Combustion mass conservation integral

Click For Summary

Homework Help Overview

The discussion revolves around a differential equation related to mass conservation in combustion, specifically integrating an expression involving surface radius, density, velocity, and a non-dimensional parameter. The participants are exploring the implications of the equation and the assumptions regarding the variables involved.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to clarify the nature of the variables, particularly whether certain parameters are constants. Questions are raised about the application of the differentiation operator and its effect on the terms in the equation. There is a focus on the integration process and the steps involved in solving the differential equation.

Discussion Status

Some participants have provided insights into the integration process and have confirmed the form of the differential equation being discussed. There is an ongoing exploration of the steps needed to solve the equation, with no explicit consensus reached on the final approach.

Contextual Notes

Participants are navigating assumptions about the dependencies of the variables involved in the equation, which may affect the integration process. There is also a mention of potential confusion regarding the differentiation and integration steps.

greggleslarue
Messages
3
Reaction score
0

Homework Statement


integrate [(rs^2)*(rhos)*(us)*(db/dr)=(d/dr)*(r^2)*(rho)*(D)*(db/dr))]

rs=radius at surface
rhos=density at surface
us=velocity at surface
r = radius
rho = density
D= diffusivity
b=spalding non-dimensional parameter


Homework Equations





The Attempt at a Solution


This is the solution after one integration based on the assumptions that r, rho, D, u are not dependent

(rs^2)*(rhos)*(us)*(b)=(r^2)*(rho)*(D)*(db/dr)+(c1)

I think this involves separation of variables to solve and integrate but I don't see the steps clearly. Can someone please show me?
 
Physics news on Phys.org
First off you mean that rs is a constant right, not r? Secondly the right hand side seems a bit awkward. The d/dr operator does it act on everything that comes after it or?

If so, ignoring the constants, is this the differential equation you're trying to solve?:

[tex] \frac{db}{dr}=\frac{d}{dr}\left(r^2\frac{db}{dr}\right)[/tex]

Edit: Looking at your attempt at a solution this must be the case.

So after one integration you're left with [itex]b=r^2 b'+c \Rightarrow b'/(b-c)=1/r^2[/itex]. You should be able to integrate this expression.
 
Last edited:
Cyosis said:
First off you mean that rs is a constant right, not r? Secondly the right hand side seems a bit awkward. The d/dr operator does it act on everything that comes after it or?

If so, ignoring the constants, is this the differential equation you're trying to solve?:

[tex] \frac{db}{dr}=\frac{d}{dr}\left(r^2\frac{db}{dr}\right)[/tex]

Edit: Looking at your attempt at a solution this must be the case.

So after one integration you're left with [itex]b=r^2 b'+c \Rightarrow b'/(b-c)=1/r^2[/itex]. You should be able to integrate this expression.

Yes you are correct that is the basic equation I am trying to solve. I am getting hung up on the first integral though. I don't understand the steps from solving this:
db/dr=d/dr*(r^2*(db/dr))

I am thinking of this as separation of variables. IE move the dr on the left over to the right side. then the left side just becomes b, but the right side becomes more complicated. What are the steps for solving the original equation?
 
not the original equation i wrote but the original equation cyosis wrote.
Thanks
 
The first integral is pretty easy. Note that both sides are just "terms" that are getting differentiated with respect to r. Therefore integrating with respect to r will cancel out the differentiation operation bar a constant.

[tex] \begin{align*}<br /> \int \frac{db}{dr} dr &=\int \frac{d}{dr}\left(r^2\frac{db}{dr}\right)dr<br /> \\<br /> b &=r^2\frac{db}{dr}+c<br /> \end{align*}[/tex]
 
Last edited:

Similar threads

  • · Replies 14 ·
Replies
14
Views
3K
Replies
0
Views
1K
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
2K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
6
Views
2K