In what way are the Concorde's engines considered efficient?

Question

The Concorde is powered by a bunch of Rolls-Royce/Snecma Olympus 593 turbojet engines. Much has been said about how efficient they are at speed, and the fact that they allow long-range cruise at Mach 2 because of how fuel-efficient they are.

If I look at the specifications, one thing immediately obvious is the compression ratio of 15.5 - quite high for a no-bypass turbojet. I'm not sure how this translates into efficiency though.

The thrust-specific fuel consumption, on the other hand, looks dismal. Wikipedia states:

1.195 lb/(lbf·h) (33.8 g/(kN·s)) cruise / 1.39 lb/(lbf·h) (39 g/(kN·s)) sl

Wait, what? The Tumansky R-25, which powered the MiG-21 and was famously fuel-hungry, had a compression ratio of only 9.5, but had a TSFC of

98 kg/(h·kN) (0.96 lb/(h·lbf)) at maximum military power

The General Electric YJ93, which powered the XB-70 and was also designed for long-range supersonic efficiency, had a TSFC of

0.700 lb/(lbf·h) or 19.8 g/(kN·s)

This doesn't seem to make sense: in what way are Concorde's engines any good? Is there something I'm missing?

Finally, I've been attempting to model the Olympus 593 in a flight simulator (Advanced Jet Engine in KSP). With the given compression ratio, though, I couldn't get the fuel efficiency to be this bad: it was around 0.9 SL and 0.85 cruise, and I had to do ridiculous things like using extremely inefficient intakes and nozzles.

Not an engineer,cant comment on math here. However concorde can travel twice faster than other airliners, so that maybe where efficiency comes from. — vasin1987, May 02 '15 at 17:52
@vasin1987 I'm comparing against military aircraft that go similarly fast, not against subsonic high-bypass engines. — ithisa, May 02 '15 at 22:32
@user54609: No fighter from 1960s is capable of sustaining supersonic speed without afterburner. And I don't think any engine had so low TSFC at Mach 2 and with afterburner. — Jan Hudec, May 04 '15 at 05:15
You are missing the speed at which the SFC is given. Comparing static values against data at flight speed is totally misleading. I don't know why @RedGrittyBrick deleted his answer, but it is the best so far. And it is true, the Olympus 593 was indeed the most efficient engine of its time and still holds up well against most of the competition. — Peter Kämpf, May 04 '15 at 16:42
Ram air in supersonic flight did improve the efficiency quite a bit, but I have no idea about these numbers. Concorde's efficiency was comically poor when it was on the ground. IIRC those beautiful 593's drank 2 tonnes of fuel for taxiing to the runway. BTW, have you been able to run the numbers on the Tu-144 KN-321 engines? I guess it would be hard because Russia, but I think it would be the best comparison. — shortstheory, May 04 '15 at 17:00

score 23 · Answer 1 · edited Apr 13 '17 at 12:59

You compare SFCs at different speeds. That is like comparing payloads for differently sized aircraft. SFC goes up with speed and, therefore, must be compared at the same speed. The work performed by an engine is thrust times distance, and higher speed means that the same thrust will perform more work per unit of time when the engine moves faster. The moving engine needs to slow down the airflow for combustion to take place, and then needs to accelerate the air by more than it has been slowed down to have positive thrust. Hence, SFC goes up in parallel with speed.

To have a meaningful comparison, we need to define efficiency. There are several, and two are of major importance for air-breathing aircraft engines: Thermal efficiency and propulsive efficiency.

Thermal efficiency

This describes how efficiently the chemical energy in the fuel $Q$ is converted into an impulse change of the air flowing through the engine. Formulated using the mass flow per unit of time $\dot{m}$, the impulse is $\dot{m}\cdot\dfrac{\Delta v^2}{2}$. Using $v_{\infty}$ for the incoming air speed and $v_{\infty} + \Delta v$ for the exit flow speed, the thermal efficiency is $$\eta_{therm} = \frac{\dot{m}\cdot \left((v_{\infty} + \Delta v)^2 - v_{\infty}^2\right)}{2\cdot Q}$$ To achieve good efficiency at high speed, a high $\Delta v$ is helpful. This explains why efficiency drops more over speed for high-bypass ratio engines and especially propellers. Since the thermal energy in fuel is the same for all engines in your question, because all run on kerosene, and we can assume a similar efficiency of combustion, we can neglect $Q$ in the comparison.

Propulsive efficiency

This describes how well the conversion is performed. Using the same variables as above, propulsive efficiency is $$\eta_{prop} = \frac{v_{\infty}}{v_{\infty} + \frac{\Delta v}{2}}$$

This equation explains the better efficiency of high-bypass ratio engines and propellers at the same speed, because propulsive efficiency is proportional to the inverse of $\Delta v$.

