any mechanical engieers that explain how entropy is used to measure steam engine performance?
The best English explanation is this: “Themodynamics” of a steam locomotive is done in terms of heat. But heat doesn’t produce TE via piston thrust; pressure does. Entropy can be thought of as a measure of ‘heat’s ability to be turned into pressure that actually does work’.
i believe entropy is actually a measure of energy that cannot be used to generate work
You asked what it was used for, not how it was defined. I’m sure you read the definition and didn’t understand before you asked.
you continue to baffel me
did I ask what it was used for?
Greg, you might find this 1889 book on the thermodynamics of the steam engine interesting.
Survivor Library Starting on page 17 is a discussion ofentropy as it pertains to piston steam engines.
Later in the book I found the section pertaining to super heat particularly interesting
I came across the Peabody book in my searches, but was reluctant to read yet another book that deals with a broad subject qualitatively rather that quantitatively. The Semmens book provides some details but they appear superficial. He interjects historical information in his technical discussions. I can’t find the same values from the steam tables used in his example and don’t understand why his enthalpy values for intake and exhaust steam have the same entropy values. If work is done, I would expect entropy to change.
i had thermo in school and believe someone more practiced can clearly explain how to select values from the steam table for a specific case and why.
i’ll give the Peabody book another look.
thanks
I remember having trouble figuring out how much energy could be derived from steam at a given temperature and pressure whenlooking at the steam tables in the CRC Handbook - this was my freshmen year. Wasn’t until taking an ME thermo class my senior year (BTW I was a EECS major) that the trick to the steam tables was following the line of constant entropy between the steam conditions at the throttle output (i.e. temp and pressure) and the steam conditions after leaving the engine or turbine and noting the difference in enthalpy. Useful work will always be less than given by an isentropic process (constamt entropy).
i think there are two modes to consider
i think what you’ve described above is the case where steam constantly enters the cyclinder, pushing the cylinder without losing pressure (ideally) and exits the cylinder at the end of the cycle. I believe in this case there is no change in the ability of the steam to do work (entropy). It’s just all wasted when it leaves the cylinder
the 2nd mode is when cutoff limits the amount of steam entering the cylinder and allowed to expand for the remainder of the cylinder cycle. i believe the entropy of the steam changes as it expands while in the cylinder.
from Entropy simplified
"Perhaps it is best defined as a non-negative thermodynamic property, which represents a part of energy of a system that cannot be converted into useful work.
I believe you are correct, but I think the effect of higher loss in long expansion may be associated more with phase-change effect than disorder. Most of the thermo discussions are written for ideal gases and leave out things like nucleate condensation that cause pronounced pressure drop while retaining the ‘latent heat’ needed to get water up to boiling point.
Personally I have always thought that entropy was defined as ‘increasing disorder’ to get other variables to have positive sign. “Work” to a themodynamicist is negative (extracted from the closed system of reference) but positive to a locomotive designer. You could always formulate S as percentage and define (1-S) as the remaining ability to ‘do work’ if the formality bothers you. That’s not really any more brain-hurting than using ‘negative’ in discussing where electrons come from in DC circuits…
presumably there are reversible processes that change the ratio of liquid and water vapor that do not change the entropy of the system.
from a video going thru a numberic example i was reminded how entropy is used to determine the change in enthalpy (btu/lb) from superheated (purely vapor) and saturated steam (mix of liquid and vapor) steam.
but i don’t understand why the entropy remains the same if work was performed. what am i missing?
presumably comparing the work performed by the cyclinder versus the change in steam state, the entropy, can account for losses (which have nothing to do with efficiency);
There are, but not in a ‘system’ that represents cylinder expansion to provide thrust. Most ‘reversible’ phase transitions involve some external manipulation of the system to remain isentropic – this may be tacit in the way the measurement is set up.
Everything I learned says that when work is extracted, the entropy of the system increases (I believe this is inherent in both the first and second “Laws” of thermodynamics).
Give us the link to the video. My expectation is that no net work was done by/on the system as the enthalpy changed if the entropy (which is a different measure from enthalpy by intent) did not.
Now, there are other effects of reversible phase change that do not work to the advantage of steam-engine designers than nucleate condensation (note this is different from ‘wall condensation’ effects; it is purely within the steam mass out of contact with any surface). One of these is that much of the condensation – both wall and nucleate – will flash back to vapor starting at the moment of exhaust release; the resulting increase in volume creates isostatic pressure so very little of it contributes to development of net exhaust flow. This doesn’t provide ‘work’ (in terms of energy that could be coupled to combustion gas in the front end) but is just a reversible phase effect that happens to make detail design difficult.
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the major processes in steam engines are certainly not reversible. irreversible processes always increase the total entropy.
entropic loss – my understanding is entropy does not go down either due to natural processes or when work is performed
as i said earlier, i believe expansion of steam in a cylinder involves a change in entropy of the steam while in the cyclinder, while if there was no expansion, the entropy changes as steam is exhausted and wasted.
no change in entropy implies no work done, right?
(Patting self on back) I’m getting better at spotting these pizzing contests in the making, and just thowing out a helpful reference and then steering clear.[^o)]
If the work was being done in a reversible manner, such as exanding gas in a frictionless and perfectly insulated cylinder, then there is no change in entropy. Same thing will happen when steam condenses in the ideal cylinder, the condensation will evaporate if the compression is done slowly enough.
You’re definitely on the right track when considering irreversible processes, which include friction, heat loss through walls, etc. The isentropic case is the ideal case, the real case is going to be worse.
The equivalent challenge for studying EE stuff is understanding the difference between and ideal voltage source and a real voltage source.
i believe i’m wrong in interpretating the entropy value in a steam table.
the Engineering Toolbox - 2nd Law of Thermodynamics page shows examples of entropy calculations using change in enthalpy values divided by the average temperature.
i am still confused by the sign of the change in entropy. If entropy is a measure of disorder, an inability to do work, why does it increase?
and what does the steam table entropy value mean?
Because, according to the ‘laws’ of thermodynamics, disorder tends to increase. In other words … it was defined that way.
It may help to think of this by looking at how some Ford designs implement their windshield-wiper delay. “Normal” haptics have speed-control and similar knobs arranged so that turning them to the right ‘turns up the volume’. On some Ford cars, turning right increases not the speed of the wipers, but the length of the delay; in other words, ‘right’ is ‘slower’, which is utterly counterintuitive unless you define ‘more delay’ as what you’re turning up … which is what a windshield-wiper ‘delay’ control could be said to be ‘increasing’.
In thermodynamics, ‘entropy’ (and some other things) were arranged so that ‘increasing disorder’ would be positive. This results in ‘less useful heat’ turning out positive. That does seem as counterintuitive as turning a knob to the right to decrease speed … but it’s how it was defined by Clausius et al., and it’s the convention that ‘everyone has agreed to use’.
of course
shouldn’t entropy go up if work is performed?