Posted by Jeremy Holmes

Andys PDF's October 07, 2009 10:47PM |
Registered: 17 years ago Posts: 1,519 |

[www.steamautomobile.com]

[www.steamautomobile.com]

-----------

More on the flash process producing work.

I don't think your valve/injector is any different than a normal valve in that it is a throttling process. Basicly presure reducing. A constant enthalpy process. The result is an admission volume in the cylander at the reduced pressure having the same initial enthalpy. The work produced is the pressure times the volume change.

As far as calculating work the flash steam and throttling process are both constant enthalpy processes. The only differance between flash steam and throttling is flash steam is initially a liquid and would not all vaporize living some lquid in the mix.

It does take some time to change phase. In the IAPWS site they talk about how well their formula extend into subcooled gas and superheated liquid states. Thoes being the unstable states during phase transition. I havn't been able to find anything on duration of thoes states other then they last long enough that their properties are important in nozel calculations.

I have to backpeddel a bit. It comes down to wether or not you have a liquid state in you injector. If it is a liquid you would have a flash steam process during admitance. But I think it would be better to have a gas that would not have any liquid content at the end of admitance.

Andy

-------------------------------------------------------------------------------

Hi Jeremy

The reasion it is less efficient to add heat during expansion comes from fact that conversion of heat energy to work when there is no heat transfer takes place along the line of constant entropy. That doesn't mean an engine doing it wont work. There are other factors can effect efficiency. Heat recovery from the exhaust steam for instance. Adding heat during expansio might benifit heat recovery and maybe wind with better over all efficiency.

The ideal cycles are static cycles. By that I mean they are not time dependent. They do not include any time dependent processes. Like heat transfer for example. A real engine is of course dynamic. Besides heat transfer being time dependent you have flow processes that are very dynamic. To analyze a dynamic cycle you have to model the processes involved in a simulation. The processes are not stady state. The valves create a palsating flow. You have pressure(sound) waves that interact with valve timing. You have heat transfer whose area exposier time varies with RPM. It a very complicated model to make. And then would have to be tested and tuned for agreement with an operatoring engine. It would take a special engine with sensors that can be used to verify the model. That has been my goal.

Heat transfer to the working fluid does work. In the clasical steam engine heat transfer takes place in the boiler at constant pressure. Work produced at constant pressure is simply the pressure times the volume change. The clasical engine utilizzes that work during admission. Analying a non-expanding cycle, one haveing admision for the full stroke, you can get around 8% efficiency. Expanding that steam along an isentropic line where you have no heat transfer and thus all energy change within the substance is converted to work, you have both pressure and volume changing during expansion so it can not simply be calculated as P * (voume change). But in theory it is equilivant to the change in internal energy.

To analyze the processes with heat transfer you break the stroke into tiny steps. Modling software does so in small time steps. Heat loss is an over all loss. Simply because energy conversion at higher temperature is more efficient than at low temperature. If you look at a PV digram of the stroke the work produced is the area under the pressure line. As heat is transfered out of the working fluid during expansion the temperature and pressure will be less then the isentropic path and less work produced. In the case of adding heat it's a lot more complicated. Looking expansion with heat transfer. For explanation say the RPM is absoloutly constant for now. During each step of the expansion the volume changes. We will say from v1 to v2. The pressure also changes from p1 to p2 during an isentropic expansion. With heat transfer we wind up at P2a. As we have a fixed volume change with constant RPM. The heat transfer will effect the pressure change. It's obvious that P2a will be greater then P2 when we add heat and less then P2 if heat is removed. In the case of adding heat you are increassing it's internal energy but not extracting it through expansion. It increases the work produced. Over all you are not getting as much internal energy conversion as you would have if all the heat add had been in the initial admission instead.

What really complicates things is that we do not have the ideal. Cylander surfaces are heat conducters. And you are going to have some heat transfer. Is it better to always have heat transfer to the working fluid? When it prevents condensation. Yes. We have stats to prove that case for steam jacketed engines using saturated steam. But it shows little if any improvement when using superheated steam. I havn't seen the stats. Did they include the heat to maintain the steam jacket? Was that counted against the engine? I would assume it was. But on the other it could have been more sails hype then fact.

There are factors other the engine efficiency that effect vehical efficiency. The botom line is MPG. A smaller less efficient more powerful engine could lead to less wind resistance and higher MPG.

Meterial stringth and cost are also factors.

As for "Flash Steam" I don't think it is correct to call it flash steam. From the formula you use, I see you are using the curent flash steam deffination. The main reasion I don't think flash steam is correct is that the curent defination of flash steam is a phase change. And there is no ware that a complete phase change of all the substance can occure having a saturated liquid heat content. In a flash steam process only part of the liquid is converted to a vapor.

Got to go now. Lunch time. Will get back to this later.

Andy

---------------------------------------------------------------------------------------------------

Hi Jeremy

First I don't think one should use old terminology to mean something different: "rejected heat of entropy" The "rejected heat" part is confusing. Rejected heat has always been used to describe the part of the heat that was not used. The heat rejected through exhaust. The part of the heat that was not converted to work etc. When you say "rejected heat of entropy" it doesn't make sense to me. The definition of entropy is the differential equation: dS=dQ/T It relates the change in entropy (dS) to the heat transferred (dQ). "dS" and "dQ" are differentials (infinitesimal smile values).

