Struggling With Internal Energy..

[Fool]

Struggling With Internal Energy.

Thermodynamics is difficult to learn mostly because it is difficult to teach, but it is hampered by 200 years of colloquial terms that sound alike, and are used misleadingly. One of those terms is the new concept of internal energy often confused with heat. Separate them in your mind so that the variables 'U' and 'Q' can be used mathematically to represent them. Please.

In the "Why a temperature difference" thread Matt Brown posted the following:

Am I still missing something here?

Yep, you're missing the most basic issue where heat equals work (ie Q=W) regardless of direction. This basic thermo tenant is what Tom struggles with where Qin=Wout and...Win=Qout. Just as the "heat of expansion" = work out, the returning piston will require work in = the "heat of compression". An ideal Stirling cycle supplies source in at a Thigh DURING expansion and sink out at Tlow DURING compression (2 distinct temperatures) where work out from expansion exceeds work in from compression proportional to the temperature differential measured via the absolute scale (deg K). Another way to view this is simply that the MEP (mean effective pressure) during expansion must exceed the MEP during compression, regardless whether the compression force comes from a flywheel or ambient pressure (aka, there's no free lunch).

An ideal Stirling cycle supplies both Thigh and Tlow isothermally (constant temperature) and whining that there's no such thing as an isothermal process is lame, since an Otto type cycle with 2 adiabatic processes will follow a similar sequence..

Taking his first comments:

"Yep, you're missing the most basic issue where heat equals work (ie Q=W) regardless of direction. This basic thermo tenant is what Tom struggles with where Qin=Wout and...Win=Qout. "

Doesn't everybody? But it gets worse:

Qin=Win applies at times to a single isothermal stroke or at other times to an entire cycle, but, and it's a big but, it never applies to an adiabatic process, single adiabatic stroke, or a complete cyclic set of adiabatic processes. Adiabatic means zero heat transfer. Q=0.0 in the equation.

The following may help.

Remember for an isolated system nothing gets in or out. That differs from a closed system which allow work and heat to enter and leave, but nothing else such as mass. I get those two confused but we are stuck with the terms, try to do your best at memorizing and separating them.

In a closed system where we are concerned with heat and work energy only entering and leaving, internal energy changes of concern are related to heat and work. Gravity, battery, speeding mass, flywheels, and other energy sources inside the system can be eliminated from the equation. They can be added back if needed for more complicated system models. So the equation becomes:

∆U=∆Q+∆W

That comes from the empirical observation of conservation of energy. If you witness a violation of this first law, you are probably fooling yourself. Check and recheck before quietly asking others privately. If they disagree it's still probably you. Rethink you experiment, ask for help. Pons and Fleishman skipped that process an got horribly slammed, and they weren't even violating, nor attempting to violate, that law.

The equation ∆U=∆W+∆Q for an adiabatic process:

∆Q is zero so the equation becomes:

∆U=∆W

During that process there is no such thing as heat. No heat loss, no heat gain, no heat conversion.

The conversion is internal energy to or from work, for an adiabatic processes. It will involve significant temperature, pressure, and volume changes.



∆U=∆W+∆Q for an isothermal process:

∆U=0.0 Internal energy doesn't change so the equation becomes:

∆W=-∆Q

Thus it is applicable to an isothermal process, zero temperature change.

Expansion: Heat comes in from Th, gets converted to internal energy (maintaining temperature and internal energy), internal energy gets converted to work going out (pressure and volume change, internal energy and temperature are maintained), no heat leaves. Colloquially 100% conversion of heat to work. One forward stroke.

Compression: Work comes in, gets converted to internal energy (maintaining temperature and internal energy), internal energy gets converted to heat going out to Tc (pressure and volume change, internal energy and temperature are maintained), no heat enters. Colloquially 100% conversion of work to heat. One reverse stroke.

If Th and Tc are the same forward work will equal reverse work. Zero net work to output. That is why it is desirable to have Th, forward work coming out, be much higher than Tc, reverse work going back in. Hotter Th, colder Tc.

The above is true for all real engines. Additional losses must also be added to be more accurate. The above is the analysis for possible maximum thermodynamic work out of a full cycle.

For a full cycle starting at U1 and ending at U1, ∆U=U1-U2=0.0

So the equation again becomes:

∆W=-∆Q
Or
W=Q

So starting at the first law the second law drops right into place:
Efficiency n=(∆Wf-∆Wr)/(∆Qin)=(Th-Tc)/Th

The second law is a consequence of the first law for the special case of a full cycle. To violate the second law is to violate the first law for the special case of a full cycle.

This means the second law doesn't apply to a single stroke, no matter how 'tricky' the path will be. It can't be violated for a tricky full cycle either. Carnot and Stirling are the target to shoot for, thermodynamically. Real engines will be worse.

