Question

Properties of stirred tank solutions a. Show that for general positive values of \(R, V, C_{i},\) and \(m_{0},\) the solution of the initial value problem $$m^{\prime}(t)=-\frac{R}{V} m(t)+C_{i} R, \quad m(0)=m_{0}$$is \(m(t)=\left(m_{0}-C_{i} V\right) e^{-R t / V}+C_{i} V\) b. Verify that \(m(0)=m_{0}\) c. Evaluate \(\lim _{t \rightarrow \infty} m(t)\) and give a physical interpretation of the result. d. Suppose \(m_{0}\) and \(V\) are fixed. Describe the effect of increasing \(R\) on the graph of the solution.

Short answer

#tag_title# Short answer: By solving the given initial valve problem and its differential equation, we find that the mass in the stirred tank will approach a constant value of \(C_iV\) as time increases, meaning the system will eventually reach equilibrium. Furthermore, as the stirring rate \(R\) increases, the mass will approach the equilibrium value more rapidly, leading to a faster stabilization of the mass in the tank. On a graph, the solution will be steeper for larger values of \(R\), indicating that the mass in the tank will stabilize more quickly.

Step-by-step solution

Solve the differential equation

To show that the given solution is correct, we need to solve the initial value problem: $$m^{\prime}(t) = -\frac{R}{V}m(t) + C_i R, \quad m(0) = m_0$$ This is a first-order linear ordinary differential equation (ODE) with constant coefficients. The general form of this ODE is \(m^{\prime}(t) + P(t)m(t) = Q(t)\). We can rewrite the given ODE in this form: $$m^{\prime}(t) + \frac{R}{V}m(t) = C_i R$$ Now, we can find the integrating factor \(\mu(t) = e^{\int \frac{R}{V} dt} = e^{\frac{Rt}{V}}\). Multiplying the ODE by the integrating factor, we get: $$e^{\frac{Rt}{V}}m^{\prime}(t) + \frac{R}{V}e^{\frac{Rt}{V}}m(t) = C_i Re^{\frac{Rt}{V}}$$ This can be recognized as the derivative of a product: $$\frac{d}{dt}\left(e^{\frac{Rt}{V}}m(t)\right) = C_i Re^{\frac{Rt}{V}}$$ We can now integrate both sides with respect to \(t\): $$\int \frac{d}{dt}\left(e^{\frac{Rt}{V}}m(t)\right) dt = \int C_i Re^{\frac{Rt}{V}} dt$$ On the left side, we get back \(e^{\frac{Rt}{V}}m(t)\), and on the right side, we can integrate using substitution: $$u = \frac{Rt}{V}, \quad dt = \frac{V}{R} du$$ And the integral becomes: $$e^{\frac{Rt}{V}}m(t) = C_i V e^{\frac{Rt}{V}} + C$$ Now, we can solve for \(m(t)\): $$m(t) = Ce^{-\frac{Rt}{V}} + C_i V$$

Verify the initial condition

To verify that the given solution satisfies the initial condition \(m(0) = m_0\), we plug in \(t = 0\) and get: $$m(0) = Ce^{-\frac{R(0)}{V}} + C_i V = C + C_i V$$ Since \(m(0) = m_0\), we can solve for \(C\): $$m_0 = C + C_i V \implies C = m_0 - C_iV$$ Thus, the solution is: $$m(t) = (m_0 - C_iV)e^{-\frac{Rt}{V}} + C_iV$$

Evaluate the limit of the solution as \(t \rightarrow \infty\)

To evaluate the limit of the solution as \(t\) tends to infinity, we consider the expression: $$\lim_{t \rightarrow \infty} m(t) = \lim_{t \rightarrow \infty} \left((m_0 - C_iV)e^{-\frac{Rt}{V}} + C_iV\right)$$ The exponential term tends to 0 as \(t \rightarrow \infty\), so we are left with: $$\lim_{t \rightarrow \infty} m(t) = C_iV$$ The physical interpretation of this result is that as time increases, the mass \(m(t)\) in the stirred tank will approach a constant value of \(C_iV\), which means that the mass will eventually reach equilibrium.

