A tank with a capacity of 1600 L is full of a mixture of water and chlorine with a concentration of 0.0125 g of chlorine per liter. In order to reduce the concentration of chlorine, fresh water is pumped into the tank at a rate of 16 L/s. The mixture is kept stirred and is pumped out at a rate of 40 L/s. Find the amount of chlorine in the tank as a function of time. (Let y be the amount of chlorine in grams and t be the time in seconds.)

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Mathematics

28 October, 11:14

A tank with a capacity of 1600 L is full of a mixture of water and chlorine with a concentration of 0.0125 g of chlorine per liter. In order to reduce the concentration of chlorine, fresh water is pumped into the tank at a rate of 16 L/s. The mixture is kept stirred and is pumped out at a rate of 40 L/s. Find the amount of chlorine in the tank as a function of time. (Let y be the amount of chlorine in grams and t be the time in seconds.)

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Nigel · Answer 1

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)