Problem 13
Question
A tank in the form of a right circular cylinder standing on end is leaking water through a circular hole in its bottom. As we saw in (10) of Section 1.3, when friction and contraction of water at the hole are ignored, the height \(h\) of water in the tank is described by $$ \frac{d h}{d t}=-\frac{A_{h}}{A_{w}} \sqrt{2 g h}, $$ where \(A_{w}\) and \(A_{h}\) are the cross-sectional areas of the water and the hole, respectively. (a) Solve for \(h(t)\) if the initial height of the water is \(H\). By hand, sketch the graph of \(h(t)\) and give its interval \(I\) of definition in terms of the symbols \(A_{w}, A_{h}\), and \(H\). Use \(g=32 \mathrm{ft} / \mathrm{s}^{2}\) (b) Suppose the tank is \(10 \mathrm{ft}\) high and has radius \(2 \mathrm{ft}\) and the circular hole has radius \(\frac{1}{2}\) in. If the tank is initially full, how long will it take to empty?
Step-by-Step Solution
Verified Answer
\( h(t) = \left(\sqrt{H} - \frac{A_h}{2A_w} \sqrt{2g} \, t \right)^2 \), \( t_{\text{empty}} \approx 227.67 \) seconds.
1Step 1: Define the given differential equation
The given differential equation for the height \( h \) of the water in the tank as a function of time \( t \) is \[ \frac{d h}{d t}=-\frac{A_{h}}{A_{w}} \sqrt{2 g h}, \] where \( A_w \) is the cross-sectional area of the tank, \( A_h \) is the cross-sectional area of the hole, and \( g \) is the acceleration due to gravity.
2Step 2: Set up the initial condition
The initial condition is given as \( h(0) = H \), indicating that at time \( t = 0 \), the height of the water in the tank is \( H \).
3Step 3: Separate variables
Separate variables to solve the differential equation: \[ \frac{d h}{\sqrt{h}} = -\frac{A_h}{A_w} \sqrt{2g} \, dt. \]
4Step 4: Integrate both sides
Integrate both sides of the separated equation. The left-hand side is \( \int \frac{d h}{\sqrt{h}} = 2\sqrt{h} \) and the right-hand side is \( \int - \frac{A_h}{A_w} \sqrt{2g} \, dt = -\frac{A_h}{A_w} \sqrt{2g} \, t + C. \)
5Step 5: Solve for the constant of integration
Using the initial condition \( h(0) = H \), solve for the constant \( C \): \[ 2\sqrt{H} = -\frac{A_h}{A_w} \sqrt{2g} \times 0 + C \Rightarrow C = 2\sqrt{H}. \]
6Step 6: Solve for height as a function of time
Substitute the constant \( C \) back into the integrated equation to get \[ 2\sqrt{h} = -\frac{A_h}{A_w} \sqrt{2g} \, t + 2\sqrt{H}. \] Solving for \( h \), we square both sides, \( h(t) = \left(\sqrt{H} - \frac{A_h}{2A_w} \sqrt{2g} \, t \right)^2. \)
7Step 7: Determine the interval of definition \( I \) for \( h(t) \)
The height \( h(t) \) is valid until the tank empties, i.e., \( h(t) = 0 \). Set \( \sqrt{H} - \frac{A_h}{2A_w} \sqrt{2g} \, t = 0 \) to find \( t \), leading to \( t = \frac{2A_w}{A_h \sqrt{2g}} \sqrt{H}. \) Therefore, the interval \( I \) is \( [0, \frac{2A_w}{A_h \sqrt{2g}} \sqrt{H}] \).
8Step 8: Substitute given values for the specific scenario
Given tank parameters: \( R=2 \) ft (convert radius of hole \( r=(1/2)/12 = 1/24 \) ft). Areas are \( A_w=\pi R^2 = 4\pi \) sq ft and \( A_h=\pi r^2 = \pi(1/24)^2 = \pi/576 \) sq ft. \[ t = \frac{2 \times 4\pi}{(\pi/576) \sqrt{2 \times 32}} \sqrt{10}. \] Simplify to calculate \( t = 72 \sqrt{10} = 227.67 \) seconds.
Key Concepts
Fluid DynamicsCylindrical TanksInitial Value ProblemsSeparation of Variables
Fluid Dynamics
Fluid dynamics, a vital branch of physics, studies the behavior of liquids and gases in motion. It is essential in understanding water flow, especially through vessels like cylindrical tanks. In this context, we explore how fluid escapes through openings, crucial for scenarios like leaking tanks.
Fluid dynamics relies on fundamental principles, including the conservation of mass and energy. For leaking tanks, Bernoulli's principle often applies, stating that an increase in fluid speed occurs simultaneously with a decrease in potential energy or pressure. This principle helps to predict how water height decreases over time in a tank with a hole.
Moreover, studying the velocity of the fluid exiting the hole involves Torricelli's Law, derived from Bernoulli's principle. This law gives the exit speed: \( v = \sqrt{2gh} \), where \( g \) is the gravitational acceleration and \( h \) is the fluid's height above the hole.
Understanding these concepts helps deduce the rate of change in water height over time in a cylindrical tank, forming the basis for differential equations used in these scenarios.
Fluid dynamics relies on fundamental principles, including the conservation of mass and energy. For leaking tanks, Bernoulli's principle often applies, stating that an increase in fluid speed occurs simultaneously with a decrease in potential energy or pressure. This principle helps to predict how water height decreases over time in a tank with a hole.
