Two large tanks, each holding 100 L of liquid, are...

Two large tanks, each holding 100 L of liquid, are interconnected by pipes, with the liquid flowing from tank A into tank B at a rate of 3 L/min and from B into A at a rate of 1 L/min (see Figure 5.2). The liquid inside each tank is kept well stirred. A brine solution with a concentration of 0.2 kg/L of salt flows into tank A at a rate of 6 L/min. The (diluted) solution flows out of the system from tank A at 4 L/min and from tank B at 2 L/min. If, initially, tank A contains pure water and tank B contains 20 kg of salt, determine the mass of salt in each tank at time $$t \geq 0$$.

Two large tanks, each holding 100 L of liquid, are interconnected by pipes, with the liquid flowing from tank A into tank B at a rate of 3 L/min and from B
into A at a rate of 1 L/min (see Figure 5.2). The liquid
inside each tank is kept well stirred. A brine solution
with a concentration of 0.2 kg/L of salt flows into tank
A at a rate of 6 L/min. The (diluted) solution flows out
of the system from tank A at 4 L/min and from tank B
at 2 L/min. If, initially, tank A contains pure water and
tank B contains 20 kg of salt, determine the mass of
salt in each tank at time $$t \geq 0$$.

Key Concepts

Conservation of Mass

This principle asserts that, within a closed system, the mass of the substance remains constant over time unless there is an addition or removal. In mixing problems, it is used to set up equations that track the amount of a solute in each compartment by accounting for inputs, outputs, and transfers between subsystems.

Differential Equations

Differential equations model the rate of change of a quantity with respect to time. In the context of mixing problems, they are used to express how the amount of salt or other solutes evolves due to incoming streams, outgoing streams, and inter-compartment flows, allowing for a dynamic analysis of the system.

Compartmental Modeling

Compartmental models divide a physical system into interconnected sections or tanks where the substance is assumed to be uniformly mixed. Each compartment’s balance is described by its own differential equation, with terms accounting for the transfer of material between compartments, as well as external inflows and outflows.

Flow Rates

Flow rates quantify how fast material enters or leaves a compartment. They are critical in setting up the differential equations for mixing problems because they directly influence the net gain or loss of the solute in each compartment, thereby affecting the overall concentration dynamics.

Initial Conditions

Initial conditions specify the starting quantities of the substance in the system and are essential for solving the differential equations uniquely. They anchor the solution in a real-world scenario by providing the baseline from which the time-dependent behavior of the system is determined.

Well-Stirred Assumption

The well-stirred or 'perfect mixing' assumption implies that the concentration of the solute is uniform throughout each compartment at any given moment. This simplification allows for the use of ordinary differential equations to model the system, since spatial variations in concentration can be neglected.

Transcript

00:01 Okay, so for this program, we need to realize some quantities at the beginning.

00:06 So, we use x to denote the mass of salt in tank a, and we use y to denote the mass of salt in tank b.

00:23 Since both tanks has, i have volume 100 liters, so the concentrate in tank a will be x over 100.

00:38 And the concentrating tank b will be y over 100.

00:44 So once we realize these quantities, we can directly write out the system for these.

00:50 So the rate of change of x equals to 1 .2 plus y over 100 times 1 minus x over 100 times 7.

01:05 And the d .y over d t equals to x over 100 times 3 .000.

01:11 3 minus y over 100 times 3.

01:18 Okay, so we use an operator d .e equals to d of dt to rewrite the ode system in the following way.

01:41 And we will do the elimination for this ode by acting an operator d plus 0 .07 to the second one.

02:08 And for the first one, we just multiply everything by minus 0 .03.

02:13 After doing this, we can eliminate the function x.

02:26 Okay, so now we are ready to eliminate the first part, and that gives us d squared plus 0 .1d plus 0 .08, acne on y equals to 0 .036, which is a non -homogeneous ode.

02:48 So the homogeneous my genius part has a characteristic quadratic equation, and we can solve this by using a quadratic formula.

03:03 So r1 equals to minus 5, minus root of 7 divided by 100.

03:10 R2 equals to minus 5 plus root of 7 divided by 100...

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