6. Energy Balance on Chemical Reactors

6. Energy balance on chemical reactors

Most of reactions are not carried out isothermally

BATCH or CSTR heated (cooled) reactors

Inlet tubing Coolant outlet Reaction Mixer mixture Double shell Coolant Outlet Internal inlet tubing heat exchanger

The balance of total energy involves: Internal energy mechanical energy (kinetic energy) potential energy R.B.Bird, W.E.Stewart, E.N.Lightfoot : .... Transport Phenomena, 2nd Edition, J.Wiley&Sons, N.Y. 2007 Transformation of various kinds of energy

Balance of total energy Input x Output

of total energy Input x Output of total energy by convective by molecular flux flux Rate of change of total energy

EUEE   Work done kin p Work done by molecular by external interactions forces Main reason to study energy balances : assesment of temperature of reacting system (reactor) Application of the 1st law of thermodynamics on the open homogeneous reacting system Rate of work done on surroundings [W] Heat flux [W]

Ee dV  V VR

Single phase reacting system

eo,e1 – specific total energy of inlet (outlet) streams [J/mol] 3 VR – volume of reaction mixture [m ] dE F e  Fe  Q  W dt oo 11

eeo ,1  molar energies of inlet and outlet streams [J/mol] Rate of work W done by the reacting system on the surroundings consists of:

• Flow work of inlet stream(s) VPFVPo o o mo o

• Flow work of outlet stream(s) VPFVP1 1   1m 1 1

• Work provided by stirrer Ws

dVR • Work done by volume change P dt

• Work done by electric, magnetic fields Wf dE dV FeVP   FeVPQP   R  WW  dto o mo o1 1 m 1 1 dt s f Neglecting potential and kinetic energies (EU  ), we have

dU dV F h  F h  Q  PR  W  W dtdt o mo11 ms f

hhmo, m1  molar enthalpies of inlet and outlet streams [J/mol]

If WWsf0,0 dU dV F h  F h  Q  P R dtdt o mo11 m

From enthalpy definition dH dU dP dV dH dP  VP  R V  F h  F h  Q dt dtR dt dt dtR dt o mo11 m Introducing partial molar enthalpies of species

N oo Fo h mo  F i H i i1 N F11 hm  F i H i i1

We have finally

NN dH dP oo VFHFHQR  i i  i i  dt dt ii11 BATCH reactor dH dP VQ dtdt R

Enthalpy is a function of temperature, pressure and composition

Total heat capacity [J/K] Partial molar enthalpy [J/mol]

HHH     N   dH  dT    dP   dni T P,, n   P  T n i1  ni jjT,, P n ji HH NN   CP dT   dP  H i dn i  m V R c P dT    dP  H i dn i PP  T,, njjii11   T n

Molar density [mol/m3 ] Molar heat capacity [J/mol/K] We know that (from thermodynamics)

 H VR VTVTR   R1  p  PT Tn, j Pn, j where the coefficient of isobaric expansion is defined as

1 VR  p   VT R Pn, j and we obtain

N dH dT dP dni mc P V R  V R1  p T   H i dt dt dti1 dt

dn Finally by substitution of i in the energy balance of the batch reactor dt NR dT dP N dn dni i VrR ki V, k mc P V R p V R T  H i  Q   dt k1 dt dti1 dt Using definition of the enthalpy of k-th reaction

N rHH k ki i i1 we have dT dP NR mc P V R p V R T  V R  r H k r V, k  Q dt dt k1 dP Isobaric reactor (  0 ) dt dT NR mc P V R V R  r H k r V, k  Q dt k1

VR  f() t we need state equation !

