water + hot gas + cilinder = ?

[PedroCG]

hi,

I've searched the foruns and find nothing similar to this... lets say a gas is compressed in a cilinder, now, if you inject water... the water would turn to steam, but how? volume would be constant, entropy? internal energy? if someone can point me in the rigth direction i'd be tankfull

 

[GregLocock]

The internal energy of the gas would drop by Cv/m degrees for each joule of energy required to warm the water up and evaporate it and warm it up some more.

I strongly recommend a thermodynamics book.




Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[BrianPetersen]

Hint: Water injection after compression causes a loss of useful energy. In simpler terms, the volume that the steam takes up, is a lot less than the shrinkage in volume of the gases as a result of their giving up the heat necessary to boil the water.

 

[ivymike]

ah, but if you used coal or wood to heat a cylinder full of water, you could use the steam to drive pistons! I wonder if anyone has ever tried moving a vehicle that way?

 

[BrianPetersen]

Maybe in oh, about 1794 ...

 

[black2003cobra]

Ah, a sense of humor!!

Pedro - The energy required to evaporate the water results in a reduction in internal energy of the hot air. Once the water starts to evaporate, the amount of energy required to change its phase is given by the product of its mass and latent heat of vaporization (Lvap), i.e.,

?Q = Lvap*m_water.

This heat energy “comes out of” the hot air and reduces its internal energy in accordance with the first law, (?U = ?Q – W = ?Q, since W = 0 for this case).

To take it a step further, this reduction of internal energy comes with a commensurate drop in its temperature. For the case as you describe, this drop in air temperature is related to its constant-volume specific heat (cv) and its mass as given by,

?T = ?Q/(cv*m_air)

Therefore the drop in air temperature from the evaporation of the water is given by,

?T = Lvap*m_water/(cv*m_air)

I agree with the above suggestion about checking out a book on thermodynamics.

 

[ivymike]

yeah, that's as good as it gets from me.

 

[GregLocock]

Bah, typo/brainfade in my post. should be * not slash.

Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[PedroCG]

thanks for the replys guys, i have a book but its one million pages long XD and i didnt know what to look for.

Greg and cobra, since pressure and temperature would not be constant Lvap would not be constant either... it could be if "delta"mwater was used i guess.

last nigth when i went to sleep i thougth of a way to aproach this, the entalpy loss of gas would be the entalpy gain of water, pressure would be the same for gas and water, this process would stop when temperature is the same for gas and water. is this rigth or should i go back to bed?

Brian, that depends on initial conditions of water (i guess)

i'll have to find time to read that million pages

 

[ivymike]

yes, heat transfer stops when the temperature is equal, and if you have no losses to or additions from the outside, total energy remains constant.

For a rough take on it:
Say you have a naturally aspirated 4cyl engine with 2.3L displacement, 90mm bore, 90mm stroke. The total mass in a cylinder after firing might be about 0.0072kg (air+fuel). Let's pick a temperature of about 4000C just for grins, and a peak firing pressure of 123kPa. TDC volume is 0.072L. (that temp is too high, but it's fine for demonstration purposes)

take your cylinder, time 0 you have a mass M0 of combustion products at temp T0 and pressure P0. You can probably get close enough for your purposes by considering the combustion products to be simply hot air. The total energy inside your cylinder is U0 = Cv.air * M0 * T0

U0 = (1.0035 kJ/kg*K) * (0.0072 kg) * (4273 K)
U0 = 30.8kJ

Add some liquid water - say 3g of it at 50C, and solve for the new temperature... total U of the system is a bit higher than it had been, since the water had some heat when it was added (4.1kJ added - new total is 34.9kJ).

first the water must be heated to boiling. in heating it, you transfer 0.63kJ from the air to the water.

next the water boils. in boiling it, you transfer 6.8kJ from the air to the water.

now you've got 3g of steam at 100C and 7.2g of air at 3246K. They'll reach equilibrium when the temperatures are equal. At 100C, it takes 2.08 kJ to raise the temperature of 1 kg of steam by 1K. If you use that value as constant, the equilibrium temperature you get is about 1915K.

Now you've got a ~0.072L container with 3g of steam and 7.2g of air inside, at ~1915K. The pressure contribution of the air would be 55kPa, and that of the steam would be 37kPa, so you would have lost about 31kPa of cylinder pressure despite having added mass and a bit of energy.

