I remember reading awhile back about the Tumansky R-15 which to the best of my knowledge has a supersonic first-stage and a pressure ratio of 4.5 : 1 at sea level.

At altitude and at Mach 2.5, the engine produces 20 tonnes of thrust!

Is the fact that the compressors operate supersonic cause it to profit so much from ram-compression even at very high altitudes?


Andrea Kent

What velocity was it capable of at seal level? I imagine slower and therefore compression is a factor of Mach number, though either I've never heard of "Tumansky R-15" or probably know it by another name.

Gimme a sec to google "Tumansky R-15"...

From a standpoint of complexity, ram compression relies on angled surfaces to reflect oblique shockwaves while a rotating system can reduce the length of the engine. I guess it would depend on the Mach number as to which is better suited.

I'm for less complexity at the possible risk of upping the R&D fund, though.

So a supersonic compressor can be shorter for the same pressure ratio?

Andrea

Quoting Blackbird (Reply 2): So a supersonic compressor can be shorter for the same pressure ratio?

How are we defining "supersonic compressor"? A compressor on an aircraft that goes supersonic or a compressor where the flow through the stages is supersonic? I would have thought the latter would be viciously inefficient due to shock losses.

A turbocompressor is certainly shorter than a ram compressor for the same pressure rise (at reasonable efficiencies) because you get the most efficient ram compression by using lots of oblique shocks, which means a long duct. You can get huge ram compression by just having a normal shock (probably the shortest possible compressor...only a few mean free paths thick) but the total pressure loss would be appalling at higher Mach numbers.

Tom.

Tom,

Actually, I'm talking about a jet engine in which the flow through at least the first stage is fully supersonic (probably low-supersonic) across the blade.

Some old russian engines have such a set-up. The engine that powered the Su-7, Su-9, MiG-21, had such an engine powering them, as did the MiG-25 (different engine, but a supersonic compressor). I don't think the first stage would have had supersonic rotation speed or something, but as the plane went to supersonic speed, the inlet would have slowed the air to a lower mach number, but still a low supersonic one.

I already knew of engines that had supersonic velocity at the tips (as an inherant result of rotational velocity), and I even knew under some circumstances it actually improved performance, but from what I read when the tip velocity exceeded Mach 1.3 to 1.4, efficency fell off. I have also heard of engines in which the airflow over parts of the blades were supercritical (without the leading edge being supersonic) much like subsonic airfoils as well, and I've been told they had good performance as well. But until the past year or two, I didn't even know it was possible to have a propeller produce thrust spinning supersonically (I now know it can be done just as a wing flies supersonic) and until I read about the MiG-21, Su-7 and Su-9, and her engines, I didn't know you could have the first compressor stage or two riding in supersonic flow and still working at all, let alone producing decent thrust.

The other reason I'm so curious about a supersonic compressor is because the MiG-25's engines operated as such and the engine which only has a pressure ratio of 4.75 : 1.0, and a maximum sea-level full-afterburning thrust on the order of around 27,000 lbf (last I checked) could produce approximately 20 tons (40,000 lbf), or 20 metric tons (~44,000 lbf) at Mach 2.5, at altitude. I'm wondering if the inlet guide-vane, and first stage compressor (and perhaps the first stage guide-vane) operating in supersonic flow act sort of inlet-like, while acting compressor-like (after all they are compressors and are spinning) at the same time producing unusual increases in thrust.

While I'm at it, do you have any rule of thumb as to determining the effects of ram-compression (increases in airspeed, mach number) combined with effects of altitude?


Andrea Kent
Should I get a heart-attack, stroke, some kind of incurable disease or cancer, disappear or die mysteriously, or under suspicious circumstances, or should I end up arrested on some kind of trumped-up bogus charge(s), you know who to blame for it.

Quoting Lehpron (Reply 1): What velocity was it capable of at seal level?

That question should perhaps be addressed to Astuteman.  

Probably about 35 knots?

I am with Tom, I cannot see much of the compressors being supersonic. Aside from inefficiency, the blades would have an unhealthy life. Viciously inefficient sounds about it.

Blackbird,

Tom is correct in what he is saying about oblique shock waves.

These shock waves are positioned within an inlet duct by moving the inlet spike (for example a English Electric lightning, SR71 Blackbird, various MIG fighters), or inlet duct "ramp doors" as found on Concorde.

Spikes move forward and backwards, and doors move up and down, which set up multiple shock waves within the inlet. Every time delivery air passes through these shock waves it is slowed to a final velocity of around 0.86 mach which I think is the optimum speed for air to be delivered to a compressor. In the Concorde, there were three shock waves present, which means the air was slowed three times before meeting the first stage of compressor blades.