In an isentropic process the temperature, enthalpy and pressure all change. But sense the entropy is constant (unchanging) there is no heat transfer to or form the substance going through the process. dS was zero through the process. Therefore dQ was also 0 through the entire process. When figuring work and other values during process having no heat transfer the term dQ/T was common. So common that they decided to give it a name. Sense it is in differential form that came up with the equation dS=dQ/T. And S = integral 1/T dQ. The thing is that entropy was invented to be able to analyze expansion and compression where no heat transfer occurs. I have one old book that talks about way entropy was invented.

In figuring what the ideal cycle should be. They first figured the most efficient heat to work conversion process and that turns out to be along a line of constant entropy. An isentropic expansion. It can be shown that adding or extracting heat during expansion reduces the percentage of heat converted to work. I went over this as best I could with out going into the math (calculus /differential equations) which is hard to do in text. Don't know your math background. I don't know if you understand differentials and integrals. I try to explain them a bit.

There is a basic problem in teaching sciences. It's done with examples. And getting the abstract idea from doing examples doesn't work very well. [arstechnica.com] I have some books that are fairly good at explaining the principals of Theo dynamics. But none really have a lot on isothermal or isenthalpic processes. It doesn't make it easy to figure when to use these processes. They are all volume change processes. I understand equilibrium to be an isenthalpic phase change process. That is where the flash steam formula come from. That isenthalpic expansion occurs when a larger volume is opened up. The example has to connected containers with a valve between them. One has a quaintly of liquid and gas in equilibrium when the valve is opened to the other container that has a vacuum. It’s much the same case as when you open your injector. What has always bothered me is why is it isentropic and not isenthalpic or vise versa. I recently found a better explanation that has cleared that up. Expansion is always isentropic. When we have an isenthalpic process it is really an isentropic process where the heat energy is converted to kinetic energy and then the kinetic energy is converted back to heat energy.

What I am getting at is that when your piston is moving you would have an isentropic process. But it won’t be pure isentropic. There will be some heat transfer and maybe some evaporation going on.

Andy

-------------------------------------------------------

> Hi Andy,

>

> Im still pretty busy with work, I had to repair a

> PLC today on a lathe that we use at work, thank

> goodness its working ok for now.

>

> Anyhow, I wanted to get back to you.

>

> For now, lets start with your first private

> message, then go forward from there.

>

> "I was at you site and see you have changed it a

> > bit."

>

> Yes I have, but not that much. Id like to build

> more content, on top of whats there. I have no

> problem admiting mistakes.

>

>

> "I looked at your Heat of rejection page and

> > have some questions and/or thoughts."

>

> "I have seen it used in some

> > thermodynamic books"

>

> For the most part, I agree with what your saying

> about heat lost from the exhaust, Andy. But if you

> have discovered something additional, Id love to

> hear more.

>

> The main reason I brought it up on my site was an

> attempt to explain, what I perceive as a necessary

> ingredient, to promote a flash steam reaction that

> successfully pushed a piston. The more I think

> about it, most likely it could be defined as

> rejected heat of entropy, what do you think of

> this? I plan to do some updating at some point on

> FlashSteam.

>

> Andy, any discussion we have here, will(in these

> pm's) be kept confidential, unless its decided

> otherwise by us both... Let me know if thats ok

> with you.

>

> I could use some help with my site, if you would

> like to contribute, you would be noted as a

> contributing author. Also, depending how deep

> things go with the grant, I would consider you for

> paid consultant.

>

> One of the formula's you discussed on the

> 'super-critical pressure' thread, that, it appears

> that you came up with on your own, interests me

> greatly.

>

>

> Best

>

> Jeremy

>

>

>

>

>

> Andy Wrote:

> --------------------------------------------------

> -----

> > Hi Jeremy

> >

> > I was at you site and see you have changed it a

> > bit. I looked at your Heat of rejection page

> and

> > have some questions and/or thoughts. Note heat

> of

> > rejection is a term sometimes applied to heat

> > rejected the heat content of the exhaust steam

> > leaving the engine. I have seen it used in some

> > thermodynamic books.

> >

> > I do all my calculations using absolute

> pressure.

> > You use psi which doesn’t indicate what you are

> > using. But from the context it appears to be

> PSIG

> > “gage pressure”. Gage being relative to local

> > atmospheric pressure. Any way using PSIA

> standard

> > atmosphere is 14.696 PSIA. At 366F the

> saturation

> > pressure (boiling point) is 164.968 PSIA or

> around

> > 150 PSIG.

> >

> > I think I see what you are saying there. But I

> > think you are wrong in your thinking. We get

> work

> > from a substance because of a change in volume.

> > There are two processes that produce work

> > involved. Expansion is one process. Heat

> transfer

> > is the other. To analyze what is going on you

> need

> > to loot at both. When no heat transfer occurs

> > steam expands or compresses along a line of

> > constant entropy. ds = dQ/T That is transferred

> > heat dQ = ds = 0. Looking at your example of

> > water at 366F I get the state point:

> > T=366F, P=164.968 PSIA, v=0.018184 ft^3/lb,

> > h=338.663 BTU/lb, s=0.523776, 0% steam.

> >

> > Where I differ with what you say is how you get

> to

> > the expanded state. Expansion work would be

> along

> > a line of constant entropy. s==0.523776. Have

> > patience I’ll get to heat transfer work.

> >

> > The ending state point would be:

> > T=212F, P=14.696 PSIA, v=3.940157 ft^3/lb,

> > h=322.311 BTU/lb, s=0.523776, 14.6495% steam

> >

> > For the ending state point to be any different

> > then heat transfer would be taking place. I do

> not

> > doubt that heat transfer takes place. If heat

> is

> > transferred to it then there would be an

> > additional increase in volume and/or pressure

> at

> > the end of expansion. More work would be

> produced.

