Chapter 6

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REACTION ENGINEERINGCKB 20104

Chapter 6 Non-isothermal Reactor

DesignDr. Kelly Yong Tau Len

Section of Chemical Engineering Technology UniKL MICETTel: 06-5512051

Email: [email protected], E-learning: CKB20104 – Kelly Yong

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E-Learning Online Laboratory Quiz

Attempts allowed: 1The quiz will be available from

Sunday, 8 November 2015, 12:00 AMThis quiz will close at

Sunday, 15 November 2015, 11:59 PMTime limit: 15 minutes

Note: Please submit the attempt before the time expired or they are not counted

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Chapter 66.1 The energy balance6.2 Algorithms for non-isothermal reaction design6.3 Equilibrium conversion6.4 Multiple steady-state

ObjectivesUpon the completion of this chapter, students are able to:o Describe the algorithm for CSTRs that are notoperated isothermally.

o Size adiabatic andnon adiabaticCSTRs.o Analyze NON-ISOTHERMALREVERSIBLE reactionso Carry out an analysis to determine the Multiple SteadyStates (MSS) in a CSTR

3

6.1 The Energy BalanceWHY WE NEED TO LEARN ENERGY BALANCE IN REACTION

ENGINEERING?

1.At times we need to calculate the volume necessary to achieve a conversion, X in a reactor.

2. For an exothermic reaction, the temperature profile vs. volume is given below

4

Reaction Temperature influenced

the Reactor Volume

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Lets say we have a CSTR and given a first order, exothermic liquid reaction carried out adiabatically (no heat loss or gained in the system).

5

1.Design equation for CSTR

VCSTR =FA0X−rA

2. Rate Law (First Order Reaction)−rA = kCA3.Because there is change of temperature during reaction therefore

k = k1 exp EaR

1T1

−1T

"

#$$

%

&''

(

)**

+

,--

4.Stochiometry for liquid phase givesCA =CA0 (1− X )

6.1 The Energy Balance

4. We combine equation 1, 2,3 and 4 to obtain

• In this equation, we have 3 unknowns (Volume, Temperature and Conversion) which need to be solved.

• Therefore we need another relationship relating X and T or T and V.

• This additional relationship can be found through energy balance equation

6

VCSTR =FA0X

k1expEaR

1T1−1T

"

#$$

%

&''

(

)**

+

,--CA0(1−X)

1stUnknown

2ndUnknown

3rdUnknown

6.1 The Energy Balance

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6.1 The Energy BalanceFirst Law Of Thermodynamics

7

Work done by system

Heat to the system

Inlet Molar Flow Rate

Inlet Enthalpy

Outlet Molar Flow Rate

Outlet Enthalpy

6.1 The Energy BalanceFirst Law Of Thermodynamics

8

General Energy Balance of An Open System

Q•

−W•

+ Eioi=1

n

∑ Fio − EiFii=1

n

∑ =d Esysdt

Heat to the system

Work done by the system Total sum or

energy in

Total sum or energy out

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6.1 The Energy Balance9

1. In terms of enthalpy (for open system), the energy balance becomes

1. And for steady state operation (accumulation = 0), the energy balance now becomes

Q•

−W•

+ Hioi=1

n

∑ Fio − HiFii=1

n

∑ =d Esysdt

Q•

−W•

+ Hioi=1

n

∑ Fio − HiFii=1

n

∑ = 0

We derive further this part

6.1 The Energy Balance 10

In : Hi0Fi0 = HA0FA0 +∑ HB0FB0 +HC0FC0 +HD0FD0 +HI 0FI 0Out : HiFi = HAFA +∑ HBFB +HCFC +HDFD +HIFI

A+ baB→ c

aC + d

aD

SPECIES INITIAL (MOL) REMAINING (MOL)

A

B

C

D

FA0FB0 = FA0ΘB

FC0 = FA0ΘC

FD0 = FA0ΘD

FA = FA0 (1− X)

FB = FA0 (ΘB −baX)

FC = FA0 (ΘC +caX)

FD = FA0 (ΘD +daX)

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Hi0Fi0 −i=1

n

∑ HiFii=1

n

∑ = HA0FA0 +HB0FB0 +HC0FC0 +HD0FD0 − (HAFA +HBFB +HCFC +HDFD )