I am not aware of inlet delivery air being supersonic in conventional gas turbine compressors.

CFM Turbo fan

Supersonic and transonic compressors most certainly do exist. In fact they seem to be commonplace, at least for military engines. Allowing the airflow in the compressor to become transonic allows a higher stage loading and so smaller and lighter compressors.

I found these course notes while googling supersonic compressors.

http://www.adl.gatech.edu/classes/propulsion/prop12.html

This is a bit different from allowing the airflow into the compressor to be supersonic, but when you add in the rotational velocity of the blades it is easy to get local flows above M=1 in the first and subsequent stages.

At high supersonic speed (as with the MiG-25) the compressor can be quite a modest pressure ratio (4.5:1 as mentioned in the OP) because most of the compression is from ram recovery.

CFMTurboFan,

Quote: These shock waves are positioned within an inlet duct by moving the inlet spike (for example a English Electric lightning, SR71 Blackbird, various MIG fighters), or inlet duct "ramp doors" as found on Concorde. Spikes move forward and backwards, and doors move up and down, which set up multiple shock waves within the inlet. Every time delivery air passes through these shock waves it is slowed to a final velocity of around 0.86 mach which I think is the optimum speed for air to be delivered to a compressor. In the Concorde, there were three shock waves present, which means the air was slowed three times before meeting the first stage of compressor blades.

I thought the English Electric Lightning's inlets were of fixed-geometry. To the best of my knowledge the radar is actually mounted in the spike...


Jetlagged,

Quote: Supersonic and transonic compressors most certainly do exist. In fact they seem to be commonplace, at least for military engines. Allowing the airflow in the compressor to become transonic allows a higher stage loading and so smaller and lighter compressors. I found these course notes while googling supersonic compressors. http://www.adl.gatech.edu/classes/propulsion/prop12.html This is a bit different from allowing the airflow into the compressor to be supersonic, but when you add in the rotational velocity of the blades it is easy to get local flows above M=1 in the first and subsequent stages.

Okay, so you often get supersonic flows when you combine incoming velocity (subsonic) with the rotational velocity? Across the whole stage (blade)? Or on the tips?

Quote: At high supersonic speed (as with the MiG-25) the compressor can be quite a modest pressure ratio (4.5:1 as mentioned in the OP) because most of the compression is from ram recovery.

Yeah, but wouldn't an engine with a pressure-ratio of 9-to-1 at the same mach-number and airspeed produce more thrust than an engine with a pressure-ratio of 4.5-to-1 as the same ram-recovery is produced in either case with a better compressor?

Is it normal for an engine with a pressure ratio of 4.5 to 1 producing 27,000 or so at sea-level producing 40,000 lbf to 44,000 lbf at altitude at Mach 2.5 or so?


Andrea Kent

This topic has a familiar ring to it.

Remembering the discussion regarding the Supersonic Through-flow Rotor and Supersonic Counter Rotating Diffuser (SSTR/SSCRD) fan engine on this forum some time ago, i did some searching for possible updates so maybe this will help:

http://gltrs.grc.nasa.gov/reports/2004/TM-2004-213139.pdf

Apparently a follow-up report of the above has been made which could be of interest but, unfortunately, is not readily available at this time.
The website states: "Do not release on a public Web site. See NPR 2810.1A "Security of Information Technology" for details." This NPR 2810.1A document does not expire until May 16, 2011.

http://gltrs.grc.nasa.gov/cgi-bin/GL.../browse.pl?all/CR-2005-213135.html


Starglider

[Edited 2008-06-24 06:32:12]

Quoting Starglider (Reply 9): Rotor and Supersonic Counter Rotating Diffuser

Neat reference! Of particular interest are the compressor maps in Figure 8 and 9 of the cited paper. They're not anything like the map of conventional compressor. Thanks!

Quoting Starglider (Reply 9): The website states: "Do not release on a public Web site. See NPR 2810.1A "Security of Information Technology" for details." This NPR 2810.1A document does not expire until May 16, 2011.

My understanding from grad school is that much of this work is still classified.


Again, thanks for the great reference, Starglider!

Quoting Blackbird (Reply 8): Okay, so you often get supersonic flows when you combine incoming velocity (subsonic) with the rotational velocity? Across the whole stage (blade)? Or on the tips?

I have no idea, but I imagine on a low aspect ratio blade the flow will be supersonic root to tip.

