Answer:
[tex]\dot Q_{out} = 104.56\,kW[/tex]
Explanation:
The steam turbine is modelled after the First Principle of Thermodynamics. Changes in potential and kinetic energy are negligible:
[tex]-\dot Q_{out} -\dot W_{out} +\dot m \cdot (h_{in}-h_{out}) = 0[/tex]
The heat rate is:
[tex]\dot Q_{out} = \dot m \cdot (h_{in}-h_{out})-\dot W_{out}[/tex]
From steam and saturated tables, specific enthalpies at inlet and outlet are found:
Inlet - Superheated vapor
[tex]P = 1600\,kPa[/tex]
[tex]T = 350^{\textdegree}C[/tex]
[tex]h = 3146.0\,\frac{kJ}{kg}[/tex]
Outlet - Saturated vapor
[tex]T = 30^{\textdegree}C[/tex]
[tex]P = 4.2469\,kPa[/tex]
[tex]h = 2555.6\,\frac{kJ}{kg}[/tex]
The loss rate is:
[tex]\dot Q_{out} = (21.4\,\frac{kg}{s} )\cdot (3146.0\,\frac{kJ}{kg}-2555.6\,\frac{kJ}{kg} )-12350\,kW[/tex]
[tex]\dot Q_{out} = 104.56\,kW[/tex]
