2. Criterio de costo descontado 9
2.5. Aproximaciones
For a floating object of known shape such as a ship or boat determination of metacentric height can be calculated as follows.
The initial equilibrium position of the object has its centre of Buoyancy, B, and the original water line is AC . When the object is tilted through a small angle the center of buoyancy will move to new
position . As a result, there will be change in the shape of displaced fluid. In the new position is the waterline. The small wedge is submerged and the wedge is uncovered. Since the vertical equilibrium is not disturbed, the total weight of fluid displaced remains unchanged.
Weight of wedge = Weight of wedge .
In the waterline plan a small area, da at a distance x from the axis of rotation OO uncover the volume of the fluid is equal to
Integrating over the whole wedge and multiplying by the specific weight w of the liquid,
Weight of wedge
Similarly,
Weight of wedge Equating Equations ( ) and (),
in which, this integral represents the first moment of the area of the waterline plane about OO , therefore the axisOO must pass through the centeroid of the waterline plane.
Computation of the Metacentric Height
Refer to Figure(), the distance is
The distance is calculated by taking moment about the centroidal axis .
The integral equals to zero, because axis symmetrically divides the submerged
portion .
At a distance x ,
Substituting it into the above equation gives
Where I 0 is the second moment of area of water line plane about . Thus,
Distance
Since,
Periodic Time of Transverse Oscillation
When an overturning moment which results an angular displacement to a floating body is suddenly removed, the floating body may be set in a state of oscillation. This oscillation behaves as in the same manner as a simple pendulum suspended at metacentre M .
Only the restoring moment sets it in a state of oscillation. So, it is equal to the rate of change of angular momentum.
Where, is the radius of gyration about its axis of rotation, and the angular acceleration.
The negative sign indicates the acceleration is in the opposite direction to the displacement. As it corresponds to simple harmonic motion, the periodic time is
From the above equation it can be observed that a large metacentric height gives higher stability to a floating object. However it reduces the time period of oscillation which may cause discomfort for passengers in a passenger ship.
Some typical metacentric heights of various floating vessels are given below Ocean going vessels : 0.3m to 1.2m.
War ship : 1m to 1.5m. River crafts : > 3.6m.
Example 1:
A wooden cylinder of length L and diameter L/2 is floating on water with its axis vertical. Find the metacentric height if the specific gravity of wood is 0.6
Given data: Length of the cylinder = L Diameter of the cylinder = L/2
Specific gravity of wood = 0.6
For flotation weight of the cylinder should be equal to weight of the water displaced. Now if the depth of immersion is h,
Weight of water displaced = Weight of the cylinder
h = 0.6
Answer: - 0.1739 L m. (The negative sign indicates the body is in unstable equilibrium.)
Example 2:
A wooden cylinder of length L and diameter D is to be floated in stable equilibrium on a liquid keeping its axis vertical. What should be the relation between L and D if the specific gravity of liquid and that of the wood are 0.6 and 0.8 respectively?
Solution :
Given data: Specific gravity of liquid = 0.6 Specific gravity of liquid = 0.8
If the depth of immersion is h
Weight of water displaced = weight of the cylinder
The depth of immersion .
Height of centre of pressure from bottom x =
Then,
For Stable equilibrium Answer: L < 0.817D.
Example 3:
A hollow cylinder closed in both end, of outside diameter 1.5 m and length of 3.8 m and specific weight 75 kN per cubic meter floats just in stable equilibrium condition. Find the thickness of the cylinder if the sea water has a specific weight of 10 kN per cubic meter.
Solution :
Given data : Outside diameter 1.5 m Length L = 3.8 m Specific weight 75 kN/m 3
Let the thickness t and immersion depth h . For flotation
Weight of water displaced = weight of the cylinder
h = 91 t
For the cylinder to be in equilibrium
Solving for t we have t = 0.0409 or 0.000829m Answer:- t = 0.83 mm
A cone is floating in water with its apex vertically downward has a vertical height H and diameter d. If the specific gravity of the material is S, find the condition for stable equilibrium.
Solution :
Given data:
Let, immersed depth be h and diameter of the cone at the water line be d 1 For floating Weight of water displaced = Weight of cone
So,
The C.G of the cone is 3/4 th of the height above the apex.
