0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Where are the terms which account for the charging up of the system to operating temperature requirement that cannot be recovered, thus reducing efficiency? $\endgroup$
    – RaSullivan
    Jan 6 '17 at 4:55
  • $\begingroup$ Can you please explain to me what is the "net" work ? Thank you. $\endgroup$
    – user141245
    Jan 6 '17 at 21:07
  • $\begingroup$ @RaSullivan-- I don't understand your question, the heat extracted Q is what maintains the system at the higher temperature, and there is also heat deposited in the lower temperature that we don't need to explicitly mention because we can always find it from energy conservation. $\endgroup$
    – Ken G
    Jan 7 '17 at 18:06
  • $\begingroup$ @Jacob-- net work just means there are two phases when work occurs, but in one of the phases, the work done is negative (the A to B phase), and in the other, it is positive (and larger). Thus the net work is the latter plus the former (where the former is negative so subtracts from the net work). "Net" just means total. $\endgroup$
    – Ken G
    Jan 7 '17 at 18:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy