# Integration help

1. Dec 1, 2015

### Triathlete

1. The problem statement, all variables and given/known data

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

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

2. Relevant equations

Initial conditions:

x=0, c1=0

3. 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!

2. Dec 1, 2015

### Staff: Mentor

What are the symbols here? What's the difference between c1 and c1(sat). Does Vc1 mean V * c1 or is it $V_{c_1}$

3. Dec 1, 2015

### Triathlete

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

4. Dec 1, 2015

### RUber

$\frac{d}{dt}Vc_1 =Ak [ c_1 (sat) - c_1 ]$
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.