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You need a good textbook. Get yourself a copy of Fundamentals of Engineering Thermodynamics by Moran et al.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.physea said:ok then which article does that, because that's what I want
Chestermiller said:You need a good textbook. Get yourself a copy of Fundamentals of Engineering Thermodynamics by Moran et al.
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 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
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.
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."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.
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 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.
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.
In an isothermal process of an ideal gas in a closed system, DH = 0 also.physea said:But in isothermal process, DU=0, not D(PV)=0.
Chestermiller said:In an isothermal process of an ideal gas in a closed system, DH = 0 also.