Relationship between work, internal energy and enthelpy.

[zzinfinity]

Hi,
I'm taking a thermodynamics class and I'm stuck on how work relates to enthalpy and internal energy.
Does work done by a system equal change in internal energy, change in enthalpy or does it depend on the situation?

The question I'm stuck on asks for the work produced by a steam turbine and gives initial and final properties for the steam. I feel like the work produced by the turbine is just equal to the ΔU for the steam, but I could also see an argument for using ΔH. Any thoughts would be appreciated.

 

[edgepflow]

Mass flowing across a control surface brings energy (neglecting potential and kinetic energy changes):

u + P $\nu$

where,

u = fluid specific internal energy
P = fluid static pressure
$\nu$ = fluid specific volume

But this is the definition of enthalpy:

h = u + P $\nu$

So enthalpy includes the internal energy and the "flow work" term P $\nu$.

So for your turbine you will use Δh as you mentioned.

 

[I like Serena]

The work done by a system depends on the situation.
The general formulas are:
dU=dQ+dW
dU=TdS-PdV
dH=TdS+VdP

I think you would use ΔH=ΔQ for a quasi-static, constant-pressure process.
And you would use ΔU=-ΔW for an adiabatic process.

I believe both processes apply in a steam turbine cycle.