$\underline{State \ 1}$
\[\begin{equation}
飽和水\left \{
\begin{array}{l}
\ p_s= 0.1\,\rm{MPa},\\
h_1=h' =417.44\,\rm{kJ/kg}&
h' '=2674.95\,\rm{kJ/kg},\\
また,\\s_1=s'=1.30256 \,\rm{kJ/(kg\cdot K)}&
s' '=7.35881 \,\rm{kJ/(kg\cdot K)}, \\
v_1=v' = 0.00104315 \,\rm{m^3/kg}
\end{array}
\right.
\end{equation}\]
$\underline{State \ 2}$
給水ポンプの仕事$l_p$は
\[
\begin{align}
l_p=v_1(p_2-p_1)
&=0.00104315 \times(10\times 10^6-0.1\times 10^6\\
&=10327.185 \,\rm{J/kg}\\
\end{align}
\]
\[
\begin{align}
\therefore h_2=h_1+l_p
&=417.44+10.327\\
&=427.767\,\rm{kJ/kg}
\end{align}
\]
$\underline{State \ 5}$
\[\begin{equation}
\left \{
\begin{array}{l}
p_5=10\,\rm{MPa}, \\
T_5=600^\circ C, \\
h_5=3625.84\,\rm{kJ/kg}, \\
s_5= 6.9045\,\rm{kJ/(kg\cdot K)}
\end{array}
\right.
\end{equation}\]
\[
s_6=s_5=6.9045\,\rm{kJ/kg}
\]
$\underline{State \ 6}$
乾き度$x$は
\[
\begin{align}
x
&=\frac{6.9045-1.30256}{7.35881-1.30256}
=0.92498
\end{align}
\]
\[
\begin{align}
\therefore
h_6
&=(1-x)h' +xh' '\\
&=(1-0.92498)\times 417.44+0.92498\times 2674.95\\
&=2505.59\,\rm{kJ/kg}
\end{align}
\]
$(1)$
理論熱効率
\[
\eta_{th}=\frac{h_5-h_6-(h_2-h_1)}{h_5-h_2}=0.347\\
\therefore
34.7 \%
\]
$(2)$
ポンプ仕事を無視した理論熱効率
\[
\eta_{th}=\frac{h_5-h_6}{h_5-h_1}=0.3491\\
\therefore
34.9 \%
\]