# Why do propellers have less theoretical range (i.e. range ignoring impatient passengers wanting to go fast) than jets, despite greater efficiency?

I was playing around with the breguet range equation recently, and came across a suprising result, that propellers have less range than jets, despite having greater efficiency.

The range equation for a prop is:

R = 550 * np * ((L/D)/(C / 3600)) * ln(Wi / Wi+1)

where C is the SFC, in lb/(hp * s), and R is the range, in feet. Note that the 550 here coincidentally is the same as the speed in the next calculation, but the 550 here is actually a conversion factor from hp to lb-f/s

According to wikipedia, the most efficient turboprop engine is the NK-12, used on the Tu-95 bomber (and Tu-114 airliner), and that engine gets a SFC of 0.36 lb/(hp*h). Assume a propeller efficiency of 0.8, which is typical.

R = 550 * 0.8 * ((L/D)/(0.36 / 3600)) * ln(Wi / Wi+1)

R = 550 * 0.8 * ((L/D)/0.0001) * ln(Wi / Wi+1)

R = 550 * 0.8 * ((L/D)/0.0001) * ln(Wi / Wi+1)

R = 4400000 * (L/D) * ln(Wi / Wi+1)

For jets:

R = (V / (C/3600)) * (L/D) * ln(Wi / Wi+1)

a SFC of 0.5 lb/(lbf*s) in cruise is considered pretty good for jets, and let's assume a 550 mph cruise speed (or 806.667 ft/s). Range is now:

R = (806.667 / (0.5/3600)) * (L/D) * ln(Wi / Wi+51)

R = 5808002.4 * (L/D) * ln(Wi / Wi+1)

The range expressions for jets and props is identical, except for the coefficient, and 5808002.4 is greater than 4400000, so therefore, jets have better range, even on paper, than props, assuming same L/D ratio and fuel ratio. This is kinda suprising, given that props are quoted to have higher efficiency than jets. I've heard people say props have less range, but I've always heard it backed up with the evidence that "props are slow, and people like to go fast". Are my calculations and data correct? If they're correct, why, intuitively, do props end up having less range?

Clarification: user3528438 has pointed out that some turboprop engines provide both a power specific fuel consumption number, as well as thrust specific. However, if you plug the thrust specific fuel consumption number for that turboprop into the jet range equation (essentially treating the prop as a jet), the resulting range is very different than the range you'll get if you plug the power specific fuel consumption number into the propeller equation -- so which one is right? I doubt my formula for propeller range is wrong (though I might be using it wrong - could someone please check) because the same formula (but without the conversion factor, because they're using metric) is found here - https://nptel.ac.in/content/storage2/courses/101104007/Module2/Lec8.pdf.

I also have my derivation of the formula on hand too, if anyone wants to check that.

• Does this answer your question? Why don't airliners have turboprop engines instead of jet engines? – Manu H Apr 27 at 20:40
• @ManuH Pretty sure it won't - 1) the question here is very specific, and 2) the old question was quite different, even if the one it was a dupe of can be useful. – Therac Apr 27 at 20:52
• Yet the longest range aircraft, the Rutan Voyager, had propellors: en.wikipedia.org/wiki/Rutan_Voyager – jamesqf Apr 28 at 4:46
• This isn't an internet forum; changing your question based on an answer breaks the principle methodology of the SE model. If you want a threaded discussion, go to a forum. If you find the question has problems, it is better to write a new, better, question. Tour, How to Ask and How to Answer provide more guidance. – KorvinStarmast Apr 28 at 13:23

Update As @Zeus has pointed out correctly: Drag coefficient stay constant in my scenario, not drag -- I've updated the explanation accordingly. Sorry for that blunder. The conclusion does not change: Engines with constant PSFC only have an advantage when flying slowly and get worse much quicker than engines with constant TSFC -- but actual engines are mostly in-between these days.

As others have pointed out already, the important thing to note is that propellers have (roughly) constant PSFC (specific fuel consumption per power output), and jets (roughly) constant TSFC (specific fuel consumption per unit of thrust).

(Second: quoting equations which include unit conversion factors is dangerous, and doing so without specifying which units are being used is doubly so.)

Power is thrust times velocity, which means that although a propeller can have a much higher propulsive efficiency (propulsive power vs shaft power), it also scales much differently with speed, even ignoring the fact that propellers have difficulty operating above Mach 0.6:

Assume we have a series of aircraft, each designed to fly at a particular speed (all well below the speed of sound), with the same glide ratio and the same weight. This means they all fly at the the same drag coefficient, and the required thrust scales with the square of speed. Now we decide which kind of engine to use:

Let's design a family of jet engines, each of which delivers a certain thrust, and all of which have the same, constant TSFC. This means 1 Newton of thrust costs the same amount of fuel per second, independent of how fast you're going. The fuel flow would of course still increase with the square of velocity because you would need more thrust the faster you're going. And that means that fuel burn per distance travelled would increase proportional to speed. Most jet airliners are actually flying a little faster than their best glide ratio (thus lowest drag coefficient) would suggest because for an airline, time is money. That's why they're burning a bit of extra fuel in order to arrive faster and do more flights with fewer aircraft.

