某一常压连续精馏塔内对苯-甲苯的混合液进行分离。原料液 组成为 0.35 (mol%, 下同), 该物系的平均相对挥发度为 2.5 , 饱和蒸汽进料。塔 顶采出率为 dfrac(q_{n, D)}(q_{n, F)}=40 % , 精馏段操作线方程 为 y_(n+1)=0.75 x_(n)+0.20 , q_(n, F)=100 mathrm(kmol) cdot h^-1 求:提馏段操作线方程式;若塔顶第一块板下降的液相组成为 0.7 , 该板的气相默夫里效率 E_(mathrm{m), 1}

(1) $ 0$.$75$=$\dfrac{R}{R+1} \quad \mathrm{R}$=$3 $$ 0$.2=$\dfrac{x_{d}}{R+1}$=$\dfrac{x_{d}}{3+1} \quad x_{\mathrm{d}}$=0.$8 $$\dfrac{q_{n, D}}{q_{n, F}}=\dfrac{x_{f}-x_{w}}{x_{d}-x_{w}}=\dfrac{0.35-x_{w}}{0.8-x_{w}}$=0.$4 \quad x_{w}$=0.$05 $\begin{array}{l}$\because \dfrac{q_{n, D}}{q_{n, F}}=40 \% $$\therefore q_{\mathrm{n}, \mathrm{D}}=0.4 \times 100=40 \mathrm{kmol} \cdot \mathrm{h}^{-1} $$q_{\mathrm{n}, W}=q_{\mathrm{n}, \mathrm{F}}-q_{\mathrm{n}, \mathrm{D}}=100-40=60 \mathrm{kmol} \cdot \mathrm{h}^{-1} $$q_{\mathrm{n} \mathrm{V}}=(\mathrm{R}+1) q_{\mathrm{n} \mathrm{D}}=$$(3+1) \times 40=160 \mathrm{kmol} \cdot \mathrm{h}^{-1} $$\delta=0 $$q_{\mathrm{n} \mathrm{V}}^{\prime}=q_{\mathrm{n} \mathrm{V}}-q_{\mathrm{n}, \mathrm{F}}=160-100=60 \mathrm{kmol} \cdot \mathrm{h}^{-1}$$q_{\mathrm{n} \mathrm{L}}^{\prime}=q_{\mathrm{n}, \mathrm{L}}=q_{\mathrm{n}, \mathrm{V}}-q_{\mathrm{n}, \mathrm{D}}$$=160-40=120 \mathrm{kmol} \cdot \mathrm{h}^{-1}$提馏段操作线方程:$y_{m+1}=\dfrac{q_{n, L}^{\prime}}{q_{n, V}^{\prime}} x_{m}-\dfrac{q_{n, W}}{q_{n, V}^{\prime}} x_{W}=2 x_{\mathrm{m}}-0.05$$\text { (2) } y_{1}^{*}=\dfrac{2.5 x_{1}}{1+(2.5-1) x_{1}}=\dfrac{2.5 \times 0.7}{1+(2.5-1) \times 0.7}=0.8573$ $\mathrm{y}_{2}=0.75 \times 0.7+0.2=0.725 $$\mathrm{y}_{1}=0.8$$E_{m, V, 1}=\dfrac{y_{1}-y_{2}}{y_{1}^{*}-y_{2}}=\dfrac{0.8-0.725}{0.8573-0.725}=0.5669$