講稿化工原理

FxFDxDFD 1000.4D0.9W100D解之得 VR1Dq將V132kmolh，D43.8kmol

FyF

1

312

xy

Fy*

R2Rmin20.8VVq1FV130kmol/hLRDqF1.6501100180kmolWxwFxFDxD1000.550yLxWxw180x2.51.38xV V yn1 ynxn1 yn11n 0.8 n1nnx*n xn1xn

n n

即

FxFDxDFD1000.4D0.9W100DD37.5kmol/hW62.5kmol/（3）

LV

V

VWV

V15062.5

62.51.417xnyw

1

2.511.5

設(shè)離開塔釜上一塊理論板液相組成為xNyw1.417xN

0.2170.0417 yn1

RR1

R

3.54.5

0.950.778

xyn10.2110.8420.211nny*n

1

312

ynyn10.8850.842n y*n

n

xnyn1ny*n

312

0.8850.822 Vy13VyD3yy

1 V0.943VyD3即yy

解之得 21（2）21

Rx

2

0.9682

n

R1

R 3 3x1

y1

11 2.51.51y2 3

0.32320.8620.3233 y1xDx1

y1 0.6151 2.51.5yRxxD30.6150.8 R1 R xW 1DFxFxWxD

2.51.5

WFD10017.182.9kmol/ ⑴y1=xD ⑵⑶xD=xF= q線:

yˊ=1.25×0.5-0.0187=0.606,代入(2) 題解 回收率=(V·y)/( FDFFxDxF 100D1000.5D0.95W解之得：

VWV

V

130.550

500.05

* yF1yF

11.25

xy

Fy*

(y6xy

1得交點坐標(biāo)為： xDye

y

R1.2Rmin1.22.68yn1

RR1

R

3.224.22

0.94

y0.76xy6xyy

提餾段操作線過點(xq、yq)和點(xw、x

yqxWxq y0.040.48x y1.52x yF11FF 0.5 F1FFFR

FFxDyFFF

y

R2Rmin21.6

DxDD0.9FyF/xD0.9100.5/0.95kmol/ (1)LDV2D290180kmol/0xx0

xD

D 1 21D對分凝器作物料衡算，可知：Vy1Lx0

Lx0V

900.818900.9（2）q5于 VVq1F1803

1200100kmol/SV100kmol

FSDFFxDxF 200100902000.5900.9WX解之得 yn1

RR1

R

1.52.5

0.95

y1xDx1

y1

11 2.251.251y20.6x10.380.60.8940.38FSDFFxDxF 即

FSDF0.5D0.95WSV2.5DDF x 1LxD=0.95xL=0.880.95

11

1 11

11.59

Vy1LxL

LDRLV0.907L0.88D

L解之得 q=1

* yF1yF

11.59

xy

Fy*

F1F2DFxF Dx12 100200D1000.62000.2D0.8W解之得：L、VLRDqF2120.51100341kmol/1VR1Dq1F3120.50361.5kmol/ DxDF1y x

