Summary • General Material Balance of Reacting System • Batch Reactor

4. The material balances for isothermal ideal reactor models

Summary

• General material balance of reacting system • Batch reactor • Continuous-flow reactors: CSTR (Continuous Stirred Tank Reactor) PFR (Plug Flow Reactor) • Steady state of CSTR and PFR • Design tasks : outlet (final conversion), given volume of reactor x volume of reactor, given outlet conversion T=const. Outlet convective molar flows (mol/s) Arbitrary volume element F1 Inlet convective molar F2 flows (mol/s) N o  AkN R0 1, F  kii . 1 i 1 o . F2 nt(i ) . . Vt( ) . FN .

o FN Molar balance of species i NR  n F0  F  r dV  c dV  i N i i ki V, k i V ki k k 1 tt V i 1 Overall molar balance NNNRNR oon Fi F i  ki r V,, k dV  F  F   k r V k dV    VV  i1 i  1 k  1 k  1 t Laboratory Pharmaceutical industry Batch reactors Special chemicals Polymers . Vconst. Mixer . T=const. . Perfect mixing N  AkN R0 1,  ki i i 1 Manometer Q P=f(t) (nt ) Heat flux i

(VtR )Volume of reaction m ixture

Molar balance of species i

NR 0 ni FFriikiV k dV  ,  V k 1 R t dn NR i   r .V  ki V ,k R dt k 1 Reactor volume constant (VR=const.)

 ni d  V NR 1 dndciiR   rccc ,,...  ki V,12 kN  VdtdtdtR k 1

Reactor pressure constant (P = konst.)

ni  ci .V R

1 dn 1 d c .V  dc d ln V NR i  i R  i  c R   r i  ki V ,k V R dt V R dt dt dt k 1

dc NR d ln V i   r  c R  ki V ,k i dt k 1 dt

To calculate P ( t ) or VR ( t ) state equation f(T,P,V,composition) = 0 is needed Example Constant volume isothermal batch reactor, ideal gas mixture

RT N Pn N  j  V R j 1  kjk j 1 dPRTRT NNNdn RN R  j VrRTr  Rkj V kk V k,, dtVdtV RRjjkk1111

Constant pressure isothermal batch reactor, ideal gas mixture

P N cy * RT ii   Vn RT Rj P j 1 dVRTVRTRTNNdn N RN R NRNR RRj dci RT  VR kj rr V,, kk V k r  c  r    ki V,, k i k V k dt Pjj11 dt kk  11 PP dtkk11 P NR dVln( ) RT NRNR R  r  k V, k r  y  r dt P k 1 ki V,, k i k V k kk11 Example Constant volume batch reactor, liquid mixture  constant pressure Second order irreversible reaction

oo A1(l) + A2(l)  A3(l) rkcV  ctccccc1 211223, 0,, , 0

dci ir V i kc12 c 2.5 dt o c1  2 mole/l 2 dc1 o c2  1 mole/l kc12 c dt o 1.5 c3  0 c dc i c1 2 mol/l kc12 c dt 1 c2 c3 dc3 0.5  kc12 c dt 0 o o o o o 0 2 4 6 ccccccccccc11222112311  ,   ,   t/min

oo c1 t dc dc  c12 c  11 kc c  coo  c    k dt  c  1 1 1 2  1 o oo oo dt o c c c kt c cc1 1c 1c 2  0 2  21 1 1  o e c1 1 1 coo c c   kt 1 2 1 o c1 Continuous-flow reactors Continuous Stirred Tank Reactor – CSTR

N  AkNR0 1,  kii i 1

(nti )

VolumeVconstR  of reactor

Molar balance of species i

NR dn FFr0  V  i iiki V k R  , k 1 dt

Overall molar balance

NR dn Fo  F   r V   k V, k R k 1 dt

We need suplementary information • state behavior of reaction mixture • start-up (shut-down) molar flow rates of individual species N Constant pressure isothermal CSTR, ideal gas mixture RTnpV   jR i 1 It arises from overall molar balance (volume of reactor, VR is constant) N dn  j NNdn NRNR ii1  FFVrFFVroo0   Rki V kRk V k ,, dtdt ii11 kk  11 NR FFVr o  Rk V k , k 1 and molar balance of species i becomes dn N RN R i Foo  y F VrVr  iiRk V kR ki V k ,,o o o dt kk11i Fii y F dndy PV NR iiR  Foo()() y y Vyr  i iRki i k V k , dt RT dt k 1 dy RTF o RT NR i =()() y o  yyri 1,  N  i iki i k V k , dt PVPR k 1 Inlet volumetric flow rate is