[Tom Booth]

....This basic thermo tenant is what Tom struggles with where Qin=Wout and...Win=Qout. "
...

No I don't.

You have a different opinion or belief or understanding about some things. You don't like or cannot tolerate anyone seeing anything differently so my unwillingness to agree with your erroneous obsolete antiquated wrong opinions that you cannot support or justify drives you crazy.

My not swallowing your horse shit is not "struggling". Sorry.

[Fool]

You do struggle. You mix single stroke rules with cycles. You cherry pick. You deny. You vituperate. Your most common logical fallacies are: ad hominem (attacking the poster), and appeal to authority. You fill page after page of science denial, in uncountable bouts of cognitive dissonance. We don't care that you think this is "horse shit", Mr potty mouth. Your post gave zero new information and after the erroneous claim of "no I don't", you proceeded with lack of any proof to support it.

You didn't even make any attempt to attack any logical flaws in my post so, we here, are all inclined to assume you accept it.

Lacking the ability to attack the logic of my post is struggling. At least struggling with something. It is also a logical fallacy.


There are many many ways to foul up logically. From Wikipedia:

https://en.m.wikipedia.org/wiki/List_of_fallacies

[Tom Booth]

Except for the fact that you inserted my name into your post and misrepresented me as "struggling" I have zero interest in whatever BS your trying to peddle here.

You've proven yourself to be dishonest and deceptive so I'm not wasting any more time trying to unravel your convoluted diatribes.

Carry on, just leave my name out of it.

[Fool]

Sheesh what a whiner. Perhaps one day you will learn the difference between, accepted consistent scientific theory backed up by rigorous mathematical principles, and your own deceptive and dishonest opinions. Until then you will struggle. It's blatantly obvious to the knowledgeable person. You prove it in every one of your ignorant logical fallacies. You prove it to everyone except yourself. I'm sorry for you, that, I'm forced to combat your belligerence so bluntly. I've tried other kinder ways, but you just get more belligerent. Besides you are way off topic with your whining and struggling.

This concept is easy. A measure of internal energy is temperature. A change in internal energy can come from work or heat flow. It is measured by how much change in temperature a single mass has experienced.

Heat flow only comes from and always flows from two separate mass temperates. Two, or more, separate internal energies and masses are involved. It is a value of energy entering or leaving a mass. It affects a mass or system similarly to, but not the same as, work.

∆U=∆W +∆Q

Q is not W is not U. There is a mathematical relationship between them. Learn this or continue to struggle.

U is considered the stationary value. ∆U is how much U has changed after at least one process.

∆U=U2-U1

U1 being U before a process. U2 being U after the process.

It is easy when taught clearly. If you'd notice, the beginning post actually defended you, that was by pointing out that everyone struggles with the concept. You would be pretty conceited to think that you were above everyone else. Is that what your whining about that you don't want to be considered an equal?

Again please attack my knowledge and logic, not whine about your own insecurities.

Work in, internal energy/temperature increase.
Work out, internal energy/temperature decrease.
Heat in, internal energy/temperature increase.
Heat out, internal energy/temperature decrease.

There is no such thing as a heat in work out conversion. There is no such thing as internal heat, or internal work. The work and heat thermodynamic relationship is governed by the above four rules. Work and heat can affect a system's internal energy/temperature simultaneously giving a colloquial appearance, an illusion, of a conversion, but internally it only happens as constrained by those four rules.

The internal and external energies (internal energies of outside masses) must balance. Work and heat don't have to ballance. They can in special circumstances, but most times not.

Getting to understand this is important.

[VincentG]

I certainly struggle. It seems odd to me that there is no consideration for pressure with internal energy.

Also seems that the relationship of work and heat could be incidental and not one and the same.

But the math is sound. Things can be computed accurately, that can't be denied.

[Tom Booth]

...
But the math is sound. Things can be computed accurately, that can't be denied.

Fools math is NOT sound. His equations are incorrect and a perversion of the first law (conservation of energy).

∆U=∆Q+∆W is not valid and not used by anyone because there is no change in heat.

This is part of fools set up for his silly number juggling where "internal energy" is viewed as internal "heat" to which more "heat" Q is added.

It is simply Q. A quantity of heat added. Not ∆Q.

Try finding any reference to Fools altered, perverted version of the first law equation which is simply ∆U=Q+W or in some applications ∆U=Q-W depending on if work is done by or on the system.

If you can find fools ∆U=∆Q+∆W version which is fundamentally wrong, with some justification given for such a variation please get back to me and provide a link, I'd like to see it.

For someone who is always going on about the strict distinction between heat and internal energy, his insertion of ∆Q in place of Q seems doubly odd and perhaps intentionally misleading.

There is no Q to add more Q to to make ∆Q and that is the basic flaw in fools earlier so-called "derivation", aside from his other chicanery and equally deceptive number juggling.

