08 energy balance

8/11/2019 08 Energy Balance

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Energy balance for chemical reactors

Gabriele Pannocchia

First Year course, MS in Chemical Engineering, University of Pisa Academic Year 20132014

Department of Civil and Industrial Engineering (DICI)University of PisaItaly

Email: [email protected]

Gabriele Pannocchia Energy balance for chemical reactors 1 / 45


2/45

Outline

1 Introduction and general balance

2 Batch reactor

3 CSTRDynamic operationSteady-state operationSteady-state multiplicity and stability

4 PFREnergy balance in transient conditionsSteady-state energy balance

PFR hot spot and runaway5 Optimized reactor arrangements

6 Summary of balance equations



3/45

General energy balance

Q

R j E m 1, E 1, c j 1m 0, E 0, c j 0

W

Arbitrary reactor volume V

Inlet: m 0 (mass rate), c j 0 (concentration of species j ), E 0 (per-mass energy)

Outlet: m 1 (mass rate), c j 1 (concentration of species j ), E 1 (per-mass energy)



4/45

General energy balance (continued)

Conservation of energy for system

rate of energy accumulation

= rate of energy entering by inow

rate of energy

entering by outow +

rate of heatadded to system

+ rate of work added to system

Balance rewritten in terms of dened variablesdE d t

= m 0 E 0 m 1 E 1 + Q + W

The total energy E is composed by several contributions, e.g. internal energy,kinetic energy, potential energy, and other types of energy:

E = U + K + +

In most chemical reactors, only changes in internal energy are relevant



5/45

Work terms

Work terms: overview

The total work W can be split into three parts: W f : the work done by ow streams while moving material into and out of thereactor volume W s : the shaft work done by stirrers, compressors, etc. W b : the work done by the reactor volume boundaries

Flow streams work termIf we assume that the ow streams enter (or leave) the reactor volume atuniform velocity, we can express W f as follows:

W f = v 0 A 0p 0 v 1 A 1p 1

where v 0 and p 0 are the inlet ow velocity and pressure, A 0 is the bounding area; likewise for v 1 , p 1 , A 1Let 0 and 1 be the densities of inlet and outlet ows respectively

We obtain: W f = Q 0p 0 Q 1p 1 = m 0p 0

0 m 1

p 1

1Gabriele Pannocchia Energy balance for chemical reactors 5 / 45


6/45

General energy balance in nal form

Enthalpy

Recall the enthalpy denition H = U + pV or per-mass basis H = U + p

Overall balance Assume that E = U + K + , we obtain:

d d t

(U + K + ) = m 0( H 0 + K 0 + 0) m 1( H 1 + K 1 + 1) + Q + W s + W b

Neglecting kinetic and potential energy terms leads todU d t

= m 0 H 0 m 1 H 1 + Q + W s + W b



7/45

Batch reactor: energy balance and derivations

The general balance

The energy balance for the batch reactor is obtained from the general balance written over the entire reaction volume ( V = V R )

Deleting the terms associated with ow streams (not present in the batchreactor) leads to: d

d t (U + K + ) = Q + W s + W b

The rate of work done by the boundaries is given by: W b = p dV R d t

Minor simplicationsUsually kinetic and potential energy terms can be neglected

Moreover, unless the reactor is treating a highly viscous mixture the rate of shaft work can be also neglected

Recall that H = U + pV R , and hence d H = dU + pdV R + V R dp , to obtain:

dU d t

+ p dV R d t

= Q d H d t

V R d p d t

= Q (1)



8/45

Manipulating the change of enthalpy

The change of enthalpy for single phase reactor

We express the change of enthalpy using ( T , p , n j ) as independent variables:

d H =H T p ,n j

dT +H p T ,n j

d p + j

H n j T ,p ,n k = j

dn j (2)

The simple termsThe rst term is by denition the total heat capacity at constant pressure:

H

T p ,n j = C p = V R C p (3)

with c P the specic heat capacity of the mixture at constant pressure

The third term is also simple because H n j T ,p ,n k = j is the molar enthalpy of

species j , denoted by H j



9/45

Manipulating the change of enthalpy (cont.)

