chemical enggineering

8/11/2019 chemical enggineering

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Temperature & Pressure

Effects in ChemicalReactors

http://blog.its.ac.id/mahfud/[email protected]

Thermodinamika vs KinetikaUnjuk kerja reaksi (konversi maksimum dan laju reaksi)

ditentukan oleh :

sifat termodinamika reaksi

sifat kinetika reaksi

a. Sifat Termodinamika Reaksi

mengunkap kelayakan reaksi kimia

menentukan konstanta kesetimbangan & konversi maximum

memberikan informasi efek panas yang timbul karena reaksi kimia

b. Sifat Kinetika Reaksi

Sifat kinetika reaksi mengungkapkan hubungan antra kondisi reaksi denganlaju reaksi

Reaksi dapat dipercepat melalui berbagai cara, yaitu :

menaikkan temperatur reaksi

menaikkan konsentrasi atau tekanan parsial reaktan

menaikkan kemungkinan tumbukan antar reaktan (

meningkatkan ketetapan orientasi tumbukan dan

menurunkan energi Aktivasi


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Thermodinamika vs Kinetika

Go, kkal./mol Kelayakan reaksi

Go < -10 sangat layak

-10 < Go = 0 layak

0 < Go < 10 dapat layak tetapi membutuhkansuhu & tekanan yang tinggi

G

o

> 10 umumnya tidak layak

TEMPERATURE EFFECT INTEMPERATURE EFFECT INTEMPERATURE EFFECT INTEMPERATURE EFFECT INCHEMICAL REACTORSCHEMICAL REACTORSCHEMICAL REACTORSCHEMICAL REACTORS

Equilibrium yield

Rate of reaction

Product distribution (multiple)

Reactions are generally accompanied by heateffects

How will they change the temperature ?

Economic considerations

Identify optimum

Reactor stability

operating point

Affected by

temperature


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Temperature and Pressure Effects

How equilibrium composition,rate of reaction, andproduct distribution are affected by changes in operatingtemperatures and pressures.to determine theoptimum temperature progression, to approximate witha real design.

Chemical reactions are usually accompanied by heateffects how these will change the temperature of thereacting mixture.to propose a number of favorablereactor and heat exchange systems-those which closely

approach the optimum. Economic considerations will select one of these

favorable systems as the best.

Temperature and Pressure Effects

Thermodynamics gives two important piecesof information :

the heat liberated or absorbed for a given extent

of reaction,

the maximum possible conversion.


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Heats of Reaction from Thermodynamics

Heat liberated or absorbed during reaction attemperature T, depends on the nature of the reacting system,

the amount of material reacting

temperature and pressure of the reacting system,

calculated from the heat of reaction Hr

Heats of Reaction from Thermodynamics

Hr calculated from thermochemical data (heats offormation Hf or heats of combustion Hc of the reactingmaterials.)

By convention we define the heat of reaction attemperature T as the heat transferred to the reactingsystem from the surroundings when a moles of Adisappear to produce r moles of R and s moles of S with

the system measured at the same temperature andpressure before and after the change.

aA + rR sS, Hr= + Reaction EndothermicHr= -Reaction Exothermic


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Heats of Reaction and Temperature.

Heats of Reaction and Temperature.


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Example 1 : Hr at various temperatures

From the Hc and Hf tables, I've calculated that the standard heat of mygas phase reaction at 25 C is as follows:

At 25 C the reaction is strongly exothermic. But t his doesn't interest me

because I plan to run the reaction at 1025 C. What is the Hr at thattemperature, and is the reaction still exothermic at that temperature?

Data. Between 25C and 1025C the average Cp values for the various

reaction components are

Solution


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Equilibrium Constants from Thermodynamics

From the second law of thermodynamics equilibrium constants, hence

equilibrium compositions of reacting systems, may be calculated. We mustremember, however, that real systems do not necessarily achieve this

conversion; therefore, the conversions calculated from thermodynamics are

only suggested attainable values.

f is the fugacity of the component at the arbitrari ly selected standard state at

temperature T, the same one used in calculating G; G is the standard freeenergy of a reacting component, tabulated for many compounds;

Standard states at given temperature are commonly chosen as follows: Gases-pure component at one atmosphere, at which pressure ideal gas behavior is closely

approximated

Solid-pure solid component at unit pressure

Liquid-pure liquid at its vapor pressure

Solute in liquid-1 molar solution; or at such dilute concentrations that theactivity is unity.

