Stability with liquid loads

Center of gravity refers to the mean location of the gravitational force acting on a body.

The center of buoyancy is the centre of the volume of water which the hull displaces.

Righting couple is a force which normally restores a ship to equilibrium once a heel has changed the relationship between her center of buoyancy and her center of gravity.

Metacentric height and conditions for buoyancy of ships and pontoons with liquid loads or ballast.

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Introduction

Tankers are designed to transport liquid loads, but most large vessels have fuel tanks needed for their propulsion.

The presence of a significant volume of liquid with a free surface has an effect on the metacentric height and hence the stability of the ship.This is because the motion of the ship, as it is displaced from its equilibrium position, causes the liquids (load or ballast) to shift or slosh continuously. This is particularly applicable to large sea tankers and bulkships. In addition, in the context of large quantities of fuel for propulsion being carried on board a ship, the constant consumption of fuel means that the load, and the corresponding metacentric height, will shift. The effect of this on the stability of ships and pontoons is considered in this section.

Note: This page should be read in conjunction with http://www.codecogs.com/reference/engineering/fluid_mechanics/ships/stability_and_metacentric_height.php "Stability and Metacentric Height"

The Metacentric Height for a Vessel with Liquid Ballast.

23287/buoyancy-with-liquid-loads-008.png

Where

$O$ $O$ is the original Waterline.
$G$ is the Centre of Gravity of the Ship and Ballast.
$B$ is the Centre of Buoyancy.

The Ship is tilted through a small angle $\theta$ clockwise giving a new waterline of $O'$ $O'$ .

Due to the movement of the water wedge the Centre of Buoyancy of the Ship ( i.e. The $C$ of $G$ of the displaced liquid) moves to $B'$ .

By the Previous Theory $BM = {\displaystyle\frac{I}{V}$ where $I$ is the second Moment of Area of the Water plane area and $V$ is the Total displacement by the Ship and it's contents.

Due to the movement of the liquid in the Ballast tank, the $C$ of $G$ of the Ship and Ballast moves to $G'$ . Now the vertical through $G'$ cuts the old vertical centre line at $N$ . The Stability of the ship depends upon whether $N$ is above or below $M$ .

The Metacentric Height is now $MN$

Consider the Liquid in the Ballast Tank.

23287/buoyancy-with-liquid-loads-009.png

Let

$I'$ = The second moment of area of the liquid surface about it's centre line. $= \frac{L'\;b'^3}{12}$
$V'$ = the Volume of liquid in the Tank.
$w'$ = the specific weight of the liquid.
$x$ = The movement of the $C$ of $G$ in the tank due to the tilting of the vessel.

The moment of the Couple due to the movement of the Wedge = The Moment of the Couple due to the movement of the $C$ of $G$ . $w'v'x = w'\times \frac{1}{2}\times \frac{b'}{2}\times \frac{b'\;\theta }{2}\times L'\times \frac{2\;b'}{3}$ $\therefore\;\;\;\;\;V'x = \frac{L'\;b'^3\;\theta }{12}$ $\therefore\;\;\;\;\;x = \frac{I'\;\theta }{V'}$

Thus the $C$ of $G$ of the ship and Contents moves from $G$ to $G'$ where $G$ $G'$ is given by:

$W\times GG' = w'\;V'\times x$ or, $w\;v\times GG' = w'\;V'\times x$ $\therefore\;\;\;\;\; GG' = \frac{w'\;V'}{w\;V}\times \frac{I}{V'}\times \theta = \frac{w'I\;\theta }{wV}$ but, $GG' = GN\times \theta$ $\therefore GN = \frac{w'\;I}{w\;V}$

Therefore the Metacentric Height: $MN = MB + BO - OG - GN$

$= \frac{I}{V} + BO - OG - \frac{w' I'}{wV}$

Thus the Metacentric Height has been reduced by the liquid ballast by an amount $\displaystyle\frac{w'I'}{wV}$

Divided Ballast Tanks

If the Ballast Tank is divided into two by a longitudinal Partition, the above equation becomes:

$\displaystyle x = \frac{w_1\;I_1}{wV}$ and $\displaystyle y = \frac{w_2\;I_2}{wV}$

23287/buoyancy-with-liquid-loads-010.png

$wV\times GG_1 = w_1V_1\times x + w_1V_2\times y$ $\therefore\;\;\;\;\;GG_1 = \frac{w_1\theta }{wV}\left(I_1 + I_2 \right)$ $\therefore\;\;\;\;\;GN = \frac{w_1}{wV}\left(I_1 + I_2 \right)$

Comparing the two separate cases:

1) Single Tank of width $b$ . $I_1 = \frac{L_1\;b_1^3}{12}$

2) Same tank divided into two equal compartments.

$I_1 + I_2 = \frac{2\times L_1}{12}\times \left(\frac{b_1^3}{2} \right)^3 = \frac{1}{4}\times \frac{L_1b_1^3}{12}$

This shows that the reduction in the Metacentric height that occurs in case two is a quarter that experienced in case one. For this reason, the tanks in Oil Tankers are divided into small compartments.

The Righting Couple with a liquid Ballast.

23287/buoyancy-with-liquid-loads-011.png

The Righting Couple due to the Buoyancy Force acting upwards through $B_1$ and $M$ and the weight of the ship acting downwards through $G_1$ and $N$ .

$F_{rc}= W\times MN\times \theta$ where $\theta$ is small.
This is the same as in the simple case without a liquid balance but it must be remembered that the Metacentric Height is now $MN$ and not $MG$ .

Example 1 [imperial]

Problem

A Pontoon 50 ft. by 20 ft. and 7 ft. deep is ballasted with 40 tons of water.
The Pontoon and its load weigh 80 tons.

23287/Buoyancy-with-liquid-loads-014.png

Find:

a)The metacentric height.
b)The angle through which the pontoon will heel if 2 tons of deck cargo are moved 10 ft. from the centre to edge.

Workings

Weight of Pontoon plus Ballast = 120 Tons.

Volume of Water displaced = $\displaystyle\frac{120\times 2240}{62.4} = 4,308\;ft\63$

Draft = $\displaystyle\frac{4308}{20\times50} = 4.308\;ft.$ $\therefore\;\;\;\;\;OB = \frac{4.308}{2} = 2.154\/ft.$

Volume of Ballast Water = $\displaystyle\frac{40\times2240}{62.4} = \displaystyle\frac{4308}{3}\;ft^3$ $\therefore\;\;\;\;\;x = \frac{4308}{3\times 50\times 20} = 1.436\;ft.$ $\therefore\;\;\;\;\;x = OG_B = \frac{x}{2} = 0.718\;ft.$

Taking Moments about the Base: $120\times OG=80\times OG_P+40\times OG_B$ $\therefore\;\;\;\;\;OG=\frac{80\times 5+40\times 0.718}{120}=3.573\;ft.$

$BM=\frac{I}{V}$ And: $GN=\frac{w_1}{w}\left(\frac{I_1+I_2}{V} \right)=\frac{2I_B}{V}$ $BM-GN=\left(\frac{50\times 20^3}{12}-\frac{2\times 50\times 10^3}{12} \right)\div 4308$ $= 5.802\;ft.$

Metacentric Height: $MN = BM + OB - OG - GN$ $= 2.154 + 5.802\;-3.573 = 4.383\;ft.$

Two Tons of Deck cargo are moved ten feet laterally.

Moment due to the movement of the load = The righting Couple. $\therefore\;\;\;\;\;2\times 10 = W\times MN\;\theta = 120\times 4.383\times \theta$ Thus: $\theta = 0.038\;rds. = 2^0\;11'$