Differential equation question

Click For Summary
SUMMARY

The discussion focuses on modeling the concentration of algae in a 1200 m³ treatment tank using a differential equation. The concentration, C, must remain between 10 gm/l and 20 gm/l for optimal algae performance. The rate of decrease of C is proportional to C, leading to the equation dC/dt = -kC. The solution involves separation of variables, yielding C = C₀e^(-kt), where C₀ represents the initial concentration. The discussion also highlights the significance of plotting ln|C| versus time to confirm the exponential decay relationship.

PREREQUISITES
  • Understanding of differential equations, specifically first-order linear equations
  • Familiarity with separation of variables technique in calculus
  • Knowledge of exponential functions and their properties
  • Basic understanding of concentration and units in chemistry
NEXT STEPS
  • Learn about first-order linear differential equations and their applications
  • Study the method of separation of variables in greater detail
  • Explore the concept of exponential decay in biological systems
  • Investigate the relationship between experimental data and theoretical models in concentration studies
USEFUL FOR

Mathematicians, environmental scientists, and chemical engineers involved in modeling biological processes and optimizing treatment systems.

lektor
Messages
56
Reaction score
0
Kinda lengthy question but bare with me please

A 1200 m3 treatment tank is used for treating effluent. The use of a particular form of algae is a major part of this process. It is important that the concentration, C, of algae not fall below 10 gm per litre as the process will not work. Above a concentration of 20 gm per litre the algae begin to die off.

Call the effective range for the concentration.
Note: We will work in terms of time in hrs, mass in gm and volume in litres.

As time passes algae are used up in the process in such a way that the rate of decrease of concentration of algae is proportional to the concentration of algae. To deal with this the algae is topped up at regular intervals of T hours. There is a special tank for growing algae from which the algae is taken.

The problem is to find the correct mass of algae to add and the time intervals at which to add it to keep the concentration as high as possible within the effective range.

1) Write the differential equation that describes the decrease in concentration of the algae. (i.e. don’t worry about the top up every T hours). If the initial concentration of algae is g /l write down the initial condition.

For this as

dC/dt is proportional to C

dc/dt = kC

2) Solve your equation using separation of variables. You MUST show every step in your solution.

Next using separation of variables i get

> dc * 1/c = k dt
> ln |C| = kt
> C = Ae^kt A = e^c

This is where I am stuck how exactly do i solve this with the information give? any help would be appreciated.
 
Physics news on Phys.org
You need an initial condition to find A. However, the description doesn't seem to give an initial C.

Therefore, put somehting like C=C_0 at t=0 which then gives the constant to be A=C_0

I presume the question goes on past part 2...
 
Since the concentration decreases, it should be dC/dt = -kt. Don't forget your constant of integration, and from this you can find out what it is since the initial concentration is g/l.
 
daveb said:
since the initial concentration is g/l.
I completely missed where it gave this in the question :redface:
 
cheers for the help guys, got C = Co*e^(-kt)

Would anyone be able to explain the mathematical reason why plotting ln|c| vs t approximating a straight line with experimental data proves this relationship.. cheers.
 
Last edited:
If you were to take the natural log of both sides, you end up with ln(c) = ln (Co) -kt. Since ln(co) is just a number (let it equal c1), if you let y = ln(c), then the equation becomes y = -kt + c1 which is a linear equation with slope -k and intercet c1. So, from experimental data, if you plot the ln(c) versus t, and you get a linear plot, then this shows that the you have exponential decay, and using interpolation, you can calculate what k and Co should be.
 

Similar threads

Replies
2
Views
2K
Replies
0
Views
1K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
5
Views
3K
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K