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Integration help

  • Thread starter Triathlete
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  • #1
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Homework Statement



I am working on a mass transfer problem and have this equation:

d/dt(Vc1) = Ak[c1(sat) - c1]


Homework Equations



Initial conditions:

x=0, c1=0

The Attempt at a Solution



I know that the result from integration should be:

c1/c1(sat) = 1 – e-(kA/V)t

But I don't understand the integration that got there. If someone could show me the steps, that would be very helpful!
Thanks!
 

Answers and Replies

  • #2
33,630
5,288

Homework Statement



I am working on a mass transfer problem and have this equation:

d/dt(Vc1) = Ak[c1(sat) - c1]
What are the symbols here? What's the difference between c1 and c1(sat). Does Vc1 mean V * c1 or is it ##V_{c_1}##
Triathlete said:

Homework Equations



Initial conditions:

x=0, c1=0

The Attempt at a Solution



I know that the result from integration should be:

c1/c1(sat) = 1 – e-(kA/V)t

But I don't understand the integration that got there. If someone could show me the steps, that would be very helpful!
Thanks!
 
  • #3
33
0
Thanks for the response, c1 is the concentration c1(sat) is the saturated concentration. It is supposed to be V*c1.
 
  • #4
RUber
Homework Helper
1,687
344
## \frac{d}{dt}Vc_1 =Ak [ c_1 (sat) - c_1 ]##
This problem is easier to do as a differential equation.

Let's call this ##V c'(t) = Ak c_{sat} - Akc(t) ##.
This gives:
##V c'(t) +Akc(t) = Ak c_{sat}\\
c'(t) + \frac{Ak}{V} c(t) =\frac{Ak}{V} c_{sat} ##
A general solution to
##c'(t) + \frac{Ak}{V} c(t)=0## is ##c(t) = N e^{- \frac{Ak}{V} t} ##.
Where N is a constant.
Using this general solution, you can solve for a particular solution that satisfies the differential form you were given and initial conditions.
Notice that you can just add ##c_{sat}## to ##c(t) ## without affecting the derivative, since it is a constant.
So a solution to
##c'(t) + \frac{Ak}{V} c(t)= \frac{Ak}{V}c_{sat}## is ##c(t) = N e^{- \frac{Ak}{V} t} +c_{sat} ##.
Now, use your initial condition to solve for N.
##c(t) = N e^{- \frac{Ak}{V} t} +c_{sat} ## with ##c(0) = 0##.
...
From there, you should see how the solution came about.
 
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