Lesson 4 · 35 min
Engines and Fuel Consumption
The second ingredient of the Breguet equations is the engine's thirst, its specific fuel consumption. Jets and propellers measure it differently, and that difference is not just units: it is why a turboprop and a jet fly the same mission at different speeds and different lift-to-drag ratios.
Learning objectives
- Define thrust-specific fuel consumption \(C\), convert it between units, and choose typical values for initial sizing.
- Define power-specific fuel consumption and propeller efficiency, and convert them to an equivalent \(C\).
- Explain, from the way \(C\) varies with speed, why jets and propeller aircraft cruise and loiter at different \(L/D\).
Jets: thrust-specific fuel consumption
For a jet engine, the fuel burned is roughly proportional to the thrust it gives. Its thrust-specific fuel consumption \(C\) is the weight of fuel burned per hour per unit of thrust, so it has units of 1/h (pounds of fuel per hour per pound of thrust, or newtons per hour per newton). Engine data sheets usually give the same consumption as a mass of fuel per unit thrust per second, \(c_T\), in mg/(N s):
Thrust-specific fuel consumption
\[ \dot W_{\text{fuel}} = C\,T, \qquad C\ (1/\text{h}) = \frac{c_T\ \bigl(\text{mg/(N s)}\bigr)}{28.33} \]Since \(1/\text{h}\) of fuel weight per unit thrust is \(\dfrac{1}{9.81}\ \text{kg/(N h)} = 28.33\ \text{mg/(N s)}\). A modern high-bypass turbofan in cruise uses about 15 to 17 mg/(N s), a \(C\) of 0.53 to 0.60.
| Engine | Cruise | Loiter |
|---|
Loiter values are lower because loiter is flown at lower speed, where a jet's \(C\) is smaller. Bypass ratio is the big lever: a high-bypass turbofan uses about half the fuel of a turbojet for the same thrust. The business jet's engines are smaller, lower-bypass turbofans than an airliner's, so this module takes \(C = 0.65\) in cruise and \(0.55\) in loiter, between the table's turbofan rows.
Propellers: power-specific fuel consumption
A piston engine or turboprop burns fuel in proportion to the power it delivers to the shaft. Its power-specific fuel consumption \(c_P\) is the mass of fuel burned per unit of shaft power per hour, in kg/(kW h) or, in US units, lb/(hp h). Data sheets often use g/(kW h): \(0.30\ \text{kg/(kW h)}\) is \(300\ \text{g/(kW h)}\). The propeller turns shaft power into thrust power with an efficiency \(\eta_p\): \(TV = \eta_p P\). The fuel burned per unit thrust then grows with speed:
Equivalent \(C\) of a propeller aircraft
\[ C = \frac{c_P\,g\,V}{1000\,\eta_p}\quad (c_P\ \text{in kg/(kW h)},\ V\ \text{in m/s}) \] \[ \text{or}\quad C = \frac{c_PV}{550\,\eta_p}\quad (\text{lb/(hp h)},\ \text{ft/s}) \] \[ \frac{V}{C} = \frac{3600\,\eta_p}{c_P\,g}\ \ \text{km} \]For a propeller aircraft \(V/C\), the propulsive factor of Breguet, does not depend on speed. \(1\ \text{lb/(hp h)} = 0.6083\ \text{kg/(kW h)}\).
Names and symbols
For a piston engine, \(c_P\) is usually called the brake-specific fuel consumption (BSFC): "brake" power is the power at the output shaft, as measured on a dynamometer brake. Turboprop data are quoted per shaft power or per equivalent shaft power, which adds the small jet thrust of the exhaust (ESFC). Raymer writes this quantity as \(C_{\text{bhp}}\), in lb/(hp h). This course uses \(c_T\) and \(c_P\) for the mass-based consumptions that engine data give, and keeps \(C\), in 1/h, for the thrust-specific form that the Breguet equations use.
| Engine | Cruise lb/(hp h) | Loiter lb/(hp h) | Cruise kg/(kW h) | \(\eta_p\) |
|---|
Example 4.1 — The \(C\) of a turboprop
A turboprop has \(c_P = 0.5\ \text{lb/(hp h)}\) and \(\eta_p = 0.8\) in cruise at \(500\ \text{km/h}\). Find its equivalent \(C\), and \(V/C\).
Show solution
Check with the speed-free form: \(3600(0.8)/(0.3041 \times 9.81) = 965\ \text{km}\). At 500 km/h the turboprop's \(C\) is about that of a high-bypass turbofan; at lower speeds it is better, at higher speeds worse.
Why the best \(L/D\) differs
Lesson 3 gave the rule; the fuel consumption explains it.
- Range is proportional to \((V/C)(L/D)\). For a jet, \(C\) is about constant, so the best range is where \(V(L/D)\) is greatest: \(L/D = 0.866\,\LDmax\), a little faster than \(\LDmax\). For a propeller aircraft, \(V/C\) is constant, so the best range is at \(\LDmax\).
