# Doubt on a shaft-turbine-generator system

Consider a shaft which connects a turbine to a generator. Initially, none of the parts are moving. Now the turbine starts to rotate and the torques act on the shaft from both the turbine and generator side. I'm interested in knowing, what happens to the torques acting on the shaft from both ends and the angular speed of rotation of the shaft, from the start of the turbine until some angular speed $$\omega$$ is reached. It makes sense to me that as the turbine starts rotating it applies a torque on the shaft from the turbine side, and because of some net unbalanced torque the speed of the shaft starts to increase. It is my guess that the shaft starts to rotate at some constant angular speed when an equal and opposite torque is developed from the generator side. I do not understand how the generator applies this torque? why the torque from the generator side increases along with the torque from the turbine side?

• Consider checking: electrical load, air resistance, bearing friction... Dec 8, 2021 at 16:24
• It's not simply a matter of speeding up until the generator torque matches the turbine torque. The speed of rotation is controlled by the turbine throttle valve, which modulates to allow more or less steam to flow through the turbine. During startup the resisting torque of the generator is fairly small because little - if any - electrical power is generated. No significant generation occurs until operating speed is reached, because it can't be synched to the power grid until it's putting out the required frequency (which is dependent on speed).
– Mark
Dec 9, 2021 at 1:48

I am not entirely certain about your question but I think, it basically boils down to this equation

$$M = I \alpha$$

where:

• M is the torque applied
• I is the mass moment inertia of all rotating masses
• $$\alpha$$ is the angular acceleration $$(\frac{d \omega}{dt})$$

In another form this is written as:

$$\Delta M = I \frac{d \omega}{dt}$$

I.e. the turbine has a load which can be translated to a torque $$M_{turb}$$ which is resisting motion. The generator is producing $$M_{gen}$$.

If:

• $$M_{gen}>M_{turb} \rightarrow M_{gen}-M_{turb}=\Delta M>0$$ then the system accelerates
• $$M_{gen}=M_{turb} \rightarrow M_{gen}-M_{turb}=\Delta M=0$$ then the system does not change its kinetic state.
• $$M_{gen} then the system decelerates

The generator start-up torque demand is the sum of the load on it, if any, its angular momentum, frictions in bearings, cogging of magnets. etc.

When a torque by the turbine is applied to the The generator starts to accelerate until the RPM of the generator is at the systems design RPM. At design RPPM the generator power is $$P= 2πNT/60$$

• N is rpm
• T= torque in Nm
• P =power in watts

So the torque of the shaft initially goes to accelerate the generator and later to have it produce power turning at a constant speed, $$\omega$$.

At all times the torque at the two ends of the shaft is equal and opposite.

## Edit

After OP's comment

Shaft unbalanced torque is a major subject in power generation plants.

At the most basic level, one can compare it to SDF torsional vibration. In reality, it is a continuous mass multi-node force function with tapering or stepping shaft diameter.

In the SDF form with damping the equation simplifies to

$$I \frac{d^2\theta}{d^2t} +C\frac{d\theta}{dt} + k\theta = \tau(t)$$

At low damping ratios frequency of the system is very near the natural frequency

$$f_n= \frac{1}{2\pi}\sqrt{\frac{k}{I}}$$.

' • Thank You. How the torque on the shaft from the generator, increases along with the torque from the turbine. Imagine the shaft is rotating at some, $\omega$, and at this time the torques are T on both ends of the shaft. The torque from turbine increases by some amount, there is a disbalance of torque on shaft, and hence it's speed increases. Is it that as speed increases the torque from generator end starts to increase? Until it matches with the torque from turbine? Dec 8, 2021 at 17:21
• In the real world, the sudden increase of torque at one end creates an imbalance that twists the shaft momentarily like a spring, and then the imbalance is absorbed by the other end. But in most engineering applications, the shaft is assumed rigid. Meaning any perturbation at one end is instantly transmitted to the other. Dec 8, 2021 at 17:34
• Sure, until the shaft breaks... Dec 8, 2021 at 17:36