工業反応装置特論
講義時間：6限
場所
：8-1A
担当：山村

1

管型反応器(Plug Flow Reactor)
Feed （原料）濃度C0Product （製品）滞留時間(residence time) t

C0pulse

入口0出口時間t 時間2管型反応器の設計方程式（１）
原料の体積流量v

FA[mol/s]
面積S0z z+ Dz

ZL

Z=z～z+Dzの微小区間で成分Aの物質収支をとる
モル数の変化DnA = 反応による生成量rASDz Dｔ+反応器への流入量FA zDｔ- 反応器からの流出量FA z+DzDｔ

3管型反応器の設計方程式（２）両辺をDtで除すと

Dn A
 rA SDz  FA z  FA z  Dz
Dt
Dt→0の極限をとればdn A
 rA SDz  FA z  FA z  Dz
dt
微小区間のモル濃度はCA=nA/DVであることに注意して両辺をDV=SDzで除すとFA z  FA z  Dz
dC A
 rA dtSDz
4管型反応器の設計方程式（３）
Dzを十分小さくとれば

FA z  Dz  FA z
DzdFAdz z

従ってdC A
1 dFA
 rA dt
S dz zdFA
 rA dV z
定常状態を考えると、dCA/dt=0なのでdFA0  rA dV5

管型反応器の設計方程式（４）dFArA dV
FAf

1
必要なPFRの体積VPFRはVPFR   dFAr
FA 0 A

6

CSTRのImpulse応答（１）
FA0

設計方程式
dC AV
 rAV  C A0v0  C Av0dtt  0でC A  C A0t  0でC A0  0, rA  （反応無）0FA
体積V[m3]dC AV
 C Av0dt
滞留時間t  V / v0より
dC At
 C Adt
7

CSTRのImpulse応答（2）
積分すると

E(t)

CAtCA
1
0 dt  t C C A dC A  t ln C A0
A0

出口CA

 exp(t / t )
C A0

t 時間

residence time distribution function, EEC A (t ) C (t )dt
A0



C A0 exp(t / t )C A0  exp(t / t )dt0

exp(t / t )t8

直列CSTRのImpulse応答（１）CA0CA2CA1体積V

N槽・・・
体積V

CAN
体積V

residence time distribution function, E
N ( Nt / t ) N 1
Eexp( Nt / t )( N  1)!
N (t / t ) N 1
N  1ではE exp(t / t )t ( N  1)!
9

直列CSTRのImpulse応答（２）
2.5

N=30
2

1.5

E
1

N=1

N=1では単槽CSTR
Nが増加するとPFRに漸近

N=10
N=5

N=20.5

0
00.5

1

1.5

2t/t
実反応器でインパルス応答実験を行い、上の曲線にfittingすれば槽数Nが決まる（槽列モデルとも呼ばれる）

10

Report 11 Chemical Process Simulator – combination of CSTR and PFR

氏名Design a chemical process with two reactors connected in series. We have single reaction, Methyl acetate+ water →acetic acid + methanol, in each reactor. We supply the feed that contains 26.75 mol/s of water,
0.25 mol/s of Methyl acetate, and 0.025 mol/s of acetic acid into the first reactor. No methanol is included inthe feed.
Q1.Show that the parameters below result in “Flow Methyl acetate” of 0.05 mol/s in stream 3.
This corresponds to the conversion of (0.25-0.05)/0.25=0.8, namely 80% of Methyl acetate in the feed isconverted into products.
Q2. Determine pairs of volumes of the CSTR/PFR reactors that satisfy “Flow Methyl acetate” of 0.05 mol/s
in stream 3. Calculate the total reactor volume (VCSTR+VPFR) against the volume of CSTR and show that thetotal volume shows a minimum at a critical VCSTR.
Q3. Switch the CSTR and PFR at the same volumes and conditions in Q2. Show that the “Flow Methylacetate” becomes larger than 0.05 mol/s in product stream, namely more unreacted ethyl acetate remains
in the system when we use PFR first and then CSTR reactors.
1. Material selection
setting-property packages-add-TEA-select-New-Model set:Peng Robinson-add-filter-Methyl acetate-OK-add-filter-waterOK-add-filter-acetic acid-OK-add-filter-methanol-OK-OK2. Reactior/Reaction selection
Reaction packages-add-CORN Reaction Package Manager-select-New-New Reaction Packages-Edit-Compounds-AddFrom material template-default-OK-check H2O-check filter-Methyl acetate-check acetic acid-OK-check methanol-OKReactions-Create-rxn1-OK-Stoich-"-1"(water)- "-1"(Methyl acetate)- "-1"(acetic acid)-"1"(methanol)-Rate-1.e-6*C(“Methylacetate")*C(“Acetic acid)-Phase-Liquid-OK3. CSTR
Insert Unit Operations-Reactors-CSTR-select-Edit unit operation-Isotherml 298.15K-Reactor-Reactor volume 1.62 m3Reactions-add-rnx1
4. PFR
Insert Unit Operations-Reactors-PFR-select-Edit unit operation-Reactor-Reactor length 1 m- Reactor diameter 1.128 =(4/p)1/2 m - Enthalpy-constant temperature 298.15K-Reactions-add-rnx1
5. FeedInsert stream-Edit/view streams-0.1013 MPa-25 C-Compound flows-Water-26.75 mol/s-Methyl acetate-0.25 mol/s-Acetic
acid 0.025 mol/s-Methanol 0 mol/s6. Stream 2 and 3
Insert stream-Edit/view streams 0.1013 MPa-25 C7. solve(F5)11