# 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

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.

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