Useful thermodynamic relations

dG = SdT + V R d p + j

j dn j (4)

d H = dG + T dS + SdT = T dS + V R dp + j

j dn j (5)

S

p T ,n j =

p T ,n j

G

T p ,n j =

T p ,n j

G

p T ,n j =

V R

T p ,n j (6)

Partial derivative of enthalpy w.r.t. pressureFrom (5) and (6) we can write:

H p T ,n j

= T S p T ,n j

+ V R = T V R T p ,n j

+ V R = V R (1 T ) (7)

= 1V R V R T p ,n j

is the isobaric coefcient of expansion of the mixture



10/45

Manipulating the change of enthalpy (cont.)

Collecting the three terms of (2)d H d t

= V R C p dT d t

+ V R (1 )d p d t

+ j

H j dn j d t

(8)

The reaction termThe reaction term appearing in (8) can be further simplied by recalling thematerial balance:

dn j d t

= R j V R =i

i j r i V R (9)

We then can write:

j H j

dn j d t

= j

H j i

i j r i V R =i j

i j H j r i V R =i

H Ri r i V R (10)

with H Ri = j i j H j the molar heat of reaction i



11/45

Final energy balances for the batch reactor

General energy balance for a batch reactorFrom (1), (10) and (8) we obtain:

V R C p dT d t

T V R dp d t

=i

( H Ri )r i V R + Q (11)

Incompressible uid or constant pressure reactorIf the mixture is incompressible, i.e. = 0, or the reactor operates at constantpressure, the balance becomes:

V R C p dT d t

=i

( H Ri )r i V R + Q (12)



12/45

Energy balance for compressible mixtures

Constant volume reactor with compressible mixture: introduction

For gas-phase mixtures, in constant volume batch reactors, it is possible torelate d p d t to the change of temperature and moles due to reaction

Derivation

Express the term dp d t

as follows (assuming constant V R ):d p d t

=p T V R ,n j

dT d t

+ j

p n j T ,V R ,n k = j

dn j d t

(13)

Inserting (13) into (11), recalling the material balance (9), we obtain:

V R C p T V R p T V R ,n j

dT d t

=i

[( H Ri )+

T j

i j p n j T ,V R ,n k = j

r i V R + Q



13/45

Energy balance for compressible mixtures (cont.)

Final result

Observe that V R C p T V R p T V R ,n j is the constant volume heat capacity,

which can be written as V R c V , and obtain:

V R C V dT

d t =

i ( H Ri ) + T V R

j i j

p

n j T ,V R ,n k = j r i V R + Q (14)

Ideal gas mixtureObserve that

T =

1,

p

n j T ,V R ,n k = j =

RT

V R

Obtain:

V R C V dT d t

=i

( H Ri + RT i )r i V R + Q (15)



14/45

Simple reaction in adiabatic conditions

Single reaction

Consider a single reaction: A

k B, with r = k (T )c n A and k (T ) = k 0 exp( E a /( RT ))

Dene x A = n A 0 n A

n A 0 , and write material and energy balance (12):

n A 0d x A

d t = r V R

V R C p dT d t

= ( H R )rV R

Obtain a relation between T and x A :dT

d x A =

( H R )n A 0V R C p

When V R is constant = ( H R )c A 0

C p is the adiabatic temperature rise, to obtain:

T = x A T = T 0 + x A



15/45


16/45

CSTR in dynamic operation: single phase system

Expanding the enthalpy change term As in the batch reactor we can write [see (8)]

d H d t

= V R C p dT d t

+ V R (1 T )d p d t

+ j

H j dn j d t

(17)

Recall the material balance:

dn j d t

= Q 0c j 0 Qc j +i

i j r i V R (18)

Combining (16), (17) and (18)V R C p dT d t + V R (1 T )

dp d t + j

H j [Q 0c j 0 Qc j + i i j r i V R ] V R dp d t = Q 0 0 H 0 Q H + Q



17/45

CSTR in dynamic operation: single phase system (cont.)

