4+-+Material+Balance+for+Chemical+Reactors

Chapter 4: Material Balance for Chemical Reactors

What do we need? Modeling and design of chemical reactors is based on

equation describing

mass transfer (Continuity equation)

heat transfer (Energy equation)

momentum transfer (Momentum equation)

chemical reaction (Rate equation)

These differential equations must be solved subject to the appropriate initial and boundary conditions

Aj

Akxy

z

Fixed coordinate

system

uj

uk

Mass transfer implies motion

u

xy

z

Moving coordinate

system

Average velocitiesLet u j be the velocity of Aj w.r.t. a fixed coordinate system

Local mass average velocity: u = ju j

j=1

n

jj=1

n

=

M jcju jj=1

n

(1)

Local molar average velocity: u* =cju j

j=1

n

cjj=1

n

=

cju jj=1

n

c

(2)

Diffusion velocities

To characterize the motion of species Aj relative to thelocal motion of the stream, we need to introduce new relative velocities:

Diffusion velocity w.r.t. u: u j u

Diffusion velocity w.r.t u*: u j u*

FluxesFluxes relative to mass average or molar average velocity:

j j= j u j u( ) (3) Units: kgm2 s

J j=c j u j u( ) (4) Units: molm2 s

j j* = j u j u

*( ) (5)J j

* =c j u j u*( ) (6)

FluxesFluxes relative to fixed coordinate system:

n j= ju j (7)

N j=c ju j (8)

Note that : u* =cju j

j=1

n

c

=N j

j=1

n

c

and u* =cjcu j

j=1

n

= x ju jj=1

n

How are these fluxes related?Ji

*=ciui ciu* = ciui

cic

cju jj=1

n

But,

Ni=ciui and xi =cic

Then

Ji* = Ni xi N j

j=1

n

(9)

Also: J j* = 0 ( Prove it! )

j=1

n

Diffusion in binary systems

Fick's First Law:JA

* =-cDABxA (10)

From eqn. (9) we have:JA

* = NA xA NA +NB( )Thus

NA = JA* + xA NA +NB( )

NA = -cDABxA + xA NA +NB( ) (11)

Things to rememberCareful!

NA = -cDABxAONLY when

NA +NB = 0

Question: Why JA* =-cDABxA

and not JA* =-DABcA ??

Things to remember

Diffusive fluxes are only defined relative to a convective flux or the total flux!

Diffusive and convective fluxes are NOT independent! They must sum to NA.

Ficks 1st Law has limitations. Everything changes for multicomponent systems!

Multicomponent systemsDiffusion in a multicomponent system of ideal gases is describedby the Stefan-Maxwell equations:

xi =1cDij

xiN j x jNi( )j=1

n

(12)For a binary system:

xA =1

cDAAxANA xANA( )+ 1cDAB

xANB xBNA( )cDABxA = xANB xBNA cDABxA = xANB 1 xA( )NA cDABxA = xANB + xANA NA NA = cDABxA + xA NB +NA( )

Multicomponent systemsFor some (but not all) systems, we may be able to define effective binary diffusivities Dim than do not depend on the other species. The Stefan-Maxwell equations then become:

xi =1

cDimxiN j x jNi( )

j=1

n

cDimxi = xiN j x jNi( )j=1

n

cDimxi = xi N j Nij=1

n

x jj=1

n

Ni = cDimxi + xi N jj=1

n

(13)

Conservation of mass

A k B


x

y

z

x

zyVolume element (xyz)

fixed in space

A k B


x

y

z

x


fixed in space

Amount of Aintroduced

per unit time

Amount of Aremoved

per unit time

Amount of Areacted perunit time

=

Amount of Aaccumulatedper unit time

A k B


x

y

z

x


fixed in space

Input of A across face at x: nAx x yz

Output of A across face at x + x: nAx x+x yz

A k B


x

y

z

x


fixed in space

Input of A across face at y: nAy y xz

Output of A across face at y + y: nAy y+y xz

A k B


x

y

z

x


fixed in space

Input of A across face at z: nAz z xy

Output of A across face at z + z: nAz z+z xy

A k B


x

y

z

x


fixed in space

Rate of production of A by chemical reactions: rA xyz

Accumulation of mass of A in volume element: At

xyz

A k B


Finally: At

+ nAxx

+nAyy

+nAzz

= rA

orAt

+nA = rA (14)

At

+AuA = rA (15)

Since this a binary mixture with 1 reaction:Bt

+BuB = rA (16)

Units?

Conservation of massBy adding (15) and (16) and noting that

rA + rB = 0

from the conservation of mass, we finally obtain:

t

+u = 0 (17)

Eulerian form of continuity equation


If we work with moles:cAt

+NA = RA* (18)

cBt

+NB = RB* (19)

Add (18) and (19) to getct

+cu* = RA* + RB

* (20)

since NA +NB = cAuA + cBuB = cu

*

??


