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Chapter 9: Problem 7

An open cylindrical vessel, having its axis vertical, is \(100 \mathrm{~mm}\) diameter and \(150 \mathrm{~mm}\) deep and is exactly two-thirds full of water. If the vessel is rotated about its axis, determine at what steady angular velocity the water would just reach the rim of the vessel.

Short Answer

Expert verified
The steady angular velocity required is approximately 19.8 rad/s.

Step by step solution

01

Understand the Problem

An open cylindrical vessel is rotating about its axis. The goal is to find the angular velocity at which water just reaches the rim of the vessel.
02

Write the Given Values

Diameter, \(D = 100 \, \text{mm} = 0.1 \, \text{m}\) Radius, \(r = \frac{D}{2} = 0.05 \, \text{m}\)Depth, \(h = 150 \, \text{mm} = 0.15 \, \text{m}\)Water fills two-thirds of the vessel: \(\text{Volume of water} = \frac{2}{3} \, h = 0.1 \, \text{m}\)
03

Determine the Free Surface Shape

When the cylindrical vessel is rotating, the water surface forms a paraboloid. The height of the paraboloid at the center is \(h_{center}\), and the height at the rim is \(0.15 \, \text{m}\).
04

Use the Equation of Paraboloid

For a paraboloid surface due to rotation: \[ h_{center} + \frac{\text{r}^2 \omega^2}{2g} = \text{h}_{rim} \] where: \(h_{center} = 0.1 \, \text{m} \) (since it's two-thirds full), \(r = 0.05 \, \text{m}\), \(g = 9.81 \, \text{m/s}^2 \), and \( \text{h}_{rim} = 0.15 \, \text{m} \)
05

Solve for Angular Velocity

Substitute the known values into the equation: \[ 0.1 + \frac{0.05^2 \omega^2}{2 \times 9.81} = 0.15 \] Simplifying: \[ \frac{0.0025 \omega^2}{2 \times 9.81} = 0.05 \] \[ \frac{0.0025 \omega^2}{19.62} = 0.05 \] \[ 0.0025 \omega^2 = 0.05 \times 19.62 \] \[ 0.0025 \omega^2 = 0.981 \] \[ \omega^2 = \frac{0.981}{0.0025} \] \[ \omega^2 = 392.4 \] \[ \omega = \sqrt{392.4} \approx 19.8 \, \text{rad/s} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

fluid mechanics
Fluid mechanics deals with the behavior of fluids (liquids and gases) at rest or in motion. In this exercise, we are mainly concerned with how water behaves when it is set into rotational motion inside a cylindrical vessel.
  • Fluids in Rest and Motion: Fluids can be found at rest (static fluids) or in motion (dynamic fluids). In our case, water starts from rest and then rotates.
  • Hydrostatic Pressure: For fluids at rest, pressure increases with depth due to the weight of the fluid above. Once the vessel begins to rotate, the distribution of water changes, affecting pressure at different points.
Fluid mechanics helps us predict and explain phenomena like the formation of a paraboloid surface in a rotating vessel. Understanding these principles is key to solving problems involving fluid motion.

angular velocity
Angular velocity, often denoted by \( \omega \), measures how rapidly an object rotates around a specific axis. It's measured in radians per second (rad/s). In this problem, we need to find the angular velocity at which water reaches the rim.
  • Definition: It’s defined as the rate of change of the angle through which an object rotates.
  • Role in Rotation: In a rotating cylindrical vessel, angular velocity affects how water is distributed within the vessel.
When the vessel rotates faster, the centrifugal force pushes the water outward, changing its surface shape. By understanding angular velocity, we can predict how fast something must spin to achieve a certain effect, like the water reaching the rim.

paraboloid surface
A paraboloid surface is the shape that fluid's surface takes when rotating. This parabolic shape happens due to the balance of forces on the fluid elements.
  • Formation: When a fluid is in a rotating vessel, each particle experiences centrifugal forces, causing the surface to curve into a paraboloid.
  • Equation: The height of the fluid at any point is given by:\[ h = h_{center} + \frac{r^2 \omega^2}{2g} \]
Understanding the formation and mathematical representation of a paraboloid surface is essential for solving fluid mechanics problems involving rotation. It allows us to relate the height at the center and rim to find the required angular velocity.

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Most popular questions from this chapter

A tall cylindrical body having an oval cross-section with major and minor dimensions \(2 X\) and \(2 Y\) respectively is to be placed in an otherwise uniform, infinite, two-dimensional air stream of velocity \(U\) parallel to the major axis. Assuming irrotational flow and a constant density, show that an appropriate flow pattern round the body may be deduced by postulating a source and sink each of strength \(|m|\) given by the simultaneous solution of the equations $$ m / \pi U=\left(X^{2}-b^{2}\right) / b \quad \text { and } \quad b / Y=\tan (\pi U Y / m) $$ Determine the maximum difference of pressure between points on the surface.

Show that the two-dimensional flow described (in metre-second units) by the equation \(\psi=x+2 x^{2}-2 y^{2}\) is irrotational. What is the velocity potential of the flow? If the density of the fluid is \(1.12 \mathrm{~kg} \cdot \mathrm{m}^{-3}\) and the piezometric pressure at the point \((1,-2)\) is \(4.8 \mathrm{kPa}\), what is the piezometric pressure at the point \((9,6)\) ?

At a speed of \(6 \mathrm{~m} \cdot \mathrm{s}^{-1}\) the resistance to motion of a rotor-ship is \(80 \mathrm{kN}\). It is propelled by two vertical cylindrical rotors, each \(3 \mathrm{~m}\) diameter and \(9 \mathrm{~m}\) high. If the actual circulation generated by the rotors is \(50 \%\) of that calculated when viscosity and end effects are ignored, determine the magnitude and direction of the rotational speed of the rotor necessary when the ship travels steadily south-east at \(6 \mathrm{~m} \cdot \mathrm{s}^{-1}\) in a \(14 \mathrm{~m} \cdot \mathrm{s}^{-1}\) north-east wind. For these conditions use inviscid flow theory to determine the positions of the stagnation points and the difference between the maximum and minimum pressures. (Assume an air density of \(1.225 \mathrm{~kg} \cdot \mathrm{m}^{-3}\).)

To the two-dimensional infinite flow given by \(\psi=-U y\) are added two sources, each of strength \(m\), placed at \((0, a)\) and \((0,-a)\) respectively. If \(m>|2 \pi U a|\), determine the stream function of the combined flow and the position of any stagnation points. Sketch the resulting body contour and determine the velocity at the point where the contour cuts the \(y\)-axis.

Which of the following functions could represent the velocity potential for the two-dimensional flow of an inviscid fluid? (a) \(x+5 y ;\) (b) \(3 x^{2}-4 y^{2} ;\) (c) \(\cos (x-y)\); (d) \(\ln (x+y)\) (e) \(\arctan (x / y) ;\) (f) \(\operatorname{arccosec}(x / y)\)
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