?tm?(80?50)?(30?20)?18.20
80?50ln30?204500?1.9?103?(80?30)So?3600?13.8m2
ko?tm
第八章 精馏
2.解：本题中,正戊烷为易发组分,以A表示，正己烷为难挥发组分，以B表示。 (1)??pApBoo?159.16?2.945 54.04oP-p101.3?54.04(2)xA?oBo??0.450 xB?0.55
pA-pB159.16?54.04px159.16?0.45yA?AA??0.707 yB?0.293
p101.3或者yA??xA2.945?0.45??0.707
1?(?-1)xA1?1.945?0.45o3. 解：(1)正戊烷用A表示 正己烷为B表示
进料液的平均摩尔质量 M??xiMi?0.4?72?0.6?86?80.4kg/km ol进料量F?5000?62.19kmol/h 80.4己知xD?0.98 xF?0.4 由?＝Fx?0.97?62.19?0.4DxD?97% 得D?F??24.62kmol/h
xD0.98FxF由F=D+W 得W=F-D=62.19-24.62＝37.57kmol/h (2)根据FxF?DxD?WxW得
xw?FxF－DXD62.19?0.4?24.62?0.98＝＝0.02 W37.575.解：原题改为y=0. 5x+0.15
(1)由精馏段操作线方程y=0.75x+0.2075可知
R?0.75 解得R?3 R?1(2)由操作线方程可得得xD?0.83
xD?0.2075 将R＝3代入 R?1(3)由进料方程 y=0.5x+1.5得
11
qq－1＝0.5 ?q??1, 过热蒸汽进料
xF?0.15 将q=－1代入得 xF＝0.3 q?1(4)由进料方程可知－
6. 解：(1)进料的平均摩尔质量M??xiMi?0.4?72?0.6?86?80.4kg/kmol 进料量 F?5000?62.19kmol/h 80.4己知xD＝0.98 xW?0.03 xF?0.4
?Fx?DxD?WxW根据?F
?F?D?W?62.19?0.4?0.98D?0.03W得： ?
62.19?D?W?解得：D=24.22 kmol/h W=37.97 kmol/h (2)精馏段操作线方程
Rxxn?DR?1R?1 2.50.98?xn??0.714xn?0.282.5?12.5?1斜率为0.714，截距为0.28 yn?1?(3)塔内第二块塔板上升蒸气组成 已知：y1?xD?0.98 由气液平衡方程 y1?0.98＝
?x1
1?(?-1)x12.92x1 解得x1=0.942
1?1.92x1由操作线方程得：
y2?0.714x1?0.28?0.714?0.942?0.28?0.953 (4)提馏段操作线方程
已知 L?RD?2.92?24.22?70.72kmol/h
F?62.19 kmol/h
W=37.97 kmol/h xW?0.03
泡点进料 q=1 提馏段操作线方程：
ym＋1'＝L＋qFWxm'?xW
L＋qF－WL＋qF－W 12
?70.72＋62.1937.97xm'??0.03
70.72?62.19?37.9770.72?62.19?37.97?1.4xm'?0.012
斜率为1.4，截距为-0.012
(5)精馏段内上升蒸气及下降液体的流量
V?(R?1)D?3.5?24.22?84.77kmol/h L?RD?2.5?24.22?60.55kmol/h
(6)提馏段内上升蒸气及下降液体的流量 L’＝L＋qF=60.55+62.19=122.74kmol/h V'?V?84.77kmol/h 8. 解：
最小回流比 Rmin?xD－yqyq－xq
泡点进料：xq＝xF?0.4
yq??xq1?(?-1)xq?2.5?0.4?0.625
1?1.5?0.4Rmin?0.96?0.6250.335??1.5
0.625?0.40.225R3??2 Rmin1.5第九章 吸收
E3.81?106＝3.76?103 2.解：m?＝P101.3?s1?103?5 H＝＝＝1.46?106EMs3.81?10?18p*＝C HC?p*H＝45?1.46?10?5＝6.57?10?4mol/L
5.解：根据NA?DP(pA1－pA2)
RTzpBm己知z＝0.004，D?2.56?10?5，P=101.3，pA2?0 101.3kPa，25℃时，pA1?3.168kPa
? pB1?101.3?3.168?98.132，pB2?101.3
pBM?pB2－pB1101.3?98.132＝?99.71 pB2101.3lnlnpB198.13213
2.56?10?5101.3? NA???(3.168-0) 8.314?298?0.00499.71?8.31?10-6kmol?m-2?s-10.005?103t??9.3h 18?8.31?10-66.解：据
111 ??KGkGHkL?S?S103E＝mp ? H????0.017
EMSmpMS32?101.3?18气膜阻力液膜阻力
11??2.89?105 -6kG3.46?10115 ??3.9?10?4HkL0.017?1.51?101? ?2.89?105?3.9?105?6.79?105
KG2KG?1.47?10?4 kmol/m?s?kPa
2KY?PKG?101.3?1.47?10?4?1.49?10-2kmol/m?s
1KG1.47?10?4kG???42.5%，即气膜阻力占总阻力的42.5％ 1kG3.46?10?6KG∴该吸收过程为气膜和液膜阻力共同控制 9.解：E?5.5?104
E5.5?104m???5.43?102
P101.3LY-Y()min?1*2 VX1-X2Y1?3%0.15%?0.031 Y2??0.15％ 1-3%1-0.15%L代入上式，得()minVY10.031?5??5.71?10 2m5.43?100.031?0.0015L2L??5.17?10?1.3()min?672.1 ,?5VV5.71?10?0X1＝*由 V(Y1-Y2)?L(X1-X2) 得 X1?V0.031-0.0015(Y1-Y2)??4.4?10?5kmol/L L672.13%?0.031 Y2?Y1(1- ? )?0.031(1-0.97)?0.00093 1?3
11.解：Y1?
LY-Y0.031?0.00093()min?1*2??0.347
0.031VX1-X20.358LL?1.6()min?1.6?0.347?0.556 VVV(Y1-Y2)?L(X1-X2)
X1?V0.031?0.00093(Y1-Y2)??0.0541 L0.556?Ym??Y1??Y2 ?Y1ln?Y2?Y1?Y1-mX1?0.031?0.358?0.0541?0.0116
?Y2?Y2?mX2?0.00093
?Ym＝0.0116?0.00093?0.00423
0.0116ln0.00093NOG?Y1?Y20.031?0.00093??7.11
0.00423?YmV0.0147??0.987m KYAA0.0149HOG?Z?HOG?NOG?7.02m
第十一章 干燥
1.解：(1) 已知101.3kPa，40℃时，水的饱和蒸气压ps?7.377kPa
H?0.622?ps20%?7.377?0.622??0.0092kg水汽/kg绝干气
P??ps101.3?20%?7.377273?t 273(2)vH?(0.772?1.244H)??(0.772?1.244?0.0092)?273?40?0.898m3/kg绝干气 273(3)?H?1?0.00921?H3 ??1.124kg/m
0.898vH(4)CH?1.01?1.88H?1.01?1.88?0.0092?1.027kJ/kg?oC (5)IH?(1.01?1.88H)t?2491H
?(1.01?1.88?0.0092)?40?2491?0.0092?64.01kJ/kg
(6) pstd??ps?20%?7.377?1.475kPa 查水的饱和蒸气压 得 td?12.6℃
2.解：(1)p?3.1kPa (2)??6% (3) td?25℃
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