Substituting the values we get,
Answer:- .
To find the metacentre of a ship of 10,000 tonnes a weight of 55 tonnes is placed at a distance of 6 m from the longitudinal centre plane to cause a heel through an angle of 3 0 . What is the metacentre height? Hence find the angle of heel and its direction when the ship is moving ahead and 2.8 MW is being transmitted by a single propeller shaft at the rate of 90 rpm.
Solution :
Given data: Weight of the ship, W = 10 7 kg Angle of heel ? = 3 0
Distance of the weight X = 6 m Weight placed w = 5.5 x 10 4 Metacentric height Answer:- 0.629 m and 0.270. Example 6:
A log of wood of 1296 cm 2 cross section (square) with specific gravity 0.8 floats in water. Now if one of its edges is depressed to cause the log roll, find the period of roll.
Solution :
Let, h be the depth of immersion and L be the length (perpendicular to the page) Since the section is square its dimension should be 0.36 m x 0.36 m
For flotation
Weight of water displaced = Weight of the log
Then, h = 0.288 m.
Time period, and we have, Answer: 5.38 second.
Example 7
A float valve controls the flow of a liquid. The spherical float has a diameter of 15cm and it is connected with a valve through a weightless link 'AOB' mounted at the hinge O. The length of the link AO is 0.2m and that of OB is 0.5m. The oil flow stops when the free surface of liquid is 0.35 m below the hinge. Now if the valve is to be pressed by a force of 10 N what should be the weight of the float? The angle AOB is given as 120 0 and the link AO should be vertical when the flow is to be stopped. The specific gravity of the liquid is given as 0.88.
Solution :
Given data: Diameter of the float 15 cm Specific gravity of the liquid 0.88
Let, the centre of the Spherical float is h m below the top of the liquid surface.
Vertical force on the float If the weight of the float is W
Net vertical force = (F W) Taking moments about the hinge
Answer:- 424.69 N
Example 8
A large iceberge, floating in seawater, is of cubical shape and its average specific gravity is 0.9. If a 20-cm -high proportion of the iceberg is above the surface of the water, determine the volume of the iceberg if the specific gravity of the seawater is 1.025.
Solution
Let the side of the cubical iceberg is h.
Then volume of the submerged portion is = ( h -20) x h 2 Total volume of the iceberg = h 3
Now,
For flotation, weight of the iceberg = weight of the displaced water
So, the side of the iceberg is 164 cm. Thus the volume of the iceberg is 4.41m3 Answer: 4.41m 3
Liquids in Rigid Body Motion
Many liquids such as water, milk and oil are transported in tankers. When a tanker is being accelerated at constant rate, the liquid within the tanker starts splashing. After that a new free surface is formed, each liquid particle moves with same acceleration. At this equilibrium stage the liquid moves as if it were a solid.
Since there is no relative motion between liquid particles the shear stress is zero throughout the liquid. At this equilibrium it is said to be liquid in rigid body motion.
Uniform linear acceleration
A liquid in a vessel is subjected to a uniform linear acceleration, a as discussed in previous section after sometime the liquid particles assumes acceleration a as a solid body.
Consider a small fluid element of dx, dy and dz dimensions as shown in figure. The hydrostatic equation (L-12.1) is applied with the acceleration component as
Note that each term of equation (L-12.1) represents respective force per unit volume.
If , the relation can be resolved into their vectorical components as
L12.2
where are the acceleration components in the x ,y ,z directions respectively. In scalar form equation (L-12.2) becomes
12.3
Special case I :- Uniform acceleration of a liquid container on a straight path.
Consider a container partly filled with a liquid, moving on a straight path with a uniform acceleration 'a'. In order to simplify the analysis the projection of the path of motion on the horizontal plane is assured to be the x-axis, and the projection on the vertical plane to be the z-axis. Note that there is no acceleration component in the y direction. i.e.
The equations ( ) of motion for acceleration fluid becomes
12.4
Therefore,
Pressure is a function of position and the total differential becomes
Substituting for the partial differentials yields
For an incompressible fluid . Pressure variations in the liquid can be computed by integration.
where c is the constant of integration. Let, at origin,
the pressure
then,
and finally the above equation becomes
ressure variation,
If the accelerated liquid has a free surface, vertical rise between two points located on the free surface is computed as follows
12.8 Note that the pressure at both points is the atmospheric pressure.
12.9 The line of constant pressure isobars are parallel to the free surface (shown in figure).
The conservation of mass of an incompressible fluid implies that the volume of the liquid remains constant before and during acceleration. The rise of the liquid level on one side must be balanced by liquid level drop on the other side.
