A tank initially contains 120 L of pure water. A mixture containing a concentration of γ g/L of salt enters the tank at a rate of 2 L/min, and the well-stirred mixture leaves the tank at the same rate. Find an expression in terms of γ for the amount of salt in the tank at any time t. Also find the limiting amount of salt in the tank as t → [infinity].

A tank contain 120L

Volume V = 120L

Solution of γ g/L of sugar

Rate of entry i.e input

dL/dt=2L/min

Let M(t) be the amount of sugar in tank at any time.

But at the beginning there was no sugar in the tank

i.e, M(0)=0, this will be out initial value problem,

The rate of amount of sugar at anytime t is

dM/dt=input amount of sugar - output amount of sugar.

Now,

Then rate of input is

2L/min × γ g/L

Then, input rate= 2γ g/mins

Output rate is

2L/mins × M(t)/120 kg/L

then, output rate = M(t)/60 g/min

So,

dM/dt=input rate -output rate

dM/dt= 2γ - M/60

Cross multiply through by 60

60dM/dt= 120γ - M

Using variable separation

60/(120γ - M) dM= dt

Integrate both sides

∫60/(120γ - M) dM= ∫dt

-60In(120γ - M)=t +C

In(120γ - M)=-t/60+C/60

C/60 is another constant, let say B

In(120γ - M)=-t/60+B

Take exponential of both side

120γ - M=exp(-t/60+B)

120γ - M=exp(-t/60)exp(B)

exp(B) is a constant let say C

-120γ - M=Cexp(-t/60)

- M=Cexp(-t/60) - 120γ

M= 120γ - Cexp(-t/60)

Now, the initial condition

a. At the start the mass of sugar in the water is 0 because it is just pure water at start.

Therefore M(0)=0,

b. Applying this to M(t)

M= 120γ - Cexp(-t/60)

M=0, t=0

0 = 120γ - Cexp(0)

0 = 120γ - C

C= 120γ

Therefore,

M= 120γ - 120γ exp(-t/60)

M =120γ[1 - exp(-t/60)]

Let know the mass rate as t tends to infinity

At infinity

exp(-∞)=1/exp(∞)=1/∞=0

Then,

The exponential aspect tend to 0

Then, M(t)=120γ as t tend to ∞

the answer is 46 mls because when you add the speed per second it increses the speed

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