領域$\rm{A}$ : 圧縮液 領域$\rm{B}$ : 湿り蒸気 領域$\rm{C}$ : 過熱蒸気 曲線$\rm{bc}$ : 飽和液線 曲線$\rm{bd}$ : 乾き飽和蒸気線 点$\rm{b}$ : 臨界点

$(a)$ 三重点 $(b)$ $273.16$ $(c)$ $610.7$ $(d)$ $0$

$(1)$ 付表2より，圧力$0.101325$の値を読み， \[ v' = 0.00104344 \,\textrm{m}^3\textrm{/kg}, \ \ v' ' = 1.67330 \,\textrm{m}^3\textrm{/kg} \\ \therefore\frac{v' ' }{v'}= \frac{1.6733}{0.00104344} = 1604\,倍 \] $(2)$ 付表2より， \[ T_s=99.974^\circ \textrm{C} \] $(3)$ 付表2より， \[ h'=418.99\,\textrm{kJ/kg}\\ h' ' =2675.53\,\textrm{kJ/kg}\\ s'=1.30672\,\textrm{kJ/kg}\\ s' '=7.35439\,\textrm{kJ/kg} \] $(4)$ \[ r=h' ' - h' より \\ r=2256.54\,\textrm{kJ/kg} \]

\[ x=\frac{v-v'}{v' '-v'} \]

\[ r=h' '-h’ \]

$(1)$ \[ \begin{equation} \left \{ \begin{array}{l} T_s=200^\circ \rm{C}\\ h' = 852.39 \,\rm{kJ/kg} & h' ' = 2792.06\,\rm{kJ/kg}\\ s' =2.33080 \,\rm{kJ/(kg \cdot K)} & s' ' =6.43030 \,\rm{kJ/(kg \cdot K)} \end{array} \right. \end{equation}\] \[ \begin{align} h &=(1-x)h' + xh' ' \\ &=0.5 \times 852.39+0.5 \times 2792.06 \\ \therefore &=1822.225 \,\rm{kJ/kg} \end{align} \] $(2)$ \[ \begin{align} s &=(1-x)s'+ xs' '\\ &=0.5 \times 2.33080 + 0.5 \times 6.43030 \\ &=4.38055 \,\rm{ kJ/(kg \cdot K)} \end{align} \]

$(1)$ \[\begin{equation} \left \{ \begin{array}{l} p_s=1.0 \,\rm{MPa} \\ h' = 762.68 \,\rm{kJ/kg} & h' ' = 2777.12\,\rm{kJ/kg}\\ s' =2.13843 \,\rm{kJ/(kg \cdot K)} & s' ' =6.58498 \,\rm{kJ/(kg \cdot K)} \end{array} \right. \end{equation}\] \[ \begin{align} h &=(1-x)h' + xh' ' \\ &=0.2 \times 762.68+0.8 \times 2777.12 \\ \therefore &=2374.232 \ \rm{kJ/kg} \end{align} \] $(2)$ \[ \begin{align} s &=(1-x)s'+ xs' '\\ &=0.2 \times 2.13843 + 0.8 \times 6.58498 \\ &=5.69567 \ \rm{kJ/(kg\cdot K)} \end{align} \]

付表３より $(1)$ \[ \therefore比容積 \ v=0.047423 \,\rm{ m^3/kg} \\ \] $(2)$ \begin{equation} \left \{ \begin{array}{l} h= 3178.18 \,\rm{ kJ/kg }\\ s= 6.5431 \,\rm{ kJ/(kg \cdot K)}\\ \end{array} \right. \end{equation} 比内部エネルギー$u$は \[ \begin{align} u &=h-p_v\\ &=3178.18 \times 10^3 + -6 \times 10^6 \times 0.047423\\ &=2.8936 \times 10^6 \,\rm{ J/kg}\\ \therefore &=2.894 \,\rm{MJ/kg} \end{align} \]

