# Integration help

## 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!

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Mark44
Mentor

## 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!

Thanks for the response, c1 is the concentration c1(sat) is the saturated concentration. It is supposed to be V*c1.

RUber
Homework Helper
## \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.

Triathlete