Problem 20 - Rotating Vessel

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

$\omega = 20 ~ \text{rad/sec}$
 

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 H²/2D when the vortex touches the bottom of the tank.)
$\dfrac{H^2}{2D} = \dfrac{6^2}{2(1.5)} = 12 ~ \text{m}$
 

Since h > H²/2D, the vortex is below the vessel. See figure below.
 

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

$x^2 = \dfrac{2gy}{\omega^2}$
 

 

At x = x₁, y = y₁
${x_1}^2 = \dfrac{2(9.81)y_1}{20^2}$

${x_1}^2 = \frac{981}{20\,000}y_1$
 

At x = x₂, y = y₁ + 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 x₁ 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