Phase Change Cylinder

[Russell Philips]

Let's start with a thought experiment: (ideal conditions)

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If an entire system was brought to 212*F, let's say for argument that room temperature was 212*F. Thus The working fluid (water) is hot (say 207*F) but is still liquid. We add a few degrees of temperature (say 217*F), paying for the heat of vaporization cost with joules of heat. The water boils to steam, expanding many times in volume. We are not in a hurry and allow time to work for us. Assume the expansion is constrained against a piston and does useful work output.

With the system expanded and liquid is now gaseous (steam), we cool the working fluid a few degrees of temperature (207*F), paying for the heat of condensation with joules of thermal energy (cooling). The steam condenses back into a liquid, contracting many times in volume. Again we are not in a hurry and allow time to work for us. We are subject to atmospheric pressure to perform useful work (keep it simple, we will increase working pressures later as we become more familiar).

The temperatures required are nominal (compared to our room temp of 212*F). The joules of energy are significant for heat of vaporization costs. However, the expansion/contraction ratios are orders of magnitude!

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[Perhaps I'm saying too little, but I want to keep it simple and light for now. I'm doing my best here. Feel free to ask questions if you are unsure about something. I believe there is enough information to begin painting the picture of the concept and theory.]


Does it appear to you that nominal temperatures could affect the desired phase changes?

I know the heat of vaporization costs are significant, but they are an order of magnitude less than the expansion ratios involved..

Chew on this rhetorical question: What might ideal outputs be per each joule of input heat/cold?

Thank you for you help and consideration on this, it is very important to me.

[Ken]

Let's look at the energy involved in this phase change.  Using a set of IAPWS-97 formulations for Excel makes things painless.  Saturated water at 207 F and 14.7 psi contains 175.15 BTU/lb.  Saturated steam at 217 F exists at 16.2 psi and contains 1152.16 BTU/lb.  So, assuming perfect insulation and heat transfer, we had to add 977.01 BTU to effect the phase change and pressure rise of 1.5 psi.    The specific volume went from 0.016680144 cubic ft/lb to 24.42856804, and increase of about 1459 times.  That's a huge increase in volume, but it can only occur with 1.5 psi pressure differential, constraining the water to develop more pressure will prevent vaporization.  A 4 inch bore and stroke piston will have a displacement of about 50.25 cubic inches (or about 0.029 cubic feet)---call it about 830 revolutions to expand the steam and about 830 more to contract it (assuming an ideal cylinder with no heat losses, leakage and so on).  Just for the heck of it, let's assume we do all this in one minute, about 830 rpm).  The Mean Effective Pressure is going to be quite a bit less than 1.5 psi, it only reaches that at 217 F, as the temperature drops, so does the pressure.  Calculating MEP can be a real bugger since the pressure/volume curves are polytropic and I would have no idea what m and n would be.  Just for the heck of it, let's say it is the average of 0 at 207 and 1.5 at 217, or about .75 psi.  Using the PLAN formula for horsepower (Pressure X Length of stroke X Area of piston X Number of strokes)/33,000 ft-lbs/min we about 0.08 horsepower; half of that being developed on the expansion and the other half on the contraction. The amount of power developed, 0.04, is purely theoretical and in real terms you would be fantastically lucky to see 25% of that.  This is why Newcomen engines are so huge for the small power output.

Now. let us assume a relatively low pressure and temperature steam engine of 150 psi and 100 degrees superheat.

At 150 psi the saturation temperature is about 358 F.  If we add 100 F superheat the temperature is now 458 and the steam possesses about 1252 BTU/lb under those conditions with a specific volume of about 3.5 cubic feet per pound.  It has about 7 times the volume of the 217 degree steam but maybe 100 times the effective pressure.  We are talking about a lot more power for a relatively small increase in the amount of energy added.  Here's the McGuffin.  We had to add 1252-175= 1077 BTU to produce the steam BUT we need to extract 977 BTU to condense the exhaust back to water inside the cylinder.  In a perfectly insulated cylinder having zero thermal mass, this means nothing, but in the real world there is a massive heat flow from the cylinder to the steam when effecting this cooling; it's easily going to bury the minute gain we got from condensing the steam.

This is the inherent down side of the Rankine cycle.  It takes enormous energy to effect phase change but no power at all is developed if the pressure remains constant, it's only by changing the pressure do we get any power....and changing the pressure changes the amount of BTU needed to change phase.  Essentially the heat of vaporization is effectively lost, especially when dealing with real world machines subject to conductive and convective heat transfers.