Overall efficiency

This is the product of thermal and propulsive efficiency, and the equation is $$\eta_{total} = \frac{T\cdot v_{\infty}}{Q}$$ where $T = \dot{m}\cdot\Delta v$ denotes the thrust. Conveniently, $\Delta v$ is eliminated in the product, allowing turbojet engines like the Olympus 593 to look much better in comparison to other engines.

Intake efficiency

This answer would be incomplete without a look at the intake of the Concorde. At cruise, it would lift the pressure of the air at the compressor face by a factor of more than six over ambient by efficiently decelerating the flow. The compressor added a compression ratio of 12, so the pressure in the combustion chamber was 80 times higher than ambient. This high pressure makes the engine so efficient, but is also needed to maintain combustion. Remember, ambient pressure in 18 km is just 76 mbar, so the absolute pressure in the combustion chamber at cruise was only 6 bar.

The full answer would be like this: The combination of intake and Olympus 593 at Mach 2.02 had a very good total efficiency, and comparisons with other engines at static conditions are misleading.

The comparison of results from a test stand on the ground would yield a very different picture, however.

Does $\dot{m}$ include the fuel mass being injected into the combustion chamber? — DrZ214, Sep 04 '17 at 05:36
@DrZ214: Normally it is neglected. To be more precise, you can add the fuel mass flow and give it zero as its initial velocity. The difference in thrust, however, is small. — Peter Kämpf, Sep 04 '17 at 07:04
I come from a rocket background so I'm not comfortable neglecting the fuel flow :-) I'm also skeptical that it will have a small change, because dodecane (C12H26) is heavy and O2 and N2 are light. But then again, I would not be surprised if the F-O ratio is less than stoichiometric for temperature reasons. I think I'll post a question on that. — DrZ214, Sep 04 '17 at 08:02
@DrZ214 indeed, jet engines run way lean of stoichiometric (even before including the bypass air) to avoid melting the turbine blades, and to take advantage of unlimited free reaction mass (i.e. the atmosphere). — pericynthion, Oct 29 '17 at 06:32

score 10 · Answer 2 · edited Jun 17 '20 at 08:28

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Wikipedia article on thrust specific fuel consumption actually uses Concorde as example, probably because it was such an extreme case. I probably should edit to make a this a real answer,but since they use your specific question as example I will just quote.

SFC varies with throttle setting, altitude and climate. For jet engines, flight speed also has a significant effect upon SFC; SFC is roughly proportional to air speed (actually exhaust velocity), but speed along the ground is also proportional to air speed. Since work done is force times distance, mechanical power is force times speed. Thus, although the nominal SFC is a useful measure of fuel efficiency, it should be divided by speed to get a way to compare engines that fly at different speeds.

For example, Concorde cruised at Mach 2.05 with its engines giving an SFC of 1.195 lb/(lbf·h) (see below); this is equivalent to an SFC of 0.51 lb/(lbf·h) for an aircraft flying at Mach 0.85, which would be better than even modern engines; it was the world's most efficient jet engine.[2][3] However, Concorde ultimately has a heavier airframe, and due to being supersonic is less aerodynamically efficient, i.e., the lift to drag ratio is far lower. In general the total fuel burn of a complete aircraft is of far more importance to the customer.

My personal interpretation as a total layperson of this is that the original air speed target was higher than the air speed of Concorde because the issues of supersonic flight were underestimated. Because of those issues the actual Concorde was built for only Mach 2 or so. The engine design was still influenced by that original airspeed target (whatever that was) and as a result had higher exhaust velocity than actually was necessary. This resulted in higher fuel consumption and noise. The reduced range and increased noise in turn limited the routes Concorde could fly and the areas where the supersonic flight could be used. Which made Concorde commercially a "limited" success and made upgrading the engines to ones optimized for the actual speed commercially impractical.

Note that the above is my speculation of the background. The important part is that the exhaust speed of the engines is faster than necessary for the Concorde. This means that despite good thermal and energy efficiency, thrust efficiency is lower than necessary

So yes, the engines were uneconomical and suffered from excessive noise and fuel consumption, but that was due to the airframe and engine being optimal for different speeds. The engines were quite efficient, best thermal efficiently achieved at the time, they just were optimized for the wrong speed that in practice was not achieved.

edited Jun 17 '20 at 08:28

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answered May 02 '15 at 23:28

