Question

A tank initially contains 50 gallons of brine, with 30 pounds of salt in solution. Water runs into the tank at 6 gallons per minute and the well-stirred solution runs out at 5 gallons per minute. How long will it be until there are 25 pounds of salt in the tank?The amount of time until 25 pounds of salt remain in the tank is " ? " minutes.Set up a differential equation and separate variables.

Answer 1

To determine the time until there are 25 pounds of salt in the tank, we set up a differential equation based on the inflow and outflow rates, separate variables, and integrate using initial conditions. The resulting equation is ln(S) = -5 ln(50 + t) + C, which can be solved to find the required time when S(t) is 25 pounds.

Step-by-step explanation:

To solve the problem of finding out how long it will take for the tank to have 25 pounds of salt, we'll set up a differential equation based on the rates of water and salt going in and out of the tank. Let's denote the amount of salt in the tank at any time t by S(t), measured in pounds, and t in minutes.

The rate of water entering the tank is 6 gallons/minute, and the rate of water leaving the tank is 5 gallons/minute. Let's assume that the concentration of salt in the tank remains uniform throughout the process due to proper stirring.

The rate of change of the amount of salt in the tank can be described by the differential equation:

dS/dt = (rate of salt in) - (rate of salt out)

Since no additional salt is being added, the rate of salt in is 0. The rate of salt out is the product of the concentration of the salt and the rate at which the solution is leaving the tank. The concentration of salt at any time is S(t) divided by the volume of the solution in the tank, which is 50 gallons plus 1 gallon per minute (because 6 gallons are entering and 5 gallons are leaving per minute).

The differential equation is:

dS/dt = 0 - (5 gallons/minute) * (S(t) / (50 + 1*t) gallons)

We can separate the variables and integrate:

dS/S = -5/ (50 + t) dt

Integrating both sides, we get:

ln(S) = -5 ln(50 + t) + C

To solve for C, we use the initial condition S(0) = 30 pounds:

ln(30) = -5 ln(50) + C

Now, we want to find the time t when S(t) = 25 pounds:

ln(25) = -5 ln(50 + t) + C

After putting in the value of C from the initial condition, we can solve for t to find out how long it takes for the tank to have 25 pounds of salt.

Answer 2

1.857 minutes

Step-by-step explanation:

From the given information:

Consider the amount of salt present in the tank as x(t) at a given time (t); &

The volume of the solution = V(t)

At t = 0 i.e. (at initial conditions) x(0) = 30; and V(0) = 50

However, the overall increase taken place for 1 gallon per minute is:

V(t) = 50 + t

The amount of salt x(t) at any given point for time (t) is;

`(x(t))/(V(t))= (x(t))/(50+t)`

After 5 gallons of solution exit per minute; the concentration of the salt solution changes at:

`(dx(t))/(dt)= -(5x(t))/(50+t)`

Taking the integral of what we have above, we get:

In x(t) = - 5 In(t + 50) + In (C)

In x(t) = In (t+ 50)⁻⁵ + In (C)

In x(t) = In C ( t + 50)⁻⁵

x(t) = C(t + 50)⁻⁵ (General solution)

To estimate the required solution; we apply the initial conditions x(0) = 30;

Thus;

x(0) = C(50)⁻⁵ = 30

⇒ C = 30 × 50⁵

Hence; x(t) = 30 × 50⁵ × (t + 50)⁻⁵

The above expression can be re-written as:

`x(t) = 25 \implies 30 * \bigg ( (50)/(t+50) \bigg ) ^5= 25`

i.e.

`\bigg ( (50)/(t+50) \bigg ) ^5= (25)/(30)`

`(50)/(t+50)= \bigg ( (25)/(30)\bigg) ^{(1)/(5)}`

`(50)/(t+50)= 0.964192504`

`{50}= 0.964192504(t+50)`

50 = 0.964192504t + 48.2096252

50 - 48.2096252 = 0.964192504t

1.7903748 = 0.964192504t

t = 1.7903748 / 0.964192504

t ≅ 1.857 minutes

We can thereby conclude that the estimated time it will require until there are 25 pounds of salt in the tank is 1.857 minutes.