ln γ 2 =
0.9595 0.9595 = 0.9595 x 2 2 x (1+ ) (1+0.5872 2 ) 2 1.634 x 1 x1
ln γ 1 = 1.634 1.634 = γ 1 =3.523 1.634 x1 2 0.0721 2 (1+ ) (1+1.703 ) 0.9595 x 2 1-0.0721
计算结果与下述实验数据比较: 乙醇
温度℃ 100 89.0 85.3 81.5 79.7 78.4 78.15 液相x1% 0 7.21 12.38 32.73 51.98 74.22 89.43 汽相y% 0 38.91 47.04 58.26 65.99 78.15 89.43
解:(1) 若当作理想溶液处理,有Py1=P1S ① Py2=P2S ② 由① 得y1= P1Sx1/P ③ ①+② 得P=P1Sx1+ P2Sx2
y1 =1-∑ y i =0.7548
2 4
已知 T, y1, y2…….., yn,求 P, x1, x2,……..xn
φ i P y i xi = P i Sφ i S γ
i
γ (初值) 仿照泡点计算,假设P, i = 1
若∑xi>1 ,则应调低P;若∑xi<1 ,则应调高P. (3)闪蒸计算 ――― 一次汽液平衡 已知:F,Zi 求V,L,xi,yi,共2N+2个未知数
,Pi S =f 3 (T )
(未知数T)
γ =f (T,x ,x ,.......x )
i 4 1 2 n
计算步骤如下:(5n+1个未知数,5n+1个独立方程,则可解)
开始 输入P, xi (已知), 估 算 温 度 T( 初 值),设所有