RTF o  Vo P Mean residence time of reaction mixture (based on inlet flow rate) is given by

VPV  RR o o VRTFo and molar balance of species i becomes finally

dy 1 RT NR i = (yo  y )  (  y  ) r i  1, N i i ki i k V, k dt o P k 1 If only one reaction takes place:

dyi 1 o RT =()() yyyriN iiiiV  1,  dtP  o 1.00

Example 0.80 N2O Start-up of an isothermal CSTR 0.60 N O  N 1/2O 2  g 2 g 2 g  yi (-) ) 4 3 o (fractional) conversion of N2O* VPR 0.1 l =10 m , T=320 C, 101 kPa 0.40 o o X1 =(F1 – F1)/F1 r= kP , k  2.028x105 mole.Pa  1 .min  1 .m -3 VNO2 o o o o 0.20 N2 O2 y11, y 2  y 3  y 4  0 (1=N 2 O, 2=O 2 , 3=N 2 , 4=He) He t0, y1  0  y 2  y 3  0, y 4  1 0.00

 o  1 0 1 2 3 4 5 t (s) Numerical solution of balance equations *) The equation is valid only at steady state Steady state

NR FFVriN o  , 1, iiRki V k  , k 1

If only one reaction takes place:

o FFVriNiiRi V  , 1,

FFo  X  jj using fractional conversion of key species j j o F j o Fj X j V R j r V() X j a) The volume of reactor for given outlet conversion

o X j Simple substitution VFRj  jrX V() j b) The outlet conversion for given volum e of reactor F o j X r( X )  f ( X )  0 Graphical or numerical (iterative) j j V j j solution V R Graphical assessment of the CSTR volume

 jVjrX() Levenspiel diagram

X j

 jVjrX() usually 1v j 

X j Graphical assessment of the outlet conversion

 jrX V() j

o F j X j V R

X j Cascade of CSTR ( 1)n  ()n F j F j

()n V R ……… ()n X j ………

n-th member of Steady state the cascade

(n ) ( n 1) ( n ) ( n ) Fi F i  V R i r V, i  1, N n  1, N C STR FFon () X ()n  jj j o F j

()(n o n ) FFXj j(1 j )

()(n o i o n ) FFFi i jX j  j F o j X()n X( n 1)  r( X ( n ) ), n  1, N ( n )  j j j V j C STR V R Graphical assessment of the outlet conversion in the cascade of CSTR

 jrX V() j

o F j (1) o XX ( 2 )  jj F j V X R (1) j V R

(1) (2) X j X j X j Homework 6 ()n ()nn V R Prove that Xk j  1  (1   o ) , o  for first order constant-volume reaction. Vo ()n  k nV In the limit ()n oR R limXej 1  , oR  n  Vo Plug Flow reactor – PFR Tubular reactors High production capacity v Catalytic reactors (e.g. ammonia synthesis) . FViR() FVVi() R R . .

z z + z VSzRR

VR VR + VR

Molar balance of species i

c(,)(,) t z1 F t z NR ii=   r (,) t z i  1, N  ki V, k t SR z k 1 Steady state dF NR i   riN 1,  ki V, k dVR k 1

If only one reaction takes place:

dFi   iV riN 1, dVR FFo  X  jj using fractional conversion of key species j j o F j

o dX j FrXjjVj  () dVR useful relations:

X j ( 2 ) dX j F F y F FVo  yci,  i  i  i jR iiNN VVm X (1)  rX() FVF j j V j j m j jj11

31 V m  molar volume (m mole ) BATCH, CSTR, PFR (PBCR), one reaction, fractional conversion of key component

BATCH 1 1 o X t j R  r ndX jV  jjdtt R  jRVjVrX 00()

1 o X t cdX j R jj CSTR dtt R rX() PFR  j 00Vj

X j CSTR X 1 oo11 j FXFXj j j j V R     r()() X11 r X j V j j V j PBCR Packed Bed Catalytic Reactor

11 PFR 11 XX XX oojj oojj F dX F dX F dX F dX j j j j V  j j  j j W    R  r r() X  r()() X r X j00 Mj M j j00 V j j V j Mean residence time

V   R V

BATCH, PFR

X 1 cdXo j jj   o  rXVj() j 0 V R  o  V o CSTR

o 1 cXjj volumetric flow rate (m3/s) of reaction V o   o mixture at inlet conditions rX()1  j Vj Gas Hourly Space Velocity

Gas volumetric flow rate(m3 /hr) GHSV  Volume of reactor (m3 )

Liquid Hourly Space Velocity

Liquid volumetric flow rate (m3 /hr) LHSV  Volume of reactor (m3 ) Tasks

1. Calculate the volume of reactor (mass of catalyst) (CSTR, PFR, PBCR) or time of reaction (BATCH) to obtain given conversion.

