Plug Flow Reactor Non Isothermal

8/10/2019 Plug Flow Reactor Non Isothermal

1/33


2/33

Plug flow is a simplified and idealized picture of the

motion of a fluid, whereby all the fluid elementsmove with a uniform velocity along parallelstreamlines.

This perfectly ordered flow is the only transportmechanism accounted for in the plug flow reactormodel.

Because of the uniformity of conditions in a crosssection the steady-state continuity equation is a verysimple ordinary differential equation.


3/33

z z + zz = 0

z

z = L

z


4/33

MASS BALANCE

VAF

VVAFVrA

0FVrFVVAAVA

dtdN A

VrFFAVAVVA

Rate of flowof A into avolumeelement

Rate of flowof A out ofthe volumeelement

Rate ofgeneration of A bychemical reactionwithin the volumeelement

Rate ofaccumulationof A withinthe volumeelement

+ =

For steady-state process: 0dt

dN A


5/33

Arz

FFA

zAzzA

0zlim

AVAVVA

rV

FF

AA r

dVdF

VrFF AVAVVA

By definition of the conversion X1FF 0AA

dXFdF 0AA


6/33

So that the continuity for A becomes:

dVrdXF A0A

A0A

rF

dXdV

(1.1)

Design equation


7/33

To design an isothermal tubular/plug-flow reactor, thefollowing information is needed:

1. Design equation

2. Rate law

(for first order reaction)

3. Stoichiometry (liquid phase)

A0A

rF

dXdV

AA Ckr

X1CC 0AA

(1.1)

(1.2)

(1.3)


8/33

Combining eqs. (1.2) and (1.3) yields:

X1Ckr 0AA

Introducing eq. (1.4) into eq. (1.1) yields:

XX

0A

X

X 0A0A0

0

X1lnCk 1X1kC dXFV

X1X1

lnkCF

X1X1

lnkCF

V 0

0A

0A

00A

0A

(1.4)

X1

X1ln

k

v

X1

X1ln

k

vV 00

0

0


9/33


10/33

T1

T1

RE

expkk1

1

(1.5)

(1.6)


(1.7)

Combining eqs. (1.1), (1.3), and (1.4) yields:

00A

0A

v X1kF X1kCdVdX

Recalling the Arrhenius equation:

T1

T1

TE

expX1vk

dVdX

10

1


11/33


12/33


13/33

For an open system in which some of the energy exchangeis brought about by the flow of mass across the system

boundaries, the energy balance for the case of only onespecies entering and leaving becomes:

Rate of

accumu-lation ofenergy

within thesystem

+ =

Rate of flowof heat tothe systemfrom the

surrounding

Rate workdone by thesystem on

thesurrounding

Rate of

energy addedto the systemby mass flow

into thesystem

Rate energyleaving

system bymass flow outof the system

outoutininsys EFEFWQ

dt

dE (1.9)


14/33

The unsteady-state energy balance for an open system

that has n species, each entering and leaving the system atits respective molar flow rate F i (mole of i per time) andwith its respective energy E i (joules per mole of i), is:

out

n

1iii

in

n

1iii

sys FEFEWQ dt

dE (1.10)


15/33

It is customary to separate the work term, , into:

flow work: work that is necessary to get the mass intoand out of the system

other work / shaft work, .

For example, when shear stresses are absent:

W

SW

S

out

n

1i

ii

in

n

1i

ii WPVFPVFW

[rate of flow work]

(1.11)

where P is the pressure and V i is the specific volume.

Stirrer in a CSTR Turbine in a PFR


16/33


outn

1iiii

in

n

1iiiiS

sys

PVEFPVEFWQ dt

dE

(1.12)

The energy E i is the sum of the internal energy (U i), thekinetic energy , the potential energy (gzi), and any

other energies, such as electric energy or light: 2u 2i

othergz2u

UE i2i

ii (1.13)

In almost all chemical reactor situations, the Kinetic,potential, and other energy terms are negligible incomparison with the enthalpy, heat transfer:

ii UE (1.14)


17/33

Recall the definition of enthalpy:

iii PVUH (1.15)

Combining eqs. (1.16), (1.15), and (1.13) yields:

out

n

1iii

in

n

1iiiS

sys HFHFWQ dt

dE (1.16)

We shall let the subscript 0 represent the inletconditions. The un-subscripted variables represent the

conditions at the outlet of the chosen system volume.

n

1iii

n

1i0i0iS

sys HFHFWQ dt

dE(1.17)


18/33

The steady-state energy balance is obtained by setting(dEsys/dt) equal to zero in eq. (1.17) in order to yield:

0HFHFWQ n

1iii

n

1i0i0iS (1.18)

To carry out the manipulations to write eq. (1.18) interms of the heat of reaction we shall use the

generalized reaction:

(1.19)DdCcBbA


19/33

The inlet and outlet terms in Equation (1.19) areexpanded, respectively, to:

0I0I0D0D0C0C0B0B0A0A0i0i FHFHFHFHFHFH

IIDDCCBBAAii FHFHFHFHFHFH

In:

Out:

(1.20)

(1.21)

We first express the molar flow rates in terms of conversion

X1FF 0AA

XbFFFXFbFF

0A

0B0A0A0BB

XbFF B0AB (1.23)

(1.22)


20/33

XcF

FFXFcFF

0A

0C0A0A0CC

XcFF C0AC (1.24)

XdFF D0AD (1.25)

I0A0A

0I0A0II FF

FFFF (1.26)


21/33


22/33

The term in parentheses that is multiplied by F A0X is calledthe heat of reaction at temperature T and is designated

HRx.

