The overall size of the engine is dictated by the diameter of the turbine and compressor wheels.

I settled on a diameter of 60mm with a blade height of 10mm for the turbine because this seems a reasonable size and I have a circular ss blank of that diameter.

That gives an annular cross-section for the gas flow through the turbine of 0.00157 m^{2}.

I aim for a low rotational speed in this design because I have no 'feeling' for the material strengths or engineering tolerances. Say, 24000 rpm - that is just above the quoted idling speed of a typical model engine (it's still 400Hz!).

That gives a peripheral velocity at the blade centre of 50mm*pi * 24000/60 = 63m/sec.

## Turbine wheel

This diagram shows the airflow between the nozzle guide vanes (NGV) and the turbine wheel.

It is important to realise that the turbine wheel is rotating at very high velocity (U). As a consequence, the apparent direction of the air entering the turbine wheel from the NGV is altered very strongly from the 'real' angle at C2.

An observer sitting on a blade of the turbine wheel and facing the NGV would find himself moving sideways so fast that the air leaving the NGV would appear to be directly in his face. In the vector diagram for the entry condition, the relative velocity is shown by the vector Wi. Note that the leading edges of the turbine blades are angled to accept air from that direction. The air enters the turbine smoothly along the face of the blade and not at right angles as you might expect from looking at the static arrangement of the blades.

As air travels through the turbine, its direction is changed by the curve of the blade and it is that change of direction (and momentum) that applies a reaction force to the blades. The force can be calculated from the change of direction and the mass flow.

When the air leaves the turbine it is moving relative to the turbine at the same angle as the blade trailing edge , but because of the rotational velocity, the final result is that the air flows directly out of the turbine with no residual 'twist'.

(Vector diagram at turbine exit shows the final velocity C3)

In the configuration shown here, the turbine wheel exit angle is the same as the NGV angle and the relative direction of entry to the turbine wheel is exactly axial, resulting in the two vector diagrams being opposite and equal triangles. This isn't the only possible configuration, but it is the most efficient.

There appears to be a general consensus that the blade angle of the turbine should be between 30 and 35 degrees. I won't argue with that. Make it 32.5, which means that the exhaust gas velocity must be 63 m/sec * Tan32.5 = 40m/sec. The volume flow must therefore be 40 m/sec * 0.00157 m^{2} = 0.06m^{3}/sec

If the turbine is to survive, the temperature at the blades shouldn't go beyond say 600 degrees C. That is 890 Kelvin and represents a temperature rise by a factor of 3 on ambient. So the gas volume has expanded by a factor of 3, which means that the density will be about 0.4kg/m^{3}. (Air is normally about 1.25kg /m^{3}). That gives a mass-flow of 0.4*0.06 = 0.024kg/sec.

From this we can calculate the turbine shaft-power. (rotational Force * rotational velocity)

The force on the turbine blades is the rate of change of momentum of the exhaust gasses in the peripheral direction as they pass through the turbine. We calculated that the exit velocity is 40m/sec. The gas enters the turbine at an angle of 31.5 degrees to the axis so it's peripheral velocity must be 40/Tan31.5 = 65m/sec. It leaves with no peripheral component (no swirl) so 65m/sec * 0.024kg/sec of momentum has converted into the turbine force = 1.56 Newtons. Multiply by the turbine peripheral velocity to get Power = 63 * 1.56 = 98 Watts.

And it also means that the volume airflow through the compressor must be 0.06/3 = 0.02 m^{3}/sec (20 litres/sec)

## The compressor

Now, this is where it all goes wrong.

There's no simple way to calculate any useful parameters for the compressor.

However, using the known shaft power of the turbine and some suspect geometry, it's possible to estimate the radial velocity. ie the speed at which air is pushed out of the compressor and into the engine. That in turn gives the flow area and hence blade height. It also gives the diffuser vane angle

In this diagram, C is the velocity/direction vector of the air as it leaves the compressor blade tip.

This can be resolved into components in two different ways.

One is to resolve with respect to the blade itself, into a component along the blade W and one in the peripheral (tangengential) direction. These form a vector sum. to give C. The angle between them is the blade tip angle (B). *NB. use of upper case B instead of greek symbol beta - similarly A for alpha in text.*

The other is to resolve radially/tangentially, giving the vectors Cm and Cu.

With a bit of geometry we can show that Cm = (U-Cu) TanB

and since Power = massflow * U * Cu (It's the same as the turbine shaft power calculation - but in reverse. ) we get that Cm = (U-P/(U*m)) TanB

U is peripheral speed of the compressor .06 * pi * 400 = 75m/sec

Assume a retro curved blade with tip angle 45 degrees.

So Cm = (75 - 98/(75 * 0.024))*Tan(45) = 30 Tan(45) = 20m/sec

So area = 0.02 / 20 = 0.001 sq m

Blade height .001/(.06*pi) = 5.5mm

Finally, alpha is the angle at which the air leaves the compressor to enter the diffuser vanes, so gives the angle for the vanes.

SinA = Cm/Cu = 20/(98/(75 * 0.024)) = 0.36

A = 21 degrees.

All of this is highly dependent on theory matching practice, which it rarely does. The efficiency of the compressor is not perfect, the gas flow is not precisely laminar, blade slip and friction losses all contribute to unfathomable variances.

I haven't allowed for blade thickness reducing the area...

So these values have to be treated as best-guess estimates and a bit of common-sense thrown in.

## 1 comment:

WOW, I HAVE THE SAME IDEA... NOT AS FAR ALONG AS YOU ARE. MOST PEOPLE I HAVE TALKED TO ARE DISCOURAGING BUT I KNOW I CAN DO IT, LIKE I KNOW YOU CAN DO IT... NICE BLOGSITE TOO BTW. MY ENGINE IS STILL A DRAWING (UNFINISHED) AND THERE IS STILL THE FUELING SYSTEM TO WORK OUT FOR INVERTED FLIGHT OR G-FORCE CALCULATIONS. THERE ARE MORE THAN A THOUSAND FACTORS.

EDEMING@ROCKETMAIL.COM

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