> > Like wise if heat were transferred from the

> > substance then there would be less work done.

> That

> > isentropic expansion with no heat transfer

> would

> > have done 16.352 BTU/lb of work. Easily

> absorbed

> > by cold cylinder walls.

> >

> > You are not necessarily expanding to

> atmospheric

> > pressure in a piston engine. You are expanding

> > between two volumes. The volume at TDC or the

> > volume of water held in your injector to the

> > ending volume at BDC or when the exhaust opens

> and

> > pressure is released. You are expanding from v1

> to

> > v2 a fixed expansion ratio of V2/V1. You would

> > have to figure you ending state from the

> initial

> > specific volume v=0.018184 *V2/V1 at constant

> > entropy. Then figure in heat transfer. If you

> > transferred say 10 BTU/lb to the steam as it

> > expanded you would wind up at a higher ending

> > pressure and that would increase the work

> > produced. But lessen the internal energy

> > conversion.

> >

> > BTUs are energy. The way you are figuring BTU

> > change does not follow the conservation of

> energy

> > law. The BTUs must either be transferred or

> > converted to work during a process for them to

> > change. You can not just figure that the ending

> > state is 180 BTUs. Study those pages on entropy

> > that you posted links for.

> >

> > A common problems is the misconception that

> heat

> > of vaporization does no work. Looking at the

> > normal steam engine processes. Water is heated

> at

> > constant pressure until it vaporizes and then

> > superheated in a boiler. There is a change in

> > volume of the water as it is heated, vaporizes

> and

> > superheated. In the engine that volume change

> does

> > work during the admittance process at constant

> > pressure. work = p*(volume change) The work

> done

> > during the expansion process is the difference

> in

> > internal energy. When you figure engine work as

> > the difference in enthalpy you are actually

> > including the constant pressure admittance work

> > p*v.

> >

> > When you heat water at constant pressure you

> are

> > increasing it's internal energy and also

> > converting some heat to work directly. The

> > increase in enthalpy accounts for both the work

> > and internal energy change. H = P*V +

> > internal_energy. While you can utilize the P*V

> > work directly. The work stored in internal

> energy

> > can only be converted by an expansion process.

> > Internal energy conversion to work is along a

> > constant entropy line. Any deviation from that

> > entropy line is produced by a heat transfer.

> That

> > heat transfer during expansion can increase or

> > decrease the work produced depending on the

> > direction of heat transfer. With a fixed

> expansion

> > ratio. Adding heat reduces the internal energy

> > conversion because you are reducing the

> internal

> > energy change by adding internal energy that

> > doesn't get converted to work. While at the

> same

> > time work out put is being increased by the

> direct

> > energy conversion of volume increase due to

> > increased heat content.

> >

> > That gets back to the reversible adiabatic

> process

> > have the highest energy conversion. The

> reversible

> > process is along the constant entropy line.

> When

> > you have heat transfer it is no longer

> reversible

> > and you have less internal energy converted to

> > work. It doesn’t matter the direction of heat

> > transfer.

> >

> > Work for a volumetric change is always p*v,

> > pressure times volume change. Work = force *

> > distance. Force = pressure * area, volume =

> > area*distance…

> >

> > See: <

> > " rel=nofollow[www.greenhills.net];;

> >

> > However during expansion pressure and volume

> are

> > changing and you must use calculus integration

> as

> > explained in the link above. The integration

> > works out to the internal energy difference.

> >

> > Those flash steam formula that you are using

> seam

> > to have more to do with natural equilibrium in

> an

> > open environment then energy conversion.

> >

> > How many BTU does it take to raise 0 PSIG water

> at

> > 180 F to 5000 PSIA if the volume is not allowed

> to

> > change.

> > P1: p=14.696 PSIA, T=180F, v= 0.016510

> ft^3/lb,

> > h= 148.02 BTU/lb, s= 0.263100

> > P2: p=5000 PSIA, T= 221.33F, v=016510 ft^3/lb,

> > h=200.56 BTU/lb, s=0.319642

> >

> > Change in enthalpy: 52.55 BTU/lb.

> >

> > Point. You are not going to get any more energy

> > out then you put in. It doesn’t take mush

> energy

> > to get a large pressure change at constant

> volume.

> > The 52.55 BTU/lb added enough heat to vaporize

> > 2.1% of the water into steam heating at

> constant

> > atmospheric pressure 14.696 PSIA. You would have

> a

> > 35:1 volume increase, constant pressure,

> heating

> > at 14.696 PSIA adding 52.55 BTU/lb

> >

> > P3: p=14.696 PSIA, T=212F, v= 0.579740

> ft^3/lb,

> > h= 200.56 BTU/lb, s= 0.3425109 2.10% vapor

> >

> > Doing an isentropic expansion from P2 to 0 PSIG,

>

> > 5000 PSIA to 14.696 PSIA:

> >

> > P4: p=14.696 PSIA, T=212F, v= 0.155785

> ft^3/lb,

> > h= 185.20 BTU/lb, s= 0.319642

> > 0.519% vapor, 9.436 expansion ratio, 15.36

> BTU/lb

> > work output.

> >

> > Andy

--------------------------------------------------------------------------------------

Hi Jeremy

I was at you site and see you have changed it a bit. I looked at your Heat of rejection page and have some questions and/or thoughts. Note heat of rejection is a term sometimes applied to heat rejected the heat content of the exhaust steam leaving the engine. I have seen it used in some thermodynamic books.