Describe the effect of increasing \(R\) on the graph of the solution

To analyze the effect of increasing \(R\) on the graph of the solution, we examine the differential equation and the solution: $$m^{\prime}(t) = -\frac{R}{V}m(t) + C_i R$$ $$m(t) = (m_0 - C_iV)e^{-\frac{Rt}{V}} + C_iV$$ As \(R\) increases, the exponential term \(e^{-\frac{Rt}{V}}\) will decay faster. Therefore, the mass \(m(t)\) in the stirred tank will approach the equilibrium value \(C_iV\) more rapidly. This means that higher values of \(R\), which represents the rate at which the solution is stirred, lead to a faster stabilization of the mass in the tank. The graph of the solution will be steeper for larger values of \(R\).

Key concepts

First-order linear ordinary differential equation

A first-order linear ordinary differential equation (ODE) is an equation that involves the derivative of a function and the function itself. These types of equations can be written in the standard form:
\[ y'(t) + P(t) y(t) = Q(t) \]where \( y(t) \) is the function we are trying to solve for, and \( P(t) \) and \( Q(t) \) are known functions.
In the specific problem given, the ODE has constant coefficients, meaning \( P(t) \) and \( Q(t) \) are constants. This simplifies the equation significantly.

• Equation: \( m'(t) + \frac{R}{V} m(t) = C_i R \)
• Initial condition: \( m(0) = m_0 \)

Solving such equations usually involves finding a special function called an "integrating factor" that helps to integrate and solve the equation.

Integrating factor

An integrating factor is a function that we multiply with the ODE to facilitate its solution. This technique turns the left-hand side of the ODE into a derivative of a product, making it easier to integrate.
For our differential equation, the integrating factor is determined by:
\[ \mu(t) = e^{\int \frac{R}{V} \, dt} = e^{\frac{Rt}{V}} \]Multiplying the entire differential equation by this integrating factor results in:

• Transformed equation: \( e^{\frac{Rt}{V}}m'(t) + \frac{R}{V}e^{\frac{Rt}{V}}m(t) = C_i Re^{\frac{Rt}{V}} \)

This can be identified as the derivative of \( e^{\frac{Rt}{V}}m(t) \), which we can then integrate over \( t \) to solve for \( m(t) \). Using the integrating factor is a powerful method that simplifies solving first-order linear ODEs.

Limit of a function

The limit of a function describes the value that a function approaches as the input approaches a certain point. In the context of this differential equation problem, we are interested in the behavior of \( m(t) \) as \( t \) approaches infinity:
\[ \lim_{t \to \infty} m(t) = C_i V \]When examining the function:

• \( m(t) = (m_0 - C_i V) e^{-\frac{Rt}{V}} + C_i V \)

As \( t \to \infty \), the term \( e^{-\frac{Rt}{V}} \) tends to zero. This simplifies the expression for \( \lim_{}t \rightarrow ingfty} m_(t) \) to \( C_i V \). Understanding limits is crucial, especially for analyzing long-term behavior in dynamic systems.

Equilibrium value

The equilibrium value in the context of differential equations refers to a stable state that the system tends towards over time. For the problem at hand, the equilibrium value of the mass function \( m(t) \) is \( C_i V \).
This signifies that as time progresses, the mass \( m(t) \) inside the tank will stabilize, reaching this constant equilibrium value.

• As \( t \to \infty \), the system no longer changes.
• Equilibrium indicates balance, where inflow rate balances out the decay rate.

Understanding equilibrium is essential for predicting steady-states in physical and engineering applications, like chemical concentrations in reactions.

Exponential decay

Exponential decay refers to a decrease at a rate proportional to the current value, which results in a rapid decrease initially that slows over time. In this exercise, the concept is encapsulated by the term \( e^{-\frac{Rt}{V}} \).
This term signifies how fast the deviation from the equilibrium value reduces.

• \( e^{-\frac{Rt}{V}} \) decreases faster with larger values of \( R \), meaning quicker stabilization.
• This pattern of fast initial decrease can be found in contexts such as radioactive decay or cooling processes.

In essence, exponential decay is crucial for understanding time-dependent changes in differential equations, indicating how fast or slow a system approaches equilibrium.