Answer:
x=0.8
Explanation:
Defining our values we have
[tex]p_1=3Mpa\\T_1=400\° c\\q_{1-2}=0[/tex]
Where [tex]p_1[/tex] is the pressure of the super heated steam
[tex]T_1[/tex] is the Temperature of the super heated steam
and [tex]q_{1-2}[/tex] is the pressure in an adiabatic process.
State 1
[tex]v_1 = 0.09936m^3/Kg\\s_1=6.9213kJ/kg.K[/tex]
State 2
[tex]s_2=s_1=6.9212[/tex]
Note that here in state 2 the process is Reversible.
At the value where [tex]T_2 = 141\°c[/tex] we have
[tex]v_2=(v_g)_{T_2}=0.4972m^3/kg[/tex]
Through this values we can calculate the Fraction of steam,
[tex]x=\frac{m_1-m_2}{m_1}\\x=1-\frac{m_2}{m_1}\\x=1-\frac{v_1}{v_2}\\x=1-\frac{0.9936}{0.4972}\\x=0.8[/tex]
