Most of reactions are not carried out isothermally - vscht.czbernauem/ark/lectures/Chapter 6.pdf ·...

6. Energy balance on chemical reactors

Most of reactions are not carried out isothermally

BATCH or CSTR heated (cooled) reactors

Inlet

tubing

Mixer

Coolant

inlet

Coolant

outlet

Reaction

mixture

Outlet

tubing

Double

shell

Internal

heat

exchanger

The balance of total energy involves: Internal energy mechanical energy (kinetic energy) potential energy .... Transformation of various kinds of energy Balance of total energy Main reason to study energy balances : assesment of temperature of reacting system (reactor)

Rate of change

of total energy

Input x Output

of total energy

by convective

flux

Input x Output of total energy

by molecular flux

Work done

by external

forces

Work done by molecular interactions

R.B.Bird, W.E.Stewart, E.N.Lightfoot : Transport Phenomena, 2nd Edition, J.Wiley&Sons, N.Y. 2007

kin pE U E E

R

V

V

E e dV

Application of the 1st law of thermodynamics on the open homogeneous reacting

system

Heat flux [W] Rate of work done on surroundings [W]

eo,e1 – specific total energy of inlet (outlet) streams [J/mol]

VR – volume of reaction mixture [m3]

Single phase

reacting

system

Rate of work done by the reacting system on the

surroundings consists of:

• Flow work of inlet stream(s)

• Flow work of outlet stream(s)

• Work provided by stirrer

• Work done by volume change

• Work done by electric, magnetic fields

1 1o o

dEF e Fe Q W

dt

W

1 1 1 1 1

o o o mo o

m

s

R

f

V P F V P

V P FV P

W

dVP

dt

W

1 1 1 1R

o o mo o m s f

dVdEF e V P F e V P Q P W W

dt dt

1, molar energies of inlet and outlet streams [J/mol]oe e

Neglecting potential and kinetic energies ( ), we have

1 1R

o mo m s f

dVdUF h Fh Q P W W

dt dt

If 0, 0s fW W

1 1R

o mo m

dVdUF h Fh Q P

dt dt

From enthalpy definition

RR

dVdH dU dPV P

dt dt dt dt

1 1R o mo m

dH dPV F h Fh Q

dt dt

E U

1, molar enthalpies of inlet and outlet streams [J/mol]mo mh h

Introducing partial molar enthalpies of species

We have finally

1

1 1

1

No o

o mo i i

i

N

m i i

i

F h F H

F h F H

1 1

N No o

R i i i i

i i

dH dPV F H F H Q

dt dt

1, , , ,

1 1, ,

j j j i

j j

N

i

iP n T n i T P n

N N

P i i m R P i i

i iT n T n

H H HdH dT dP dn

T P n

H HC dT dP H dn V c dT dP H dn

P P

BATCH reactor

R

dH dPV Q

dt dt

Enthalpy is a function of temperature, pressure and composition

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

3Molar density [mol/m ] Molar heat capacity [J/mol/K]

We know that (from thermodynamics)

, ,

1

j j

RR R p

T n P n

VHV T V T

P T

,

1

j

Rp

P nR

V

V T

where the coefficient of isobaric expansion is defined as

1

1N

im P R R p i

i

dndH dT dPc V V T H

dt dt dt dt

and we obtain

Finally by substitution of in the energy balance of the batch reactor

1

Ni

m P R p R i

i

dndT dPc V V T H Q

dt dt dt

,

1

NRi

R ki V k

k

dnV r

dt

idn

dt

Using definition of the enthalpy of k-th reaction

we have

1

N

r k ki i

i

H H

,

1

NR

m P R p R R r k V k

k

dT dPc V V T V H r Q

dt dt

Isobaric reactor ( ) 0dP

dt

,

1

( )

NR

m P R R r k V k

R

k

dTc V V H r Q

dt

V f t

we need state equation !