Ken

[Max Kennedy]

Good explanation Ken, I was trying to figure out how to say that simply.  The use of a concrete example was the ticket.

Max

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It can be done

[Russell Philips]

Thanks to all for the quick responses and support!

I do not hear arguments against the base principle so much as that outputs are very limited.

Ken, Thank You for crunching the numbers, you seem to be able to make quick work of it.

Would you please look up the value of saturated water at 172 F at 146 psi to see how many BTU/lb it contains?
And also, Saturated steam at 252 F at 250 psi to see how many BTU/lb it contains? 
(172  & 252  are +/- 40 F from 212.)

Are these tables available online?

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Date: Wed, 5 Feb 2014 04:58:40 -0800
From: ken...@aol.com
To: open-sou...@googlegroups.com
Subject: Re: Phase Change Cylinder

[Ken]

Hi Russell,

 

I spent about 90 minutes putting together a thoughtful reply, and Google swallowed rather than posting it.  This effort may be a bit more brief.

 

I really can't give you the values you are asking for, it has to do with the definition of saturated.  Saturated water and steam exist at whatever combination of temperature and pressure produces boiling, at that temperature you can have pure steam (100% quality), pure water (0%)), half and half (50%) or any other ratio.  That said, it is possible to find the enthalpy of water at the pressures and temperatures you state.

 

172 F at 146 psi =  140.325 BTU/ lb

252 F at 250 psi =  221.104 BTU/ lb

 

You can tell at a glance that both samples are beneath the saturation curve (the curve which describes the temperatures and pressures at which water vaporizes)., the heat content is just way too low.  At 146 psi water vaporizes at 356 F and at 250 it is 400 F.

 

I use a set of functions that are designed to work in Excel, they are add-ins that act as though they were regular Excel functions except that they calculate a variety of water and steam variables.  The original author's website is gone but I found you can download the same IAPWS 97 formulations at http://ascentengineering.com/Documents/XSteam_Excel_v2.6_US.xls.  A less involved method is to use the online Sarco Spirax steam table calculators at http://www.spiraxsarco.com/us/resources/steam-tables.asp

 

Let me clear up a misapprehension, the basic premise of condensing the working fluid inside the engine is not good.  Ideally, steam expansion would be isentropic, neither giving nor receiving energy from the cylinder.  This is impossible in any practical form when condensing in the cylinder, it is necessary to somehow extract heat from the steam and this is only going to be accomplished by either cooling the cylinder or by introducing a coolant into the cylinder.  Either method will inevitably cool the cylinder itself, which in turn will cause the cylinder to absorb heat from the steam on the next stroke.  Stumpf developed the uniflow to keep the cylinder hot at the admission end and cool at the exhaust end, this produced a heat gradient from one end of the cylinder to the other and in turn reduced transfer of heat to and from the walls.  It's easy to see how the heat drops gradually on expansion.  Since the uniflow exhausts out the cylinder side walls at the bottom of the stroke, it must compress whatever remnant steam is still in the cylinder on the upstroke.  This compression can return the small amount of steam to around the original admission pressure and often to higher temperatures (the compression ratio is necessarily higher than the expansion ratio).  The compression heating also helps maintain the temperature gradient which in turn raises efficiency,.

 

It could be argued that a Newcomen engine produced work by condensing in the cylinder, so this should also apply to a more modern steam engine.  The reason this is not so is that the Newcomen is NOT really a steam engine, it is an atmospheric pressure engine.  All the power comes from the atmosphere working against a piston under which a partial vaccuum is maintained.  You can't rob the expansion steam of heat energy simply because there IS NO expansion occurring.  Remember, however, the Watt engine was far more efficient than the Newcomen simply by addiing a condenser; within 20 years the Watt was superceded by increasingly more efficient steam prime movers until you reach the central electrical generating plants of today....  All of which work to retain heat inside the expander housing to the greatest degree possible.

 

Regards,

 

Ken

[Russell Philips]

Ken, I am sorry for your lost reply - that is so frustrating!
Thank you for keeping in our minds the aspects that will allow us to be successful in steam.
Thank you for providing the information to continue in our suspension of disbelief.

We need to set room temperature to 356 F, where water saturates and begins to boil at 146 psi.

By sealing the system at this working pressure, we raise the ability for atmosphere to do work on the contraction end of the cycle.
As we create a lower pressure in the cylinder the atmospheric pressure pushes in with greater forces than if we were working at normal room pressure. We raise the amount of work potential the contraction stroke can do.

