Reactor Design and Kinetics of Reactions

Click For Summary
SUMMARY

The discussion focuses on the application of reaction kinetics in mass conservation equations, specifically in Continuous Stirred-Tank Reactors (CSTRs). Participants clarify that the consumption term should be expressed as -kVCn rather than -V(dC/dt), emphasizing that the latter is only valid for batch reactors. The conversation highlights the confusion stemming from textbooks that incorrectly equate these terms, which can mislead students in chemical engineering. A recommended resource for further understanding is "Chemical Reaction Engineering" by Octave Levenspiel.

PREREQUISITES
  • Understanding of mass conservation laws in chemical engineering
  • Familiarity with reaction kinetics, specifically rate laws
  • Knowledge of Continuous Stirred-Tank Reactor (CSTR) dynamics
  • Basic calculus, particularly differentiation and integration
NEXT STEPS
  • Study the derivation of mass balance equations in CSTRs
  • Learn about the differences between batch reactors and flow reactors
  • Explore the implications of reaction rate expressions in chemical engineering
  • Read "Chemical Reaction Engineering" by Octave Levenspiel for in-depth concepts
USEFUL FOR

Chemical engineering students, particularly those studying reaction kinetics and reactor design, as well as professionals involved in process engineering and wastewater treatment.

CivilSigma
Messages
227
Reaction score
58

Homework Statement



My problem deals with understanding why we substitute in the reaction kinetics equivalent into mass conservation equation instead of dealing with differentials

Homework Equations


From the conservation of mass law about an envelope:
$$\frac{dm}{d t} = \sum m_{in} - \sum m_{out} + Generation - Consumption$$

Assuming there is a consumption of some compound, then:
$$Consumption = -\frac{dc}{dt} = kC^n$$

The Attempt at a Solution


[/B]

Why is that when we go to solve the mass balance equation, usually for the unknown concentration that varies with time (in this case for a complete mix reactor), we make the following substitution:

$$\frac{dm}{d t} = \sum m_{in} - \sum m_{out} - kVC^n$$

Why can't we do this:

$$\frac{dm}{d t} = \sum m_{in} - \sum m_{out} + -\frac{dc}{dt} \cdot V$$
$$ V \cdot \frac{dc}{dt} = \sum m_{in} - \sum m_{out} -\frac{dc}{dt} \cdot V$$

and then isolate for the differential terms, and integrate to get the final concentration??

Thank you.
 
Physics news on Phys.org
Hi,

1. check dimensions. In your first equation you don't have ##m_{\rm in}## but ##{dm_{\rm in}\over dt}## (usually written as ##\,\dot m_{\rm in}## ).
2. Consumption is not -dc/dt but dcV/dt -- as you use in your third equation
3. In your last equation you seem to take V as constant. It's not.
 
The problem is with this equation: $$Consumption = -\frac{dc}{dt} = kC^n$$
It is only valid for a batch reactor. I know it is often common to write down an expression like this, but it always struck me as just plain stupid when referring to flow reactor. So, in the case of a flow reactor, dc/dt is not equal to ##-kC^n##. In a flow reactor, the rate of consumption of a species is equal to the reactor volume times the reaction rate: ##VkC^n##.
 
  • Like
Likes   Reactions: CivilSigma
Yes I agree @Chester.

Also, I forgot to state my assumption, sorry:

Assume that we are dealing with a CSTR ( complete mixing ) and that the volume of the tank is constant.

But, in the mass balance equation, why do we explicitly equate consumption to

$$-kVC^n$$

but not simply use
$$-V\cdot \frac{dc}{dt}$$

in the equation?

I know they are equal, but I have an urge to keep the differential form of consumption in the equation and solve it.
 
sakonpure6 said:
Yes I agree @Chester.

Also, I forgot to state my assumption, sorry:

Assume that we are dealing with a CSTR ( complete mixing ) and that the volume of the tank is constant.

But, in the mass balance equation, why do we explicitly equate consumption to

$$-kVC^n$$

but not simply use
$$-V\cdot \frac{dc}{dt}$$

in the equation?

I know they are equal, but I have an urge to keep the differential form of consumption in the equation and solve it.
In a cstr, they are not equal. Read my post again.
 
  • Like
Likes   Reactions: CivilSigma
@Chestermiller, that's so weird, in my textbook, they say they are equal and transition from dc/dt to kC
 

Attachments

  • 20170708_005319.jpg
    20170708_005319.jpg
    29.4 KB · Views: 424
sakonpure6 said:
@Chestermiller, that's so weird, in my textbook, they say they are equal and transition from dc/dt to kC
This is why, in my first post, I said that it is "just plain stupid." The author of your textbook (and many others in their textbooks) has done you a disservice by writing the equation in this way and confusing you. Unfortunately, this is not the only thing in textbooks that confuse students. Have you had a course in thermodynamics yet? Thermodynamics texts are chock full of confusing and incorrect statements.
 
  • Like
Likes   Reactions: CivilSigma
Chestermiller said:
This is why, in my first post, I said that it is "just plain stupid." The author of your textbook (and many others in their textbooks) has done you a disservice by writing the equation in this way and confusing you. Unfortunately, this is not the only thing in textbooks that confuse students. Have you had a course in thermodynamics yet? Thermodynamics texts are chock full of confusing and incorrect statements.

Oh wow, I've been cheated :(. No , I will have not and am not required to take any thermodynamics courses (Civil Engineering Student). My original question arose from a Waste Water Process Engineering course I am currently taking.

Can you please recommend a reference that I can read to help learn about more this?
 
sakonpure6 said:
Oh wow, I've been cheated :(. No , I will have not and am not required to take any thermodynamics courses (Civil Engineering Student). My original question arose from a Waste Water Process Engineering course I am currently taking.

Can you please recommend a reference that I can read to help learn about more this?
I like Chemical Reaction Engineering by Octave Levenspiel. This book is available as a pdf on line. However, even Levenspiel makes this same boo boo with the time derivative. In his defense, he no longer continues to do this when he gets to sections of the book where he describes how to analyze CSTRs and plug flow reactors.
 
  • Like
Likes   Reactions: CivilSigma
  • #10
thanks
 

Similar threads

  • · Replies 27 ·
Replies
27
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
2
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
19
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 16 ·
Replies
16
Views
2K