i) In a cylinder fitted with piston ii) In a Turbine

i) In a cylinder fitted with piston: for isentropic expansion

The initial volume before expansion can be found out by

$v_1=\frac{RT_1 m}{P_1}=\frac{(0.4651×(184.2+273)×1}{1000}=0.2126m^3$

$p_1 v_1^k=p_2 v_2^k$

$10×0.2126^1.327=1×v_2^k$

$v_2=1.2057m^3$

The final temperature and the boundary work are determined as

k for steam is 1.327,R=0.4651 Kj/KgK(from property table)

$T_2=\frac{P_2 V_2}{mR}=\frac{100×1.2057}{(1×0.4651)}=259.23K$

$W_b=\frac{P_2 V_2-P_1 V_1}{1-k}=\frac{100×1.2057-1000×0.2126}{1-1.327}=281.43KJ$

ii) In a turbine:

$s_1=s_2$

From steam table for 10 bar

$h_1=h_g=2777.1KJ/Kg,s_1=s_g=6.5850KJ/Kg$

For end of expansion we can calculate

$s_2=s_1=6.5850KJ/KgK,h_f2=417.5KJ/Kg,s_fg=6.056KJ/KgK,$

$h_fg2=2257.4KJ/Kg,s_f=1.3028 KJ/KgK$

$s_1=s_2=s_f+x×s_fg x=\frac{s_2-s_f}{s_fg} =\frac{6.585-1.3028}{6.056}=0.8722$

Condition of steam, x= 87.22 dry

$work=h_1-h_2$

$where,h_2=h_f2+xh_fg2=417.5+0.8722×2257.4=2386.4KJ/Kg$

work=2777.1-2386.4=390.69KJ/Kg