Q. What is Ventuary Effect?
The Venturi effect is the reduction in fluid pressure associated with increased velocity that results when a fluid flows through a constricted section of a pipe. The Venturi effect is named after its discoverer, Giovanni Battista Venturi. Thus, whenever there is relative movement of water between ship and a constricted passage an increase flow of water and fall of water level will result.
Q. What is Bank Suction?
Water stream through constricted passage moves at an increased speed. This reduces the pressure (- – -) causing a drop in water level. On the other side however, the level does not drop thus maintaining a positive pressure (+++) or higher level of water level compared to the bankside level. The higher level on outer side is responsible for bank suction.
Q. What is bow cushion?
1. Positive pressure pushing the vessel to bank effectively pushes the ship over the middle & after body rather than in the forward part. This is because the ship’s body is more full in middle & aft part rather than in the fore part. The push therefore, has an effect of pivoting the ship at pivot point that is one third the length of ship from forward.
2. The water that enters area between the stem and the bank has slight wedging effect deflecting the stem outwards.
3. Low pressure at stern may cause attraction of stern towards the bank.
Q. What is smelling the ground?
In a river or a narrow channel, shallowing in its course, may cause a sudden constriction between the ship and the underwater shoal. The ship is then likely to take a sudden sheer. The sheer is first towards the shallow, then violently away from it. AB is likely to complain immediately. This is due ventuary effect.
Q. What action must be taken?
If such effect is evident, then hard over rudder on opposite side and a healthy ‘kick ahead’ are essential to regain control. Reducing speed or stopping, when faced with the ship sheering the wrong way and apparently failing to respond to progressively larger angles of helm would often not help.
Q. What else may be observed as unusual behavior?
The constriction can occur sideways or from underneath. If the ships pass from deep to shallow water, at any time during the maneuver, the forces will increase drastically and extreme caution should be exercised.
Q. What precautions are important to take while navigating in narrow channel?
While navigating in shallow waters, the vessel must try and steer in geometric center of the channel rather than apparent middle of channel width at surface. This way a continuous caution against the bank effect is taken. Speed should be low to reduce the interactive forces. There is then, plenty of reserve power for corrective ‘kicks ahead’.
Q. What is Squat?
Squat is reduction of under keel clearance caused in shallow waters. When a vessel runs into shallow water, (depth less than twice the draft), a restricted passage exists for the water, which must pass from underneath or sides. Water filling up the virtual vacuum behind the ship from underneath is at higher velocity than that coming from sides. This results in squatting of ship. This also results in build-up of water ahead of the ship causing increased longitudinal resistance pushing, the pivot point back. This reduces steering lever.
The sinkage is associated with the change of trim. This is because, there is increase of underwater volume, though the displacement actually does not increase. The phenomenon may be considered similar to the one where ship goes to less dense water. The centre of buoyancy will move forward or aft, wherever the COF is. The ship is lifted forward or aft respectively. In the following example the COF is aft of COB. The change of trim is forward
Q. What factors influence squat?
1. Velocity is the most important factor. Squat varies almost as the square of velocity.
2. Squat is directly proportional to the block coefficient.
3. Blockage factor: In a canal the squat can get much more than calculated and sometimes, it can even doubled.
Q. What is the formula for squat that allows the blockage factor?
Squat = 0.05 x Cb x V2.08BF 0.81……… [BF is the blockage factor]
Q. If the breadth and depth of a channel is 72m 30 m resp. Ship’s breadth & draft being 28m & 10 m resp. Cb = 0.72 & speed 16 kn. What will be the squat.
= 0.05 x 0.72 x 319.6 x 0.19 = 2.19 m.
Q. What if the width of the channel is very wide, say 2 miles or so?
In such cases the effective width of channel is found from the formula: Effective breadth = [7.7 + 20 (1-Cb)2]b. Alternately, the artificial canal width may be taken as 8.35b for loaded tankers, 9.5b for a general cargo ship & 11.7b for container ships.
Q. Wide shallow channel of average depth of 12m. Ship’s breadth & draft being 28m & 10 m resp. Cb = 0.72 & speed 16 kn. What will be the squat.
Effective breadth is given by breadth = [7.7 + 20 (1-Cb)2]b = [7.7+ (20 x 0.282)] x 28 = 259.5 m.
Squat = 0.05 x Cb x V2.08BF 0.81 = 0.05 x 0.72 x 162.08 0.0.09 0.81= 0.05 x 0.72 x 319.6 x 0.142 = 1.63 m.
Q. Do you know any empirical formula to find the wind thrust?
Wind thrust = K x A x V2
K = 0.52 x 10-4 for beam wind & 0.39 x 10-4 for F & A wind. A is area in m2. V is wind speed in m/s.
Thus, if the area in side elevation is 2000m2, wind speed is 18 kn, the wind thrust due beam wind will be 0.52 x 10-4 x 2000 x 9.262 = 8.92 t.
There is another formula, which gives wind thrust for 1000m2. Using another formula i.e. thrust
Q. How will you calculate bollard pull required to tow a vessel?
An approximate formula for required bollard pull in case of ship-shaped tow is given by the following formula:
Bollard pull in tonnes = 1.1161 x 10-5 x wet area in ft2 x v2 x factor 2 or 3. (V = Towing speed in knots). Wet area in m2 is given by A = 1.025 x LBP in m (Cb x Breadth in m + 1.7 x Draft in m) m2. To convert m2 to feet² it is multiplied by 10.764262.
Q. Length of tow 115 m, block coefficient = 0.72, breadth = 22m, draft = 7m. Find bollard pull to tow at 6kn. Take weather factor as 3.
Wet area = 1.025 x 115 (0.72 x 22 + 1.7 x 7) = 1.025 x 115 (15.84 + 11.9) = 3269.85m2.
Bollard pull = 1.1161 x 10-5 x (3269.85 x 10.764262) x 36 x 3 = 44.13 t.
Q. Is there any other formula you know?
There is another approximate formula to find required Bollard Pull
Q. The displacement of ship (tow) being 20,000 tonnes is towed at a speed of 6kn. Assuming the tug to be with Fixed pitch propeller and kort-nozzle, calculate the BHP required of tug.
Q. What are the three critical moments when a ship is met on a reciprocal course in a narrow / shallow channel?
In the situations of adverse hydrodynamic interaction, there exists the possibility of losing control and sheering violently out of a channel. Counter helm to correct the swing may be sluggish. On completion of the maneuver each ship should regain the centre of the channel as quickly as possible to avoid any furtherance of bank effect. The three phases of caution are as shown in the diagram.
Q. What are the three critical moments when a ship is overtaking other in a narrow / shallow channel?
The stem of overtaking vessel gets pulled towards the stern of the vessel being overtaken as shown in the first diagram (phase 1). In the second diagram, the sterns tend to come closer and in the third phase of overtaking, the head of the vessel being overtaken tends to come close to the stern of the overtaking vessel. This can become dangerous when the vessel being overtaken is much smaller in comparison with the size and power of overtaking vessel.
The smaller of two ships, are likely to be the most seriously affected. Large ships should be aware of this and adjust their speed accordingly.
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