Isentropic Compression of Saturated Water: Workdone Calculation

[ajayguhan]

Workdone= integral Pdv
a kg of saturated water was compressed isentropically from 1 bar to 10 bar.
I solved it in the following logic: Since water is incompressible dv=0 , work =0

But my answers was wrong.

The solution was integral Vdp where V is V_f at 1 bar

My doubt is , is workdone =integral {(Pdv) or (Vdp)}...? Or both?
If both, how work=integral pdv and vdp ?
If it is equal to integral Vdp, then how is it?

 

[haruspex]

If water were completely incompressible then you would be right that no work is done, but it is very slightly compressible.
Beyond that, I agree with you that work done is PdV. Without knowing the compressibility coefficient there is no way to deduce the work done. Using VdP makes no sense to me.

 

[jack action]

The relationship between work and energy:

[...] work is the energy associated with the action of a force, [...]

Wikipedia​

​

In this regard:

\delta W = d\left(pV\right) = pdV + Vdp
The first part is know as boundary work, where there is a variation of volume. The second part is any other form of work (often called shaft work) where there is no variation of volume.

The shaft work may not do actual work on the system (i.e. displacement), but it changes the energy level of the system. If the process is reversible and adiabatic, then work had to be done somewhere for this to happen. So, in a sense, «work done» is still technically correct.

But, if someone wants to argue, an isentropic process is not necessarily reversible and adiabatic. :p

 

[Chestermiller]

Was it saturated liquid water or saturated water vapor? Either way, the work is PdV.

Chet

 

[ajayguhan]

Saturated liquid water

 

[Chestermiller]

Then your original answer is definitely correct, and it's not vdp.

Chet

 

[ajayguhan]

But the textbook solution is vdp only.

 

[haruspex]

But the textbook solution is vdp only.

Can you post the whole of the textbook solution? That might shed more light on it.

 

[Chestermiller]

vdp is definitely incorrect. But there is a more accurate solution to this problem than simply saying that dw = pdv = 0. This solution involves the coefficwient of volumetric thermal expansion $\alpha$ and the bulk compressibility $\beta$. If you are interested, I can lead you through how to obtain the correct solution.

Chet

 

[jack action]

@ Chestermiller:

I don't understand why you say that it is incorrect. Here's an example of what is taught at Ohio university about steam power plants:

In order to determine the enthalpy change Δh of the cooling water (or in the feedwater pump which follows), we consider the water to be an Incompressible Liquid, and evaluate Δh as follows:

[...]

We now consider the feedwater pump as follows:

 

[Chestermiller]

@ Chestermiller:

I don't understand why you say that it is incorrect. Here's an example of what is taught at Ohio university about steam power plants:

Hi Jack Action,

The original problem statement didn't say anything about an open system (flow process). In fact, it seems to imply a closed system of 1 kg.

For an open system, your description is certainly appropriate, with the change in enthalpy per unit mass of water passing through the system (at steady state) equal to minus the shaft work. For a closed system, of course, the development you provided does not apply.

It is not clear whether this homework assignment was designed to give the student experience with applying the open system version of the first law.

In any event, in my previous post, I though it might be interesting to develop the closed system solution for compression of 1 kg of water at constant entropy. Any interest?

Chet

 

[ajayguhan]

It is the isentropic compression by pump in RANKINE CYCLE ...I'll post the problem statement with solution soon. (Unable to post now due to some error)

 

[Chestermiller]

Now, don't you think it would have been pretty important to let us know that from the outset?

Chet

 

[ajayguhan]

the solution is integral Vdp where V is Vf at 1 bar.
dp= p2-p1 , where p1= 1 bar and p2=10bar.