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Enthaply calculation for a closed system undergoing reversible adiabatic compression

  • #1

Homework Statement



a fluid undergoes a reversible adiabatic compression from .5 MPa, .2 m3 to .05 m3 according to the law PV1.3 = constant. determine the change in enthalpy, internal energy and work transfer during the process.

Homework Equations



P1V11.3 = P2V21.3

dH = T dS + V dP
dS = 0, due to reversible - adiabatic process.
dH = ∫V dP
dU = dH - (P2V2 - P1V1)
W = dU

The Attempt at a Solution



when i solve the integral equation of enthalpy, i found its value H2-H1 = 223.3 kJ
U2-U1 = 171.77 kJ.

i didn't understand the conceptual part of the problem.

gas is being compressed by doing work on the system. As it is being compressed, the work done on the system is completely transformed into it internal energy (increase in energy stored by the system).

enthalpy rise = (internal energy rise) + Δ(PV)
= 171.77 kJ + 51.53 kJ
i understood the part that enthalpy of gas is increased due to the rise in internal energy and also the rise internal energy is due to the work done on the system.

but i didn't understand the part, how enthalpy value got increased due to the PV contribution? i mean there is no other energy transfer to the system apart from work transfer and all the energy due to work transfer has gone into rising it's internal energy.

doesn't change in internal energy, enthalpy requires energy transfer (heat ot work) in case of closed system?
 
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Answers and Replies

  • #2
1,197
0


The fluid is being compressed. That is what supplies the work that changes the internal energy and enthalpy. Your problem statement does not specify whether mass is crossing a control surface. When it does, the pv term arises.
 
  • #3


what if it was a closed system?
lets assume it was a piston cylinder problem.
what would be its enthalpy change then?
 
  • #4
19,677
3,987


The system you studied indeed is a closed system. The enthalpy is simply a defined thermodynamic function that is convenient to use in certain applications, particularly flow problems. It is a function of state, and depends only on the initial and final states of a closed system. For your problem, even though its change can be calculated, this change is less relevant than in other types of problems.
 
  • #5


for flow problems, it's very easy to understand the concept of enthalpy change when we are dealing with open system energy balance equation.

from what you said does that mean for problems involving closed systems undergoing thermodynamic processes like polytropic or adiabatic, there is no physical interpretation for enthalpy change?
 
  • #6
19,677
3,987


for flow problems, it's very easy to understand the concept of enthalpy change when we are dealing with open system energy balance equation.

from what you said does that mean for problems involving closed systems undergoing thermodynamic processes like polytropic or adiabatic, there is no physical interpretation for enthalpy change?
For a closed system in which heat is added while the pressure is held constant, the enthalpy change is equal to the heat added.
 
  • #7
vela
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Science Advisor
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One way to look at enthalpy is that it's the total energy required to create the system from scratch. Not only do you have to supply the internal energy for the contents of the system, you also have to make room in the environment in which to place the system. If a system is to occupy a volume V and the environment exerts constant pressure P, you have to do work in the amount of PV to create the empty space into which the system will go.

I don't see a satisfying way how this picture relates to this problem, though, other than calculating PV for the initial and final states and seeing there's a difference.
 
  • #8
19,677
3,987


One way to look at enthalpy is that it's the total energy required to create the system from scratch. Not only do you have to supply the internal energy for the contents of the system, you also have to make room in the environment in which to place the system. If a system is to occupy a volume V and the environment exerts constant pressure P, you have to do work in the amount of PV to create the empty space into which the system will go.

I don't see a satisfying way how this picture relates to this problem, though, other than calculating PV for the initial and final states and seeing there's a difference.
I never though of it this way, but it's a very interesting way of looking at it, particularly for an ideal gas where the internal energy is independent of the pressure.
 
  • #9


For a closed system in which heat is added while the pressure is held constant, the enthalpy change is equal to the heat added.
yes! i totally agree with that statement. only in the case of constant pressure process the enthalpy change concept makes sense. but how about in any general polytropic process? what happens then?
 

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