[Fool]

It seems odd to me too. Still thinking about it.

The problem with heat and work is that neither really exist. Their early existence in Caloric theory was discarded. Both are a mathematical tally for energy being transferred. It's easy to visualize work. Heat and it's influence is harder, as a lot of its influence is associated with internal energy. Heat of a fire is felt after your skins internal energy rises get hotter.

True. Science boils down to the mathematics. No mathematics, no theory. Lots of people and scientists struggle and are in denial of the mathematics equals science equivalence.

A lot of this seems to depend on viewpoint. The mathematics doesn't, but it depends on meticulous rigor.

[Fool]

"Fools math is NOT sound. His equations are incorrect and a perversion of the first law (conservation of energy)."

Oh my gosh, Mr name unmentionable, someone else disagrees with you. Someone with a good mathematical background. Stop attacking the poster with a ludicrous opinion. Put out your 'superior' mathematics. Please.

"If you can find fools ∆U=∆Q+∆W version"

I would like to see your fundamentally correct version.

I suppose you might want to write:

∆U=W+Q

Doesn't really change anything but convention, as how the numbers are entered into the variables that matters not the names given.

Could have written:
= +

[Tom Booth]

...
Doesn't really change anything
....

It changes the entire perspective from heat added or Q as a finite quantity of energy in Joules over to your obsolete "Calorics" nonsense where heat is viewed as a substance being added to the that already present in the working fluid.

Your changing the equation is just a tell tale sign of your entire wrong headed perspective.

I'd like to see either you or VincentG find some reference to justify such an alteration.

If I'm wrong, I'm perfectly happy and willing to be corrected, so prove me wrong and provide a reference.

[VincentG]

IMO, it would start with defining "U" as a static value.

[Fool]

In the section of Wikipedia:

https://en.m.wikipedia.org/wiki/Internal_energy

Internal energy of a closed thermodynamic system

It provides the equation:

dU=dQ-dW

The minus sine is from positive work out of the system, where I defined both work and as being positive into the system, meaning if the work is coming out it would have a negation.

The 'd's are small infinitesimal ∆, for standard Integral Calculus. I used ∆'s because I know how to get them on my scratch and sniff phone screen/entry pad.

Please get a real point. You are still in the quagmire of opinion. In other words, "prove me wrong".

VincentG do you mean "static" or a 'state' variable? It is a state variable.

[Tom Booth]

In the section of Wikipedia:

https://en.m.wikipedia.org/wiki/Internal_energy

Internal energy of a closed thermodynamic system

It provides the equation:

dU=dQ-dW

The minus sine is from positive work out of the system, where I defined both work and as being positive into the system, meaning if the work is coming out it would have a negation.

The 'd's are small infinitesimal ∆, for standard Integral Calculus. I used ∆'s because I know how to get them on my scratch and sniff phone screen/entry pad.

Please get a real point. You are still in the quagmire of opinion. In other words, "prove me wrong".

Well, thanks for clarifying, you just proved me right, not my fault you don't know how to find "d" on your phone.

Calling an infinitesimal heat input or work output a "delta" however is still wrong. A finite quantity is not a difference or delta, is it?

[VincentG]

From the Wiki page...

The internal energy cannot be measured absolutely. Thermodynamics concerns changes in the internal energy, not its absolute value.

 It is a state variable, a thermodynamic potential, and an extensive property.

The unit of energy in the International System of Units (SI) is the joule (J). The internal energy relative to the mass with unit J/kg is the specific internal energy. The corresponding quantity relative to the amount of substance with unit J/mol is the molar internal energy.[4]

I mean that it would help if there was some base unit for thermodynamic potential at STP.

Air at 14.7psi and 300k has a certain specific internal energy, but no potential to do work on earth. Air at 30psi and 600k has double the specific internal energy, but infinitely more potential to do work on earth.

And I apologize for using the overly philosophical "infinitely more" phrase, but where that difference may just be a rounding error for very high delta U systems(like ICE), it is a massive error for systems that operate near STP.

[Fool]

"not my fault you don't know how to find "d" on your phone."

You really know how to get to the main points!

It's not a matter of not knowing. It is a matter of not wanting to spend the time looking for obscure symbols.

The letter for an infinitesimal is the lower case delta. I just know that anyone knowledgeable in calculus would know the substitution of capital delta and lower case delta. Lower case delta looks like a "d", with a squiggly back.

The Wikipedia link I gave uses the correct delta symbol, pronounced "dee", in mathematics.

The difference between ∆ and "dee", a change and an infinitesimal change is just the size and numbers to add up. ∆x is often changed to do if allowable in calculus.

The point here is internal energy is vibrations, temperature. Heat is not. All equations clearly separate them, even when you quote them.