9
'(1)(1)(2.921)42.99168.52/V R D q F kmol h =+--=+⨯= (泡点进料:q=1) ' 2.9242.991121.53238.06/L RD qF kmol h =+=⨯+⨯= ④求操作线方程
精馏段操作线方程为
10.7490.244211D
n n n x R
y x x R R +=+=+++
提馏段操作线方程为
'
1'' 1.4120.092m m w m L W
y x x x V V +=-=-
(2)逐板法求理论板 又根据min (1)1
[]11d D
F f
x x R x x α-=-α-- 可解得
α=2.475 相平衡方程为: 2.4751(1)1 1.475x
x
y x x αα==+-+
1D y x = = 0.957 11
11111(1) 2.475(1)
y
y x y y y y ==+α-+-=0.901
211110.7450.24420.915d
x
R y x R R x =+++=+= 2
2220.813(1)y x y y =
=+2.475-
320.7450.24420.850y x =+= 3333(1)
y x y y ==+2.475-0.696
430.7450.24420.763y x =+= 44440.565(1)
y x y y ==+2.475-
540.7450.24420.665y x =+= 55550.420(1)
y x y y ==+2.475-
650.7450.24420.557y x =+= 66660.337(1)
y x y y ==+2.475-