Moreover, studying the velocity of the fluid exiting the hole involves Torricelli's Law, derived from Bernoulli's principle. This law gives the exit speed: \( v = \sqrt{2gh} \), where \( g \) is the gravitational acceleration and \( h \) is the fluid's height above the hole.
Understanding these concepts helps deduce the rate of change in water height over time in a cylindrical tank, forming the basis for differential equations used in these scenarios.
Cylindrical Tanks
Cylindrical tanks are commonly used to store and manage fluids, thanks to their simple shape and efficient use of space. These tanks are characterized by a circular base and a consistent height, making volume calculations straightforward.
The geometry of cylindrical tanks directly influences how they are analyzed in fluid dynamics. For instance, the cross-sectional area of a vertical cylindrical tank with radius \( R \) is given by \( A_w = \pi R^2 \). This plays a significant role in determining the flow rate of fluids through any openings.
When dealing with problems involving fluid flow in these tanks, knowing the cross-sectional area of both the opening and the tank is critical. The ratio of these areas helps determine the speed at which the fluid exits. In particular, if a cylindrical tank has a small hole at the bottom, the flux of water is determined by the ratio \( \frac{A_h}{A_w} \), where \( A_h \) is the area of the hole.
This relation also affects the time it takes for the tank to empty completely, providing a real-world application of these mathematical concepts.
The geometry of cylindrical tanks directly influences how they are analyzed in fluid dynamics. For instance, the cross-sectional area of a vertical cylindrical tank with radius \( R \) is given by \( A_w = \pi R^2 \). This plays a significant role in determining the flow rate of fluids through any openings.
When dealing with problems involving fluid flow in these tanks, knowing the cross-sectional area of both the opening and the tank is critical. The ratio of these areas helps determine the speed at which the fluid exits. In particular, if a cylindrical tank has a small hole at the bottom, the flux of water is determined by the ratio \( \frac{A_h}{A_w} \), where \( A_h \) is the area of the hole.
This relation also affects the time it takes for the tank to empty completely, providing a real-world application of these mathematical concepts.
Initial Value Problems
Initial value problems (IVPs) are a type of differential equation problem where the solution must satisfy specific initial conditions. They are crucial in modeling real-world phenomena, such as predicting fluid behavior over time in cylindrical tanks.
In this context, the differential equation describes how the water level changes, and the initial condition specifies the water height at the start. Our problem, for instance, starts with the condition \( h(0) = H \), indicating the height initially. By imposing this condition, we find a particular solution that describes the water height at any future time.
Solving an IVP typically involves integrating the differential equation and applying the initial condition to solve for the constant of integration. This results in a specific solution tailored to the problem's scenario, such as determining how long it takes for a fluid to leak from a cylindrical tank.
Understanding IVPs allows us to predict system behavior across various applications, from engineering to environmental science, utilizing specific initial scenarios.
In this context, the differential equation describes how the water level changes, and the initial condition specifies the water height at the start. Our problem, for instance, starts with the condition \( h(0) = H \), indicating the height initially. By imposing this condition, we find a particular solution that describes the water height at any future time.
Solving an IVP typically involves integrating the differential equation and applying the initial condition to solve for the constant of integration. This results in a specific solution tailored to the problem's scenario, such as determining how long it takes for a fluid to leak from a cylindrical tank.
Understanding IVPs allows us to predict system behavior across various applications, from engineering to environmental science, utilizing specific initial scenarios.
Separation of Variables
Separation of variables is a powerful technique for solving differential equations. It involves rearranging the equation so that each variable and its differential appear on opposing sides. This allows each side of the equation to be integrated independently.
To illustrate with our tank problem, the equation \( \frac{d h}{d t}=-\frac{A_{h}}{A_{w}} \sqrt{2 g h} \) is separated into \( \frac{d h}{\sqrt{h}} = -\frac{A_h}{A_w} \sqrt{2g} \, dt \). This enables the integration of each side separately, simplifying the solution process.
Once separated, we integrate both sides. The left side \( \int \frac{d h}{\sqrt{h}} = 2\sqrt{h} \) and the right side \( -\frac{A_h}{A_w} \sqrt{2g} \, t + C \) yields a solution in terms of \( h(t) \). Next, we use initial conditions to solve for constants and find the specific solution.
Separation of variables is instrumental in problems involving fluid dynamics, as it allows for the direct integration of otherwise complex differential equations, paving the way for solutions to real-world problems like fluid leaks in tanks.
To illustrate with our tank problem, the equation \( \frac{d h}{d t}=-\frac{A_{h}}{A_{w}} \sqrt{2 g h} \) is separated into \( \frac{d h}{\sqrt{h}} = -\frac{A_h}{A_w} \sqrt{2g} \, dt \). This enables the integration of each side separately, simplifying the solution process.
Once separated, we integrate both sides. The left side \( \int \frac{d h}{\sqrt{h}} = 2\sqrt{h} \) and the right side \( -\frac{A_h}{A_w} \sqrt{2g} \, t + C \) yields a solution in terms of \( h(t) \). Next, we use initial conditions to solve for constants and find the specific solution.
Separation of variables is instrumental in problems involving fluid dynamics, as it allows for the direct integration of otherwise complex differential equations, paving the way for solutions to real-world problems like fluid leaks in tanks.
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