any mechanical engieers that explain how entropy is used to measure steam engine performance?
greg - Philadelphia & Reading / Reading
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
gregci believe entropy is actually a measure of energy that cannot be used to generate work
Entropy simplified
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
OvermodYou asked what it was used for
did I ask what it was used for?
gregchow entropy is used to measure steam engine performance?
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
Convicted OneGreg, you might find this 1889 book on the thermodynamics of the steam engine 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).
Erik_Magthe 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.
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.
Convicted One gregc i believe entropy is actually a measure of energy that cannot be used to generate work Entropy simplified
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.
gregcI believe the entropy of the steam changes as it expands while in the cylinder.
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...
Overmod but I think the effect of higher loss in long expansion may be associated more with phase-change effect than disorder.
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);
gregcpresumably there are reversible processes that change the ratio of liquid and water vapor that do not change the entropy of the system.
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).
... from a video going thru a numeric 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?
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.
... presumably, comparing the work performed by the cylinder...
Just to be sure: you mean via the piston, right?
versus the change in steam state, the entropy, can account for losses (which have nothing to do with efficiency);
I was under the impression that purely entropic loss in steam expansion was relatively slight and, more specifically, couldn't describe any 'reversible' process by definition.
What might help would be to see if you can find some published data for high-pressure steam used in the first expansion stage of a marine engine, where at least nominally the pressure and superheat may give a closer approximation to an ideal gas, and see what the equations produce. I agree with you that anything that "shows" no entropic change when actual nonreversible work is extracted is set up wrong somehow.
the major processes in steam engines are certainly not reversible. irreversible processes always increase the total entropy.
OvermodI was under the impression that purely entropic loss in steam expansion was relatively slight and, more specifically, couldn't describe any 'reversible' process by definition.
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.
OvermodGive 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.
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.
gregc but i don't understand why the entropy remains the same if work was performed. what am i missing?
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?
gregcIf entropy is a measure of disorder, an inability to do work, why does it increase?
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'.
Overmod gregc If entropy is a measure of disorder, an inability to do work, why does it increase? Because, according to the 'laws' of thermodynamics, disorder tends to increase. In other words ... it was defined that way.
gregc If entropy is a measure of disorder, an inability to do work, why does it increase?
of course
shouldn't entropy go up if work is performed?
gregcshouldn't entropy go up if work is performed?
Yes, I think so.
In addition, I have always thought 'naively' that work extracted from a thermodynamic system 'acts' just like other sources of irremediable loss (it being unimportant whether the change is 'reversible' until it is actually reversed in practice) so you could treat it as part of the 'change in ability to do further work' that the idea of 'entropy' encompasses for theory.
So with that extension you would always find entropy 'increasing' when work is abstracted from the system. Whether it 'would' decrease if, say, you subsequently did work on the piston to recompress the steam is interesting (as some descriptions of what entropy 'describes' about the physics of the steam involve irreversible changes rather than reversible effects) but operationally this would only come into discussions of things like post-exhaust-cutoff compression.
If you have not read or worked through the old ICS course material on 'steam', see if you can find it. There are some practical issues in discussion there that I don't remember seeing 'as such' in more sophisticated and theoretical books on scientific and even engineering thermo.
OvermodSo with that extension you would always find entropy 'increasing' when work is abstracted from the system. Whether it 'would' decrease if, say, you subsequently did work on the piston to recompress the steam is interesting ... but operationally this would only come into discussions of things like post-exhaust-cutoff compression.
but operationally this would only come into discussions of things like post-exhaust-cutoff compression.
in a classic example from Halliday and Resnick, a gas in a compressed and thermally insulated container is released to fill a similar adjacent container. the pressure obviously drops and the ability of perform work decreases (hence entropy increases, right?).
but if the volume of the adjacent contained is then reduced (piston) and forced entirely back into the original container, the state is reversed and the ability to do work restored (entropy decreases, right?)
isn't allowing the gas to enter the adjacent container analogous to steam entering the cylinder. But unlike the example described above, isn't the steam performing work by moving the piston, (ignoring cutoff) equivalent to the slight decrease in pressure (ignoring heat loss) due to the increased volume of the steam, the entire boiler volume + cylinder volume?
of course exhauting the steam from the cylinder further reduces pressure (increases entropy?) and wasting the remaining ability of the steam to do work (hence compounded engines)?
Greg,
Think of gas in in a piston/cylinder system as a spring. When the cylinder is filled with high pressure gas, that is the equivalent of a compressed spring. With an ideal spring, all of the energy (work) used to compress the spring can be recovered in letting the spring decompress under load - in this case, the entropy of the spring has not changed. In the real world, the spring will not return all of the energy used to compress it and entropy will increase.
In an ideal heat engine (Carnot cycle), the entropy of the working fluid (gas) does not change in either the compression part of the cycle or the expansion part of the cycle. The entropy does change in the heating portion of the cycle and cooling portion of the cycle.
