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-