Homework 8: Energy balance of ideal gas isobaric batch reactor

Isochoric reactor ( ) 0RdV

dt

Homework 9: Energy balance of ideal gas isochoric batch reactor

,

1

( )

NRp

m R V R r k k V k

k T

dTV c V H T V

d

P f

Qt

t

r

,

the coefficient of isobaric expansion the coefficient of isothermal compressibility

1 1

the volume variation due to k-th

j

R Rp T

P n TR R

V V

V T V P

1 , ,

, ,

chemical reaction

, where is the partial molar volume of species i

the specific heat capacity at

j i

j j

N

k ki i i

i i T p n

RV P P R

V n P n

VV V V

n

VP PC C T C TV

T T T

constant volume

2

, j

p

p P R

V n T

C TV

Variation of the pressure can be derived from total differential of volume

1,

1

,

1j

j

NiR

iNiP n p i

i

iR T R T

T n

dnV dTV

T dt dt dndP dTV

Vdt dt V dt

P

Summary of energy balance of BATCH reactor

0dP

dt ,

1

( )

NR

m P R R r k V k

R

k

dTc V V H r Q

dt

V f t

0RdV

dt

,

1

( )

NRp

m R V R r k k V k

k T

dTV c V H T V

d

P f

Qt

t

r

If liquid (condensed) systems

Rate of change Rate of heat generation Rate of heat

of reaction mixture by chemical reactions loss (input)

enthalpy

Heat flux :

0,p p Vc c ,

1

NR

m P R R r k V k

k

dTc V V H r Q

dt

2 1

2

the overall (global) heat transfer coefficient [W.m .K ]

the heat exchange area [m ]

temperature of external co

(

oling (heating) f

)

luid

H

H

e

eQ S T T

S

T

Isothermal

reactor

Adiabatic

reactor

Limiting cases

,

1

0NR

m P R R r k V k

k

dTc V V H r Q

dt

,

1

0NR

m p R R r k V k

k

dTQ c V V H r

dt

Example Adiabatic reactor with 1 reaction, constant heat capacities

Energy balance on adiabatic BATCH reactor

Molar balance of key component

m p R R r V

j jo

j j V

dTc V V H r

dt

dc dXc r

dt dt

1

1 1 1

1 1 1

,

Assuming that

( )

N

m R p i pi

i

N N No oi

m R p i pi i pi i j j pi

i i i j

o oN N Nj j j jo o

i pi i pi i pi p

i i ij j

pi p

o

j r o j

o

o o

j i pi j p

V n c y c

V c n y c n c n n X c

n X n Xn c c n c c

c const c const

y H T XT T

v y c y c X

1

1

( ) the adiabatice rise of temperature

o j jN

j

i

o

j r o

j No o

j i pi j p j

i

T X

y H T

v y c y c X

Trajectories of T(t) and Xj(t)

To

T(t)

Xj(t)

T(t)

Xj(t)

t

Xj = 1

1

( )

o

j r o

j No o

j i pi j p j

i

y H T

v y c y c X

Exothermal reaction

0r H

Homework 10

The reversible reaction

A1 + A2 A3

is carried out adiabatically in a constant-volume BATCH reactor. The

kinetic equation is

Initial conditions and thermodynamic data

Calculate X1(t),T(t).

1/2 1/2

1 2 3

3 1

1

5 1

2

(373 K)=2x10 s E 100 kJ/mol

(373 K)=3x10 s E 150 kJ/mol

f b

f

b

r k c c k c

k

k

3

1 1

3

2 2

298 3

0.1 mol/dm 25 J/mol/K

0.125 mol/dm 25 J/mol/K

40 kJ/mol 40 J/mol/K = 373K

o

p

o

p

o o

r p

c c

c c

H c 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

Data

In neglecting variation of heat capacities with temperature,

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

-1 -1 -1 1 3

1

1

0.3 mol.l , 3.8 kJ.kg .K , 209kJ.mol , =1070 kg.m

.S =200W.K 300 K, 0, ,

initial mass fractions, - mass heat capacities (kJ.

o

pM r

pio o

H e p pM i pMi pMi

i i

o

i pMi

c c H

cT T c c w c c

M

w c

-1 -1 1

1 1

kg .K ), - molar weight (kg.mol )

8.31446 J.mol .

iM

R K

46500

7 3 1

12.14 10 mol.m .minRTVr e c

0.00

0.20

0.40

0.60

0.80

1.00

1.20

304

306

308

310

312

314

316

0 10 20 30 40 50 60

t [min]

T [K] X1

By numerical integration we get

1

11

1 465000.1 209E3 2.14E7xexp - 300 (1 ) 200 (300-T)

(1070 3.8E3 0.1) (8.31446 )

465002.14E7 exp - (1 )

(8.31446 )

dTX

dt T

dXX

dt T

Continuous (perfectly) stirred reactor (CSTR)

