Confused with the energy equation in thermodynamics

[physea]

Hello

I am confused with the energy conservation equation in thermodynamics.

Is it U2 - U1 = Q - W ? This is what most books say.

Or is it H2 - H1 = Q - W ?

Which should I use to solve problems? For example we have a gas with U1 and P1V1 (which makes H1). After a process, we have Q transferred in or out and W transferred in or out. How can I write the equation in this situation?

Thanks

 

[gleem]

How do you understand the meaning of the quantities in the two equations?

 

[physea]

U is the internal energy, the energy due to the molecular motion of the gas
H is the internal energy plus the pressure energy PV
Q is the heat transferred in/out of the gas
W is the work transferred in/out of the gas

What are you trying to say? Please be clear

 

[gleem]

That W and PV both have units of energy but are not the same. Work is not a state variable and thus depends on the process that produced it.

 

[physea]

So which of the two equations are correct? If both, how can that be?

 

[gleem]

U₂ - U₁ = Heat added to the system - Work done by the system

H₂ - H₁ = U₂ - U₁ + P₂V₂ - P₁V₁

You use the one appropriate to the problem.

 

[physea]

So the equation H2 - H1 = Q - W is not correct?

Also, in U2 - U1 = Q - W, you say that Q is heat ONLY added to the system and W only DONE by the system. What about heat substracted by the system and work added to the system?

 

[gleem]

Yes W ≠ PV₂ - PV₁

The first law is written as U2 - U1 = Heat added to the system - Work done by the system =Q - W where both Q and W are positive quantities. A negative Q will mean heat removed from the system and a negative W is the work done ON the system.

You must always watch out if you are adding or removing heat to a system and whether work is done on the system versus having the system do work.

 

[physea]

So, there is no situation where there is no change in U, there is no heat exchanged and there is work done on or by the system? Ie. in an adiabatic expansion or compression, there is ALWAYS dU≠0?

 

[gleem]

An adiabatic process is one that is thermally insulated so ΔQ = 0. So dU = work done by( du < 0) / on (du>0) the system.

 

[physea]

So you are saying that there is no possibility for work to be done adiabatically in the system and not to increase U, but rather increase pressure energy PV ?

 

[Chestermiller]

Hello

I am confused with the energy conservation equation in thermodynamics.

Is it U2 - U1 = Q - W ? This is what most books say.

Or is it H2 - H1 = Q - W ?

Which should I use to solve problems? For example we have a gas with U1 and P1V1 (which makes H1). After a process, we have Q transferred in or out and W transferred in or out. How can I write the equation in this situation?

Thanks

The first equation is always correct. The second equation is correct only for a process carried out at a constant pressure, equal to the original pressure (say by adding heat at constant pressure).

Do you have a specific problem that we can use as an example.

Here is a Physics Forums Insights article that should be helpful to you. Although the title refers to the second law of thermodynamics, the first section covers the first law of thermodynamics: https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/

 

[gleem]

In the case of constant pressure H₂ - H₁ = Q

 

[Chestermiller]

In the case of constant pressure H₂ - H₁ = Q

Oops. Thanks. I didn't notice the W there.

 

[physea]

This NASA webpage says something different than what you say:
https://www.grc.nasa.gov/www/k-12/airplane/thermo1f.html
They say that dH=Q-W (if you ignore kinetic energy). To me this is more complete, because it takes into account the change in pressure energy of the system, which can be caused by either Q or W.

 

[physea]

I don't have a specific problem, but if I am asked this:
We know U1, U2, Q, W, P1, V1, V2 and they ask me P2, how do I answer?

I do NOT want to learn equations that have limited application, ie. when P=constant or V=constant or whatever. I want to know one complete energy conservation equations that applies in EVERY situation and that I can calculate every parameter from it, if I know the rest parameters.

