4.1 4.1

P.133

4-1

Logic

vs.

Memorization

Level 1:

Mole balance

Level 2:

Design equation

Level 3:

-rA = f(X) ??

Level 4:

Rate law

Level 5:

Stoichiometry

Level 6:

-rA = f(X) L4 + L5

P.135

4-2

Example:

PFR reactor volume

for a first-order

gas-phase reaction

4.2 ( CSTR)

P.136

4.2.14.2.1

AA

1 dNr

V dt =

0

0

( / )1 1 AA A AA

d N VdN dN dCr

V dt V dt dt dt= = = =

(4-1)

(4-2)

P.137

AA

dCr

dt− = −

A B→

A0 A 0

dXN r V

dt= − (2-6)

(will be used in Chapter 5)

P.137

2A Ar kC− =

A A0 (1 )C C X= −

2A0 (1 )

dXkC X

dt= −

A02(1 )

dXkC dt

X=

−

(4-4)

(3-29)

(4-3)

t = 0, X = 0 k

(tR)

P.137

20 0

A0

1

(1 )

t X dXdt

kC X=

−∫ ∫

A0

1

1

Xt

kC X = −

(4-5)

P.138

4-1

tR

A B→

90%

k = 10-4 s-1

k = 10-4 s-1 90% 6.4 h

P.138

R

1 1 1 1 2.3ln ln

1 1 0.9t

k X k k= = =

− −

R 4 1

2.323,000 s 6.4 h

10 st − −= = =

kCA0 = 10-3 s-1

kCA0 99% 27.5 h

P.138

RA0 A0 A0

1 0.9 9

1 (1 0.9)

Xt

kC X kC kC= = =

− −

R 3 1

99000 s 2.5 h

10 st − −= = =

P.138

4-2

90% conversion (from CA0 to 0.1CA0 )

order of magnitude

(characteristic reaction time, tR)

(total cycle time)

P.139

t f e c Rt t t t t= + + +

4-3

(5~60)

P.139

44444444--------11111111

k

CSTR 6200 10 lb / year×

k

C STR

40 C° 55 C°

2 4

2

H SO2 2 2 2

O CH OH|

CH CH H O CH OH

A B C

+ →

+ →觸媒

─／ ＼─ ─

500 mL 2 M 500 mL

0.9 wt% 55 C°

( E4-1.1)

P.140

44444444--------11111111

P.140

44444444--------11111111

CD-ROM www.engin.umich.edu/~problemsolving

(A G)

A. Ak

B.

P.141

44444444--------11111111C.

C1.

AA

dNr V

dt=

A A Ar k C− =

C2.

A B,C C CC

t

C3.

1. AC =

2. Ak =

3. V =

C4.

C5.

P.141

44444444--------11111111D.

1.

2.

3.

4.

5.

1. ( B B0C C≅ )

E.

F. 1 3

G. 4-1 4-2

P.141

44444444--------11111111解： 1.

AA

1 dNr

V dt= (E4-1.1)

2.

A Ar kC− = (E4-1.2)

t

B B0( )C C≅

3. 0V V= ( E4-1.2)

P.142

44444444--------11111111

BΘ B A B0B

A0

N

N

Θ =

即

A AA

0

A 0A A

0

( / )1

N NC

V V

d N VdN dC

V dt dt dt

= =

= =

4.

AA

dCkC

dt− = (E4-1.3)

5. k (E4-1.3)

A

A0

A

0 0A

C t t

C

dCk dt k dt

C− = =∫ ∫ ∫

A A00 ,t C C= = A 31.0 kmol / m

(1 mol/L)

結合莫耳均衡速率定律與化學計量

P.142

44444444--------11111111

A0

A

lnC

ktC

= (E4-1.4)

t

A A0ktC C e−= (E4-1.5)

C A0 A0 A

A B C

N N X N N

+ →= = −

0V V=

C CC A0 A A0

0

(1 )ktN NC C C C e

V V−= = = − = − (E4-1.6)

A0 C

A0

lnC C

ktC

− = −

AC

CC

t

t

P.143

44444444--------11111111E4-1.3

t

(min)

CC

/ 3(kmol m )

−C C

C

A0 C

A0

−

C C

C

A0 C

A0

ln

0.0 0.000 1.000 0.0000

0.5 0.145 0.855 −0.1570

1.0 0.270 0.730 −0.3150

1.5 0.376 0.624 −0.4720

2.0 0.467 0.533 −0.6290

3.0 0.610 0.390 −0.9420

4.0 0.715 0.285 −1.2550

6.0 0.848 0.152 −1.8840

10.0 0.957 0.043 −3.1470

A0 C A0ln[( ) / ]C C C− t k− E4-1.1

E4-1.3 Excel A0 C A0ln( ) /C C C− t

A0 C A0ln[( ) / ]C C C− t k Excel

E4-4.1

由分批反應之濃度 -時間關係計算比反應速率

P.143

44444444--------11111111

E4-1.1 Excel

1

1

0.311 min

0.311 min

k

k

−

−

= − = −

=

斜率

1A A0.311 minr C−− =

CSTR k

4 CD-ROM 3 Excel

4.3 4.3 (CSTR)(CSTR)

