Confused with the energy equation in thermodynamics

In summary, the first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted between different forms. There are different equations that represent this law, such as U2 - U1 = Q - W and H2 - H1 = Q - W, which are used depending on the specific situation. In an adiabatic process, where no heat is exchanged, dU will always equal the work done by or on the system. In a steady flow energy equation, dH = Q - W, where H refers to the enthalpy change between inlet and outlet streams in an open system. In a closed system, the equation dU = Q - W is used. The work in the steady flow
  • #36
physea said:
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.
 
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  • #37
OK I will read it, but I need quickly a concise explanation as I am taking exams soon!
Chestermiller said:
You need a good textbook. Get yourself a copy of Fundamentals of Engineering Thermodynamics by Moran et al.
 
  • #38
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
 
  • #39
physea said:
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.
 
  • #40
Chestermiller said:
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.
 
  • #41
physea said:
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."
 
  • #42
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?
 
  • #43
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.
 
  • #44
physea said:
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.
 
  • #45
Chestermiller said:
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.
 
  • #46
physea said:
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.
 
  • #47
Chestermiller said:
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?
 

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