Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of

chemical reactions and the design of the reactors in which they take place.

Lecture 19

Today’s lecture   Gas Phase Reactions   Trends and Optimums

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User Friendly Equations Relate T and X or Fi 1. Adiabatic CSTR, PFR, Batch, PBR achieve this:

Last Lecture

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2. CSTR with heat exchanger, UA(Ta-T) and a large coolant flow rate

T Ta

User Friendly Equations Relate T and X or Fi

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3. PFR/PBR with heat exchange

FA0 T0

Coolant Ta

User Friendly Equations Relate T and X or Fi

3A. In terms of conversion, X

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User Friendly Equations Relate T and X or Fi 3B. In terms of molar flow rates, Fi

4. For multiple reactions

5. Coolant Balance

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Reversible Reactions

endothermic reaction

exothermic reaction

KP

T

endothermic reaction

exothermic reaction

Xe

T

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Heat Exchange

Example: Elementary liquid phase reaction carried out in a PFR

FA0 FI

Ta

Heat Exchange Fluid

The feed consists of both inerts I and Species A with the ratio of inerts to the species A being 2 to 1.

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Heat Exchange a)   Adiabatic. Plot X, Xe, T and the rate of disappearance as

a function of V up to V = 40 dm3.

b)   Constant Ta. Plot X, Xe, T, Ta and Rate of disappearance of A when there is a heat loss to the coolant and the coolant temperature is constant at 300 K for V = 40 dm3. How do these curves differ from the adiabatic case.

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Heat Exchange c)   Variable Ta Co-Current. Plot X, Xe, T, Ta and Rate of

disappearance of A when there is a heat loss to the coolant and the coolant temperature varies along the length of the reactor for V = 40 dm3. The coolant enters at 300 K. How do these curves differ from those in the adiabatic case and part (a) and (b)?

d)   Variable Ta Counter Current. Plot X, Xe, T, Ta and Rate of disappearance of A when there is a heat loss to the coolant and the coolant temperature varies along the length of the reactor for V = 20 dm3. The coolant enters at 300 K. How do these curves differ from those in the adiabatic case and part (a) and (b)? 10

Example: PBR A ↔ B

5) Parameters….

Gas Phase Heat Effects

•  For adiabatic:

•  Constant Ta:

•  Co-current: Equations as is

•  Counter-current:

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Reversible Reactions

Mole Balance:

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Reversible Reactions

Rate Law:

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Reversible Reactions

Stoichiometry:

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5( ) CA = CA 0 1− X( )y T0 T( )

€

6( ) CB = CA 0Xy T0 T( )

€

dydW

=αyFTFT0

TT0

⎛

⎝ ⎜

⎞

⎠ ⎟ = −

α2y

TT0

⎛

⎝ ⎜

⎞

⎠ ⎟

W = ρV

dydV

= −αρb2y

TT0

⎛

⎝ ⎜

⎞

⎠ ⎟

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Reversible Reactions

Parameters:

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Reversible Reactions

Example: PBR A ↔ B

3) Stoich (gas):

€

v = v0 1+εX( ) P0

PTT0

5( ) CA =FA 0 1− X( )v0 1+εX( )

PP0

T0

T=CA 0 1− X( )

1+εX( )y T0

T

6( ) CB =CA 0X1+εX( )

y T0

T

7( ) dydW

=−α2y

FTFT 0

TT0

=−α2y

1+εX( ) TT0

Gas Phase Heat Effects

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Reversible Reactions

€

KC =CBe

CAe

=CA 0XeyT0 T

CA 0 1− Xe( )yT0 T

8( ) Xe =KC

1+KC

Example: PBR A ↔ B

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Gas Phase Heat Effects Reversible Reactions

Example: PBR A ↔ B

Gas Phase Heat Effects

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Exothermic Case:

Xe

T

KC

T

KC

T T

Xe ~1

Endothermic Case:

Reversible Reactions

Adibatic Equilibrium Conversion on Temperature

Exothermic ΔH is negative

Adiabatic Equilibrium temperature (Tadia) and conversion (Xeadia) X

Xeadia

Tadia T 19

X2

FA0 FA1 FA2 FA3

T0 X1 X3 T0 T0

Q1 Q2

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X

T

X3

X2

X1

T0

Xe XEB

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Gas Phase Heat Effects

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€

dTdV

=−rA( ) −ΔHRx( ) −Ua T −Ta( )

∑FiCPi

€

∑FiCPi= FA 0 ∑ΘiCPi

+ ΔCPX[ ]Case 1: Adiabtic and ΔCP=0

€

T = T0 +−ΔHRx( )X∑ΘiCPi

(16A)

Additional Parameters (17A) & (17B)

€

T0, ∑ΘiCPi= CPA

+ΘICPI

Example A ↔ B

Heat effects:

€

dTdW

=−ra( ) −ΔHR( ) −Ua

ρbT −Ta( )

FA 0 θ iCPi∑ 9( )

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Case 2: Heat Exchange – Constant Ta

Gas Phase Heat Effects Example A ↔ B

Case 3. Variable Ta Co-Current

Case 4. Variable Ta Counter Current

Guess Ta at V = 0 to match Ta0 = Ta0 at exit, i.e., V = Vf

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Example A ↔ B

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Effects of Inerts In The Feed

What happens when we vary

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Endothermic

As inert flow increases the conversion will increase. However as inerts increase, reactant concentration decreases, slowing down the reaction. Therefore there is an optimal inert flow rate to maximize X.

First Order Irreversible

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T

X

Adiabatic T

and Xe

T0

exothermic

T

X

T0

endothermic

Gas Phase Heat Effects Trends: -Adiabatic:

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Adiabatic: •  As T0 decreases the conversion X will increase, however the

reaction will progress slower to equilibrium conversion and may not make it in the volume of reactor that you have.

•  Therefore, for exothermic reactions there is an optimum inlet temperature, where X reaches Xeq right at the end of V. However, for endothermic reactions there is no temperature maximum and the X will continue to increase as T increases.

Gas Phase Heat Effects

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X

T

Xe

T0

X

T

X

T

Adiabatic:

Gas Phase Heat Effects

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Effect of adding Inerts

X

T

V1 V2 X

T T0

Xe

X

€

X =T −T0( ) CpA +θ ICpI[ ]

−ΔHRx

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Exothermic Adiabatic

As θI increase, T decrease and

€

dXdV

=k

υ0 Hθ I( )

k

θI

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Adiabatic

Endothermic

Exothermic

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Heat Exchange

Endothermic

Exothermic

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Derive the Steady State Energy Balance (w/o Work)

Differentiating with respect to W:

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Mole Balance on species i:

Enthalpy for species i:

Derive the Steady State Energy Balance (w/o Work)

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Differentiating with respect to W:

Derive the Steady State Energy Balance (w/o Work)

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Final Form of the Differential Equations in Terms of Conversion:

A:

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Final Form of terms of Molar Flow Rate:

B:

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End of Lecture 19

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Example: PBR A ↔ B

2) Rates:

1) Mole balance:

Gas Phase Heat Effects

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