Questions: A tank contains 80 kg of salt and 1000 L of water. A solution of a concentration 0.04 kg of salt per liter enters a tank at the rate 6 L / min. The solution is mixed and drains from the tank at the same rate. (a) What is the concentration of our solution in the tank initially? concentration = 0.08 (kg / L) (b) Set up an initial value problem for the quantity y, in kg, of salt in the tank at time t minutes. dy/dt = (kg / min) y(0) = (kg) (c) Solve the initial value problem in part (b). y(t) = (d) Find the amount of salt in the tank after 3 hours. amount = (kg) (e) Find the concentration of salt in the solution in the tank as time approaches infinity. concentration = (kg / L)

A tank contains 80 kg of salt and 1000 L of water. A solution of a concentration 0.04 kg of salt per liter enters a tank at the rate 6 L / min. The solution is mixed and drains from the tank at the same rate.

(a) What is the concentration of our solution in the tank initially?
concentration = 0.08 (kg / L)

(b) Set up an initial value problem for the quantity y, in kg, of salt in the tank at time t minutes.
dy/dt = (kg / min) y(0) = (kg)

(c) Solve the initial value problem in part (b).
y(t) =

(d) Find the amount of salt in the tank after 3 hours.
amount = (kg)

(e) Find the concentration of salt in the solution in the tank as time approaches infinity.
concentration = (kg / L)

Transcript text: A tank contains 80 kg of salt and 1000 L of water. A solution of a concentration 0.04 kg of salt per liter enters a tank at the rate $6 \mathrm{~L} / \mathrm{min}$. The solution is mixed and drains from the tank at the same rate. (a) What is the concentration of our solution in the tank initially? concentration $=$ $\square$ 0.08 $(\mathrm{kg} / \mathrm{L})$ (b) Set up an initial value problem for the quantity $y$, in kg , of salt in the tank at time t minutes. $\frac{d y}{d t}=$ $\square$ $(\mathrm{kg} / \mathrm{min}) \quad y(0)=$ $\square$ $(\mathrm{kg})$ (c) Solve the initial value problem in part (b). $y(t)=$ $\square$ (d) Find the amount of salt in the tank after 3 hours. amount $=$ $\square$ $(\mathrm{kg})$ (e) Find the concentration of salt in the solution in the tank as time approaches infinity. concentration $=$ $\square$ $(\mathrm{kg} / \mathrm{L})$

Solution

Solution Steps

Step 1: Initial Concentration Calculation

The initial concentration of the salt in the tank is calculated by dividing the initial amount of salt, 80 kg, by the initial volume of the solution, 1000 L. This gives us an initial concentration of 0.08 kg/L.

Step 2: Differential Equation Formulation

The differential equation representing the rate of change of the amount of salt in the tank over time is given by: $\frac{dS}{dt} = R_{in}C_{in} - R_{out}\frac{S}{V}$. Here, $R_{in}C_{in}$ represents the rate at which salt enters the tank, and $R_{out}\frac{S}{V}$ represents the rate at which salt leaves the tank, assuming the volume, V, remains constant over time.

Step 3: Solving the Differential Equation

By solving the differential equation with the given initial conditions, the amount of salt in the tank after 180 units of time is approximately 53.58 kg.

Step 4: Long-term Behavior

As time approaches infinity, the concentration of salt in the tank approaches a steady state where the inflow and outflow rates of salt are equal. This steady-state concentration is 0.04 kg/L.

Final Answer:

The initial concentration is 0.08 kg/L, the amount of salt after 180 units of time is 53.58 kg, and the steady-state concentration as time approaches infinity is 0.04 kg/L.

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