I do all my calculations using absolute pressure. You use psi which doesn’t indicate what you are using. But from the context it appears to be PSIG “gage pressure”. Gage being relative to local atmospheric pressure. Any way using PSIA standard atmosphere is 14.696 PSIA. At 366F the saturation pressure (boiling point) is 164.968 PSIA or around 150 PSIG.

I think I see what you are saying there. But I think you are wrong in your thinking. We get work from a substance because of a change in volume. There are two processes that produce work involved. Expansion is one process. Heat transfer is the other. To analyze what is going on you need to loot at both. When no heat transfer occurs steam expands or compresses along a line of constant entropy. ds = dQ/T That is transferred heat dQ = ds = 0. Looking at your example of water at 366F I get the state point:

T=366F, P=164.968 PSIA, v=0.018184 ft^3/lb, h=338.663 BTU/lb, s=0.523776, 0% steam.

Where I differ with what you say is how you get to the expanded state. Expansion work would be along a line of constant entropy. s==0.523776. Have patience I’ll get to heat transfer work.

The ending state point would be:

T=212F, P=14.696 PSIA, v=3.940157 ft^3/lb, h=322.311 BTU/lb, s=0.523776, 14.6495% steam

For the ending state point to be any different then heat transfer would be taking place. I do not doubt that heat transfer takes place. If heat is transferred to it then there would be an additional increase in volume and/or pressure at the end of expansion. More work would be produced. Like wise if heat were transferred from the substance then there would be less work done. That isentropic expansion with no heat transfer would have done 16.352 BTU/lb of work. Easily absorbed by cold cylinder walls.

You are not necessarily expanding to atmospheric pressure in a piston engine. You are expanding between two volumes. The volume at TDC or the volume of water held in your injector to the ending volume at BDC or when the exhaust opens and pressure is released. You are expanding from v1 to v2 a fixed expansion ratio of V2/V1. You would have to figure you ending state from the initial specific volume v=0.018184 *V2/V1 at constant entropy. Then figure in heat transfer. If you transferred say 10 BTU/lb to the steam as it expanded you would wind up at a higher ending pressure and that would increase the work produced. But lessen the internal energy conversion.

BTUs are energy. The way you are figuring BTU change does not follow the conservation of energy law. The BTUs must either be transferred or converted to work during a process for them to change. You can not just figure that the ending state is 180 BTUs. Study those pages on entropy that you posted links for.

A common problems is the misconception that heat of vaporization does no work. Looking at the normal steam engine processes. Water is heated at constant pressure until it vaporizes and then superheated in a boiler. There is a change in volume of the water as it is heated, vaporizes and superheated. In the engine that volume change does work during the admittance process at constant pressure. work = p*(volume change) The work done during the expansion process is the difference in internal energy. When you figure engine work as the difference in enthalpy you are actually including the constant pressure admittance work p*v.

When you heat water at constant pressure you are increasing it's internal energy and also converting some heat to work directly. The increase in enthalpy accounts for both the work and internal energy change. H = P*V + internal_energy. While you can utilize the P*V work directly. The work stored in internal energy can only be converted by an expansion process. Internal energy conversion to work is along a constant entropy line. Any deviation from that entropy line is produced by a heat transfer. That heat transfer during expansion can increase or decrease the work produced depending on the direction of heat transfer. With a fixed expansion ratio. Adding heat reduces the internal energy conversion because you are reducing the internal energy change by adding internal energy that doesn't get converted to work. While at the same time work out put is being increased by the direct energy conversion of volume increase due to increased heat content.

That gets back to the reversible adiabatic process have the highest energy conversion. The reversible process is along the constant entropy line. When you have heat transfer it is no longer reversible and you have less internal energy converted to work. It doesn’t matter the direction of heat transfer.

Work for a volumetric change is always p*v, pressure times volume change. Work = force * distance. Force = pressure * area, volume = area*distance…

See: < " rel=nofollow[www.greenhills.net];;

However during expansion pressure and volume are changing and you must use calculus integration as explained in the link above. The integration works out to the internal energy difference.

Those flash steam formula that you are using seam to have more to do with natural equilibrium in an open environment then energy conversion.

How many BTU does it take to raise 0 PSIG water at 180 F to 5000 PSIA if the volume is not allowed to change.

P1: p=14.696 PSIA, T=180F, v= 0.016510 ft^3/lb, h= 148.02 BTU/lb, s= 0.263100

P2: p=5000 PSIA, T= 221.33F, v=016510 ft^3/lb, h=200.56 BTU/lb, s=0.319642

Change in enthalpy: 52.55 BTU/lb.

Point. You are not going to get any more energy out then you put in. It doesn’t take mush energy to get a large pressure change at constant volume. The 52.55 BTU/lb added enough heat to vaporize 2.1% of the water into steam heating at constant atmospheric pressure 14.696 PSIA. You would have a 35:1 volume increase, constant pressure, heating at 14.696 PSIA adding 52.55 BTU/lb

P3: p=14.696 PSIA, T=212F, v= 0.579740 ft^3/lb, h= 200.56 BTU/lb, s= 0.3425109 2.10% vapor

Doing an isentropic expansion from P2 to 0 PSIG, 5000 PSIA to 14.696 PSIA:

P4: p=14.696 PSIA, T=212F, v= 0.155785 ft^3/lb, h= 185.20 BTU/lb, s= 0.319642

0.519% vapor, 9.436 expansion ratio, 15.36 BTU/lb work output.

Andy

-------------------------------------------------------------

Hi All,

I thought these notes, and PDF's, by Andy P. / SteamerAndy

Hold relevance, I refer to them, from time to time, so I wanted to make it easier for me to access the files. And I made this thread, thanks for your work on this Andy...

