Continuing with the previous exercise, let Tank 2 be another...

Continuing with the previous exercise, let Tank 2 be another tank filled with $V_{2}$ gal of water. Assume that the dye solution from Tank 1 empties into Tank 2 as in Figure $5,$ mixes instantaneously, and leaves Tank 2 at the same rate $R$ . Let $c_{2}(t)$ be the dye concentration in Tank 2 at time $t .$ (a) Explain why $c_{2}$ satisfies the differential equation $$ \frac{d c_{2}}{d t}=\frac{R}{V_{2}}\left(c_{1}-c_{2}\right) $$ (b) Use the solution to Exercise 39 to solve for $c_{2}(t)$ if $V_{1}=300$ , $V_{2}=200, R=50$ , and $c_{0}=10$ . (c) Find the maximum concentration in Tank 2. (d) Plot the solution.

Continuing with the previous exercise, let Tank 2 be another tank filled with $V_{2}$ gal of water. Assume that the dye solution from Tank 1 empties into Tank 2 as in Figure $5,$ mixes instantaneously, and leaves Tank 2 at the same rate $R$ . Let $c_{2}(t)$ be the dye concentration in Tank 2 at time $t .$
(a) Explain why $c_{2}$ satisfies the differential equation
$$
\frac{d c_{2}}{d t}=\frac{R}{V_{2}}\left(c_{1}-c_{2}\right)
$$
(b) Use the solution to Exercise 39 to solve for $c_{2}(t)$ if $V_{1}=300$ , $V_{2}=200, R=50$ , and $c_{0}=10$ .
(c) Find the maximum concentration in Tank 2.
(d) Plot the solution.

Key Concepts

Initial Conditions in Differential Equations

Specifying the initial concentration (or state) is essential to finding a unique solution for a differential equation. Initial conditions tie the general solution down to the particular scenario being modeled and are key to solving real-world problems.

Plotting Solutions of Differential Equations

Graphical representation of the solution to a differential equation provides insight into the dynamic behavior of the system over time. Plots help in visualizing how the solution approaches equilibrium, its rate of change, and other transient features of the process.

Instantaneous Mixing Assumption

This assumption means that as soon as the dye enters the tank, it is uniformly distributed throughout the entire volume. This simplifies the modeling by allowing the concentration throughout the tank to be represented by a single variable rather than a spatially varying function.

Mixing Problems in Differential Equations

This concept involves modeling the concentration of substances in tanks or reactors where fluid flows are involved, specifically accounting for inputs, outputs, and mixing between different regions. The differential equations represent how the concentration changes over time due to these processes.

First-Order Linear Differential Equations

These are equations of the form dc/dt + p(t)c = q(t), which have well-known solution techniques such as the integrating factor method. They are widely used to describe processes with a rate of change proportional to the current state, which is typical in mixing and decay problems.

Steady-State Analysis

In mixing problems, steady-state analysis involves determining the equilibrium condition where the inflow and outflow concentrations become identical, leading to no further change in concentration as time tends to infinity. This analysis is crucial for understanding the long-term behavior of the system.

Transcript

00:02 Okay, part a -rex asked to explain why we c2 satisfy the differential equation.

00:08 D -c2 over dt equals r over 32, c1 minus c2.

00:13 Well, if you let g2 of t be the amount of grams of dye in tank 2 at time t.

00:27 And then the, at time t, the concentration of die in tank 1, and thus the concentration of die in 2 tank 2 is given by g2, prime t is equal to the rate in which is the grams of diet in minus outs but that's is equal to c1 of t times r minus c2 times r well if we factor out at r we get that r times c1 t minus c2 t and since g2 time t is equal to b2 c prime 2 2 what can we do here well we can plug in this portion over here so we get to e2 prime t times v2 is equal to r c1 minus c2 i divided by v2 on both sides we get the c2 prime t is equal to r b2 c1 minus c2 at time t okay now for part b we ask to use to use the solution of exercise 39 to solve for c2 t if v1 is equal to 300 v2 is equal to 200 r is equal to 50 and c not is equal to 10 okay so based on parts 1 we can say that let's see we have c0 b2 is equal to 200 yeah based on part a we can say that c1 of t is equal to 10 minus or times e negative t over 6 and since v2 is equal to 200 c2 prime of t is equal to well r which is 50 over v2 which is 200 that's this 1 over 4 times c20 which is 10 e to the negative t over 6 minus c2 of t and let's rewrite this in a form we know so let's add this portion over here so we get c2 prime t plus one -fourth of c2 t is equal to five over two e to the negative t over six okay well our integrating factor alpha of t is equal to e to integral of one -four d t so then this gives me e one -fourth t now multiplying by e one -fourth t on both sides over here we get that e one -fourth t times c -2 t is equal to let's see one fourth t one nuk minus one over six five over two e to the t over 12 okay now if we integrate on both sides we get that e one fourth t times c2 of t is equal to dirty et over 12 plus c2 of t is equal to 30 e negative t over 6 plus c2 of t is equal to 30 e negative t over 6 plus c e negative t over 4.

04:33 Now since tank 2 starts out entirely filled with water, we have c2 of 0 is equal to 0.

04:44 And plugging that in, we get 30 e0, which is 1 plus c.

04:51 So c is equal to negative 30.

04:55 So our equation becomes c2 of t is equal to 30.

05:01 E2.

05:02 Negative t over 6 plus or an x minus e to the negative t over 4.

05:09 Okay, now for part c, we're asked to find the maximum concentration in time 2...

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