Erik_MagIf the gas in the cylinder goes through reversible adiabatic expansion or compression, then there will be no change in entropy.
after re-re-reading Halliday and Resnick, expansion in an thermally isolated container (T is constant) is irreversible and heat needs to be added in a reversible process to re-compress the gas and lower the entropy. Both expansion and compression change the entropy.
see no point in discussing reversible processes regarding a steam locomotive or any practical engine
greg - Philadelphia & Reading / Reading
Backshop Meanwhile, in the real world...
Meanwhile, in the real world...
.
gregcsee no point in discussing reversible processes regarding a steam locomotive or any practical engine
As we keep saying, in one form or another.
The only point of discussing 'reversible' anything is to understand what things might not necessarily exhibit non-increasing entropy. So that you might get a better idea for yourself about how 'entropy' is used in discussing steam-engine output, and not how entropy is used academically in irrelevant examples.
A related issue, which is masked somewhat by the terms that get used, is why many attractive-looking ORC cycles, notoriously including the original ether 'bottoming' cycle for marine compound engines circa 1850, prove out so wretched in practice. Ether has a nifty high vapor pressure and it boils low, so let's bottom the exhaust steam in an 'ether boiler', let it power-expand against a cylinder, and then condense it in cold river water... voila! a better Rankine efficiency!
The problem is that the 'high' pressure drops dramatically on even slight expansion, e.g. when you try moving a piston to do useful work on the paddle or propeller shaft. This is related to the 'heat content' of the ether which is something that only peripherally shows up in 'steam' tables, but is essential in picking a useful working fluid for a practical transportation engine...
gregc after re-re-reading Halliday and Resnick, expansion in an thermally isolated container (T is constant) is irreversible and heat needs to be added in a reversible process to re-compress the gas and lower the entropy. Both expansion and compression change the entropy.
Greg,
You are confusing work (mechanical energy) and heat (thermal energy). Recompressing gas in an adiabatic cylinder requires meachanical work, adding heat will increase the pressure if the gas volume is held constant.
Once the steam is admitted to the cylinder and the steam admission valve is closed, and before the exhaust valve is opened, the action of the expanding steam is reasonably well described as an isentropic process. The actions of the throttle valve, pressure loss in the pipes and valve gear are not reversible - but these are topics in fluid mechanics (look up nozzles and diffusers).
Another way to put it is that entropy puts a hard physical limit on how much work you can get out of steam, and the goal of the ME's designing the engine is to get as close to that limit as reasonably possible.
Big Jim,
Entropy is a scientific term most commonly used in thermodynamics. The science of thermodynamics was originally developed to help improve steam engines and thus most certainly belongs in the Steam and Preservation section.
BigJim Backshop Meanwhile, in the real world... ..."Entropy" sounds like a medical term, so, this thread serves no useful purpose on the "Steam & Preservation" forum!
..."Entropy" sounds like a medical term, so, this thread serves no useful purpose on the "Steam & Preservation" forum!
Erik_MagYou are confusing work (mechanical energy) and heat (thermal energy). Recompressing gas in an adiabatic cylinder requires meachanical work, adding heat will increase the pressure if the gas volume is held constant.
gregcheat needs to be added in a reversible process to re-compress the gas and lower the entropy.
my mistake. heat needs to be removed. from halliday and resnick
Let us now return the gas to ins initial state by a reversible process of slow compression in which we keep the temperature constant at T by removing an amount of heat Q. This reversible process brings the gas back to a pressures p1, volume V1, temperature T and entropy S1 and involves an interaction of the system with its environment. In this process the entropy of the has changed from S2 to S1 in a measurable way; we removed a quantity of heat Q at the temperature T from the gas so that the entropy of the gas decreased by an amount Q / T from S2 to S1. Hence, the entropy of the gas after free expansion is greater than its entropy before free expansion.
i'm focused on this H&R example because i think it clearly describes a change in state and change in value of entropy and how it can be lowered, to restore the gas' ability to do work.
i don't understand why compressing the gas (reducing the volume) doesn't just increase the pressure (PV/T right). Temperature didn't change when it expanded
(while it makes sense work is performed to lower the entropy, increasing the ability of the gas to do work, it's ironic to me that heat needs to be removed. Since neither work was performed nor heat lost during expansion, I wonder if the work performed during compression is the equivalent of the heat removed?)
Erik_MagOnce the steam is admitted to the cylinder and the steam admission valve is closed, and before the exhaust valve is opened, the action of the expanding steam is reasonably well described as an isentropic process.
i understand any thermo dynamic system is complicated. I tried to describe a very simple step
gregcisn'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?
by "ignoring cutoff" i meant the admission valve is not closed until exhaust.