Equilibrium Constants from Thermodynamics


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Equilibrium Conversion.

The equilibrium composition, as governed by the equilibriumconstant, changes with temperature, and fromthermodynamics the rate of change is given by

Equilibrium Conversion

These expressions allow us to find the variation of the equilibriumconstant, hence, equilibrium conversion, with temperature.


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Thermodynamics.1. The thermodynamic equilibrium constant (K) is unaffected by the

pressure of the system, by the presence or absence of inerts, orby the kinetics of the reaction, but is affected by the temperatureof the system.

2. Though the thermodynamic equilibrium constant is unaffected bypressure or inerts, the equilibrium concentration (XAe) of materialsand equilibrium conversionof reactants can be influenced by these

variables.

Thermodynamics.3. K >> 1 indicates that practically complete conversion may be

possible and that the reaction can be considered to be irreversible.

K


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Example 2 : Equilibrium conversion at differenttemperatures

(a) Between 0C and 100C determine the equilibriumconversion for the elementary aqueous reaction

Present the results in the form of a plot of temperatureversus conversion.

(b) What restrictions should be placed on the reactoroperating isothermally if we are to obtain a conversion of

75% or higher?

Solution(a) With all specific heats alike, Cp = 0. Then the heat of

reaction is independent of temperature and is given by

Putting T values, then K into Eq. aboves gives the changing equilibriumconversion as a function of temperature in the range of O C to 100C


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(b) From the graph we see thatthe temperature must stay below

78C if conversion of 75% orhigher may be expected.

General Graphical Design Procedure

Temperature, composition, and reaction rate are uniquelyrelated for any single homogeneous reaction, and this may berepresented graphically in one of three ways.

The first of these, the composition-temperature plot, is themost convenient so we will use it throughout to representdata, to calculate reactor sizes, and to compare designalternatives.


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Example 3 : Construction of the rate-conversiontemperature chart from kinetic data

With the system of Example 2 and starting with an R-freesolution, kinetic experiments in a batch reactor give 58.1%

conversion in 1 min at 65 C, 60% conversion in 10 min at25C. Assuming reversible first-order kinetics, find the rateexpression for this reaction and prepare the conversion

temperature chart with reaction rate as parameter.


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Solution


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Optimum Temperature Progression Optimum temperature progression - progression which minimizes

V/FAofor a given conversion of reactant. This optimum may be anisothermal or it may be a changing temperature: in time for a batch reactor,

along the length of a plug flow reactor, or

from stage to stage for a series of mixed flow reactors.

The optimum temperature progression in any type of reactor is asfollows: At any composition, it will always be at the temperature wherethe rate is a maximum. The locus of maximum rates is found byexamining the r(T, C) curves shows this progression : For irreversible reactions, the rate always increases with temperature at

any composition, so the highest rate occurs at the highest allowabletemperature. This temperature is set by the materials of construction or bythe possible increasing importance of side reactions.

For endothermic reversible reactions a rise in temperature increasesboth the equilibrium conversion and the rate of reaction. Thus, thehighest allowable temperature should be used.

For exothermic reversible reactions the situation is different, for here twoopposing factors are at work when the temperature is raised-the rate offorward reaction speeds up but the maximum attainable conversiondecreases. Thus, in general, a reversible exothermic reaction starts ata high temperature which decreases as conversion rises.


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Operating lines for minimum reactor size.

Heat Effects When the heat absorbed or released by reaction can markedly

change the temperature of the reacting fluid, this factor must be

accounted for in design.

