Answer:
y(t) = 20 [1600^(-5/3)] x (1600-24t)^ (5/3)
Step-by-step explanation:
1) Identify the problem
This is a differential equation problem
On this case the amount of liquid in the tank at time t is 1600−24t. (When the process begin, t=0 ) The reason of this is because the liquid is entering at 16 litres per second and leaving at 40 litres per second.
2) Define notation
y = amount of chlorine in the tank at time t,
Based on this definition, the concentration of chlorine at time t is y/(1600−24t) g/ L.
Since liquid is leaving the tank at 40L/s, the rate at which chlorine is leaving at time t is 40y/(1600−24t) (g/s).
For this we can find the differential equation
dy/dt = - (40 y)/ (1600 -24 t)
The equation above is a separable Differential equation. For this case the initial condition is y(0)=(1600L )(0.0125 gr/L) = 20 gr
3) Solve the differential equation
We can rewrite the differential equation like this:
dy/40y = - (dt)/ (1600-24t)
And integrating on both sides we have:
(1/40) ln |y| = (1/24) ln (|1600-24t|) + C
Multiplying both sides by 40
ln |y| = (40/24) ln (|1600 -24t|) + C
And simplifying
ln |y| = (5/3) ln (|1600 -24t|) + C
Then exponentiating both sides:
e^ [ln |y|]= e^ [(5/3) ln (|1600-24t|) + C]
with e^c = C , we have this:
y(t) = C (1600-24t)^ (5/3)
4) Use the initial condition to find C
Since y(0) = 20 gr
20 = C (1600 -24x0)^ (5/3)
Solving for C we got
C = 20 / [1600^(5/3)] = 20 [1600^(-5/3)]
Finally the amount of chlorine in the tank as a function of time, would be given by this formula:
y(t) = 20 [1600^(-5/3)] x (1600-24t)^ (5/3)
