Help With Gas Law Equations

[ckurtz]

I'm interested in doing a simple mathematical analysis of pressure changes when compressing air beginning at ambient conditions and then introducing heated air, such as run through a passive solar heater. I'm interested at looking at pressure boosts due to heat of compression factors when a given condition is that no heat is lost but is retained in the compressed mass. I'm only interested in starting and ending values of set increments such as 1.2 kg of air per step. I apologize that when dealing with pressures I work in pounds per square inch as while the English system of measurements is a completely illogical hash so is my sorry brain when it comes to thinking about pressures - mea culpa. Maybe also referring to pressure in terms of bar or atmospheres might help...

For example let's say we start with ambient air more or less at standard conditions of 15 degrees C with a mass of 1.2 kg per cubic meter; the pressure vessel volume is constant at 1 cubic meter; air pressure is 1 bar | 14.7 psia. All additions of air are in 1.2 kg amounts. For instance: Step One is the addition of a 1.2 kg mass at the above standard conditions; Step Two is the addition of an another 1.2 kg mass which has been run through a passive solar heater and heated to 65 degrees C. The Heat of Compression factor (R) for this range of temps is basically 1.4. For this example the pressure tank is perfectly insulated and all normal operating losses are ignored.

The results I get for Step One is a simple doubling of pressure to 2 bar | 29.4 psia on a volumetric basis then by multiplying by the heat of compression R factor of 1.4 boosts the pressure to 2.8 bar | 41.16 psia. Is that correct??

Could someone be so kind as to demonstrate the calculations showing what the pressure figures would be at the end of Step Two? I thought it would be simple, however, the results I got weren't pretty. Any help would be greatly appreciated.

[wrtner]

This is pretty standard thermodynamics stuff. I would send an email to the Professor of Mechanical Engineering at Bristol University and see if he will ask a student to reply.

All Mechanical Engineering staff people
.

[ckurtz]

Thanks for the suggestion. It does seem pretty standard which is why it seemed reasonable to think there are forum members here who can figure it out without much effort - they can't all be dummies like me now, can they..?

[wrtner]

Thanks for the suggestion. It does seem pretty standard which is why it seemed reasonable to think there are forum members here who can figure it out without much effort - they can't all be dummies like me now, can they..?

There are some VERY smart people indeed on this board, and the other two (along with the goons, wide eyed enthusiasts and troublemakers). But I have never noticed anyone discussing about thermodynamics.

[I_Like_Science]

For example let's say we start with ambient air more or less at standard conditions of 15 degrees C with a mass of 1.2 kg per cubic meter; the pressure vessel volume is constant at 1 cubic meter; air pressure is 1 bar | 14.7 psia. All additions of air are in 1.2 kg amounts. For instance: Step One is the addition of a 1.2 kg mass at the above standard conditions; Step Two is the addition of an another 1.2 kg mass which has been run through a passive solar heater and heated to 65 degrees C. The Heat of Compression factor (R) for this range of temps is basically 1.4. For this example the pressure tank is perfectly insulated and all normal operating losses are ignored.

The results I get for Step One is a simple doubling of pressure to 2 bar | 29.4 psia on a volumetric basis then by multiplying by the heat of compression R factor of 1.4 boosts the pressure to 2.8 bar | 41.16 psia. Is that correct??

So, let me get this straight, you have one cubic meter that is 'magically' receiving an extra 1.2kg of air? If it really is magical, and temperature is not allowed to rise in the first case, you are right that the pressure will be double.

P=rho*R*T

rho is double for you, R is a gas constant (287 metric can't be bothered with units ) and T should be obvious. If you don't allow for an increase in temperature (aka Magic) then pressure doubles.

Your second scenario is less trivial, however.

Energy of the ideal gas at constant volume can be determined from...

E = cv*m*T

cv is specific heat (kJ/(kg*K)
m is mass (kg)
T is obvious (K)

Original energy => E1 = 0.718*1.2*300 = 258.48kJ
Added energy => E2 = 0.718*1.2*350 = 301.56kJ
E1 + E2 = 560kJ (close enough)

New temp = 560/(0.718*2.4) = 325K

New density = 2.4kg/m^3 (by definition)

P = 2.4*287*325 = 223860Pa = 32.5 psi

However, this, by your defining, takes into account very little of reality, so I'd take these numbers as a very rough ballpark estimate.

But I have never noticed anyone discussing about thermodynamics.

Because thermodynamics debunks all over-unity/perpetual energy theories, most people here tend to turn a blind eye to it.

[aljhoa]

Because thermodynamics debunks all over-unity/perpetual energy theories, most people here tend to turn a blind eye to it.

