Given: $T_3= 500^oC, T_5= 400^oC, T_6= 40^oC, x_6 = 0.85$

Solution:

At $40^oC$, from Steam tables,

$s_6 = sf_6 + x_6 × sfg_6$

= 0.572 + 0.85(7.686)

= 7.1051kJ/K

$h_6 = hf_6 + x_6 × hfg_6$

= 167.5 + 0.85(24069)

= 2213.365k/kg

Process 5-6 is isentropic,

$s_5 = s_6 = 7.1051kJ/kgK$

At $400^oC$, from Superheated Steam tables,

By Interpolation,

$\frac{20 - P_5}{22 - P_5} = \frac{7.130 - 7.1051}{7.082 - 7.1051}$

$P_5 = 21.03 bar$

Therefore, $$P_4 = P_5 = 21.03 bar$$

Similarly by Interpolation,

At 21.03bar, from Steam tables,

By Interpolation,

$h_4 = 2798.227 kJ/kg$

$s_4 = 6.31846kJ/kg$

Process 3-4 is isentropic,

$s_3 = s_4 = 6.3184kJ/kgK$

At $500^oC$, from Superheated Steam tables,

By Interpolation,

$\frac{150 - 156.943}{160 - 156.943} = \frac{33106 - h_3}{3297.1 - h_3}$

$h_3 = 3301.226kJ/kg$

At $40^oC$, from Steam tables,

$h_1 = h_2 = (hf)_40 = 167.5kJ/kg$

Work Output is given as

$W = h_3 - h_4 + h_5 - h_6 = 3301.226 - 2798.227 + 3247.052 - 2213.365 = 1536.686 kJ/kg$

Now, Efficiency is given as,

$η = \frac{W}{h_3 - h_2 + h_5 - h_4} = \frac{1536.686}{3301.226 - 167.5 + 3247.052 - 2798.227} = 0.4289 = 42.89$%

The Mean Effective Pressure is given as,

$Steam Rate = \frac{3600}{W} = \frac{3600}{1536.686} = 2.3427kJ/kWh$