Homework 8: Energy balance of ideal gas isobaric batch reactor dV Isochoric reactor ( R  0 ) dt NR dT  p mV R c V V R   r H k  T  V kr V, k  Q dt k1  T  Pf ()t

the coefficient of isobaric expansion the coefficient of isothermal compressibility

11VVRR    pT      VTVP RR P, nj   T the volume variation due to k-th chemical reaction N V VVVk  ki i, where i   is the partial molar volume of species i i1 ni T,, p n ji the specific heat capacity at constant volume  2 PP VR   p CCTCTVVPPR        pCTV P R TTT     Vn, j Pn, j  Vn, j T Variation of the pressure can be derived from total differential of volume

N VR dT dni  Vi   N TPn, dti1 dt  dPj p dT 1 dni     Vi dtVR  dt Vi1 dt  TRT P Tn, j Homework 9: Energy balance of ideal gas isochoric batch reactor Summary of energy balance of BATCH reactor

dT NR dP  c V V  H r  Q  0 m P Rdt R r k V, k dt k1 V f() t R

dT NR   V c V  H  Tp  V r  Q dVR m R V R  r k k V, k  0 dt k1  T  dt Pf ()t

dT NR mc P V R V R  r H k r V, k  Q If  p0,cc p V dt k1 liquid (condensed) systems Rate of change Rate of heat generation Rate of heat of reaction mixture by chemical reactions loss (input) enthalpy Heat flux : QSTT H ( e )   the overall (global) heat transfer coefficient [W.m21 .K ] 2 SH  the heat exchange area [m ]

Te  temperature of external cooling (heating) fluid Limiting cases

dT NR Isothermal  mc P V R0  V R  r H k r V, k  Q dt  reactor k1

Adiabatic dT NR Q0  c V  V  H r reactor m p R R r k V, k dt k1

Example Adiabatic reactor with 1 reaction, constant heat capacities

Energy balance on adiabatic BATCH reactor dT  c V V  H r m p Rdt R r V Molar balance of key component dc dX jj cro  dtj dt j V N mV R n, c p y i c pi i1 NNN   Vcnyc  nc  noo i nXc  m R p i pi  i pi  i j j pi i1 i  1 i  1  j NNNoo oonj X j n j X j ni c pi   i c pi   n i c pi   c p i1 j i  1 i  1  j

Assuming that cpi const   c p  const yo  H() T X TT j r o j TX o N o j j oo vj y i c pi y j c p X j i1 yo  H() T    j r o the adiabatice rise of temperature j N oo vj y i c pi y j c p X j i1 Trajectories of T(t) and Xj(t)

Xj = 1 Exothermal reaction T(t) Xj(t) r H 0 Xj(t)

yo  H() T    j r o j N oo vj y i c pi y j c p X j T(t) i1

t Homework 10 The reversible reaction

A1 + A2  A3 is carried out adiabatically in a constant-volume BATCH reactor. The kinetic equation is

1/2 1/2 r kfb c1 c 2 k c 3 k (373 K)=2x1031 s E 100 kJ/mol f 1 51 kb (373 K)=3x10 s E2  150 kJ/mol

Initial conditions and thermodynamic data

cco 0.1 mol/dm3 25 J/mol/K 11p o 3 cc220.125 mol/dm p 25 J/mol/K

oo rpH298  40 kJ/mol c 3  40 J/mol/K T = 373K

Calculate X1(t),T(t). Example Acetic anhydride reacts with water

(CH3CO)2O + H2O  2CH3COOH

in a BATCH reactor of constant volume of 100 l. Kinetics of reaction is given by 46500  r2.14 1073 eRT c 1 mol.m .min V 1

Data

o -1 -1 -1 1 3 c1 0.3 mol.l , cpM  3.8 kJ.kg .K ,  H r   209kJ.mol ,  =1070 kg.m c .S =200W.K1 T Too  300 K,  c  0, c  w c , c  pi H e p pM i pMi pMi M i i wco  initial mass fractions, - mass heat capacities (kJ.kg-1 .K -1 ),M - molar weight (kg.mol 1 ) i pMi i RK 8.31446 J.mol11 .

In neglecting variation of heat capacities with temperature,

calculate T(t) and X1(t) for an non-adiabatic and adiabatic case.

