Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

A tank of capacity 100 gallons is initially full of pure alcohol. The flow rate of the drain pipe is 5 gallons per minute; the flow rate of the filler pipe can be adjusted to \(c\) gallons per minute. An unlimited amount of \(25 \%\) alcohol solution can be brought in through the filler pipe. Our goal is to reduce the amount of alcohol in the tank so that it will contain 100 gallons of \(50 \%\) solution. Let \(T\) be the number of minutes required to accomplish the desired change. (a) Evaluate \(T\) if \(c=5\) and both pipes are opened. (b) Evaluate \(T\) if \(c=5\) and we first drain away a sufficient amount of the pure alcohol and then close the drain and open the filler pipe. (c) For what values of \(c\) (if any) would strategy (b) give a faster time than (a)? (d) Suppose that \(c=4\). Determine the equation for \(T\) if we initially open both pipes and then close the drain.

Step-by-Step Solution

Answer

(a) 13.33 minutes; (b) 20 minutes; (c) Faster if \(c > 15\); (d) \(T = 25\) minutes.

Step 1: Understand the Problem

We need to reduce the amount of alcohol in the 100-gallon tank such that it ends up being a 50% alcohol solution. Initially, the tank contains pure alcohol (100% alcohol). The filler pipe brings in a 25% alcohol solution at a rate of \(c\) gallons per minute, while the drain pipe removes liquid at 5 gallons per minute. Our task is to determine the time \(T\) in different scenarios.

Step 2: Formulate Equations for Scenario (a)

In scenario (a), both pipes are opened, with \(c = 5\). The filler and drain pipes ensure no net volume change, thus no change in volume, only concentration over time. The rate of alcohol reduction is the difference between the draining of pure alcohol and the filling of alcohol from the 25% solution. At a steady state:\[ \frac{dA}{dt} = -5 + 0.25 \times 5 = -3.75 \text{ (gallons of alcohol per minute)} \] where \(A\) is the alcohol content in the tank. We need \(A = 50\) to achieve a 50% solution.

Step 3: Calculate Time for Scenario (a)

Starting with 100 gallons of pure alcohol, we need to reduce it to 50 gallons:\[100 - 3.75T = 50 \3.75T = 50 \T = \frac{50}{3.75} \T \approx 13.33 \text{ minutes}\]

Step 4: Determine Time for Scenario (b)

In scenario (b), we first drain and then fill. We need to drain enough to reach 50 gallons of alcohol because after reaching 50 gallons, the closed drain ensures reaching the required solution directly by adding 25% alcohol solution until the tank is full again:\[ t_1 \text{ (draining time) } =\ \frac{50}{5} = 10 \text{ minutes} \] We then fill the remaining 50 gallons with the 25% solution at \(c=5\), takes another 10 minutes. Thus, \[ T = 10 + 10 = 20 ext{ minutes}\]

Step 5: Compare Strategies for Scenario (c)

Strategy (b) will be faster if the combined time (for draining till half and filling to the top) is less than 13.33 minutes. Since strategy (b) involves 10 minutes draining, adding so that the filling should be within the remaining timespan \(x = 3.33 \) minutes, requiring \(c = \frac{50}{x} = \frac{50}{3.33} \approx 15 \) gallons per minute. Hence \(c > 15\) is needed.

Step 6: Solve for Equation for Scenario (d)

Initially opening both pipes with a filler rate of 4 gallons per minute, results in a net drain rate of 1 gallon per minute. So initially: \[ \frac{dA}{dt} = -5 + 0.25 \times 4 = -4 ext{ gallons per minute} \]Let \(t_1\) be the time where we close the drain, then \(100 - 4t_1 = 50\) thus \(t_1 = 12.5 \) minutes.For filling after closing drain, the equation \[ \text{100} = 4(t - t_1) \Rightarrow \text{50} = 4(t - 12.5) \] solving gives \(t = 25\) minutes.Thus, \(T = 25\) minutes.

Key Concepts

Concentration Change

Concentration change is a vital concept when dealing with solutions, especially in problems involving differential equations. In this scenario, we begin with a 100-gallon tank filled entirely with pure alcohol; thus, the concentration is initially 100% alcohol. By adjusting the flow rates of the filler and drain pipes, the concentration of alcohol in the tank is modified. The goal is to reach a 50% alcohol concentration at the end.

The concentration change occurs as the 25% alcohol solution enters the tank through the filler pipe, while the pure alcohol is removed via the drain pipe. Over time, the input and output flow rates affect the alcohol concentration, until the desired 50% concentration is achieved. This involves calculating how much pure alcohol needs to be drained and how much diluted alcohol solution should be added to reach this state.

Alcohol Solution

An alcohol solution refers to a mixture where alcohol is diluted with another substance, typically water. In this exercise, the tank initially contains pure alcohol, which constitutes a 100% concentration. The challenge is to bring in an external 25% alcohol solution via the filler pipe to adjust the concentration.

Having a 25% solution means that for every gallon, 0.25 gallons is pure alcohol, and 0.75 gallons is the diluting substance (water). By managing the inflow and outflow, we can gradually decrease the alcohol concentration in the tank.

• This controlled mixture process essentially allows adjusting the solution's composition until the target concentration of 50% is reached.
• It exemplifies how different solution percentages can be used effectively in real-world applications of chemical processing.

Flow Rates

Flow rates are crucial for controlling how quickly solutions enter or leave a system, in this case, the tank. The rates are measured in gallons per minute (GPM), with the filler and drain pipes having adjustable rates.

In the given problem, the drain pipe operates at 5 GPM, reducing the volume of alcohol in the tank. At the same time, the filler pipe can be adjusted to different rates, denoted by \(c\), to add the 25% solution.

• When both pipes are open, achieving a balance between inflow and outflow is key to controlling the effective change in alcohol concentration. This indicates how fast the mixture reaches the desired 50% alcohol solution.
• Selecting different values of \(c\) allows exploring alternate strategies to reach the same end-point in varying spans of time.

This manipulation of flow rates to achieve desired concentrations is a practical illustration of fluid dynamics in action.

Time Calculation

Time calculation is essential in determining how long it will take to achieve the target concentration. Here, the variable \(T\) represents time in minutes. Calculating \(T\) involves considering how quickly alcohol is drained from the system compared to how the 25% solution is added.

Starting with pure alcohol, the time required to reach the 50% solution depends on several factors:

• The flow rate of both the drain and filler pipes.
• The initial and target alcohol concentrations.

In different scenarios, such as opening both pipes simultaneously or draining first and then filling, the time to reach the desired concentration varies.
Ultimately, calculating \(T\) involves setting up equations where volumes of alcohol and mixture change over time, solving for \(T\) using algebraic manipulation. Understanding these calculations showcases the practical implications of differential equations in predicting how long processes take in real-world applications.