A tank contains 3,000 L of brine with 15 kg of dissolved salt. Pure water enters the tank at a rate of 30 L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate.
How much salt is in the tank after t minutes?

Step-by-step explanation:

Volume of tank is 3000L.

Mass of salt is 15kg

Input rate of water is 30L/min

dV/dt=30L/min

Let y(t) be the amount of salt at any time

Then,

dy/dt = input rate - output rate.

The input rate is zero since only water is added and not salt solution

Now, output rate.

Concentrate on of the salt in the tank at any time (t) is given as

Since it holds initially holds 3000L of brine then the mass to volume rate is y(t)/3000

dy/dt= dV/dt × dM/dV

dy/dt=30×y/3000

dy/dt=y/100

Applying variable separation to solve the ODE

1/y dy=0.01dt

Integrate both side

∫ 1/y dy = ∫ 0.01dt

In(y)= 0.01t + A, .A is constant

Take exponential of both side

y=exp(0.01t+A)

y=exp(0.01t)exp(A)

exp(A) is another constant let say C

y(t)=Cexp(0.01t)

The initial condition given

At t=0 y=15kg

15=Cexp(0)

Therefore, C=15

Then, the solution becomes

y(t) = 15exp(0.01t)

At any time that is the mass.

⊂⊃⊂⊃ω

step-by-step explanation:

i hope this : )

pi x^2

step-by-step explanation:

pi times x squared