Ville Niemi

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I was comparing with other Mach 2 engines, though, not with low-speed engines. To be fair though, none of the planes using those engines were able to reach Mach 2 without afterburner, which would significantly increase fuel consumption. – ithisa May 03 '15 at 04:37
@user54609 Yes, it does. I think the big part here is that Concorde had high exhaust velocity and once you take that into account it was very efficient. A proposed upgrade would have had lower exhaust velocity resulting in less noise and fuel consumption, but since super sonic travel was not a commercial success it wasn't built. – Ville Niemi May 03 '15 at 07:54
"Yes it does"? What does what? Lol – ithisa May 03 '15 at 20:50
@user54609 Afterburner, increase fuel consumption significantly. Straight follow up on the comment being replied to. – Ville Niemi May 03 '15 at 21:50
@user54609 Edited in my interpretation what the quote and data means to the answer below the quote. Not sure if it helps, but at least it looks more like a proper answer now! – Ville Niemi May 04 '15 at 11:46
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The engines were initially designed for Mach 2.2 and redesigned for Concorde, and are stated as getting the 43% efficiency in actual cruise. Concorde was designed to be a Mach 2 turbojet aircraft from the start. – fooot May 06 '15 at 02:53
@fooot That is debatable. There is lots of stuff about the issues of supersonic drag being worse than people expected, so I think the actual design process was more complex than is accounted by that. Still you are right that the way I worded that makes it look Concorde is a failed Mach 3 aircraft while what I actually meant was that relative lack of experience with supersonic flight complicated the development process and left some parts badly optimized. I guess I'll reword the answer to focus on that instead of my speculation on the original goals of the project. Thanks. – Ville Niemi May 06 '15 at 09:47
@fooot Or at least add a note that I don't mean that the actual Concorde was designed for higher speeds... – Ville Niemi May 06 '15 at 09:48
@fooot Made the edit, hopefully less misleading. – Ville Niemi May 06 '15 at 09:58
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The comparison between different speeds is due to the fact that the SFC figures are per hour, not per kilometer. Concorde @ Mach 2.2 simply flew 2.5 times further per hour than a subsonic plane at Mach 0.85. Thus the "equivalent" SFC figure of 0.5 is just dividing the SFC of 1.195 by the speed ratio's. That's the root cause of the engine efficiency: the faster Concorde covered the same routes in less time. The wikipedia comment correctly notes that the engines are efficient at producing thrust, but the heavy airframe requires a lot of lift and the supersonic wings have a poor lift/drag ratio. – MSalters Aug 18 '15 at 09:38

score 7 · Answer 3 · edited Jun 17 '20 at 08:28

In what way are the Concorde's engines considered efficient?

TSFC/Speed

The Wikipedia article, "Thrust specific fuel consumption", referred to in the question, says

although the nominal SFC is a useful measure of fuel efficiency, it should be divided by speed to get a way to compare engines that fly at different speeds.

For example, Concorde cruised at Mach 2.05 with its engines giving an SFC of 1.195 lb/(lbf·h) (see below); this is equivalent to an SFC of 0.51 lb/(lbf·h) for an aircraft flying at Mach 0.85, which would be better than even modern engines; it was the world's most efficient jet engine.[2][3]

I think what they are saying is perhaps that the amount of thermodynamic work being produced per unit quantity of fuel was high. Concorde was cruising at Mach 2 and had a range of 7000 km. There probably aren't many aircraft around that need to do that. It's engines were producing a lot more work than a typical high-bypass turbofan attached to a widebody jet at Mach 0.85 does.

Thermal Efficiency

They are considered to have high "thermal efficiency" at Mach 2.

They are considered inefficient at lower speeds.

The Rolls-Royce Olympus 593 Mk 610 installed in Concorde STILL remains the most efficient jet engine in the world at Mach 2, as far as thermal efficiency is concerned. This is due to the design of the engine itself of course, but mainly down to the intake, and to a lesser extent the individual nozzle designs. ... (As efficient as the OLY 593 is at Mach 2 and about, at slower speeds it uses fuel like it’s going out of fashion, hence the need for a minimum of low speed flying with Concorde).

From a bunch of Concorde enthusiasts.

Specific Impulse

Another way to measure engine efficiency is specific impulse

enter image description here
^{Graph by Kashkhan}

XB-70 is Mach 3, the question gets a further argument actually — Trebia Project., May 02 '15 at 19:36
@Trebia: I can't find sources for the YJ93 TSFC in an XB-70 at Mach 3. Wikipedia says the the XB-70 had structural failures at mach 3 and was then limited to Mach 2.5. It also seems to have had problems reaching it's intended range. It's puzzling. Also I can't find if the XB-70 needed afterburners in cruise like the SR-71 did (YJ93 unsurprisingly has 1.8 TSFC with afterburners) — RedGrittyBrick, May 02 '15 at 19:53
in order YJ93 to have the same efficiency compared with Olympus 593, that fuel consumption should be set at Mach 1,37 in XB-70. Very likely is higher, but without the actual proper reference, anything is especulation. — Trebia Project., May 02 '15 at 19:59

score 4 · Answer 4 · answered Dec 29 '15 at 15:19

Efficiency is an energy thing.

