Answer:
a) [tex]\dot m = 6.878\,\frac{kg}{s}[/tex], b) [tex]T = 104.3^{\textdegree}C[/tex], c) [tex]\dot S_{gen} = 11.8\,\frac{kW}{K}[/tex]
Explanation:
a) The turbine is modelled by means of the First Principle of Thermodynamics. Changes in kinetic and potential energy are negligible.
[tex]-\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0[/tex]
The mass flow rate is:
[tex]\dot m = \frac{\dot W_{out}}{h_{in}-h_{out}}[/tex]
According to property water tables, specific enthalpies and entropies are:
State 1 - Superheated steam
[tex]P = 7000\,kPa[/tex]
[tex]T = 500^{\textdegree}C[/tex]
[tex]h = 3411.4\,\frac{kJ}{kg}[/tex]
[tex]s = 6.8000\,\frac{kJ}{kg\cdot K}[/tex]
State 2s - Liquid-Vapor Mixture
[tex]P = 100\,kPa[/tex]
[tex]h = 2467.32\,\frac{kJ}{kg}[/tex]
[tex]s = 6.8000\,\frac{kJ}{kg\cdot K}[/tex]
[tex]x = 0.908[/tex]
The isentropic efficiency is given by the following expression:
[tex]\eta_{s} = \frac{h_{1}-h_{2}}{h_{1}-h_{2s}}[/tex]
The real specific enthalpy at outlet is:
[tex]h_{2} = h_{1} - \eta_{s}\cdot (h_{1}-h_{2s})[/tex]
[tex]h_{2} = 3411.4\,\frac{kJ}{kg} - 0.77\cdot (3411.4\,\frac{kJ}{kg} - 2467.32\,\frac{kJ}{kg} )[/tex]
[tex]h_{2} = 2684.46\,\frac{kJ}{kg}[/tex]
State 2 - Superheated Vapor
[tex]P = 100\,kPa[/tex]
[tex]T = 104.3^{\textdegree}C[/tex]
[tex]h = 2684.46\,\frac{kJ}{kg}[/tex]
[tex]s = 7.3829\,\frac{kJ}{kg\cdot K}[/tex]
The mass flow rate is:
[tex]\dot m = \frac{5000\,kW}{3411.4\,\frac{kJ}{kg} -2684.46\,\frac{kJ}{kg}}[/tex]
[tex]\dot m = 6.878\,\frac{kg}{s}[/tex]
b) The temperature at the turbine exit is:
[tex]T = 104.3^{\textdegree}C[/tex]
c) The rate of entropy generation is determined by means of the Second Law of Thermodynamics:
[tex]\dot m \cdot (s_{in}-s_{out}) + \dot S_{gen} = 0[/tex]
[tex]\dot S_{gen}=\dot m \cdot (s_{out}-s_{in})[/tex]
[tex]\dot S_{gen} = (6.878\,\frac{kg}{s})\cdot (7.3829\,\frac{kJ}{kg\cdot K} - 6.8000\,\frac{kJ}{kg\cdot K} )[/tex]
[tex]\dot S_{gen} = 11.8\,\frac{kW}{K}[/tex]
