Energy conversion efficiency of the steam engine.

Question

The path followed by water in steam engines can be described as the following:

From A to B: Liquid water is compressed in pump, it receives a work $ W_{A \rightarrow B} $.

From B to C: Water is heated in a boiler where it receives heat $ Q $ without any work, and from where it gets out as vapor.

From C to D: In the turbine, water gives the amount of work $ W_{C \rightarrow D} $ without gaining or losing heat.

From D to A: Water is cooled in a condenser, without work transfer, where it goes back to its original state and properties, to go the pump again to get compressed and so on...

Q: What is the energy conversion efficiency $ \eta $ of the steam engine?

We know that it is the ratio of the useful output energy and the input energy.

In this case, the output energy is the work $ W_{C \rightarrow D} $ done by the water during the phase $ C \rightarrow D $ , and the input is the work of compression $ W_{A \rightarrow B} $ and the heat Q.

In the solution of the problem, the energy conversion efficiency is expressed as: $$ \eta = \frac{W_{A \rightarrow B} + W_{C \rightarrow D} }{Q} $$

Can you explain to me why it is not: $ \eta = \frac{W_{C \rightarrow D} }{Q + W_{A \rightarrow B} } $ ?

Thank you.

Answer 1

The first thing to realize is that W_AB is a negative work, so putting it in the numerator correctly reduces the efficiency, but putting in the denominator would incorrectly increase the efficiency. Maybe you want to think of it as a positive number and put it in the denominator, but that's just not how efficiency is defined, it is defined as the net work obtained (so the sum of the positive and negative W in your example), divided by the heat extracted, Q.

Answer 2

A engine is a device which converts some form of energy into work. So you get energy from diesel (input) and move your car (work).

In steam engine, input is heat it gets from boiler. Q in denominator should be clear enough.

Thinking about work is more tricky. Think about how we propel ourselves while swimming. When we move our hands back we try to push as much water as possible. This is the positive work. But we need to bring our hand back in front to complete the cycle. We do it gently without pushing water (by bringing hands outside water). If we apply equal push on water while bringing our hand in forward direction, net work will be zero and we not move!

Now imagine yourself inside the steam engine. You push the turbine with great force ($W_{CD}$). But the problem is we end up with low pressure steam. So it has to be cooled (DA) and compressed again (gently) (AB) to complete the cycle. Negative work done in this phase is added to work. (actually subtracted due to minus sign).

It would be easier if you imagine the process starting from B and finishing at A. So process BC will be the first step.

Answer 3

Old fashioned Carnot cycle efficiency... and Thermodynamic conversion... Chemical energy (Wood/coal/oil/gas) ---> Thermal energy (Heat) ---> *boils water conducting heat thru another fluid medium (Heat)---> Boiler pressurized steam thru piston (Mechanical energy) --->energy lost in transmission conversion