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