= HA0FA0 −HAFA +HB0FB0 −HBFB +HC0FC0 −HCFC +HD0FD0 −HDFD

= HA0FA0 −HAFA0 (1− X)+HB0ΘBFA0 −HBFA0 ΘB −baX

$

%&

'

()

+HC0ΘCFA0 −HCFA0 ΘC +caX

$

%&

'

()+HD0ΘDFA0 −HDFA0 ΘD +

daX

$

%&

'

()

= HA0FA0 −HAFA0 +HAFA0X +HB0ΘBFA0 −HBΘBFA0+HBFA0baX

+HC0ΘCFA0 −HCΘCFA0 −HCFA0caX +HD0ΘDFA0 −HDΘDFA0 −HDFA0

daX

= FA0 HA0 −HA( )+ΘBFA0 HB0 −HB( )+ΘCFA0 HC0 −HC( )+ΘDFA0 HD0 −HD( )

+HAFA0X +HBFA0baX −HCFA0

caX −HDFA0

daX

= FA0[ HA0 −HA( )+ΘB HB0 −HB( )+ΘC HC0 −HC( )+ΘD HD0 −HD( )]

−daHD +

caHC −

baHB −HA

$

%&

'

()FA0X


Hi0Fi0 −i=1

n

∑ HiFii=1

n

∑ =

= FA0[ HA0 −HA( )+ΘB HB0 −HB( )+ΘC HC0 −HC( )+ΘD HD0 −HD( )]

−daHD +

caHC −

baHB −HA

$

%&

'

()FA0X

FAO Θi (Hio −H i )i=1

n

∑

ΔHRx(T )FAOX

1. From the original energy balance equation

1. With manipulation of the molar flow rates, we obtain the energy balance

Q•

−W•

+ Hioi=1

n

∑ Fio − HiFii=1

n

∑ = 0

Q•

−W•

+ FAO Θi (H io−H i )i=1

n

∑ −ΔHRx(T )FAOX = 0

Heat of Reaction

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4. From equation 3 previously

5. We know the enthalpy term is defined as following

6. Therefore the energy balance now becomes

Q•

−W•

+ FAO Θi (Hio −Hi )i=1

n


Hio −Hi = ΔHi (T ) = Hf,ioo + CP dT

TR

To

∫$

%&&

'

())− Hf,i

o + CP dTTR

T

∫$

%&&

'

())

Hio −Hi = − CPi dTTo

T

∫ = −CPi (T −To )

Q•

−W•

−FAO ΘiCPi (T −To )i=1

n


Reference T

Entrance T Reaction T



7. We know the heat of reaction, HRx term is defined as following

9. Therefore the energy balance now becomes

Q•

−W•

− FAO ΘiCPi (T −To )i=1

n


ΔHRx(T ) = ΔHRx(TR )+ΔCP (T −TR )

Q•

−W•

− FAO ΘiCPi (T −To )i=1

n

∑ − FAOX ΔHRx(TR )+ΔCP (T −TR )%

&'(= 0

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8. Work, W can be separated into flow work and other work, Ws. The term Ws(shaft work) could be produced from STIRRER in CSTR or turbine in PFR. Flow work are usually negligible compared to other terms therefore it is omitted.

9. Therefore the FINAL energy balance is

Q•

−W•


n

∑ −FAOX ΔHRx(TR )+ΔCP (T −TR )%& '(= 0


Q•

−Ws −FAO ΘiCPi (T −To )i=1

n

∑ −FAOX ΔHRx (TR )o +ΔCP (T −TR )%& '(= 0

Equation are given in Exam!

Now we learn to apply the Energy

Balance equation

16

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Q•

−WS−FAO ΘiCPi (T −To )i=1

n


6.1 The Energy Balance(Heat Capacity, Cpi)

17

1. Typical unit for heat capacity, Cpi of a reactant is represented as J/mol.K

2. The terms could be calculated for ALL REACTANT

ONLY. 3. Recall that Θi for product is zero.