1 1

N No o

R i i i i

i i

dH dPV F H F H Q

dt dt

1

1 1

1N

im P R R p i

i

N No o

R i i i i

i i

dndH dT dPc V V T H

dt dt dt dt

dPV F H F H Q

dt

,

1

, 1,NR

oii i R ki V k

k

dnF F V r i N

dt

Energy balance on CSTR

Total differential of enthalpy

Molar balance on CSTR

,

1 1

1 1

N NR

o

m P R p R i i i R ki V k

i k

N No o

i i i i

i i

dT dPc V TV H F F V r

dt dt

F H F H Q

We get

,

1

1

NR

m P R R p R r k V k

k

No o

i i i

i

dT dPc V V T V H r

dt dt

F H H Q

1

N

r k ki i

i

H H

Enthalpy

change for

k-th reaction

,

1 1

+NR N

o o

m P R R r k V k i i i

k i

dTc V V H r F H H Q

dt

The CSTR usually works at constant pressure (no pressure drop)

0dP

dt

( )T mQ S T T

Steady state

0idndT

dt dt

,

1 1

+ 0NR N

o o

R r k V k i i i

k i

V H r F H H Q

1

o

o

T

o o

i i i i pi

T

T

o

r k r k p kT

N

p ki piki

H H H H c dT

H H c dT

c c

Assuming ideal mixture, i.e. , we have i iH H

,

1 1

,

1

0

0, 1,

o o

T TNR No o

R r k p V k i pikk iT T

NRo

i i R ki V k

k

V H c dT r F c dT Q

F F V r i N

Molar and enthalpy

balances give

N+1 unknown

variables T, Fi

N+1 unknown variables in N+1 non linear algebraic equations

1 2, , ,.... 0, 1, 1i NG T F F F i N

Issues:

• multiple solutions

• slow convergence (divergence)

Example Adiabatic CSTR with 1 reaction, constant heat capacities

1

1

1

1

( ) ( )

0

( )

N No o o o o o o

R r p V i pi i pi

i i

o

j j R j V

o Nj jo o o o

r p i pi

o

j j

V

R j

o o

r j jo

No o

j

ij

i pi p j j

i

o

j j

F

V H c T T r F c T T F T T y c

F X V r

y XH c

Xr

V

H y XT T

y c c y X

T X

T T T T y c

cf. adiabatic const.-volume BATCH

2 nonlinear algebraic equations for

unknown T and Xj

0

0.2

0.4

0.6

0.8

1

310 330 350 370 390

Xj

T (K)

f2(T)

f1(T)

Steady state 1

Steady state 2

Steady state 3

Multiple steady states of adiabatic CSTR (exothermal reaction)

1 + ( , ) =0 (T) o

j j j V j R jF X r X T V X f

2 ( )o j j jT T X X f T

11 10.001exp[ ( )] min

298

10 kcal/mol

V CO

a

a

r kc

Ek

R T

E

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)

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.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 500 1000 1500 2000 2500 3000 3500 4000 4500

X1(T)

T/K

1

11

1 1

Adiabatic enthalpy balance

( )

( )

No o

i pi

i

o o o o

r p

y c T T

XH y c y T T

1 1

1 1

2

1

Molar balance of CO

(2 )( )( )

2 1

(2 ) 4

2

R

o

V P X Xk Tg T

RT F X

g gX

3 steady states

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:

rV = A.exp(-E/RT).cA1 (kmol/m3/s) Data

A = 5.1017 s-1

E = 132.3 kJ/mol

VR = 2 m3

To = 310 K

= 800 kg.m-3

3.33 l/soV

3

1 2 /o

Ac kmol m

,298 100 /rH kJ mol

-1 -14.19 kJ.kg . Ko

pC

z + z

VR + VR

z

VR

1R

N

i i

i V

F H

1

R R

N

i i

i V V

F H

v

R RV S z

Balance of enthalpy on Plug Flow Reactor (PFR)

1

1

1

1 1

in the steady state

0=

R R R R

R R R

N N

R i i i i

i iV V V V

N N

i i i i

i i

N

i i

iRV V RV

H dPV F H F H Q

t dt

dF H F H Q

dQF H

dV dV

RV

H

t

Q

,

1 1 1

,

1 1

,

1 1

ideal mixture ( , , ) ( )

( ) ( )( ) ( )

i i

N N NRi i i

i i ki V k i i

i i kR R R

NR N

r k V k i pi

k iR R

NR Ni

ki V k ki i

k iR

H T P composition h T

dF dh T dh T dTh T F r h T F

dV dV dT dV

dT dQH r Fc

dV dV

dFr h

dV

, 21

1

2

,

1

1

( )

1 4( )

4

4( )

