Problem 1 - Aerodynamics for Engineers

Question

A truck carries an open tank, that is \(6 \mathrm{~m}\) long, \(2 \mathrm{~m}\) wide, and \(3 \mathrm{~m}\) deep. Assuming that the driver will not accelerate or decelerate the truck at a rate greater than \(2 \mathrm{~m} / \mathrm{s}^{2}\), what is the maximum depth to which the tank may be filled so that the water will not be spilled?

Step-by-Step Solution

Answer

1.224 meters

Step 1: Identify the Problem

We need to calculate the maximum depth of water in an open tank carried by a truck, given that the tank may tilt during acceleration. The tank must not spill water when the truck accelerates up to a maximum of 2 m/s².

Step 2: Sketch the Tank Scenario

The tank has a rectangular surface with length 6 m and width 2 m. During acceleration, the water surface will tilt. We'll focus on the critical condition, which occurs when the tank is filled to a depth that causes the water nearly to spill at one end.

Step 3: Determine the Critical Angle

If the truck accelerates at 2 m/s², that causes a horizontal force that tilts the water. The angle θ formed between the water surface and the horizontal plane is defined by the tangent function: \( \tan \theta = \frac{F_{\text{horizontal}}}{F_{\text{gravity}}} = \frac{a}{g} \) where \( a = 2 \, \text{m/s}^2 \) and \( g = 9.8 \, \text{m/s}^2 \).

Step 4: Calculate Tan(θ)

Using \( \tan \theta = \frac{2}{9.8} \approx 0.204 \). This represents the ratio of height change between the ends of the tank.

Step 5: Relate Water Depth and Tilt to Tank Length

When the tank is tilted, the change in water depth across the length 6 m is crucial. If filled to a maximum depth \( d \), the incline could cause the water line to drop by \( d_l = \tan \theta \times 6 \, \text{m} \).

Step 6: Set the Critical Condition

Ensure the water does not spill by equating the two critical conditions: \( d - d_l = 0 \). Solve: \( d - (\tan \theta \times 6 \mathrm{~m}) = 0 \).

Step 7: Solve for Maximum Water Depth

Substitute the known value of \( \tan \theta \) as previously calculated: \( d - (0.204 \times 6) = 0 \). Calculate \( d = 1.224 \, \text{m} \).

Key Concepts

Acceleration Effects

When a truck accelerates, it experiences forces that can affect any load it carries. In this case, a significant force arises from the horizontal acceleration of 2 m/s². The acceleration results in a dynamic effect where the water in the tank tends to move towards the rear of the truck. This not only shifts the center of mass but also tilts the water surface within the tank.
Such acceleration effects can lead to a tilting phenomenon. This tilt affects how water distributes itself in the tank, creating a slope between the front and back ends. Understanding these acceleration effects is necessary to prevent water spillage while the truck is in motion.

Open Tank Dynamics

The open tank has a defined shape and size, which play a crucial role in its dynamics under motion. It is 6 meters long, 2 meters wide, and 3 meters deep. Such tanks often face challenges with fluid dynamics, especially when in motion, due to their open top and interaction with acceleration.
During are acceleration, the tank's dynamics change as the water tilts. This means the water can quickly shift and create uneven pressure points on the tank walls. To address these dynamics, one must understand how the tank's dimensions interact with any applied forces. Proper considerations ensure stability and prevent unforeseen movements that can result in water spillage.

Water Spillage Prevention

Preventing water spillage in a moving truck involves understanding the critical conditions under which spillage occurs. Key to prevention is ensuring that the water's surface remains stable enough during the truck's acceleration.
First, knowing the tank’s maximum fill capacity ensures it is filled only to a level that allows room for water tilting without reaching the tank’s edge. Calculations must take into account the truck's maximum acceleration and how it influences water movement.

• Calculate the angle of tilt using acceleration and gravitational forces.
• Ensure that this tilt is within safe limits to prevent water from reaching the tank's edge.

These steps help in keeping the water secure, minimizing potential losses and hazards associated with water spillage.

Critical Angle Calculation

The critical angle is the angle at which the tilt causes the water to be on the verge of spilling over. It is calculated using the tank's maximum acceleration and incorporates gravitational force. To find this angle:

• Use the tangent of the angle, derived from the equation \( an \theta = \frac{a}{g}\), where \(a = 2 \text{ m/s}^2\), the acceleration, and \(g = 9.8 \text{ m/s}^2\), the gravitational force.

After calculating, \(\tan \theta\) gives a ratio of approximately 0.204. This indicates the degree of tilt across the 6-meter length of the tank. Using this angle helps in determining how much of the tank can be filled safely, ensuring water levels don't surpass the critical point beyond which spillage would occur.