By numerical integration we get

dT 146500  0.1  209E3  2.14E7xexp - 300  (1 X1 )  200  (300-T) dtT (1070 3.8E3  0.1) (8.31446 )

dX1 46500 2.14E7  exp - (1  X1 ) dt(8.31446 T )

316 1.20

T [K] 314 1.00 X1

312 0.80

310 0.60

308 0.40

306 0.20

304 0.00 0 10 20 30 40 50 60 t [min] Continuous (perfectly) stirred reactor (CSTR)

Energy balance on CSTR NN dH dP oo VFHFHQR  i i  i i  dt dt ii11

Total differential of enthalpy

N dH dT dP dni mc P V R  V R1  p T  H i  dt dt dti1 dt NN dP oo VFHFHQR  i i  i i  dt ii11

Molar balance on CSTR

NR dni o Fi  F i  V R ki r V, k , i  1, N dt k1 We get

dT dP N NR c V  TV  H Fo  F  V r  Enthalpy mPRdt pR dt  ii iR kiVk, ik11change for NN oo k-th reaction FHFHQi i  i i  ii11 N rHH kki i  i1

dT dP NR mc P V R V R p T  V R  r H k r V, k  dt dt k1 N oo  FHHQi i  i   i1 dP  0 dt The CSTR usually works at constant pressure (no pressure drop)

NR N dT oo QSTT Tm() mPRc V V R  rkVk H r, + F ii H  H i   Q dt ki11 Steady state

dT dn i 0 dt dt

NR N oo VR r H k r V , k +0  F i H i H i  Q ki11

Assuming ideal mixture, i.e. HH ii  , we have T Hoo H  H  H   c dT i i i i pi To T o rH k   r H k   c p  dT  k Molar and enthalpy To N balances give cc  pk  ki pi N+1 unknown i1 variables T, Fi NRTT N V  Hoo   cdTr  FcdTQ   0 R r k pk V, k i pi ki11 TToo NR o Fi F i  V R ki r V, k 0, i  1, N k1 N+1 unknown variables in N+1 non linear algebraic equations

GiN T, F12 , F ,.... F  0, i  1, N  1

Issues: • multiple solutions • slow convergence (divergence) Example Adiabatic CSTR with 1 reaction, constant heat capacities

NN o o o o o o o VR r H   cTTr p  V  FcTT i pi()()   FTT  yc i pi ii1 1 o o FjjX Fj X j V R j r V 0  rV  VRj o yX N 2 nonlinear algebraic equations for Hco  T  T ojj  () T T o y o c  r p   i pi unknown T and Xj  j i1  Hoo y X o  r j j TT N  oo  j  yi c pi c p y j X j i1 o  TX jj cf. adiabatic const.-volume BATCH Multiple steady states of adiabatic CSTR (exothermal reaction)

o Fj X j + j r V ( X j , T ) V R =0 X j f1 (T)

0.8 f1(T)

0.6 Steady state 1

j f2(T) X Steady state 2 0.4

0.2 Steady state 3

0 310 330 350 370 390 T (K)

T To  j X j  X j  f2 ( T ) Example You are to consider an irreversible gas-phase reaction in an adiabatic CSTR at constant pressure (101 kPa). The gas phase reaction is:

CO(g) + 3 H2(g)  CH4(g) + H2O(g)

rV kc CO E 11 k 0.001exp[ a (  )] min1 RT 298

Ea 10 kcal/mol

The feed to the CSTR consists of CO and H2 at the following (stoichiometric) concentrations:

CCO(in) = 0.0102 mol/liter CH2(in) = 0.0306 mol/liter The heat of reaction at 298 K is equal to –49.0 kcal/mol.

The heat capacities of CO, H2, CH4 and H2O are all constant and are equal to 7 cal/mol/K. The temperature of the feed stream is equal to 298 K, the pressure is equal to 101 kPa, the volumetric flow rate of feed is 8 liter/min and volume of reactor is 0.5 l. The gas mixture behaves as ideal gas.

A. Use the energy and molar balance to calculate the CO conversion and temperature of the effluent stream from the adiabatic CSTR. B. Calculate the composition in molar fraction of the outlet stream. C. Calculate the volume of the adiabatic CSTR required to achieve the desired CO fractional conversion equal to 0.99.