Best

Jeremy

[www.steamautomobile.com]

-----------

More on the flash process producing work.

I don't think your valve/injector is any different than a normal valve in that it is a throttling process. Basicly presure reducing. A constant enthalpy process. The result is an admission volume in the cylander at the reduced pressure having the same initial enthalpy. The work produced is the pressure times the volume change.

As far as calculating work the flash steam and throttling process are both constant enthalpy processes. The only differance between flash steam and throttling is flash steam is initially a liquid and would not all vaporize living some lquid in the mix.

It does take some time to change phase. In the IAPWS site they talk about how well their formula extend into subcooled gas and superheated liquid states. Thoes being the unstable states during phase transition. I havn't been able to find anything on duration of thoes states other then they last long enough that their properties are important in nozel calculations.

I have to backpeddel a bit. It comes down to wether or not you have a liquid state in you injector. If it is a liquid you would have a flash steam process during admitance. But I think it would be better to have a gas that would not have any liquid content at the end of admitance.

Andy

-------------------------------------------------------------------------------

Hi Jeremy

The reasion it is less efficient to add heat during expansion comes from fact that conversion of heat energy to work when there is no heat transfer takes place along the line of constant entropy. That doesn't mean an engine doing it wont work. There are other factors can effect efficiency. Heat recovery from the exhaust steam for instance. Adding heat during expansio might benifit heat recovery and maybe wind with better over all efficiency.

The ideal cycles are static cycles. By that I mean they are not time dependent. They do not include any time dependent processes. Like heat transfer for example. A real engine is of course dynamic. Besides heat transfer being time dependent you have flow processes that are very dynamic. To analyze a dynamic cycle you have to model the processes involved in a simulation. The processes are not stady state. The valves create a palsating flow. You have pressure(sound) waves that interact with valve timing. You have heat transfer whose area exposier time varies with RPM. It a very complicated model to make. And then would have to be tested and tuned for agreement with an operatoring engine. It would take a special engine with sensors that can be used to verify the model. That has been my goal.

Heat transfer to the working fluid does work. In the clasical steam engine heat transfer takes place in the boiler at constant pressure. Work produced at constant pressure is simply the pressure times the volume change. The clasical engine utilizzes that work during admission. Analying a non-expanding cycle, one haveing admision for the full stroke, you can get around 8% efficiency. Expanding that steam along an isentropic line where you have no heat transfer and thus all energy change within the substance is converted to work, you have both pressure and volume changing during expansion so it can not simply be calculated as P * (voume change). But in theory it is equilivant to the change in internal energy.

To analyze the processes with heat transfer you break the stroke into tiny steps. Modling software does so in small time steps. Heat loss is an over all loss. Simply because energy conversion at higher temperature is more efficient than at low temperature. If you look at a PV digram of the stroke the work produced is the area under the pressure line. As heat is transfered out of the working fluid during expansion the temperature and pressure will be less then the isentropic path and less work produced. In the case of adding heat it's a lot more complicated. Looking expansion with heat transfer. For explanation say the RPM is absoloutly constant for now. During each step of the expansion the volume changes. We will say from v1 to v2. The pressure also changes from p1 to p2 during an isentropic expansion. With heat transfer we wind up at P2a. As we have a fixed volume change with constant RPM. The heat transfer will effect the pressure change. It's obvious that P2a will be greater then P2 when we add heat and less then P2 if heat is removed. In the case of adding heat you are increassing it's internal energy but not extracting it through expansion. It increases the work produced. Over all you are not getting as much internal energy conversion as you would have if all the heat add had been in the initial admission instead.

What really complicates things is that we do not have the ideal. Cylander surfaces are heat conducters. And you are going to have some heat transfer. Is it better to always have heat transfer to the working fluid? When it prevents condensation. Yes. We have stats to prove that case for steam jacketed engines using saturated steam. But it shows little if any improvement when using superheated steam. I havn't seen the stats. Did they include the heat to maintain the steam jacket? Was that counted against the engine? I would assume it was. But on the other it could have been more sails hype then fact.

There are factors other the engine efficiency that effect vehical efficiency. The botom line is MPG. A smaller less efficient more powerful engine could lead to less wind resistance and higher MPG.

Meterial stringth and cost are also factors.

As for "Flash Steam" I don't think it is correct to call it flash steam. From the formula you use, I see you are using the curent flash steam deffination. The main reasion I don't think flash steam is correct is that the curent defination of flash steam is a phase change. And there is no ware that a complete phase change of all the substance can occure having a saturated liquid heat content. In a flash steam process only part of the liquid is converted to a vapor.

Got to go now. Lunch time. Will get back to this later.

Andy

---------------------------------------------------------------------------------------------------

Hi Jeremy

First I don't think one should use old terminology to mean something different: "rejected heat of entropy" The "rejected heat" part is confusing. Rejected heat has always been used to describe the part of the heat that was not used. The heat rejected through exhaust. The part of the heat that was not converted to work etc. When you say "rejected heat of entropy" it doesn't make sense to me. The definition of entropy is the differential equation: dS=dQ/T It relates the change in entropy (dS) to the heat transferred (dQ). "dS" and "dQ" are differentials (infinitesimal smile values).

In an isentropic process the temperature, enthalpy and pressure all change. But sense the entropy is constant (unchanging) there is no heat transfer to or form the substance going through the process. dS was zero through the process. Therefore dQ was also 0 through the entire process. When figuring work and other values during process having no heat transfer the term dQ/T was common. So common that they decided to give it a name. Sense it is in differential form that came up with the equation dS=dQ/T. And S = integral 1/T dQ. The thing is that entropy was invented to be able to analyze expansion and compression where no heat transfer occurs. I have one old book that talks about way entropy was invented.