 

[Chestermiller]

This NASA webpage says something different than what you say:
https://www.grc.nasa.gov/www/k-12/airplane/thermo1f.html
They say that dH=Q-W (if you ignore kinetic energy). To me this is more complete, because it takes into account the change in pressure energy of the system, which can be caused by either Q or W.

This equation applies only to the open system version of the first law, in which mass is flowing into and out of the system. And H refers to the enthalpy change between the inlet and outlet streams of the flow system, not the change in enthalpy within the system. And, did you notice the subscript S on the W. That is not the total work done. It is only the "shaft work". So the NASA document does not apply to a closed system that you have been asking about.

 

[physea]

Mmmm now it starts to make sense.

So there IS dH=Q-W, and it is the Steady Flow Energy Equation!

BTW, when did I say I am asking for closed systems?

So can you elaborate now, why in SFEE we have dH=Q-W (ignoring kinetic and potential energy) and in closed systems we have dU=Q-W ?
And what is the Work in the SFEE that is only "part"of the work done?

 

[Chestermiller]

I don't have a specific problem, but if I am asked this:
We know U1, U2, Q, W, P1, V1, V2 and they ask me P2, how do I answer?

I do NOT want to learn equations that have limited application, ie. when P=constant or V=constant or whatever. I want to know one complete energy conservation equations that applies in EVERY situation and that I can calculate every parameter from it, if I know the rest parameters.

That's covered in my Physics Forums article.

Regarding the specific question you asked, is this the actual question they asked you? What other information is given?

 

[physea]

As I told you I don't have a specific question. I just got confused between dU=Q-W and dH=Q-W.

 

[Chestermiller]

Mmmm now it starts to make sense.

So there IS dH=Q-W, and it is the Steady Flow Energy Equation!

BTW, when did I say I am asking for closed systems?

So can you elaborate now, why in SFEE we have dH=Q-W (ignoring kinetic and potential energy) and in closed systems we have dU=Q-W ?
And what is the Work in the SFEE that is only "part"of the work done?

You need to look up the derivation of the SFEE. It focuses on an open control volume, and part of the work goes into pushing material into and out of the control volume. The shaft work does not include this work. That is why there is a $\Delta H$, not a $\Delta U$.

Regarding your comment "BTW, when did I say I am asking for closed systems?", what were we supposed to think? We can't read your mind. You need to provide us with all the information that is relevant.

 

[Chestermiller]

As I told you I don't have a specific question. I just got confused between dU=Q-W and dH=Q-W.

Does what I said about the SFEE help?

 

[physea]

Mmm, I read a bit about derivation of SFEE but I can't figure out much. In few words, I understand that in a closed system, P2V2-P1V1=W. In an open system it seems that P2V2-P1V1 is not W. How is the work done to the system or produced by the system in SFEE conditions categorized? And what is its relations to the difference in pressure energy?

 

[Chestermiller]

Mmm, I read a bit about derivation of SFEE but I can't figure out much. In few words, I understand that in a closed system, P2V2-P1V1=W.

This is totally incorrect. You need to go back and review closed systems so that you can understand them properly before you can move on to open systems.

 

[physea]

This is totally incorrect. You need to go back and review closed systems so that you can understand them properly before you can move on to open systems.

OK, W=d(PV), which means when we have constant P, we can write W=PdV and when we have constant V, we can write W=VdP.

But can anyone explain me in few words why in SFEE, for energy conservation, we consider pressure energy and in closed systems we don't?

 

[Chestermiller]

OK, W=d(PV), which means when we have constant P, we can write W=PdV and when we have constant V, we can write W=VdP.

Totally incorrect again (except at constant pressure). You'll never be able to understand SFEE if you first don't understand the closed system version of the first law. You obviously haven't read my Physics Forums Insights article that I referred you to.

Chet

 

[physea]

I read the article! (the first law part). It says only that W=PdV when P=const. Obviously when V=const there is no W. Yet, in that situation, pressure energy increases! How is that depicted?

