Ram Compression on Turbojet/Turbofan Thrust

[Jane-DoH]

Does anybody have a formula for determining the effects of ram-compression on thrust levels for turbojet/turbofan engines at subsonic and supersonic speed?

Also, when it comes to air compression -- is there a point where air will reach a pressure at which it won't proportionately compress right (i.e. you have a 9:1 pressure ratio, the inlet has increased the atmospheric pressure to 10 atmospheres, so when the air flows through the compressor after coming off the inlet, it is about 90 atmospheres. Is there a point where if you were flying at high supersonic speeds where the compression off the inlet would be so high that when you run the air though the compressor, it wouldn't increase the pressure 9-fold?)?

[engine-eer]

Jane,

You need a set of mach tables for standard atmosphere to calculate the pressure at the compressor face. A proper set of mach tables has the total pressure ratio and total temperature already calculated. I couldn't find a set on the web, but I did find a link to a NASA calculator that gives you what you are looking for. Plug in your flight conditions (and push the enter key with the cursor in either the altitude or the speed box) and this will tell you what you want. The units are funky, you have to calc the pressure ratio after you convert the total pressure back from lbs/sq ft back to psi, but it is all there.

AtmosModeler Simulator - Version 1.2a (http://www.grc.nasa.gov/WWW/K-12/airplane/atmosi.html)

What you want to remember is that there is a pressure loss in any inlet, so the calculated inlet pressure has to be multiplied by the inlet recovery. Low speed inlets (like on a big turbofan) are typically about 98% efficient or better, but long high mach inlets like on a fighter can be as bad a 92 or 93%.

The faster you go, the higher the inlet pressure, but also, the higher the inlet temperature becomes. As you go to high mach numbers (above three) the inlet temperature will be closing in on 800 degrees F. The pressure is there, but if you try to compress it much the temperature coming out of the compressor and into the combustor gets so high that you can't add much heat. This is why high mach turbines don't have much of a compressor pressure ratio. Engines running at high mach are typically seeing pressure ratios of 4 or 5 at those conditions. The J58 has a pressure ratio of 6, but when the inlet is hot it doesn't make that much becaise this higher temperature reduces the corrected speed of the compressor, which means that to do work on the air you need to run higher mechanical speed or you have to give up pressure ratio. Also, hotter air takes more work to compress, so your are taking more work out of the flowpath to drive the compressor so the efficiency of the engine suffers more. At some point a ram jet makes more sense than a turbojet.

[Jane-DoH]

engine-eer

You need a set of mach tables for standard atmosphere to calculate the pressure at the compressor face. A proper set of mach tables has the total pressure ratio and total temperature already calculated. I couldn't find a set on the web, but I did find a link to a NASA calculator that gives you what you are looking for. Plug in your flight conditions (and push the enter key with the cursor in either the altitude or the speed box) and this will tell you what you want. The units are funky, you have to calc the pressure ratio after you convert the total pressure back from lbs/sq ft back to psi, but it is all there.

It doesn't seem to make any mention of being Mac compatible...

What you want to remember is that there is a pressure loss in any inlet, so the calculated inlet pressure has to be multiplied by the inlet recovery. Low speed inlets (like on a big turbofan) are typically about 98% efficient or better, but long high mach inlets like on a fighter can be as bad a 92 or 93%.

So that's why very high speed aircraft like the XB-70 and SR-71 tended to have a slower initial acceleration rate than you would expect?

The faster you go, the higher the inlet pressure, but also, the higher the inlet temperature becomes. As you go to high mach numbers (above three) the inlet temperature will be closing in on 800 degrees F.

I thought stagnation temps tended to produce around 600-F at Mach 3?

[engine-eer]

Sorry about the Mac. Maybe you can find a set of tables somewhere on the web. I just did a short search and didn't find any. Used to be Allison and Pratt produced a little pocket sized book with Isentropic shock tables and Standard atmosphere tables in it. They used to give them out at trade shows and every gas turbine engineer had one (I still have mine and use it frequently). Here is a link to a used one on the web:

Overstock Auctions AERONAUTICAL POCKET HANDBOOK 1966 Pratt & Whitney Item 46008598 (http://auctions.overstock.com/AERONAUTICAL-POCKET-HANDBOOK-1966-Pratt-Whitney/Collectibles/item/46008598)

Probably the best $8 you will ever spend if you are interested in propulsion. Pratt also sells it new for $34, again, a bargin at twice the price.

So that's why very high speed aircraft like the XB-70 and SR-71 tended to have a slower initial acceleration rate than you would expect?

The reason that these aircraft have slow initial acceleration is that they have turbojets. Turbojets have higher velocity, lower mass flow rate exhausts, and at low speeds the propulsive efficiency is poor. Think of it this way, bigger fan, better low speed thrust, prop, even bigger low speed thrust. The early turbojets took forever to accelerate. Those really long runways for the B47, B52 and the early fighters were necessary because the early turbojets didn't produce much low speed thrust.

I thought stagnation temps tended to produce around 600-F at Mach 3?

Depends on the altitude. Total temperature at Mach 3 is 2.69 x the absolute temperature in the free stream. Higher altitudes, above 45k it can be that low, but at 20,000 ft it will be almost 800. On the deck at Mach 3 where we were designing some missile engines, we were seeing almost 1,000F.

[Jane-DoH]

engine-eer

The reason that these aircraft have slow initial acceleration is that they have turbojets. Turbojets have higher velocity, lower mass flow rate exhausts, and at low speeds the propulsive efficiency is poor. Think of it this way, bigger fan, better low speed thrust, prop, even bigger low speed thrust.

That I understand actually, it's more efficient to accelerate a large volume of air a small amount, than accelerate a small volume of air a large amount (though the faster you go, you need a higher exhaust speed). For this reason propeller-driven fighters often had more thrust/weight than early jet fighters, though as the speed increased, many of the jets had them beat (ram-compression was another variable).

The early turbojets took forever to accelerate.

Yes, but it wasn't due entirely to a fast-exhaust velocity; some of them were quite underpowered: The P-59 had a T/W of 0.32/1.0 to 0.36/1.0; the FH-1 at 0.27/1.0 to 0.32/1.0; the F2H at 0.32/1.0 to 0.36/1.0 (which are comparable to modern commercial airliners).

Those really long runways for the B47, B52 and the early fighters were necessary because the early turbojets didn't produce much low speed thrust.

As I understand the B-47's weighed around 200,000 and had 6 x J-47's which produced around 6,500 to 7,500 pounds of thrust, IIRC, and with that being said, that gives you a T/W of 0.20/1.0 to 0.23/1.0; The B-52's weighed around 420,000 pounds fully loaded with 8 x J-57's producing around 11,000 to 13,500 pounds of thrust yielding a T/W ratio between 0.21 to 0.27

Regardless, the point I was making was that two aircraft with the same T/W ratio, one with a subsonic inlet (short, round-lipped, bell-mouthed), the other with a supersonic inlet (long, sharp-lipped, covergent-divergent); the one with the longer inlet tends to accelerate more poorly.