发布时间 : 星期三 文章分离工程习题指导更新完毕开始阅读097875c1d5bbfd0a795673c4
所以?xi??yiKi?0.0516.5?0.103.2?0.550.43?1.614
选异丁烷为参考组分
K42?K41??xi?0.43?1.614?0.694
由25℃,K42=0.694查得P=560KPa,查得各组分的Ki值 求得?xi?0.990?1故混合物在25℃时的露点压力为560KPa
序号 组成 组成 1000KPa Ki xi 560KPa Ki xi 1 2 3 4
甲烷 乙烷 0.05 16.5 0.003 27.5 0.10 3.2 0.031 5.20 0.002 0.019 丙烷 0.30 1.0 0.30 1.70 0.176 异丁烷 0.55 0.43 1.28 0.694 0.793 1.614 0.990 ?6 含有80%(mol)醋酸乙酯(A)和20%(mol)乙醇(E)的二元物系。液相活
度系数用Van Laar方程计算,AAE=0.144,AEA=0.170。试计算在101.3kPa压力下的泡点温度和露点温度。 解:
由Vanlaar方程得:
AAE(1?xAEAAExEAAEA)2lnrA??(1?0.1440.144?0.80.17?0.2)2,得rA=1.0075
lnrE?(1?AEAxEAAEAxAEAAE)2?(1?0.1700.170?0.20.144?0.8)2, 得rB=1.1067
因为低压气体可视为理想气体,故
ripixipspyi?ripixis,得
yi?
(1) 泡点温度时,设T=348.15K,由安托尼方程得
pA=94.377kPa, pESs=88.651kPa
ss故
?yi?yA?yE?rApAxAprEpExE+
p1.0075?94.377?0.8=
101.3?1.1067?88.651?0.2101.3
=0.945<1, 可知所设温度偏低,重设T=349.82K: 此时
pAS=99.685kPa,
rApAxApspEs=94.819kPa
rEpExEs?yi?yA?yE?+
p1.0075?99.685?0.8=
101.3?1.1067?94.819?0.2101.3=1.00033≈1
故泡点温度为349.82K
pyipiris(2) 求露点温度,此体系可视为理想气体,由设T=349.8K 由安托尼方程得
pASpyi?ripixisxi?,得
=99.620kPa,
pEs=94.743kPa,
101.3?0.8故?偏低
xi?xA?xB=99.620?1.0075?101.3?0.294.743?1.1067=1.4>1,故所设温度
重设T=350.1K时?xi?xA?xB=0.992≈1
故露点温度为350.1K
解1:⑴泡点温度
此时xA?0.8,xE?0.2 ln?A?AAE?Ax?1?AEA?AEAxE?????2?0.1440.144?0.8???1??0.170?0.2??2 ?0.0075 ?A?1.0075 ln?E?AEA?Ax?1?EAE?AAExA?????2?0.1700.170?0.2???1??0.144?0.8??2?0.1013
?E?1.1067 设T=350K
lnPAPASS?21.0444?2790.50?T?T?57.15??11.516
?100271PaS
lnPEPES?23.8047?3803.98?41.68??11.4669
?95505Pa KA??APAPS?0.9972S
KE?
?EPEPi?1.0434
?K?Kxi?KAxA?KExE?0.9972?0.8?1.0434?0.2?1.0064 xi?1
i所以泡点温度为350K。
⑵露点温度
此时yA?0.8,yE?0.2 设T=350K,
lnPAPASS?21.0444?2790.50?T?T?57.15??11.516
?100271PaS 设?lnPEPES?23.8047?3803.98?41.68??11.4669
?95505PaEA?1,??1
KA??APAPS?0.9898 4 KE??EPEPS?0.94279
xA?yAKAyEKE'?0.8082
xE? ln??0.2121
AAEA??Ax?1?AEA?AEAxE?????2?0.1440.144?0.8082???1??0.170?0.2121??2?0.00806
?
'A?1.0081'E
AEA?0.1700.170?0.2121???1??0.144?0.8082??2ln???Ax?1?EAE?AAExA?????2?0.09908
?E'?1.1042 KA?'?'APASP?1.0081?1002711013001.1042?95505101300yEKE'?0.99785
KE?
'?'EPESP?yAK'A??1.04103
?yiKi'??0.8017?0.1921?0.9938
?yiKi?1
所以露点温度为350K。
解2:(1)计算活度系数:
lnrA?(1?AAEAAExAAEAxE}2?0.144(1?0.144?0.80.17?0.2)2?0.0075
rA=1.0075