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Parabolic Free Surface

For a liquid rotating as a rigid body in an open vessel, the free surface becomes a paraboloid.

$$z=\frac{\omega^2 r^2}{2g}$$$$\Delta h=\frac{\omega^2 R^2}{2g}$$

$\Delta h$ is the rise from the center to the wall for a cylindrical vessel of radius $R$.

Volume Conservation Before Spill

Before spill occurs, the average liquid depth stays equal to the original depth. In a cylindrical vessel, the wall rises by $\Delta h/2$ and the center drops by $\Delta h/2$.

$$h_{wall}=h_0+\frac{\Delta h}{2}$$$$h_{center}=h_0-\frac{\Delta h}{2}$$

Speed Just Reaching Rim

An open cylindrical tank has radius 0.75 m, liquid depth 0.90 m, and tank depth 1.10 m. Find the angular velocity when liquid just reaches the rim.

$$h_{wall}-h_0=1.10-0.90=0.20\text{ m}$$$$\frac{\Delta h}{2}=0.20 \Rightarrow \Delta h=0.40\text{ m}$$$$0.40=\frac{\omega^2(0.75)^2}{2(9.81)}$$$$\omega=3.73\text{ rad/s}$$

Answer: $\omega=3.73\text{ rad/s}$, about 35.6 rpm.

Closed Vessel Pressure Field

For a closed rotating vessel, use a known pressure point and add both elevation head and rotation head. There is no exposed free surface to apply volume conservation.

$$\frac{p_2-p_1}{\gamma}=(z_1-z_2)+\frac{\omega^2(r_2^2-r_1^2)}{2g}$$

Problem: Volume of Water Spilled After Rotation Starts

A cylindrical container has radius 0.60 m and is initially filled with water to a depth of 1.20 m. The container is open at the top with walls 1.50 m tall. It is set rotating about its vertical axis at 6.0 rad/s. Determine the height of the paraboloid at the wall, the new center depth, and the volume of water spilled.

$$\Delta h = \frac{\omega^2 R^2}{2g} = \frac{(6.0)^2(0.60)^2}{2(9.81)} = \frac{12.96}{19.62} = 0.660 \text{ m}$$ $$h_{wall,no-spill} = h_0 + \frac{\Delta h}{2} = 1.20 + 0.330 = 1.530 \text{ m}$$

Since $h_{wall} = 1.530 \text{ m} > 1.50 \text{ m}$ (tank wall height), water spills at the rim.

After spill, water surface touches the rim at r = R. Center depth = wall depth − Δh. But the paraboloid height equals Δh, and after spill the rim is fixed at the wall top.

$$h_{center,after} = 1.50 - \Delta h = 1.50 - 0.660 = 0.840 \text{ m}$$ $$V_{remaining} = \pi R^2 h_{center} + \frac{\pi R^2 \Delta h}{2} = \pi(0.60)^2\left(0.840 + \frac{0.660}{2}\right)$$ $$V_{remaining} = \pi(0.36)(1.170) = 1.324 \text{ m}^3$$ $$V_{original} = \pi R^2 h_0 = \pi(0.36)(1.20) = 1.357 \text{ m}^3$$ $$V_{spilled} = 1.357 - 1.324 = 0.033 \text{ m}^3$$

Answer: Wall level would reach 1.530 m without spill, but since the wall is only 1.50 m, spilling occurs. Volume spilled is approximately 0.033 m³.

Problem: RPM Required to Reach a Given Center Depth

A cylindrical open tank has radius 0.50 m and initial water depth 0.80 m. The tank walls are 1.00 m tall. Find the angular velocity (in rpm) at which the water surface at the center just touches the bottom of the tank. Also find the depth at the wall at this condition.

When the paraboloid just touches the bottom, the center depth = 0. Since no water spills before this point (check required), the total water volume is conserved. The average depth remains 0.80 m, meaning the rise at the wall = Δh/2 = 0.80 m, so Δh = 1.60 m.

$$\Delta h = \frac{\omega^2 R^2}{2g}$$ $$1.60 = \frac{\omega^2(0.50)^2}{2(9.81)} = \frac{0.25\omega^2}{19.62}$$ $$\omega^2 = \frac{1.60(19.62)}{0.25} = 125.6 \Rightarrow \omega = 11.21 \text{ rad/s}$$ $$N = \frac{60\omega}{2\pi} = \frac{60(11.21)}{2\pi} = 107.1 \text{ rpm}$$ $$h_{wall} = h_0 + \frac{\Delta h}{2} = 0.80 + 0.80 = 1.60 \text{ m}$$

But wall height is 1.00 m — water spills before the vortex reaches the bottom. A revised analysis with spill is needed in practice.

