Problem Statement
An open cylindrical tank with a 0.5 m diameter and 1 m height is completely filled with water and rotated about its axis at 240 rpm. Determine the radius up to which the bottom will be exposed and the volume of water spilled out of the tank.
Solution
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Given:
Diameter \( = 0.5\,\text{m} \) ⇒ Radius, \( r = 0.25\,\text{m} \)
Height \( = 1\,\text{m} \)
Rotation speed \( N = 240\,\text{rpm} \)The angular velocity is computed as:
\( \omega = \frac{2\pi N}{60} = \frac{2\pi \times 240}{60} \approx 25.13\,\text{rad/s} \) -
Determine the Free Surface Depression at the Tank Wall:
For a rotating liquid, the depression at the wall is given by:
\( z = \frac{r^2\,\omega^2}{2g} \)Substituting \( r = 0.25\,\text{m} \), \( \omega \approx 25.13\,\text{rad/s} \), and \( g = 9.81\,\text{m/s}^2 \):
\( z = \frac{(0.25)^2 \times (25.13)^2}{2 \times 9.81} \approx 2.01\,\text{m} \) -
Determine the Depth by Which the Bottom is Exposed:
Since the tank is only 1 m high, the extra depression beyond the tank is:
\( z_1 = z – 1 = 2.01 – 1 = 1.01\,\text{m} \) -
Find the Radius \( r_1 \) Up to Which the Bottom is Exposed:
At the point where the bottom is just exposed, the free surface depression is:
\( z_1 = \frac{r_1^2\,\omega^2}{2g} \)Solving for \( r_1 \):
\( r_1 = \sqrt{\frac{2g\,z_1}{\omega^2}} = \sqrt{\frac{2 \times 9.81 \times 1.01}{(25.13)^2}} \approx 0.18\,\text{m} \) -
Calculate the Volume of Water Spilled:
The volume of water spilled is the difference between the volume of the full paraboloidal segment and the volume that remains inside the tank.
\( V = \frac{1}{2}\pi r^2 z – \frac{1}{2}\pi r_1^2 z_1 \)Substituting the known values:
\( V = \frac{1}{2}\pi (0.25)^2 \times 2.01 – \frac{1}{2}\pi (0.18)^2 \times 1.01 \approx 0.146\,\text{m}^3 \)
Explanation
When the tank rotates, centrifugal forces cause the free surface to form a paraboloid. The depression at the tank’s wall is calculated using \( z = \frac{r^2\,\omega^2}{2g} \). Since this depression (≈ 2.01 m) exceeds the tank height, the bottom becomes partially exposed by an amount \( z_1 = 2.01 – 1 = 1.01\,\text{m} \). The radius \( r_1 \) where the free surface meets the bottom is found by relating the local depression to the parabolic profile. The volume of water spilled is then determined by subtracting the volume of water that remains in the tank (corresponding to the exposed part) from the total volume of the paraboloid.
Physical Meaning
This problem demonstrates how high rotational speeds can significantly alter the free surface of a fluid. In this case, the intense centrifugal forces create a depression that not only changes the fluid level but also leads to water spillage as the free surface extends beyond the container’s confines. The analysis highlights the balance between centrifugal effects and gravity in shaping the fluid’s surface.
