Question

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. A \(2000-\) Leank 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

Answer: The mass of sugar in the tank at any given time t is given by the expression: M(t) = 100 * 200 * e^((1/200)t) + 76000 * e^(-(1/200)t)

Step-by-step solution

Define the given parameters

We are given the following parameters: - Tank volume: 2000 L - Initial sugar concentration: 40 g/L - Flow rate of incoming sugar solution: 10 L/min with a concentration of 10 g/L - Flow rate of outgoing mixed solution: 10 L/min Now let's define: - M(t): mass of sugar in the tank at time t - V: volume of the tank (constant)

Write a differential equation for the change in mass

We will write a differential equation to represent the change in mass M(t) in the tank. The mass change results from the incoming and outgoing flows. The mass increase due to the incoming flow is given by: rate_in = 10 (L/min) * 10 (g/L) = 100 g/min The mass decrease due to the outgoing flow is proportional to the sugar concentration in the tank at each moment and given by: rate_out = 10 (L/min) * (M / V) (g/L) = 10 * M / 2000 g/min Thus, the net rate of change of the mass is equal to the input rate minus the output rate: dM/dt = rate_in - rate_out = 100 - 10 * M / 2000 This is a differential equation representing the initial value problem for the mass of the substance in the tank.

Solve the initial value problem

We need to solve the first-order linear differential equation: dM/dt + (1/200) * M = 100 We can solve this using the method of integrating factors. The integrating factor is given by: IF = e^(∫(1/200)dt) = e^((1/200)t) We multiply the differential equation by the integrating factor and integrate both sides: ∫(e^((1/200)t)dM/dt + (1/200)e^((1/200)t)M) dt = ∫(100e^((1/200)t)dt) Integrate both sides: e^((1/200)t)M = 100 * 200 * e^((1/200)t) + C Where C is the constant of integration. Now, find C by using the initial condition, M(0) = 40 * 2000: C = 40 * 2000 - 100 * 200 = 76000 Now, solve for M(t): M(t) = 100 * 200 * e^((1/200)t) + 76000 * e^(-(1/200)t)

Final solution

We have successfully obtained the initial value problem and solved it. The mass of sugar in the tank at any given time t is now given by the following expression: M(t) = 100 * 200 * e^((1/200)t) + 76000 * e^(-(1/200)t)

Key concepts

Differential Equations

Differential equations are powerful mathematical tools that describe how a particular quantity changes over time or spatial dimensions. They encapsulate relationships between functions and their derivatives, essentially capturing the essence of dynamic systems and phenomena.

In the stirred tank reaction example, we constructed a differential equation to track the mass of sugar within the tank over time. The mass of the sugar changes due to the inflow and outflow of the sugar solution. A differential equation, such as the one formed \(dM/dt = rate_{in} - rate_{out}\), serves as a mathematical model that helps us quantify this change.

Differential equations come in various forms, and they can be linear or nonlinear, and ordinary or partial, among other classifications. The equation in our example is a first-order linear ordinary differential equation, meaning it involves first derivatives and does not display any products or powers of the unknown function or its derivatives.

Understanding how to set up and solve differential equations is a vital skill in many fields, ranging from physics to engineering and ecology, as it allows one to predict and understand behaviors of systems over time.

Integrating Factors

An integrating factor is a clever mathematical technique used to solve certain types of differential equations, particularly first-order linear equations. When the ordinary methods of integration by separation of variables do not work, integrating factors come to the rescue.

For a differential equation in the form \(\frac{dM}{dt} + P(t)M = Q(t)\), where \(P(t)\) and \(Q(t)\) are functions of \(t\), the integrating factor is generally expressed as \(e^{\int P(t) dt}\). Multiplying the entire differential equation by this integrating factor transforms it into a form that can be easily integrated.

In our sugar tank problem, \(P(t)\) corresponds to \(1/200\) and the integrating factor is \(e^{(1/200)t}\). The process of multiplying the equation by the integrating factor and then integrating both sides effectively 'undoes' the differentiation, leading us to an expression for \(M(t)\), the mass of sugar in the tank at any time \(t\).

Integrating factors are a testament to the ingenuity of mathematicians, simplifying complex differential equations to a form where solutions are accessible and practical applications, such as predicting the concentration of a substance in a mixer, are made possible.

Exponential Functions

Exponential functions are mathematically expressed as \(f(x) = a^{bx}\), where \(a\) and \(b\) are constants, and \(x\) is the variable. These functions are unique in that they grow (or decay) at a rate proportional to their current value, which makes them incredibly important in the realms of growth processes, decay, and many areas within calculus.

In the context of our initial value problem for the stirred tank reaction, exponential functions appear as the solution to the differential equation involving the mass of sugar in the tank \(M(t)\). As the integrating factor involves an exponential term, \(e^((1/200)t)\), it indicates that the change in mass over time has an exponential characteristic.

Exponential functions oftentimes emerge in natural processes, such as population growth, radioactive decay, and compound interest in finance. Recognizing and understanding the behavior of exponential functions is essential not only in solving differential equations but also in interpreting the solutions and their implications for real-world problems.