In figuring what the ideal cycle should be. They first figured the most efficient heat to work conversion process and that turns out to be along a line of constant entropy. An isentropic expansion. It can be shown that adding or extracting heat during expansion reduces the percentage of heat converted to work. I went over this as best I could with out going into the math (calculus /differential equations) which is hard to do in text. Don't know your math background. I don't know if you understand differentials and integrals. I try to explain them a bit.

There is a basic problem in teaching sciences. It's done with examples. And getting the abstract idea from doing examples doesn't work very well. [arstechnica.com] I have some books that are fairly good at explaining the principals of Theo dynamics. But none really have a lot on isothermal or isenthalpic processes. It doesn't make it easy to figure when to use these processes. They are all volume change processes. I understand equilibrium to be an isenthalpic phase change process. That is where the flash steam formula come from. That isenthalpic expansion occurs when a larger volume is opened up. The example has to connected containers with a valve between them. One has a quaintly of liquid and gas in equilibrium when the valve is opened to the other container that has a vacuum. It’s much the same case as when you open your injector. What has always bothered me is why is it isentropic and not isenthalpic or vise versa. I recently found a better explanation that has cleared that up. Expansion is always isentropic. When we have an isenthalpic process it is really an isentropic process where the heat energy is converted to kinetic energy and then the kinetic energy is converted back to heat energy.

What I am getting at is that when your piston is moving you would have an isentropic process. But it won’t be pure isentropic. There will be some heat transfer and maybe some evaporation going on.

Andy

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> Hi Andy,

>

> Im still pretty busy with work, I had to repair a

> PLC today on a lathe that we use at work, thank

> goodness its working ok for now.

>

> Anyhow, I wanted to get back to you.

>

> For now, lets start with your first private

> message, then go forward from there.

>

> "I was at you site and see you have changed it a

> > bit."

>

> Yes I have, but not that much. Id like to build

> more content, on top of whats there. I have no

> problem admiting mistakes.

>

>

> "I looked at your Heat of rejection page and

> > have some questions and/or thoughts."

>

> "I have seen it used in some

> > thermodynamic books"

>

> For the most part, I agree with what your saying

> about heat lost from the exhaust, Andy. But if you

> have discovered something additional, Id love to

> hear more.

>

> The main reason I brought it up on my site was an

> attempt to explain, what I perceive as a necessary

> ingredient, to promote a flash steam reaction that

> successfully pushed a piston. The more I think

> about it, most likely it could be defined as

> rejected heat of entropy, what do you think of

> this? I plan to do some updating at some point on

> FlashSteam.

>

> Andy, any discussion we have here, will(in these

> pm's) be kept confidential, unless its decided

> otherwise by us both... Let me know if thats ok

> with you.

>

> I could use some help with my site, if you would

> like to contribute, you would be noted as a

> contributing author. Also, depending how deep

> things go with the grant, I would consider you for

> paid consultant.

>

> One of the formula's you discussed on the

> 'super-critical pressure' thread, that, it appears

> that you came up with on your own, interests me

> greatly.

>

>

> Best

>

> Jeremy

>

>

>

>

>

> Andy Wrote:

> --------------------------------------------------

> -----

> > Hi Jeremy

> >

> > I was at you site and see you have changed it a

> > bit. I looked at your Heat of rejection page

> and

> > have some questions and/or thoughts. Note heat

> of

> > rejection is a term sometimes applied to heat

> > rejected the heat content of the exhaust steam

> > leaving the engine. I have seen it used in some

> > thermodynamic books.

> >

> > I do all my calculations using absolute

> pressure.

> > You use psi which doesn’t indicate what you are

> > using. But from the context it appears to be

> PSIG

> > “gage pressure”. Gage being relative to local

> > atmospheric pressure. Any way using PSIA

> standard

> > atmosphere is 14.696 PSIA. At 366F the

> saturation

> > pressure (boiling point) is 164.968 PSIA or

> around

> > 150 PSIG.

> >

> > I think I see what you are saying there. But I

> > think you are wrong in your thinking. We get

> work

> > from a substance because of a change in volume.

> > There are two processes that produce work

> > involved. Expansion is one process. Heat

> transfer

> > is the other. To analyze what is going on you

> need

> > to loot at both. When no heat transfer occurs

> > steam expands or compresses along a line of

> > constant entropy. ds = dQ/T That is transferred

> > heat dQ = ds = 0. Looking at your example of

> > water at 366F I get the state point:

> > T=366F, P=164.968 PSIA, v=0.018184 ft^3/lb,

> > h=338.663 BTU/lb, s=0.523776, 0% steam.

> >

> > Where I differ with what you say is how you get

> to

> > the expanded state. Expansion work would be

> along

> > a line of constant entropy. s==0.523776. Have

> > patience I’ll get to heat transfer work.

> >

> > The ending state point would be:

> > T=212F, P=14.696 PSIA, v=3.940157 ft^3/lb,

> > h=322.311 BTU/lb, s=0.523776, 14.6495% steam

> >

> > For the ending state point to be any different

> > then heat transfer would be taking place. I do

> not

> > doubt that heat transfer takes place. If heat

> is

> > transferred to it then there would be an

> > additional increase in volume and/or pressure

> at

> > the end of expansion. More work would be

> produced.

> > Like wise if heat were transferred from the

> > substance then there would be less work done.