Example 1:
An open rectangular open tank 6m x 4.5m x 3m high containing water up to a level of 2m is accelerated at 3m/ s 2A)horizontally along the longer side. B) vertically downwards and C) vertically upwards D) in
30 0 inclination with horizontal along the longer side. Find in each case the shape of the free water surface and the pressure on the bottom and on the side walls.
Solution Case a) Length = 6m. Answer: 2.918m and 1.082m. Force on bottom 529.74 kN Leading face 25.84 kN Trailing face 187.94 kN
Case b) Downward acceleration,
Fig.Ex1: case (c) Fig.Ex1: case (d)
Case c) Upward acceleration Case d) Inclined acceleration
Answer: 2.69 m and 1.31 m.
Uniform rotation about a vertical axis
When a liquid in a container is rotated about its vertical axis at constant angular velocity, after sometime the liquid will move like a solid together with the container. Since every liquid particle moves with the same angular velocity: no shear stresses exit in the liquid. This type of motion is also known as forced vortex motion.
As shown in figure a cylindrical coordinate system with the unit vector in the radial direction and in the vertical upward direction, is selected.
A fluid particle p' rotating with a constant angular velocity ' ' has a centripetal acceleration direct radially toward the axis of rotation (-ve direction). By substituting the acceleration component the pressure equation ( ) for the fluid particle becomes
Expanding equation ( )
The scalar components are
Since , the total differential is
L-13.4
L-13.5 where c is the constant of integration.
The equation for the surface of constant pressure (for example free surface) is
L13.6
where and this equation indicates that the isobars are paraboloids of revolutions. Special case : Cylinder liquid-filled container
Let, the point (1) on the axis of rotation is at height from the origin. Since the pressure at point (1) is at atmospheric pressure, we can neglect the effect of the pressure.
Substituting pressure and position of (1) the equation ( ) gives
The equation of the free surface becomes
Consider a cylinder element of radius r , free surface height z and thickness dr. The volume of the element is
The volume of paraboloid generated by the free surface is
Since the liquid mass is conserved and incompressible this volume must be equal to the initial volume of the liquid before rotation.
The initial volume of fluid in the container is
Equating these two volumes we get
In the case of a closed container with no free surface or with a partly exposed free surface rotated about the vertical axis an imaginary free surface based on equation(##) can be constructed.
Example 1:
An open cylindrical container 0.5m in diameter and 0.8m in height ,filled with oil upto 0.5 m and rotating about its vertical axis. Find the speed at which the liquid will start to spill over and also the speed at which the point of the bottom centre will just exposed. The specific gravity of liquid is 0.88.
Solution
Given data: Diameter of the cylinder 0.5 m Height of the cylinder 0.8 m
Case a) The liquid will start to spill over when the maximum height at the periphery becomes 0.8 m.
z 0 = 0.5 m
Case b) The bottom of the centre will expose when z 0 will be zero and z max will be still 0.8 m while some oil may spill over.
thus z 0 = 0 m Ans : 13.72 rad/s , 15.84 rad/s.
An upright manometer of limbs 1m high and 0.5 m apart are filled with water upto 0.5 m. now if it starts rotating about a vertical axis 0.2m apart from one limb at the rate of 10 rad/s what would be the levels of liquid in the two limbs ?
Solution
Given data: The radius of rotation of the two limbs are r 1 = 0.2 m and r 2 = 0.3 m
Minimum ref. point of the parabolic shape being Z min
Also (Since the total volume of liquid is constant) Answer: 0.627 m and 0.373 m.
Example 3:
A conical vessel with the base open is filled completely with water and is rotated at 60 rpm. If only 0.0142 m 3 of water is left then calculate the ratio of radius and height of the cone.
Solution
The depth of the paraboloid,
Now,
Volume of water left
We have Note that the side of the cone is tangential to the parabola. From the above two equations, we get the ratio R/H = 0.73