And it doesn't mention anything about SFEE.

 

[Chestermiller]

I read the article! (the first law part). It says only that W=PdV when P=const. Obviously when V=const there is no W.
It doesn't mention anything about SFEE.

It doesn't say that W=PdV only when P=const.

 

[physea]

It doesn't say that W=PdV only when P=const.

What is that then:
w˙(t)=P(t)V˙(t)

Reference https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/

You mean that ALWAYS W=PdV? I can understand that.

 

[physea]

any update?

will anyone explain me why in closed systems we have dU=Q-W and why in open steady flow systems we have dH=Q-W?

thanks!

 

[Chestermiller]

What is that then:
w˙(t)=P(t)V˙(t)

You left out the subscript I on the P indicating that it is the force per unit area at the interface where work is being done.
$$\frac{dw}{dt}=P_I(t)\frac{dV}{dt}$$

You mean that ALWAYS W=PdV? I can understand that.

I mean $dw = P_IdV$. $P_I$ is sometimes referred to in the literature as $P_{ext}$, the externally applied pressure. This is the same as the gas pressure at the interface.

 

[Chestermiller]

any update?

will anyone explain me why in closed systems we have dU=Q-W and why in open steady flow systems we have dH=Q-W?

thanks!

I don't think anyone should even try to explain this until you have mastered the application of the first law to closed systems. You need to get some significant practice solving closed system first law problems.

I won't be responding any further to this thread.

 

[physea]

I don't think anyone should even try to explain this until you have mastered the application of the first law to closed systems. You need to get some significant practice solving closed system first law problems.

I won't be responding any further to this thread.

OK but your article doesn't help much then.

 

[Chestermiller]

OK but your article doesn't help much then.

The objective of my article was not to distinguish between the open system and closed system versions of the first law.

 

[physea]

The objective of my article was not to distinguish between the open system and closed system versions of the first law.

ok then which article does that, because that's what I want

 

[Chestermiller]

ok then which article does that, because that's what I want

You need a good textbook. Get yourself a copy of Fundamentals of Engineering Thermodynamics by Moran et al.

 

[physea]

OK I will read it, but I need quickly a concise explanation as I am taking exams soon!

You need a good textbook. Get yourself a copy of Fundamentals of Engineering Thermodynamics by Moran et al.

 

[physea]

this article explains the situation:
http://www.tech.plym.ac.uk/sme/mech225/steadya.pdf

so basically the P2V2-P1V1 is only the work to overcome the pressure to enter new mass in the system (or the work done by the system if P2V2-P1V1<0 as it seems that way it would facilitate the flow)

it is a bit tricky to understand that basically we need some work to overcome the pressure energy of the inlet, PLUS then accelerate the fluid to attain the kinetic energy

so the kinetic energy of the system is not the reason that new mass enters the system, if it was the reason, it would be reduced, which is not. so there is needed an additional work to maintain the flow

which work is totally different than the work done to or by the system!

This is what I need in all my questions! The right resource that explains in few lines my fair questions! Yet I get responses from people, go read a textbook or go search online, etc

 

[Chestermiller]

this article explains the situation:
http://www.tech.plym.ac.uk/sme/mech225/steadya.pdf

so basically the P2V2-P1V1 is only the work to overcome the pressure to enter new mass in the system (or the work done by the system if P2V2-P1V1<0 as it seems that way it would facilitate the flow)

it is a bit tricky to understand that basically we need some work to overcome the pressure energy of the inlet, PLUS then accelerate the fluid to attain the kinetic energy

so the kinetic energy of the system is not the reason that new mass enters the system, if it was the reason, it would be reduced, which is not. so there is needed an additional work to maintain the flow

which work is totally different than the work done to or by the system!

This is what I need in all my questions! The right resource that explains in few lines my fair questions! Yet I get responses from people, go read a textbook or go search online, etc

Yet, in the end, that's exactly what you successfully did. That's what we were encouraging you to do. So, all you did was prove us correct.

