Flow Work vs PdV Work: Is PdV Valid for Both Closed & Open Systems?

[Soumalya]

Hello friends,
Referring to the attachment it can be clearly understood that flow work in an open system is a form of the PdV work defined as the boundary work or displacement work for a closed system.

Does that mean PdV work is valid for both a closed system and an open system?

 

[Chestermiller]

Hi Soumalya,

If you are having trouble understanding the extension of the first law to open systems, maybe consulting a different reference will help:

Smith and Van Ness: Introduction to Chemical Engineering Thermodynamics

They do a very nice job.

Chet

 

[Soumalya]

Hi Soumalya,

If you are having trouble understanding the extension of the first law to open systems, maybe consulting a different reference will help:

Smith and Van Ness: Introduction to Chemical Engineering Thermodynamics

They do a very nice job.

Chet

I went through the book searching for a clarification to my query but couldn't find any specific illustration about what I am looking for even though the book is incredibly good for building up concepts!Thanks for the excellent reference

Regarding my doubt I would summarize what I have understood so far:

1.In an open flow system the 'flow work' is itself a form of 'PdV' work crossing the control surface for a control volume even though the system boundary(control surface)for an open system is fixed.While in a closed system we focus on a fixed quantity of matter(control mass) enclosed by an imaginary boundary that is either fixed or free to move,the work of expansion or compression is always accompanied by the movement of system boundary.On the other hand in an open system the system boundary(control surface) is immovable but again we have a form of 'Pdv' work entering or leaving the system boundary as flow work.So is it appropriate to render 'flow work' as Pdv work for an open system?

2.Other than 'flow work' for a control volume in an open flow system there cannot be any form of 'PdV' work as the control surface is fixed thereby not allowing any expansion or compression i.e, dV=0.

Am I correct?

 

[Chestermiller]

I went through the book searching for a clarification to my query but couldn't find any specific illustration about what I am looking for even though the book is incredibly good for building up concepts!Thanks for the excellent reference

Regarding my doubt I would summarize what I have understood so far:

1.In an open flow system the 'flow work' is itself a form of 'PdV' work crossing the control surface for a control volume even though the system boundary(control surface)for an open system is fixed.While in a closed system we focus on a fixed quantity of matter(control mass) enclosed by an imaginary boundary that is either fixed or free to move,the work of expansion or compression is always accompanied by the movement of system boundary.On the other hand in an open system the system boundary(control surface) is immovable but again we have a form of 'Pdv' work entering or leaving the system boundary as flow work.So is it appropriate to render 'flow work' as Pdv work for an open system?

Yes. There are really two systems that we are considering: 1. The open system within the boundaries of the control volume and 2. The closed system formed by the material within the control volume at time t, plus the material about to flow into the system between time t and time t + Δt. At time t + Δt, this same closed system consists of the material now within the control volume, plus the material that has just flowed out of the system between time t and time t + Δt. So there is p-V work that has been done on this closed system, and we can apply the closed system version of the first law to it. This leads to the open system version of the first law. However, for the open system within the control volume, there is no p-V work done on the part of the fixed control where no material is entering or leaving.

2.Other than 'flow work' for a control volume in an open flow system there cannot be any form of 'PdV' work as the control surface is fixed thereby not allowing any expansion or compression i.e, dV=0.

Am I correct?

Yes. Well said!

Chet

 

[Soumalya]

Yes. There are really two systems that we are considering: 1. The open system within the boundaries of the control volume and 2. The closed system formed by the material within the control volume at time t, plus the material about to flow into the system between time t and time t + Δt. At time t + Δt, this same closed system consists of the material now within the control volume, plus the material that has just flowed out of the system between time t and time t + Δt. So there is p-V work that has been done on this closed system, and we can apply the closed system version of the first law to it. This leads to the open system version of the first law. However, for the open system within the control volume, there is no p-V work done on the part of the fixed control where no material is entering or leaving.

Chet

Thanks for clarifying Chet!

Are you familiar with the equation for the reversible work output for a steady flow device?

W_rev= -∫VdP-Δ(k.e.)-Δ(p.e.)

I am not very convinced about the way they use the second TdS relation i.e, TdS=dh-VdP for a closed system into the differential form of the steady flow energy equation to arrive at that conclusion.

Are they considering unit mass of fluid in flow both as part of the control volume for the open system and also as a closed system itself such that they apply the S.F.E.E. and the first law for a closed system?

δq_rev=du+d(k.e.)+d(p.e.)+d(pv)+δw_rev and

δq_rev=du+pdV

such that,du+pdV=du+d(k.e.)+d(p.e.)+d(pv)+δw_rev

or,du+pdV=du+d(k.e.)+d(p.e.)+pdV+Vdp+δw_rev

or,w_rev= -∫VdP-Δ(k.e.)-Δ(p.e.)

Is it the way I am thinking?

 

[Chestermiller]

Thanks for clarifying Chet!

Are you familiar with the equation for the reversible work output for a steady flow device?

W_rev= -∫VdP-Δ(k.e.)-Δ(p.e.)

I am not very convinced about the way they use the second TdS relation i.e, TdS=dh-VdP for a closed system into the differential form of the steady flow energy equation to arrive at that conclusion.

This relationship TdS=dh-VdP is independent of whether some system is open or closed. It is a fundamental physical property relationship for the particular material under consideration.

Are they considering unit mass of fluid in flow both as part of the control volume for the open system and also as a closed system itself such that they apply the S.F.E.E. and the first law for a closed system?

What does SFEE stand for?

They are considering what happens to a unit mass in its transition between inlet to the control volume and the exit from the control volume.

δq_rev=du+d(k.e.)+d(p.e.)+d(pv)+δw_rev and

δq_rev=du+pdV

such that,du+pdV=du+d(k.e.)+d(p.e.)+d(pv)+δw_rev

or,du+pdV=du+d(k.e.)+d(p.e.)+pdV+Vdp+δw_rev

or,w_rev= -∫VdP-Δ(k.e.)-Δ(p.e.)

Is it the way I am thinking?

This is basically correct. If you want to read more about this, consult Smith and Van Ness' section on the Bernoulli equation p 217.

Also note, that, in this equation, w_rev represents the reversible shaft work. Examples of shaft work include work done by pumps, turbines, and compressors on the fluid as it passes through the control volume.

Chet