A brine solution of salt flows at a constant rate of 99 L/min into a large tank that initially held 100100 L of brine solution in which was dissolved 0.20.2 kg of salt. The solution inside the tank is kept well stirred and flows out of the tank at the same rate. If the concentration of salt in the brine entering the tank is 0.040.04 kg/L, determine the mass of salt in the tank after t min. When will the concentration of salt in the tank reach 0.030.03 kg/L?

Question

Dale Stout

Mathematics

24 April, 09:34

A brine solution of salt flows at a constant rate of 99 L/min into a large tank that initially held 100100 L of brine solution in which was dissolved 0.20.2 kg of salt. The solution inside the tank is kept well stirred and flows out of the tank at the same rate. If the concentration of salt in the brine entering the tank is 0.040.04 kg/L, determine the mass of salt in the tank after t min. When will the concentration of salt in the tank reach 0.030.03 kg/L?

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Yahir Fritz · Answer 1

Step-by-step explanation:

Let m (t) represent mass of the salt at any time (t)

Given that the rate of flow is 99 L/mins

Then dV/dt=99 L/min

Given that the input concentration of brine is 0.04 kg/L

dM/dV = 0.04Kg/L

Then, the rate of the input mass into the tank.

dM/dt = dM/dV * dV/dt

dM/dt = 0.04*99

dM/dt = 3.96Kg/min

Now, output rate.

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

Since it holds initially holds 100L of brine then the mass rate is m (t) / 100

dM/dt = dV/dt * M (t) / 100

dM/dt = 99*M (t) / 100

dM/dt = 0.99M

Given the initial value condition

M (0) = 0.2Kg

Then, the rate of mass rate is given as the input mass rate minus output mass rate

dM/dt = 3.96-0.99M

Using variable separation

1 / (3.96-0.99M) dM = dt

Integrate both side

∫1 / (3.96-0.99M) dM = ∫dt

-0.99In (3.96-0.99M) = t+C

Divide through by - 0.99

In (3.96-099M) = -1.01t-1.01C

-1.01C is still a constant, let say D

In (3.96-099M) = -1.01t+D

Take exponential of both side

3.96-0.99M=exp (-1.01t+D)

3.96-0.99M=exp (-1.01t) exp (D)

exp (D) is a constant, let say A

3.96-0.99M=Aexp (-1.01t)

-0.99M=-3.96+Aexp (-1.01t)

Divide through by - 0.99

-0.99M=-3.96-1.01Aexp (-1.01t)

-1.01A is still a constant, let say B

M (t) = 4+Bexp (-1.01t)

Using the initial condition M (0) = 0.2kg

0.2=4+Bexp (0)

0.2-4=B

B=-3.8

Then, the equation becomes

M (t) = 4-3.8exp (-1.01t)

2. When will salt concentration be 0.03kg/L

The mass is salt concentration * volume of liquid.

Mass=0.03kg/L * 100L

The mass is 3kg

M (t) = 4-3.8exp (-1.01t)

3=4-3.8exp (-1.01t)

3-4=-3.8exp (-1.01t)

-1=-3.8exp (-1.01t)

-1/-3.8=exp (-1.01t)

0.2632=exp (-1.01t)

Take In of both side

In (0.2632) = -1.01t

t=In (0.2632) : -1.01

t=1.32seconds