This does not have a negative effect on how hard it is to heat - expansion still occurs similarly. The same few degrees in temperature change still have a great effect upon volume - just at a higher temperature base or zero point. The ability to choose your zero point is powerful!

As students of life, as inventors, or in this case as system engineers, we will be bound by a temperature range.  This is given to us by the potential of the thermal differentials that we will apply to the cylinder. The lowest hanging fruit first shows us that we will be seeking in the lower end of the differentials. The possibility for our success only occurs IF IT EXISTS -and- IF WE LOOK FOR IT. 

Each set of thermal differentials requires changes to the cylinder and perhaps working fluid.

Working inside a temperature range of +/- 22 F, we add 44 to 356 F. Ken has provided that 400 F is 250psi. Our pressure for output work is the difference between 250 psi and 146 psi or 104 psi. Let's compare this to our atmospheric maximum limitation and see where we stand.

If the two equal then great. If the limitation affects us then we can only get a percentage (contraction %) of work output. If the pressure is inside of the limitation then we are not fully harvesting our thermal potential. The closer they equal the better.

Where does this stand currently?

What is the ideal potential from a pressurized system working at 146 psi +/-22 F?

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Date: Thu, 6 Feb 2014 08:09:39 -0800

[Ken]

Hi Russell,

 

I think the problem here is that you are trying to impose what you think is a "common sense" solution to a thermodynamics problem.  At first blush we can see that condensing the vapor is a non-reversible thermodynamic process, if we add water or chill the cylinder to condense the vapor we can't get the heat to flow back into the cylinder or the water to separate out from the vapor.  By contrast, if we leave the steam in the cylinder we can compress and expand it back and forth to nearly the original states.  Reversible thermodynamic reactions are much more desireable than non-reversible.

 

You really can't analyze such problems verbally as you are trying to do, you need to apply mathematics and the correct variables for the specified states to get a meaningful answer.  Basically you are trying to invent a thermodynamic cycle but trying to avoid doing any real thermodynamics.   People have known that the heat of vaporization is lost in the Rankine Cycle since the 1800s and development of the cycle is ongoing today, this isn't a topic which has received scant attention.

 

Taking your initial condtion of steam at 146 psi, you are assuming that the way to extract the energy is to condense the steam. That is hideously inefficient.   As the pressure goes up, the relative volume decreases, so there is actually less relative volume from which to extract work..   Expanding the steam as much as possible in the cylinder extracts energy much more efficiently and nearly to the same pressure.  In fact, I've worked with turbines that expand the steam down beneath 28 inches of mercury...a pretty high grade vaccuum.  Rather than trying to collapse the volume, the volume has been maximized to the highest degree possible and energy extracted from the kinetic energy produced by expansion.  This huge expansion was possible because the steam had condensed after leaving the turbine, thus there was no back pressure to limit the expansion.  Thpugh the energy was wrung out down to almost no pressure, the heat of vaporization was NOT recovered from the steam.  This can be explained by understanding reversible vs non-reversible processes and the desirability of isentropic processes.

 

Expanding and compressing in the cylinder is considered to be nearly isentropic, meaning the entropy does not change.  There are any number of definitions of entropy but for our purposes the one that best fits is that entropy is a measure of the amount of energy within the working fluid that is NOT available for work...increasing entropy means that we have a potential to wring less energy from the steam.  Expansion and compression are considered isentropic (well, nearly so) because not that much energy is transferred out of the steam to the environment.  By contrast, the condensation you are suggesting is designed to raise entropy directly, the energy is transferred out of the working fluid.  With higher entropy, less energy is available to perform work.  As mentioned before, the expansion is a reversible process while the condensation isn't, condensation leaves less energy remaining to harvest.

 

This really isn't a topic amenable to discussion in any great detail in a forum like this but something that requires some study of thermodynamic principles.  Very few people ever manage to develop a new, viable, thermodynamic process despite many attempts...I'm reminded of the Stirling engine using steam rather than air as a working fluid; for example.  If you want, go to Google patents and search for D-Cycle Associates, they patented a number of different thermal cycles during the time of the first energy crisis....none of which ever caught on.

