I finally got a handle on the diameter and blade length of the direct drive
turbine needed to work well with a rotary. A 19,000 RPM Formula 1 engine
would
require a smaller turbine. This correlates well with the geared turbine used
on the R3350 TC engine. The 11.3 inch diameter R3350 turbine speed is 16,000
RPM
at cruise power and 19,000 RPM at takeoff power. Gear ratio is 6.52:1.
Engine
RPM is 2900.

A stationary blow down turbine with air flowing over the blades generates
maximum torque. As the turbine accelerates the torque reduces in a straight
line. As the turbine blades reach the speed of the air flowing over them no
torque is generated. This will be the maximum theoretical RPM of the
turbine.
The maximum HP generated by the turbine will be at a RPM that is roughly
half of
the peak RPM. Torque at mid RPM will be roughly half of peak torque. Since
HP is
RPM times torque peak HP is generated at half peak RPM and half peak torque
roughly speaking.

If the total projected blade area of the turbine is the same as the cross
sectional area of the exhaust pipe feeding it the velocity of the gas
flowing
over the blades will be the same as the exhaust pipe gas velocity. The
projected
area is that area between the inner diameter of the blades and the outer
diameter. Not the total blade surface area.

The part of the exhaust system closest to the blades is called the nozzle.
It is
usually possible to neck down the nozzle area to speed the air flowing over
the
blades at the modest expense of increasing the back pressure on the engine.
This
is analogous to changing the A/R ratio of a turbo charger. Usually there is
an
overall increase in net HP of the turbo compound system because gas velocity
is
more important than back pressure. The blades respond to the gas velocity
squared just like lift on a wing responds to an increase in forward speed.
The
force the blades exert on the hub is the density of the gas times the
surface
area of the blades times the lift coefficient times the gas velocity
squared.
Just like a wing. Lift coefficients on the highly cambered blades would be
between one and one and a half.

Since you need to keep the projected cross sectional area of the blades more
or
less the same as the exhaust pipe area the length of the blades must change
as
the overall diameter of the turbine changes. In general a large diameter
turbine
with short blades will generate more torque and peak HP and will be
generated at
a lower RPM. The up side to this is the gear ratio required between turbine
and
crankshaft is reduced. The draw back is leakage around the ends of the
blades.
The aspect ratio of the blades in wing terms is low. A smaller diameter
turbine
with longer blades will operate at a higher RPM but the blades aerodynamic
losses may be lower due to a more favorable aspect ratio. What also must be
considered is the centrifugal loads on the blades. Longer blades put more
stress
on the roots of the blades which is the critical area. Another factor of
course
is metal loses it strength as the temperature of the metal increases. There
is
an area of operation bounded on one side by the turbine inlet temperature
and on
the other side by the RPM. these two lines eventually converge so one must
keep
the turbine operating in the safe area between the two curves. Turbine
design is
best left up to people with experience. One of those companies is Barber
Nichols
in the US.

I used QB (below) to help with the calcs. A spread sheet would also work.
The results are in the attached jpg. The theoretical free speed of the 10
inch
turbine is 23,000 RPM. Of course the turbine would not live at that speed.
This
turbine will be generating max HP around half of that or about 11,500 RPM.
As
long as it is connected to the e-shaft (crankshaft) it will not see any RPM
higher than the engine.  This is a racing application. For an aircraft
engine
the turbine will need to be a bit bigger.

Check my logic and math.

Paul Lamar

------------------------------------------------------------------
CLS

PRINT : GasVel = 1000'FPS

PRINT "Gas  Vel  FPS ="; GasVel

PipeR = 1.5: AreaPipe = PipeR ^ 2 * 3.1416

PRINT "Area Pipe sq in ="; AreaPipe
PRINT

R1 = 5: R2 = 4.77

PRINT "Turbine Dia OD in ="; R1 * 2

PRINT "Turbine Dia ID in ="; R2 * 2: PRINT

FlowArea = R1 ^ 2 * 3.1416 - R2 ^ 2 * 3.1416

PRINT "Flow Area sq in ="; FlowArea

Circum = R1 * 2 * 3.1416

PRINT : PRINT "Circum Inches ="; Circum

PRINT : PRINT "Circum Feet ="; Circum / 12

PRINT : PRINT "Rev per second ="; GasVel / (Circum / 12)

PRINT : PRINT "Rev per minute ="; (GasVel / (Circum / 12)) * 60


-----------------------------------
Paul,

I thought we talked about a larger diameter, slower turning turbine some
time ago.  This looks very good!  I need one for around 5500 continuous
engine rpm.

Steve Brazil

------------------------------------------------
I think we finally have a handle good enough to recommend a direct drive turbine
diameter and blade configuration as a starting point.

I'll run through the numbers for you. Such a low engine RPM may not be practical
for a direct drive turbine. So don't get your hopes up too high.

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To get closer in I need the exhaust pipe diameter and a measurement of the
dynamic pressure in the pipe using a SS pito tube.
The dynamic pressure will be in the range of 500 to 1000 pounds per square
foot or 3 to 7 pounds per square inch.

Paul Lamar

--

Paul

In measuring the dynamic pressure via a stainless pito tube does the pito
tube itself introduce an error in the reading?  I was under the
impression Mazda used some type of high temp. piezo electric sensor
inserted into a bung welded on the manifold before the turbine.

Doug in Japan

Not significantly. I think Mazda was measuring static pressure in the exhaust.
It is the combination of exhaust gas density and exhaust gas velocity you are
after. That is what generates dynamic pressure. Dynamic pressure is what
generates lift on a wing or a turbine blade. As you climb and the air density is
reduced you have to fly faster to get the same lift on the wing. If you heat the
air the density also goes down. That is were the term density altitude comes
into play.

-------------------------------------------------------------------------

I am just thinking out loud here. Given the dynamic pressure (we can measure
that) and exhaust gas temperature (we can measure that too) what is the gas
velocity? Anybody have any ideas on how to calculate this?

I suspect. It has something to do with the ratio of the temperature expressed
in Kelivin. 100 F is 311 K.  1700 F is 1200 K or four times higher.
Does that mean air density at 1700 F is one quarter of what it is at 100 F?

Dynamic pressure in pounds per square foot at sea level density
and standard temperature is .0026 (AD) time velocity squared.
D.P. = A.D. X V^2

-----------------------------------------------

Here is something else to thing  about. The rotor face is 3 inches wide
and 7.5 inches long. The face moves from top dead center to bottom dead center
in 1/2 revolution. The rotor face moves in and out 2.5 inches. 11,000 RPM is 183 revolutions per second. One revolution in .005 seconds. Half a revolution in .0025 seconds. So the rotor face speed is 2.5/.0025 or 1000 inches per second
in and out. If there were no rotor housing that would be the speed of the column of exhaust gas moving in an out with the face of the rotor.

What happens is all of this column of exhaust gas goes through the exhaust port which has an area of only 2 square inches. The speed of the exhaust gas going out the port is speeded up. So the speed of the exhaust gas going out the port is proportional to the ratio of the areas or (7.5 X 3)/2 or 11 times the speed air moving in and out with the rotor. Namely 11000 inches per second or 917 feet per second. I guessed 1000 feet per second which is pretty close. The question remains. What is density of the exhaust gas?


Paul Lamar