(CSTR)

P.144

CSTR

FA0= 0CA0 (2-13)

0 X

CSTR CSTR

P.144

A0

A exit( )

F XV

r=

−

0 A0

A

C XV

r

υ=

−

A0

0 A

C XV

rτ

υ= =

−

(2-13)

(4-6)

(4-7)

4.3.1 4.3.1 CSTRCSTR

(3-29)

(4-7) (3-29)

P.144

A Ar kC− =

A A0 (1 )C C X= −

1

1

X

k Xτ = −

(3-29)

(3-29) (4-8) A CA

P.145

1

kX

k

ττ

=+

A0A 1

CC

kτ=

+

(4-8)

(4-9)

CSTR

Damköhler number (Da) AA (dimensionless)

Damköhler

Damköhler

P.145

A0

A0

A ADa

A A

r V

F

−= = =在入口處的反應速率 “ 之反應速率”

之流入速率 “ 之對流速率”

A0 A0

A0 0 A0

Dar V kC V

kF C

τυ

−= = =

2A0 A0

A0A0 0 A0

Dar V kC V

kCF C

τυ

−= = =

CSTR (4-8) Damköhler number

P.145

Da 0.1 0.1

Da 10 0.9

X

X

< <> >

若 ，則 ；

若 ，則 。

Da

1 DaX =

+

( = 0)CSTRs ( 4-3)

P.146

4.3.24.3.2 CSTRsCSTRs ((CSTRsCSTRs in Series)in Series)

4-3 CSTRs

CSTR A (4-9)

1=V1 / 0 CSTR

CA2 CSTR

P.146

A0A1

1 11

CC

kτ=

+

0 A1 A2A1 A22

A2 2 A2

( )C CF FV

r k C

υ −−= =

−

A0A1A2

2 2 2 2 1 11 (1 )(1 )

CCC

k k kτ τ τ= =

+ + +

( 1= 2 = )(k1 = k2 = k)

n CSTRs ( 1= 2=…= n=

i= (Vi / 0))(k1 = k2 = … = kn= k)

P.146

A0A2 2(1 )

CC

kτ=

+

A0 A0A (1 ) (1 Da)n n n

C CC

kτ= =

+ +(4-10)

CAn

n CSTRs

n A

P.147

A0A0 (1 )

(1 Da)nC

C X− =+

1 11 1

(1 Da) (1 )n nX

kτ= − ≡ −

+ +(4-11)

A0A A (1 )n n n

Cr kC k

kτ− = =

+

P.147

4-4 Damköhler

CSTRs( 4-5)

i

P.148

4.3.34.3.3 CSTRsCSTRs ((CSTRsCSTRs in Parallel)in Parallel)

A0A

ii i

i

XV F

r

= −

(4-12)

P.148

4-5 CSTRs

P.148

1 2 nX X X X= = = =⋯

A1 A2 A Anr r r r− = − = = − = −⋯

i

VV

n=

(4-12)

P.149

A0A0i

FF

n=

A0

A

i

i

F XV

n n r

= −

A0 A0

A A

i

i

F X F XV

r r= =

− −(4-13)

(4-12)

CSTR

= 0

CA=CA0(1-X) FA0X= 0CA0X

0

P.149

4.3.44.3.4 CSTRCSTR

A0 A02

A A

F X F XV

r kC= =

−

0 A02 2A0 (1 )

C XV

kC X

υ=

−

20 A0 (1 )

V X

kC Xτ

υ= =

−

(4-14)

(4-15)

(4-15) X

1

P.149

2 2A0 A0 A0

A0

A0 A0

A0

(1 2 ) (1 2 ) (2 )

2

(1 2 ) 1 4

2

kC kC kCX

kC

kC kC

kC

τ τ ττ

τ ττ

+ − + −=

+ − +=

(1 2 Da) 1 4 Da

2 DaX

+ − +=

(4-16)

P.150

4-6 CSTRDamköhler ( kCA0)

Da 10 , X: 0.67�0.88CSTR operates at the lowest reactant concentration

P.150

44444444--------22222222

CSTR /× 6200 10 lb year

2000 (ethylene glycol, EG) 912.2 10 tons×

26

88% 12% 2004 EG $0.28/ lb

6200 10 lb / year× EG 31 lb mol / ft EO

( E4-2.1)

0.9 wt% 4-1 10.311 min− Mukesh

Warsteel

(a) 80% CSTR

(b) 800 gal

(c) 800 gal

P.150

44444444--------22222222

P.151

44444444--------22222222

解： EG

EG lb mol/min

8C

lb 1 yr 1day 1 h 1 lb mol lb mol2 10 6.137

yr 365 days 24 h 60 min 62 lb minF = × × × × × =

C A0F F X=

(EO)

CA0

6.137 lb mol7.67 (58.0 g mol /s)

0.8 min

FF

X= = =

(a) 80% CSTR CRE

1.

