C_L = √(3C_D0/K).
• Endurance is proportional to √ρ, so it is best at low altitude (sea level).
• The speed for best endurance is about 0.76 times the speed for best range.

Assumptions and limits. Constant η_p and c; parabolic polar; steady level flight with the fuel weight being the only change; no allowance for takeoff, climb, descent, reserves or wind. Real range calculations use these equations segment by segment with reserves added. Dimension check: η_p/c has units of metres, so range comes out in metres directly. In the endurance formula the group √(2ρS)·W^(−1/2) has units of s/m, so with W in N, ρ in kg/m³, S in m² and c in 1/m, endurance comes out in seconds.

Comparison with jets. A jet's fuel flow scales with thrust, not power, so its optimum conditions shift: the jet flies for best range at the Carson speed and has best endurance at (L/D)_max (next topic).

Formulas

c = BSFC × g / (3.6 × 10⁶) (kg/(kW·h) to 1/m)

dW/dt = −c·P_s = −c·D·V / η_p

R = (η_p / c) · (L/D) · ln(W₀ / W₁) (Breguet range, propeller)

R_max = (η_p / c) · (L/D)_max · ln(W₀ / W₁)

E = (η_p / c) · (C_L^(3/2) / C_D) · √(2ρS) · (W₁^(−1/2) − W₀^(−1/2)) (Breguet endurance, propeller)

(C_L^(3/2)/C_D)_max at C_L = √(3·C_D0/K), C_D = 4·C_D0

Symbols: R range (m); E endurance (s); η_p propeller efficiency; c specific fuel consumption on weight basis (N of fuel per W per s = 1/m); BSFC (kg/(kW·h)); g = 9.81 m/s²; L/D lift-to-drag ratio; W₀ initial weight (with fuel), W₁ final weight (N); ρ density (kg/m³); S wing area (m²); C_L, C_D lift and drag coefficients; P_s shaft power, P_R power required (W); D drag (N); V true airspeed (m/s).

Worked examples

Example 1 (standard): maximum range. Given: η_p = 0.85, BSFC = 0.25 kg/(kW·h), (L/D)_max = 15, W₀ = 20 000 N, fuel weight 3000 N (W₁ = 17 000 N).

1. Convert c: c = 0.25 × 9.81/3.6 × 10⁶ = 6.8125 × 10⁻⁷ m⁻¹.
2. η_p/c = 0.85/6.8125 × 10⁻⁷ = 1.2477 × 10⁶ m.
3. ln(W₀/W₁) = ln(20 000/17 000) = 0.16252.
4. R = (η_p/c)·(L/D)·ln(W₀/W₁) = 1.2477 × 10⁶ × 15 × 0.16252 = 3.042 × 10⁶ m = 3042 km.

Example 2 (GATE level): maximum endurance. Given: the same aircraft and fuel, S = 20 m², drag polar C_D = 0.025 + 0.04·C_L², flying at sea level (ρ = 1.225 kg/m³).

1. Minimum-power point: C_L = √(3 × 0.025/0.04) = 1.3693; C_D = 4 × 0.025 = 0.100.
2. C_L^(3/2)/C_D = 1.3693^1.5/0.100 = 16.02.
3. √(2ρS) = √(2 × 1.225 × 20) = 7.000 (units kg^½/m^½).
4. W₁^(−1/2) − W₀^(−1/2) = 1/√17 000 − 1/√20 000 = 0.0076696 − 0.0070711 = 0.00059855 N^(−1/2).
5. E = (η_p/c)·(C_L^(3/2)/C_D)·√(2ρS)·(W₁^(−1/2) − W₀^(−1/2)) = 1.2477 × 10⁶ × 16.02 × 7.000 × 0.00059855 = 83 770 s = 23.3 h. Sanity check: the mean power required is about 45 kW, so fuel flow ≈ 0.25 × 45/0.85 ≈ 13 kg/h, and 306 kg of fuel lasts about 23 h.

Common mistakes

• Using BSFC in kg/(kW·h) directly. Convert to a weight-per-watt-per-second basis (1/m) first, or the units do not cancel.
• Using the jet range formula (with V/c_t) for a propeller aircraft.
• Flying for best range at minimum power or for best endurance at (L/D)_max — that is the jet case reversed.
• Writing ln(W₁/W₀) (negative) or using fuel weight in place of the weight ratio.
• Thinking a propeller aircraft gains range by flying higher. In the ideal equation altitude cancels; only endurance (∝ √ρ) changes, and it gets worse with height.

For GATE AE

Expect direct Breguet-range numericals with unit conversion of specific fuel consumption, ratio questions (effect of improving η_p, L/D or fuel fraction), endurance calculations at the minimum-power condition, and conceptual questions on which C_L and altitude are best for range and endurance of propeller versus jet aircraft. Practise the derivation from dW/dt so you can adapt it to other flight programmes.

Quick check

1. At which aerodynamic condition is a propeller aircraft's range a maximum?
2. At which aerodynamic condition is its endurance a maximum?
3. Does the ideal propeller range depend on altitude?
4. Convert 0.3 kg/(kW·h) to the weight basis in 1/m.

Answers: 1. (L/D)_max. 2. Maximum C_L^(3/2)/C_D (minimum power). 3. No (only through η_p and c in practice). 4. 0.3 × 9.81/3.6 × 10⁶ = 8.18 × 10⁻⁷ m⁻¹.