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Turbine Blades

One of the main objectives of the IMPRESS project is to produce low mass, intermetallic turbine blades for jet aircraft engines. The lower mass leads to a lower fuel consumption and so less green house gas emission.

The design of a jet turbine (or ‘turbofan’) is very complex but a basic overview is set out here.

 Typical Aero Turbine Engine

The first section is the single set of large fan blades. These draw in and push back huge amounts of air, around one ton per second for a Rolls-Royce Trent engine.

Behind the fan blade is the compressor stage. A small proportion of the air drawn in by the fan blades is drawn in to the ‘core’ of the engine where it first meets the compressor blades. Here the air is compressed into a small volume with a corresponding rise in temperature.

Next fuel is mixed with the air and combustion occurs. The temperature soars but the exhaust gases cannot expand and so the pressure becomes very high.

The next two stages are the high pressure and the low pressure turbine stages. Here the exhaust gases gradually expand, whilst forcing the turbine blades to rotate.

By the time the exhaust gases leave the engine the turbine blades have extracted much of their thermal energy and used it to rotate a drive shaft which in turn drives the fan blades and the compressor blades. The (still hot) exhaust is surrounded by a sheath of cool air from the fan blades, reducing the noise that the engine generates.

Most of the air that passes through the fan blades is pushed directly backwards and never enters the rest of the engine. However the momentum given to this bypass air provides the bulk of the thrust of the engine. It also gives the engine design its name, the high bypass turbofan.

Very similar engines (without the fan) are used in gas fired power stations. But what are the conditions that rotating turbine blade has to survive?

A turbine blade is a very complex shape consisting of a ‘root’ at the base of the blade, an aerofoil surface that extracts the thermal energy from the hot exhaust gases and a ‘shroud’ at the top. The root of each blade is attached to a disc and a single disc with around a hundred or so blades attached is called a ‘stage’ and there are several stages in each section of the engine. 

 An IMPRESS Turbine Blade

For our purposes we can use a simplified model blade made of three rectangular blocks shown below this is a very basic model of a low pressure turbine blade:

 Simplified Turbine Blade

Traditionally these blades are made of ‘nickel superalloy’ with a density of around 9.0 g cm-3. An intermetallic of titanium aluminide would have a density of around 4.5 g cm-3 and so would be far lighter.

What would the moments of inertia of the two blades be for the axis shown on the diagram?

When installed in an engine the blades are mounted on a disc, shown below (a single blade is shown). The disc is made of the same material as the blades. Calculate the moment of inertia of the disc about its axis of rotation and so determine the moment of inertia of a complete stage comprising a disc and the 80 blades mounted on it.


 Simplified Disc and Blade

Calculate the constant angular acceleration, and so the torque, required to accelerate the stage up to its full speed of 3000 rpm, from rest in a period of 30 seconds.

What is the kinetic energy of a single blade when operating at this speed? Why might this figure be important in the event of a ‘bird strike’ on the engine?

So far we’ve looked at circular motion, acceleration, energy and the associated ‘rotational forces’ or torques involved but what about the forces inside an object? We know that to keep an object moving in a circular path we need a centripetal force. For an object like our turbine blade the only thing that can provide these forces are the internal cohesive forces that give rise to its tensile strength. The spinning turbine blade will be in tension but how large is that tension and how does it vary along the length of the blade?

 Tension in a rotating Blade: Simplified Schematic

The figure shows a blade, mounted on a disc and rotating at an angular velocity of ω. Highlighted is a section of the blade, of thickness δr and mass m at a distance r from the axis of rotation. Since the size of m will depend on where in the blade we decide to take the section we should write it as m(r) (mass, as a function of r). To keep it moving in its circular path it requires a force δF that is directed towards the centre. The size of the force is just the familiar formula for centripetal force, which when written with our chosen symbols becomes:


The surface immediately towards the centre of our section must provide this provide this force in the form of a tension in order to keep the section attached to the rest of the blade. However our chosen section must support the next section out which in turn supports the next section and so on until we reach the end of the blade. As a result the tension in the blade at point x can be written as:


Clearly the tension will depend on the angular velocity, the distance from the centre and, importantly, the mass distribution along the blade. We have chosen to use a simplified blade made from three rectangular blocks.

 Tension in a rotating Object

If the block has a mass M, dimensions A, B, C and is positioned a distance D from the centre of rotation, then so long as D is large compared to A then:


So for the shroud, where D is the distance of the inside edge of the shroud from the centre of rotation and B is length of the shroud:


This is increases as x decreases and so the tension in the shroud is a maximum at its inside edge (x = D).


Whenever you meet an unfamiliar equation such as this it is always a good idea to check it to see that it makes sense. Does it produce the right units, or even better does it reduce to a familiar equation? For this equation we could take the extreme case where the distance, D to the object is far greater than its size, B.


Since D >> B:


And so our new equation reduces to the familiar form of


And so we can be confident that it is correct. Applying the equation to the shroud of the blade:


For the TiAl blade in the previous question the mass of the shroud was 189 g and the angular velocity was 314 rad s-1, so:


Next for the aerofoil section we can repeat the calculation but this section must also provide the force needed by the shroud, so:


For the TiAl blade the mass of the aerofoil was 338 g (the angular velocity must of course be the same), so:


Similarly for the root, which has a mass of 189 g:


This is the equivalent of hanging over three and a half tonnes off the end of the blade, and all this when it is operating at a temperature of around 700 °C. Many real blades in functioning aircraft engines must support even greater forces, the weight of a double decker bus is frequently used as an example to describe the strength required by an aero-engine blade.


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