I need to calculate the efficiency of a Carnot machine, in which, some of its extreme points are in the region of liquid transition to steam with the works, differences of energy and heats of each section. I'm not very sure how start this, so any help is welcome!

  • $\begingroup$ Can you describe your particular Carnot cycle in more detail? $\endgroup$ – Chet Miller Oct 8 '18 at 0:45
  • $\begingroup$ You are asking questions about a Carnot cycle as a two phase steam power cycle. I suggest you google "the Carnot cycle as a two phase power cycle" and click on the mit.edu hit. Then, if you still don't understand it, come on back. Good luck. $\endgroup$ – Bob D Oct 8 '18 at 3:33

A Carnot engine is defined as a heat engine process that absorbs heat isothermally at two temperatures, $T_{1}>T_{2}$, with the transitions between those two temperatures being adiabatic (so that no heat is exchanged). The efficiency $\eta$ is defined to be the fraction of the heat $Q_{1}$ the system takes in while at $T_{1}$ that is converted to work $W$; that is $\eta=W/Q_{1}$. By energy conservation (the First Law of Thermodynamics), the total heat taken in at $T_{1}$ must be equal to the work $W$ plus the heat expelled at the lower temperature $T_{2}$. (Remember that while the system is transitioning between the two temperatures, no heat flows in or out.) Therefore $Q_{1}=W+Q_{2}$, and the efficiency can be related to the ratio of heat flows $\eta=1-Q_{2}/Q_{1}$.

What is particularly amazing about a Carnot engine is that this efficiency can be expressed in terms of just the temperatures $T_{1}$ and $T_{2}$, rather than the exchanged heats $Q_{1}$ and $Q_{2}$. This is very helpful because it is generally much easier to measure the temperature at which the system is operating than the heat transfer. This allows you to avoid the difficulties of dealing with the equation of state, provided you know the upper and lower operating temperatures. (There is a complication, since the equation of state also determines what the adiabatic paths between $T_{1}$ and $T_{2}$ are, but there are ways to make the paths very nearly adiabatic without detailed calculations; the system can be thermally isolated, so that no heat transfer is possible; or the adiabatic paths may be traversed very quickly, so that there is no time for appreciable heat transfer.)

For an isothermal process, the heat exchanged is (by the Second Law of Thermodynamics) $\Delta Q=T\Delta S$, in terms of the entropy $S$. No entropy enters of leaves the system except during the isothermal processes at $T_{1}$ and $T_{2}$; and no entropy is ever generated, since the whole Carnot cycle is reversible. Therefore, the entropy changes $\Delta S_{1}$ and $\Delta S_{1}$ appearing in $Q_{1}=T_{1}\Delta S_{1}$ and $Q_{2}=T_{2}\Delta S_{2}$ are equal: $\Delta S_{1}=\Delta S_{2}$. An amount of entropy is absorbed by the system at the high temperature $T_{1}$, and an equal amount is expelled at the low temperature $T_{2}$. This allows us to express the key quantity $Q_{2}/Q_{1}$ that appears in the efficiency in terms of the relatively easily measured temperatures; and the final formula is $\eta=1-T_{2}/T_{1}$. The efficiency of the Carnot engine depends only on its two operating temperatures, not on anything else.

| cite | improve this answer | |

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.