THTHbTHcTHdH ABCDRx (1.28)

All of the enthalpies (e.g., H A, HB) are evaluated at thetemperature at the outlet of the system volume, andconsequently, [ HRx(T)] is the heat of reaction at thespecific temperature ip: The heat of reaction is alwaysgiven per mole of the species that is the basis ofcalculation [i.e., species A (joules per mole of A reacted)].


23/33

Substituting eq. (1. 28) into (1. 27) and reverting to

summation notation for the species, eq. (1. 28) becomes

XFHHHFFHFH 0ARxn

1ii0ii0A

n

1iii

n

1i0i0i (1.29)

Substituting eq. (1.29) into (1.18) yields:

0XFHHHFWQ 0ARx

n

1i i0ii0AS (1.30)


24/33

The enthalpy changes on mixing so that the partial molalenthalpies are equal to the molal enthalpies of the purecomponents.

The molal enthalpy of species i at a particular temperatureand pressure, H i, is usually expressed in terms of anenthalpy of formation of species i at some referencetemperature T R, Hi (TR), plus the change in enthalpy that

results when the temperature is raised from the referencetemperature to some temperature T, H Qi

QiR0ii HTHH (1.31)


25/33

The reference temperature at which H i is given is usually25C. For any substance i that is being heated from T 1 to T 2 inthe absence of phase change

2

1

T

TPQi dTCH (1.32)

A large number of chemical reactions carried out inindustry do not involve phase change. Consequently, weshall further refine our energy balance to apply to single-phase chemical reactions. Under these conditions the

enthalpy of species i at temperature T is related to theenthalpy of formation at the reference temperature T R by

T

T

piR0ii

R

dTCTHH (1.33)


26/33


27/33

The heat of reaction at temperature T is given in eq. (1.28):

THTHbTHcTHdH ABCDRx (1.28)

where the enthalpy of each species is given by eq. (1.33):

T

TpiR

0ii

R

dTCTHH (1.33)

If we now substitute for the enthalpy of each species, we have

R0AR

0BR

0CR

0DRx THTbHTcHTdHH

T

T

pApBpCpDR

dTCbCcCdC (1.37)


28/33

The first set of terms on the right-hand side of eq. (1.37) isthe heat of reaction at the reference temperature T R ,

R0AR

0BR

0CR

0DR

0Rx THTHbTHcTHdTH (1.38)

The second term in brackets on the right-hand side of eq.

(1.37) is the overall change in the heat capacity per mole of Areacted, C p,

pApBpCpDp CbCcCdCC (1.39)

Combining Equations (1.38), (1.39), and (1.37) gives us

T

TpR

0RxRx

R

dTCTHTH (1.40)


29/33

The heat flow to the reactor, Q , is given in terms of the overallheat-transfer coefficient, U, the heat-exchange area, A, andthe difference between the ambient temperature, T a, and thereaction temperature, T.

When the heat flow vanes along the length of the reactor,such as the case in a tubular flow reactor, we must integratethe heat flux equation along the length of the reactor toobtain the total heat added to the reactor,

V

0a

A

0a dVTTUadATTUQ

where a is the heat-exchange area per unit volume of reactor.

(1.41)


30/33

The variation in heat added along the reactor length (i.e.,volume) is found by differentiating with respect to V :

TTUadVQ d

a (1.42)

For a tubular reactor of diameter D, a = D/4For a packed-bed reactor, we can write eq. (1.43) in terms of

catalyst weight by simply dividing by the bulk catalystdensity

TTUa

dV

Q d1a

BB

(1.43)

Recalling dW = B dV, then

TTUa

dW

Q da

B


31/33

Substituting eq. (1.40) into eq. (1.36), the steady-state energy

balance becomes

0XFdTCTHdTCFWQ 0AT

TpR

0Rx

n

1i

T

Tpii0AS

R0i

(1.44)

For constant of mean heat capacity:

n

1i0ipii0A0ARpR

0RxS TTC

FXFTTC

THWQ

(1.45)


32/33

EXAMPLE 1.1Calculate the heat of reaction for the synthesis of ammoniafrom hydrogen and nitrogen at 150C in kcal/mol of N

2reacted.

SOLUTION

Reaction: N2

+ 3H2 2NH

3

R0NR

0HR

0NHR

0Rx THTH3TH2TH 223

= 2 ( 11.02) 3 (0) 0 = 20.04 kcal/mol N 2

K.Hmolcal992.6C 2p2H

K.Nmolcal984.6C 2p2N

K.NHmolcal92.8C 3p 3NH


33/33

Documents

Plug Flow Reactor Non Isothermal