Naval Ship Architecture - Lab Report Example

NAVAL ARCHITECTURE Student’s name Code+ course name Professor’s name University name City, State Date Question 1a The metacentric height (GM) is used to measure the initial static stability of a floating vessel. It is computed as the distance between the ship’s metacenter and its centre of gravity. A lager GM indicates that the vessel has a better initial stability against capsizing. The commonly accepted method to obtain an accurate assessment of the vessel’s initial GM is the inclining experiment, which is carried out after the vessel is fully built[Ree10]. In general, the GM of the vessel plays a critical role in setting the stability and loading capacity of the ship. Metacenter and metacentric height Figure 1: Simple Metacentric Stability Consider a vessel which was at equilibrium and is then inclined by an angle θ When the vessel is heeled by an angle, its center of buoyance will change from B to B1 (see figure 2 below). When vertical lines are drawn from point B and B1, they will intersect at a point referred to as metacenter of the vessel. From Figure 2, the metacentric height (GM) is measured as the distance between the metacenter and centre of gravity of the ship, and it helps determine the stability of the ship. Requirements for Carrying out the Inclining Experiment There are a number of that should be met before the inclining experiment is carried out. They include: The experiment should be done when the vessel is fully built or when major structural changes have been carried out. The experiment should be done with an empty vessel or the vessel should be near to empty as possible. The vessel have to be in upright position The vessel should be placed in calm waters and sheltered The gangway have to be lifted and the mooring ropes ought to be slacked The density and draught of the water should be accurately noted Any tank in the vessel should be empty or completely pressed up to minimize free surface effect Figure 2: Heeling the Vessel Anyone not conducting the experiment should leave the vessel The additional weights on board must not go beyond 20 percent of the lightweight Boiler and cargo cooling, lube and fuel oil, sanitary, firefighting, cooling water or hydraulic systems must be filled up to operational conditions The test ought to be done on a calm weather day so that the current and wind do not have an effect on the free movement of the vessel during the experiment The heeling weights should be in such a way that they do not cause inclination angles of more than two degrees. Two movements should be used to carry out the inclinations to each side and zero points ought to be marked in the protocol. A tool referred to as the stabilograph is used to conduct the inclining experiment[Mol11]. A stabilography tool has a heavy metal pendulum that is balanced on a knife edge and mounted to a pointer to register the readings of the heel angle (see Figure 3). A minimum of two stabilographs are normally used and are arranged at a maximum distance from one another. In other words, one is placed in forward and the other one at the aft. On each side of the mid vessel, four masses are placed on the vessel deck. This placement is done in such a way that they are away from the centre line as shown in Figure 4. These masses are then moved each at a time till all of them are on the same side, then all of them are moved to the other side, and finally, two on either side (see Figure 5). Figure 3: A Stabilography The deflection on the two stabilographs is noted/recorded for each and every movement of the four masses. The average of these readings is used to calculate the metacentric height. Figure 4: The Effect of Heeling the Vessel on the Centre of Buoyancy and Metacentre Taking θ to be the angle of heel and G1 as the new or moved position of the centre of gravity after inclination; so using the trigonometry rules, GG1 = GM tanθ GG1 can also be given by m*d/Δ Where, d = distance moved by the mass Δ = displacement of the vessel in water m = mass moved Therefore, GM = m * d/Δtanθ The value of GM is the metacentric height, and the value of tanθ can be calculated using the readings obtained from the stabilograph[Mol11]. Figure 5: Layout and Movements of the Four Masses Question 1b Station Half area (m2) Waterplane half-breaths (M) Multiplier Volume Function Area Function Lever Product 1st Mnt Product 2nd Mnt 0 0.00 0.0 1 0 0 5 0 0 1 3.04 1.7 4 12.16 6.8 4 48.64 194.56 2 4.90 2.6 2 9.8 5.2 3 29.4 88.2 3 5.02 3.4 4 20.08 13.6 2 40.16 80.32 4 4.62 3.4 2 9.24 6.8 1 9.24 9.24 5 4.21 3.4 4 16.84 13.6 0 127.44 Aft - 6 3.80 3.4 2 7.6 6.8 1 7.6 7.6 7 3.38 3.4 4 13.52 13.6 2 27.04 54.08 8 2.93 3.4 2 5.86 6.8 3 17.58 52.74 9 1.79 2.9 4 7.16 11.6 4 28.64 114.56 10 0.00 0.00 1 0 0 5 0 0 Ʃfvol = 102.26 ƩfA = 84.8 80.86 Fwd 601.3 Difference 46.58 Aft i. Mass Displacement Displacement refers to the weight () or volume () of water of water displaced by the body of the ship (hull). The volume displaced by the hull can be conceived as the volume of the hole in the water used by the ship quantified in cubic metres. Displacement as a weight refers to the weight of the water displaced by the vessel. This is equivalent to the volume of the water displaced