- Endurance is proportional to \((1/C)(L/D)\). For a jet, that is greatest at \(\LDmax\). For a propeller aircraft, \(C \propto V\), so endurance is greatest where \((L/D)/V\) is, at minimum power, which is at \(0.866\,\LDmax\), slower than \(\LDmax\).
Derivations: the best \(L/D\) for range and endurance, and the range ratio
Drag and \(L/D\) against speed
Start from Module 1's parabolic drag polar, \(\CD = \CDz + K\CL^2\), in level flight, where lift equals weight and \(\CL = 2W/(\rho V^2 S)\). Call \(V_{md}\) the speed at which \(K\CL^2 = \CDz\), where the drag is least and \(L/D = \LDmax\), and measure speed by \(u = V/V_{md}\). The zero-lift drag grows with \(V^2\) and the induced drag falls with \(1/V^2\), and at \(u = 1\) each is half the minimum drag:
\[ D_0 = \tfrac12 D_{\min}\,u^2, \qquad D_i = \tfrac12 D_{\min}\,u^{-2} \] \[ \frac{D}{D_{\min}} = \frac{u^2 + u^{-2}}{2}, \qquad \frac{L/D}{\LDmax} = \frac{2}{u^2 + u^{-2}} \]Range and endurance of a segment
The engines burn fuel weight at \(C\) times the thrust, and in level flight the thrust equals the drag, \(W/(L/D)\):
\[ \frac{\dd W}{\dd t} = -\frac{C\,W}{L/D} \]Divide by \(W\) and integrate over the segment, holding \(V\), \(C\) and \(L/D\) constant as Breguet does. Time gives the endurance, and distance, \(\dd R = V\,\dd t\), the range:
\[ E = \frac{1}{C}\,\frac{L}{D}\,\ln\frac{W_{i-1}}{W_i} \] \[ R = \frac{V}{C}\,\frac{L}{D}\,\ln\frac{W_{i-1}}{W_i} \]For a given fuel load the logarithm is fixed. The best endurance maximizes \((1/C)(L/D)\), and the best range maximizes the range factor \((V/C)(L/D)\) of Module 2.
Jets: \(C\) about constant
Endurance. Maximize \(L/D\) itself: \(u = 1\), at \(\LDmax\).
Range. Maximize \(V(L/D)\). In terms of \(u\),
\[ V\,\frac{L}{D} \;\propto\; u\cdot\frac{2}{u^2 + u^{-2}} = \frac{2}{u + u^{-3}} \]and the denominator is least where
\[ \frac{\dd}{\dd u}\left(u + u^{-3}\right) = 1 - 3u^{-4} = 0 \] \[ u = 3^{1/4} = 1.316, \qquad \frac{L/D}{\LDmax} = \frac{2}{\sqrt3 + 1/\sqrt3} = \frac{\sqrt3}{2} = 0.866 \]At that speed \(\CL\) is \(1/\sqrt3\) of its value at \(\LDmax\), so the induced drag is a third of the zero-lift drag.
Propellers: \(C\) proportional to \(V\)
With \(C = c_P\,g\,V/(1000\,\eta_p)\) from the propeller section above, \(V/C\) does not depend on speed.
Range. \((V/C)(L/D)\) is proportional to \(L/D\): \(u = 1\), at \(\LDmax\).
Endurance. With \(C\) proportional to \(u\),
\[ \frac{1}{C}\,\frac{L}{D} \;\propto\; \frac{L/D}{u} = \frac{2}{u^3 + u^{-1}} \]which is greatest where the power required, \(DV\), is least:
\[ \frac{\dd}{\dd u}\left(u^3 + u^{-1}\right) = 3u^2 - u^{-2} = 0 \] \[ u = 3^{-1/4} = 0.760, \qquad \frac{L/D}{\LDmax} = \frac{2}{1/\sqrt3 + \sqrt3} = 0.866 \]Now the induced drag is three times the zero-lift drag. The two \(0.866\) points sit on either side of the peak of the \(L/D\) curve, a factor of \(\sqrt3\) apart in speed.
| Best | Maximizes | \(D_i/D_0\) | \(V/V_{md}\) | \(L/D\) |
|---|---|---|---|---|
| Jet range | \(\CL^{1/2}/\CD\) | 1/3 | 1.316 | \(0.866\,\LDmax\) |
| Jet endurance, propeller range | \(\CL/\CD\) | 1 | 1 | \(\LDmax\) |
| Propeller endurance | \(\CL^{3/2}/\CD\) | 3 | 0.760 | \(0.866\,\LDmax\) |
The "Maximizes" column is how performance texts state the same results. At a fixed weight \(V\) is proportional to \(\CL^{-1/2}\), so \(V(L/D)\) is proportional to \(\CL^{1/2}/\CD\) and \((L/D)/V\) to \(\CL^{3/2}/\CD\).