Answer: Without spill constraint, the center vortex touches bottom at ω = 11.21 rad/s (107.1 rpm). The wall would be at 1.60 m — since the tank wall is only 1.00 m, water spills before this speed is reached.

Problem: Pressure Distribution in a Closed Rotating Cylinder

A closed cylindrical vessel is completely filled with water and rotated about its vertical axis at 10 rad/s. The radius is 0.40 m. Point A is at the center of the base (r = 0, z = 0) where the gage pressure is 50 kPa. Find the gage pressure at: (a) point B on the outer wall at the base (r = 0.40 m, z = 0), and (b) point C at the center of the top lid (r = 0, z = 0.80 m).

$$\frac{p_B - p_A}{\gamma} = \frac{\omega^2(r_B^2 - r_A^2)}{2g} + (z_A - z_B)$$ $$\frac{p_B - 50}{9.81} = \frac{(10)^2[(0.40)^2 - 0]}{2(9.81)} + 0$$ $$\frac{p_B - 50}{9.81} = \frac{100(0.16)}{19.62} = 0.815 \text{ m}$$ $$p_B = 50 + 9.81(0.815) = 50 + 8.0 = 58.0 \text{ kPa}$$
$$\frac{p_C - p_A}{\gamma} = \frac{\omega^2(0 - 0)}{2g} + (0 - 0.80)$$ $$p_C = 50 + 9.81(-0.80) = 50 - 7.85 = 42.15 \text{ kPa}$$

Answer: Point B (outer base) = 58.0 kPa; Point C (center top) = 42.15 kPa. Rotation increases pressure radially outward; elevation reduces pressure going up.

Problem: Paraboloid Geometry and Volume of Dry Core

A cylindrical tank of radius 0.80 m and height 2.0 m is initially full of water (open top). The tank is rotated at angular velocity ω = 8.0 rad/s. Assuming no water has spilled yet is incorrect — find whether spill occurs. If it does not spill, compute the height of the paraboloid and the depth of the dry cone at the center. If it does spill, find the volume spilled.

$$\Delta h = \frac{\omega^2 R^2}{2g} = \frac{(8.0)^2(0.80)^2}{2(9.81)} = \frac{40.96}{19.62} = 2.088 \text{ m}$$ $$h_{wall,trial} = h_0 + \frac{\Delta h}{2} = 2.0 + 1.044 = 3.044 \text{ m}$$

Since $h_{wall} = 3.044 \text{ m} > 2.0 \text{ m}$ (full tank height), spilling occurs.

After spilling with rim fixed at 2.0 m, let center depth drop by $\Delta h$ from the rim:

$$h_{center} = 2.0 - \Delta h = 2.0 - 2.088 = -0.088 \text{ m (below base)}$$

The paraboloid vertex goes below the tank bottom, so the dry volume forms a paraboloid of height $y_0$ within the tank.

$$V_{spilled} \approx \pi R^2 \frac{\Delta h}{2} = \pi(0.80)^2\frac{2.088}{2} = 2.104 \text{ m}^3$$

Answer: Water spills. The volume spilled is approximately 2.10 m³. The central dry vortex (paraboloid apex) would extend slightly below the floor of the tank at this speed.

Problem: Radial and Vertical Pressure Gradient in a Rotating Vessel

A sealed cylindrical vessel (radius 0.30 m, height 0.50 m) is completely filled with oil (SG = 0.85) and rotated at 15 rad/s about its vertical axis. The pressure at the axis of the bottom (r = 0, z = 0) is 80 kPa gage. Find the pressure at: (a) the outer rim of the bottom (r = 0.30 m, z = 0), and (b) the outer rim of the top lid (r = 0.30 m, z = 0.50 m).

$$\gamma_{oil} = 9.81(0.85) = 8.339 \text{ kN/m}^3$$ $$\frac{p_B - p_A}{\gamma} = \frac{\omega^2(r_B^2 - r_A^2)}{2g} = \frac{(15)^2(0.30)^2}{2(9.81)} = \frac{20.25}{19.62} = 1.032 \text{ m}$$ $$p_B = 80 + 8.339(1.032) = 80 + 8.61 = 88.61 \text{ kPa}$$
$$\frac{p_C - p_B}{\gamma} = \frac{\omega^2(r_C^2 - r_B^2)}{2g} + (z_B - z_C)$$ $$= 0 + (0 - 0.50) = -0.50 \text{ m (elevation change only, same r)}$$ $$p_C = 88.61 + 8.339(-0.50) = 88.61 - 4.17 = 84.44 \text{ kPa}$$

Answer: Outer rim of bottom = 88.61 kPa. Outer rim of top = 84.44 kPa. Centrifugal effect raises pressure radially outward; elevation reduces pressure upward.