multiplied by the density of water, which is a constant. The density constant in fresh water is equal to 1000 kg/m3, while that of salt water is 1025 kg/m3[Tup01]. Therefore, weight displacement in fresh water Weight displacement in sea water Mass or weight displacement is equivalent to the total weight of the ship while the vessel is at rest in still water and at equilibrium. The waterline spacing or the common interval h can be obtained as follow h= water length/ 10 The water length provided is 50 m. Therefore, h = 50/10 = 5m Volume displacement = 2 x 1/3 X h X Ʃfvol = 2 x 1/3 X 5 X 102.26 = 340.867 m3 Mass displacement = volume displacement X salt water density = 340.867 x 1025 = 349388.33 kg = 349.388 tones ii. Waterplane Area Waterplane area refers to the area at the waterline of the vessel and the surface of the water. This area may vary depending on the draft of the ship[Tup13]. The symbol of the waterplane area is AW. Waterplane area AW = 2 X 1/3 X h X ƩfA = 2 X 1/3 X 5 X 84.8 = 282.667 m2 iii. Longitudinal Centre of Buoyancy (LCB) The center of buoyancy (B) refers to the point in which the total force of buoyancy can be assumed to act. In other words, it is the centroid of the underwater form of the vessel. It can also be referred to as the geometric centre of the underwater volume of the ship. The position of this point is defined by Longitudinal Centre of Buoyancy (LCB) and KB[Tup13]. LCB refers to the longitudinal distance determined either from amidships or FP or AP. On the other hand, KB refers to the vertical distance that is above the base or rather the height of the centre of buoyancy. KB is also known as the VCB; it is measured from the reference point K, and hence, the initials KB. The position of centre of buoyancy (B) is as illustrated in Figure 6 below: Figure 6: Position of B LCB = = 46.58/102.26 X 5 = 2.278 m Aft amidships iv. BM BM refers to the metacentric radius which is measured from B (centre of bouyancy). There are two types of BM initialized as BML and BMT. BML refers to the longitudinal metacentric radius which is measured from B, while BMT refers to the metacentric radius which is measured from B[Bri131]. BM = second moment of area/volume displacement Second Moment of area = 2 X 1/3 X 53 X601.3 = 50108.33 m4 = 50108.333/340.867 = 147.003 m v. Tonnes per centimetre Immersion (TCP) Tonnes per centimeter refer to the mass required to decrease or increase the mean draft of a ship by 1 cm. In other words, it is the mass that should be deducted from or added to a ship so as to change the ship’s draught by 1 cm. Its value varies depending on the waterplane area (AW). On the other hand, the waterplane may vary depending on the draft of the ship, and for this reason, the TPC may vary depending on the draft of the ship[Nev11]. TCP = AWPρ/100 = 282.667 X 1.025/100 = 2.897 tonnes vi. Initial KG Before calculating the initial KG, it is good to know it defines the centre of gravity (G). Centre of gravity refers to the point through which the ship’s total weight can be assumed to act (see Figure 7 below). G is defined by LCG and KG. LCG refers to the longitudinal distance which is measured from either amidships or FP or AP. On the other hand, KG refers to the vertical distance above the base. Therefore, G can also be defined as the point through which the gravitational force is exerted vertically downwards; or the point through which a pivot may be positioned and keep the body balanced; or a uniform body’s geometric centre. In the case of ship, G refers to the point through which the force generated by the mass of the ship is exerted vertically downwards. The G’s position changes depending on the loading conditions of the ship[Pap142]. KG is also referred to as the height of the centre of gravity. KG varies according to the vertical distribution of the mobile masses of the ship, such as ballast, fuel, and cargo. It is calculated from the reference point K, and hence, identified as the KG of the ship. Figure 7: Centre of Gravity, G KG = Total moment = 601.3 Volume Displacement = 340.867 = = 1.764 m vii. Initial GM GM refers to the metacentric height which is measured from the centre of gravity. That is, it is the distance between G and M as shown in Figure 8 below. Figure 8: Righting Moment with a Decreased GM The position of G with reference to M is very important as it is concerned with the ability of the shup to right itself. Normally, G ought to be below M. In such a case, the GM is said to be positive. The bigger the distance between G and M, the greater the positive GM is; also, the bigger the GM, the greater the righting lever. When the G is approaching M, the righting lever will decrease, and this causes the righting moment to be weak[Tup13]. When the GM is zero, which means that G is coinciding with M, then the righting lever does not exist. When an external force changes the heel of the ship to a small angle, without the righting moment, the vessel will remain heeled at this angle. When G is above M, GM will be negative and the righting lever will be non-existent and it will also become a capsizing moment. Applying a light external force will cause the ship to incline sharply, and according to the hull’s shape, the ship can even capsize completely. A