Thus we need to use both the material and energy balanceexpressions..

if the reaction is exothermic and if heat transfer is unable to

remove all of the liberated heat, then the temperature of thereacting fluid will rise as conversion rises.

for endothermic reactions the fluid cools as conversion rises. Let

us relate this temperature change with extent of conversion.


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Adiabatic Operations

Consider either a mixed flow reactor, a plug flow reactor, or a section ofplug flow reactor, in which the conversion is X, with limiting reactant A taken

as the basis.

Let Subscripts 1, 2 refer to temperatures of entering and leaving streams.

With T1 as the reference temperature on which enthalpies and heats of

reaction are based.


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Nonadiabatic Operations

For the adiabatic operating line of Fig. 9.7 to more closely approach theideals of Fig. 9.5, we may want deliberately to introduce or remove heat

from the reactor. In addition, there are heat losses to the surroundingsto account for.

Let us see how these forms of heat interchange modify the shape of the

adiabatic operating line.

Let Q be the total heat added to a reactor per mole of entering reactantA, and let this heat also include the losses to the surroundings. Then,

the energy balance about the system, is modified to


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Sketch of the energy balance equation showingthe shift in adiabatic line caused by heat exchangewith surroundings.

Ways of approaching the ideal temperature profile with heat

exchange: (a) and (b) exothermic reaction; (c) endothermic reaction.


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Best location for the adiabatic operating line. For plug flow,

a trial and error search is needed to find this line; for mixedflow, no search is needed.

Conversion in a mixed flow reactor as a functionof T and r, from the material balance equation


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Three types of solutions to the energy and material balances

for exothermic irreversible reactions.

Non-isothermal CSTRv

CA0T0

VT

v

CA

Cooling water

h, A, TC

A B irreversible

- Heat transfer coefficient h

- AreaA

- Temperature TC- Kinetics: (-rA)= kCA- Enthalpy change of reaction = H

- Assume: CpA = CpB = const.

Mass balance:

( ) ( )A

AA

A

AA

rCC

rXC

vV

=

==

00

Subst. kinetics

v

kV

CC

Ck

CC

v

V AA

A

AA

+=

=

1

0

1

0


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Insert Arrhenius

expression: ( )[ ]

+

=

v

VRT

Ek

CC AA

exp1

0

0

Energy balance:Heat generated (exothermic) - ( ) ( )HVkCQ Ag =

Rate of reaction

per unit volume

Heat evolved per

mole reactantArrhenius - ( ) HVCRTEkQ Ag = /exp0

Substitute for CA from above and rearrange:

Thus in the limit - ( )RTEVk

v

HvCQ Ag

/exp10

0

+

=

0,0 gQT

0

0

1

,

Vk

v

HvCQT Ag

+

Heat removed -

( ) ( ) ) )CppCpr hATTCvThACvTThATTCvQ ++=+= 00

AT STEADY STATE: Qr= Qg

Qr 1 2 3Qg

Qg, Qr

T

Cases 1, 3 - single steady states

Case 2 - multiple steady states

Case 2 has three possibilities - which steady state

do we actually get ?


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Stability of steady statesStability of steady statesStability of steady statesStability of steady states

Definition: a steady state is stable if, when subjected

to a perturbation, it returns to the steady state.

Qr

T

QgQg,

Qr

Ta Tb Tc

a

b

c Steady state a

-If Ta increases to Ta + T

Qr > Qg thus T reduces

-If Ta decreases to Ta T

Qr < Qg thus T increases

System acts to eliminate the perturbation

- thus steady state a is stable.

Steady state b

- If Tb increases to Tb + T

Qr < Qg T increases further until steady state c is reached- If Tb decreases to Tb - T

Qr > Qg thus T decreases further to Ta

Steady state c

as per case a, i.e. stable

Implications for operability

T

Qg

Ti

QrQr,

QgStart up temperature, TsIf Ts>Ti, Qg>Qr and the reactor goes to the

upper steady state (high T implies high

conversion). This is usually the desired

steady-state.

This may not be desired due to:-

high T encourages side reactions, or the

material of the reactor may not stand high

T, or the rate may become dangerously high.

