Example 10.3 For The System Methanol (1) / Methyl Acetate
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Example 10.3 For the system methanol (1) / methyl acetate (2) ln γ 1 = Ax22
,
ln γ 2 = Ax12
ln P1sat = 16.59158 −
where ,
3643.31 T − 33.424
A = 2.771 − 0.00523T
ln P2sat = 14.25326 −
2665.54 T − 53.424
Assuming the validity of (10.5) , calculate : P , xi ; T = 318.5 K , y1 = 0.6 (b) DEW POINT calculation : for T = 318.5 K , y1 = 0.6 •
Star with assuming P=
γ1 = γ 2 = 1 1
•
Calculate P by
•
Calculate x1 by
•
Substitute xi into the activity coff. Eq ( γ , γ ) and
y1
x1 =
γ 1 P1
sat
y1 P γ 1 P1sat
+
y2
γ 2 P2sat
and
x2 = 1 − x1
1
compare with the assumed value • Return to the first step. • Another data: P1sat : ln P1sat = 16.59158 −
3643.31 318.5 − 33.424
;
P1sat = 44.51kPa
2
:
P2sat
ln P2sat = 14.25326 −
2665.54 318.5 − 53.424
A
:
A = 2.771 − 0.00523(318.5) = 1.107
γ1
:
2 γ 1 = exp ( 1.107 ) ( x2 )
γ2
:
2 γ 2 = exp ( 1.107 ) ( x1 )
;
P2sat = 65.64kPa
ɤ1(ne ɤ2(ne ɤ1
P/kPa
ɤ2
x1
x2
w)
w)
51.08 0.688 0.311 1.113 1.690 1
1
828
676
324
261
496
1.113 1.690 63.63 0.770 0.229 1.059 1.929 261
496
977
593
407
989
676
1.059 1.929 62.99 0.801 0.198 1.044 2.034 989
676
139
076
924
779
777
1.044 2.034 62.90 0.811 0.188 1.040 2.073 779
777
411
612
388
069
413
1.040 2.073 62.89 0.815 0.184 1.038 2.086 069
413
377
153
847
549
676
1.038 2.086 62.89 0.816 0.183 1.038 2.091 549
676
261
331
669
05
122
1.038 2.091 62.89 0.816 0.183 1.037 2.092 05
122
248
722
278
885
6
1.037 2.092 62.89 0.816 0.183 1.037 2.093 885
6
247
851
149
831
09
1.037 2.093 62.89 0.816 0.183 1.037 2.093 831
09
247
894
106
813
252
1.037 2.093 62.89 0.816 0.183 1.037 2.093 813
252
247
908
092
807
306
1.037 2.093 62.89 0.816 0.183 1.037 2.093 807
306
247
913
087
805
324
1.037 2.093 62.89 0.816 0.183 1.037 2.093 805
324
247
915
085
804
33
1.037 2.093 62.89 0.816 0.183 1.037 2.093 804
33
247
915
085
804
332
1.037 2.093 62.89 0.816 0.183 1.037 2.093 804
332
247
915
085
804
332
1.037 2.093 62.89 0.816 0.183 1.037 2.093 804
332
247
915
085
804
332
∴ P = 62.89274kPa, x1 = 0.816915, x 2 = 0.183085, γ 1 = 1.037804, γ 2 = 2.093332Q.E.D.
Navapon Pittayaporn 50290015 CHE
,
ln γ 2 = Ax12
ln P1sat = 16.59158 −
where ,
3643.31 T − 33.424
A = 2.771 − 0.00523T
ln P2sat = 14.25326 −
2665.54 T − 53.424
Assuming the validity of (10.5) , calculate : P , xi ; T = 318.5 K , y1 = 0.6 (b) DEW POINT calculation : for T = 318.5 K , y1 = 0.6 •
Star with assuming P=
γ1 = γ 2 = 1 1
•
Calculate P by
•
Calculate x1 by
•
Substitute xi into the activity coff. Eq ( γ , γ ) and
y1
x1 =
γ 1 P1
sat
y1 P γ 1 P1sat
+
y2
γ 2 P2sat
and
x2 = 1 − x1
1
compare with the assumed value • Return to the first step. • Another data: P1sat : ln P1sat = 16.59158 −
3643.31 318.5 − 33.424
;
P1sat = 44.51kPa
2
:
P2sat
ln P2sat = 14.25326 −
2665.54 318.5 − 53.424
A
:
A = 2.771 − 0.00523(318.5) = 1.107
γ1
:
2 γ 1 = exp ( 1.107 ) ( x2 )
γ2
:
2 γ 2 = exp ( 1.107 ) ( x1 )
;
P2sat = 65.64kPa
ɤ1(ne ɤ2(ne ɤ1
P/kPa
ɤ2
x1
x2
w)
w)
51.08 0.688 0.311 1.113 1.690 1
1
828
676
324
261
496
1.113 1.690 63.63 0.770 0.229 1.059 1.929 261
496
977
593
407
989
676
1.059 1.929 62.99 0.801 0.198 1.044 2.034 989
676
139
076
924
779
777
1.044 2.034 62.90 0.811 0.188 1.040 2.073 779
777
411
612
388
069
413
1.040 2.073 62.89 0.815 0.184 1.038 2.086 069
413
377
153
847
549
676
1.038 2.086 62.89 0.816 0.183 1.038 2.091 549
676
261
331
669
05
122
1.038 2.091 62.89 0.816 0.183 1.037 2.092 05
122
248
722
278
885
6
1.037 2.092 62.89 0.816 0.183 1.037 2.093 885
6
247
851
149
831
09
1.037 2.093 62.89 0.816 0.183 1.037 2.093 831
09
247
894
106
813
252
1.037 2.093 62.89 0.816 0.183 1.037 2.093 813
252
247
908
092
807
306
1.037 2.093 62.89 0.816 0.183 1.037 2.093 807
306
247
913
087
805
324
1.037 2.093 62.89 0.816 0.183 1.037 2.093 805
324
247
915
085
804
33
1.037 2.093 62.89 0.816 0.183 1.037 2.093 804
33
247
915
085
804
332
1.037 2.093 62.89 0.816 0.183 1.037 2.093 804
332
247
915
085
804
332
1.037 2.093 62.89 0.816 0.183 1.037 2.093 804
332
247
915
085
804
332
∴ P = 62.89274kPa, x1 = 0.816915, x 2 = 0.183085, γ 1 = 1.037804, γ 2 = 2.093332Q.E.D.
Navapon Pittayaporn 50290015 CHE
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