ΘiCPii=1

n

∑

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For example given a reaction of A + B à C + D with A as the basis of calculations, therefore:

withPIIPBPPi CCCCBAiΘ+Θ+=Θ∑

~~~

0

0

A

BB FF

=Θ

POSITIVE SIGN

REMEMBER:Only take the REACTANT

into consideration including INERT if its in the

reaction.

Q•


n


6.1 The Energy Balance(Heat Capacity, Cpi)

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Q•


n


6.1 The Energy Balance(Temperature Definition & Difference)

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1. T = Reaction temperature in which the reaction is conducted2. T0 = Entrance/Inlet Temperature of the reactant

3. TR = Reference Temperature in which the heat capacity at constant pressure, Cp or the heat of reaction, ΔHoRxwere obtained (usually 25oC) unless stated other value.

Q•


n


6.1 The Energy Balance(Heat of Reaction at reference temperature ΔHoRx (TR))

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1. Heat of reaction at reference temperature (usually 25oC (unless stated other value) are specific depending on the reaction.

2. It is either given directly or can be tabulated based on the enthalpies of formation, Hof(TR) for all reactants and products involved in the reaction.

3. The enthalpies of formation of many compounds, Hof(TR) are usually tabulated at 25oC and can readily be found in any handbook.

4. ΔHo Rx (TR) is to indicate that in the formula, the heat of reaction ΔHo Rx is calculated at the reference temperature.

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For example given a reaction of A + bBà cC + dDand enthalpies of formation, Hof(TR) of reactants (A and B), and products (C and D) are given, therefore:

ΔHRx (TR )

o =daHO

D(TR )+caHO

C(TR )−baHO

B(TR )−HO

A(TR )#

$%

&

'(

21

Positive for product Negative for reactantREMEMBER: REACTANT AND PRODUCT SHOULD BE TAKEN INTO

CONSIDERATION

Q•


n



22

EXAMPLECalculate ΔHo Rx (TR) when given the reaction as following:

SOLUTION: For the above reaction, the ΔHo Rx (TR) is as following:

From reference, it is found that the heat of formation of hydrogen, (Ho H2,(TR)) and nitrogen (Ho N2,(TR)) are zero at 25oC, and Ho NH3 (TR) = –11,020 cal/mol therefore

ΔHRx (TR )o = 2HO

NH 3(TR )−3HO

H 2(TR )−HO

N 2(TR )( )

ΔHRx (TR )o = 2(−11,020)−3(0)− 0 = −22,040cal /mol


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Q•


n


6.1 The Energy BalanceOverall Change In The Heat Capacity, ΔCp

23

1. The overall change in the heat capacity, ΔCp is based on per mole of A reacted during the reaction.

2. It can be obtained based on the heat capacities for all reactant and product, Cpi

3. These values can be found from any handbook.

1. It can be calculated based on the following method2. For example given a reaction of A + bBà cC + dDwith heat capacities, Cpi of reactants (A and B), and products (C and D) are given, therefore:

ΔCP =daCPD

+caCPC

−baCPB

−CPA

24

REMEMBER: REACTANT AND PRODUCT SHOULD BE TAKEN INTO CONSIDERATION

Q•


n


Positive for product Negative for reactant


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25

Calculate ΔCpwhen given the reaction as following:

SOLUTION:


Q•


n


6.1 The Energy BalanceHeat Added to the Reactor

26

1. Q is HEAT added to the reactor when heating up a reaction is required.

2. For CSTR, the Q value can be calculated from a specified equation.

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6.1 The Energy BalanceHeat added to CSTR with heat exchanger, Q

27

TO, FAO T, FA

Ta1, mc Ta2CSTR reactor

Heat exchanger

Heat transfer fluid enters the exchanger at a mass flow rate mc (kg/s) at a temperature Ta1 and

leaves at a temperature Ta2.

For exothermic reactions (T>Ta2>Ta1)

For endothermic reactions (Ta1>Ta2>T)

28

3. For moderate to low coolant rates:

4. For large coolant rates, Ta1 (Heat transfer fluid inlet temperature) = Ta2 (Heat transfer fluid outlet temperature)

U = Overall heat transfer coefficient of jacketA = Heat exchange area (steam jacket area)Cp,c = Heat capacity of the heat transfer fluidTa1 = Heat transfer fluid inlet temperature (jacket steam saturation temperature)T = Reaction temperature

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ −−−=

c

cpc

apc CmUATTCmQ exp1)( 1

..

Q.=UA(Ta1 −T )

Usually we use this unless indicated otherwise

6.1 The Energy BalanceHeat added to CSTR with heat exchanger, Q

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6.2 Algorithms for non-isothermal reaction design (CSTR Algorithm for non-adiabatic System)

29

X is given, calculate V and T T is given, calculate X and V

1.Design equation for CSTR: VCSTR =FA0X−rA

2. Rate Law (Elementary): − rA = kCA3. Stochiometry for liquid phase: CA =CA0 (1− X )

4. Change of temperare, k = k1 exp EaR

1T1

−1T

"

#$$

%

&''

(

)**

+

,--

4. Combine all to give: VCSTR =FA0X

k1 exp EaR

1T1

−1T

"

#$$

%

&''

(

)**

+

,--CA0 (1− X )1st Unknown

2ndUnknown

3rdUnknown

Q•


n

∑ −FAOX ΔHRx (TR )

o +ΔCP (T −TR )%&

'(= 0

30

X is given, calculate T first using energy balance equation

Calculate new k value

k = k1 exp ER

1T1

−1T

"

#$$

%

&''

(

)**

+

,--

Calculate VCSTR

VCSTR =FA0X

kCAO (1− X )

Using the T above to calculate the k

Scheme 1: X is given, calculate V and T


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T is given, Calculate X and V

Calculate X from Energy Balance Equation

Q•

−WS −FAO ΘiCPi (T −To )i=1

n


Calculate VCSTR

VCSTR =FA0X

kCAO (1− X )

Using the X above to calculate the V


32

Given the following reaction in a non adiabatic CSTR.

(V)(A)

(Ta)


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33

(CPA)

(FA0)(T0)

(CPB) (CPC)

Given the following reaction in a non adiabatic CSTR.


34

1. Reaction is non-adiabatic, CSTR reactor, irreversible liquid phase, elementary reaction.

2. CSTR volume, V = 125 gal and conversion, X = 1.0 (proceeds to completion) so find the reaction temperature, T.

3. From question,

U

Ta

FAO

ΔHRx(TR )o


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35

4. From the energy balance equation,

ΔCP =caCPC −

baCPB −CPA = 2(47.5)− 44.0−51.0 = 0

PCΔ

Q•


n


(CPA) (CPB) (CPC)


36

5. From the energy balance equation

Θi CPi =∑ CPA +ΘBCPB ,ΘB =10lbmol / hr10lbmol / hr

=1

Therefore Θi CPi =∑ CPA +CPB = 51.0+ 44.0 = 95.0 Btu / lb mol.oF

PiC

Q•

−W.s−FAO ΘiCPi (T −To )

i=1

n


(CPA) (CPB) (CPC)


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Calculate the heat added into the reactor by the heat exchanger

Q•

−W.s−FAO ΘiCPi (T −To )

i=1

n


37

Q.

=UA.

(Ta −T )

= (150 Btu / hr. ft 2.oF )(10 ft 2 )(365.9−T ) oF=1500(365.9−T )Btu / hr


38


[ ] 0)()()(1

.=−Δ+Δ−−Θ−− ∑

=

•

RPRo

AO

n

iioPiiAOs TTCTHXFTTCFWQ

Rx

W.

s = 25hp× 9.486×10−4Btu / s

1.341×10−3hp×3600s1hr

= 63664Btu / hr


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39

8. From the energy balance equation, substitute in all the values into the equation to obtain T as

Q•

−W.

s− FAO ΘiCPi (T −To )i=1

n

∑ − FAOX ΔHRx (TR )

o +ΔCP (T −TR )%&'

()*= 0

where Q.

=1500(365.9−T ) Btuhr

, Ws = 63664 Btuhr

where Θi CPi =∑ 95.0 Btulb mol.oF

,ΔHRx(TR )

o =20,000 Btulb mol

,ΔCP = 0

where FA0 =10 lbmolhr

,T0 = 80 oF ,TR = 365.9 oF ,X =1(into completion)

Substitute all into equation1500(365.9−T )−63664− (10 )(95.0 )(T −80)− (10)(1)[20,000+0]= 0

T =147.4 oF


40

FOR ADIABATIC SYSTEM;; Q = 0,

The energy balance for CSTR with no heat exchange Q= 0

−Ws − FAO ΘiCPi (T −To )i=1

n


o +ΔCP (T −TR )%&'

()*= 0

6.2 Algorithms for non-isothermal reaction design (CSTR Algorithm for Adiabatic System)