A0

A

F XV

r=

− (E4-2.1)

P.151

44444444--------22222222

2.

A Ar kC− = (E4-2.2)

3. 0( )υ υ=

A0AA A0

0 0

(1 )(1 )

F XFC C X

υ υ−= = = − (E4-2.3)

4.

A0 0

A0(1 ) (1 )

F X XV

kC X k X

υ= =− −

(E4-2.4)

5.

A 3A01 1 lb mol / ftC =

3

A0A0 3

A01

7.67 lb mol / min ft7.67

min1 lb mol / ft

F

Cυ = = =

P.151

44444444--------22222222

B0 A0υ υ=

3

B0 B0 B01 3

ft lb 1 lb mol lb mol7.67 62.4 26.6

min 18 lb minftF Cυ

= = ⋅ × =

3

30 A0 B0

ft15.34 (7.24 dm /s)

minυ υ υ= + =

10.311mink −= (E4-2.4)

3

301

ft 0.815.34 197.3 ft

(1 ) min (0.311 min )(1 0.8)

XV

k X

υ−= = =

− −

80% 5 ft 10 ft

(density of water)

(= 1480 gal = 5.6 m3)

P.152

44444444--------22222222

(b) CSTRs 800 gal CSTRs ( E4-2.2 )3

07.67 ft / min ( / 2)υ (E4-2.4)

P.152

44444444--------22222222

0 1

V Xk k

Xτ

υ= =

−

1

kX

k

ττ

=+

(E4-2.5)

3

30

1 ft 1800 gal 13.94 min

/ 2 7.48 gal 7.67 ft / min

Vτυ

= = × × =

Damköhler

0.311

Da 13.94 min 4.34min

kτ= = × =

(E4-2.5)

Da 4.34

0.811 Da 1 4.34

X = = =+ +

CSTR 81%

P.152

44444444--------22222222

(c) CSTRs 800 gal

[ (E4-2.5) ]

11

11

kX

k

ττ

=+

(E4-2.6)

3

13

01

1ft 1800 gal 6.97 min

7.48 gal 15.34 ft / min

Vτυ

= = × × =

Damköhler

1 1

1

0.311Da 6.97 min 2.167

min2.167 2.167

0.6841 2.167 3.167

k

X

τ= = × =

= = =+

1 2 01 02 0,V V V υ υ υ= = = =

1 2τ τ τ= =

第一個CSTR

P.153

44444444--------22222222

� � �

A1 A2 A2

0

0F F r V

− + =

− + =

流入 流出 生成

1

A1 A0 1 A2 A0 2(1 ) (1 )F F X F F X= − = −及

A1 A2 2 1A0

A2 A2

F F X XV F

r r

− −= =− −

A0 2A2A2 A2 A0 2

0 0

(1 )(1 )

kF XFr kC k kC X

υ υ−− = = = = −

[(2-24) ]

A0 2 1 A0 0 2 1 0 2 1

A2 A0 2 2

( ) ( )

(1 ) 1

F X X C X X X XV

r kC X k X

υ υ − − −= = = − − − (E4-2.7)

P.153

44444444--------22222222

2X

1 12

Da 0.684 2.1670.90

1 Da 1 1 2.167

X X kX

k

ττ

+ + += = = =+ + +

(4-11)

2 2

1 11 1 0.90

(1 ) (1 2.167)nX

kτ= − = − =

+ +

800 gal 3(3.0 m ) 6200 10 lb / year× EG

第二個CSTR

P.154

44444444--------22222222

0 CSTRs ( E4-2.3)

CSTRs

串連之轉化率大於並連之轉化率串連之轉化率大於並連之轉化率串連之轉化率大於並連之轉化率串連之轉化率大於並連之轉化率，，，，由第由第由第由第2章反應器的討論章反應器的討論章反應器的討論章反應器的討論，，，，即可預測串連之轉即可預測串連之轉即可預測串連之轉即可預測串連之轉化率較高化率較高化率較高化率較高。。。。

4.4 4.4

(plug flow)

P.155

1-9

PFR

P.155

A0 A

dXF r

dV= −

A00

A

X dXV F

r=

−∫

A →產物

(2-15)

(2-16)

2A Ar kC− =

= 0PFR

k

P.156

2A

A0

kCdX

dV F=

A A0 (1 )C C X= −

A0 02 2

0A0A0 1(1 )

xF dx XV

kC XkC X

υ = = −− ∫

化學計量化學計量化學計量化學計量 ( 液相液相液相液相 )

結結結結 合合合合

X 0

Da2 Damköhler ( )

P.156

A0 2

A0 2

Da

1 1 Da

kCX

kC

ττ

= =+ +

A0kCτ

PFR

P.156

化學計量化學計量化學計量化學計量 ( 氣相氣相氣相氣相 )