4

tr k e

R R

NRt R

r k V k eNk RR R R

i pi

i

NRR

r k V k eNk R

i pi

i

dSdQH T T

dV dV

dS d dzdTH r T T

ddV d dVFc dz

ddTH r T T

dz dFc

Circular cross

section

2

2 2

4( )

4

. .( , ) ( , )

4 4

0, , 0

Rr V eo o

pM R

jj jR RV j V jo o

j j

o j

ddTH r T T

dz F c d

dX d dr X T r X T

dz F F

z T T X

One reaction, constant heat capacity of reaction mixture. Profiles of conversion and temperature are given by following

equations:

Limiting cases 1.Isothermal reactor

2.Adiabatic reactor

4 4

0 ( ) ( )r V e r V e

R

o

R

dTH r T T H r T T

dz

T T

d d

2 (

( )

)( )

4

o

j r

o jo

j p

o

j r jRr Vo o

M

o opM j pM

F H dXddTH r

dz F c dzF

T T

c

y HX

c

, ,

1 1

,

1

N N

i p i i p i

i i

No o o o

i p i pM

i

Fc Fy c

F y c F c

2

,

1

2 2

4( )

4

. .( , ) ( , )

4 4

0, , 0

Rr V e

oNRjo o

i p i p j

i j

jj jR RV j V jo o

j j

o j

ddTH r T T

dz dyF y c c X

dX d dr X T r X T

dz F F

z T T X

One reaction, constant heat capacity of species. Profiles of conversion and temperature are given by following

equations:

Limiting cases 1.Isothermal reactor

4 4

0 ( ) ( )r V e r V e

R

o

R

dTH r T T H r T T

dz

T T

d d

, ,

1 1

,

1

ji ii V

R j R

o

j j

j o

j

o o oi ii i j j j j

j j

N No oi

i p i p i i j j

i i j

oNjo o

i p i p j

i j

dFdFr

dV dV

F FX

F

F F F F F X

Fc c F F X

yF y c c X

2

, ,

1 1

,

1

,

1

( )( )

4

( )

( )o

o

j r jRr V

o oN Nj jo o o o

i p i p j j i p i p j

i ij j

o o o

j r p j

No o

j i p i j p j

i

Tj

No o ooj r pT

j i p i j

i

F H dXddTH r

dz dzy yF y c c X F y c c X

y H c T T dX

dzy c y c X

dXdT

y H c T Ty c y

,

1

0

( )

j

o o

j r jo

No o

j i

X

o

p

p

j

i j p j

i

y H XT T

y c y c X

c X

2.Adiabatic reactor

Cf. adiabatic BATCH and CSTR

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0

100

200

300

400

500

600

700

0.0 0.2 0.4 0.6 0.8 1.0

XSO2 [-] T [oC]

V [m3]

Conversion of SO2

Temperature

„hot“ point

Exercise: reactor for oxidation of SO2 to SO3

Numerical method: • Euler method „Stiff“ solvers MATLAB www.netlib.org www.athenavisual.com http://wxmaxima.sourceforge.net/wiki/index.php/Main_Page

Balance of mechanical energy in PFR Profile of overall pressure (P(z))

z=0

VR=0

z

VR

dR

(0)P ( )P z

2

2

(0) ( )

f R

f

R

P P z zv

d

dPv

dz d

Bernoulli equation

density of fluid(kg/m3) friction coefficient(-)

fluid mean velocity

f

(Re, / )w Rf d

v

Catalytic PFR Profile of pressure is calculated using Ergun equation:

22 2

1 22 3 3

1 1150 1.75

f fb o o o obf f f f f f

p b p b

dPv v A v A v

dz d d

f - fluid dynamic viscosity (Pa.s)

f - fluid density (kg/m3)

o

fv - superficial fluid mean velocity (m/s)

b - bed porosity (-)

pd - catalyst particle diameter (m)

2

2 2

2

1 2

4( )

4

. .( , ) ( , )

4 4

0, , 0,

Rr V eo o

pM R

jj jR RV j V jo o

j j

o o

f f f f

o j o

ddTH r T T

dz F c d

dX d dr X T r X T

dz F F

dPA v A v

dz

z T T X P P

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

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:

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)

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

1 2

4 1 3 1

( )

115148.9( ) 3.2 10 exp (mol )

Vr k T c c

k T m sRT

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:

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.

1 2

4 1 3 1

( )

115148.9( ) 3.2 10 exp (mol )

Vr k T c c

k T m sRT