$\underline{State \ 1}$ \[\begin{equation} 飽和水\left \{ \begin{array}{l} \ p_s= 0.1\,\rm{MPa},\\ h_1=h' =417.44\,\rm{kJ/kg}& h' '=2674.95\,\rm{kJ/kg},\\ また，\\s_1=s'=1.30256 \,\rm{kJ/(kg\cdot K)}& s' '=7.35881 \,\rm{kJ/(kg\cdot K)}, \\ v_1=v' = 0.00104315 \,\rm{m^3/kg} \end{array} \right. \end{equation}\] $\underline{State \ 2}$ 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 \[ \eta_{th}=\frac{h_5-h_6}{h_5-h_1}=0.3491\\ \therefore 34.9 \% \]