Using propellers with reciprocating engines, though, although it's converting more of the shaft power to thrust, the fuel flow through the engine scales with power, not thrust, and power needed for constant thrust scales with velocity. So Doubling speed at equal thrust requires twice as much fuel per second, but since thrust increasing with the square of velocity, fuel flow actually scales with the cube of velocity. So fuel consumption per distance is scaling with the square of velocity. That's why the most efficient propeller aircraft would be flying fairly slow, with optimal speed being a trade-off between engine efficiency and the speed at which the wing can comfortable generate enough lift. Going faster gets expensive quick.

So however efficient your propeller is at some (low) speed, as you get to faster and faster aircraft, it will eventually be worse off than a jet because its fuel per distance quadruples when doubling speed, but the jet's only doubles. At whichever velocity our fictional jet engine and piston/propeller get equal fuel economy, accelerating just bit with the propeller aircraft will be twice as expensive as with the jet. That's also why piston engines with propellers are the weapon of choice for long-endurance flights, where distance covered is less important than time afloat, or where costs are more important than speed.

However, turboprops do not have constant PSFC, since they have a turbojet core providing the shaft power, which benefits from the increased pressure of air in the intake at higher velocities. And modern high-bypass turbofan engines don't have constant TSFC, either, since they have more losses in the bypass duct at higher velocities at constant thrust. In fact, an extremely high-BPR turbofan starts to approach the characteristics of a comparably low-BPR turboprop (except at large Mach numbers, but we're still ignoring that here).

Conclusion

It's dangerous to confuse TSFC and PSFC. An engine with constant PSFC (old-fashioned piston-driven propeller) might fly a lot more efficiently at low speeds but gets worse quicker than engines with constant TSFC.

Because nobody (except for this guy -- which is amazing) wants to take days to finish their intercontinental flight, and because propellers don't work so well at higher Mach numbers, jet engines (that is: Turbofans) are dominating commercial long-range flights. For high-efficiency flying over long distances, where speed is not that important, however, Propellers (mostly in the form of turboprops) are still popular. See for example auxiliary engines for sailplanes, or most military transport aircraft. For the latter, range is everything, speed (and noise...) are secondary. However, since those are built to military specifications, they're not that useful as civil cargo aircraft, and since the civil cargo aircraft market is very small, most of it is covered by converted passenger aircraft.

Aside Another issue with piston engines is that their performance reduces the higher the fly, as the air thins. This issue can be reduced by adding a turbocharger, but a jet engine already is essentially a very big turbocharger with a turbine attached. The other advantage of jet engines is their power-to-weight ratio. This is why e.g. most helicopters don't use piston engines and have turbines instead.

• "Doubling speed at equal thrust requires twice as much fuel per second." Formally correct in the context, but how do you do that? By slashing the drag coefficient 4 times? – Zeus Aug 5 at 0:57
• @Zeus: Yikes! You're right! c_D stays constant but drag increases of course ... Will correct the scenario asap – Zak Aug 6 at 9:05

For more background, this question is about the usage of Breguet Range Equation (I find this explanation better suited for this question)

There are many problems with your calculation:

1. Time unit of speed should be the same as the time unit of SFC. For your first calculation, since you used mph for speed and lb/(hp*h) for SFC, your SFC term shouldn't have 1/3600. For your second calculation, since your speed is fps and your SFC is also in lb/(lbf*s), there's no need to have 1/3600 either.

2. For both calculations, end result should be the same unit of length. Your first calculation uses mph and the second uses fps, so the the end result would be in miles and feet so not comparable. But since you made the mistake #1 you end with numbers still on the same order of magnitude.

3. For the propeller case you are using PSFC (Power specific fuel consumption) rather than TSFC (Thrust-specific fuel consumption). Breguet Range Equation requires TSFC to work. If you look at the Breguet Range Equation you see the only difference between prop case and jet case is TSFC. So to compare relative range differences, you only need to compare TSFC and avoid all other complications.

Take Cruise TSFC from some modern examples:

(1/0.38)/(1/0.545) - 1 = 43% so with everything else being the same, prop should have about 40% more range than jet.