Energy is force times distance.

Differentiate, you get power is force times speed. It turns out you don't even need to know the cruise thrust, you can calculate the efficiency on the TSFC (which is pretty much why it is used).

Although Concorde's engine were generating less force per unit flow, the vehicle was going 2.5 times further each second. If you divide the useful power (thrust times speed) into the power in the fuel (fuel flow times energy per unit mass of fuel) you can calculate the energy efficiency of Concorde's engines.

So let's do that. Here's the basic numbers I found on the web, and I convert them into SI base units:

Cruise speed = 2,124 km/h = 590m/s TSFC @ cruise = 33.8 g/(kN·s) = 33.8e-6 kg/N.s Specific energy of aviation fuel = 43.15 MJ/kg

Energy per N m/s = power per Newton = force times speed per Newton = 590 m/s/Newton = 590 Joules per second per Newton Fuel energy used by engine per N s = 43.15e6 MJ/kg x 33e-6 kg/N.s = 1458 joules per second per Newton

So dividing one by the other, Newtons cancel and we get 40% efficiency.

Bearing in mind that's in an aircraft, which is made as light as possible, for any aircraft, that's amazingly good; better than most electric generating power plants which are bolted to the ground, but some diesel engines can reach over 50% in very, very large ships, and you might exceed that in cars with cast-iron engines which would be far too heavy to be flown.

Let's take a 747-400 at long range cruise with a CF6 engine:

Cruise speed 907km/h = 251 m/s TSFC 17.1 g/(kN·s) = 17e-6 kg/N.s So it's generating 251 J/s.N and burning 17e-6 x 43.15e6J/kg = 733 J/s

I make that 34% engine efficiency.

That's it beating a high bypass turbofan that's used for long distance subsonic cruise. So it's not just that these engines were good for the time; they're still world-class.

score 2 · Answer 5 · answered May 04 '15 at 16:21

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I think you are seeing a difference here in where the SFC is being reported. The engines on the Concorde are optimized for supersonic cruise, and SFC is reported there. The fighter is not just optimized for supersonic cruise, so you can't assume the SFC value is for cruise. The YJ93 never saw much use, and the only value I can find for SFC is the one included in the question, but this is probably not at cruise.

Compare the 33.8 g/(kNs) for the Concorde engines to other similar engines, using only values for supersonic cruise. The J-58 (SR-71) at Mach 3.2 was 53.8 g/(kNs) (wet), and the RD-36 (Tu-144) at cruise was 35 g/(kNs).

answered May 04 '15 at 16:21

fooot

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Doesn't SFC usually decrease with speed though? – ithisa May 06 '15 at 02:06
I don't think so. – fooot May 06 '15 at 02:55
SFC decreases with the exhaust velocity of the engine because the energy contained by the fuel stays the same but the kinetic energy of the exhaust increases with the square of its velocity. If the exhaust velocity stays the same, ie. it was higher than needed for the previous speed, increasing air speed improves SFC. But if gaining the higher air speed requires increasing exhaust velocity SFC goes down. (Bit complex to explain in comment, hope the idea, the link between exhaust velocity, achievable air speed, and SFC was clear.) – Ville Niemi May 06 '15 at 10:11
@user54609 Thrust- and Power- Specific Fuel Consumption goes down with speed, but that's not the same as SFC. – habu May 06 '15 at 14:25
What is SFC if not a short form of TSFC? Sorry, am a real noob on planes. – ithisa May 06 '15 at 22:23
Wikipedia: "Thrust specific fuel consumption (TSFC) or sometimes simply specific fuel consumption, SFC, is an engineering term that is used to describe the fuel efficiency of an engine design with respect to thrust output." – ithisa May 06 '15 at 22:24

score -1 · Answer 6 · answered Aug 18 '15 at 11:35

I don't think it was the engines alone that allowed supercruise (supersonic cruise without reheat). On Concorde, the engines engaged reheat in order to accelerate in level flight from subsonic to supersonic. One of the main reasons why Concorde had so much more range than the TU144 (which had to stay in reheat to stay supersonic), is the design of the wing. The Concorde wing evolved into a more complex shape during its development (I think between the first prototype and the production Concordes) specifically for this reason. If you look at the Concorde from the front the wing tips kind of sweep downwards. See http://www.concordesst.com/wing.html for more information. Even the great Kelly Johnson didn't do it this way for the SR-71, which stays in reheat the entire time it is supersonic.