> That

> > isentropic expansion with no heat transfer

> would

> > have done 16.352 BTU/lb of work. Easily

> absorbed

> > by cold cylinder walls.

> >

> > You are not necessarily expanding to

> atmospheric

> > pressure in a piston engine. You are expanding

> > between two volumes. The volume at TDC or the

> > volume of water held in your injector to the

> > ending volume at BDC or when the exhaust opens

> and

> > pressure is released. You are expanding from v1

> to

> > v2 a fixed expansion ratio of V2/V1. You would

> > have to figure you ending state from the

> initial

> > specific volume v=0.018184 *V2/V1 at constant

> > entropy. Then figure in heat transfer. If you

> > transferred say 10 BTU/lb to the steam as it

> > expanded you would wind up at a higher ending

> > pressure and that would increase the work

> > produced. But lessen the internal energy

> > conversion.

> >

> > BTUs are energy. The way you are figuring BTU

> > change does not follow the conservation of

> energy

> > law. The BTUs must either be transferred or

> > converted to work during a process for them to

> > change. You can not just figure that the ending

> > state is 180 BTUs. Study those pages on entropy

> > that you posted links for.

> >

> > A common problems is the misconception that

> heat

> > of vaporization does no work. Looking at the

> > normal steam engine processes. Water is heated

> at

> > constant pressure until it vaporizes and then

> > superheated in a boiler. There is a change in

> > volume of the water as it is heated, vaporizes

> and

> > superheated. In the engine that volume change

> does

> > work during the admittance process at constant

> > pressure. work = p*(volume change) The work

> done

> > during the expansion process is the difference

> in

> > internal energy. When you figure engine work as

> > the difference in enthalpy you are actually

> > including the constant pressure admittance work

> > p*v.

> >

> > When you heat water at constant pressure you

> are

> > increasing it's internal energy and also

> > converting some heat to work directly. The

> > increase in enthalpy accounts for both the work

> > and internal energy change. H = P*V +

> > internal_energy. While you can utilize the P*V

> > work directly. The work stored in internal

> energy

> > can only be converted by an expansion process.

> > Internal energy conversion to work is along a

> > constant entropy line. Any deviation from that

> > entropy line is produced by a heat transfer.

> That

> > heat transfer during expansion can increase or

> > decrease the work produced depending on the

> > direction of heat transfer. With a fixed

> expansion

> > ratio. Adding heat reduces the internal energy

> > conversion because you are reducing the

> internal

> > energy change by adding internal energy that

> > doesn't get converted to work. While at the

> same

> > time work out put is being increased by the

> direct

> > energy conversion of volume increase due to

> > increased heat content.

> >

> > That gets back to the reversible adiabatic

> process

> > have the highest energy conversion. The

> reversible

> > process is along the constant entropy line.

> When

> > you have heat transfer it is no longer

> reversible

> > and you have less internal energy converted to

> > work. It doesn’t matter the direction of heat

> > transfer.

> >

> > Work for a volumetric change is always p*v,

> > pressure times volume change. Work = force *

> > distance. Force = pressure * area, volume =

> > area*distance…

> >

> > See: <

> > " rel=nofollow[www.greenhills.net];;

> >

> > However during expansion pressure and volume

> are

> > changing and you must use calculus integration

> as

> > explained in the link above. The integration

> > works out to the internal energy difference.

> >

> > Those flash steam formula that you are using

> seam

> > to have more to do with natural equilibrium in

> an

> > open environment then energy conversion.

> >

> > How many BTU does it take to raise 0 PSIG water

> at

> > 180 F to 5000 PSIA if the volume is not allowed

> to

> > change.

> > P1: p=14.696 PSIA, T=180F, v= 0.016510

> ft^3/lb,

> > h= 148.02 BTU/lb, s= 0.263100

> > P2: p=5000 PSIA, T= 221.33F, v=016510 ft^3/lb,

> > h=200.56 BTU/lb, s=0.319642

> >

> > Change in enthalpy: 52.55 BTU/lb.

> >

> > Point. You are not going to get any more energy

> > out then you put in. It doesn’t take mush

> energy

> > to get a large pressure change at constant

> volume.

> > The 52.55 BTU/lb added enough heat to vaporize

> > 2.1% of the water into steam heating at

> constant

> > atmospheric pressure 14.696 PSIA. You would have

> a

> > 35:1 volume increase, constant pressure,

> heating

> > at 14.696 PSIA adding 52.55 BTU/lb

> >

> > P3: p=14.696 PSIA, T=212F, v= 0.579740

> ft^3/lb,

> > h= 200.56 BTU/lb, s= 0.3425109 2.10% vapor

> >

> > Doing an isentropic expansion from P2 to 0 PSIG,

>

> > 5000 PSIA to 14.696 PSIA:

> >

> > P4: p=14.696 PSIA, T=212F, v= 0.155785

> ft^3/lb,

> > h= 185.20 BTU/lb, s= 0.319642

> > 0.519% vapor, 9.436 expansion ratio, 15.36

> BTU/lb

> > work output.

> >

> > Andy

--------------------------------------------------------------------------------------

Hi Jeremy

I was at you site and see you have changed it a bit. I looked at your Heat of rejection page and have some questions and/or thoughts. Note heat of rejection is a term sometimes applied to heat rejected the heat content of the exhaust steam leaving the engine. I have seen it used in some thermodynamic books.

