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Equation describing the contraction of a rotation shaft

  1. Dec 8, 2008 #1
    It's a fact that a shaft in a turbine is expanding when the temperature is increasing. A simple equation gives that the expansion can be described as: deltaX=(T1-T0)*alpha*x0, where deltaX is the expansion, T0 is the initial temperature, T1 is the final temperature and alpha is a coefficient (approximately equal to 11.5*10^-6 /°C). At least this fomula is accurate when the temperature is in the range 0- 100 °C.

    However, how can the contraction (in length) due to rotation be calculated? Until very recently I was not even aware that this was something that could be measured if we for instance is talking about a shaft that is 25 m long and 0.6 m i diameter rotating in 3000 rpm. I've been told that this effect is the same as the temperature expansion which means the length of the shaft is not changed at all when the turbine is running on full power if it's designed correctly.

    Any ideas?
  2. jcsd
  3. Dec 8, 2008 #2


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    It's going to depend on how the shaft is constrained within the turbine, i.e. bearing and lock nut locations (if used). Without looking over the shaft layout it's impossible to say. You have a lot of things other than centrifugals to worry about.

    If all else fails, use conservation of mass. Use the standard shaft dimensions to calculate a volume. Then use the new OD and ID with the volume you just calculated to calculate a new length. Again, this assumes that the shaft is completely unconstrained and is allowed to move over its entire length.
  4. Dec 8, 2008 #3
    Thanks Fred for your reply.

    I understand it's a complicated problem and that there's a lot of unknown details. However, at the moment I'm only looking for a figure that is somewhere near the truth. The same goes for the formula I mentioned earlier, concerning the expansion due to increasing temperature.

    I don't understand what you mean by OD and ID. Can you clearify?

    The shaft has one fixed point, the "axial" bearing (maybe not correct English, but it's just a point of the shaft that cannot move in the direction of the shaft. Beside from that point the shaft is free "grow" och "shrink".
  5. Dec 8, 2008 #4
    ID = Inner Diameter (assuming it has one)
    OD = Outer Diameter

    I think what Fred is getting at, is that you cant simply use a simple formula to accurately describe the expansion of a turbine shaft. If you assume an ideal shaft that is not constrained and assumed to be "free floating" you could determine such an equation. I would probably approach the problem using Castigliano's theorem where ultimately the expansion of the shaft will be a function of its geometry, angular speed, and modulus of elasticity of its material.
  6. Dec 9, 2008 #5
    I know, but my objective was just to get a hint about the "true" answer. Is it close to 1 mm? 10 mm? 20 mm? Etc.

    The thing is that we have a sensor reading the differential expansion of the turbine shaft and right now it's showing 10 mm too little according to some people but I doubt that from a metrology perspective. I would simple like to know if there is any possibility that the rotation COULD explain at least a major part of the "error".

    I guess I will have to study the Castigliano theorem to get a deeper understanding of the phenomenon.
  7. Dec 16, 2008 #6
    We can hope are looking at 1 mm on the length. If it is much larger than that, all the turbine seals are likely to be wiped because of it.
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