Quoting Blackbird (Reply 8): Yeah, but wouldn't an engine with a pressure-ratio of 9-to-1 at the same mach-number and airspeed produce more thrust than an engine with a pressure-ratio of 4.5-to-1 as the same ram-recovery is produced in either case with a better compressor? The problems will probably be easier to deal with if it was, rather than having a mixed flow. Is it normal for an engine with a pressure ratio of 4.5 to 1 producing 27,000 or so at sea-level producing 40,000 lbf to 44,000 lbf at altitude at Mach 2.5 or so?

At Mach 2.5 the adiabatic ram recovery ratio is just over 17:1. Factor in the 4.5:1 compressor pressure ratio and that is a mighty 76:1 overall ratio. More than plenty I would have thought.

The Olympus Tubo Jets Fan Blades are Super Sonic but to operate it requires Subsonic Air at the Face of the Engine.

If Supersonic Airflow is allowed to enter the Engine it will lead to an Unstart / Compressor Stall leading to an Expensive Mess for an Engineering team to clean up if they are Lucky if unlucky there could be a Crash and another kind of Mess to Clean up.

If Supersonic Intakes go wrong as has Happend on a Prototype Concorde the resulting Pressure inside the Ducting Caused at least one Intake Ramp to Be SPAT OUT of the Intake which is not a good thing as the Intake Ramps are rather Heavy and Quite Large.

The idea that gas turbines can't work with supersonic airflow is well entrenched but incorrect. I remember being shown examples of supersonic compressor blades while at university back in the 70s.

High bypass turbofans deal with supersonic air on takeoff every day.

For a supersonic aircraft, it makes sense to slow down the intake airflow as much as possible before it gets to the engine, and so take advantage of ram recovery, of course.

Quoting Jetlagged (Reply 13): High bypass turbofans deal with supersonic air on takeoff every day.

There's a big difference between the blade being supersonic locally (which happens all the time) and the bulk flow through the engine being supersonic. In the former case you have shocks off the blades but no bulk supersonic flow. The flow in the stators, inlet, and combustor is all subsonic so you don't have choking, expansion fans, and all those other fun features.

Tom.

Jetlagged,

Quote: I have no idea, but I imagine on a low aspect ratio blade the flow will be supersonic root to tip.[/i] Wow, so even with subsonic flow, the rotational velocity would kick the speed supersonic across the whole diameter of the blade? [quote]At Mach 2.5 the adiabatic ram recovery ratio is just over 17:1. Factor in the 4.5:1 compressor pressure ratio and that is a mighty 76:1 overall ratio. More than plenty I would have thought

I didn't think you could just multiply the inlet compression by the pressure-ratio. I kind of figured that past a point when you'd build air up past a certain pressure you'd get a point of diminishing returns thing where the compressor couldn't compress the air as proportionately and (ultimately, although well above pressures ever encountered in any jet-engine) eventually the pressure would get so high that it simply couldn't compress anymore ultimately...

While we are at it though, what is the adiabatic ram-recovery ratios for the following...

-M = 0.25 (~Takeoff speed of a large airliner)
-M = 0.85 (Speed at which many jetliners fly at)
-M = 1.00
-M = 2.00
-M = 2.20 (Concorde Maximum-Speed)
-M = 2.35 (Cruise speed MiG-25, Tu-144, and MiG-31)
-M = 2.70 (Boeing SST Design-Speed)
-M = 2.85 (MiG-25 Maximum Listed Safe Mach Number)
-M = 3.00
-M = 3.20 [i](Lower Estimate, MiG-25 Max-Speed)
-M = 3.40 (Upper Estimate, MiG-25 Max-Speed)
-M = 3.50
-M = 4.00
-M = 4.35
-M = 5.00
-M = 6.00
-M = 8.00
-M = 10.00
-M = 12.00

And if possible...
-M = 20.00
-M = 22.00
-M = 25.00


BlackProjects,

Quote: The Olympus Tubo Jets Fan Blades are Super Sonic but to operate it requires Subsonic Air at the Face of the Engine.

When the engine is sitting still and spinning at 100% RPM, would the incoming air and rotational velocity already get the blade into the supersonic zone? Or would that only happen when you get the incoming flow to Mach 0.30 to 0.85 or something?

Quote: If Supersonic Intakes go wrong as has Happened on a Prototype Concorde the resulting Pressure inside the Ducting Caused at least one Intake Ramp to Be SPAT OUT of the Intake which is not a good thing as the Intake Ramps are rather Heavy and Quite Large.

What caused such a disastrous reaction?

BTW: If you have a compressor that's
1.) Not designed to have supersonic or transonic flow and it is exposed to full supersonic flow, what happens then?
2.) Not designed to have supersonic flow (not supersonic straight through flow, but a combination of subsonic flow and high rotational velocities), but transonic flow and is exposed to full transonic flow...