\[\begin{equation} \left \{ \begin{array}{l} T_s=5^\circ \rm{C}\\ h' = 21.02 \,\rm{kJ/kg} & h' ' = 2510.07 \,\rm{kJ/kg}\\ \end{array} \right. \end{equation}\] \[\begin{equation} \left \{ \begin{array}{l} T_s=10^\circ \rm{C}\\ h' = 42.02 \,\rm{kJ/kg} & h' ' = 2519.23 \,\rm{kJ/kg}\\ \end{array} \right. \end{equation}\] 内挿法により，$T_s=8^\circ \rm{C}$では， \[ \begin{align} h' &=21.02+\frac{42.02-21.02}{10-5}\times (8-5) \\ &=33.62 \,\rm{kJ/kg} \end{align} \] \[ \begin{align} h' ' &=2510.07+\frac{2519.23-2510.07}{10-5}\times (8-5) \\ &=2515.57 \,\rm{kJ/kg} \end{align} \]

$(1)$ $(a)$ \[ 170.41^\circ \rm{C} \] $(b)$ \[ 2046.5\,\rm{kJ/(kg \cdot K)} \] $(c)$ \[ \begin{align} v &=v' +x(v' ' - v') \\ &=0.00111498 + 0.6 (0.240257 - 0.00111498) \\ &=0.144600 \end{align} \] \[ \therefore 0.1446\,\rm{m^3/kg} \] $(d)$ \[ \begin{align} h &=720.932 \times 0.6 \times (2046.5) \\ &=1948.8 \end{align} \] \[ \therefore 1949\,\rm{kJ/kg} \] $(e)$ \[ \begin{align} s &=2.04572+0.6 \times (6.65960-2.04572) \\ &=4.814 \end{align} \] \[ \therefore 4.814\,\rm{kJ/kg \cdot K} \] $(2)$ \[ \begin{align} v_1 &= v'_1 + x(v_1' ' - v_1') \\ &=0.00113858+0.9 \times (0.1632 - 0.001138580) \\ &=0.146994 \end{align} \] \[ \begin{align} x_2 &=\frac{v_1 - v'}{v' ' - v'} \\ &=\frac{0.146994 - 0.00109284}{0.374676-0.00109284} \\ &=0.3905 \end{align} \] $(3)$ $(f)$ \[ 419.7 \,\rm{kJ/kg} \] $(g)$ \[ 3478.3 \,\rm{kJ/kg} \] $(h)$　 \[ \begin{align} Q &=m(h_2 - h_1) \\ &=5 \times (3478.2 - 419.7) \\ &=15292.5 \,\rm{kJ} \\ &=15.29 \,\rm{MJ} \end{align} \] $(4)$ \[ 3204.9 + \frac{3248.7 - 3204.9}{2}=3226.8\\ \\ \therefore 3227 \,\rm{kJ/kg} \]

$(1)$　 $0.1 \,\rm{m^3/kg}$ $(2)$　 $2800 \,\rm{ kJ/kg}$ $(3)$　 $1.5 \,\rm{ MPa}$ $(4)$　 $0.5 \,\rm{m^3/kg}$ $(5)$　 $0.9$

%=image:/media/2015/02/03/142289847198710800.png:

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

%=image:/media/2015/02/03/142289863990369200.png:

%=image:/media/2015/02/03/142289893357120400.png:

\[\begin{equation} \left \{ \begin{array}{l} h_1=400\,\rm{kJ/kg} \\ h_2=413\,\rm{kJ/kg}\\ h_3=230\,\rm{kJ/kg} \\ h_4=230\,\rm{kJ/kg} \end{array} \right. \end{equation}\] $(1)$ \[ \begin{align} COP_H &=\frac{h_2-h_3}{h_2-h_1}\\ &=\frac{413-230}{413-400}\\ &=\frac{183}{13}\\ &=14.1 \end{align} \] $(2)$ \[ \begin{align} COP_R &=\frac{h_1-h_4}{h_2-h_1}\\ &=\frac{400-230}{413-400}\\ &=\frac{170}{13}\\ &=13.1 \end{align} \] $(2)$の別解 同一サイクルにおいて冷凍機の成績係数はヒートポンプの成績係数より１小さいので， \[ \begin{align} COP_R &=COP_H-1 \\ &=14.1-1 \\ &=13.1 \end{align} \]