Question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate. Find the time that will elapse before the concentration of dye in the tank reaches \(1 \%\) of its original value.

Short answer

To summarize, in this mixing problem we defined a function Q(t) to represent the amount of dye in the tank at a given time t. We then set up a differential equation to model the rate of change of this quantity and solved it using integration. Finally, we determined the time it will take for the concentration of dye in the tank to reach 1% of its initial value. The result is approximately 460.52 minutes.

Step-by-step solution

Define variables

Let's define \(Q(t)\) as the amount of dye in the tank at time \(t\) (in grams), and \(V(t)\) as the volume of the solution in the tank at time \(t\) (in liters).

Write the differential equation

Now let's write a differential equation to express the rate of change in the amount of dye in the tank with respect to time, using the input and output flow of water. Note that the concentration of dye in the output flow is equal to the concentration of dye in the tank at any given time. $$ \frac{dQ(t)}{dt} = -\frac{2}{V(t)}Q(t) $$The negative sign in the equation denotes that the amount of dye is decreasing over time.

Investigate the volume of the solution

The volume of the solution in the tank stays constant since the inflow rate and outflow rate are the same, thus \(V(t) = 200\) liters. We can plug this into the differential equation:$$ \frac{dQ(t)}{dt} = -\frac{2}{200}Q(t) $$Now we have a first-order linear differential equation.

Solve the differential equation

To solve this differential equation, we can use the method of integration of variables. From the equation, we have:$ \frac{dQ(t)}{Q(t)} = -\frac{2}{200} dt \(Integrating both sides, we have:\) \int \frac{dQ(t)}{Q(t)} = -\frac{2}{200} \int dt \(After integrating this yields:\) \ln{Q(t)} = -\frac{2}{200}t + C, $ where C is the constant of integration. Then, we write the solution using the exponential function:$ Q(t) = Ce^{-\frac{1}{100}t}. $

Determine the initial condition and find C

Since the initial concentration of dye is 200 liters multiplied by 1 g/liter, we have \(Q(0) = 200\) grams. Now we can use this initial condition to find the value of C: $$ 200 = Ce^{-\frac{1}{100}(0)} \implies C = 200. $$So the function that describes the amount of dye in the tank at time t is $$ Q(t) = 200e^{-\frac{1}{100}t}. $$

Calculate the time for the concentration to reach \(1 \%\) of its initial value

To find the time when the concentration of dye reaches \(1\%\) of its initial value, we need to find the time \(t\) when the ratio of the amount of dye at time \(t\) to the initial amount of dye is equal to \(0.01\). Thus, we have the equation:$$ \frac{Q(t)}{200} = 0.01. $$Substituting the expression for \(Q(t)\), we get:$$ \frac{200e^{-\frac{1}{100}t}}{200}=0.01. $$Solving for \(t\), we have:$$ e^{-\frac{1}{100}t}=0.01. $$Taking the natural logarithm of both sides, we get:$$ -\frac{1}{100}t = \ln{0.01}. $$Now, isolate t to find the time:$$ t = -100\ln{0.01} = 100\ln{100} \approx 460.52 \text{ minutes}. $$So it will take approximately 460.52 minutes for the concentration of dye in the tank to reach \(1\%\) of its initial value.

Key concepts

First-Order Linear Differential Equation

A first-order linear differential equation is a fundamental concept in mathematics, often used to model real-world phenomena. The general form of these equations is \(\frac{dy}{dt} + P(t)y = Q(t)\), where \(P(t)\) and \(Q(t)\) are functions of \(t\). In our exercise, the differential equation is simplified as it involves the decay of dye in a tank being rinsed with water: \[\frac{dQ(t)}{dt} = -\frac{2}{200}Q(t). \]This equation models a situation where the amount of dye decreases over time, proportional to its current amount. This key characteristic of the equation showcases why it's termed as "linear"; the rate of change of \(Q(t)\) is directly proportional to \(Q(t)\) itself.Understanding first-order linear differential equations helps to interpret diverse scenarios, with solutions often providing insight into the temporal dynamics of a system. The solution to our example introduces the concept of exponential decay, where integration and rearranging lead to an expression involving an exponential function. The mathematical tools involved in solving these types of equations are powerful and widely applicable.

Exponential Decay

Exponential decay describes a process where a quantity decreases at a rate proportional to its current value. This is a common pattern observed in natural systems and is essential in fields such as physics, chemistry, and even finance.The equation derived from our exercise showcases exponential decay: \[Q(t) = 200e^{-\frac{1}{100}t}.\]In this context, the initial amount of dye is 200 grams, and \(e^{-\frac{1}{100}t}\) represents how it diminishes over time due to the continuous rinsing with fresh water. The term "exponential" comes from the appearance of the constant \(e\), which is a mathematical constant approximately equal to 2.718. The factor of \-\frac{1}{100}\ represents the decay rate.A characteristic feature of exponential decay is its consistency over equal time intervals. In simple terms, a constant percentage of the dye amount is lost over each equal period of time. This principle allows for easier modeling and prediction of such systems and highlights the importance of understanding exponential functions when analyzing rates that change over time.

Hydrodynamic Models

Hydrodynamic models use mathematical equations to describe the motion of fluids, such as liquids and gases. These models are pivotal in engineering, environmental studies, and physical science, offering insight into fluid behavior under various conditions. In the given exercise, we deal with a straightforward hydrodynamic model. The tank experiment involves a continuous input and output of water, with the dye uniformly distributed in the solution. By keeping the volume constant—in this case, 200 liters—the model simplifies to a scenario where we only need to consider the dye's rate of change. Hydrodynamic models like this help simulate real-world processes. They allow us to predict how a system will evolve over time. Specifically, this model is illustrative of how substances are flushed from a system, a common requirement in cleaning processes or chemical reactions. By applying these models, engineers can design systems that efficiently monitor and manage the distribution of substances within fluid environments.