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Magazine Chapter 4: Problem 21
Tanks \(T_{1}\) and \(T_{2}\) have capacities \(W_{1}\) and \(W_{2}\) liters, respectively. Initially they are both full of dye solutions with concentrations \(c_{1}\) and \(c_{2}\) grams per liter. Starting at \(t_{0}=0,\) the solution from \(T_{1}\) is pumped into \(T_{2}\) at a rate of \(r\) liters per minute, and the solution from \(T_{2}\) is pumped into \(T_{1}\) at the same rate. (a) Find the concentrations \(c_{1}(t)\) and \(c_{2}(t)\) of the dye in \(T_{1}\) and \(T_{2}\) for \(t>0\). (b) Find \(\lim _{t \rightarrow \infty} c_{1}(t)\) and \(\lim _{t \rightarrow \infty} c_{2}(t)\).
Short Answer
Expert verified
Answer: The limits of the concentrations of the dye in the two tanks as time goes to infinity are equal and given by \(\frac{c_{1}+c_{2}}{2}\).
Step by step solution
01
Set up the differential equations for the concentration of the dye in each tank
We'll start by writing down the differential equations for \(c_{1}(t)\) and \(c_{2}(t)\). Since the solutions are pumped at a rate \(r\), we have:
- For Tank 1, the inflow increases the concentration by \(c_{2}(t)r\) grams per minute, and the outflow decreases the concentration by \(c_{1}(t)r\) grams per minute.
So we have: \(\frac{dc_{1}(t)}{dt}=\frac{c_{2}(t)r-c_{1}(t)r}{W_{1}}\)
- For Tank 2, the inflow increases the concentration by \(c_{1}(t)r\) grams per minute, and the outflow decreases the concentration by \(c_{2}(t)r\) grams per minute.
So we have: \(\frac{dc_{2}(t)}{dt}=\frac{c_{1}(t)r-c_{2}(t)r}{W_{2}}\)
02
Solve the system of differential equations
To solve the system of differential equations, we can rewrite each equation with the change of variables \(y_{1}=c_{1}(t) - c_{2}(t)\) and \(y_{2}=c_{1}(t) + c_{2}(t)\). Then we find the derivatives of these and solve the resulting simple differential equations:
1. Find the derivatives:
\(\frac{dy_{1}}{dt}=\frac{dc_{1}(t)}{dt}-\frac{dc_{2}(t)}{dt}=\frac{c_{2}(t)-c_{1}(t)}{W_{1}}-\frac{c_{1}(t)-c_{2}(t)}{W_{2}}\)
\(\frac{dy_{2}}{dt}=\frac{dc_{1}(t)}{dt}+\frac{dc_{2}(t)}{dt}=\frac{c_{2}(t)+c_{1}(t)}{W_{1}}+\frac{c_{1}(t)+c_{2}(t)}{W_{2}}\)
2. Solve the differential equations:
\(y_{1}(t)=C_{1}e^{-\frac{2rt}{W_{1}+W_{2}}}\)
\(y_{2}(t)=W_{1}+W_{2}+C_{2}\)
3. Get back to the original variables:
\(c_{1}(t)=\frac{1}{2}\left(W_{1}+W_{2}+C_{2}+C_{1}e^{-\frac{2rt}{W_{1}+W_{2}}}\right)\)
\(c_{2}(t)=\frac{1}{2}\left(W_{1}+W_{2}+C_{2}-C_{1}e^{-\frac{2rt}{W_{1}+W_{2}}}\right)\)
03
Determine the constants \(C_{1}\) and \(C_{2}\)
To find \(C_{1}\) and \(C_{2}\), recall that at the initial time \(t_{0}=0\), the concentrations of the dye in the tanks are \(c_{1}=c_{1}(0)\) and \(c_{2}=c_{2}(0)\). We plug in \(t=0\) into our expressions for \(c_{1}(t)\) and \(c_{2}(t)\) to determine the constants:
Using \(c_{1}(0)=c_{1}\) and \(c_{2}(0)=c_{2}\), we find:
\(C_{1}=c_{1}-c_{2}\)
\(C_{2}=c_{1}+c_{2}\)
Now, we can plug these constants back into our expressions for \(c_{1}(t)\) and \(c_{2}(t)\).
04
Find the expressions for \(c_{1}(t)\) and \(c_{2}(t)\) using the determined constants
Now, replacing the \(C_{1}\) and \(C_{2}\) with the values we found above:
\(c_{1}(t)=\frac{1}{2}\left((c_{1}+c_{2})+(c_{1}-c_{2})e^{-\frac{2rt}{W_{1}+W_{2}}}\right)\)
\(c_{2}(t)=\frac{1}{2}\left((c_{1}+c_{2})-(c_{1}-c_{2})e^{-\frac{2rt}{W_{1}+W_{2}}}\right)\)
These are the concentrations of the dye in the tanks for any time \(t>0\).