Again, i'm trying to keep things analagous (and of course not iso-whatever) and intentionally ignoring other parts of the system and other causes of heat and mechanical loss to keep things simple because i don't think they are significant to the discussion and detract from it.
i am aware that after cutoff steam expands and pressure decrease due to the expansion within the cylinder.
and in both cases, i believe this is analagous to the Halliday & Resnick example of an expanding gas in a thermally isolated container (of course the cylinder isn't) ...
... but the difference is that in the cylinder, work is performed during the expansion and i believe entropy increases.
of course, the steam goes thru a final state change during exhaust. I see no point in following the steam after it leaves the cylinder
while even above, if clearly stated and understood, is interesting, the significance is H&R makes entropy calculations which quantify it's change. In the video i posted on Apr 24, 12:12, when using the steam table, the steam table entropy value (Joule/lb*T) is kept constant !
My understanding is this is not the entropy of the steam (even in terms if 1/lb) and even though it's value is kept constant in the video example, my understanding is that it does not imply that there was no change in entropy (wasn't work performed?). I don't understand the significance of that value in the steam tables
i'm interested in this to better understand the value in the Stemmens book
gregci don't understand why compressing the gas (reducing the volume) doesn't just increase the pressure (PV/T right). Temperature didn't change when it expanded
Go back to the H&R example again. In real-world expansions the temperature ALWAYS falls when the pressure does, just as the temperature and pressure both increase when 'compression' work is done naively. You have to cheat to get the 'reversibility' to completely reverse. I suspect the example will say something like that the sample was very, very slowly expanded. Artificial tinkering so that PV=nRT only shows reciprocal change between terms on the left side of the equation. For demonstration purposes; theoretical-demonstration purposes...
Overmod Go back to the H&R example again. In real-world expansions the temperature ALWAYS falls when the pressure does, just as the temperature and pressure both increase when 'compression' work is done naively. You have to cheat to get the 'reversibility' to completely reverse.
Go back to the H&R example again. In real-world expansions the temperature ALWAYS falls when the pressure does, just as the temperature and pressure both increase when 'compression' work is done naively. You have to cheat to get the 'reversibility' to completely reverse.
Caveat to the underlined text is that is true when the gas does work (e.g. pushing a piston or driving a turbine) when expanding. I distinct remember my freshman chem prof's discussion of gas laws explaining that hydrogen gas can actually warm up when expanding through an orifice. This was presumably due to heating from gas flow equivalent of friction.
I suspect the example will say something like that the sample was very, very slowly expanded.
I would expect a more rigorous qualification is that the expansion is much slower than the sonic velocity of the gas (which is a function of temperature). The other critical item is that there is no heat transfer to the cylinder or piston. The latter is from a gas spring demonstration with thin aluminum sheets were placed in the gas to increase damping, which is an irreversible process.
H&R??? The text for my statistical mechanics class (Physics 4E) used Volume 5 of the Berkeley Physics Course (go Bears!). One of the more amusing memories was working through a couple of problems with nuclear quadrupole interactions - then 20 years later doing work on nuclear quadrupole resonance...
Erik_MagI distinct remember my freshman chem prof's discussion of gas laws explaining that hydrogen gas can actually warm up when expanding through an orifice.
That's not 'ideal gas law' assumptions, though; that's Joule-Thomson inversion which for hydrogen is somewhere around 200K which is not a 'real-world' temperature in most situations.
Your mentioning Berkeley reminded me that this was the base text I wish I could remember the volume offhand) for relativity we used in 'honors' spring freshman physics (106) instead of H&R. Remarkable how different those two were in how they went about explaining things.
Hydrogen was the one exception to general rule about gases cooling off with expansion. IIRC, making the gas do a bit of work will cool the gas more than just letting it expand willy nilly.
The Berkeley Physics Course volumes were: 1 Mechanics (Phys 4A); 2 E&M (Phys 4B); 3 Waves (Phys 4C); 4 Relativistic and Quantum Mechanics (Phys 4D); 5 Statistical Mechanics (Phys 4E). My prof for 4D used Tipler as he thought it was much better than the BPC book. I started with 4A the winter quarter of my freshman year, and the UCB Physics department switched to Physics 5A-E using H&R the fall quarter of my sophomore year, with the last Phys 4E classes taught the fall quarter of my Junior year - had one friend who took that session as he was interning spring quarter of our sophomore year.