negative GM should be avoided at whatever cost. Figure 9: G = 0, Neutral Equilibrium Figure 10: Negative GM, Capsizing Moment GM = KB + BM – KG = 0.737 + 147.003 - 1.764 = 145.969 m Question 1c Original KG = 1.764 m Distance 3.6 tonnes (A&B) from original G = 145.969 – 4.4 = 1.41.569 m GG’ = = 1.444m KG’ = KG + GG’ = 1.764 + 1.444 = 3.208m Distance 10 tonnes is from G = 3m GG’’ = = 0.085m KG’’ = KG’ + GG’’ = 3.208 + 0.085 = 3.293 m Therefore, the shift of G is 1.444 + 0.085 = 1.529 m and new KG is 3.293 m. Question 1d KN Curves The KN curves, also referred to as the cross curves of stability, is a group of curves in which one can read the KN values for a group of constant heel-angle values at any given displacement (see Figure 11). Therefore, we can have a curve 10o heel angle, 20o, 30o and so on[Pap142]. In order to determine the values of KN for a particular displacement volume, one has to draw a vertical line and the values of where the vertical line crosses the curves are read. In general, the KN curves enable us to present stability for various heeling angles, displacements, and an initial trim, both in waves and still water. Figure 11: KN Curves As the location of g (centre of gravity) changes with the ship’s loading, it is not applicable to determine the actual righting arm for each loading condition. Rather, an assumption is made that the g is zero and fall together with the keel. This results to a value presented as KN*sin(θ)[Pap142]. KN curves can be determined to either side of the vessel for a wide range of heeling angles, trims, and displacements. The KN curves only depend on the geometry of the vessel and not it’s loading. Therefore, they apply to each condition a ship may be operating in. Because g is a function of the loading condition, the basis of the KN curves is taken as a fixed point. Therefore, the righting arm is determined using the following formula: GZ = KN – KG sinθ A number of assumptions have been made while preparing the KN curves as follows: i. Irrespective of the heel angle, the ship’s g remains fixed at the pole point, or assumed g. ii. The hull of the ship is assumed to be perfectly watertight iii. The deckhouses and superstructures above the water deck are assumed as non-watertight. iv. Adjustments have to be made to account for the moments and volumes of the immersed appendages such as sonar domes, propellers, rudder, among others, and freely flooding spaces. GZ Curve When a ship heels, the part under water changes behavior; this indicates that the centre of buoyance (B) varies according to the heel of the ship. As heeling changes, the GZ value also changes. Figure 12 below indicates that as the GZ value increases, the ship heels more. The GZ value reaches a maximum at some point, and at that point the ship has a maximum righting force. After the maximum point, the GZ value decreases and when it reaches zero, the ship will capsize. At different degrees of heeling, the GZ values can be drawn to form the GZ curve as shown in Figure 12 below. The GZ curves provide an impression of the ship and its stability. There are different GZ curves for different types of ships. Figure 12: GZ Curve Drawing the GZ curve Mass displacement = 349.388 tones KG = 1.764 m KS = 3.193 m SG = KS – KG = 3.193 – 1.764 = 1.429 Heel (θ) KN Sin θ KG sinθ GZ = KN - KGsinθ 0o 0 0 0 0 10o 0.5 0.174 0.307 0.193 20o 0.9 0.342 0.603 0.297 30o 1.27 0.500 0.882 0.388 40o 1.8 0.643 1.134 0.666 50o 2.1 0.766 1.351 0.749 60o 2.26 0.866 1.528 0.732 70o 2.3 0.940 1.658 0.642 80o 2.27 0.985 1.738 0.532 90o 2.20 1.00 1.764 0.436 Figure 13: GZ Curve As seen from Figure 13, the maximum GZ value is 0.749 m. Question 1e MCT1cm MCT1cm refers to moment to change trim 1cm. it is used to measure the change in trim when loading, shifting and unloading of a mass. MCT1cm = = = In this case, we will use the formula MCT1cm = Where, = the ship’s displacement in tonnes GML= the longitudinal metacentric height in metres L= the Ship’s length in metres The given values are: = 6300 tonnes GML = 104 m L = 100 m Therefore, MCT1cm = = 65.52 tonnes m/cm Draught Forward and Aft In order to calculate new draught forward and aft, we have to find the trimming moment. A trim refers to the difference between the draughts aft and forward. If the draught aft is bigger than the draught forward, it is referred to as trim by the stern. On the other hand, when the draught forward is bigger than draught aft, it is referred to as trim by the bow or head. Trimming moment = w x d Where, w = weight in tonnes d = distance in metres Therefore, trimming moment = 60 x 50 = 3000 tonnes m Change of trim = = = 45.788 cm Change of draft aft in cm = x change of trim in cm Where, I = distance of centre of floatation from aft in metres L = Length of the ship in metres Therefore, Change of draft aft = x 45.788 = 22.894 cm Change of draft forward = x 45.788 = 22.894 cm Original drafts were 6m A and 6m F. Therefore, new draft will be: New draught A = 6 – 0.229 = 5.771 m A New draught F = 6 + 0.229 = 6.229 m F Reference List Ree10: , (Reed, 2010), Mol11: , (Molland, 2011), Tup01: , (Tupper & Rawson, 2001), Tup13: , (Tupper, 2013), Bri131: , (Brian & Pulido, 2013), Nev11: , (Neves, et al., 2011), Pap142: , (Papanikolaou, 2014), Read More