結結結結 合合合合

A0A AA A0

0 0

(1 ) (1 )

(1 ) (1 ) (1 )

F XF F XC C

X X Xυ υ ε υ ε ε− −= = = =

+ + +

2

A0 2 20

A0

(1 )

(1 )

X XV F dX

kC X

ε+=−∫

CA0

k

A.1

P.156

恆溫反應之恆溫反應之恆溫反應之恆溫反應之k為為為為常數常數常數常數

二級氣相反應二級氣相反應二級氣相反應二級氣相反應的反應器體積的反應器體積的反應器體積的反應器體積

2A0

2 20

A0

(1 )

(1 )

XF XV dX

kC X

ε+=−∫

220

A0

(1 )2 (1 ) ln(1 )

1

XV X X

kC X

υ εε ε ε += + − + + −

(4-17)

P.156

4-7 (using (4-17))

0

A0kC

υ

V (m3)

= 0 ( =0) , 0 ( 0) 0 ( 0)

( A→ B) =0 , = 0 ( = 0)

( 2A → B)

P.157

0(1 0.5 )Xυ υ= −

(3-45)

0(1 )Xυ υ ε= + (4-17)

( 0 , 0 )( A → 2B)

P.157

0(1 )Xυ υ= +

4-8

44444444--------33333333

P.158

/× 6300 10 lb year

2000 1050 10 lb× $0.27

65% 20% 16%

5% 5%

10300 10 lb / year× 乙烯 PFR

1100 K 6 atm

80%

解：

2 6 2 4 2C H C H H→ +

2 6 2 4 2A C H , B C H , C H= = =

A B C→ +

經 濟

用 途

44444444--------33333333

P.158

SI

6B

lb 1 year 1day 1h lb mol300 10

year 365 days 24 h 3600 s 28 lb

lb mol g mol0.340 154.4

s s

F = × × × × ×

=

0.34 lb mol/s 80%

A0F

B A0

6A0

0.34 lb mol0.425 (402 10 lb / yr)

0.8 s

F F X

F

=

= = ×

1.

A0 AdX

F rdV

= − (2-15)莫耳均衡

44444444--------33333333

P.159

A00 A

X dXV F

r=

−∫ (E4-3.1)

2.

A Ar kC− = 1000 K 10.072 sk −= (E4-3.2)

82 kcal/g mol

3.

T0 0

T0

(1 )F

XF

υ υ υ ε= = +

A0AA A0

0

(1 ) 1

(1 ) 1

F XF XC C

X Xυ υ ε ε− − = = = + +

(E4-3.3)

A0C (1 )

C XC

Xε=

+ (E4-3.4)

速率定律

化學計量

44444444--------33333333

P.159

4. (E4-3.1) (E4-3.3)

A0 A0

0 0A0 A0

A0

0A0

(1 )

(1 ) /(1 ) (1 )

(1 )

(1 )

X X

X

dX X dXV F F

kC X X kC X

F X dX

C k X

εε

ε

+= =− + −

+=−

∫ ∫

∫

(E4-3.5)

5. k A.1

A0 A0

0A0 A0

(1 ) 1(1 ) ln

1 1

XF FX dXV X

kC X kC X

ε ε ε+ = = + − − − ∫ (E4-3.6)

6.

A0 0A0 A0 T0 3

0

33

6 atm(1)

(0.73 ft atm / lb mol R) (1980 R)

lb mol0.00415 (0.066 mol / dm )

ft

y PC y C

RT

= = =

⋅ ⋅ ° × °

=

A0 (1)(1 1 1) 1yε δ= = + − =

結合設計方程式速度定律與化學計量

解析解

計 算

44444444--------33333333

P.159

1000 K 1100 K

2 11 2

2 11

1 2

1

1 1( ) ( )exp

( )exp

0.072 82,000 cal / g mol(1100 1000) Kexp

s 1.987 cal /(g mol K)(1000 K)(1100 K)

3.07 s

Ek T k T

R T T

T TEk T

R T T

−

= −

−=

−= ⋅

=

(E4-3.7)

(E4-3.6)

3

0.425 lb mol / s 1(1 1) ln (1)

1(3.07 /s)(0.00415 lb mol / ft )

133.36 ft 2ln

1

V XX

XX

3

= + − −

= − −

(E4-3.8)

0.8X =

3

3 3 3

133.36 ft 2ln 0.8

1 0.8

80.7 ft (2280 dm 2.28 m )

V = − −

= = =

44444444--------33333333

P.160

2-in. 80 40 ft 80

CA 20.0205 ft

3

2

80.7 ft98.4

(0.0205 ft )(40 ft)n = = (E4-3.9)

(E4-3.8) V

CA

C

Vz

A= (E4-3.10)

(E4-3.9) 2C 0.0205 ftA = (E4-3.8) (E4-3.3)

