分离工程试题库 - 图文

A关?L0.5544??0.99K关V关0.56log(N?所以 理论板数为

A??0.99?0.9)log()1??1?0.9?1??1?9.48logAlog0.99

AiN?1?AL③它组分吸收率公式 Ai?,?i?N?1

VKiAi?1计算结果如下: 组分 进料量 相平衡常数Ki 76.5 4.5 3.5 17.4 3.75 1.3 0.56 0.4 0.18 0.144 0.056 — Ai ?i被吸收量 塔顶尾气 数量 74.05 3.834 2.009 0.250 0.045 0.0 0.0 0.0 组成 0.923 0.048 0.025 0.003 0.0006 0.0 0.0 0.0 CH4 C2H6 C3H8 0.032 0.032 2.448 0.148 0.148 0.668 0.426 0.426 1.491 0.99 0.90 2.250 4.455 1.500 2.500 4.500 19.810 i-C4H10 2.5 n-C4H10 4.5 i-C5H12 1.5 n-C5H12 2.5 n-C6H14 4.5 合计 100.0 1.386 0.99 3.08 3.85 9.9 — 1.00 1.00 1.00 — 80.190 以CH4为例:

L0.5544??0.032AVK17.4 i=i

0.0329.48?1?0.99?0.329.48?1i0.032?1=

?V1(CH4)=(1-

y(1CH4)?V(1CH4)V1??i)VN+1=(1-0.032)?76.5=74.05

74.05?0.92376.5④塔内气体平均流率:

v?100?80.190?90.102Kmol/h

L0?(L0?19.81)?L0?9.9052塔内液体平均流率:L=

由v=0.5544;?L0=40.05Kmol/h

215. 某1、2两组分构成二元系,活度系数方程为ln?1?Ax2,ln?2?Ax12,端值常数

l与温度的关系:A=1.7884-4.25?10-3T (T,K)

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4050SlnP?16.0826?1T 蒸汽压方程为 4050SlnP2?16.3526?T (P:kPa:T:K)

假设汽相是理想气体,试问99.75Kpa时①系统是否形成共沸物?②共沸温度是多少? 解:设T为350K

则A=1.7884-4.25?10-3?350=1.7884-1.4875=0.3009

SlnP1?16.0826?40504050lnP2S?16.3526?350;350

?

P1S=91.0284;P2S=119.2439

因为在恒沸点

S?1P?1P2S1 由?12? ?1得 ??2P1S?2P2SS?1P221ln?lnS?ln?1?ln?2?A(x2?x1)?A(1?2x1)?P22?

ln91.0284?0.3009(1?2x1)119.2439

?x1=0.9487 x2=0.0513

2ln?2?0.3009?0.9487

2?ln?1?0.3009?0.0513

?1=1.0008 ?2=1.3110

?xiPiS?iP==1.0008?0.9487?91.0284+1.3110?0.0513?119.2439=95.0692

?99.75

设T为340K

则A=1.7884-4.25?10-3?340=0.3434

SlnP1?16.0826?40504050lnP2S?16.3526?340;340

?

P1S=64.7695;P2S=84.8458

lnSP1lnS?A(1?2x1)P2

64.7695?0.3434(1?2x1)84.8458

?x1=0.8931 x2=1-0.8931=0.1069

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?ln?1?0.3434?0.1069

22ln?2?0.3434?0.8931

?1=1.0039

S?2=1.3151

P=??ixiPi=1.0039?0.8931?64.7695+1.3151?0.1069?84.8458=69.9992

?99.75

设T为352K

则A=1.7884-4.25?10-3?352=0.2924

SlnP1?16.0826?40504050;lnP2S?16.3526? 352352?

P1S=97.2143;P2S=127.3473

SP1由lnS?A(1?2x1)

P2ln97.2143 ?0.2924(1?2x1)127.3473?x1=0.9617

x2=1-0.9617=0.0383

ln?2?0.2924?0.96172

?ln?1?0.2924?0.03832

?1=1.0004

?2=1.3105

P=??ixiPiS=1.0004?0.9617?97.2143+1.3105?0.0383?127.3473=99.9202

?99.75

16. 在101.3Kpa压力下氯仿a.-甲醇b.系统的NRTL参数为: ?12=8.9665J/mol,

?12=-0.83665J/mol,?12=0.3。试确定共沸温度和共沸组成。

安托尼方程(PS:Pa;T:K)

S2696.79氯仿:lnP 1?20.8660?(T?46.16)甲醇:lnP2S?23.4803?3626.55 (T?34.29)S2696.79则lnP 1?20.8660?(326.65?46.16)解:设T为53.5℃

.55 lnP2S?23.4803?3626(326.65?34.29)P1S=76990.1 P2S=64595.6

(??ij?ij)由Gij?exp,?ij=?ji

G12?exp(??12?12)exp(?0.3?8.9665)==0.06788

G21?exp(??21?21)(0.3?0.8365)=exp=1.2852

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2ln?1?x2???12G12?22?(x?xG)(x?xG)?2112??1221??2?21G21=

8.9665?0.06788?(1?x1)??22?[x?(1?x)?1.2852]([1?x)?0.06788x]??1211??=??1.38170.60862?(1?x1)??22?(1.2852?0.2852x)(1?0.93212x)?11???

)?1.285222?(?0.8365ln?2?2x1?

2?x1??21G21??22?(x?xG)(x?xG)?1221??2112??2?12G12=

28.9665?0.06788?0.8365?1.2852???2(x1)[x1?1.2852(1?x1)]2??1?x1?0.06788??=

?0.04131?1.075072?x1??22?(x1)(1.2852?0.2852x1)?1?0.93212???lnP1SP2S

ln?1ln?2-==

ln76990.164595.6=0.1755

求得x1=0.32 ?1=1.2092 ?2=0.8971

?xiPiS?i?x1P1S?1?x2P2S?2=0.32?76990.1?1.2092?0.68?64595.6?0.8971 =69195.98Pa?101.3kPa 设T为60℃

S2696.79lnP1?20.8660?(333.15?46.16)则

.55lnP2S?23.4803?3626(333.15?34.29)

P1S=95721.9 P2S=84599.9

ln?1ln?2lnP1SP2S-==

ln95721.984599.9=0.1235

设T为56℃

S2696.79则lnP 1?20.8660?(329.15?46.16).55lnP2S?23.4803?3626(329.15?34.29) P1S=83815.2 P2S=71759.3

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