# Watt's Governor - Equation

• phiby
In summary, the equation is used to calculate the torque delivered by the engine in response to a change in load, and it's based on the assumption that the torque produced by the engine is proportional to its RPM.

#### phiby

I was reading up on the mathematical modelling of Watt's governor from Brennan's Book on Differential Equations.

The book has 2 equations one for the governor itself which I understand.
However, there is an equation for the steam engine flywall & Drive shaft which I don't understand. This is the equation

J * dΩ/dt = -β* Ω + τ

This is the text following the equation
J is the moment of Inertia of the flywheel
Ω - rotational speed of the engine
First term on the right is the torque due to the load & the second term on the right is the steam generated torque referred to the drive shaft.

What exactly is β in the equation? What's the unit of β?

All the terms in the equation must be in the same units.

$\tau$ is obviously a torque. So is $J\, d\Omega/dt$.

So $\beta\, \Omega$ must also be a torque.

$\Omega$ is the rotational velocity, so $\beta$ is a measure of the "size" of the engine, in other words how much torque it produces at 1 rad/sec.

If $\beta$ is a constant, your equation is assuming that the torque produced by the engine is proportional to its RPM. That seems a strange assumption to make for a steam engine, but then your book is about differential equations, not engineering.

AlephZero said:
All the terms in the equation must be in the same units.

$\tau$ is obviously a torque. So is $J\, d\Omega/dt$.

So $\beta\, \Omega$ must also be a torque.

$\Omega$ is the rotational velocity, so $\beta$ is a measure of the "size" of the engine, in other words how much torque it produces at 1 rad/sec.

If $\beta$ is a constant, your equation is assuming that the torque produced by the engine is proportional to its RPM. That seems a strange assumption to make for a steam engine, but then your book is about differential equations, not engineering.

I don't think β is a measure of the size of the engine. This is because of 2 reasons
1) βω part is preceded by -minus sign - if it was the torque produced by the engine, it would be positive in the equation, not negative
2) Also the text I quoted in my original post says "First term on the right is the torque due to the load" - i.e. it's the negative torque which acts on the system because of the load driven by the engine. The quoted text also says that τ is the torque produced by the engine.

EDIT:
Looking a few more places, it seems that β is load damping co-efficient (unit N-m/rad/s).
However, it's not clear what exactly this is.

Last edited:
I took "torque due to the load" in your OP as meaning the torque required to drive the load. That is not the same as the "torque delivered by the engine" if the speed is changing.

With that interpretation, one sign on the right hand side should be + and the other -, as they are in your equation, though I did think they were using a strange sign convention!

But OK, if you are now saying $\tau$ is the NET torque delivered by the engine, (i.e. the torque acting on the flywheel, which is the difference between the total engine torque and the torque that is driving the load) then it would make sense if $\beta\, \Omega$ was a "damping" force proportional to speed.

In "real life" there would be many causes of the energy lost from the system, for example friction viscous forces in the oil lubrication, air resistance to the moving parts, etc. Lumping all those things together and assuming they produce a "torque" proportional to linear or angular velocity is a good way to make the math fairly simple, but you can't easily relate it to the detailed physics of what is going on.

But this is often used as a modelling technique, because (as you will probably see when your book discusses what the solution to the equation looks like) you can find the value of beta for a particular system by measuring how the system responds to a sudden change in load, for example, and that is much easier than trying to model all the "real physics" involved.

This modelling technique of including some "general" parameters (like $\beta$ in your example) in the math model to represent the "global" effect of a lot of details is often used, and the process of finding numerical values for the parameters from the behaviour of the real-world system is called "system identification". You can also use this process the other way round: play with the math to find a "good" value of $\beta$ to get the response you want, and then go back to the physics and design something to produce that value of $\beta$ in real life.

Last edited:

## 1. What is Watt's Governor?

Watt's Governor is a type of centrifugal governor used to regulate the speed of steam engines. It was invented by James Watt in the late 18th century and is still used in various forms today.

## 2. How does Watt's Governor work?

Watt's Governor works by utilizing centrifugal force to regulate the flow of steam to the engine. As the engine speeds up, the centrifugal force increases, causing the governor balls to move outwards and close the valve. This reduces the flow of steam and slows down the engine. Conversely, as the engine slows down, the centrifugal force decreases, allowing the balls to drop and open the valve, increasing the flow of steam and speeding up the engine.

## 3. What is the equation for Watt's Governor?

The equation for Watt's Governor is F = mω²r, where F is the centrifugal force, m is the mass of the governor balls, ω is the angular velocity, and r is the distance from the center of rotation to the center of mass of the governor balls.

## 4. What are the advantages of using Watt's Governor?

Watt's Governor is a simple and effective way to regulate the speed of steam engines. It is also self-governing, meaning it does not require external power to function. Additionally, it is relatively low-cost and easy to maintain.

## 5. Are there any limitations to Watt's Governor?

While Watt's Governor is a reliable speed regulator, it does have some limitations. It is primarily designed for steady-state operation and may not respond quickly enough to sudden changes in speed. It also requires careful adjustment to ensure optimal performance, which can be time-consuming and labor-intensive.