For the compression and exansion portions of real heat engines (e.g. gas turbines), you will come across references to isentropic efficiency. For turbines, this refers to how much work can be extracted with a real turbine as compared to an ideal turbine - large turbines can have 95% isentropic efficiencies. For compressors, the efficiency is how much work is needed to compress the fluid in an ideal compressor divided by the work needed to get the same pressure increase with a real compressor. Typical result is that the compressed fluid from a real compressor is warmer than what it would be from an ideal compressor.
For our next lesson, we will get into negative temperatures. This actually exists when dealing with nuclear or electron spins in magnetic resonance.
As it happens we used an earlier edition of Halliday and Resnick in the introductory physics courses, so this is a somewhat familiar and nostalgic thing...
gregcisn't allowing the gas to enter the adjacent container analogous to steam entering the cylinder.
Not quite... the moment you admit steam to the cylinder, you tacitly include some part of the admission tract, potentially right back up through the elements to the dry pipe and boiler, in the 'closed system' that has to be considered.
This could be 'approximated' as a transfer between volumes, as in the example, with the initial 'container' holding the admission charge in the admission-cutoff volume, and the 'combined' container volume being that of the final stage. This removes any potential confusion with steam admission, wire-drawing, throttling or shrouding, or other problems inherent in real-world 'admission to the cylinder', but it neatly preserves 100% of where the 'work done by the piston' is concerned in the expansion.
If the steam engine were frictionless (in other words there were no machine losses) and were, say, being tested in the air or on a test plant so that the wheels were free to revolve, you'd have the effect of the 'free expansion' as the piston moved from one point in the cylinder to another. It would of course do this 'spontaneously' (as the steam "followed the laws of thermodynamics" to reach a point where the free paths of molecules are longer, density is less, etc. etc. etc.) and, as in the example, putting the steam back into its original volume is 'reversible' (given the assumptions).
Note that there is no net 'work' done here; more precisely no energy has been abstracted from the overall 'closed system' that is the thermodynamic reference frame in the example.
But unlike the example described above, isn't the steam performing work by moving the piston, (ignoring cutoff) equivalent to the slight decrease in pressure (ignoring heat loss) due to the increased volume of the steam, the entire boiler volume + cylinder volume?
Theoretically you clould do the calculation using the 'whole boiler volume' but it would be much more complex to analyze, and many of the factors involved necessarily involve empirically-determined factors and constants (see Fry's boiler discussion from 1922 and look at the derivation of the proportionality factors and constants, for example) so I'd model the system conservatively, if not as above in some way that eliminates the effect of admission shock or throttling losses that would be difficult if not impossible to reverse in practice.
...of course exhausting the steam from the cylinder further reduces pressure (increases entropy?)...
In my opinion, the whole portion of expansion where the exhaust port is open is uncharacteristic of any reversible process for meaningful engineering analysis, and should not be considered, as I think you're suggesting, as though it were.
Now, what happens to the mass of steam in the cylinder after the instant of effective exhaust valve CLOSURE is far more interesting, as it now constitutes a reversible process just as above that does demonstrable practical work at a very important point of a reciprocating locomotive's cycle. In my opinion there is value in modulating compression in such a way that the cylinder pressure just at the moment the intake valve opens is a reasonable high percentage of admission pressure, which among other things limits wiredrawing losses and greatly mitigates issues with nominal dead space ... this can be studied in conjunction with reversible compression effects net of functional 'cushioning' of inertial rod force.
... and wasting the remaining ability of the steam to do work (hence compounded engines)?
Erik_MagFor our next lesson, we will get into negative temperatures. This actually exists when dealing with nuclear or electron spins in magnetic resonance.
I want to see what happens when we get into temperatures measured in volts*. That should be good for head-scratchin'!
*Well, technically in electron-volts ... but the volt part has comparable sense...
not very specific .... baffling
Erik_MagThink of gas in in a piston/cylinder system as a spring.
gregcbut if the volume of the adjacent contained is then reduced (piston) and forced entirely back into the original container, the state is reversed and the ability to do work restored (entropy decreases, right?)
isn't that what i said?
Meanwhile, in the real world...
gregc Erik_Mag Think of gas in in a piston/cylinder system as a spring. gregc but if the volume of the adjacent contained is then reduced (piston) and forced entirely back into the original container, the state is reversed and the ability to do work restored (entropy decreases, right?) isn't that what i said?
Erik_Mag Think of gas in in a piston/cylinder system as a spring.
gregc but if the volume of the adjacent contained is then reduced (piston) and forced entirely back into the original container, the state is reversed and the ability to do work restored (entropy decreases, right?)
Yes and no.
The no part: If the gas in the cylinder goes through reversible adiabatic expansion or compression, then there will be no change in entropy. A change in entropy would come from deviations from reversible (e.g. friction) and heat transfer in or out of the cylinder. You might reduce entropy with sufficient heat loss from the cylinder, but that involves throwing away energy that could be use to perform useful work. Given initial gas temperature and pressure, and a final temperature or pressure, the isentropic line defines the maximum amount of work that can be provided by the expansion of gas.
The exchange of thermal energy does involve changing the entropy of the working fluid.
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