Answer:
Δu=1300kJ/kg
Explanation:
Energy at the initial state
[tex]p_{1}=200kPa\\t_{1}=300^{o}\\u_{1}=2808.8kJ/kg(tableA-5)[/tex]
Is saturated vapor at initial pressure we have
[tex]p_{2}=200kPa\\x_{2}=1(stat.vapor)\\v_{2}=0.8858m^3/kg(tableA-5)[/tex]
Process 2-3 is a constant volume process
[tex]p_{3}=100kPa\\v_{3}=v_{2}=0.8858m^{3}/kg\\u_{3}=1508.6kJ/kg(tableA-5)[/tex]
The overall in internal energy
Δu=u₁-u₃
We replace the values in equation
Δu=u₁-u₃
[tex]=2808.8kJ/kg-1508.6kJ/kg\\=1300kJ/kg[/tex]
Δu=1300kJ/kg