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41

Given the following reaction in an adiabatic CSTR.

FAO FBO


42

Given the following reaction in an adiabatic CSTR.

You are feeding 2500 lb mol/h of A and equal molar flowrate of B to the reactor. The feed stream consist of 46.62 ft3/hr of A and 233.1 ft3/hr of B. The temperature of both feed stream is 58oF prior to mixing but there is an immediate 17oF temperature rise upon mixing of the 2 feed streams caused by the heat of mixing. The entering temperature of all feed streams is thus taken to be 75oF.For the conditions similar to those, it is found that that the reaction is first order with respect to A and zero order with respect to B with the specific reaction rate given as

k = Ae-Ea/RT = 16.96x1012(e–32,400Btu/lb mol/RT)/hrCalculate the exit temperature at which the reaction is conducted and subsequently the CSTR volume when given the conversion as 10%. Assume the work done by stirrer is negligible.


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43

Following data are given as following:


44

1. In this question, the exit conversion is given, X therefore we need to find the temperature, T at which the reaction is conducted and subsequently the volume of the CSTR, V

2. Based on the algorithm for adiabatic system,

X is given, calculate T first using energy balance equation

Calculate T from Energy Balance

Q−Ws −FAO ΘiCPi (T −To )i=1

n



'(= 0

Because adiabatic, Q = 0 and stirring work negligible, Ws = 0Therefore equation becomes


n



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45

5. From the energy balance equationCalculate T


n



'(= 0

We know ΔHRxo (TR ) = c

aHC(TR )

o −baHB(TR )

o −HA(TR )o

Therefore ΔHRx(TR)

0 = HC (TR )o −HB(TR )

o −HA(TR )o

= −226,000− (−123,000)− (−66,600) = −36,400 Btu / lb mol

Reference Temperature, TR


46


4. Given FAO = 2500 lb mol/hr of A to the reactor and equal molar flowrate of B

Calculate T


n



'(= 0

Θi CPi =∑ CPA +ΘBCPB ,ΘB =25002500

=1

Therefore Θi CPi =∑ CPA +CPB = 35.0+18.0 = 53.0 Btu / lb mol.oF

To = 75oFTR = 68oF

From Question:You are feeding 2500 lb mol/h of A and equal molar flowrate

of B


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Calculate T


n



'(= 0

47

pAppp CCabC

acC know We

BC−−=Δ

Therefore ΔCp =CpC −CpB −CpA = 46−18−35 = −7 Btu / lb mol. oF



8. Substitute in all the values we calculate previously to determine the reaction temperature, T when given the conversion, X = 0.1

Calculate T


n



'(= 0

48


n


o +ΔCP (T −TR )%&'

()*= 0

where ΔHRx(TR)0 = −36,400 Btu

lbmol, Θi CPi =∑ 53.0 Btu

lbmol.oF,ΔCp = −7 Btu

lbmol.oF

where FAO = 2500 lbmolhr

,T0 = 75oF ,TR = 68oF ,X = 0.1

Substitute in the equation,− 2500(53.0)(T −75)− (2500)(0.10)[−36,400+ (−7)(T −68)]= 0T = 144.9 oF


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49

9. From the logarithm, we have solved T value. 10. Therefore next is to find the new k value when

reaction temperature, T is 144.9 oF. Given from the question that

k = Ae-Ea/RT = 16.96x1012(e-32,400Btu/lb mol/RT)/hr

k =16.96×1012e−32,400 Btu/lb mol

RT

=16.96×1012e−32,400 Btu/lb mol

(1.987Btu/lb mol. oR)(144.9+459.69 oR) / hr = 32.84 / hr


50

SOLUTION11.Next we obtain the volume of the CSTR, V.12.To find the initial concentration of A, we know FA0 = 2500 lb

mol/hr and the initial volumetric flowrate, as given in question as 46.62 ft3/hr of A and 233.1 ft3/hr of B. Therefore υ0,TOTAL = 46.62 + 233.1 ft3/hr = 279.72 ft3/hr

VCSTR =FA0X

kCAO (1− X )=

υ0Xk(1− X )

=279.72 ft

3

hr(0.1)

32.84hr

(1−0.1)"

#$

%

&'

= 0.95 ft3


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6.3 Equilibrium ConversionQUESTION: What Happen if its NON-ISOTHERMAL REVERSIBLE reactions? 1. The highest conversion can be achieved in reversible reactions is