 

[PedroCG]

thanks ivymike thats what i've dreamed about XD, just one thing, you are using Cp not Cv, so the result is H0 (entalpy) not U0 (internal energy). wich i think is what should be used since entalpy is total energy.

now, if the work needed to compress the gas is bigger than the work produced by the expansion... how is the crower six stroke engine supossed to have better efficiency?
is there so much heat on the piston to compensate for this?

 

[ivymike]

we're using a constant volume, and a constant mass (if you slightly redefine the "problem" as follows: a cylinder initially contains 7.2g of air at 4000C and 3g of water at 50C. Initial pressure is 123kPA. Find the equilibrium pressure and temperature.)

Correcting my error above (Cv = 0.718 kJ/kg.K), we find that you get an equilibrium temperature of 1489K and an equilibrium pressure of 72kPa, for a loss of 51kPA.

 

[ivymike]

regarding the "crower six stroke engine," perhaps you'll need to explain what it is and what it supposedly does before anyone answers that. I'm sure the information is out there on the internet somewhere, but I don't feel like searching for it.

 

[ivymike]

and one final add-on post, for little benefit:

http://www.roymech.co.uk/Related/Thermos/Thermos_Specific_heat.html

Enthalpy
In many thermodynamic fluid process analyses the sum of the internal energy (U) and the product of pressure (P) and volume(V) is present. The combination (U + PV) is called the enthalpy of the fluid. H is a thermodynamic fluid property but it does not have an absolute value (because it includes internal energy U) and therefore enthalpy changes are generally applied or enthalpy values are identified relative to a fixed state e.g. water at 273 deg.K . It is important to note that enthalpy is simply a combination of properties ..it is not a form of stored energy although for certain applications it can be treated as energy.

 

[PedroCG]

here you go,

http://www.autoweek.com/apps/pbcs.dll/article?AID=/20060227/FREE/302270007/1023/THISWEEKSISSUE,

it has 4 normal strokes than he added another compression stroke folowed by water injection and expansion.

 

[ivymike]

"The engine has yet to operate against a load on a dyno, but his testing to date encourages Crower to expect that once he gets hard numbers, the engine will show normal levels of power on substantially less fuel, and without overheating. "

That article is 3 years old. Hasn't it been run in a test cell yet? What were the results?

http://en.wikipedia.org/wiki/Crower_six_stroke

Apparently people have been dinking around with the concept since 1915. For some reason it hasn't caught on. (Must be some kind of conspiracy!)

 

[GregLocock]

I wouldn't mind betting that steam cleaning all the oil off the cylinder walls at very frequent intervals poses a bit of a problem.



Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[patprimmer]

Isaac

The relative cost of fuel vs the cost of corrosion resistant materials of desired strength and the cost of producing more complex designs may have changed over the years to a point that the 6 stroke may become viable.

The mentioned disadvantage of having to carry double he fuel load due to the water tank is a bit dubious as if it gets a significant fuel economy increase, for the same range,it carries less fuel which partly offsets the weight and volume of the water carried.

Bruce Cower is a smart inventive engineer, but he retired a few years ago and I believe (from anecdotal evidence) his company was taken over by his daughter who has financial and marketing type rather than an engineering background.

I don't know if this is his retirement project and if so, what resources he has or is prepared to invest for a non patentable development.

He did develop and offer for sale a rang of camshafts that had VERY late inlet closing an a recommended much higher compression ratio, like 13:1 on pump gas I think. As far as I know it was not a commercial success.

He certainly had an excellent reputation and was a commercial success with automotive after market and racing engine components. His racing valve train, con rods, cranks and mechanical fuel injection components.


Regards
Pat
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[ivymike]

the "6 stroke" doesn't sound like one of his good ideas to me. I'm not the least bit surprised that we haven't heard any recent "success stories" about it.

The cam you describe is something I always wished I could get for my Honda, because I believed at the time that I'd be able to trade power for fuel economy, which I was willing to do (even from a starting point of 105hp). I didn't know that someone was offering them at the time, and I never bothered calling a cam company to see if they'd make one for me.

 

[black2003cobra]

PedroCG - "Greg and cobra, since pressure and temperature would not be constant Lvap would not be constant either"

I was assuming the change in temperature and pressure would not be too great, so that considering Lvap to be a constant would be a reasonable estimate. Similarly, I also assumed cv to be constant since it is a very weak function of temp.

Looks like ivymike has filled in the holes for you.

Best regards.