The range ratio
How much does flying at another speed cost? The range ratio is the range flown on the same fuel at speed \(u\) divided by the best range, \(R/R_{\text{best}}\). The logarithm is the same for both, so it is simply the ratio of the range factors. The endurance ratio \(E/E_{\text{best}}\) is defined the same way. (Do not confuse either with the weight ratio \(W_{i-1}/W_i\) inside the logarithm.) From the expressions above:
\[ \text{Jet:}\quad \frac{R}{R_{\text{best}}} = \frac{4}{3^{3/4}\,(u + u^{-3})}, \qquad \frac{E}{E_{\text{best}}} = \frac{2}{u^2 + u^{-2}} \] \[ \text{Propeller:}\quad \frac{R}{R_{\text{best}}} = \frac{2}{u^2 + u^{-2}}, \qquad \frac{E}{E_{\text{best}}} = \frac{4}{3^{3/4}\,(u^3 + u^{-1})} \]These are the percentages Figure 4.2 shows as you move its speed. The optimum is flat. A jet flying 10% faster than its best-range speed keeps 98.7% of its best range, and 10% slower, 98.2%. Airlines use that flatness: their "long-range cruise" speed, about 9% faster than the best-range speed, keeps 99% of the best range and saves time. Flying far from the optimum is another matter: a jet cruising at \(\LDmax\) keeps only 87.7% of its best range.
The model holds \(C\) constant and ignores compressibility. A jet airliner's real best-range speed is also limited by the rise in drag near its drag-divergence Mach number (Module 9).
The same argument is clearer on a graph. Plot the engine's fuel flow (kg per hour) against speed, as in Figure 4.2:
- Endurance, hours per kilogram of fuel, is greatest where the fuel flow is least: the bottom of the curve.
- Range, kilometres per kilogram of fuel, is speed divided by fuel flow. A line from the origin to a point on the curve has slope fuel flow over speed, so range is greatest where that line is shallowest: where it just touches the curve.
A jet's fuel flow follows its thrust, which in steady level flight equals the drag. A propeller aircraft's follows its power, which is drag times speed. Multiplying by \(V\) tilts the curve up on the right, and both points move to a slower speed by the same factor, \(3^{1/4} = 1.32\). With the parabolic drag polar, the speed for least drag \(V_{md}\) is where \(L/D = \LDmax\); the jet's best range is at \(1.32\,V_{md}\), and the propeller's best endurance at \(V_{md}/1.32 = 0.76\,V_{md}\), both at \(L/D = 0.866\,\LDmax\).
(a) Jet: fuel flow follows thrust, \(T = D\)
(b) Propeller: fuel flow follows power, \(P = DV\)
Does this apply to turboprops?
Yes, to a first approximation. A turboprop is a gas turbine driving a propeller, and its fuel flow follows shaft power, so for sizing it is a propeller aircraft: cruise at \(\LDmax\), loiter at \(0.866\,\LDmax\). Several things blur the picture in practice:
- Part-power penalty. Engines burn more fuel per unit power at the low power of a loiter, which is why the loiter values of \(c_P\) in the table above are higher than the cruise values. A gas turbine suffers more than a piston engine, because its efficiency depends strongly on running near its design power, so a turboprop gains less from slowing to the minimum-power speed than the simple rule suggests. Maritime patrol turboprops such as the P-3 Orion often shut down an engine on station so that the others run at a more efficient power setting.
- A little jet thrust. The exhaust gives a few percent of the thrust directly (manufacturers quote it as an equivalent shaft power), a small jet-like part on top of the propeller.
- Propeller efficiency varies. \(\eta_p\) falls at low speed and above about Mach 0.6, so the true optimum shifts toward the speeds where the propeller is most efficient.
- Time costs money. Turboprop airliners usually cruise well above their best-range speed, near maximum cruise power, because crew, maintenance and the number of trips a day are paid for by the hour, and the turbine is most efficient near its rated power. Jets do the same, less dramatically: airlines fly a "long-range cruise" a little faster than the best-range speed for 99% of the best range.
For a first sizing, use the typical values in the tables above. If the requirement fixes the cruise speed, the \(L/D\) at that speed is what belongs in the Breguet equation; Module 9 shows how to find it.
Example 4.2 — An hour on station
A piston patrol aircraft with \(\LDmax = 12\) loiters for one hour at \(180\ \text{km/h}\), with \(c_P = 0.5\ \text{lb/(hp h)}\) and \(\eta_p = 0.8\). Find the loiter weight fraction.
Show solution
Under 2% of the weight for an hour aloft: slow propeller aircraft are very good at loitering, which is why maritime patrol aircraft have propellers.
Check your understanding
Key takeaways
- Jets: \(C\) in 1/h, about 0.9 (turbojet), 0.8 (low-bypass) and 0.5 (high-bypass) in cruise; \({C = c_T[\text{mg/(N s)}]/28.33}\).
- Propellers: \(C = c_PgV/(1000\,\eta_p)\) with \(c_P\) in kg/(kW h); \(V/C = 3600\,\eta_p/(c_Pg)\) km.
- Jets: cruise \(0.866\,\LDmax\), loiter \(\LDmax\). Propellers: cruise \(\LDmax\), loiter \(0.866\,\LDmax\).
- Next, Lesson 5 combines \(L/D\) and \(C\) into the mission's weight fractions.