Open Cylindrical Tank Rotation Analysis Determining Minimum Depth And Spillage

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This article delves into the analysis of an open cylindrical tank containing water when subjected to rotation around its vertical axis. This is a classical problem in fluid mechanics, often encountered in engineering applications involving rotating machinery or storage tanks. Understanding the behavior of fluids under rotation is crucial for designing safe and efficient systems. The primary focus will be on determining the minimum depth of the tank required to prevent water spillage at a given rotational speed. We will explore the underlying principles of fluid mechanics, including centrifugal forces and the formation of a parabolic water surface, and apply them to solve a practical engineering problem. This exploration aims to provide a comprehensive understanding of the concepts involved and their application in real-world scenarios, making it a valuable resource for students, engineers, and anyone interested in the dynamics of rotating fluids. This article will serve as a valuable resource for understanding the interplay between rotational dynamics and fluid behavior, providing insights relevant to various engineering disciplines.

Problem Statement

Consider an open cylindrical tank with a diameter of 1 meter, initially filled with water to a depth of 3 meters. Our investigation revolves around two key scenarios:

  • (a) Determining the minimum tank depth: When the tank rotates at a speed of 100 revolutions per minute (rpm) about its vertical axis, what is the minimum depth required to prevent any water from spilling out?
  • (b) Analyzing the rotation at 150 rpm: If the tank's rotation is increased to 150 rpm, how much water will spill out, assuming the tank has the minimum depth calculated in part (a)?

Theoretical Background: Fluid Mechanics of Rotating Tanks

Centrifugal Force and Free Surface Formation

When a cylindrical tank containing a liquid is rotated about its vertical axis, the liquid experiences a centrifugal force acting radially outward. This force is proportional to the mass of the liquid element, the square of the rotational speed, and the distance from the axis of rotation. This centrifugal force causes the free surface of the liquid to deform into a parabolic shape. The height of the parabola increases with both the rotational speed and the radial distance from the axis. This phenomenon is a direct consequence of Newton's laws of motion and is fundamental to understanding the behavior of rotating fluids. The interplay between gravity and centrifugal force dictates the shape of the free surface, a critical aspect in the design of rotating equipment and storage tanks.

Equation of the Parabola

The equation describing the free surface of the rotating liquid is a parabola, given by:

y = (ω^2 * r^2) / (2 * g)

Where:

  • y is the height of the water surface above the lowest point (vertex) of the parabola.
  • ω is the angular velocity in radians per second.
  • r is the radial distance from the axis of rotation.
  • g is the acceleration due to gravity (approximately 9.81 m/s²).

This equation is derived from the equilibrium condition where the pressure gradient within the fluid balances the combined effects of gravity and centrifugal force. The parabolic shape is a natural consequence of this equilibrium, reflecting the increasing centrifugal force with radial distance. Understanding this equation is crucial for predicting the behavior of fluids in rotating containers and designing systems that can handle these effects.

Volume Conservation

A crucial principle in solving these types of problems is the conservation of volume. The volume of water in the tank remains constant regardless of the rotation speed. This means the volume of the air space created by the parabolic depression must equal the volume of the water that rises above the initial water level. This principle allows us to relate the initial water depth to the final water profile under rotation. The conservation of volume provides a vital constraint in analyzing the fluid dynamics of rotating systems, enabling accurate predictions of liquid levels and potential spillage.

Solution Approach

Part (a): Determining Minimum Tank Depth at 100 rpm

  1. Convert rpm to radians per second: The rotational speed is given as 100 rpm. To use it in the parabola equation, we need to convert it to radians per second (rad/s).

    ω = (100 rpm) * (2π rad/rev) * (1 min/60 s) ≈ 10.47 rad/s

  2. Determine the radius of the tank: The diameter of the tank is 1 meter, so the radius (r) is 0.5 meters.

  3. Calculate the height of the parabola at the edge of the tank: Using the parabola equation, we can find the height (y) at the edge of the tank (r = 0.5 m).

    y = (ω^2 * r^2) / (2 * g) = (10.47^2 * 0.5^2) / (2 * 9.81) ≈ 1.39 meters

    This value represents the height difference between the lowest point of the water surface (vertex of the parabola) and the water level at the edge of the tank.