Useful observationsRecall that:

j i i j H j = i j i j H j = i

H Ri

Notice that: Q 0 0 H 0 = j

Q 0c j 0 H j 0 j

Qc j H j = Q H

Final balance

V R C p dT d t

V R T d p d t

=i

( H Ri )r i V R + Q 0 j

c j 0( H j 0 H j ) + Q (19)

Reasonable approximation for liquid mixturesThe heat capacity is reasonably constant with temperature and = 0. Hence:H j 0 H j = C p j (T 0 T ) j c j 0( H j 0 H j ) = 0 C p (T 0 T ):

V R C p dT d t

=i

( H Ri )r i V R + Q 0 0 C p (T 0 T ) + Q (20)



18/45

Steady-state operation

General steady-state material and energy balances

The steady-state energy balance comes from (19) by setting dT d t = 0 and dp d t = 0

In general material and energy balances must be solved together:

Q 0c j 0 Qc j +

i

i j r i V R = 0 j = 1,. . . , n s (21)

i ( H Ri )r i V R + Q 0

j c j 0( H j 0 H j ) + Q = 0 (22)

Steady-state material and energy balances for liquid mixtures

Q 0(c j 0 c j ) +i

i j r i V R = 0 j = 1,. . . , n s (23)

i ( H Ri )r i V R + Q 0 0 C p (T 0 T ) + Q = 0 (24)



19/45

Simple reaction in adiabatic conditions

Single reaction

Consider a single reaction: A k B, with r = k (T )c n A and k (T ) = k 0 exp( E a /( RT ))

Dene x A = N A 0 N A

N A 0 , and write material and energy balance (12):

Q 0c A 0x A = r V R

Q 0 C p (T T 0) = ( H R )r V R

Obtain a relation between T and x A :

T T 0 =( H R )c A 0

C p x A

Let = ( H R )c A 0

C p , observe that in general:

T = x A


d l l


20/45

Steady-state multiplicity

IntroductionThe coupling of energy and material balances can result in a surprisingly complex behavior even for simple kinetic schemes

This is true even in steady-state operation, in particular because there may exist multiple steady states

ExampleConsider a rst order reaction:

A k B, r = k (T )c A , k (T ) = k 0 exp( E a /( RT ))

Adiabatic conditions in liquid phase. Recall that = V R / Q 0 and x A = c A 0 c A

c A 0 :

x A =k

1 + k (25)

C p

(T T 0) = k (1 x A )c A 0( H R ) (26)

Examine ( x A ,T ) vs. for different H R Gabriele Pannocchia Energy balance for chemical reactors 20 / 45

S d l i li i l


21/45

Steady-state multiplicity: example

Temperature vs. residence time

260

280

300

320

340

360

380

400

420

440

460

1 10 100 1000 10000 100000

T ( K )

(min)

H R = 300 H R = 200 H R = 100

H R = 50 H R = 0

H R = 50


S d l i li i l ( )


22/45

Steady-state multiplicity: example (cont.)

Conversion vs. residence time

0

0.2

0.4

0.6

0.8

1

1 10 100 1000 10000 100000

x A

(min)

H R = 300 H R = 200 H R = 100

H R = 50 H R = 0

H R = 50


St d t t lti li it i iti d ti ti i t


23/45

Steady-state multiplicity: ignition and extinction points

Temperature vs. residence time for H R = 300kJ/mol

280

300

320

340

360

380

400

420

440

460

0 5 10 15 20 25 30 35 40

T ( K )

(min)

A

E

B

D

F

C

G


Stead state stabilit : heat generation and remo al


24/45

Steady-state stability: heat generation and removal

Heat generation and removalIn the energy balance (26), we replace x A from (25) to obtain:

C p

(T T 0)

Q r =

k 1 + k

c A 0( H R )

Q g Q r is the heat removal rate, associated to the different between output andinput stream enthalpies

Q g is the heat generation rate, associated with the reaction rate

Necessary condition for stability

An equilibrium is said to be stable if the system returns to that steady stateafter any sufciently small perturbation

A necessary condition for stability is:

d Q r dT

>d Q g dT


Steady state stability: heat generation and removal (cont )


25/45

Steady-state stability: heat generation and removal (cont.)

Heat generation and removal for = 1.79 min

-100

0

100

200

300

400

500

300 350 400 450 500

H e a t ( M J / m

3 m i n )

T (K)

A

E (extinction)

RemovalGeneration


Steady state stability: heat generation and removal (cont )


26/45

Steady-state stability: heat generation and removal (cont.)