Laws of Thermodynamics

The Laws of Thermodynamics are considered part of the bedrock of physics (and among physicists, about as immobile as bedrock). But there are two problems.

The first is essentially a clarification. The Laws of Physics (including specifically the Laws of Thermodynamics) are, in reality, not laws of nature, but instead man-made theories or informed guesses of how man believes nature works and how she apparently limits herself. They are a theory, and no amount of propaganda or public relations will convert them magically into fundamental laws which cannot be breached. It takes, as a matter of fact, only one contrary example to shake the most basic bedrock of theory. And the current problem is that we’re all living in an earthquake zone!

The second problem with the Law of Conservation of Energy (aka the First Law of Thermodynamics) is that one of the Assumptions on which it is based is often neglected in the mathematical treatment of the law and the results which are derived from it. No rational physicist would argue that fundamental to the Law of Conservation of Energy is the restraint or assumption that we are dealing with a closed system! If the system is not closed, then the law is not strictly applicable, and thus there is no violation of the law.

Laws of Thermodynamics

Al

[I_Like_Science]

Laws of Thermodynamics

I read most of that linked article.

While the first point you chose to quote is technically all true, the main point to dwell on is this.

It takes, as a matter of fact, only one contrary example to shake the most basic bedrock of theory.

Incredibly true. But such a contrary example to the laws of thermodynamics has never been found, at least not on a scale useful to us mortals. Think on that for a bit before you believe those who would identify centuries of research as merely theory.

On the second point, yes, the laws of conservation of mass and energy do assume a closed systems. However, this is because, to date, we have not found evidence to the contrary of the assumption that the Universe is closed, again, on a level of usefulness. And, no evidence has been brought to bear that current ideal closed systems are in fact not closed, as this author is suggesting, meaning there is no 'hidden energetic connectedness' that has ever been observed.

Even assuming the Universe is an open system still doesn't lend itself to over-unity/perpetual energy devices. Just because a system is open does not mean energy is required to be flowing through it, it only means that it can. In all the peer reviewed work to date, such an accidental flow of magic energy into a system has yet to be demonstrated.

Really, the author can play semantics with science all they want, but until someone comes up with evidence of a contrary theories validity, it's all just semantics and theorizing.

EDIT: Oh yeah, and sorry for taking this way off-topic

[wrtner]

Because thermodynamics debunks all over-unity/perpetual energy theories, most people here tend to turn a blind eye to it.

Every tiny corner of this Rolls Royce engine,which has three concentric shafts, was designed with and by thermodynamics.

[ckurtz]

So, let me get this straight, you have one cubic meter that is 'magically' receiving an extra 1.2kg of air? If it really is magical, and temperature is not allowed to rise in the first case, you are right that the pressure will be double.

I admit to finding the field of fluid dynamics rather magical; however, I apologize if I somehow gave the impression that temperature was to remain constant. In Step One I thought the temperature rise was addressed by the use of the heat of compression factor of 1.4 which boosted the pressure to the finale figure of 2.8 bar | 41.16 psia. Was that view incorrect?

Step Two was meant to build on Step One which ended with an air mass of 2.4 kg, by adding another air mass of 1.2 kg at a temperature of 65 degrees C. for a total air mass of 3.6 kg.

The only restriction for either Step One or Step Two is in regards to maintaining a constant volume; while I am mainly interested in the pressure, the final temperature would be of interest as well.

When I've looked at examples of ideal gas law calculations, especially those of constant volume, it seemed most ignored the method by which the additional mass was supplied, and since my interest is in seeing the effect that the heat of compression can have the conditions were set to reflect that interest.

Thanks for showing how to approach the problem. Give me a chance to try and digest the information you've provided, on first look it seems pretty logical now that it's all laid out, and if I have any questions I'll do another post. Appreciate your taking the time to set the basic calculations out; it is very helpful even for a non-mathematical type as myself.

Oh yeah, and sorry for taking this way off-topic.

Hell, if you'll take the time to try and answer my questions, go off on any topic you want. I stick to terms such as 'renewable energy' myself since as long as the sun shines, the breezes blow, and the rivers flow it can't be all bad.

[ckurtz]

There are some VERY smart people indeed on this board, and the other two (along with the goons, wide eyed enthusiasts and troublemakers). But I have never noticed anyone discussing about thermodynamics.