 

Regards,

 

Ken

[Russell Philips]

Hello Ken,
I believe there may be a problem in these previous calculations:

Let's look at the energy involved in this phase change.  Using a set of IAPWS-97 formulations for Excel makes things painless.  Saturated water at 207 F and 14.7 psi contains 175.15 BTU/lb.  Saturated steam at 217 F exists at 16.2 psi and contains 1152.16 BTU/lb.  So, assuming perfect insulation and heat transfer, we had to add 977.01 BTU to effect the phase change and pressure rise of 1.5 psi.    The specific volume went from 0.016680144 cubic ft/lb to 24.42856804, and increase of about 1459 times.  That's a huge increase in volume, but it can only occur with 1.5 psi pressure differential, constraining the water to develop more pressure will prevent vaporization.  A 4 inch bore and stroke piston will have a displacement of about 50.25 cubic inches (or about 0.029 cubic feet)---call it about 830 revolutions to expand the steam and about 830 more to contract it (assuming an ideal cylinder with no heat losses, leakage and so on).  Just for the heck of it, let's assume we do all this in one minute, about 830 rpm).  The Mean Effective Pressure is going to be quite a bit less than 1.5 psi, it only reaches that at 217 F, as the temperature drops, so does the pressure.  Calculating MEP can be a real bugger since the pressure/volume curves are polytropic and I would have no idea what m and n would be.  Just for the heck of it, let's say it is the average of 0 at 207 and 1.5 at 217, or about .75 psi.  Using the PLAN formula for horsepower (Pressure X Length of stroke X Area of piston X Number of strokes)/33,000 ft-lbs/min we about 0.08 horsepower; half of that being developed on the expansion and the other half on the contraction.

We add 977.01 BTU input energy and get .08 hp output work in an ideal calculation.
1 HP  is roughly 42.4 BTU.               
.08 hp is roughly 3.4 BTU, so how does 977 = 3.4?

[Ken]

[Ken]

EXACTLY the point, you can't effectively recover the heat of vaporization during the condensation phase.  You're making the assumption that you can recover those  977 BTU --- as though they were tied up in a reversible process like compression.expansion --- but they aren't.  It is lost energy and the biggest single problem with the basic Rankine cycle. It's the classic thermodynamic example of why you avoid non-reversible processes. If you look at reports of the first Otto engines fitted to boats circa the 1880s you find that the IC engines were essentially modified steam engines and that the engines were desirable not only because they were so much less complicated to run, but also because they were so much more economical.  Ais isn't normally condensible, so there is no latent heat loss. 

What you CAN do is to use a feed water to transfer some of the heat from the exhaust into the cold water entering the boiler; this preservation of energy reduces the fuel that must be burned and enhances efficiency somewhat.

Very simply put, there is NOTHING for the steam to work against when it condenses.  Vaccuum produces no work, it is substances working against the vaccuum that produce work.  If we were on the surface of the moon, the cylinder would still produce expansive work when pressurized steam was admitted at the top of the stroke and allowed to expand as the piston descends.  If we condensed the steam, however, absolutely no work is performed.  A Newcomen engine on the moon produces no work.  Expansion produces work in and of itself, condensation doesn't.   Think about it, suppose we had a fluid that required only 100 BTU to vaporize to the same volume as steam and another that required 2000 BTU to vaporize to the same volume.  Condensing both of these vapors in pistons along side a piston filled with steam creates the same pressure differential (14.7 psi) and in all three cases the piston is displaced the same distance; since force times distance equals work, the work produced in all three cases is independant of the BTU that was shed to effect condensation.  The heat of vaporization was irrelevant to the work produced, in all three cases that work is still pretty small compared to the energy lost to condense the steam.

 

If you're still not convinced, I'd suggest going to "Google Books" and referencing some old thermodynamics texts...preferably some after 1910 or so.  The lost energy certainly isn't something that engineers have simply overlooked for 200 years, they are painfully aware of it.  The physics simply don't give you a way to recover it and this is why most vehicles operate on cycles using non condensible gasses. In large scale there are potentials to boost steam engine efficiencies greatly, these involve heat exchanges throughout various parts ot the cycle to recover losses and some impressive efficiencies can be attained.  Even then, you have to start getting exotic to go much further than than what a small Diesel engine or very efficient large Brayton (jet engine) can do.

 

Regards,

 

Ken

[Eerik Wissenz]

Thanks Ken for a good explanation. I'm no expert but have learned a few things.

I think a point to add is that condensation does no "work" on the piston as you said, but rather releases "heat". Efficient condensation requires a large surface area which by nature dissipates the heat making it a poor heat source. To make things worse you also need a tmp diff to condense meaning your condensed liquid is quickly at the cold end of your gradient (ie ambient), and so has no difference to work on. I.e. a second heat engine powered by the condensed heat doesn't work well since for the condenser to be efficient must be as close to ambient, whereas the second heat engine requires higher than ambient heat source by definition.