Heat generation and removal for = 15 min

-10

0

10

20

30

40

50

60

300 350 400 450 500

H e a t ( M J / m

3 m i n )

T (K)

B

D (unstable)

F

RemovalGeneration



27/45

PFR: energy balance in transient conditions


28/45

PFR: energy balance in transient conditions

PFR scheme and thin disk volume element

q (z ) z D

Q (z + z )

c j (z + z ) V R j

Q (z )

c j (z )

Q , c j , H Q 0, c j 0, H 0

H (z ) H (z + d z )

q (z )

Energy balance: transient conditions in V = A c z

( U V )t

= (m H ) z (m H ) z + z + q D z


PFR: energy balance in transient and steady state


29/45

PFR: energy balance in transient and steady stateconditions

Obtaining the nal balanceDivide by V and take the limit as V 0:

( U )t

= (Q H )

V +

4D

q

Rewrite H

= j c i

H j and obtain:

( U )t

= (Q j c j H j )

V +

4D

q

q is the heat ux (i.e. heat ow per unit of transfer area), usually expressed as:q = U 0(T a T ) where T a is the external temperature (e.g. jacket temperature)

Steady-state balance

d (Q j c j H j )

dV =

4D

q (27)


PFR: steady-state balance rearrangement


30/45

PFR: steady state balance rearrangement

Expand the l.h.s. termRecall the material balances:

d (Qc j )dV

=i

i j r i

Expand the d (Q j c j H j )dV as follows:

d (Q j c j H j )

dV =

j Qc j

d H j dV

+ H j d (Qc j )

dV

d (Q j c

j H

j )

dV = Q

j c j

d H j

dV +

i H Ri r i (28)


PFR: steady-state balance rearrangement (cont.)


31/45

PFR: steady state balance rearrangement (cont.)

Expanding the enthalpy change term

j c j d H j

Let H = j c j

H j be the enthalpy per volume of reacting mixture. We have:

d H = j

H j dc j + c j d H j (29)

As in previous cases, its change can be expressed as:

d H = H T p ,n j

dT + H p T ,n j

d p + j

H n j T ,p ,n k = j

dn j (30)

From previous derivations, (30) is re-written as:

d H = C p dT + (1 T )d p + j

H j dc j (31)

Finally, by comparison of (29) and (31) we obtain:

j c j d H j = C p dT + (1 T )d p (32)


PFR: steady-state nal balance


32/45

PFR: steady state nal balance

General balance

From (27), (28) and (32) we obtain:

Q C p dT dV

+ Q (1 T )d p dV

=i

( H Ri )r i +4D

q (33)

Ideal gas or neglect pressure drop

Q C p dT dV

=i

( H Ri )r i +4D

q (34)

Incompressible uid

Q C p dT dV

+ Q d p dV

=i

( H Ri )r i +4D

q (35)


PFR hot spot and runaway


33/45

p y

IntroductionFor exothermic (gas-phase) reactions in a PFR, the heat release can lead to theformation of a hot spot

When reactions are highly exothermic, small changes in operating parameters(e.g., inlet concentration, owrate, temperature) can result in very changes of

the temperature proleThis is known as the PFR parametric sensitivity , and a sudden increase intemperature prole in known as run away

Example: exothermic reaction in a jacketed PFR

Irreversible rst order reaction in gas phase: A k B

Observe ( N A ,T ) as function of z for different inlet stream temperatures T 0


PFR parametric sensitivity


34/45

p y

Reactant molar ow

0

0.0002

0.0004

0.0006

0.0008

0.001

0.0012

0.0014

0.0016

0 0.2 0.4 0.6 0.8 1 1.2 1.4

N A

( k m o l / s )

z (m)

T 0 = 615 K T 0 = 625 K T 0 = 631 K T 0 = 635 K



35/45

Optimized reactor arrangements: introduction


36/45

p g

Motivations and rationaleIn reactor design, it is often desirable to minimize the overall reaction volumeThis goal is made possible by arranging a number of reactors (e.g. in series) tofollow a path at maximum reaction rate

To follow such optimized paths, heat transfers are required between reactors

A number of typical arrangements are described and analyzed next

A reversible reaction: kinetic data analysisSingle reversible reaction:

A k 1

k 1

B

Kinetic data can be conveniently expressed graphically in the plane ( T , x A )

