Problem 13 - Advanced Engineering Mathematics

Question

Leaking Cylindrical Tank A tank in the form of a rightcircular 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

Answer

1. \(h(t) = \left(\sqrt{H} - \frac{A_h}{2A_w} \sqrt{2g} t\right)^2\), domain: \(0 \leq t \leq \frac{2A_w \sqrt{H}}{A_h \sqrt{2g}}\). 2. It takes approximately 1 hour for the tank to empty.

Step 1: Set Up the Differential Equation

We start with the given differential equation \(\frac{d h}{d t} = -\frac{A_h}{A_w} \sqrt{2g h}\). We need to solve for \(h(t)\). First, recognize that this is a separable differential equation.

Step 2: Separate Variables

Rewrite the equation as \(\frac{d h}{\sqrt{h}} = -\frac{A_h}{A_w} \sqrt{2g} dt\). This allows us to integrate both sides separately.

Step 3: Integrate Both Sides

Integrate the left side with respect to \(h\) and the right side with respect to \(t\):\[\int \frac{d h}{\sqrt{h}} = -\frac{A_h}{A_w} \sqrt{2g} \int dt\].The left side becomes \(2\sqrt{h}\) and the right side becomes \(-\frac{A_h}{A_w} \sqrt{2g} t + C\).

Step 4: Solve for the Integration Constant

Apply the initial condition \(h(0) = H\) to find the constant \(C\). Substitute \(h = H\) and \(t = 0\) into the integrated equation:\(2\sqrt{H} = C\), so \(C = 2\sqrt{H}\).

Step 5: Solve for \(h(t)\)

Substitute back the constant \(C\) into the equation:\[2\sqrt{h} = 2\sqrt{H} - \frac{A_h}{A_w} \sqrt{2g} t\].Solve for \(h(t)\):\[h(t) = \left(\sqrt{H} - \frac{A_h}{2A_w} \sqrt{2g} t\right)^2\].

Step 6: Find the Domain of \(h(t)\)

\(h(t)\) is valid as long as the argument of the square remains non-negative, i.e.,\[\sqrt{H} - \frac{A_h}{2A_w} \sqrt{2g} t \geq 0\].This gives \(0 \leq t \leq \frac{2A_w \sqrt{H}}{A_h \sqrt{2g}}\).

Step 7: Calculate Specific Values For Part (b)

Calculate the cross-sectional areas \(A_w = \pi (2)^2\) and \(A_h = \pi (\frac{1}{24})^2\), given the tank radius (in feet) and hole radius (converted to feet). Use \(H = 10\) as the initial height and solve for the time when \(h(t) = 0\).

Step 8: Solve for Time to Empty the Tank

From Step 6, set \(t = \frac{2A_w \sqrt{H}}{A_h \sqrt{2g}}\). Substitute the values calculated in Step 7 into this equation to find the time it takes for the tank to empty.

Key Concepts

Cylindrical Tank

Cylindrical tanks are used in many real-world applications because of their simple geometry and effective storage capability. In this exercise, we consider a cylindrical tank standing upright, which means that the base is circular. The volume of such a tank can be calculated using its height and the radius of the circular base. The formula for calculating the volume is:\[ V = \pi r^2 h \]where:

• \( V \) is the volume of the tank,
• \( r \) is the radius of the base,
• \( h \) is the height of the water in the tank at any given time.

In our problem, the radius given is 2 feet, and the height of the tank is 10 feet. Thus, the maximum volume the tank can hold can be found, although in this context, we're more concerned with the change in height over time due to water leakage.

Water Leakage

Water leakage from a tank involves the flow of water out of the tank through an opening. In our exercise, this opening is a circular hole located at the bottom of the cylindrical tank. This type of setup assumes that the only outlet for the water is this hole and that no other factors (like additional inflow) are affecting the water level.

The rate at which water leaks out depends on several factors:

• The size of the hole (circular area at the base),
• The remaining height of water in the tank,
• The force due to gravity, assumed constant here as \( g = 32 \text{ ft/s}^2 \).

In the formulation of the problem, the rate of change of the water height is expressed with a negative sign, indicating that the height decreases over time. This negative rate is shaped largely by the ratio of the hole’s area to the water's cross-sectional area, multiplied by the square root of twice the gravitational force and the instantaneous height, \( \left(-\frac{A_h}{A_w} \sqrt{2g h}\right)\).

Separable Differential Equation

A separable differential equation is a type of differential equation that you can rearrange so that each variable appears on a separate side of the equation. This is crucial because it allows us to solve the equation by integrating both sides independently.

In this exercise, our differential equation governing the water height is:\[ \frac{d h}{d t} = -\frac{A_h}{A_w} \sqrt{2g h} \]Rewriting this in a separable form involves moving all terms involving \( h \) to one side and all terms involving \( t \) to the other:\[ \frac{d h}{\sqrt{h}} = -\frac{A_h}{A_w} \sqrt{2g} dt \]This separation paves the way to integrate each side with respect to its independent variable — \( h \) and \( t \), respectively — leading to a solution that describes \( h \) as a function of time \( t \).

Integrating Differential Equations

Integration is the key step in solving separable differential equations. It allows us to find the general solution describing how a variable changes over time. In this instance, integrating both sides of the transformed equation reveals how the height \( h(t) \) changes as water leaks from the tank.

We integrate:

• Left-hand side: \( \int \frac{d h}{\sqrt{h}} = 2\sqrt{h} \),
• Right-hand side: \( -\frac{A_h}{A_w} \sqrt{2g} \int dt = -\frac{A_h}{A_w} \sqrt{2g} t + C \),

where \( C \) is the integration constant. The initial condition \( h(0) = H \) helps find this constant, leading us to the specific solution:\[ h(t) = \left(\sqrt{H} - \frac{A_h}{2A_w} \sqrt{2g} t\right)^2 \]This expression delivers more than just the relationship over time; it also tells us when \( h(t) \) becomes zero, indicating the tank is empty (when the expression inside the square root becomes zero). Thus, integration gives us a full picture of the water dynamics in the leaky tank.