Molar balance of CO 0.9

VR PXkT() X 11(2 ) o gT() RT21 FX11 0.8 (2gg )  4  2 X  1 2 0.7 Adiabatic enthalpy balance N oo 1  yi c pi () T T 0.6 X  i1 1 o o o o rpH y11   c y() T  T

0.5 X1(T)

0.4 3 steady states

0.3

0.2

0.1

0 0 500 1000 1500 2000 2500 3000 3500 4000 4500 T/K Remarks: . Unrealistic temperature of the 3rd steady state  backward reaction will occur . The 2nd steady state is unstable  carefull temperature control has to be used . The dynamic behavior of reactor should be studied

Homework 11

Determine steady states of adiabatic CSTR in which the exothermal liquid state reaction takes place

A1  A2 Reaction rate:

3 rV = A.exp(-E/RT).cA1 (kmol/m /s)

Data

17 -1 A = 5.10 s E = 132.3 kJ/mol

3 o VR = 2 m T = 310 K

 = 800 kg.m-3 o 3 cA1  2/ kmol m

o o -1 -1 V  3.33 l/s Cp  4.19 kJ.kg . K

Hr,298  100 kJ / mol Balance of enthalpy on Plug Flow Reactor (PFR)

v N N  FHii  FHii i1 V i1 VV R H RR  t VR

z z + z VSRR z

VR VR + VR Q H dP NN  VFHFHQR  i i  i i   t dt ii1 1 VR VVVRRR in the steady state NN d N dQ 0= FHFHQi i i i     FHii ii11 dVi1 dV VVRRRV R R ideal mixture  Hii ( T , P , composition ) h ( T ) N N NR dFi dh i()() T   dh i T dT  hi( T ) F i      ki r V, k h i ( T )  F i   i1dVRRR dV  i  1  k  1 dT dV  NRdT N dQ  rH k r V, k  Fc i pi  ki11dVRR dV dF NR N dQ dS Circular cross i t section kirh V, k ki i rHTT k  ( e  ) dVR ki11 dVRR dV NR dT 14 dSt  dR dz N  rH k r V, k ( T e  T )  2 dVRRRk 1 d dV dR  Fci pi  dz i1 4 2 NR dT  dR 4 N  rH k r V, k () T e  T dzk 1 dR 4 Fci pi i1 One reaction, constant heat capacity of reaction mixture. Profiles of conversion and temperature are given by following

equations: NN

2 Fci p,, ii p i Fy c dT  dR 4  ii11 oo () Hr r V  T e  T  dz4 F cpM d R N o o o o 22F yi c p, i F c pM dX jj..dd  j  RR i1   oorV(,)(,) X j T r V X j T dz44 FFjj

z0, T  Toj , X  0

Limiting cases dT 44 1.Isothermal 0 (Hr ) r V  T e  T  ( H r ) r V  T e  T  dz dd reactor R R TT o 2.Adiabatic 2 Fo ( H) dX dT  dR j r j reactor () HrrV  dz4 Foo c F ooc dz pM j pM yHo () TTjrX ojo  jpc M One reaction, constant heat capacity of species. Profiles of conversion and temperature are given by following equations: dF  dFj 2 ii r  iV dT  dR 4 dVR j dV R    ()Hr r V T e T  dzd N o FFo  ooy j R jj X j  4F yi c p, i   c p X j  F o  j i1 j oii o o FFFFFXi i  j  j   j j 22  dX ..dd  j jj jj RRr(,)(,) X T  r X T ooV j V j NN dz44 F F Fc c Foo i F X  jj i p,, i p i i j j ii11 j

N o ooy j F yi c p, i   c p X j z0, T  Toj , X  0  i1 j Limiting cases dT 44 0 (Hr ) r V  T e  T  ( H r ) r V  T e  T  1.Isothermal dz ddR R reactor TT o