2. Calculate the outlet conversion for given volume of reactor.

3. Calculate the reaction rate in laboratory reactor to obtain kinetic law and estimate the kinetic parameters. NN A 0 = NR = 1 iii j – subscript of key component ii11

BATCH FLOW

  nnnnXooo  i . FFrdVFFXooo  ..i iiiijj iiiVRijj   j  j

NN oooo  oo nnnnnX  .. FFFX jj. iijj  ii11  j j

oo  oooo  nn X i . xxoo X i . FF Xxxii X.. ijj  ijj ijjijj  nikj  Fi jj xi  xi  n ooo  F ooo  nn Xx X jjjj.1. FF Xx X jjjj.1. jj jj

  ooi xoo i x. X xi x j. X j i j j n n x 1  F F x 1  j i i i j c i  i  i  ci     i V n. V V V  o VFVVV.  o mmm1. xX mmm1. xX jj  jj  j j

V m - molar volume of reaction mixture (m3 /m ol)

Gas phase - Ideal state behavior RT Vm  P   xoo i x. X xoo i x. X i j j i j j xPiiP j FFPi i i P j ci    ci     RT RTRT  o VRTRTRT  o 1. xXjj F. 1 xjj . X P  j P  j

nnii FFii ci   ci    V T Po  o V  Vx X (1 ) T Po  o oj j V1  x X TP  o j j oj TP oj  oii  oo o n n.. Xc c X o o   o o  i  j ji j j Fii F.. X c c X TTPP1  i j j   i j j  oojj    TTooPP1 jj      T Po VT oo  P oo 11x Xx X TPVTPoo     j jj j o o o  11xj X j x j X j jj      jj   

Homework 7 Calculate volume of PFR to produce 150 kt ethylen/year. Reaction of the 1st order

C2H6(g)  C2H4(g) + H2(g) takes place at 1100 K and 0,6 Mpa. Final conversion of ethan is 80 %. Reaction rate is given by

rV = k. cA k (1000 K) = 0,072 s-1 E = 343,6 kJ/mol

Pure ethan is fed into the reactor.

Assumptions: Ideal gas, molar weight of ethylen is 28,054 kg/kmol.

K. J. Laidler and B. W. Wojciechowski: Kinetics and Mechanisms of the Thermal Decomposition of Ethane. I. The Uninhibited Reaction, Proceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences, Vol. 260, No. 1300 pp. 91-102 Homework 4 (due after Chapter 4)

o Calculate volumes of CSTR a PFR working at 150 C and 300 kPa to produce 1 t COCl2/day with CO conversion equal to 95 %. A mixture of CO and Cl2 (molar ratio 1:1) is fed at 300 kPa and 150 oC. Data k(423 K) = 0.07 (m3mol-1)3/2.s-1

MCOCl2 = 98.92 kg/kmol.

Answer:

3 3 VCSTR = 0.053 m VPFR = 0.0021 m  o ooi F   F FFFiijj X i i  CO(g) + Cl2(g) → COCl2(g) j steady state; j - key component IN OUT o o CO (1) F1 FFX1 11 1  o Cl2 (2) FFX2 11 1  o COCl2 (3) 0 FFX311 oo o  FF 2 1 FFX11 2  RT 3 ideal gas V m [m /mol] P o FFFP PP F11 XX   c cc 1 c  1  1 1  11 [m ol/m3 ] C OC l 12o 2 VVFRTFRTFXRTXm 1 22 11  

5 / 2 5 / 2 P 1  X r kc c3 / 23 k 1 [m ol/m /s] V C O C l2  RTX2  1 6 F 1 10 /98.92/ 24 3600  F o 3  0.123163 m ol/s 1 1 X 1 0.95 FXFXoo11 0.123163 0.95 V 1 1  11 CSTR 1 5 / 25 / 2  rX( ) 1 5 / 2 1 3 5 / 2 11V P 1  X   300 10 1 0.95 k 1 0.07  1    RTX2  1 8.3145 423 2 0.95  0.0503 m 3 1 5 / 2 o X j oo0.950.95 5 / 2 FdXjjFdXFX RT 2  VdX  1111 PFR  5 /2 5 / 2  1  rVj( XkPX )11   1 j 000 P 1  X 1  k  RTX 2  1 5 / 2 0.123163 8.3145 423  79.42692225 2.08 103 m 3 3 0.07300 1  0 Summary