Dene x A = c A 0 c A

c A 0 and recall:

r (T , x A ) = k 1(T )c A 0(1 x A ) k 1(T )c A 0x A


Optimized reactor arrangements: kinetic data


37/45

Kinetic data

Kinetic data report, in ( T , x A ), a number lines of constant reaction rateEach line is dened by the implicit relation:

a = k 1(T )c A 0(1 x A ) k 1(T )c A 0x A

where a > 0 species the reaction rate

The line at a = 0 denes the (implicit) equilibrium relation x Ae (T )

Locus of maximum ratesFor each given conversion, there exists a temperature at which the reactionrate is at maximum

This is formally dened as:r (T , x A )

T x A =

dk 1dT

c A 0(1 x A ) dk 1dT

c A 0x A = 0


Kinetic data for an exothermic reaction


38/45

Parametric plot of normalized reaction rate r / c A 0

0

0.10.2

0.3

0.4

0.50.6

0.7

0.8

0.9

1

60 80 100 120 140

x A

T (o C)

0.002

0.004

0.01

0.02

0.04

0.1

0.2

0.41.0

EquilibriumIsokinetic

Maximum rate


Kinetic data for an endothermic reaction


39/45

Parametric plot of normalized reaction rate r / c A 0

0

0.1

0.2

0.3

0.4

0.50.6

0.7

0.8

0.9

1

60 70 80 90 100 110 120 130 140

x A

T (o C)

0.0020.004 0.01 0.02 0.04 0.1 0.2 0.4 1.0



Optimized CSTR arrangements


40/45

Scheme 1: Preheating and inter-cooling Cooler

x A 2

T 2 T 2 T 3

x A 3

T 0 T 0 T 1 T 1

x A 1

CoolerHeater

0

0.10.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

50 60 70 80 90 100 110 120 130 140 150

x A

T (o C)


Maximum rate


Optimized CSTR arrangements (cont.)


41/45

Scheme 2: Auto-thermal preheating

x A 1T 0 T 1

T 1 T 2

T 0

x A 2

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

50 60 70 80 90 100 110 120 130 140 150

x A

T (o C)


Maximum rate


Optimized CSTR arrangements (cont.)


42/45

Scheme 3: Preheating and feed quenching

x A 3

T 3T 0 T 0 T 1

x A 1

Heater

T 1 T 2

x A 2

T 2

x A 2x A 1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.70.8

0.9

1

50 60 70 80 90 100 110 120 130 140 150

x A

T (o C)


Maximum rate



43/45

Summary of material and energy balances


44/45

Balance equations for the batch reactor

Case Material balances Energy balance

Single phased (V R c j )

d t =

i i j r i V R V R C p

dT d t

TV R dp d t

=i

( H Ri )r i V R + Q

Constant pressured (V R c j )

d t =

i i j r i V R V R C p

dT d t

=i

( H Ri )r i V R + Q

Constant volume(ideal gas)

dc j d t = i i j r i V R

C v

dT

d t = i (

H Ri + RT i )r i V R + Q

Liquid phasedc j d t

=i

i j r i V R C p dT d t

=i

( H Ri )r i V R + Q

Steady-state balance equations for the CSTR


Single phase Q 0c j 0 Qc j +i

i j r i V R = 0 j

Q 0c j 0( H j 0 H j ) +i

( H Ri )r i V R + Q = 0

Liquid phase Q 0(c j c j 0) +i

i j r i V R = 0 Q 0 0 C p (T 0 T ) +i

( H Ri )r i V R + Q = 0


Summary of material and energy balances (cont.)


45/45

Steady-state balance equations for the PFR


Single phased (Qc j )

dV =

i i j r i Q C p

dT dV

+ Q (1 T )dp dV

=i

( H Ri )r i +4D

q

Constant pressureor ideal gas

d (Qc j )dV

=i

i j r i Q C p dT dV =

i ( H Ri )r i + 4D

q

Liquid phase Q dc j dV

=i

i j r i Q C p dT dV

+ Q d p dV

=i

( H Ri )r i +4D

q

Liquid phase,constant pressure

Q dc j

dV =

i i j r i Q C p

dT

dV =

i ( H Ri )r i +

4

D q


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