One fun memory from about that time was one quarter I would be heading back to the dorm from a class and almost invariably pass by Glenn Seabord heading down to the campus from LBL.
Oh, gee. This thread is out-of-control.
The way you do calculations based on thermodynamics based on steam tables is that you decide on a thermodynamic variable that is conserved.
For example, if you have a "process" (such as expansion of steam in a cylinder), you figure out what kind it is. Such as "adiabatic-isentropic." Adiabatic means that you are not allowing heat to be transfered either into or out of the cylinder. Isentropic means that the process is "reversible." An example of a reversible process is an expansion steam in a cylinder where if you reversed the expansion into a compression, whatever work was done during the expansion is applied to the piston during the compression, and the "state" of the steam (temperature, pressure, volume, etc.) at the end of the compression is exactly where you left off.
A reversible process generally doesn't involve friction, expansion of a gas by just venting it instead of having it do work against an expander, and so on.
So what does this have to do with anything. If you start with the a steam engine with a certain cutoff, you start with the cylinder right at cutoff and then expand it. So you know the volume and temperature of the steam at the start of the expansion phase. You know the volume at the end of the expansion phase because you know the cutoff as a fraction of the cylinder volume. So how do you know the end state of temperature and pressure and steam "quality" (percent that is condensed or perhaps amount of superheat remaining if not condensed?
Because this is an undergrad engineering homework problem, or maybe a steam locomotive enthusiast trying to get a handle on the upper limits of steam locomotive thermal efficiency, you assume the expansion is adiabatic-reversible, that is, adiabatic-isentropic. So what state variable is conserved?
Entropy!
You don't have to get into all of these angels dancing on the head-of-a-pin arguments, the state variable is entropy-end-of-story-and-you-don't-need-to-know-what it means. You know the initial entropy of the steam charge from the boiler conditions from its temperature and pressure filling the cylinder along with its specific volume. At the end of the expansion you know its final specific volume, and you know its final entropy -- same as its initial -- from the adiabatic-isentropic assumption. From its final volume and entropy, you know the final temperature and pressure. From the difference between initial and final enthalpy (gosh, I love that word!), you find the work done on the piston.
Now I know that Overmod will be "all over this" because of heat transfer to the cylinder wall causing condensation and pressure drop in the valves and this or the other condition where the cylinder expansion is far from ideally adiabatic-isentropic.
But that's it. You find out what thermodyamic state variables are conserved (entropy in isentropic, enthalpy in adiabatic), start with the initial conditions, run the process, find out what has changed (such as the specific volume in a piston, a pressure drop in a turbine), use the conserved quantities to use the partial end state to find the full end state, and boom, turn in your answer to the homework problem. (voice of Alec Guiness -- Use the steam tables to look up state values, Luke, don't try to solve these problems by path integration).
Now here is a homework problem for ya. All of the thermo textbooks teach enthalpy difference between inlet and outlet to figure out the power extracted by a turbine, a flow device. How to you figure out the work done in a steam cylinder over the interval prior to cutoff? How much work is done if you expand at 100% cutoff? The work is not zero, and the efficiency will be poor, but it isn't zero, but you cannot solve this like a steam turbine problem, and if anyone wants to discuss this with me, send a message through the Trains Forum message board.
If GM "killed the electric car", what am I doing standing next to an EV-1, a half a block from the WSOR tracks?
Great Course that covers much of this....."IF" you could find a copy:"The Study of Steam Power" by Tom ? & T Parkinson PHD. I took the course and it was VERY thorough. Was taught at NAMES and later at our Live STeam Club NEOLS. Tom ? (Sorry-can't recall his last name-my bad!) originally taught it...and later T Parkinson took over. Great Course taught in Laymans terms.....
It seems the whole of this has disappeared as far as the Internet is concerned.
Very old contact information for Barbara and T Parkinson:
Telephone (440)285-2327
bpnames@yahoo.com
Perhaps Vince can get Mike to look into this and find actual current references or sources.
here are the specific examples from Semmens describing steam locomotive efficiency that i'm trying to understand or more precisely how he goes about determining it (assuming 1st order effects)
his initial example determines the BTUs to raise water from 50F 1 atmosphere to 200 psi and compares it to the work performed when the steam is expanded to 5 psi. He determines the BTU of water at 50 F as 18 BTU (pt G on graph), the BTU of 200 psi as 1201 BTU (pt A) and the BTU of the exhaust steam as 1026 (pt B).
he determines the efficiency as 14.8% == 175 / 1183 = (1201 - 1026) / (1201 - 18)
pts G, B and A all correspond to the same steam table entropy value. I don't understand why the steam table entropy value is kept constant. When work is done by the steam its entropy must have increased. (i don't know how to relate the steam table entropy value, BTU/lbT, to the entropy, BTU/lb, of the steam)
in a 2nd example he determine the efficiency when the boiler pressure is raised to 250 psi as 16.1% == 191 / 1185 (pts C and D)
a 3rd example consider efficiency when steam at 200 psi is superheated by 200 F as 16.0% == 207 + (1183 + 113) (pts A, E and F). In this example, if i interpret it correctly, the steam table entropy value change due to super-heating from pt A to E. why?