If Ts < Ti, Qg < Qr and the reactor goes to the lower steady-state (low conversion).

May need to pre-heat reactants at start-up even if reaction is exothermic.


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Solution of energy and material balances forreversible exothermic reaction.

Example 4 : Performance for the optimal

temperature progression

Using the optimal temperature progression in a plug f low reactor for thereaction of Examples 2 and 3,

(a) calculate the space time and volume needed for 80% conversion of

a feed of FAo= 1000 mol/min where CAo= 4 mol/liter.

(b) plot the temperature and conversion profile along the length of thereactor. Let the maximum allowable operating temperature be

95C.

Note that Fig. E9.3 was prepared for CAo = 1 mol/liter, not 4 mol/liter.

Solution :(a) Minimum Space-Time. On the conversion-temperature graph,

draw the locus of maximum rates. Then, remembering thetemperature restriction, draw the optimum path for this system (line

ABCDE )and integrate graphically along this path to obtain :


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(b) T and X, Profiles Through the Reactor. Let us take 10%increments through the reactor by taking 10% increments in areaunder the curve of Fig. b. This procedure gives XA = 0.34 at the 10%

point, XA = 0.485 at the 20% point, etc. The correspondingtemperatures are then 362 K at XA = 0.34 (point C), 354 K at XA =

0.485 (point D), etc.

In addition we note that the temperature starts at 95"C, and at XA =0.27 (point B) it drops. Measuring areas in Fig. b we see that this

happens after the fluid has passed 7% of the distance through thereactor.

In this manner the temperature and conversion profiles are found.

The result is shown in the following figure :


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Example 5 : Optimum mixed flow reactorperformance

A concentrated aqueous A-solution of the previous examples (CAo = 4

mollliter, FAo = 1000 mol/min) is to be 80% converted in a mixed flowreactor.(a) What size of reactor is needed?

(b) What is the heat duty if feed enters at 25C an d product is to be withdrawn at this temperature?

Cp = 1000 cal /kg .K = 250 cal/mol A.KSolution :

(a) Reactor Volume. For CAo = 4 mol/liter we may use the XA versus T

chart of Fig. E9.3 as long as we multiply all rate values on this chart by 4.Following figure,the mixed flow operating point should be located where

the locus of optima intersects the 80% conversion line (point C).Here the reaction rate has the value

-rA = 0.4 mol A converted/min . liter

From the performance equation for mixed flow reactors, the volumerequired is given by


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(b) Heat Duty. Of course we can use joules in our calculations;however, since we are dealing with aqueous solutions it is simpler touse calories. Let us use calories. Then the slope of the energy balance

line is

Drawing this line through point C (line BCD) we see that the feed must

be cooled 20(from point A to point B) before it en ters and reactsadiabatically. Also, the product must be cooled 37C (from point C to

point E). Thus the heat duty is


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Example 6 : Adiabatic plug flow reactor performance

Find the size of adiabatic plug flow reactor to react the feed of Example5 (FAo = 1000 mol/min and CAo = 4 mol/liter) to 80% conversion.SOLUTION

Following the procedure of Fig. 9.9 draw trial operating lines with aslope of 1/72 (from Example 5), and for each evaluate the integral

to find which is smallest. The following figures show this procedure for

lines AB and CD. Line CD has the smaller area, is in fact closer to theminimum, and is therefore the desired adiabatic operating line. So


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8.Temperature Effects With Multiple Reactions8.Temperature Effects With Multiple Reactions8.Temperature Effects With Multiple Reactions8.Temperature Effects With Multiple Reactions

Product DistributionProduct DistributionProduct DistributionProduct Distribution

(a) Parallel reactions

A

R desired

S unwanted

(1)

(2)

Assume reactions are of the same

Order. We want: k1 large

k2 small

k = k0exp(-E/RT)

If E1>E2, (1) increases with T more rapidly than (2) - Use high T

If E1E2 use high T

If E1E2>E3 - use high T

(ii) E1>E2, E1


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SEKIAN

Terima Kasih

chemical enggineering

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