2nd order differential equation for Catalyst

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SUMMARY

The discussion centers on deriving the maximum temperature equation for a catalyst pellet in a reactor, specifically Tmax = Ts + [(-H)*(D*Cas)]/k. The user successfully developed an energy balance equation and derived the temperature equation using boundary conditions. They also formulated a mole balance equation for concentration, d^2Ca/dr^2 + (2/r)*dCa/dr - (k/D)*Ca = 0, but faced challenges in solving it. Ultimately, the user confirmed their temperature equation and sought guidance on solving the concentration equation.

PREREQUISITES
  • Understanding of second-order differential equations
  • Familiarity with energy and mole balance concepts in chemical engineering
  • Knowledge of boundary conditions in differential equations
  • Proficiency in calculus, particularly integration techniques
NEXT STEPS
  • Study the method of solving second-order differential equations with variable coefficients
  • Learn about boundary value problems in the context of chemical reactors
  • Explore numerical methods for solving differential equations, such as the finite difference method
  • Investigate the implications of heat transfer coefficients on reaction rates in catalytic processes
USEFUL FOR

Chemical engineers, students studying reaction engineering, and researchers focused on catalyst behavior in reactors will benefit from this discussion.

FaNgS
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Homework Statement



There's a catalyst pellet in a reactor and I'm supposed to prove an equation for the maximum temperature which is:

Tmax=Ts +[(-H)*(D*Cas)]/k

Tmax= max temp,
Ts = temp at catalyst surface,
T=temperature,
H=enthalpy of rxn,
D=diffusivity,
Cas= concentration at catalyst surface,
Ca=concenration
k= heat transfer coefficient,
r= radius,
R=radius of catalyst sphere,
Rxn= rate of reaction

2. The attempt at a solution
First thing I did was develop an Energy Balance across the spherical catalyst and I got the following equation:

(1/r^2)*d/dr(r^2*k*dT/dr) + (-H)*(-Rxn) = 0

expanding i get:

d^2T/dr^2 + (2/r)*dT/dr + (-H)*(-Rxn)/k = 0

Boundary Conditions: at r=0, dT/dr =0 AND at r=R, T=Ts (at the catalyst surface i.e. r=R the temperature T = Ts (catalyst surface temperature))

Using the boundary conditions and integrating factor I got
T=Ts + [(-H)(-Rxn)*(R^2-r^2)]/(6*k) ...(eqn 1)

Now for the mole balance across the catalyst I got (where Ca is the concentration):

d^2Ca/dr^2 + (2/r)*dCa/dr - (k/D)*Ca =0 ... (eqn 2)

Boundary Condition: at r=R, Ca=Cas

Can someone confirm the equation I got for temperature (eqn 1) and also for the concentration (eqn 2) how do I go about to solve it??

I tried another way to solve it, as suggested by my instructor, which is by using the un-expanded forms of the energy and mole balance equations which are:

(1/r^2)*d/dr(r^2*k*dT/dr) + (-H)*(-Rxn) = 0

(1/r^2)*d/dr(r^2*D*dCa/dr) + (Rxn) = 0

So here I have the "Rxn" term common in both equations and I combined and got the following after some simplifications:

d^2Ca/dr^2 + 2*dCa/dr = [k/(H*D)]* { r*d^2T/dr^2 + 2*dT/dr}

But in this case it looks way more complicated and I'm not sure how to deal with this type of an equation, since I have 2 derivates on both sides of the equation one with respect to the Concentration Ca and one with respect to temperature T.

:confused: not sure which way to proceed and how to proceed
 
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FaNgS said:

Homework Statement



There's a catalyst pellet in a reactor and I'm supposed to prove an equation for the maximum temperature which is:

Tmax=Ts +[(-H)*(D*Cas)]/k

Tmax= max temp,
Ts = temp at catalyst surface,
T=temperature,
H=enthalpy of rxn,
D=diffusivity,
Cas= concentration at catalyst surface,
Ca=concenration
k= heat transfer coefficient,
r= radius,
R=radius of catalyst sphere,
Rxn= rate of reaction

2. The attempt at a solution
First thing I did was develop an Energy Balance across the spherical catalyst and I got the following equation:

(1/r^2)*d/dr(r^2*k*dT/dr) + (-H)*(-Rxn) = 0

expanding i get:

d^2T/dr^2 + (2/r)*dT/dr + (-H)*(-Rxn)/k = 0
Since it was already partially solved for you don't expand!
d(r^2k(dT/dr)= -HRxn r^2 dr
gives a first integral of
r^2k (dT/dr)= -(HRxn/3)r^3+ C
and then
k dT= (-(HRxn/3)r+ Cr^{-2}) dr
integrates to
kT= -(HRxn/6)r^2- C/r+ D


Boundary Conditions: at r=0, dT/dr =0
The fact that dT/dr exists at r= 0 means that C= 0

AND at r=R, T=Ts (at the catalyst surface i.e. r=R the temperature T = Ts (catalyst surface temperature))
so kTs= -(HRxn/6)R^2+ D and D= kTs+ (HRxn/6)R^2

Using the boundary conditions and integrating factor I got
T=Ts + [(-H)(-Rxn)*(R^2-r^2)]/(6*k) ...(eqn 1)

Now for the mole balance across the catalyst I got (where Ca is the concentration):

d^2Ca/dr^2 + (2/r)*dCa/dr - (k/D)*Ca =0 ... (eqn 2)

Boundary Condition: at r=R, Ca=Cas

Can someone confirm the equation I got for temperature (eqn 1) and also for the concentration (eqn 2) how do I go about to solve it??

I tried another way to solve it, as suggested by my instructor, which is by using the un-expanded forms of the energy and mole balance equations which are:

(1/r^2)*d/dr(r^2*k*dT/dr) + (-H)*(-Rxn) = 0

(1/r^2)*d/dr(r^2*D*dCa/dr) + (Rxn) = 0

So here I have the "Rxn" term common in both equations and I combined and got the following after some simplifications:

d^2Ca/dr^2 + 2*dCa/dr = [k/(H*D)]* { r*d^2T/dr^2 + 2*dT/dr}

But in this case it looks way more complicated and I'm not sure how to deal with this type of an equation, since I have 2 derivates on both sides of the equation one with respect to the Concentration Ca and one with respect to temperature T.

:confused: not sure which way to proceed and how to proceed
 
so here you used D as the integration constant after the second integration and got D=kTs+ (H*Rxn/6)*R^2, if i substitute that back into the equation won't everything cancel out??

Edit 1: nevermind, what i said earlier...ok so now this confirms the first equation.

can you give me any hints on how to proceed to solve the equation for the concentration?


Edit 2: sorry for being jumpy, i figured it out and got the same equation as required (Tmax)
thanks a lot for your help i really appreciate it, I've been stuck on this for some time

thank you sir!
 
Last edited:

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