• I believe you're misinterpreting the 550 in the 1st calculation, the 550 is a conversion factor from horsepower (which is never used in calculations) to ft-lb/s, which is the one used (imperial is weird in that lots of units are used in data, but not used in calculations, e.g. mass is written in lbs, but in calculation, you use slugs, power is written in hp, but in calculation, you use ft-lb/s). Sorry for picking 550 mph as the speed, that was confusing, my bad. – Anonymous Person Apr 27 at 22:11
• I didn't see that some turboprops also have TSFC data, I guess I'll plug those into the range equation, and the results you get make more intuitive sense (props SHOULD have more theoretical range). I just redid my derivation of the range equation for propellers and the results still don't match yours, but I guess finding the error in that derivation is another can of worms, more fit for the physics stackexchange. – Anonymous Person Apr 27 at 22:31

I think there are some confusion here, in particular, as @user3528438 said, with units used in formulas. I'll also try to highlight problem with another point of view.

1. Formulas comparison

First of all, be really careful when using the first formula. At first, I thought it was not consistent regarding range R units. The way you make calculation formulates the range on « ft2.s-2 ». After taking a look on the demonstration of the paper you shared, I realised that the SFC is here defined as « Newton of fuel » / « Power x time ». It seems it does not provide any error here cause your SFC uses lbs and it is equivalent to pound force lbf.

Then, it is important to notice that both formulas say exactly the same thing ! There is not a « jet range equation » or method « treating a turboprop as a jet » :

• One consider Power SFC whereas the second use Thrust SFC (Pwr = T.V and so PSFC = TSFC/V).
• An other difference is the efficiency in the first equation. Not sure about it, but it seems to comes from the SFC definition again. In the paper, the author define the power as the power used by « reciprocating engine » which mean available input power. It could be electrical power or chemical power in burner. That's why they need to include the efficiency.
• Note that, it is not just the propeller efficiency but the overall system efficiency (including thermal and propulsion efficiency for a gas turbine) as it is defined like « input power » / « aircraft power ». For me, it is more convenient to include it in SFC data (at least, I guess it is in data we can find online).

2. SFC definition

Here we have 2 definition TSFC and PSFC, introduced in the previous section, both sometimes called Specific Fuel Consumption:

• TSFC: 1 lbs/(lbf*h) of SFC means the engine will consume 1 lbs/h fuel for each pound force of thrust provided.

• PSFC: 1 lbs/(hp*h) of SFC means the engine will consume 1 lbs/h fuel for each horsepower (or 550 lbf.ft/s) provided.

According to wikipedia, a power is the rate of a system doing a work or providing energy by units of time. In our problem, it's a measurement of how fast a mechanical work is given to the aircraft by the engine (e.g. T x V).

1. In your specific case, you compare 2 engines: the first one is a turboprop with a SFC specified in power definition = (c/eta) = 0.36/0.8 = 0.45 lbs / (hp * h). The second is a turbofan with 0.5 SFC in thrust used during a steady flight of 806.667 ft/s speed. So we can calculate its PSFC = 0.5 * 550/806 = 0.34 lbs / (hp * h). The quotient 0.45/0.34 is what you got (e.g. 5808002.4/4400000). Several things to note:
• data used for turboprop SFC does not match with first formula definition (at least, I'm pretty sure that eta is already include in the value 0.36 lbs / hp*h.

• your turbofan as a better PSFC and provide more power with same fuel or need less fuel to provide same power that's why the range is bigger.

1. Remember that the data you find SFC = 0.36 lbs/(hp * h) is equivalent to a SFC = 0.36 lbs / (lbf * h) for a flight speed of 550 ft/s. Now, let's try to put the turboprop in same conditions than the turbofan (e.g. 806 ft/s).
• if we keep TSFC = cst: we can compare range directly and we got about 40% more range for the turboprop. Is what's @user3528438 made in his answer.

• if we keep PSFC = cst: we can calculated the TSFC at this specific flight point (806 * 0.36 / 550 = 0.527 lbs / (lbf * h)). Here, the turboprop provide a slight increase of range.

So which case is the true one ? Probably none of these.. Intuitively, I want to say a turbofan has more chance to keep TSFC constant when changing flight condition while is PSFC for a turboprop. Indeed, you can try to reduced little by little the speed condition and find the limit where it's preferable to have a turboprop rather than a turbofan. But it is huge assumptions here so I'm not confident with results. Nevertheless, it could be interesting to compare it with flight speed of most aircraft equipped with turboprop.

However, this provide a first explanation why turboprop is usually operating at low speed. The other reason is a propeller need to have large diameter to be really effective and give enough thrust. Therefore we got an aircraft speed limit at the point where the flow become supersonic at blades tip. You can still reduce the rotational speed but it could be difficult to reduce it even with gearbox without affecting overall gas turbine efficiency.

Conclusion

In conclusion, SFC is an important parameter of engine performance. But it is good to keep in mind that it is calculated for specific design points (e.g. flight conditions) thus dependent of the aircraft mission. SFC is specially used for comparing 2 engines designed for same aircraft.