**Problem 1**

A closed cylindrical vessel 3 m. in diameter and 6 m high is filled with water to a height of 4.5 m. The rest is filled with air, the pressure of which is 105 kPa. If the vessel is rotated at 191 rpm about its axis, determine the maximum and minimum inside pressure at the base.

**Solution 1**

$\omega = 191 ~ \dfrac{\text{rev}}{\text{min}} \times \dfrac{2\pi ~ \text{rad}}{\text{rev}} \times \dfrac{1 ~ \text{min}}{60 ~ \text{sec}}$

$\omega = 20 ~ \text{rad/sec}$

TIP: Multiply rpm by π/30 for fast conversion to rad/sec. Notice that the above procedure is actually a multiplication of this amount.

$y = \dfrac{\omega^2x^2}{2g}$

When x = r = 1.5 m, y = h

$h = \dfrac{\omega^2r^2}{2g} = \dfrac{20^2(1.5^2)}{2(9.81)}$

$h = 45.871 ~ \text{m}$

Determine the position of the vortex:

(Note: The height of paraboloid is equal to H^{2}/2D when the vortex touches the bottom of the tank.)

$\dfrac{H^2}{2D} = \dfrac{6^2}{2(1.5)} = 12 ~ \text{m}$

Since h > H^{2}/2D, the vortex is below the vessel. See figure below.

$y = \dfrac{\omega^2x^2}{2g}$

$x^2 = \dfrac{2gy}{\omega^2}$

At x = x_{1}, y = y_{1}

${x_1}^2 = \dfrac{2(9.81)y_1}{20^2}$

${x_1}^2 = \frac{981}{20\,000}y_1$

At x = x_{2}, y = y_{1} + 6

${x_2}^2 = \dfrac{2(9.81)(y_1 + 6)}{20^2}$

${x_2}^2 = \frac{981}{20\,000}(y_1 + 6)$

Volume of air

$V_{air} = \frac{1}{4}\pi (3^2)(1.5) = \frac{27}{8}\pi ~ \text{m}^3$

Final volume of air = Initial volume of air

$\frac{1}{2}\pi {x_2}^2 (y_1 + 6) - \frac{1}{2}\pi {x_1}^2 y_1 = V_{air}$

$\frac{1}{2}\pi [ \,\frac{981}{20\,000}(y_1 + 6) \, ](y_1 + 6) - \frac{1}{2}\pi (\frac{981}{20\,000}y_1) y_1 = \frac{27}{8}\pi$

$\frac{981}{20\,000}(y_1 + 6)^2 - \frac{981}{20\,000}{y_1}^2 = \frac{27}{4}$

$(y_1 + 6)^2 - {y_1}^2 = \frac{15\,000}{109}$

$({y_1}^2 + 12y_1 + 36) - {y_1}^2 = \frac{15\,000}{109}$

$12y_1 = \frac{11\,076}{109}$

$y_1 = 8.468 ~ \text{m}$

The minimum pressure at the base occurs at all points within the circle of radius x_{1} and is equal to the original air pressure.

$p_{min} = 105 ~ \text{kPa}$ *answer*

The maximum pressure will occur anywhere along the circumference of the base.

$p_{max} = 105 + \gamma_w(h - y_1) = 105 + 9.81(45.871 - 8.468)$

$p_{max} = 471.92 ~ \text{kPa}$ *answer*

## Recent comments