Naval Architecture and Ships Structure - Assignment Example

Naval Architecture Name: Course: Instructor: Institution: Date of Submission: ASSIGNMENT 3 Q1a. Describe in detail (using appropriate diagrams/drawings) a common accepted method (experiment) you would use to obtain an accurate assessment of the vessel’s initial GM. The adjustable immersion hydrometer is a tool, centered on the Archimedes Law. The tool according to the law assists to regulate the liquid density (Dokkum, 2013). The hydrometer utilized to calculate the water density where the ship floats is typically made through materials that are not easily corroded. The materials also mainly have a subjective bulb with a thin rectangular stem that transmits a scale for gauging densities between 1000 and 1025 kilograms per cubic metre, i.e. 1.000 and 1.025t/m3. On the stem provided or developed according to hydrometer, the marks are positioned as follows. The first step involves letting the hydrometer as presented in figure 1 to float in the fresh water upright at the part marked X. To weight the water accurately, one procures the hydrometer from the water and records the measurements. To record effectively, the mass is selected as Mx kilograms (Dokkum, 2013). Once the hydrometer has been attained from the water, it is replaced with the bulb until one can perceive it hovering where the position marked Y, is the higher close of the stem, in the waterline used. Then one assesses the hydrometer once more and perceives the mass now to be measured as My kilograms. The mass is important for developing the initial GM of the vessel. Figure: 1 The water mass of that is displaced by the stem is recorded amid X and Y, which is hence equivalent to My - Mx kilograms. About 1000kg of the water (fresh) used inhabit a cubic of a metre, which assists measure the stem volume. The stem (figure 1) volume is measured between X and Y as presented in figure one above, which is equal to My – Mx /1000 cu. m. In the following equation, L is the length of the stem provided. That is; the distance between X and Y is the stem length. On the other hand, letter A in the equation is a representative of the stems’ cross sectional part. A = Volume / Length = (My – Mx / 1000L) sq m Now let the hydrometer float in water of density ρ kg/m3 with the waterline ‘x’ metres below Y: Volume of water displaced = MY / 1000 – xa = MY /1000 – x{ (MY - MX) /1000L} ……………………(i) But Mass of water displaced Volume of water displaced = Density of water displaced = My / 1000 ρ …………………………….. (ii) Equate (I) and (II).’ My / 1000 ρ = My / 1000 - x{ (MY - MX) /1000L} Or ρ = My / MY - x {(MY - MX) /L} In this equation, My, Mx and L are known constants whilst ρ and x are variables. As a result, to spot the gauge it is now simply essential to select numerous standards of ρ and to calculate the corresponding values of x. Q1b Using the following information provided below, calculate the following measurements for this vessel? Mass Displacement. Waterplane area. LCB (longitudinal Centre of Buoyancy). BM. TPC (Tons per centimeter Immersion) Initial KG Initial GM The vessel is in salt water. ST ½(A)m2=Y ½ breadth(m) SM A*SM X ASMX Y Y*SM 0 0 0 1 0 0 0 0 0 1 3.04 1.7 4 12.16 5 60.8 4.913 19.652 2 4.90 2.6 2 9.8 10 98.0 17.576 35.152 3 5.02 3.4 4 20.08 15 301.2 39.304 157.216 4 4.62 3.4 2 9.24 20 184.8 39.304 76.608 5 4.21 3.4 4 16.84 25 421.0 39.304 157.216 6 3.80 3.4 2 7.60 30 228 39.304 76.608 7 3.38 3.4 4 13.52 35 473.2 39.304 157.216 8 2.93 3.4 2 5.86 40 234.4 39.304 76.608 9 1.79 2.9 4 7.16 45 322.2 24.389 97.556 10 0 0 1 0 50 0 0 0 ∑Y= 33.69 ∑ASM= 102.26 ∑ASMX= 2323.6 ∑Y*SM= 853.83 Where; SM is Simpson’s multiplier Using Simpson’s first rule, Area of waterplane = 1/3 * CI * ∑Y * 2 where CI=h/d, 50/10 = 1/3 * 5 * 33.69 * 2 =112.3m2 Under water volume = 1/3 * CI * ∑ASM * 2 = 1/3 * 5 * 102.26 * 2 = 340.867m3 Mass displacement = volume under water * sea water density = 340.867 * 1.025 = 349.389 tons Waterplane area Area of waterplane = 1/3 * CI * ∑Y * 2 where CI=h/d, 50/10 = 1/3 * 5 * 33.69 * 2 =112.3m2 LCB = ∑ (ASMX) / ∑ (ASM) = 2323.6 / 102.26 =22. 722m from bow BM First moment of inertia, I will be calculated I =1/12 (Cwp) 2 * LB3 where Cwp =Awp / LB = 1/12 (0.612)2 * (50 * 6.83) = 474.79m4 BM = I / V = 474.79 / 340.867 = 1.3929 m TPCsw = ( Wpa * RD ) / 100 = (112.3 * 1.025) / 100 =1.151 tonnes/m Initial KG = moment of inertia / weight displaced Moment of inertia, ICL = 2/3 * CI * ∑YSM = (2/3 *5 *853.832) / (349. 