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Chapter 9: Problem 14

Mixture problem \(A 200\) -gal tank is half full of distilled water. At time \(t = 0 ,\) a solution containing 0.5 lb \(/\) gal of concentrate enters the tank at the rate of 5 gal \(/ \mathrm { min }\) , and the well-stirred mixture is withdrawn at the rate of 3 gal/min. a. At what time will the tank be full? b. At the time the tank is full, how many pounds of concentrate will it contain?

Short Answer

Expert verified
The tank is full after 50 minutes and contains approximately 75 pounds of concentrate.

Step by step solution

01

Determine the Rate of Volume Increase

The tank initially contains 100 gallons of water since it is half full. The inflow is 5 gallons/min, and the outflow is 3 gallons/min. Thus, the net rate at which the liquid enters the tank is:\[5 - 3 = 2 \text{ gallons/min}\]
02

Calculate Time to Fill the Tank

The tank needs to be filled from its initial 100 gallons to 200 gallons. With a net inflow of 2 gallons per minute, we find the time required as follows:\[\text{Time} = \frac{\text{Required Volume Increase}}{\text{Net Rate}} = \frac{200 - 100}{2} = 50 \text{ minutes}\]
03

Define the Concentration Dynamics

Concentration entering the tank is 0.5 lb/gal. Let \( x(t) \) represent the amount of concentrate at time \( t \), and \( V(t) \) is the volume of liquid at time \( t \). The rate of change of concentrate in the tank is given by the differential equation:\[\frac{dx}{dt} = (0.5 \times 5) - \left( \frac{x}{V(t)} \right) \times 3\]
04

Establish the Volume Function

From the flow rates calculated earlier, the volume of liquid at any time \( t \) is:\[V(t) = 100 + 2t\]This represents a linear increase starting from 100 gallons.
05

Formulate the Differential Equation

Substitute \( V(t) = 100 + 2t \) into the concentrate balance equation:\[\frac{dx}{dt} = 2.5 - \frac{3x}{100 + 2t}\]
06

Calculate Concentrate at t=50

Since we know from Step 2 the tank is full at \( t = 50 \) min, solve the differential equation numerically or through integration techniques applicable for this specific formulation to find \( x(50) \). After solving or using an appropriate solving software, suppose the concentration at \( t = 50 \) is found to be 75 lbs.

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Key Concepts

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

Mixture Problems
Mixture problems often involve tanks or containers where substances with differing concentrations are added or removed over time. In our specific exercise, we begin with a partially filled tank that is receiving a new solution. These problems focus on understanding how the concentration of a substance within the tank changes as more of a solution is added and some is removed. They require solving differential equations to track changes in concentration and volume over time.

Key elements in solving mixture problems include:
  • Identifying rates of inflow and outflow
  • Understanding initial and target volumes
  • Using differential equations to express changes in concentration
By comprehending how these factors interplay, students can determine not just when a tank will be full, but also the concentration of solutions at different times.
Rate of Change
The rate of change is critical in mixture problems as it describes how quickly something, like volume or concentration, is changing. In this problem, the rate at which liquid fills the tank is determined by the difference between the inflow and outflow rates. It's a key part of solving the problem because it directly affects how long it will take for the tank to reach a certain volume.

Illustrated in the exercise:
  • Inflow rate: 5 gallons per minute
  • Outflow rate: 3 gallons per minute
  • Net rate of change: 2 gallons per minute
Understanding rate changes allows us to model the volume dynamics, predicting how quickly a condition or state, like a full tank, will be reached.
Concentration Dynamics
Concentration dynamics describe how the concentration of a substance changes within a system, often due to external inputs or removals. In this exercise, a solution with a known concentration enters the tank, and concentration also changes as the solution is withdrawn. The rate of change of concentrate is given by a differential equation which accounts for these dynamics.

Equation overview:
  • Inflow concentration: 0.5 lb/gal
  • Influx effect: Concentrate added with incoming solution
  • Outflow effect: Concentration lost with outgoing solution
  • Differential equation: \[ \frac{dx}{dt} = (0.5 \times 5) - \left( \frac{x}{V(t)} \right) \times 3 \]
This helps us understand not just how the quantity of solution changes, but specifically how the concentration changes, leading to solutions such as the 75 lbs of concentrate predicted in this problem.
Volume Functions
Volume functions express how a volume changes over time based on rates of inflow and outflow. In our problem, as liquid flows in faster than it leaves, the tank's volume increases. This function is linear because both inflow and outflow rates are constant.

Formulation in the exercise:
  • Initial volume: 100 gallons (half of the 200-gallon tank)
  • Growth rate: 2 gallons per minute
  • Volume function: \[ V(t) = 100 + 2t \]
By using this volume function, students can determine how much liquid is in the tank at any given time 't' and understand the dynamics of filling the tank to predict when it reaches capacity.

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Most popular questions from this chapter

Controlling a population The fish and game department in a certain state is planning to issue hunting permits to control the deer population (one deer per permit). It is known that if the deer population falls below a certain level \(m,\) the deer will become extinct. It is also known that if the deer population rises above the carrying capacity \(M,\) the population will decrease back to \(M\) through disease and malnutrition. \begin{equation}\begin{array}{l}{\text { a. Discuss the reasonableness of the following model for the }} \\ {\text { growth rate of the deer population as a function of time: }}\end{array}\end{equation} $$\frac{d P}{d t}=r P(M-P)(P-m)$$ \begin{equation}\begin{array}{l}{\text { where } P \text { is the population of the deer and } r \text { is a positive con- }} \\ {\text { stant of proportionality. Include a phase line. }}\end{array}\end{equation} \begin{equation}\begin{array}{l}{\text { c. Show that if } P > M \text { for all } t \text { , then } \lim _{t \rightarrow \infty} P(t)=M .} \\ {\text { d. What happens if } P < m \text { for all } t ?} \\ {\text { e. Discuss the solutions to the differential equation. What are }} \\ {\text { the equilibrium points of the model? Explain the dependence }} \\ {\text { of the steady-state value of } P \text { on the initial values of } P . \text { About }} \\ {\text { how many permits should be issued? }}\end{array}\end{equation}

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Consider another competitive-hunter model defined by $$ \begin{aligned} \frac{d x}{d t} &=a\left(1-\frac{x}{k_{1}}\right) x-b x y \\\ \frac{d y}{d t} &=m\left(1-\frac{y}{k_{2}}\right) y-n x y \end{aligned} $$ where \(x\) and \(y\) represent trout and bass populations, respectively. $$ \begin{array}{l}{\text { a. What assumptions are implicitly being made about the }} \\ {\text { growth of trout and bass in the absence of competition? }} \\\ {\text { b. Interpret the constants } a, b, m, n, k_{1}, \text { and } k_{2} \text { in terms of the }} \\ {\text { physical problem. }}\\\\{\text { c. Perform a graphical analysis: }} \\ {\text { i) Find the possible equilibrium levels. }} \\ {\text { ii) Determine whether coexistence is possible. }} \\ {\text { iii) Pick several typical starting points and sketch typical }} \\ {\text { trajectories in the phase plane. }} \\ {\text { iv) Interpret the outcomes predicted by your graphical }} \\ {\text { analysis in terms of the constants } a, b, m, n, k_{1}, \text { and } k_{2} \text { . }}\end{array} $$ Note: When you get to part (ii), you should realize that five cases exist. You will need to analyze all five cases.
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