That's an interesting observation, which I guess is an indication of where most members' interest lie, such as with radiant or electro-magnetic phenomena. I'll admit to an interest in most any approach to renewable energy however most of my time has been spent looking at solar for third world or off-grid applications. Initially that was in the field of collector design and energy storage but eventually morphed into more indirect approaches. Spent the past year or so looking at developing a specific alternative to high temperature solar power extraction designs, and believe it or not the question here is related to that as I've been seduced by the magic of fluid dynamics for some time now, especially in viewing the atmosphere as the most practical solar energy reservoir we have access to. Expect to be posting on this subject over at the energyscience forum in a few days. Thanks for taking the time to post your observations.

[I_Like_Science]

I admit to finding the field of fluid dynamics rather magical; however, I apologize if I somehow gave the impression that temperature was to remain constant. In Step One I thought the temperature rise was addressed by the use of the heat of compression factor of 1.4 which boosted the pressure to the finale figure of 2.8 bar | 41.16 psia. Was that view incorrect?

If you do allow for the temperature to rise, you are indeed correct in saying the new pressure will be 2.8 bar. Out of curiosity, what equation did you use?

Just for reference, this is the equation I used for temperature...

(T2/T1) = (rho2/rho1)^(gamma-1)

Where rho is density, and gamma is the ideal gas constant of 1.4. Then plug rho2 and T2 into the ideal gas equation for P2.

I think you can now take care of the second step with the process I laid out already, but feel free to PM me any questions, though I'm not religiously on this site.

I stick to terms such as 'renewable energy' myself since as long as the sun shines, the breezes blow, and the rivers flow it can't be all bad.

100% agreed.

Good luck.

[ckurtz]

All right, after contemplating things for awhile... well, after banging my head against the wall a few times and then whimpering about it, here is my attempt at answering mine own question along the line suggested by I_Like_Science. All the original conditions remain the same: T1 = 288K; P1 = 14.7psia; density = 1.2kg/cubic meter; there are NO heat losses; Volume is constant at 1 cubic meter. The sign 'Y' = gamma = cp/cv = 1.4; the gas constant 'R' = cp - cv = 0.287; rho = density.

A) Step One (re-stated a little): Compress 2 cubic meters of air (2.4 kg) to 1 cubic meter.

P2 = P1(V1/V2)^Y = 14.7psia(2/1)^1.4 = 14.7psia*2.639 = 38.79 psia
T2 = T1(rho2/rho1)^Y-1 = 288K(2.4/1.2)^0.4 = 288K*1.3195 = 380K
(temperature calculation check: T2 = T1(P2/P1)^(Y-1)/Y = 288K(38.79/14.7)^0.286 = 288K*1.32 = 380K)
E1 = cv*m*T = 0.718*2.4kg*380K = 654.82kj/kgK

B) Step Two: Add 1.2kg of air at 338K to the above; determine final pressure(Pf) and temperature(Tf).

E2 = 0.720 * 1.2kg * 338K = 292kj/kgK
Tf = (E1 + E2)/(cv3*rho3) = 946.85kj/kgK/(0.719*3.6) = 365.80K
Pf = rho3*R*Tf = 3.6kg*287*365.80K = 377,945 Pascal/(6895Pa/psi) = 54.8psia

*(note: I think the 287 in the Pf equation above is the 0.287 R value wrote large to adjust for a very good reason... just don't ask me to specify what very good reason that would be exactly - maybe the Pascal thing; but after all a 'monkey see, monkey do' form of involvement can only take understanding so far...)
---

Thoughts of a mathematical illiterate: If you look at changing spherical volumes they are suggestive of the values expressed in the cp and cv specific heat coefficients and all of the incestuous relationships spun off from them. For instance the surface area of a 1 cubic meter sphere is around 4.836 square meters, while that for 0.5 cubic meters is about 3.046 square meters. From that it can be seen that decreasing the volume by half may increase the frequency of impacts per surface area by about 1.58 times, or inversely decrease it to about 63 per cent, affecting the pressure and temperature values similar to what actually happens if not exactly to the same degrees. I mention this because trying to use anonymous values without any apparent connection to the other values given makes it exceedingly difficult, for me anyway, to figure out which values are appropriate when faced with the multitude of variations on a theme which pop-up when dealing with this subject of changing pressures, volumes, and temperatures. It probably would have been best to start with formula dealing with moles which directly deal with the actual molecules that make up a gaseous mass. That would have given a direct connection to thoughts about the number of molecules that impinge on the surface area of a sphere and their effect on temperature and pressure values, and how these values change once the number and density of molecules and the mean distance they travel before a collision changes, or the volume and surface area size change.

If I messed this up, and it will come as no surprise if I did, please let me know.

Thanks to I_Like_Science for his help and to NASA for their page on compression and expansion.
wright.nasa.gov/airplane/compexp.html.

Alright, I'm taking a break - it's about time for a banana or two...