x120.50.81000.60.943xV V

Ly1xD x1 1LF，VV

5.84.8 L

F

FxF V V D 0.6D

WW

5.8xW14.8xW

Fxx 1000.3x D

xD

0.6

0.61xW

0.69.50.6xWxW0.396（舍去（2）xW0.09143 yn1

RR1

R

6.57.5

0.70.867

xdxFyd0.867xd0.09330.8670.350.0933xW，即ym1xWxmxW

ydxWxd ym10.20.397xm ym11.313xmy1xDx1

y1

11 2.51.51y20.867x10.09330.8670.4830.0933x2

y2

0.29621 2.51.52y31.313x20.06271.3130.2960.0627 Wx3 0.162W1

2.51.5

2100%5 由DxD

/

FqDR531

xFxFDxD

3.030

xxD

/

/

/LRDqF2.532.13

/yLx

190.33x67.870.0161.55x

V V

(1)yexeyy3

23

yy

1xF=0.45，2.5yy3

2

yy

q3R

xDye

ye

R1.5Rmin1.51.556DxD

DxDWxw

D0.95xF0.950.45

Wxw1

L

RD

RDq

V R1Dq

R1Dq

WxwV

R1Dq

F

3.3340.4523

yLxWxw1.47xV V (yq，液相組成為xq1F3F31

2

xDyq

y

R1.5Rmin1.51.64yn1

RR1

R

2.463.46

0.96

y1xDx1

y1

11 2.51.51y20.71x10.2770.710.9060.277 Fxx D xD

由V0.8L可得 R（2）

yW

1可 0.08

11

x0

y0 1 yRxxD30.920 R1 R x1

y1 1

ye

40.4513 xDye0.900.766 y 0.766 R2Rmin20.424ym1

R1

R

提餾段操作線方程：據(jù)題意，VLVW1

V V V

LV

V

y1xDx1

0.6921 43y20.4590.6920.487x2

y2 0.5081 43y31.5x20.5xW1.50.5080.5xW0.762xW

y3

0.762

0.7621

30.7620.5xW

1.714W 1.5xW

0.762

xW

1000.45D0.90W100D Lx1

10000.5

D500q0yexFxe

0.251 32 xDye

0.9

y

0.5R2Rmin21.6VR1D4.2500

VVq

yn1

R1

R

3.24.2

0.9

y1xDx1

0.751 32y20.762x10.2140.7620.750.214

1D

R、VFD，L

RD

V

DxWxDRLxDV2， x

D

WdxyxdWyxy=1.5x+0.05

0.5x

dx

1dWx0 dxW1dWx040.5x FlnF

2ln0.50.40.050.50.1

FWD

0.1yD

6.250.41

yD xn1

n n

nnx*nn

yn

11 1nnnLxn1xnVynyn1nn 1000.4x100ynnyn0.7 xD=1。此時，

F

0.41

qxFxeye

1

2.511.5

y

xxD3x將xe=0.3ye=0.517

R

R D

x F

0.3yn

34

0.750.75x4 FD0.5F0.8D DxD0.620.8

LRDqF30.62F1F

LV

V

2.862.48

0.380.01

yexDxe

ye 0.6151 2.51.5 LRminDqF0.62Rmin1FVRmin1D0.62Rmin1FLV

0.62Rmin10.62Rmin1LyexW

V xe

0.62Rmin11.3060.62Rmin1解之

(xD 或 xD=y1=0.8, y2=0.628,x2=0.360;y3=0.470, { 由式 代入式(2)得:D=120.5kmol/hL=RD=2×120.5=241kmol/hV=L+D=241+120.5=361.5kmol/h在兩進料間和塔頂進行物料衡算,L＂、V＂,塔板序S。y＂=(L＂/V＂)x''+(DxD–F1xf1)/V＂ FDWFFxDxWxF 即

102W100.520.98W0.05W4.89kmol/、F

yn1

RR1

R

89

9

LLDLLVyn1LxnDxD

LV

DxDVLLRD823.1112.89kmol/sVVR1D9218kmol/sy

n

12.89

20.983.110.90.716

y1xDx1

y1

11 2.51.511

y8x0.9880.9510.98 9 2y2x2

0.892x212

2.51.5 ( （２） ｙ＇＝（Ｌ＇／Ｖ）ｘ＇－（Ｗ／Ｖ）ｘw（ｘD／（１－ｘD（１－ｘw）／ｘw0.)（(1-0.)/0.1∴完成上述分離任務(wù)所需ＮTmin≥５（包括塔釜 x1

y1 0.6941 2.51.5y

xxD

R0.6940.850.694R R1

R

R

R

Rx

R 0.694R0.852 2.51.52

R

0.694R解之得

于 yRx0.856.690.50.85 R1

R

3 2.51.5 2.51.53L V

wxWV

RD

x3

FxFR1DqRDqFR1D

FxFR6.69D11000.3251000.5D3.024D17.5

2.5xW

Fxx DxD

1000.5xW

3

x50

42

y125

51 0.96 ｘ1=又∵ 故 22 2222 FSD

10060DFF

0.4D0.95W塔頂輕組分回收率DxD36.90.95

xe

ye 20.41 2ee

、，則

SLW

ye

W

xe

FSD

10060DF F

0.40.95D0.6解之得

/ LL114.29