What happens?


Andrea Kent

Quoting Blackbird (Reply 15): I didn't think you could just multiply the inlet compression by the pressure-ratio. I kind of figured that past a point when you'd build air up past a certain pressure you'd get a point of diminishing returns thing where the compressor couldn't compress the air as proportionately and (ultimately, although well above pressures ever encountered in any jet-engine) eventually the pressure would get so high that it simply couldn't compress anymore ultimately...

I don't see why not. If you have compressed the air "as much as you can" increasing pressure ratio would be pointless. If a compressor has a pressure ratio of X, the output is X times the inlet pressure. That is the freestream total pressure, which includes ram recovery.

Quoting Blackbird (Reply 15): While we are at it though, what is the adiabatic ram-recovery ratios for the following...

The formula for total pressure ratio Pt/Ps at the inlet is:

(1 + 0.2 * (M) ^2)^3.5

Multiply static pressure by the result of this formula to get total pressure. Obviously due to shock wave the recovery will not be ideal, but this gives you an idea of the numbers involved.

How do you determine static pressure?

None of the calculators I got can multiply something to the 3.5th power...


Andrea Kent

Quoting Blackbird (Reply 15): I didn't think you could just multiply the inlet compression by the pressure-ratio. I kind of figured that past a point when you'd build air up past a certain pressure you'd get a point of diminishing returns thing where the compressor couldn't compress the air as proportionately and (ultimately, although well above pressures ever encountered in any jet-engine) eventually the pressure would get so high that it simply couldn't compress anymore ultimately...

For the pressures we're talking about, it's straight multiplicative. The ideal gas law is valid for almost everything below hypersonics. If the pressure gets *really* high you start to go out of the ideal gas range and the stuff starts behaving vaguely liquid-like and most of the equations go out the window, but that's a very long way from the pressure typically seen in jet engines.

Quoting Blackbird (Reply 15): 1.) Not designed to have supersonic or transonic flow and it is exposed to full supersonic flow, what happens then?

Transonic won't necessarily kill it, since the supersonic areas may be contained to the front side of the compressor blades. Supersonic, I would expect, would cause a compressor stall due to adverse shock interactions and separation.

Quoting Blackbird (Reply 17): How do you determine static pressure?

Static port on the side of a smooth and flat part of the aircraft.

Quoting Blackbird (Reply 17): None of the calculators I got can multiply something to the 3.5th power...

Most graphing calculators can do this. Excel can too. Alternatively, just raise it to the 7th power then take the square root, which gives you the same answer.

Tom.

Quoting Blackbird (Reply 17): How do you determine static pressure? None of the calculators I got can multiply something to the 3.5th power...

Static pressure = ambient pressure

Even the windows calculator can do it in scientific view (x^y button)

Jetlagged,

Quote: Static pressure = ambient pressure Even the windows calculator can do it in scientific view (x^y button)

Unfortunately, I have a Mac... who's calculator on the dashboard is a primitive basic calculator.


Andrea Kent

In which case follow Tom's advice above.

Or consider http://magicnumbermachine.googlepages.com/ or one of the other supplemental calculators for Mac.

PITIngres,

Thank you, I downloaded it and have used it...


Jetlagged,
Okay... I still have a problem, which may have to do with the fact that I haven't taken math in years...

Step 1: (1 + 0.2 * (M) ^2)^3.5
Step 2: (1 + 0.2 * (2.5)^2)^3.5
Step 3: (1 + 0.2 * (6.25)^3.5)
Step 4: (1 + 0.2 * (~610.4))
Step 5: (1.2 * (~610.4))
Step 6: 732.42

732.42 / 14.70 (atmospheric pressure)

Ram Pressure = ~49.825 x atmospheric @ M 2.5

That should be it, but I thought the Ram Recovery was 17:1 @ Mach 2.5

How did I end up 42 atmospheres off?


Andrea Kent

Quoting Blackbird (Reply 23): How did I end up 42 atmospheres off?

By making an order-of-operations mistake in step 3.

Quoting Blackbird (Reply 23): Step 2: (1 + 0.2 * (2.5)^2)^3.5 Step 3: (1 + 0.2 * (6.25)^3.5)

The power of 3.5 applies to the entire term in the outer set of brackets, not just the 2.5^2 term.

You should have gotten:
Step 3: (1+0.2*6.25)^3.5
Step 4: (2.25)^3.5
Step 5: 17.1

Exactly as Jetlagged predicted.

Tom.

Tom,

Thanks for the assistance. As I said, it's been awhile since I've done even minimally complex mathematics.


Andrea Kent