FWIW, the D&H 1403 is widely thought to have had the highest thermal efficiency of any reciprocating steam locomotive in the US. The one figure I saw put the BTU of coal to drawbar hp-hr efficiency at 9%. Data from Kratville's book on the UP Big Boy suggests a figure of a bit over 4% when pulling trains uphill east of Ogden.
Note that steam cycle thermal efficiency for these locomotives is significantly better as the figures for overall thermal efficiency, which are hurt by not so great boiler efficiency. You would be lucky to get 70% of the heat content of the fuel going into the firebox ending up heating the water in the boiler.
Oh yeah, "How Steam Locomotives Really Work."
To correct my earlier remarks on the subject, an adiabatic-reversible thermodynamic process takes place at constant entropy. This diagram is assuming that steam expansion in a cylinder takes place at constant entropy, hence the vertical lines.
A couple of critiques.
1. You don't have the steam in the cylinders at boiler pressure owing to pressure drops through the superheater tubes and "wire-drawing" through the valves.
2. The "bottom" part of the expansion is suspect. At most usable cutoff values, the pressure at exhaust release is considerable higher than 5 PSI. You might have 5 PSI backpressure acting on the opposite face of the piston owning to not working the locomotive very hard, which I guess, was British practice. Wardale, I believe (he insists on metric units that I am second guessing whether I did the conversions correctly), claims as much as 30 PSI backpressure running, say, a Pennsy Q2 locomotive working flat out. The back pressure from the exhaust system acts like a parastiic back force on the piston.
3. The vertical axis better be enthalpy rather than internal energy, because the change in that quantity represents work.
4. This chart doesn't account for the work done on the piston before cutoff. When you are filling the cylinder with steam, whatever the pressure is below the boiler pressure from valve and other losses, you are doing P times V (pressure times volume) work. My suggestion on how to quantify this on a steam consumption of pounds/per hour is to multiply pounds/per hour steam consumption times (enthalpy/pound - internal energy/pound) to get the P times V work prior to cutoff.
You are getting some power, although at low thermal efficiency, even if you work steam non-expansively, as Semmens attributes to Trevithick's original locomotive. This diagram would suggest you are getting zero work if you don't use the steam expansively, which is certainly not the case.
5. If you don't have enough superheat that you don't have any superheat left after expansion, you are going up "wall" and "nucleate" condensation effects that can rob power. The British, especially, went with minimal levels of superheat on the theory that if the steam isn't wet after expansion, you are wasting energy, but such is probably not true on account of condensation effects -- Overmod can weigh in on this topic.
Other than all of that, Mrs. Lincoln, how did you enjoy the play? At least the graph you show is a starting point for a "sharp pencil" calculation to get a handle on steam locomotive thermal efficiency.
Paul MilenkovicTo correct my earlier remarks on the subject, an adiabatic-reversible thermodynamic process takes place at constant entropy.
so since a steam locomotive cylinder is doing work and entropy must increase how can the process be adiabatic-reversible?
Paul MilenkovicThis diagram is assuming that steam expansion in a cylinder takes place at constant entropy, hence the vertical lines.
if work is done and entropy increases, shouldn't the lines move down and right toward > entropy?
Greg: What follows doesn’t address your question about steam engines but is intended to help with some of the underlying conceptual aspects.
gregc don't understand why compressing the gas (reducing the volume) doesn't just increase the pressure (PV/T right). Temperature didn't change when it expanded
don't understand why compressing the gas (reducing the volume) doesn't just increase the pressure (PV/T right). Temperature didn't change when it expanded
When the gas expanded into the larger volume, the energy of the gas remained constant and therefore so did its temperature, assuming a perfect gas. Compressing the gas requires work to be done on the gas, increasing the energy of the gas and thus its temperature.
gregc (while it makes sense work is performed to lower the entropy, increasing the ability of the gas to do work, it's ironic to me that heat needs to be removed
Work done on a system can never reduce its entropy. In this case, the entropy of the gas is reduced by the heat transfer to the environment which also restores the gas to its original temperature.
gregc shouldn't entropy go up if work is performed?
shouldn't entropy go up if work is performed?