2.Adiabatic reactor

2 Fo () H dX dT  dR j r j () HrrV   dzNNyyoo    dz 4Fo y o c jj  c X  F o  y o c   c X  ipi,, pj j ipi pj ii11jj    o o o yj r H   c p () T  T  dX j  N oo dz  j y i c p, i y j c p X j i1 X T dT j dX  j  o o o  N o  T yj r H   c p () T  T  0 o o  j y i c p, i y j cXp j i1 yoo() H X o j r j Cf. adiabatic BATCH and CSTR TT N oo  ji y cp,i y j c p X j i1 Exercise: reactor for oxidation of SO2 to SO3

Numerical method: „hot“ point • Euler method 700 Conversion of 1 SO2 0.9 „Stiff“ solvers 600 MATLAB 0.8 www.netlib.org 500 0.7 0.6 www.athenavisual.c 400 Temperature o T [ C] 0.5 om XSO2 [-] 300 http://wxmaxima.so 0.4 urceforge.net/wiki/i 200 0.3 ndex.php/Main_Pag 0.2 100 e 0.1

0 0 0.0 0.2 0.4 0.6 0.8 1.0 V [m3] Balance of mechanical energy in PFR Profile of overall pressure (P(z))

P(0) Pz()

z=0 z

VR=0 VR

Bernoulli equation

P(0) P ( z ) z 2   v  f density of fluid(kg/m3)  d fR  friction coefficient(-) dP  f 2  fd(Re,wR / )  v dz dR v fluid mean velocity

Catalytic PFR Profile of pressure is calculated using Ergun equation:

2 22 dP ff1b  o1b o o o  1502 3vf  1.75 3  v f  A12 f v f  A f v f  dz dp b d p b

 f - fluid dynamic viscosity (Pa.s)

 f - fluid density (kg/m3)

o v f - superficial fluid mean velocity (m/s)

b - bed porosity (-)

d p - catalyst particle diameter (m)

PFR model for one reaction with constant heat capacity of reaction mixture

2 dT  dR 4 oo () Hr r V  T e  T  dz4 F cpM d R dX 22  jj..ddRRj  oorV(,)(,) X j T  r V X j T dz44 Fjj F

2 dP oo  A12f v f  A f v f  dz 

z0, T  To , X j  0, P  P o Example A gas phase reaction between butadiene and ethylene is conducted in a PFR, producing cyclohexene:

C4H6(g) + C2H4(g)  C6H10(g) A1 + A2  A3

The feed contains equimolar amounts of each reactant at 525 oC and the total pressure of 101 kPa. The enthalpy of reaction at inlet temperature is -115 kJ/mol and reaction is second-order:

rV  k() T c12 c

41 3 1 115148.9    k( Tm ) s3.2  10 exp  (mol ) RT

Assuming the process is adiabatic and isobaric, determine the volume of reactor and the residence time for 25 % conversion of butadiene.

Data: Mean heat capacities of components are as follows (supposing that heat capacities are constant in given range of temperature)

-1 -1 -1 -1 -1 -1 cp1 = 150 J.mol .K , cp2 = 80 J.mol .K , cp3 = 250 J.mol .K

Project 15 A gas phase reaction between butadiene and ethylene is conducted in a PFR, producing cyclohexene:

C4H6(g) + C2H4(g)  C6H10(g) A1 + A2  A3

The feed contains equimolar amounts of each reactant at 525 oC and the total pressure of 101 kPa. The enthalpy of reaction at inlet temperature is -115 kJ/mol and reaction is second-order:

rV  k() T c12 c

41 3 1 115148.9    k( Tm ) s3.2  10 exp  (mol ) RT 1. Calculate temperature and conversion profiles in adiabatic PFR. 2. Assuming the process is adiabatic and isobaric, determine the volume of reactor and the residence time for 25 % conversion of butadiene.

Data: Heat capacities of components will be taken from open resources [1,2] 1. http://webbook.nist.gov/chemistry/ 2. B. E. Poling, J.M.Prausnitz, J.P.O’Connell, The Properties of Gases and Liquids, Fifth Edition, McGraw-Hill, N.Y. 2001.