E4-3.1 100 6300 10 lb× /

E4-3.1

並連PFR的管數

44444444--------33333333

P.160

4.5 4.5

(3-46) i

v ( vA= -1 , vB= -b /a ) P/P0

V W

P.161

4.5.14.5.1

0A0

01i i

i

v X TPC C

X TPεΘ + = +

0A0

A0

,ii

Fy

Fε δΘ = =

(4-18)

P.161

若 P ≠ P0則必須使用PFR/PBR設計方程式的微分型式微分型式微分型式微分型式

2A B C→ +

A0 A

dXF r

dW

′= − ⋅

克莫耳

克莫耳 分

2A Ar kC′− =

(4-17)

(4-19)

( 3-5)

(4-20) (2-17) (T = T0)

P.161

A0 0A

0

(1 )

1

C X TPC

X P Tε−

=+

2

A0 0A

0

(1 )

1

C X TPr k

X P Tε −′− = +

22

A0A0

0

(1 )

1

C XdX PF k

dW X Pε − = +

(4-20)

FA0( 0CA0 )

(T = T0)

P

P.162

22A0

0 0

1

1

kCdX X P

dW X Pυ ε − = +

1( , )dX

F X PdW

= (4-21)

Ergun(Ergun equation) (see BSL, 2002, p.191)

P.162

4.5.24.5.2

�Term 2Term 1

3pc p

150(1 )11.75

dP GG

Ddz g D

φ µφρ φ

−− += −

�����

(4-22)

Dominant for

laminar flow

Dominant for

turbulent flow

P.163

P = 2flb / ft (kPa)

φ = = =空隙體積

床體總體積

1 φ− = 固體體積

床體總體積

cg = 2m f32.174 lb ft / h lb ( )⋅ ⋅ 單位轉換因子

8 2m f4.17 10 lb ft / h lb= × ⋅ ⋅ ( c 1.0g = )

pD = ft (m)

µ = mlb / ft h (kg / m s)⋅ ⋅

z = ft (m)

u = (superficial velocity)

= ÷ ft/h (m/s)

ρ = 3 3lb / ft (kg / m )

G = uρ = 2 2mlb / ft h (kg / m s)⋅ ⋅

(kg/s) ( )

(3-41)

P.163

mɺ 0mɺ

0

0 0

m m

ρ υ ρυ==

ɺ ɺ

00

0 0

T

T

P FT

P T Fυ υ

=

0 0 00 0

0

T

T

T FP

P T F

υρ ρ ρυ

= =

(3-41)

(4-23)

(4-22) (4-23)

0

P.163

03

p 0 00 c p

(1 ) 150(1 )1.75 T

T

P FdP G TG

dz D P T Fg D

φ φ µρ φ

− −= − +

00

0 0

T

T

P FdP T

dz P T Fβ

= −

0 3P0 c P

(1 ) 150(1 )1.75

GG

Dg D

φ φ µβρ φ

− −= +

(4-24)

(4-25)

Wz z

Ac (bulk density) b ( / )

C ( 1- φ )

P.164

c c(1 )W A zφ ρ= − ×

= ×

����� ����� �����

催化劑 固 體 固 體

重 量 體 積 密 度

b c(1 )ρ ρ φ= −

c c(1 )dW A dzφ ρ= −

(4-26)

(4-26) W z Ergun

y = P / P0 (dP = P0 dy)

P.164

0 0 T

c c 0 T0(1 )

P FdP T

dW A P T F

βφ ρ

= − −

0 T

0 0 T02 /

P FdP T

dW T P P F

α = −

T

0 T02

Fdy T

dW y T F

α= −

0

c c 0

2

(1 )A P

βαρ φ

=−

(4-27)

(4-28)

(4-29)

FT0

ε

(FT / FT0 ) (4-28)

P.164

A0T T0 A0 T0

T0

1F

F F F X F XF

δ δ

= + = +

T

T0

1F

XF

ε= +

A0A0

T0

Fy

Fε δ δ= =

0

(1 )2

dy TX

dW y T

α ε= − +

(3-43)

(4-30)

(3-35)

(4-30)

(4-21)

P.165

2 ( , )dP

F X PdW

=

1( , )dX

F X PdW

=

(4-31)

(4-21)

Two coupled first-order ODEs � must be solved simultaneously

22A0

0 0

1

1

kCdX X P

dW X Pυ ε − = +

(1 )2

dyX

dW y

α ε= − +

ε = 0 1 >> ε X ε X(4-30)

ε = 0 (4-30)

y

P.165

ε = 0 2

dy

dW y

α−=

2dy

ydW

α= −

2dy

dWα= −

(4-32)

y =1 ( P = P0 ) , W = 0 y = y , W = W

ε 0

z (4-26) W (4-33)

P.166

2( ) 1y Wα= −

1/2

0

(1 )P

y WP

α= = −

0

c c 0

2

(1 )A P

βαφ ρ

=−

1/2

0

0 0

21

zPy

P P

β = = −

(4-33)

