Answer:
The work of the turbine is done at the expense of the enthalpy of the fluid.
Explanation:
Ideally speaking, a turbine is a steady-state devices that has no heat interactions and generates works at the expense of enthalpy. By the First Law of Thermodynamic, we have the following definition for steady-state conditions: (all energy components are measured in joules)
[tex]E_{in} - E_{out} + E_{gen} = 0[/tex] (1)
Where:
[tex]E_{in}[/tex] - Inlet energy.
[tex]E_{out}[/tex] - Outlet energy.
[tex]E_{gen}[/tex] - Generated energy.
Generated energy is associated with endothermic and exothermic chemical reactions and fission processes in radioactive elements and isotopes. Since fluids used in turbines are usually inert, the generated energy is zero and each remaining componets of the expression are defined:
[tex]E_{in} = U_{g,in}+K_{in}+H_{in}[/tex] (2)
[tex]E_{out} = U_{g,out}+K_{out}+H_{out}+W_{out}[/tex] (3)
Where:
[tex]U_{g,in}[/tex], [tex]U_{g,out}[/tex] - Inlet and outlet gravitational potential energies.
[tex]K_{in}[/tex], [tex]K_{out}[/tex] - Inlet and outlet translational kinetic energies.
[tex]H_{in}[/tex], [tex]H_{out}[/tex] - Inlet and outlet enthalpies.
[tex]W_{out}[/tex] - Energy generated by the turbine.
Usually, turbines at steady state does not report significant changes in translational and gravitational potential energies and (2) and (3) can be reduced into this form:
[tex]E_{in} = H_{in}[/tex] (2b)
[tex]E_{out} = H_{out}+W_{out}[/tex] (3b)
And we find the following equation of work by substituting on (1):
[tex]W_{out} = H_{in}-H_{out}[/tex] (4)
In a nutshell, the work of the turbine is done at the expense of the enthalpy of the fluid.
