Why is Flow Work defined as pV and not pdV?

[omberlo]

In the chapter relative to First law applied to flow processes, my book explains the "Flow Work" which appears in the energy balance, as work needed to push a volume of fluid into (and out of) the control volume. It also says that it's equal to pV, and then it is added to the internal energy U to become enthalpy H. My question is, why is such work pV, which in infinitesimal form would become (PdV + Vdp) and not just PdV?

 

[Wolfgang2b]

As far as I know, the total work done is actually PV. As you said, in the infinitesimal form dW = PdV + VdP. Usually, we consider systems where pressure is kept constant, hence dP =0. That is how you get dW=PdV. But this is not always the case.

 

[I like Serena]

Hi omberlo!

Work done is not generally PV.
It really is dW = PdV.
You can see this if you consider that work is force times displacement.

Enthalpy is defined as H = U + PV
In infinitesimal form this becomes dH = dU + PdV + VdP.

For a reversible process this becomes dH = (dQ - PdV) + PdV + VdP = dQ + VdP.
This is a handy result, since it means that in a reversible process where pressure is constant, we have dH=dQ.

 

[omberlo]

Hi omberlo!

Work done is not generally PV.
It really is dW = PdV.
You can see this if you consider that work is force times displacement.

Enthalpy is defined as H = U + PV
In infinitesimal form this becomes dH = dU + PdV + VdP.

For a reversible process this becomes dH = (dQ - PdV) + PdV + VdP = dQ + VdP.
This is a handy result, since it means that in a reversible process where pressure is constant, we have dH=dQ.

Hi!

I understand what enthalpy is, but the book is trying to justify the use of enthalpy in the first law applied to flow processes by saying that in the energy balance internal energy of the fluid entering/exiting the control volume must be accounted for as well as the flow work, which is the work done by the upstream fluid to push the fluid into/out of the control volume.
Going by this explanation, it would make more sense to me to use (u + Pdv) in the energy balance equation, rather than enthalpy, since work is really PdV. I doubt the equation is wrong, hence I suppose there is a deeper meaning into this "push work" that justifies the use of d(pv) rather than PdV. Can anyone give me the intuition on this?

 

[I like Serena]

Ah yes, I see what you mean.

The volume is not compressed or expanded, but it is displaced.
The work done is pressure (assumed constant) times volume displaced.

The flow work done on a unit mass that is displaced into the control volume is Pv.

Pdv would be something different.
This would be the work done by the fluid to expand.
But the fluid does not expand, it only gets displaced.
This happens to be Pv.
If the fluid would also expand there would be additional work being Pdv.

 

[256bits]

Omberlo
You have to go back to the definition of control volume which is any arbitrarially chosen closed surface through which heat, work, mass can cross AND the contol volume can be fixed, move or expand.

Conservation of mass is considered in the analysis of the control volume where an infinitisimal mass, \deltam enters or leaves the control volume. That specific mass has a specific volume and that entering can be different than that leaving.

The infinitisimal work for the control volume is stated as
\deltaW = \deltaWcv + [ ( P\upsilon\deltam ) out - ( P \upsilon\deltam ) in ]

The last 2 terms are the flow work.
The first term would be shaft work, electrical work or fluid expansion work.

 

[Studiot]

The infinitisimal work for the control volume is stated as
.........

This was not the question asked.

Why is Flow Work defined as pV and not pdV?

Omberlo,
Some (more advanced) texts use a differential control volume as above but many use a finite one as in my extract from Joel - a great book at this level.
In fact a differential volume is uneccessary the control volume may include a whole pipe or other well defined volume.

If you use a differential volume the work/energy equations will obviously need to be cast in this form.

go well

 

[omberlo]

Thanks everyone for the answers so far.
From what I gather from all of you, flow work seems to be defined as pv because pressure is assumed to be constant.
But isn't this assumption inaccurate for unsteady-flow processes?

 

[I like Serena]

Actually, flow work is $W=\int_0^L F ds$,
where F is the force with which fluid is pushed into the control volume, A is the area of the cross section and L is some distance that the fluid is pushed.

For a small amount, this simplifies to $W=F L$ or $\frac F A \times A L = P \Delta V$,
where $\Delta V$ is the displaced amount of fluid (and in particular not the increase in volume of an amount of fluid).

This does not really mean that P has to be constant, but for a unit mass, this is Pv, which happens to coincide with the term in enthalpy.

 

[Studiot]

But isn't this assumption inaccurate for unsteady-flow processes?

This gets much more difficult.

If pressure is not constant then you have to do a piecewise calculation adding up all the different pressures. If you can derive an analytic expression for the time variation of the pressure you can integrate this.

 

[omberlo]

Actually, flow work is $W=\int_0^L F ds$,
where F is the force with which fluid is pushed into the control volume, A is the area of the cross section and L is some distance that the fluid is pushed.

For a small amount, this simplifies to $W=F L$ or $\frac F A \times A L = P \Delta V$,
where $\Delta V$ is the displaced amount of fluid (and in particular not the increase in volume of an amount of fluid).

This does not really mean that P has to be constant, but for a unit mass, this is Pv, which happens to coincide with the term in enthalpy.

What do you mean that for a small amount, it simplifies?
By following your math I get

$W=\int_0^L F ds$,
introducing pressure
$W=\int_0^L \frac F A A \times ds$
which in terms of pressure and volume is
$W=\int_0^V P dV$

this is equal to $P \Delta V$ only if pressure is constant

if pressure is not constant, de + PdV ≠ dh

If I'm missing something obvious I'm sorry :(.

This gets much more difficult.

If pressure is not constant then you have to do a piecewise calculation adding up all the different pressures. If you can derive an analytic expression for the time variation of the pressure you can integrate this.

Yes, but I'm still not sure whether the correct integral would involve just PdV or (PdV + Vdp)

 

[Studiot]

Yes, but I'm still not sure whether the correct integral would involve just PdV or (PdV + Vdp)

Neither.

In order to obtain the integral you have to write one variable in terms of the other.
You can only do this if you have an equation of state connecting them, or if one is constant.

For example you cannot integrate y²dx unless you know y in terms of x.

 

[I like Serena]

What do you mean that for a small amount, it simplifies?
By following your math I get

$W=\int_0^L F ds$,
introducing pressure
$W=\int_0^L \frac F A A \times ds$
which in terms of pressure and volume is
$W=\int_0^V P dV$

this is equal to $P \Delta V$ only if pressure is constant

if pressure is not constant, de + PdV ≠ dh

If I'm missing something obvious I'm sorry :(.

What you write is entirely correct and I guess I'm not explaining this very well.
Sorry.

Let me try again.

In thermodynamics dV usually denotes the amount a volume V expands.
In this case P is a function of V: when V expands, P will decrease (disregarding the effect of temperature for now).
Furthermore the work is done by the fluid itself, which means it is energy lost.

In a flow process dV would be the infinitesimal volume that is displaced by an external force on the fluid.
In this case P is not a function of V: dV is simply the amount displaced by some independent pressure P.
And since this is externally applied, it is energy gained.

These are different types of work, which will add up if applicable.