# How tds=du+pdv valid for open flow system?

1. Aug 5, 2015

### hemant singh

I have been having this doubt for long. As by the first law dq=du+dw (neglecting KE and PE change). And we know that the work done by the system can be utilized in things (eg. boundary work, shaft work, shear work, paddle wheel work..etc).
For open flow system replacing dq=tds, we get,
tds=du+ dw (process should be reversible)
if we replace dw= pdv+(shaft work..etc), then we get,
tds=du+pdv+(shaft work..etc)

Why isn't the above equation valid, instead we use tds=du+pdv. Why are we considering only boundary work not other works?

2. Aug 5, 2015

### Staff: Mentor

You are aware that u, v, and s are functions of state, independent of any process or kind of equipment arrangement used the bring the material comprising the system to that thermodynamic equilibrium state, correct?

Chet

3. Aug 5, 2015

### hemant singh

I know that u v s are state properties. But in open flow system KE PE both are present if that is taken into consideration then tds=dE+pdv?
where E is total macroscopic and microscopic energy? Why do we use tds= du+ pdv ?

4. Aug 5, 2015

### Staff: Mentor

As you said, u,v,and s are state properties. That means that they each are unique functions of T and P only. So they are actually physical properties of the material being processed, and do not depend on the nature of any process. Also, it doesn't matter whether the material is traveling at 10 km/hr or 0 km/hr, or whether the material is on top of mount Everest or at the bottom of Death Valley. This does not affect its u, v, and s. For example, regarding kinetic energy, to an observer traveling with the same velocity as the material, the material is standing still. Since the properties of a material cannot depend on the frame of reference of the observer, u, v, and s are independent of its Kinetic energy.

Imagine two closely neighboring thermodynamic equilibrium states of a material at temperatures and pressures (T,P) and (T+dT ,P+dP). The equation Tds=du+Pdv describes the unique relationship between ds, du, and dv between these two thermodynamic equilibrium states for the material. So the equation only applies to differentially separated thermodynamic equilibrium states.

Hope this helps. Something else that might help is the following link to my recent Physics Forums Insights article: https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/

Chet

5. Aug 6, 2015

### hemant singh

Thanks a lot sir.. doubt cleared.