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Chapter 8: Problem 25

For each of the following stirred tank reactions, carry out the following analysis. a. Write an initial value problem for the mass of the substance. b. Solve the initial value problem and graph the solution to be sure that \(m(0)\) and \(\lim _{t \rightarrow \infty} m(t)\) are correct. A 2000 -L tank is initially filled with a sugar solution with a concentration of \(40 \mathrm{g} / \mathrm{L} .\) A sugar solution with a concentration of \(10 \mathrm{g} / \mathrm{L}\) flows into the tank at a rate of \(10 \mathrm{L} / \mathrm{min} .\) The thoroughly mixed solution is drained from the tank at a rate of \(10 \mathrm{L} / \mathrm{min}\).

Short Answer

Expert verified
Answer: As time approaches infinity, the mass of sugar in the tank will be 20,000 g.

Step by step solution

01

Write an initial value problem for the mass of the substance

Let \(m(t)\) represent the mass of sugar in the tank after \(t\) minutes. Since the solution of 10g/L flows in at a rate of 10L/min, the mass of sugar flowing into the tank per minute is \((10 \frac{g}{L})(10 \frac{L}{min})=100 g/min\). As the well-mixed solution drains from the tank at a rate of 10L/min, the concentration of sugar in the drained solution is the same as the concentration in the tank. To find the mass of sugar drained, we can write the concentration as \(\frac{m(t)}{2000}\) g/L, and the mass drained per minute is then \(\left(\frac{m(t)}{2000}\frac{g}{L}\right)(10\frac{L}{min})=\frac{m(t)}{200} \frac{g}{min}\). So, the initial value problem for the mass of the substance can be written as: \[\frac{dm}{dt}=100-\frac{m}{200}\] The initial condition of the mass in the tank is: \[m(0)=40\frac{g}{L}(2000 L)=80000 g\]
02

Solve the initial value problem and verify the solution

To solve the initial value problem, we rewrite it as a separable differential equation and integrate both sides: \[\int \frac{dm}{m-20000}=\int \frac{1}{200} dt\] This can be further written as: \[ln|m-20000|=\frac{1}{200} t + C\] Solve the above equation for \(m(t)\): \[m(t)=20000+Ce^{\frac{t}{200}}\] Applying the initial condition, \(m(0)=80000\) to find the value of \(C\): \[80000=20000+Ce^{0}\] Solving for \(C\), we get: \[C=60000\] So, the equation for the mass of the substance in the tank with respect to time is: \[m(t)=20000+60000e^{\frac{-t}{200}}\] Now, we find \(\lim _{t \rightarrow \infty} m(t)\): \[\lim _{t \rightarrow \infty} m(t)=\lim _{t \rightarrow \infty} (20000+60000e^{\frac{-t}{200}}) = 20000\] As per the initial value problem and its solution, \(m(0)=80000\) and \(\lim _{t \rightarrow \infty} m(t)=20000\). The initial mass and the mass as time approaches infinity are correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Initial Value Problem
An **initial value problem** involves finding a function that satisfies a differential equation and meets specific initial conditions. In the example of the tank mixing problem, we're tasked with modeling the mass of sugar in a 2000 L tank over time.

Initially, the tank contains a sugar concentration of 40 g/L, giving us a total initial mass of 80,000 g. This serves as our initial condition, expressed as \(m(0) = 80000\). The inflow and outflow rates of the solution remain balanced, both at 10 L/min. By accounting for the sugar entering and leaving the tank, we derive the differential equation:
  • The rate of sugar inflow: 100 g/min.
  • The rate of sugar outflow: \(\frac{m(t)}{200}\) g/min.
  • Resulting differential equation: \(\frac{dm}{dt} = 100 - \frac{m}{200}\).
This mathematical model allows us to predict the mass at any future time given the initial state.
Separable Differential Equation
A **separable differential equation** can be rewritten to allow separation of variables, enabling integration with respect to two distinct variables. The obtained equation in the tank problem is \(\frac{dm}{dt} = 100 - \frac{m}{200}\). By treating terms involving \(m\) and \(t\) separately, we arrive at: \[\int \frac{dm}{m-20000} = \int \frac{1}{200} dt\] The integration of these terms yields a logarithmic relationship: \[\ln|m-20000| = \frac{1}{200} t + C\] Upper or lower integration can be used to solve for \(m(t)\), evolving into an exponential form: \[m(t) = 20000 + Ce^{\frac{-t}{200}}\] Using the initial condition \(m(0) = 80000\), we determine \(C = 60000\). Consequently, the solution \(m(t) = 20000 + 60000e^{\frac{-t}{200}}\) effectively describes the system.
Tank Mixing Problem
The **tank mixing problem** is a classical example of an application of differential equations in real life. It models the concentration of substances within a mixing tank that has inputs and outputs. In this case, we are considering the sugar solution in a tank.

Here are the essentials:
  • Inflow: solution with a concentration of 10 g/L at 10 L/min.
  • Outflow: solution concentration equals current tank concentration \(\frac{m(t)}{2000}\) g/L, also at 10 L/min.
Despite constant volume due to balanced flow rates, the concentration of sugar in the tank changes differently due to varying mass content. Solving involves understanding how inflow and outflow rates impact the overall substance concentration. This offers predictions about future changes in mass \(m(t)\) throughout the process.
Limit Evaluation
The concept of **limit evaluation** gives vital insights into the system鈥檚 behavior as time progresses towards infinity, denoted by \(t \rightarrow \infty\). In our differential equation analysis for the tank, we ended with the function \[m(t) = 20000 + 60000e^{\frac{-t}{200}}\]Taking the limit of \(m(t)\) as \(t\) approaches infinity helps us understand the final state of the system. As time elapses, the influence of the exponential term \(60000e^{\frac{-t}{200}}\) diminishes due to the negative exponent coupled with \(t\).

Eventually, this term trends towards zero, simplifying our solution to a stable state at:
  • \(\lim_{t \rightarrow \infty} m(t) = 20000\)
This limit indicates how the mass of sugar stabilizes at 20,000 g, aligning with physical expectations given continuous inflow and outflow dynamics.

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Most popular questions from this chapter

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