露点温度与压力的计算

含有80%(mol)醋酸乙酯(A)和20%乙醇(E)的二元物系，液相活度系数用Van Laar 方程计算，AE A =0.144，EA A =0.170。试计算在101.3kPa 压力下的露点温度。

安托尼方程为：

醋酸乙酯：()15.5750.27900444.21ln --=T P S A

乙醇：()

68.4198.38038047.23ln --=T P S E (P S ：Pa ；T ：K) 解1：露点温度

此时8.0=A y ，2.0=E y

设T=350K ，

()Pa P T P S A S A 100271516.1115.5750.27900444.21ln ==--=

()Pa

P T P S E S

E 955054669.1168.4198.38038047.23ln ==--= 设1=A γ，1=E γ

9898

4.0==P P K S

A A A γ 94279.0==

P P K S E E E γ 8082.0==A

A A K y x 2121.0==E

E E K y x 00806.02121.0170.08082.0144.01144.01ln 22'=⎪⎭⎫ ⎝⎛⨯⨯+=⎪⎪⎭⎫ ⎝⎛+=E EA A AE AE

A x A x A A γ

0081.1'=A γ

09908

.08082.0144.02121.0170.01170.01ln 22'=⎪⎭⎫ ⎝⎛⨯⨯+=⎪⎪⎭⎫ ⎝⎛+=A AE E EA EA

E x A x A A γ 1042.1'

=E γ

99785.0101300

1002710081.1''=⨯==P P K S

A A A γ 04103.1101300955051042.1''=⨯==

P P K S E E E γ 9938

.01921.08017.0'''

=+=+=∑E E A A i i

K y K y K y ∑≈1i

i K y

所以露点温度为350K 。

解2：(1)计算活度系数：

0075.0)2.017.08.0144.01(144.0}1(ln 22=⨯⨯+=+=E EA A AE AE A x A x A A r

A r =1.0075

10137.0}8.0144.02.017.01(17.0}1(ln 22=⨯⨯+=+=A AE E EA EA E x A x A A r E r =1.107

(2)计算露点温度

设T=353.15K(80℃)

∴9867.0894.01037.12=⨯=A K

)15.575.27900444.21exp(0075.110013.19867.0522

--=⨯⨯==T r P K P A A S A 解得T 2=349.67K(76.52℃)

∴s A P ln =11.505

Pa P s A 41092.9⨯= s E P ln =11.454

Pa P s E 410425.9⨯= 9866.0013.1992.00075.1=⨯=A K 03.1013.19425.0107.1=⨯=E K

1005.103.12.09866.08.0≈=+=∑i x

故露点温度为76.52℃