Letting the temp of the confense liquid rise simply renders the cindenser and so first engine less efficient, making more overall losses with addef cost of "second" engine.

Using the condensed liquid as feed provides some gain simply because it is always hotter than ambient so might as well, but since to cindense it must transfer to heat sink, there's not much to recover, and I believe it would always be better to cool all the way to heat sink tmp if that were possible (otherwise why condense at all).

For all reasons mentioned by ken this is why the super efficient turbines today are super critical. At arround 400 bar the density of "steam" meets the "density" of water at same tmp and pressure, and so they become the "same". Result, no phase change and all the advantages mentioned by ken. But even so this is not recovering phase change but rather just the initial energy to heat liquid to saturation tmp (ie work done to heat water to boiling point).

The efficient non-super-critical turbines get their gains by having pretty high temp diff nonetheless (ie good ratio between energy for phase change and energy heating the liquid/gas), and then very good implantation of super heating before each stage precisely to reduce condensation in the turbine as much as possible. 

Unfortunately it's difficult to scale these things down. 

Best on small scale is either:

1. Simple reciprocating (counter flow) but with some use for waste heat, like grain processing. Not so efficient but cheap and cost effective if associated with process that needs a lot of heat anyways.

2. Monoflow which increases efficiency by avoiding condensation in the engine by not heating and cooling parts alternatively as in case of counter flow. But monoflow needs condenser to have real advantage which adds cost and maintenance complexity. 

The real advantage of small scale steam is not efficiency, but rather it is compatible with solar thermal as a heat source, and IC is not. With a "free" high temp heat source efficiency is less relevant and use for the waste heat can always be found when close to point of use (agricultural processing,  drying various things,   space heating,  water pasteurisation).

Solar thermal/simple steam engine can be very cheap in capital but is high maintenance,  so is suited to "developing" countries. At least until tech is mature and "developed" countries realize they're quite poor for the most part.

Limiting factor not steam tech but rather high temp solar but still cheap to build.

This is my analysis anyhow,  but not an expert on the thermo stuff so am welcome to correcting any points. It is how I visualize the points of ken which are obvious when looking at the math but difficult to retain if thermo is not ones training.

Best to all

Eerik

www.decent-domocracy.org

www.solarfire.co

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Eerik Wissenz

[Eerik Wissenz]

Sry, sentence about recovering heat to boiling point should a paragraph above not about super-critical. ( smart phone woes).

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Eerik Wissenz

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Eerik Wissenz

[Russell Philips]

Ken,

In the original example:

Saturated water at 207 F and 14.7 psi contains 175.15 BTU/lb.
Saturated steam at 217 F exists at 16.2 psi and contains 1152.16 BTU/lb.
So, assuming perfect insulation and heat transfer, we had to add 977.01 BTU to effect the phase change and pressure rise of 1.5 psi.
The specific volume went from 0.016680144 cubic ft/lb to 24.42856804, and increase of about 1459 times.

A 4 inch bore and stroke piston will have a displacement of about 50.25 cubic inches (or about 0.029 cubic feet)

(24.42856804 - 0.016680144) / 0.029 = 841.79 (not sure how you got 830 RPM?)

PLAN Formula (using MEP):
(.75 x 4.0 x 12.5625 x 841.79)/33000 = .96 HP (not sure how you got .08 HP?)

[Ken, If I made mistakes in the above process, please show the corrected steps.]

Let's look closer at the math. It will require breaking the cycle into the expansion math, HP (and equivalent BTU output). Plus the contraction math, HP (and equivalent BTU output), separately.

Let's enlarge the piston and stroke - fitting it to the expansion volume with a single stroke.

Keeping the expansion volume of 24.412 cubic ft (24.42856804 - 0.016680144).
A 48 inch piston has an area of 150.8 square inches (48 x 3.1416).
Our stroke equals 280 inches (24.412 (12 x 12 x 12) / (150.8)).

PLAN Formula (using MEP):
(.75 x 48 x 150.8 x 1)/33000 = .16 HP

.16 HP is roughly 6.8 BTU (1 HP  is roughly 42.4 BTU)(.16 x 42.4)

Our input energy is 977.01 BTU.
Our output energy is 6.8 BTU.

WHERE DID 970 BTU VANISH TOO?