I do all my calculations using absolute pressure. You use psi which doesn’t indicate what you are using. But from the context it appears to be PSIG “gage pressure”. Gage being relative to local atmospheric pressure. Any way using PSIA standard atmosphere is 14.696 PSIA. At 366F the saturation pressure (boiling point) is 164.968 PSIA or around 150 PSIG.

I think I see what you are saying there. But I think you are wrong in your thinking. We get work from a substance because of a change in volume. There are two processes that produce work involved. Expansion is one process. Heat transfer is the other. To analyze what is going on you need to loot at both. When no heat transfer occurs steam expands or compresses along a line of constant entropy. ds = dQ/T That is transferred heat dQ = ds = 0. Looking at your example of water at 366F I get the state point:

T=366F, P=164.968 PSIA, v=0.018184 ft^3/lb, h=338.663 BTU/lb, s=0.523776, 0% steam.

Where I differ with what you say is how you get to the expanded state. Expansion work would be along a line of constant entropy. s==0.523776. Have patience I’ll get to heat transfer work.

The ending state point would be:

T=212F, P=14.696 PSIA, v=3.940157 ft^3/lb, h=322.311 BTU/lb, s=0.523776, 14.6495% steam

For the ending state point to be any different then heat transfer would be taking place. I do not doubt that heat transfer takes place. If heat is transferred to it then there would be an additional increase in volume and/or pressure at the end of expansion. More work would be produced. Like wise if heat were transferred from the substance then there would be less work done. That isentropic expansion with no heat transfer would have done 16.352 BTU/lb of work. Easily absorbed by cold cylinder walls.

You are not necessarily expanding to atmospheric pressure in a piston engine. You are expanding between two volumes. The volume at TDC or the volume of water held in your injector to the ending volume at BDC or when the exhaust opens and pressure is released. You are expanding from v1 to v2 a fixed expansion ratio of V2/V1. You would have to figure you ending state from the initial specific volume v=0.018184 *V2/V1 at constant entropy. Then figure in heat transfer. If you transferred say 10 BTU/lb to the steam as it expanded you would wind up at a higher ending pressure and that would increase the work produced. But lessen the internal energy conversion.

BTUs are energy. The way you are figuring BTU change does not follow the conservation of energy law. The BTUs must either be transferred or converted to work during a process for them to change. You can not just figure that the ending state is 180 BTUs. Study those pages on entropy that you posted links for.

A common problems is the misconception that heat of vaporization does no work. Looking at the normal steam engine processes. Water is heated at constant pressure until it vaporizes and then superheated in a boiler. There is a change in volume of the water as it is heated, vaporizes and superheated. In the engine that volume change does work during the admittance process at constant pressure. work = p*(volume change) The work done during the expansion process is the difference in internal energy. When you figure engine work as the difference in enthalpy you are actually including the constant pressure admittance work p*v.

When you heat water at constant pressure you are increasing it's internal energy and also converting some heat to work directly. The increase in enthalpy accounts for both the work and internal energy change. H = P*V + internal_energy. While you can utilize the P*V work directly. The work stored in internal energy can only be converted by an expansion process. Internal energy conversion to work is along a constant entropy line. Any deviation from that entropy line is produced by a heat transfer. That heat transfer during expansion can increase or decrease the work produced depending on the direction of heat transfer. With a fixed expansion ratio. Adding heat reduces the internal energy conversion because you are reducing the internal energy change by adding internal energy that doesn't get converted to work. While at the same time work out put is being increased by the direct energy conversion of volume increase due to increased heat content.

That gets back to the reversible adiabatic process have the highest energy conversion. The reversible process is along the constant entropy line. When you have heat transfer it is no longer reversible and you have less internal energy converted to work. It doesn’t matter the direction of heat transfer.

Work for a volumetric change is always p*v, pressure times volume change. Work = force * distance. Force = pressure * area, volume = area*distance…

See: < " rel=nofollow[www.greenhills.net];;

However during expansion pressure and volume are changing and you must use calculus integration as explained in the link above. The integration works out to the internal energy difference.

Those flash steam formula that you are using seam to have more to do with natural equilibrium in an open environment then energy conversion.

How many BTU does it take to raise 0 PSIG water at 180 F to 5000 PSIA if the volume is not allowed to change.

P1: p=14.696 PSIA, T=180F, v= 0.016510 ft^3/lb, h= 148.02 BTU/lb, s= 0.263100

P2: p=5000 PSIA, T= 221.33F, v=016510 ft^3/lb, h=200.56 BTU/lb, s=0.319642

Change in enthalpy: 52.55 BTU/lb.

Point. You are not going to get any more energy out then you put in. It doesn’t take mush energy to get a large pressure change at constant volume. The 52.55 BTU/lb added enough heat to vaporize 2.1% of the water into steam heating at constant atmospheric pressure 14.696 PSIA. You would have a 35:1 volume increase, constant pressure, heating at 14.696 PSIA adding 52.55 BTU/lb

P3: p=14.696 PSIA, T=212F, v= 0.579740 ft^3/lb, h= 200.56 BTU/lb, s= 0.3425109 2.10% vapor

Doing an isentropic expansion from P2 to 0 PSIG, 5000 PSIA to 14.696 PSIA:

P4: p=14.696 PSIA, T=212F, v= 0.155785 ft^3/lb, h= 185.20 BTU/lb, s= 0.319642

0.519% vapor, 9.436 expansion ratio, 15.36 BTU/lb work output.

Andy

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Hi All,

I thought these notes, and PDF's, by Andy P. / SteamerAndy

Hold relevance, I refer to them, from time to time, so I wanted to make it easier for me to access the files. And I made this thread, thanks for your work on this Andy...

Best

Jeremy

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