 

[physea]

Yet, in the end, that's exactly what you successfully did. That's what we were encouraging you to do. So, all you did was prove us correct.

I am not sure how I proved you correct, since you never explained why in steady flow the 1st law is different than in closed system.

And it's not always simple to go read a textbook or online, I was just lucky to find that brilliant article, as I read other resources that weren't so clear. They used maths/calculus, without explaining the real sense and meaning behind the numbers.

 

[Chestermiller]

I am not sure how I proved you correct, since you never explained why in steady flow the 1st law is different than in closed system.

And it's not always simple to go read a textbook or online, I was just lucky to find that brilliant article, as I read other resources that weren't so clear. They used maths/calculus, without explaining the real sense and meaning behind the numbers.

The analysis in your "brilliant article" is almost a carbon copy of the analyses I have seen in many textbooks, including the reference I provided you in post #36, Fundamentals of Engineering Thermodynamics by Moran et al. Another book with a similar treatment is Introduction to Chemical Engineering Thermodynamics by Smith and van Ness. I have never seen a treatment SFEE in any thermo textbook that differs significantly from that in your "brilliant article."

 

[Chestermiller]

Let's review the historical sequence in this thread. Take a look at your original post. Is there any indication there that you were wondering about the difference between the closed system- and open system (SFEE) versions of the first law? Did you say that you thought that W=Δ(PV)? Was it not until post #18 that you indicated that you were talking about the open system version of the first law? What was I supposed to think? My conclusion was that you did not even understand the closed system version of the first law, so how could you possibly understand the open system version. That's when I started suggesting that go back to your textbooks to solidify your knowledge base. Please read the thread over, and try to put yourself in my place, trying to "read between the lines" as to what you were asking, and trying to figure out how to advise you. Based on my determination, right or wrong, that you did not even understand the closed system version of the first law, I made the determination that it would be a waste of effort to try to explain the closed system version until you got a better understanding. That's why I recommended going back to your textbooks. Can you possibly understand how I came to that determination?

 

[physea]

OK maybe you're right, but when I said dH=Q-W you should recognize that that's SFEE.
And yes, it seems the explanation is in many books, even the lecture slides of my course, but that lecture slides were never presented to me.

 

[Chestermiller]

OK maybe you're right, but when I said dH=Q-W you should recognize that that's SFEE.
And yes, it seems the explanation is in many books, even the lecture slides of my course, but that lecture slides were never presented to me.

Thanks for being so understanding. I really appreciate it. Regarding whether I should have recognized that dH = Q - W implies SFEE, there are cases for closed systems where $\Delta (PV)$ is equal to zero, such that $\Delta U=\Delta H$ (e.g., isothermal expansion of an ideal gas). Also, when you wrote that dH = Q - W, I just thought you were referring to a closed system and didn't know what you were talking about. Apparently that was not the case.

 

[physea]

Thanks for being so understanding. I really appreciate it. Regarding whether I should have recognized that dH = Q - W implies SFEE, there are cases for closed systems where $\Delta (PV)$ is equal to zero, such that $\Delta U=\Delta H$ (e.g., isothermal expansion of an ideal gas). Also, when you wrote that dH = Q - W, I just thought you were referring to a closed system and didn't know what you were talking about. Apparently that was not the case.

But in isothermal process, DU=0, not D(PV)=0.

 

[Chestermiller]

But in isothermal process, DU=0, not D(PV)=0.

In an isothermal process of an ideal gas in a closed system, DH = 0 also.

 

[physea]

In an isothermal process of an ideal gas in a closed system, DH = 0 also.

OK, I agree. So under any circumstances, U is only depended on T? In either closed and open systems?

Also, if we have a SFEE system, and we put some work on it, is there a way to determine if that work is used to increase U or PV or both and at which percentage?