Calculating Work Out for a Sterling Engine

[ofathens]

I was reading a list of science experiment ideas, and came across one that intrigued me: calculating the efficiency of a sterling engine.

We can say that the efficiency n_th is:

n_th = W_out / Q_in

Assuming we are heating the engine with a beaker of hot water,

Q_in = m * c * \DeltaT

But how would we calculate work out? If the engine is making a wheel spin, as most sterling engines do, then we could calculate the work out as a rotational analog of the definition of work:

W_out = \tau\Delta\theta

Where \tau is the torque on the wheel, which we could substitute with

\tau = \alphaI

Where \alpha is the angular acceleration, and I is the moment of inertia of the wheel.

Thus, we must be able to calculate the angular acceleration of the wheel, as well as its moment of inertia. Is this the easiest way to do this? How else could we measure the effieciency of a sterling engine?

 

[Lojzek]

Calculating work put into kinetic energy of a wheel would almost certainly give very poor results, because:

1. You need a long time to measure input heat accurately (you need to asume that input heat is much larger than the change of internal energy of the engine). However you can't accelerate a wheel with the full engine's power for a long time, because of the friction and air resistance.

2. Moving parts inside the engine are also accelerating, so wheel does not get the full power.

I would rather chose this method:

Let the engine propel the wheel that has very little friction, but a rope carriing a weight with mass m is wrapped several times around the wheel. On one side rope should be hanging free (but strained with weigth force), on the other side it should be fixed. If the rope is wrapped so many times, that the force on the fixed point is negligible, then the force sloving down the wheel is:

F=m*g

And power is: P=m*g*v=m*g*2*Pi*r*rotation frequency

You would probably want to maximize power by selecting a proper weight mass (since the rotation frequency will be dependent of the brake force)