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I made a few very conservative educated guesses about the conditions inside the exhaust pipe. These can be easily verified by measurement's which is the next step.

Gas velocity in the 3 inch diameter exhaust pipe is assumed to be 1000 FPS or 681 MPH. A variable nozzle would increase that at the expense of some back
pressure.

Air density in the pipe is assumed to be half sea level or .0013 for the MPH constant. That affects the dynamic pressure directly. Not to be confused with static pressure.

I assumed the lift coefficients on the blades was 1.5.

There were 31 blades at about one inch apart. It is possible to build
a turbine with blades much closer together so that is going to affect the
results directly.

Bottom line I estimated  37 HP from the turbine at 11,000 RPM which is slightly over 12.3% more power from a 300 HP P-port rotary no increase in fuel burn. This would be about the same for a BSFC improvement. NASA in a paper study with a super charged rotary obtained 80 HP additional. The DD15 diesel turbo compound
obtained 11% and diesel's have much less waste energy in the exhaust.

The actual results are going to vary according to the real numbers. There are a  lot of factors here. This is probably a good turbine to start with.

I think the theory and calculation method is good. We will not know the actual numbers until we build the hardware and test it. I think this method will get us in the ball park. I won't bore everybody with the equations so if you want
them let me know.

Paul Lamar

Paul,

I think we can get into mass flow with HP, BSFC and air/fuel ratio.

let's assume a 360 hp rotary on 75% cruise, BSFC as 0,45 lb/HPh, A/F as
leaned to 1:16.

75% cruise is 275 hp, 122 lb/h. Converting that to grams/sec = 15,3
grams/sec fuel flow. Air/fuel was 16:1, so required air is 245 grams/sec.

Air density on 15 deg C, std. pressure, is 1,225 kg/m3.
That translates to 1,225 grams/litre. So our intake air volume is 200
litres/sec. That translates to 414 CFM.

So, a 360 hp rotary on 75% with 0,45 lb/HPh BSFC and 16:1 A/F gives:

Mass flow: 15,3 (fuel) + 245 (air) = 260 grams/sec or 0,574 lb/sec.
Intake air volume: 200 ltr/sec or 414 CFM

That agrees quite nicely on (3-rotor equivalent) 4 ltr engine table
value - on 6500 rpm/80% VE/ 14:1 A/F  = 350 CFM.

Putting that on kinetic energy equation gives us:

305 m/s^2 * 0,26 kg/sec * 0,5 = 12 kJ/sec = 12 kW = 16,1 HP = 4,5%
340 m/s^2 * 0,26 kg/sec * 0,5 = 15 kJ/sec = 15 kW = 20,1 HP = 5,6%
380 m/s^2 * 0,26 kg/sec * 0,5 = 19 kJ/sec = 19 kW = 25,1 HP = 7,0%

Percents as against 360 SHP. And this is total kinetic energy, before
conversion.

Thus, in order to have 20% power recovery, gas speed should be at least
643 m/s, ie. 2.5 Mach... not happening.

Something's not right here. As much as i can see, mass flow should be
quite right, but total energy and it's recovery should be calculated
some other way...

This is quite basic case, so there must be a myriad of calculated
examples on textbooks..

Hot gas should have stored energy as kinetic, heat and pressure. Any
ideas? I'll give up for now, need to sleep...

Cheers,

Petri