Problem Statement
A cylindrical vessel of 0.5 m diameter and 0.6 m height is completely filled with water under a pressure of 9.81 KN/m2. It is rotated at 300 rpm about its vertical axis. Determine the pressure at a point adjacent to the wall of the vessel.
Solution
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Given:
P = 9.81 KN/m2 (i.e., \( P = 9810\,\text{Pa} \))
Diameter = 0.5 m ⇒ \( r = 0.25\,\text{m} \)
Vessel Height = 0.6 m
\( N = 300\,\text{rpm} \)
\( g = 9.81\,\text{m/s}^2 \) \( \gamma = 9810\,\text{N/m}^3 \) -
Angular Velocity:
The angular velocity is given by \( \omega = \frac{2\pi N}{60} \). Substituting \( N = 300 \):
\( \omega = \frac{2\pi \times 300}{60} = 31.41\,\text{rad/s} \) -
Virtual Height Difference (z):
Under rotation, the free surface of the water forms a paraboloid. The difference in height between the center and the wall is given by \( z = \frac{r^2\,\omega^2}{2g} \). Substituting \( r = 0.25\,\text{m} \) and \( \omega = 31.41\,\text{rad/s} \):
\( z = \frac{(0.25)^2 \times (31.41)^2}{2 \times 9.81} \approx 3.14\,\text{m} \) -
Determining the Virtual Rise (h):
The liquid that appears above the original level occupies the volume of a paraboloid. Equate the volume of the paraboloid to the volume of liquid that has risen:
\( \pi r^2 h = \frac{1}{2}\pi r^2 z \)Canceling \( \pi r^2 \) gives: \( h = \frac{z}{2} \). Substituting \( z = 3.14\,\text{m} \):
\( h \approx 1.57\,\text{m} \) -
Effective Liquid Height and Pressure at the Wall:
The effective height of the water column at the wall is the sum of the vessel height and the virtual rise:
\( h_1 = 0.6\,\text{m} + 1.57\,\text{m} = 2.17\,\text{m} \)Hence, the pressure at the point adjacent to the wall is:
\( P_C = P + \gamma\,h_1 = 9810\,\text{Pa} + 9810 \times 2.17 \approx 31098\,\text{Pa} \)
Explanation
When the vessel is rotated, centrifugal forces cause the water’s free surface to form a paraboloid. The difference in height between the center and the wall is given by \( z = \frac{r^2\,\omega^2}{2g} \). Since the vessel is completely filled and under pressure, the liquid appears to rise above the original level by a virtual height \( h \), determined by equating the volume of the paraboloid to the additional volume of water. This virtual rise is then added to the vessel’s height to yield the effective water column at the wall, which is used to compute the pressure.
Physical Meaning
This problem demonstrates how rotation deforms the free surface in a completely filled vessel. The increase in effective liquid height at the wall leads to a higher hydrostatic pressure at that point. Such analyses are important in the design and operation of rotating containers, where understanding the distribution of pressure is critical.
