Problem 14 - Advanced Engineering Mathematics

Question

When the height of the water is \(h,\) the radius of the top of the water is \(\frac{2}{5}(20-h)\) and \(A_{w}=4 \pi(20-h)^{2} / 25 .\) The differential equation is \\[ \frac{d h}{d t}=-c \frac{A_{h}}{A_{w}} \sqrt{2 g h}=-0.6 \frac{\pi(2 / 12)^{2}}{4 \pi(20-h)^{2} / 25} \sqrt{64 h}=-\frac{5}{6} \frac{\sqrt{h}}{(20-h)^{2}} \\] Separating variables and integrating we have \\[ \frac{(20-h)^{2}}{\sqrt{h}} d h=-\frac{5}{6} d t \quad \text { and } \quad 800 \sqrt{h}-\frac{80}{3} h^{3 / 2}+\frac{2}{5} h^{5 / 2}=-\frac{5}{6} t+c \\] Using \(h(0)=20\) we find \(c=2560 \sqrt{5} / 3,\) so an implicit solution of the initial-value problem is \\[ 800 \sqrt{h}-\frac{80}{3} h^{3 / 2}+\frac{2}{5} h^{5 / 2}=-\frac{5}{6} t+\frac{2560 \sqrt{5}}{3} \\] To find the time it takes the tank to empty we set \(h=0\) and solve for \(t .\) The tank empties in \(1024 \sqrt{5}\) seconds or 38.16 minutes. Thus, the tank empties more slowly when the base of the cone is on the bottom.

Step-by-Step Solution

Answer

The tank empties in 1024√5 seconds or 38.16 minutes.

Step 1: Identify the Initial Condition

We begin by noting that at time \(t = 0\), the height of the water \(h\) is given as 20. This initial condition is crucial for finding the constant \(c\) after integrating the differential equation.

Step 2: Separate Variables in the Differential Equation

Given the differential equation \( \frac{d h}{d t}=-\frac{5}{6} \frac{\sqrt{h}}{(20-h)^{2}} \), we rewrite it in a manner to separate variables: \( \frac{(20-h)^{2}}{\sqrt{h}} \, d h=-\frac{5}{6} \, d t \). This allows us to integrate both sides independently.

Step 3: Integrate Both Sides

Integrate the left side with respect to \(h\) and the right side with respect to \(t\):\[\int \frac{(20-h)^{2}}{\sqrt{h}} \, d h=-\int \frac{5}{6} \, d t\]This will provide us with an expression involving \(h\) and \(t\).

Step 4: Solve the Integration

After integrating, we get the equation: \( 800 \sqrt{h} - \frac{80}{3} h^{3/2} + \frac{2}{5} h^{5/2} = -\frac{5}{6} t + c \). This expression relates the height \(h\) at any time \(t\) and includes the integration constant \(c\).

Step 5: Determine the Integration Constant

Using the initial condition \(h(0) = 20\), we substitute to solve for \(c\):\[800 \sqrt{20} - \frac{80}{3} (20)^{3/2} + \frac{2}{5} (20)^{5/2} = \frac{2560 \sqrt{5}}{3}\]Thus, \(c = \frac{2560 \sqrt{5}}{3}\).

Step 6: Find the Time to Empty the Tank

To find when the tank empties, set \(h = 0\) and solve for \(t\) using the implicit solution.The equation becomes:\[0 = -\frac{5}{6} t + \frac{2560 \sqrt{5}}{3}\]Solving for \(t\) gives: \(t = 1024 \sqrt{5}\) seconds.

Key Concepts

Fluid Dynamics

Fluid dynamics is a fundamental area in physics and engineering that explores how fluids (liquids and gases) move. In problems such as the draining of a tank, understanding fluid dynamics is crucial. Here, we look at how the water level changes over time in a tank with a varying radius, highlighting the relationship between fluid behavior and geometry.
Fluid dynamics is governed by several laws, including Bernoulli's principle, which relates the speed of the fluid and its pressure, and the continuity equation, which ensures mass conservation in fluid flow.

• Involves studying the forces causing fluid motion.
• Uses principles like continuity and energy conservation to predict fluid behavior.
• Requires considering variables like density, velocity, and viscosity.

In this exercise, the dynamics of fluid draining from a tank are encapsulated in a differential equation that considers changes in water height and radius, modeling the rate at which water is draining.

Initial Value Problem

An initial value problem (IVP) is a differential equation coupled with a known value of the function at a certain point, often used to find a specific solution. Here, the initial condition is the height of the water at time zero, given as 20. This condition is pivotal as it allows us to solve for the integration constant within the equation.
Solving an IVP can be broken down into key steps:

• Identify the differential equation that models the situation.
• Use the initial condition to find any constants after integrating.
• Apply the initial condition to get a particular solution.

By utilizing the given initial condition, it's possible to ensure the solution is consistent with the physical constraints of the problem, such as ensuring the water starts at a height of 20 meters.

Variable Separation

Variable separation is a technique used to solve differential equations by separating the variables on different sides of the equation. This method is especially useful for equations where the variables are multiplicatively separable, like the one presented here.
The steps involved in separating variables include:

• Rewriting the differential equation to isolate derivatives on one side and integrals on opposing sides.
• Integrating both sides independently to arrive at a general solution.

For example, in our exercise, rewriting the differential equation allows us to separate the variable dependent on time from the one dependent on height. This facilitates solving the equation through integration, ultimately leading to a solution that models the fluid dynamics of the system.

Integration Constant

The integration constant, often denoted as 'c', arises when integrating a differential equation. It represents the unknown constant that must be determined using initial or boundary conditions. In the exercise, once the differential equation has been integrated, we obtain an expression with a constant.
To find the value of the integration constant:

• Substitute the initial or boundary condition into the integrated equation.
• Solve for the constant, thereby obtaining a particular rather than general solution.

In our exercise, the initial condition that the height is 20 at time zero allows us to find the exact value of the integration constant. This accurately and uniquely determines the behavior of the fluid, making it possible to predict how long it takes for the tank to empty.