Please prove that dW=-vdP where v is specific volume

[benny_91]

Homework Statement

I came across this equation denoting the work done by an open system (e.g. turbine or compressor). I wonder how they arrived at such an equation.

Homework Equations

Differential form of the steady flow energy equation for an infinitesimally small control volume neglecting kinetic and potential energy changes[/B]
u+pv+dQ-dW-(p+dp).(v+dv)=u+du
du+pdv+vdp=dQ-dW ...(neglecting the term dp.dv)

The Attempt at a Solution

If we use dQ=du+pdv in the above equation we shall get the required equation for work done.
But how can we use this relation of dQ=du+pdv which is valid for close systems having only one kind of work i.e. expansion work.
If you feel the above method is incorrect please provide the appropriate method to prove this relation.

 

[stevendaryl]

I am not sure I understand the situation in which you're supposed to compute work. But typically, the work done by a system as it exands is dW = P dV (or the negative of that if you are computing the work done ON the system) In the special case of constant temperature, P V is cconstant, so P dV = - V dP

 

[benny_91]

I am not sure I understand the situation in which you're supposed to compute work. But typically, the work done by a system as it exands is dW = P dV (or the negative of that if you are computing the work done ON the system) In the special case of constant temperature, P V is cconstant, so P dV = - V dP

I believe this equation is true not just for isothermal processes but other open system processes as well.

 

[Chestermiller]

You are aware that, for an open system, the VdP corresponds to the shaft work in irreversible operation (not the total work), correct? If not, what does the steady state version of the first law reduce to for an open system when the change in kinetic energy and potential energy from inlet to outlet of the control volume are negligible?

 

[benny_91]

You are aware that, for an open system, the VdP corresponds to the shaft work in irreversible operation (not the total work), correct? If not, what does the steady state version of the first law reduce to for an open system when the change in kinetic energy and potential energy from inlet to outlet of the control volume are negligible?

Thanks Chet for replying. If you have a look at the derivation I have posted, you will see that I have done it using the steady state energy equation neglecting changes in kinetic and potential energy. If you can explain to me the proof of this relation or provide any link which shows its derivation I would be very grateful!

 

[Chestermiller]

Thanks Chet for replying. If you have a look at the derivation I have posted, you will see that I have done it using the steady state energy equation neglecting changes in kinetic and potential energy. If you can explain to me the proof of this relation or provide any link which shows its derivation I would be very grateful!

$$dh=dQ-dW_s$$
where all quantities are per unit mass. But,
$$dh=Tds+vdP$$
So,
$$Tds+vdP=dQ-dW_s$$
If the operation is reversible, then $Tds=dQ$. So, $$dW_s=-vdP$$
This is also derived in Moran, et al, Fundamentals of Engineering Thermodynamics (a wonderful book IMHO).

 

[benny_91]

$$dh=dQ-dW_s$$
where all quantities are per unit mass. But,
$$dh=Tds+vdP$$
So,
$$Tds+vdP=dQ-dW_s$$
If the operation is reversible, then $Tds=dQ$. So, $$dW_s=-vdP$$
This is also derived in Moran, et al, Fundamentals of Engineering Thermodynamics (a wonderful book IMHO).

I shall try to explain where I find the equation difficult to understand. In the 2nd step of the derivation you have stated this equation:

$$dh=dQ-dW_s$$
where all quantities are per unit mass. But,
$$dh=Tds+vdP$$
So,
$$Tds+vdP=dQ-dW_s$$
If the operation is reversible, then $Tds=dQ$. So, $$dW_s=-vdP$$
This is also derived in Moran, et al, Fundamentals of Engineering Thermodynamics (a wonderful book IMHO).

I shall try to explain where I find the equation difficult to understand. In the 2nd step of the derivation you have stated this equation:
$$dh=Tds+vdP$$
Though I don't know how this equation is actually derived I have given it a try.
Consider a infinitesimally small control volume with properties of the fluid within it denoted by the usual notations. Let the flow be reversible.
Now if we use closed system analysis here we can come up with the following,
$$dQ=du+pdv$$
Adding and subtracting the term vdp in the RHS we get
$$dQ=du+pdv+vdp-vdp$$
$$dQ=d(u+pv)-vdp$$
$$dQ=dh-vdp$$
Now my question is how can we write $$dQ=du+pdv$$ in the first hand. It means that the entire heat that was supplied was utilized in increasing the internal energy of the fluid within the control volume and also to do a closed system kind of work (or simply increase its specific volume by expanding against the pressure within the control volume). But why can't it be possible that some of the heat supplied was utilized for doing the shaft work as well. So why is that not taken into consideration? Thanks in advance!

 

[Chestermiller]

The equation in question follows from the fundamental relationship between differential changes in internal energy u, entropy s, and volume v between two closely neighboring thermodynamic equilibrium states:
$$du=Tds-Pdv$$