(4-29)

(4-34)

D = cmu = cm/sf = G = u , g / 2 s

P.166

4.5.34.5.3

22dP du f GG

dL dL Dρ= − − (4-35)

G u G /(4-23) FT (4-35)

L = 0 , P = P0 f

P.166

22

00

20

P dP dP f GG

P dL P dL Dρ − + =

2 220 0 0

0

2 ln2

P P P PLG f

D Pρ− = +

1/221/2

p0 0 0 c

41 (1 )

P f G VV

P P A Dα

ρ

= − = −

2

pc 0 0

4 f G

A P Dα

ρ=

(4-36)

44444444--------44444444

P.167

60 ft 12

1 in. 40 14

in. 104.4 lb/h

260 C° 45%

10 atm

解： (4-34) z L=

1/ 2

0

0 0

21

LP

P P

β = −

(E4-4.1)

0 3pc 0 p

(1 ) 150(1 )1.75

GG

Dg D

φ φ µβρ φ

− −= +

(4-25)

c

mG

A=ɺ

(E4-4.2)

12

1 in. 40 2c 0.01414 ftA =

m m2 2

104.4 lb / h lb7383.3

0.01414 ft h ftG = =

⋅

計算各個壓降參數

44444444--------44444444

P.167

260 C° 10 atm

m

30 m

0.0673 lb / ft h

0.413 lb / ft

µρ

= ⋅

=

3 3m0 3

0 m

104.4 lb / h252.8 ft / h (7.16 m / h)

0.413 lb / ft

mυρ

= = =ɺ

1p 4

8 mc 2

f

in. 0.0208 ft , 0.45

lb ft4.17 10

lb h

D

g

φ= = =

⋅= ×⋅

　

(4-25)

2

m0 8 2 3 3

m f m

7383.3 lb / ft h(1 0.45)

(4.17 10 lb ft / lb h )(0.413 lb / ft )(0.0208 ft)(0.45)β

⋅ −= × ⋅ ⋅

m m2

150(1 0.45)(0.0673 lb / ft h) lb1.75(7383.3)

0.0208 ft ft h

− ⋅× + ⋅ (E4-4.3)

44444444--------44444444

P.168

Term 1 Term 2

f m f0 2 3

m

lb h lb lb0.01244 [(266.9 12,920.8)] 164.1

ft lb ft h ftβ ⋅= + =

⋅ ⋅

��� �����

(E4-4.4)

2 1

2

f0 3 2 2

f

lb 1 ft 1atm164.1

ft 144 in. 14.7 lb / in.β = × × (E4-4.5)

0atm kPa

0.0775 25.8ft m

β = =

1/ 20.1551/ 2

0

0 0

2 2 0.0775 atm / ft 60 ft1 1

10 atm

β × ×= = − = −

�����LP

yP P

(E4-4.6)

0

0

0.265 2.65 atm (268 kPa)

10 2.65 7.35 atm (744 kPa)

P P

P P P

= =∆ = − = − =

(E4-4.7)

P υ (4-34)

00 ,T Tε = =

0 00

P

P y

υυ υ= = (E4-4.8)

44444444--------44444444

P.168

(4-34) (E4-4.8) E4-4.1

E4-4.1 P υ

(ft)z 0 10 20 30 40 50 60

(atm)P 10 9.2 8.3 7.3 6.2 4.7 2.65 3(ft / h)υ 253 275 305 347 408 538 955

3

c 120 lb / ftρ =

03 2

c c 0

2 2(0.0775) atm / ft

(1 )A 120 lb / ft (1 0.45)(0.1414 ft )10 atmP

βαρ φ

= =− −

1 10.00165 lb 0.037 kgα − −= =

44444444--------44444444

P.169

E4-4.1 E4-4.1

E4-4.1 P υ

4-9

P.169

4.5.44.5.4

(4-33)

1. (2-17)

2. (4-19)

P.169

A B→

A0 A

dXF r

dW′= −

2A Ar kC′− =

3. (3-45)

(4-33) (4-37)

4.

5.

P.170

AA A0 (1 )

FC C X y

υ= = −

1/2(1 )y Wα= −

(4-37)

(4-33)

1/2A A0 (1 )(1 )C C X Wα= − −

22 1/2 2A0

A0

(1 ) [(1 ) ]kCdX

X WdW F

α= − −

A02 2A0

(1 )(1 )

F dXW dW

kC Xα= −

−

Only for ε = 0

FA0 = CA0 0 , W = 0 , X = 0

(4-38)

P.170

0

A0

11 2

X WW

kC X

υ α = − −

A0

0

A0

0

12

1 12

kC W W

XkC W W

αυ

αυ

− = + −

1/20 A01 {1 [(2 ) / ][ / (1 )]}kC X X

Wυ α

α− − −

= (4-39)

