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Equilibrium Reaction In Plug Flow Reactor

plug flow equilibrium constant activation energy pre exponential factor methane water co hydrogen rate law

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#1 TonyD

TonyD

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Posted 20 August 2015 - 04:03 AM

Hi

 

I have to model the equilibrium reaction

 

CH4 + H2O <-> CO + 3 H2

 

in a plug flow reactor (isotherm) with a katalytic reaction.

 

I have following data given:

 

The Inlet stream:

 

y(CH4) = 1,715*10^-3

y(H2O) = 0,366

y(CO) = 0

y(H2) = 0,448

y(N2) = 0,184285

 

T = const = 650 °C

p = const = 5 bar

 

RATE LAW: r = -kCH4 p ( y(CH4) - y(CH4,Equilibrium) )

 

I have calculate the equilibrium constant to be KEq = 2,7724

and the y(CH4,Equilibrium) concentration to be 0,00052935

 

k0,CH4 = 90,2 mol/(kg*Pa*s)

 

EA,CH4 = 63,5 kJ/mol

 

kCH4 = 2,30264*10^-5 kmol /(kg*Pa*s)

 

Is there a way to simulate this in a Plug Flow reactor?

 

Thank you very much!


Edited by TonyD, 20 August 2015 - 04:06 AM.


#2 Lucian Gomoescu

Lucian Gomoescu

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Posted 21 October 2015 - 11:20 AM

Hello,
 

Yes, there is a way to simulate this in a PFR within Aspen Plus.

I. What i understood about the problem is described in the following lines:

1. The reactor is of plug flow type and it is filled with solid catalyst particles.
2. The equilibrium reaction mentioned above describes the overall reaction.
3. The inlet stream is fully defined, except flow rate (but which is irelevant for this topic).
4. The process rate, rp [kmol/kg_cat-Pa-s], within reactor should respect this simple law:

 rp = k * (P_CH4 - P_CH4,ch_eq)  equivalent to rp = k * P *(y_CH4 - y_CH4,ch_eq)

, where:

k = k0 * exp(-Ea/RT);

k0 = 90.2 mol/kg_cat-Pa-s;
Ea = 63.5 kJ/mol;

y_CH4,ch_eq  is the methane mole fr. @ chemical equilibrium of overall reaction;

y_CH4,ch_eq = 0,00052935 => partial pres. of CH4 @ ch_eq is 265 Pa (considering total pressure constant & equal to 5 bar(a) ).

        

II. Possible solution in A+:

 

Use a reaction type 'general' class 'powerlaw' with checked box for reversible.
Specify the kinetic factor :

k is k0, n=0, E is Ea, T0 to be empty (because it is not considered inn kinetic factor expression);

Specify the proper basis and units (i.e. partial pressure in Pa, catalyst wt, rate in kmol/kg_cat-s, reacting phase: vapor).

Specify the driving force, i.e. (in our case) : (P_CH4^1 - P_CH4_ch_Eq).

 

It's general form is : K1 * Prod(P_i ^ expon_i) - K2 * Prod(P_i ^ expon_j).

 

For forward reaction rate:

K1 is 1 => A, B, C, D are zero.

exponents are 1, 0 // 0, 0.

So, the first term becomes 1 * P_CH4 * 1.

 

For reverse reaction rate:

K2 is 265 Pa => A=ln(265)=5.5785

exponents are 0, 0 // 0, 0

So, the 2nd term is 265 Pa * 1

 

III. Observations and suggestions:

1. The mass of catalyst is very important for the reactor performance therefore it should be defined correctly acc reactor geometry and catalyst bulk density.

2. The process rate, r, law is very rigid, i.e. for a different feed P or/and composition, the driving force reverse reaction rate param. A must be changed based on new yCH4_ch_eq and feed P, otherwise the law would not be respected.

3. A calculator block for the calculation from points (1) & (2) would solve this issue.

 

 

 

I hope this answers your question.
 

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