Problem Statement
A cylindrical vessel closed at the top and bottom has an inner diameter of 300 mm and is 1 m long. It contains water up to a depth of 0.8 m, and the air above the water is at a pressure of 60 kPa. If the vessel is rotated at 250 rpm about its vertical axis, determine the pressure head at the bottom of the vessel at the center and at the edge.
Solution
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Given:
Air pressure, \( P = 60\,\text{kPa} = 60000\,\text{Pa} \)
Head due to pressure, \( h = \dfrac{P}{\gamma} = \dfrac{60000}{9810} \approx 6.11\,\text{m} \)
Inner diameter = 300 mm, so radius \( r = 0.15\,\text{m} \)
Tank length = 1 m, water depth = 0.8 m → air gap = \( 1 – 0.8 = 0.2\,\text{m} \)
Rotational speed, \( N = 250\,\text{rpm} \) -
Determine the Angular Velocity:
\( \omega = \dfrac{2\pi N}{60} = \dfrac{2\pi \times 250}{60} \approx 26.16\,\text{rad/s} \) -
Calculate the Free Surface Depression at the Center:
The depression at the center is given by \( z_1 = \dfrac{r^2 \omega^2}{2g} \).
\( z_1 = \dfrac{(0.15)^2 \times (26.16)^2}{2 \times 9.81} \approx 0.785\,\text{m} \) -
Determine the Free Surface Elevation at the Edge:
Let \( z_2 \) be the elevation of the free surface at the edge relative to the original water level. In a rotating fluid, the elevation at a radial position \( r_1 \) (with \( r_1 \) less than \( r \)) is given by \( z_2 = \dfrac{r_1^2 \omega^2}{2g} \). In this problem, \( z_2 \) is expressed as \( z_2 = 34.88\,r_1^2 \) (a) (since \( \dfrac{\omega^2}{2g} \approx 34.88\,\text{(for } \omega=26.16\,\text{rad/s)} \)).
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Volume Balance for the Trapped Air:
The volume of air originally present above the water (of height 0.2 m) is \( V_{\text{air}} = \pi r^2 \times 0.2 \). This air is compressed when the free surface deforms into a paraboloid with its lower boundary shifted upward. Equate the initial air volume to the volume of the paraboloidal section:
\( \pi (0.15)^2 (0.2) = \dfrac{1}{2}\pi r_1^2 z_2 \)Canceling \( \pi \) gives:
\( (0.15)^2 \times 0.2 = \dfrac{1}{2} \, r_1^2 \, z_2 \)Numerically, \( 0.15^2 \times 0.2 = 0.0225 \times 0.2 = 0.0045 \). Thus,
\( r_1^2 z_2 = 0.009 \) (b) -
Solving Equations (a) and (b):
From (a), \( z_2 = 34.88\,r_1^2 \). Substitute into (b):
\( r_1^2 (34.88\,r_1^2) = 34.88\,r_1^4 = 0.009 \)Thus,
\( r_1^4 = \dfrac{0.009}{34.88} \approx 0.000258 \quad\Longrightarrow\quad r_1 \approx (0.000258)^{1/4} \approx 0.13\,\text{m} \)Then, using (a):
\( z_2 = 34.88 \times (0.13)^2 \approx 34.88 \times 0.0169 \approx 0.59\,\text{m} \) -
Determine the Pressure Heads at the Bottom:
The total pressure head at the bottom is the sum of the head due to the trapped air and the additional head from the water column deformation.
At the center, the water surface is depressed; the additional head is
\( \text{Additional head at center } = QS = 1 – z_2 = 1 – 0.59 = 0.41\,\text{m} \)Thus, the pressure head at the center is:
\( \text{Head}_{\text{center}} = h + QS = 6.11 + 0.41 = 6.52\,\text{m} \)At the edge, the effective water column is increased by the depression difference:
\( AD = 1 + (z_1 – z_2) = 1 + (0.785 – 0.59) = 1 + 0.195 = 1.195\,\text{m} \)Hence, the pressure head at the edge is:
\( \text{Head}_{\text{edge}} = h + AD = 6.11 + 1.195 = 7.305\,\text{m} \)
Explanation
The air above the water exerts an initial pressure head of approximately \( 6.11\,\text{m} \) of water. When the vessel rotates, centrifugal forces deform the water’s free surface into a paraboloid. The depression at the center, \( z_1 \), is calculated using \( z_1 = \frac{r^2 \omega^2}{2g} \). A further elevation \( z_2 \) at the edge (relative to the deformed surface at a reduced radius \( r_1 \)) is determined by a volume balance between the original air space and the volume of the paraboloidal segment. Solving these relations gives \( r_1 \approx 0.13\,\text{m} \) and \( z_2 \approx 0.59\,\text{m} \). These values are then used to compute the additional water head at the center (a reduction by \( 0.41\,\text{m} \)) and an increased head at the edge (an addition of \( 1.195\,\text{m} \)).
Physical Meaning
This problem illustrates how the combined effects of trapped air pressure and the centrifugal deformation of the water free surface influence the pressure distribution at the bottom of a rotating, closed cylindrical vessel. The pressure head at the center is lower than that at the edge due to the depression of the free surface, while the trapped air pressure contributes a base head that adds to the overall pressure at both locations.