P.170

44--55

4-4

2A B C→ +

20 m 121 40 4-4

SI 0P 10 atm 1013 kPa= =

3 30 7.15 m / h (252 ft / h)υ =

1p 4

0.006 m ( in.)D = 約

3 3c 1923 kg / m (120 lb / ft )ρ =

12

1 in. 40 2C 0.0013 mA =

0 25.8 kPa / mβ =

20 mL =

P.171

44--55

k

12

(a)

(b)

(c) (b) (b) 30.1 kmol / m k

612 m

kmol kg cat hk =

⋅ ⋅

解： (4-38)

A0

0

A0

0

12

1 (1 )

kC W W

XkC W

W

αυ

αυ

− =

+ − (4-38)

P.171

44--55

3b c(1 ) (1923)(1 0.45) 1058 kg / mρ ρ φ= − = − =

20 m 12

1 in. 40

2c b 3

6A0

3 30

kg(0.0013 m ) 1058 (20 m)

m27.5 kg

12 m kmol 27.5 kg0.1 4.6

kmol kg cat h m 7.15 m / h

W A L

W

kC W

ρ

υ

= =

=

= ⋅ ⋅ =⋅ ⋅

(a) 0P∆ = ( 0α = )

A0

0

A0

0

4.60.82

1 4.61

kC W

XkC Wυ

υ

= = =++

(E4-5.1)

0.82X =

P.171

44--55 (b) (4-29) b c(1 ) 1058ρ φ ρ= − =3kg / m

0

20 c b3

kPa2 25.8

2 mkg

(1013 kPa)(0.0013 m ) 1058m

P A

βαρ

= =

(E4-5.2)

10.037 kg−=

(0.037)(27.5)

1 1 0.492 2

Wα − = − =

(E4-5.3)

A0

0

A0

0

12 (4.6)(0.49) 2.36

1 (4.6)(0.49) 3.261 1

2

kC W W

XkC W W

αυ

αυ

− = = =

+ + −

(E4-5.4)

0.693X =

82.2% 69.3%

P.172

44--55 (c) Rober 2

Ergun (E4-4.5) 2 1

p

150(1 )1.75G

D

φ µ−>> (E4-5.5)

(4-25)

0 3p0 c p

(1 ) 150(1 )1.75

GG

Dg D

φ φ µβρ φ

− −= +

2

0 30 c p

1.75 (1 )G

g D

φβρ φ

−= (E4-5.6)

(E4-4.4) 0β

0P

1~

Dβ

P

1~

Dα

P.172

44--55 (b) 2 1p p2D D=

1

2

p 12 1

p

1

1(0.037 kg )

2

0.0185 kg

D

Dα α −

−

= =

=

(E4-5.7)

α (E4-5.4)

2

0.0185(27.5)(4.6) 1

3.4320.0185(27.5) 4.43

1 (4.6) 12

0.774

X

X

− = = + −

=

10

12

( 10-6)

P~ 1/k D [ (12-35) ] k

P.173

44--55

E4-5.1 P4-23

(1)(2)

G P

(1) (channel) (2) ( )

P.173

44--66

P.174

1997 97 10 lb× $0.58

9$4.0 10× 60% (30%)

(30%) (10%) (5%)

60%

12 4 2 2 22

12

O

C H O CH CHA B C

+ →+ →

／ ＼─

260 C°

0.30 lb mol/s 10 atm 10

100 12

1 in. 40

43 10 lb mol / s−×

14

in. 3120 lb / ft 0.45

經 濟

用 途

44--66

P.174

1/3 2/3A A B lb mol / lb cat hr kP P′− = ⋅

lb mol

0.0141 260 Catm lb cat h

k = °⋅ ⋅

在

解： 1.

A0 AdX

F rdW

′= − (E4-6.1)

2.

1/3 2 /3 1/3 2 /3A A B A B( ) ( )r kP P k C RT C RT′− = = (E4-6.2)

1/3 2/3A BkRTC C= (E4-6.3)

3. 0 0(1 )( / )X P Pυ υ ε= +

A0 A0AA

0 0

(1 ) (1 )

1 1

C X C X yF P PC y

X P X Pυ ε ε − −= = = = + +

其中 (E4-6.4)

A0 BBB

( / 2)

1

C XFC y

Xυ εΘ −

= =+

策 略

44--66

P.174

B0B

A0

1

2

F

FΘ = =

A0B

(1 )

2 (1 )

C XC y

Xε−=

+

(4-30)

(1 )2

dyX

dW y

α ε= − + (E4-6.6)

4.

