NON-ISOTHERMAL REACTOR DESIGN

of
>process
where temperature a system changed as it
progresses .

> to describe such a

system ,
heat balance/energy balanced is required to solve a non-

isothermal system .

> Consider a non-isothermal PFR :

A> B

The design equation :

-
CAo-CA dX -
-

A ①
SA dV FAo

Let the reaction is first order :

-

A >B
-

rA =
k CA ②
↑
where
changes of K is
governed by PFR :

Arkennius
eq
.

dX ①
As E /Re A
-
* I

59
↳ =

E [I - dV FA
=Ar ,
⑭

Assuming rate of reaction to be first order ;

From conversion :

X =
270-CA

CA0 rA =
KCA &

CA =
CAL1-X) ⑤
-

rA =

kCA(1-x) ⑥ Conversion ;

Substitute & into ⑭ :

X =

CA0-CA
-

ra
=
As
*
I- 5 c- ( -

x) ⑦
CA
CA =
CA01 -x)
Substitute ⑦ into ① : dX = -
kCA(1-X)
)
ArEa[
-

dV FAS
(1-x)
,

dX ny
Arhenius
=

W
dV 0
Vo - -

kA011-X) equation ;

where : Nt *RT
vo =

Go (mom3) = mol/s -
=
k(1 -

x)
= As G

Fo (m/s) Vo - "It

The final
equation :

[ ( -

5)5
dX -Ae 11 x)
-

Energy balance dV
equation :

conversion
Vo
initial
X temp /
W

7 :

To + 1 AAR) X -

⑥
I

CPA
Heat
capacity

IN > > OUT L

when the rate constant

Fr
2
changes the rate also
->

I
↳

change ->
change the
as volume of reactant conversion of the
changes along the length reaction .

of the tube >T also changed

↳ when T
changes , V
changes -

the rate constant also change .

 * steps to tackle non-is othermal

he I calculate -
given Calculate Determine Determine the
7
+ rate constantis FAo/VA

EXAMPLE

Calculate the heat of reaction for the synthesis of NAs from hydrogen and
nitrogen at 150 % in

kcal/mol of Nagas reacted and k3/mol of

in
hydrogen reacted -

N2 + 3 He - 2 NH3
① What ?
is
required
Heat of formation

HNH3 : -
11020 Keal/mol NHs 25°
HH2 = 0

HN2 =
0

② Equation needed

Atm =
2AR (product) -

2A (reactants)

③ Heat capacity :

CPA2 :

6 992
.
cal/mol H2 K .

(pNa 6 984 cal(molNc K

:
.
.

(PNAs :
8 .
92 cal/moINAs .
K

AGp =2Cp(product) 2CpCreactants)

-

Determine the enthalpy change of reaction :

A AR 2 HNHs -[3HN2 + HN2 J

=

ne

2- 11028) -
0
=-22040 cal/mol N2

Determine the min , heat

capacity :

&
(p 2
CPNHs- 13 CPNH +(pHc)
=

=-10 12 .

cal/molNe . K

The heat of reaction at 150° [

:
reaction
reaction initial
AHR (423) AHR (298) +
DCpLT- Top
=

- -
22040 + 1-10-12) (423-298)
--23 310 cal/mol Na
,

= -
97 5 -
15/mol Ne

For mol Ne consumed the heat -97

54]A
is
every
.

For consumed
hydrogen ,

>
=> 97 5k] -

X
/N2 =- 32 -

5 k3/mol H2
#
Ne 3 mol He
NON-ISOTHERMAL REACTORS

ENERGY BALANCES FOR

COMMON REACTORS

Starts from the First law of

Thermodynamics

dE =

Sc-SW < 8 -> not exact differential

OR of the state function

dE -
dQ- dWa work done
1
E by the
changes
total
in Heat flow
into the
system on the
energy surrounding
of the system system

dEsys= -i +FinEin-Foutout Rate of energy diff

dt ~ of the
by mass system
flow (in and out)

Eitoof
component
↑ n

dEsys= G-y+
L

In Es Fit
*

i
It put

upon substituting with ;