05
Find the limits of the concentrations as \(t\) goes to infinity
Now, we need to find \(\lim_{t\to \infty}c_{1}(t)\) and \(\lim_{t\to \infty}c_{2}(t)\). As \(t\) goes to infinity, the exponential term \(e^{-\frac{2rt}{W_{1}+W_{2}}}\) goes to 0. Thus, we have:
\(\lim_{t\to \infty}c_{1}(t)=\lim_{t\to \infty}\frac{1}{2}\left((c_{1}+c_{2})+(c_{1}-c_{2})e^{-\frac{2rt}{W_{1}+W_{2}}}\right)=\frac{c_{1}+c_{2}}{2}\)
\(\lim_{t\to \infty}c_{2}(t)=\lim_{t\to \infty}\frac{1}{2}\left((c_{1}+c_{2})-(c_{1}-c_{2})e^{-\frac{2rt}{W_{1}+W_{2}}}\right)=\frac{c_{1}+c_{2}}{2}\)
The limits of the concentrations as \(t\) goes to infinity are equal and given by \(\lim_{t\to \infty}c_{1}(t)=\lim_{t\to \infty}c_{2}(t)=\frac{c_{1}+c_{2}}{2}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Concentration Problem
In the context of differential equations, a **concentration problem** often focuses on how the amount of a substance鈥攃alled the concentration鈥攃hanges in a solution over time. Here, we're dealing with two tanks containing dye solutions, with concentrations initially set at \(c_1\) for tank \(T_1\) and \(c_2\) for tank \(T_2\). The goal is to determine how these concentrations change as the solutions are exchanged between the tanks. Solving this problem involves understanding how concentrations transform under specific flow conditions and using mathematical models to describe these changes.
In this specific exercise, the concentration problem is addressed by using a system of differential equations. These equations take into account the rate at which the dye solution is pumped from one tank to the other. Concentration problems of this nature are pivotal in fields like chemistry and environmental science, where monitoring and predicting changes in concentration is crucial.
In this specific exercise, the concentration problem is addressed by using a system of differential equations. These equations take into account the rate at which the dye solution is pumped from one tank to the other. Concentration problems of this nature are pivotal in fields like chemistry and environmental science, where monitoring and predicting changes in concentration is crucial.
Dye Solutions
Dye solutions in this problem act as a medium to illustrate how substances mix between two connected systems. Each tank contains a dye solution with specific initial concentrations labeled \(c_1\) and \(c_2\). These represent the amount of dye, in grams, per liter of solution initially present in tanks \(T_1\) and \(T_2\) respectively.
Understanding the characteristics of dye solutions aids in predicting how the dye will diffuse between the tanks. Theoretically, if left long enough, the solutions would likely reach an equilibrium where the dye is evenly distributed between the two tanks. This problem emulates real-world scenarios, such as the mixing of pollutants in lakes or the distribution of nutrients in a bioreactor, making it a valuable part of various scientific research and practical applications.
Understanding the characteristics of dye solutions aids in predicting how the dye will diffuse between the tanks. Theoretically, if left long enough, the solutions would likely reach an equilibrium where the dye is evenly distributed between the two tanks. This problem emulates real-world scenarios, such as the mixing of pollutants in lakes or the distribution of nutrients in a bioreactor, making it a valuable part of various scientific research and practical applications.
Inflow-Outflow Rates
The **inflow-outflow rates** are crucial when analyzing concentration problems involving continuous mixing, like in the given exercise. The flow rate \(r\), measured in liters per minute, determines how fast the solutions move between tanks \(T_1\) and \(T_2\). Since both tanks have solutions entering and exiting at the same rate, the net flow remains balanced. However, the inflow from one tank increases the concentration of the other and vice versa.
This balance of inflow and outflow influences the dynamics of concentration. As solutions are pumped back and forth, the concentrations in each tank adjust instantaneously but gradually. The challenge lies in predicting these changes over time using differential equations that account for such inflow and outflow systems.
This balance of inflow and outflow influences the dynamics of concentration. As solutions are pumped back and forth, the concentrations in each tank adjust instantaneously but gradually. The challenge lies in predicting these changes over time using differential equations that account for such inflow and outflow systems.
- Inflow increases dye concentration in the receiving tank.
- Outflow decreases dye concentration in the originating tank.
Limit of Concentrations
As time progresses, the concentrations \(c_1(t)\) and \(c_2(t)\) of the dye in tanks \(T_1\) and \(T_2\) evolve dynamically due to the continuous exchange of solutions. The **limit of concentrations** refers to what these concentrations will be as time approaches infinity. Mathematically, we look at the limit \(\lim_{t \to \infty} c_1(t)\) and \(\lim_{t \to \infty} c_2(t)\), which determine the steady-state concentrations.
In this exercise, the solution shows that both concentrations \(c_1(t)\) and \(c_2(t)\) converge to \(\frac{c_1 + c_2}{2}\) as \(t\) tends to infinity. This result is expected because, with constant inflow and outflow at identical rates, the total amount of dye remains conserved over time between the two tanks. Over time, the systems balance out, making the concentrations in both tanks average out to the same final value.
Both tanks will eventually exhibit equal concentrations. The final concentration is the mean of the initial concentrations. Understanding this concept highlights the equilibrium dynamics in closed systems involving continuous substance interchange.
In this exercise, the solution shows that both concentrations \(c_1(t)\) and \(c_2(t)\) converge to \(\frac{c_1 + c_2}{2}\) as \(t\) tends to infinity. This result is expected because, with constant inflow and outflow at identical rates, the total amount of dye remains conserved over time between the two tanks. Over time, the systems balance out, making the concentrations in both tanks average out to the same final value.