2/31/3

A0 A0A 0

(1 ) (1 )( ) ( )

1 2(1 )

C X C Xr kRT y y

X Xε ε − − ′− = + +

(E4-6.7)

2/312( ) 0.63= A0 A0 0P C RT= (E4-6.7)

A1

1

Xr k y

Xε− ′ ′− = +

(E4-6.8)

2/31A0 A02( ) 0.63k kP kP′ = =

結合步驟可利用下列方法計算1. 解析法2. 圖解法3. 數值法4. 使用套裝軟體

44--66

P.175

5. ( 1000)

4A0

4B0

4 22 I

2

4I

0 A0 B0 I

A0A0

0

A

3 10 lb mol / s 1.08 lb mol / h

1.5 10 lb mol / s 0.54 lb mol / h

0.79 mol NI N 1.5 10 lb mol / s

0.21 mol O

5.64 10 lb mol / s 2.03 lb mol / h

3.65 lb mol / h

1.080.30

3.65

T

T

F

F

F

F

F F F F

Fy

F

yε

−

−

−

−

= × =

= × =

= = = × ×

= × == + + =

= = =

=

乙烯：

氧：

惰性氣體 ：

合計：

10 2

A0 A0 0

2 / 31A0 2

(0.3)(1 1) 0.15

3.0 atm

lb mol lb mol( ) 0.0141 3 atm 0.63 0.0266

atm lb cat h h lb cat

P y P

k kP

δ = − − = −= =

′ = = × × =⋅ ⋅ ⋅

α

0

c c 0

2

(1 )A P

βαφ ρ

=−

計算壓降參數

44--66

P.175

G

A0

B0

I0

lb mol lb1.08 28 30.24 lb / h

h lb mol

lb mol lb0.54 32 17.28 lb / h

h lb mol

lb mol lb2.03 28 56.84 lb / h

h lb mol

m

m

m

= × =

= × =

= × =

ɺ

ɺ

ɺ

0

02 2

c

lb104.4

h104.4 lb / h lb

7383.30.01414 ft h ft

T

T

m

mG

A

=

= = =⋅

ɺ

ɺ

4-4

4-4 0β α

0

02 3

c c 0

5

atm0.0775

ft2 (2)(0.0775) atm / ft

(1 ) (0.01414 ft )(0.55)(120 lb cat / ft )(10 atm)

0.0166( 3.656 10 /g cat)

lb cat

A P

β

βαφ ρ

α −

=

= =−

= = ×

44--66

P.176

6. (E4-6.1) (E4-6.8)

A0

1

1

dX k Xy

dW F Xε′ − = +

(E4-6.9)

(1 )

2

dy X

dW y

α ε+= − (E4-6.10)

lb mol

0.0266h lb cat

k′ = (E4-6.11)

A0lb mol

1.08 lbh

F = (E4-6.12)

0.0166

lb catα = (E4-6.13)

0.15ε = − (E4-6.14)

0 , 0W X= = f1.0 , 60 lby W= = 60 lb

60% 60%

44--66

P.176

Polymath Polymath (E4-6.9)

(E4-6.10) [(4-6.11) (4-6.14) ]

E4-6.1 E4-6.2 (E4-6.9) (E4-6.10)

Polymath CD-ROM

MATL AB Polymath ASPE N

CD-ROM Polymath CD-ROM

(E4-6.9) (E4-6.14)

(3-45)

0 0 00

0 0

(1 )( / )(1 )

/

P X T TTX

P T P P

υ ευ υ ε += + = (3-45)

44--66

P.177

f υ 0υ

(3-45)

0

1 Xf

y

υ ευ

+= = (E4-6.15)

E4-6.1 Polymath

體積流率隨壓降之增加而增加

0

0

f

Py

P

υυ

=

=

44--66

P.177

E4-6.2 0 0, / , /X y P P f υ υ= =

0

( E4-6.2) Polymath (

CD) 60% 44.5 lb

44--66

P.178

E4-6.2 65% 66% 1%

(3.5 lb) 1% (0.41 lb) 8.5

5% 3.8 atm 2.3 atm

44.5 lb 5 atm

(E4-6.9) 1y =

A0 1(1 )ln

1

FW X

k Xε ε = + − ′ −

(E4-6.16)

1.08 1

(1 0.15) ln ( 0.15)(0.6)0.0266 1 0.6

= × − − − − (E4-6.17)

35.3 lb / ( ) (16 kg / )= 催化劑 管 忽略壓降 管

E4-6.2 35,300 lb (35.3 lb/ ) 53%

加觸媒對轉化率的影響

忽略壓降導致不良的設計(53% 對60% 的轉化率 )

4.5.54.5.5

PBRG = / AC (4-22) G

P.178

mɺ

4.64.6

P.179

= −− −

利潤 產品售價 反應物成本

操作成本 分離成本

P.179

4-10 (1/2)

P.179

4-10 (2/2)

400×106 lb / year 200×106 lb / year

P.180

8 8

6

$0.38 lb $0.04 lb2 10 4 10

lb year lb year

$0.043 lb2.26 10 $8,000,000

lb year

$76,000,000 $16,000,000 $54,000 $8,000,000

$

−

× ×= − × ×

− −

− × × −

= − − −≅

��������� ���������

����������� �����

乙二醇成本 乙烷成本

利潤

操作成本硫酸成本

52 million