Using equation (11) [ Fick's Law]:

cAt

+ cDABxA + xA NA +NB( ) = RA*

cAt

+ cDABxA( )+ xAcu*( ) = RA*

cAt

+cAu* + cDABxA( ) = RA* (21)

Conservation of massNote that: nA = cDABxA + gA nA + nB( )

nA = cDABxA + gAunA = cDABxA + Au (22)

From (22) and (14)At

+Au+ DABgA( ) = rA (23)And since jA = DABgA (24)

At

+Au+ jA = rA (25)

Dividing by MA we obtaincAt

+cAu+JA = RA* (26)

Multicomponent systemscjt

+cju+J j = Rj* (27)

And since J j = Djmcj (28)

cjt

+ cjux( )x

+ cjuy( )y

+ cjuz( )z

=

= x

Djmcjx

+ y

Djmcjy

+ z

Djmcjz

+ Rj

* (29)

Assumptions The only contribution to mass flux Ji comes from

concentration gradients.

Additional contributions may come from

pressure gradients (centrifuge);

external forces acting unequally on different species (ionic solutions);

temperature gradients (thermal diffusion or Soret" effects).!These are usually small but can be enhanced by steep temperature gradients

Assumptions

The continuity equation assumes perfectly ordered flow.

Turbulent flow introduces an additional flux that can also be expressed as proportional to the concentration gradient - eddy diffusivity.

The two fluxes are summed to obtain an effective diffusivity.

When we have highly turbulent flow, the molecular diffusion is negligible and the effective diffusivity is the same for all species.

Simplifications

Tubular Reactors

Empty tubes

Flow in one direction

Under isothermal conditions, major gradients exist only in the axial direction

Then, we may use average values of variables over the cross section of the reactor: !!

= 1

d

Tubular reactorsAssuming

1. turbulent flow and2. constant diffusivity

cjt

+ cjuz( )z

= Dj ,m2cjz2

+ Rj* (1)

Additional assumptions:1. Steady state2. Diffusion term is negligible

d cju( )dz

= Rj* (2)

Tubular reactors

Let Ac be the constant cross sectional area of the tube. Then:Volumetric flow rate: Q = uAc

d cju( )dz

=d cjQ( )Acdz

=d cjQ( )

dV= Rj

* (4.66)

and since cjQ = N j :dN jdV

= Rj* (4.67)

Units?

Gas phase reactionsIdeal Gas Law:

cjj

! = PRT "N j

j!

Q=

PRT

"

Q = RTP

N jj

! (4.69)

cj =N jQ

=P

RTN j

N jj

!

#

$

%%

&

((

(4.70)

Liquid phase reactionsTotal mass flow: M = N j M j

j!

M = ! Q (4.71)

(4.67) " M jdN jdV

= M j Rj* "

d M j N j( )dV

= M j Rj*

Sum over j:

d M j N jj

!#

$%&

(

dV= M j Rj

*

j! "

dMdV

= 0 and M 0( ) = M feed

Liquid phase reactionsTotal mass flow is constant w.r.t. axial position!

Chemical reaction cannot change mass flow!

Mass flow rate: Q =M f!

(4.72)

If the liquid density is constant: ! z( ) = ! fQ = Qf (4.73)

Time spent in reactor: " =V Q

If Q is constant:d cjQ( )

dV= Q

dcjdV

=dcjd!

= Rj* (4.74)

Well-mixed reactors

VRj

Q1cj1

Q2cj 2

S2

S1

Well-mixed reactorsStart with the continuity equation:

! cj! t

+ " #cju + " #J j = Rj*

Concentrations are uniform in well-mixed reactor!

Integrate over the entire volume of the reactor:

! cj! t

dVV$ = % " #cju( )dV

V$ % " #J j( )dV

V$ + Rj* dV

V$

From Gauss divergence theorem:

ddt

cj dVV$ % cjumn dS

S$ = % cjun dS% j in dS

S$

S$ + Rj* dV

V$

Surface of control volume may move with time

Well-mixed reactorsd cjV( )

dt= ! cj un ! umn( )dS

S! " j in dS

S! + Rj* dV

V! =

= cj1u1S1 " cj 2u2S2( ) ! j in dSS1+S2

" + Rj*V

where

S1,S2: open end surfaces of control volume

u1,u2: average velocities over cross sectional areas

(1)

The integral term (2) is usually negligible compared to term (1).

(2) (3)

Well-mixed reactorsd cjV( )

dt= cj1u1S1 ! cj 2u2S2( )+ Rj*V

Batch reactor: No fluid enters or leaves the reactor

d cjV( )dt

= Rj*V (4.5)

CSTR: cj1u1S1 = cj , fQf Inlet molar flow rate

cj 2u2S2 = cjQj Outlet molar flow rate

d cjV( )dt

= cj , fQf ! cjQj + Rj*V (4.36)

4+-+Material+Balance+for+Chemical+Reactors

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