Ken, Where is the conservation of mass and energy?
We have not included the contraction cycle, so that can not be the loss.
We must be able to make an accounting with the expansion only.
Something is amiss!
            

[Ken]

Russell,  

Forget the math for a moment, it isn't important right now, it's a break down in conception at this point. 

The reason you can't get the heat of vaporization back is entropy.  Yes, the energy is still there as per the Law of Conservation of Energy - but that doesn't matter.  Entropy is the process whereby concentrations of energy gradually disperse to a uniform state; eventually entropy will wear the universe down.  The energy will all still be there, but once it is evenly distributed it is functionally the same as being non-existent...it is the differential in energy concentration that allows us to do work. 

In the thermodynamic tables entropy is basically a measure of how much energy is NOT available for work whereas enthalpy is a measure of the total thermal energy in the material.  The entropy of steam entering the cylinder is assumed to be constant throughout the stroke in an ideal (frictionless, perfectly insulated) cylinder and piston.  At exhaust, the entropy of the steam jumps (you can see it on the tables) and the same amount of energy is no longer available to do work. 

Look at it another way. 

The Carnot Cycle is the most theoretically efficient thermodynamic cycle possible, one that can't be met in reality.   There is a lot of fancy math involved when examining theoretical work out because you have to examine the energy content of the working fluids; but the Carnot efficiency is a simple ratio of the temperatures, not the total energy, because it is concentration of energy that determines what kind of output you will see.  The Carnot efficiency formula is E = (Th - Tl)/Th  with Th being the highest absolute temperature in the cyle (admission steam) and Tl being the lowest absolute temperature in the cycle.  It's readily apparent that the efficiency is a function of the temperature differential and has nothing to do with the total amount of energy....which is quite logical.  The temperature at 212 F is  equal to 373 K, Kelvin being the measure of absolute temperature. Now, let's apply this recovering the heat of vaporization.  The saturated steam temperature at atmospheric is 373 K and the saturated water temperature is 373 K ..plugging this into Carnot gives us E =(373-373)/373 = 0.  Zero efficiency is equal to no work output.  The energy is still there, the enthalpy of the steam is about 1140 BTU per pound and the water enthalpy is about 180 BTU per pound...so the Law of Conservation of Energy is still satisfied.  But Carnot says that the efficiency with which we can exploit this energy is proportional to the difference in absolute temperatures, which is zero.   

The energy is still there, but we can't tap it.  No one said the universe was fair, and the universe sure seems to take full advantage of that!  :-)

It IS possible to use the hot condensate to vaporize a lower vapor pressure fluid in a bottom cycle and extract energy from that, but the exhaust temperature will still necessarily be higher than the ambient temperature, so the efficiency of this bottom cycle will be significantly less. 

Regards, 

Ken

[Ken]

Hi Eerik

 

You're quite welcome, and thank you.

 

I have to admit that I wish I had more formal thermodynamics training, as it is I learned thermodynamics in much the same way I learned about sex...a combination of furtive reading and out on the streets from low companions!

 

You are dead on in your analysis that it is energy concentration that matters.  The one point on which I would disagree is that one should subcool the condensate back down to ambient temperature.  When I was in the navy we referred to the difference between saturation temperature and condensate temperature as "condensate depression".  You need a certain amount of condensate depression so that fluctuations in pressure won't allow the condenser contents to flash back into vapor, but you try to keep the depression as low as possible.  The hotter the condensate is, the less energy it takes to vaporize it when it is fed back into the boiler.  This applies for single cycle powerplants.  In a combined bottoming cycle you want to transfer as much energy as you can into the secondary working fluid...but this transfer is going to be limited to the condensing temperature of the lower working fluid.  If it condenses at 135 F you still want to avoid condensate depression in that system as well, so 135 is the theoretical lowest temperature you can take your primary working fluid down to.  In reality, heat exchangers aren't perfectly efficient so it won't go down that far.

 

The general rule of thumb is that you do not want to reject any more heat into the environment than you absolutely must else you will need to resupply that heat later.  This is why some stationary and marine gas turbines (and Stirling engines)  have recuperators.

 

Regards,

 

Ken

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Eerik Wissenz

[Russell Philips]

As we seek the dynamics and ideal work potential of a cylinder expansion from liquid at 146 Psi and 356 F to steam at 250 Psi and 400 F,

Please limit the scope of discussion to expansion equation and math.

Let's use a piston size of 1 square inch and a stroke of 36 inches.

What is the fluid volume at 356 F that expands to 36 cubic inches at 400 F?