389 * 9.81) = 0.8304m Initial GM GM = KB + BM – KG = 0.737 + 1.3929 – 0.8304 = 1.2995m Q1c. On the same vessel. 2 x weights A&B (1.8t each) are placed 2.2m each side of the LCB (v/l in equilibrium) Weight A is moved to the position of weight B giving a pendulum movement of 75mm. Pendulum length 3m (mounted at the LCB) If the weight A is moved back to the original position and a new weight of 10t is loaded amidships with a CG of 3m. What will be the shift of G and the new KG.? Solution From 1b LCB is 22.72, when weight A & B are added to the vessel it remains at equilibrium, when weight A is shifted to weight B LCB shifts by 3m to 19.72 When weight A moves back to its original position, it has a distance d1 from the new LCB The new weight of 10t has a distance d2 from LCB Weight B that remained at its original position has a distance d2 from the new LCB The shift will be GG = GG1 + GG2 + GG3 Using GG = w * d / W Where w is the weight at a distance d, W is the total weight The total weight W = 349.39 +1.8 + 1.8 + 10 = 363tonnes GG1 = 1.8 * 5.18 / 363 = 0.026m GG2 = 10 * 5.28 / 363 = 0.145m GG3 = 1.8 * 0.8 / 363 = 0.004m Therefore, the shift of G will be 0.026 + 0.145 + 0.004 = 0.175m Thus the new KG will be (.8304 – 0.175) = 0.6554m Q1d Explain how a KN (SZ) curve is used to determine the vessel’s stability data? The processof developing a cross curve of the stability of the ship occurs through one plotting the assumed height with the righting levers. The height is determined by the gravity center that occurs beyond the keel (Barrass & Derrett, 2011). The constructed curves are also assumed in other cases a KG of zero. The developed curves are identified as KN curves where KN is perceived as the righting lever that is usually determined from the keel. The displacement and KG values righting levers are obtained where the KN value is first attained through the inspection of the curves displacement value. When one subtracts the KN values the appropriate righting levers are also determined, which is an equal correction of the sin heel and the KG product. In Figure b, let KN signify the ordinate acquired from the curves. In addition, let the ship’s centre of gravity be at G so that KG characterizes the genuine elevation of the centre of gravity beyond the keel and GZ embodies the distance (L) of the righting lever. Now GZ = XN = KN - KX or GZ = KN - G sin θ Thus, to determine the righting lever one must use the KN and KG. That is; the ordinate of the KN should be subtracted from the KG sin heel correction equal. Fig. b Then using the vessel’s calculated displacement from 1b and the KN (SZ) curve provided below for this small commercial work boat. (KS is 3.193m) Draw the vessel’s GZ curve of statical stability and state the maximum GZ value? KN cross curves of stability Solution The following data on the table below was obtained from at 349.4 tonnes on the KN curves GZ = KN – G sinθ Taking G as 0.8304 Angle (O) 10 20 30 40 50 60 70 80 90 KN (m) 0.52 0.98 1.42 1.80 2.08 2.26 2.34 2.32 2.20 GZ 0.38 0.70 1.00 1.27 1.44 1.54 1.56 1.50 1.37 The maximum GZ value is 1.56 m at 65 degrees Q1e A different vessel with the following dimensions:- Waterline length of 100 metres Displacement of 6300 tonnes. Mean draught of 6 metres, Longitudinal metacentre = 104m Longitudinal Centre of Floatation (LCF) = Amidships A weight of 60 tonnes is moved from forward to aft over a distance of 50 metres. Calculate the MCT 1 cm and the new draughts forward and aft for this vessel. Solution MCT1cm MCT1cm = (W * BMl) / 100L BML = L2 /12d = 1002 / 12 *6 = 138. 89m The distance BG should be recorded, which is small when correlating it to the GML and BML distances on the stem. Therefore, for this motive, there is a high probability that BML has an appreciable error that is occasionally replaced for GML in the method developed of calculating MCT 1 cm. = 6300 * 138.89 / 100 * 100 = 87.5 tonnes / cm Change of trim = w * d / MCT1cm = 60 * 50 / 87.5 =34. 2857 cm by the stern Change of draft aft = I/L * change of trim Where; I is the distance of the centre of floatation from the aft in metres L is the ships length in metres = 50/100 * 34.2857 = 17.1429cm Original draft 6.0000 A 6.0000 F Change due to trim + 0.17143 - 0.17143 New drafts 6.1714 m A 5.829 m F References Barrass, B. & Derrett, C. D., 2011. Ship stability for masters and mates. New York: Butterworth-Heinemann. Dokkum, K., 2013. Ship knowledge: ship design, construction and operation. New York: DOKMAR. Read More