2nd order differential equation for Catalyst

In summary, the student is trying to find an equation for the temperature at the catalyst surface, using the energy and mole balance across the catalyst sphere. They get the equation for temperature and concentration, but are stumped on how to solve for the concentration.
  • #1
FaNgS
91
0

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
 
Physics news on Phys.org
  • #2
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!
[tex] d(r^2k(dT/dr)= -HRxn r^2 dr[/tex]
gives a first integral of
[tex]r^2k (dT/dr)= -(HRxn/3)r^3+ C[/tex]
and then
[tex]k dT= (-(HRxn/3)r+ Cr^{-2}) dr[/tex]
integrates to
[tex]kT= -(HRxn/6)r^2- C/r+ D[/tex]


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 [itex]kTs= -(HRxn/6)R^2+ D[/itex] and [itex]D= kTs+ (HRxn/6)R^2[/itex]

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
 
  • #3
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:

1. What is a 2nd order differential equation for Catalyst?

A 2nd order differential equation for Catalyst is a mathematical equation that describes the rate of change of a catalyst in a chemical reaction. It takes into account the concentration of the catalyst, the reaction rate, and the rate of change of the reaction rate.

2. Why is a 2nd order differential equation used for Catalyst?

A 2nd order differential equation is used for Catalyst because it allows for a more accurate prediction of the behavior of a catalyst in a chemical reaction. It takes into account not only the initial concentration of the catalyst, but also how the reaction rate changes over time.

3. How is a 2nd order differential equation solved for Catalyst?

A 2nd order differential equation for Catalyst can be solved using various mathematical techniques, such as separation of variables, substitution, or using a system of equations. The specific method used will depend on the specific form of the equation.

4. What factors can affect the accuracy of a 2nd order differential equation for Catalyst?

Several factors can affect the accuracy of a 2nd order differential equation for Catalyst, including temperature, pressure, and the presence of other substances in the reaction. Other factors such as the purity of the catalyst and any experimental errors can also impact the accuracy of the equation.

5. How can a 2nd order differential equation for Catalyst be applied in real-world situations?

A 2nd order differential equation for Catalyst can be applied in real-world situations by using it to predict the behavior of a catalyst in a chemical reaction. This can be useful in industries such as pharmaceuticals, where understanding the behavior of catalysts is crucial for the development of new drugs. It can also help in optimizing reaction conditions and identifying potential issues or limitations in a reaction process.

Similar threads

Replies
0
Views
462
  • Calculus and Beyond Homework Help
Replies
15
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
136
  • Calculus and Beyond Homework Help
Replies
1
Views
294
Replies
4
Views
509
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
993
  • Calculus and Beyond Homework Help
Replies
7
Views
294
  • Calculus and Beyond Homework Help
Replies
2
Views
920
  • Calculus and Beyond Homework Help
Replies
3
Views
575
Back
Top