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Observe the picture below, taken from this link.

enter image description here

Now compare the figure above with the same shaft (being rotated by the turbine) with a torsional moment, but now the other end is fixed to the wall (instead of being attached to the generator with bearings in between). For this case, I can understand that a net reaction torque must exist on the shaft face which is attached to the wall, because the wall is exerting equal but opposite torque on that face, to ensure an overall equilibrium system. But for the figure attached here, the shaft can rotate without any restriction or resistance from the generator, because they have ball bearings in between.

So can someone just tell me that is the system shown in the figure in equilibrium or not? Because the figure shows presence of reaction torque but the rod still rotates without any resistance. Plus, why does the reaciton torque even exist in this system?

Rameez Ul Haq
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  • What does the magnetism do? – Solar Mike Dec 31 '21 at 11:46
  • I'm not sure what you think ball bearings have to do with the problem but the generator will present a mechanical load (reaction torque) to the turbine proportional to the electrical power output of the generator. $P_{out} = P_{in} \times efficiency$. – Transistor Dec 31 '21 at 12:10
  • The reaction torque is due to the resistance to rotation by the generator when providing energy to a load. (Try spinning a bicycle wheel with a hub generator with the terminals open-circuit and again with the terminals short-circuited. You could do the same with a DC motor.) – Transistor Dec 31 '21 at 13:06
  • A system is considered in equilibrium when the input energy equals the output work plus, in statics - no connect parts is moving; in the rotary machine - no broken parts. – r13 Dec 31 '21 at 22:32

3 Answers3

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general

A motor or a generator (i.e. something with a rotating mass), will accelerate if the sum of torsional moments are greater than zero, and it will decelerate if the sum is less than zero. The angular acceleration is determined by the equation

$$\Delta M = I \cdot a$$

Where:

  • $\Delta M$ is the algebraic sum of torsional moments
  • $I$ is the mass moment of are of the rotating masses
  • $\alpha$ is the angular acceleration (usually in rad/s^2).

In the steady state, angular acceleration is zero, and therefore $\Delta M =0$

At any moment the power going through the shaft is

$$P = M\cdot \omega$$ where:

  • M is the torsional moment
  • $\omega$ is the angular velocity in rad/s

Motors

When the shaft of a motor turns (e.g. the electric motor of an EV ) the opposing load (or load for short) is provided by the resistance of the car (which is mainly due to the aerodynamic drag).

generators

A shaft that rotates a generator provides power to the generator. The power that it provides is equal to $P= M\omega$. Depending on the efficiency of the generator, that power is converted to electric energy and distributed to many households.

If at any point the power required is more that the power provided, then the shaft will start to decelerate. (That deceleration is due to the magnetic fields in the coil of the generator.) See this question for more details.

conclusion

so the bearing are not there to take the torsional moment. The function of the bearings is to allow the shaft to move.

NMech
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  • what I meant is that the presence of bearing between the shaft and generator causes the rotational motion of the shaft to not get resisted at all. This is what confuses me then why there exist a reaction torque. – Rameez Ul Haq Dec 31 '21 at 21:21
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Try to imagine the shaft was really narrow, say a bicycle spoke, would it work?

If the generator had no load on it and was designed to turn idle without any resistance, then there wouldn't be a torque in the shaft between the turbine and the generator. The turbine would produce a torque but the shaft would turn idle, with no torque.

The shaft is just a mechanical coupling between the generator and the turbine.

  • If the generator load is removed, the torque on the shaft would be zero.

  • if the generator load is smaller or larger than the turbine torque by $\Delta \tau$, the whole system will accelerate by $\alpha=\frac{\Delta \tau}{I_{turbine+generator}}, $ meaning Part of the turbine torque will drive the generator and the excess torque will accelerate the whole system.

  • If the sudden demand on the generator is too big its torque will break the shaft.

kamran
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0

You can image a simply supported beam with links in between, in which $\sum F = 0$. The system is in equilibrium provides the input/applied torque produces an equal but opposite reaction (output work).

enter image description here

r13
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