  4. Determine the drop in water level at the center: The volume of the air space formed by the rotation is a paraboloid. The volume of a paraboloid is half the volume of the cylinder that encloses it. Therefore, the drop in water level at the center is half the rise at the edge. The drop in water level at the center is half the rise at the edge, which means the water level at the center drops by half the height of the parabola at the edge relative to the initial water level.

    Drop in water level at the center = 1.39 m / 2 = 0.695 m

  5. Calculate the minimum tank depth: The initial water depth was 3 meters. The water level at the center drops by 0.695 meters, so the minimum depth of the tank to prevent spillage must be the initial depth plus the rise at the edge.

    Minimum Tank Depth = Initial Water Depth + Rise at Edge - Drop at Center Minimum Tank Depth = 3 m + 0.695 m = 3.695 m However, a simpler calculation is Initial Depth + y = 3 + 1.39 = 4.39 meters

    To ensure no spillage, the tank depth must be such that even at the edge, the water level doesn't exceed the tank's height. The water level at the edge rises, forming the parabolic shape. The key is to calculate the total height the water reaches at the tank's edge and ensure the tank is deep enough to accommodate it.

    The lowest point of the water surface will be 0.695 meters below the initial level. The highest point (at the edge) is 1.39 meters above this lowest point. Therefore, the height difference between the initial water level and the highest point is 1.39 - 0.695, which equals 0.695 meters from the initial level plus 3 meters initial. The correct method involves finding how much the water level drops at the center (vertex of the parabola) and ensuring the tank's height accommodates both this drop and the parabolic rise.

    Minimum tank depth = Initial water depth + Drop at center + Height of parabola = 3 m + 0.695 m + 1.39 = 4.39 meters

  6. The minimum depth required to prevent water spillage is 4.39 meters. This calculation ensures that the tank is sufficiently deep to contain the water even with the parabolic rise due to rotation.

Part (b): Water Spillage at 150 rpm

  1. Convert rpm to radians per second: The rotational speed is now 150 rpm.

    ω = (150 rpm) * (2π rad/rev) * (1 min/60 s) ≈ 15.71 rad/s

  2. Calculate the height of the parabola at the edge of the tank: Using the parabola equation with the new angular velocity.

    y = (ω^2 * r^2) / (2 * g) = (15.71^2 * 0.5^2) / (2 * 9.81) ≈ 3.15 meters

  3. Determine the drop in water level at the center: Drop in water level at the center = 3.15 m / 2 = 1.575 m

  4. Calculate the water level at the edge relative to the initial level: The total height of the water at the edge above the tank bottom is the drop at the center plus the parabolic height, meaning that The water level at the edge relative to the tank bottom = Drop at Center + Height of Parabola = 1.575 m + 3.15 m = 4.725 m

  5. Determine the amount of spillage: The tank depth is 4.39 meters (from part a). The water level at the edge is 4.725 meters. Therefore, the amount of spillage is:

    Spillage = Water Level at Edge - Tank Depth = 4.725 m - 4.39 m = 0.335 m (Height Difference)

    Now we calculate the volume of the paraboloid above the tank's edge which represents the volume spilled out. The area of parabolic segment is (2/3) * base * height. Base is the radius of the tank, 0.5 m. Height is the spillage height 0.335 m.

    Area of Spilled Segment = (2/3) * 0.5 m * 0.335 m = 0.1117 m²

    Since this parabolic segment is rotated around the vertical axis, its volume is the volume of revolution. The volume of a paraboloid is (1/2)πr²h where r is radius (0.5m) and h is the height(spillage) of the segment.

    Volume of Spilled Water = (1/2) * π * (0.5 m)² * 0.335 m = 0.1317 m³

    Therefore, approximately 0.1317 cubic meters of water will spill out.

Conclusion

This analysis has demonstrated the application of fluid mechanics principles to a practical engineering problem involving a rotating cylindrical tank. We successfully determined the minimum tank depth required to prevent water spillage at a rotational speed of 100 rpm and calculated the amount of water spillage at 150 rpm, considering the minimum depth requirement. The key takeaways from this study include the importance of understanding centrifugal forces, the parabolic shape of the free surface in rotating liquids, and the principle of volume conservation. These concepts are fundamental in various engineering applications, including the design of rotating machinery, storage tanks, and other fluid handling systems. Further research could explore more complex scenarios, such as partially filled tanks, non-cylindrical geometries, or the effects of fluid viscosity and surface tension.