the equilibrium conversion.2. For endothermic (need heat) reactions, the equilibrium conversion

increases with increasing temperature up to a maximum of 1.0.3. For exothermic (release heat) reactions, the equilibrium

conversion decreases with increasing temperature.

51

Equilibrium Conversion

Reaction Temperature

Example for Exothermic reaction

6.3 Equilibrium Conversions (Reversible, Exothermic and Adiabatic Reactions)

52

1. To determine the maximum conversion, X that can be achieved in an EXOTHERMIC reaction carried out ADIABATICALLY (Q=0) in a REVERSIBLE REACTION, we find the intersection of the equilibrium-temperature plot with temperature-conversion relationships from the energy balance equation.

Equilibrium Conversion, Xe

Reaction Temperature, T

Equilibrium PlotPlot obtained from Energy balance equation

ToAdiabatic temperature

Inlet Temperature

T

XMaximum Conversion

Q•


n


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6.3 Equilibrium Conversions (Reversible, Exothermic and Adiabatic Reactions)

53

1. If inlet temperature is increased from T0 to T01, the energy balance line will be shifted to the right and parallel to the original line, as shown by the dashed line.



Equilibrium PlotPlot obtained from Energy balance

equation

ToNew Inlet Temperature

Inlet Temperature

T01

Q•


n


6.3 Equilibrium Conversions (Reversible, Exothermic and Adiabatic Reactions)How to improve conversion for exothermic reaction?1. By connecting reactors in series with interstage cooling

54



Equilibrium PlotPlot obtained from Energy balance

equation

ToInlet Temperature

X1X2

Cooling down

Cooling down

X3Conversion

increases with interstage

cooling to Inlet Temperature

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6.3 Equilibrium Conversions (Reversible, Endothermic and Adiabatic Reactions)

How to improve conversion for ENDOTHERMIC reaction?1. By connecting reactors in series with interstage heating

55



Inlet TemperatureTo

Plot obtained from Energy

balance equation

Equilibrium Plot

X1 Heating up

Heating upX2X3Conversion

increases with interstage

heating to Inlet Temperature

6.4 Multiple Steady-state56

1. Lets say in our reaction, we don’t have both data of Conversion (X) and Reaction Temperature (T) but we know the Volume of Reactor (V)

2. If we consider a first order irreversible liquid phase reaction with single steady state in a CSTR we can obtain 2 separate plots of Conversion, X vs. Reaction Temperature

3. First plot is obtain when we calculate Conversion, XMBvs. Temperature based on Mass Balance Equation.

VCSTR =FA0XMB

kCA0 (1− XMB )=

v0XMB

k(1− XMB )

Rearrange we get XMB =VCSTRk

v0 +VCSTRk

When T changes, k = Ae−EaRT also changes

When k changes, XMB also changes

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1. Second plot is obtain when we calculate Conversion, XEB vs. Temperature based on Energy Balance Equation.

Q•


n



'(= 0


1. By combining this 2 plot we can find the Reaction Conversion, X by determining its intersection.

Conversion, X


Conversion plot from Mass Balance Equation, XMB

Conversion plot from Energy Balance Equation, XEB

X

To

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1. If a reaction parameter were changed slightly, the XEB line could move slightly to the left and there might be more than one intersection between both plots.

2. When more than one intersection occurs, there is more than one set of conditions that satisfy both the energy balance and mole balance

3. This is called as multiple steady states at which the reaction may operate

Conversion, X


Conversion plot from Mass Balance Equation, XMB

Conversion plot from Energy Balance Equation, XEB

X1

ToTo1

X2a

X2b

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