Hi =
Ui+ PV ;

E R I
&+ Z T
dEsys=
*

Fictio -
ilti
=1
M4
*

shatn
in

represent inlet

stirring turbine t &

Pump

compressor

Consider the
system at
steady state :
(CAs become of

dEsys =
0

At

c -us + Fictio-Filti= 0
if "*
Heat of reaction at
Enthalpy change of
species involved temp ,
I
in the reaction Atte (x) T
=>

Creactants products FNOXBHRXCT)

=>

, ,

inerts]
=>

Fr 6 (Hio-Hi)
it
&
Fio
Fo

-
8-Ws+FA26 (Hio-Hi)-AArxCt)FA0X =
0

E
5
Initial enthalpy
i Hi (T) =
Nox
1
Ped
 I-is-FA
20Cp :

[TioT-AAnxt) Frox =
8

AARX CTR) + ACp CT-TR3

By neglecting the shaft work ,

is :

a IT-Tio) -[AHRx (In) (T-tn) J FroX

-FAziCp : +ACp =
0

For adiabatic
an
system :

Q =

Upon rearrangement ,

XEz =

20:Cp: (T-ToS
-
[AHx (TR) + ACp <T-Tref J
DHRx(25°)
commonly used
Ex :

X
I
> inert

i XEB <CPA+ 01(pe) (T-To)

I equation
=

⑧
-

BHRx LTR>

Can also obtain the temperature at the reactor outlet

;

[ BHRx (TR) / + 20
7 x
:Gi To +XACpTR
= -

20 :
Gi+XDG
used
t
X commonly

i T =
To + I- AHRx (TR) JX

(px +

01CpI
I equation

Mance
/
IMB - GA0-CA
CA

r =

kCA
=KCA
kTr)e- )
A
/

k =

CONVERSION

I I
Liquid Gas

CA =
CA01 x) -

i
(A =

<A0(ix)" (I)
↑
/
;*
48/aaactants
 ENERGY BALANCE WITH REACTION + HEAT EXCHANGER * Refer Autorial 1

Reactor

Batch Continuos/Fed-Batch
?
Feed temp feed exit ↓
da=
&Hi FtCp(Tf-Te) FAAR(x-xe) MeGdT +AHR(rV)
+
=

E
MtCpdT +AAR(rV)
It
Exit temp
↳ Ah CTs-T) dt
=>
hoAhlTs-T) dt =

mtCpCTs-7) +AAR(rV) At At non-steady state

,

MtCp &T =

FtCpCTf-te) +UoAh(Ts-Te)-AAm(GyCf)
At
-

0 Lat S -5) (rV)

->

YoAhITe-Ts) =-
AAR(QfCf) +
FtCp (Tf-Te)
reason for NOT the rate reaction
considering of -
:

=>
when
temp control is CRITICAL the reaction rate is often
,

very FAST with

respect to the rate of heat transfer
->
the factor that determine the design are (i) the rate of energy
with the (ii) the feed rate
exchange surroundings
only energy balance is
required I!
=>

=> the reaction may be assumed to be at

thermodynamics equilibrium
the rate
expression and mass balance are NOT
required

From
example

UoAh (Te-Ts)=-AHR (GOfeed) NIs +

FECp(Ifeed-Te)
of Mass

Given that :

BHR=-2 .

23x103/kg4
In term of mol :
-

product
=

N4(CH2)6
Relative =140
Atr 23x1063 molecular mass
=-2 -

x 140
Eg 4 mol sbbnal I molshy
*

AHR =-78 050k3/kmol ,

14
:An
=

686x10-3 0 1146614 19)(25-100)

-)(1 -
+
- .

0 .

483(100 25) -

The total mass of components at the end of the reaction

->

mass of NHs fed + mass of formalin already in the reactor

NH3
(x10m3
[0 11466
91 3 : min
605] + 0 901m 1 10
3
I /
I
1619 248
=

X
#
x x
- -

x
~
=> .

x
.
.

~ 1000g
-