Problem 57 - Differential Equations and Linear Algebra

Question

A solution containing 3 grams/L of a salt solution pours into a tank, initially half full of water, at a rate of 6 L/min. The well-stirred mixture flows out at a rate of \(4 \mathrm{L} / \mathrm{min}\). If the tank holds \(60 \mathrm{L}\), find the amount of salt (in grams) in the tank when the solution overflows.

Step-by-Step Solution

Answer

When the tank overflows, there will be \(180 \textrm{grams}\) of salt in it.

Step 1: Define the variables given in the problem

Let's define the given values and units: - Inflow rate: \(6 \frac{\textrm{L}}{\textrm{min}}\) - Outflow rate: \(4 \frac{\textrm{L}}{\textrm{min}}\) - Salt concentration: \(3 \frac{\textrm{g}}{\textrm{L}}\) - Tank capacity: \(60 \textrm{L}\)

Step 2: Find the time it takes for the tank to overflow

Since the tank is half full initially and the inflow rate is greater than the outflow rate, it will eventually overflow. To determine when this happens, we need to find the time it takes for the tank to become full. We can use the net flow rate to find this time: Net flow rate = Inflow rate - Outflow rate = \(6 \frac{\textrm{L}}{\textrm{min}} - 4 \frac{\textrm{L}}{\textrm{min}} = 2 \frac{\textrm{L}}{\textrm{min}}\) Now, considering the tank is half full initially, we have \(30 \textrm{L}\) of empty space remaining. Dividing this by the net flow rate, we can find the amount of time it takes to fill the tank: Time to fill the tank = \(\frac{\textrm{remaining volume}}{\textrm{net flow rate}} = \frac{30 \textrm{L}}{2 \frac{\textrm{L}}{\textrm{min}}} = 15 \textrm{minutes}\)

Step 3: Calculate the amount of salt in the tank when it overflows

Now that we know the tank will overflow after 15 minutes, we can calculate the total salt quantity in it by considering the salt inflow and outflow: Salt Inflow: Inflow rate × Salt concentration × Time = \(6 \frac{\textrm{L}}{\textrm{min}} × 3 \frac{\textrm{g}}{\textrm{L}} × 15 \textrm{minutes} = 270 \textrm{g}\) Salt Outflow: To find the salt outflow, we need to consider that the salt in the tank will gradually increase, so it's not going to be a constant rate. We can find the average salt concentration in the tank, as the inflow and outflow are linear: Average Salt concentration = \(\frac{0 + 3}{2} = 1.5 \frac{\textrm{g}}{\textrm{L}}\) Now, we can find the salt outflow with this average concentration: Salt Outflow = Outflow rate × Average salt concentration × Time = \(4 \frac{\textrm{L}}{\textrm{min}} × 1.5 \frac{\textrm{g}}{\textrm{L}} × 15 \textrm{minutes} = 90 \textrm{g}\) Finally, we can find the total amount of salt in the tank when it overflows: Total Salt in the tank = Salt Inflow - Salt Outflow = \(270 \textrm{g} - 90 \textrm{g} = 180 \textrm{g}\) So, there are \(180 \textrm{grams}\) of salt in the tank when the solution overflows.

Key Concepts

Salt Solution Dynamics

When studying salt solution dynamics, we focus on how salt concentration changes as a liquid solution is added to and removed from a tank. Imagine a mixing scenario, where a saltwater solution, continuously poured into a tank, is well-stirred and then discharged at a different rate. This process alters the salt concentration over time.
The salt concentration at any moment depends on the total volume of liquid in the tank, the amount of salt introduced, and the mixing efficiency. In our problem, a salt solution flows into a tank, initially half-filled with water, at a rate of 6 liters per minute, with a concentration of 3 grams per liter. The tank's contents are constantly mixed, ensuring uniform salt distribution.

• The inflow adds concentration to the solution but is balanced by the outflow.
• As the volume of solution increases or decreases, so does the concentration until steady state or overflow occurs.

Grasping the dynamics requires understanding these movements, mixing rates, and concentration shifts.

Rate of Change

Rate of change describes how one quantity changes in relation to another over a period, often seen in differential equations. In the context of our tank problem, this refers to the change in volume and concentration over time.
To analyze this, consider the net rate of liquid inflow against outflow. Here, the rate of change for the solution's volume is the difference between the inflow rate (6 L/min) and the outflow rate (4 L/min). This net inflow rate of 2 L/min dictates how quickly the tank fills.

• Concentration rates: A rise in volume not equally matched in concentration changes leads to dilution.
• Important: When inflow brings a higher concentration, the rate of change impacts the salt's total volume in the tank over time.

In summary, understanding the rate of change allows us to anticipate the time taken for the tank to fill and overflow, as well as the expected salt content during this process.

Tank Overflow Problem

Tank overflow problems involve determining at what point a liquid volume will exceed the container's capacity, taking into account inflows and outflows. The scenario in the current exercise involves understanding how the tank initially half-full with water, gradually fills to its limit.

• The tank starts with 30 L of space remaining (half of its 60 L capacity).
• With a net inflow of 2 L/min (since inflow is greater than outflow), we calculate the overflow time by dividing the available space by this net rate. This results in 15 minutes to full capacity.

Once the overflow time is known, calculating the concentration of salt at overflow involves tracking how much salt enters the tank minus what has left. Proper mixing assumptions dictate an average concentration over time, influencing the outflow's salt. Identifying these points clarifies how much salt remains at the moment of overflow, guiding solution decisions and safety measures in practical applications.