No. A system performing work does not experience an increase in entropy solely by virtue of performing work on its surroundings. I could be wrong, but my sense is you are interpreting the situation as follows:
1) A system produces some work.
2) The system then “has less work left in it.”
3) Since it can’t do as much work as it could before, its entropy must have increased.
This is not the case. You were on the right track when you wrote the quote below.
gregc i believe entropy is actually a measure of energy that cannot be used to generate work
i believe entropy is actually a measure of energy that cannot be used to generate work
Let’s say we have a system with a certain pressure, temperature, volume, etc. Assume that at these conditions, the energy of the system is 100 units. Let’s also assume the system can produce a maximum of 80 units of work, leaving 20 energy units that cannot be converted into work. The entropy of the system corresponds to the 20 units.
Now imagine the system undergoes a reversible process (which is the best you can do) and produces 80 units of work with 20 energy units remaining. The entropy of the system remained unchanged because there are still 20 units of energy that cannot be converted into work.
In the real world, there are losses (friction, etc.), resulting in the system producing less than the maximum 80 units of work. Let’s say the real system produced 70 units of work with 10 units wasted due to friction. Now there are 30 energy units (10 plus the original 20) that cannot be converted to work. The entropy of the system increased.
I hope this helps.
AnthonyV Let’s say we have a system with a certain pressure, temperature, volume, etc. Assume that at these conditions, the energy of the system is 100 units. Let’s also assume the system can produce a maximum of 80 units of work, leaving 20 energy units that cannot be converted into work. The entropy of the system corresponds to the 20 units. Now imagine the system undergoes a reversible process (which is the best you can do) and produces 80 units of work with 20 energy units remaining. The entropy of the system remained unchanged because there are still 20 units of energy that cannot be converted into work. In the real world, there are losses (friction, etc.), resulting in the system producing less than the maximum 80 units of work. Let’s say the real system produced 70 units of work with 10 units wasted due to friction. Now there are 30 energy units (10 plus the original 20) that cannot be converted to work. The entropy of the system increased.
The meaning of entropy has to be consistent with the Carnot cycle and the efficiency of the reversible heat engine. The amount of work you can extract depends on the temperature difference between the heat source and the cooling sink. If you expand the working fluid down to absolute zero, you can extract every last bit of the internal energy that the fluid started with into work and your heat rejection to the cold sink will be zero.
For entropy to be an intrinsic property of the fluid at that start of the expansion and maintained through the expansion, which it is, it cannot be the 20 energy units that cannot be converted to work because you in theory convert all the energy into work.
The key has to be delta S = delta Q/T. The change in entropy brought about by a change in heat Q is proportional to the reciprocal of temperature, so if you chilled the system to a fraction of a degree above absolute zero, 1/T is very large, and a tiny amount of heat produces whatever change in entropy you see. If you warm the system up, it takes a lot of heat to make the change in entropy.
It is that 1/T factor that makes entropy so hard to understand. If entropy is a measure of "disorder", if you have a messy, cluttered house, a gas explosion isn't going to change the condition very much. But if you have very clean, organized house, one person tracking their muddy boots over the carpet will cause a big difference.
This is only a metaphor, people, try to keep your gas-fired appliance maintained that they don't explode, because you really don't want that to happen.
Anthony
thanks for the comments
AnthonyVgregc shouldn't entropy go up if work is performed? No. A system performing work does not experience an increase in entropy solely by virtue of performing work on its surroundings. I could be wrong, but my sense is you are interpreting the situation as follows: 1) A system produces some work. 2) The system then “has less work left in it.” 3) Since it can’t do as much work as it could before, its entropy must have increased. This is not the case. You were on the right track when you wrote the quote below. gregc i believe entropy is actually a measure of energy that cannot be used to generate work
don't understand the distinction
The Parkinsons moved from Ohio to Georgia about 10 years ago to escape the Ohio Winters wear and tear on their bodies and be nearer their Son as they aged. They were driving forces in our Live Steam Club NEOLS
The distinction is between a reversible process and irreversible process. A low friction piston compressing gas in a well insulated cylinder is close to being a reversible process, i.e. most of the energy expended in compressing the gas can be extracted by letting the gas pressure push the piston back out. This would have low increase in entropy. A pinhole leak in a compressed air tank is an example of an irreversible process as there no chance of recovering the energy used in compressing the air. This would have a high increase in entropy.
A humorous description of entropy is nature's way of enforcing the ban on perpetual motion machines.
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