Ship Stability for Masters and Mates 5 Episode 7 ppt

The following worked example shows the effect of subdivisions in slack tanks in relation to free surface effects (FSE): Question: A ship has a displacement of 3000 tonnes. On the vessel is a rectangular double-bottom tank 15 m long and 8 m wide. This tank is partially ®lled with ballast water having a density of 1.025 t/m 3 . If the GM T without free surface effects is 0.18 m calculate the virtual loss in GM T and the ®nal GM T when the double bottom tank has: (a) no divisional bulkheads ®tted; (b) one transverse bulkhead ®tted at mid-length; (c) one longitudinal bulkhead ®tted on c L of the tank; (d) two longitudinal bulkheads ®tted giving three equal divisions. Answer FSEvirtual loss in GM T or rise in G  I  r SW W  l  b 3 1  r SW 12  W see FigX 20X4a ; virtual loss in GM T  15  8 3  1X025 3000  12  0X02187 m 4 ; GM T finally  0X1800 À0X2187 À 0X0387 m 4 i.e. unstable ship!! Calculating the effect of free surface of liquids (FSE) 199 Fig. 20.4(a) FSEvirtual loss in GM T or rise in G  2 @ l 2  b 3 1 12  W  r SW (see Fig. 20.4(b)) ; virtual loss  2  7X5 Â8 3  1X025 12  3000  0X2187 m 4 This is same answer as for part (a). Consequently it can be concluded that ®tting transverse divisional bulkheads in tanks does not reduce the free surface effects. Ship is still unstable!! FSEvertical loss in GM T or rise in G 2 @ l 1 b 3 2 12  W r SW ; virtual loss in GM T  2Â15Â4 3 Â1X025 12  3000 see FigX 20X4c 0X0547 m 4 iXeX 1 4 of answer to part a Hence final GM T  0X1800 À 0X0547 m 0X1253 m Ship is stable. 200 Ship Stability for Masters and Mates Fig. 20.4(b) Fig. 20.4(c) GM T is now ve, but below the minimum GM T of 0.15 m that is allowable by D.Tp. regulations. FSEvirtual loss in GM T or rise in G  3 d l 1  b 3 3 12  W  r SW (see Fig. 20.4(d)) ; Virtual loss in GM T  3  15  8 3  3 Â1X025 12  W 0X0243 m 4 i.e. 1 9 of answer to part (a) Hence final GM T  0 X1800 À 0X0243 0X1557 m ship is stable. Ship is stable and above D.Tp. minimum GM T value of 0.15 m. So longitudinal divisional bulkheads (watertight or wash-bulkheads) are effective. They cut down rapidly the loss in GM T . Note the 1/n 2 law where n is the number of equal divisions made by the longitudinal bulkheads. Free surface effects therefore depend on: (I) density of slack liquid in the tank; (II) ship's displacement in tonnes; (III) dimensions and shape of the slack tanks; (IV) bulkhead subdivision within the slack tanks. The longitudinal divisional bulkheads referre d to in examples in this chapter need not be absolutely watertight; they could have openings in them. Examples on board ship are the centreline wash bulkhead in the Fore Peak tank and in Aft Peak tank. Calculating the effect of free surface of liquids (FSE) 201 Fig. 20.4(d) 202 Ship Stability for Masters and Mates Exercise 20 1 A ship of 10 000 tonnes displacement is ¯oating in dock water of density 1024 kg per cu. m, and is carrying oil of relative density 0.84 in a double- bottom tank. The tank is 25 m long, 15 m wide, and is divided at the centre line. Find the virtual loss of GM due to this tank being slack. 2 A ship of 6000 tonnes displacement is ¯oating in fresh water and has a deep tank (10 m Â15 m Â6 m) which is undivided and is partly ®lled with nut oil of relative density 0.92. Find the virtual loss of GM due to the free surface. 3 A ship of 8000 tonnes displacement has KG 3.75 m, and KM  5.5 m. A double-bottom tank 16 mÂ16m Â1 m is subdivided at the centre line and is full of salt water ballast. Find the new GM if this tank is pumped out until it is half empty. 4 A ship of 10 000 tonnes displacement, KM 6 m, KG 5.5 m, is ¯oating upright in dock water of density 1024 kg per cu. m. She has a double bottom tank 20 m Â15 m which is subdivided at the centre line and is partially ®lled with oil of relative density 0.96. Find the list if a mass of 30 tonnes is now shifted 15 m transversely across the deck. 5 A ship is at the light displacement of 3000 tonnes and has KG 5.5 m, and KM 7.0 m. The following weights are then loaded: 5000 tonnes of cargo KG 5 m 2000 tonnes of cargo KG 10 m 700 tonnes of fuel oil of relative density 0.96. The fuel oil is taken into Nos. 2, 3 and 5 double bottom tanks, ®lling Nos. 3 and 5, and leaving No. 2 slack. The ship then sails on a 20-day passage consuming 30 tonnes of fuel oil per day. On arrival at her destination Nos. 2 and 3 tanks are empty, and the remaining fuel oil is in No. 5 tank. Find the ship's GM's for the departure and arrival conditions. Dimensions of the tanks: No. 2 15Â15 m Â1m No. 3 22 m  15 m Â1m No. 5 12 m  15 m Â1m Assume that the KM is constant and that the KG of the fuel oil in every case is half of the depth of the tank. 6 A shi p's displacement is 5100 tonnes, KG 4 m, and KM 4.8 m. A double-bottom tank on the starboard side is 20 m long, 6 m wide and 1 m deep and is full of fresh water. Calculate the list after 60 to nnes of this water has been consumed. 7 A ship of 6000 tonnes displacement has KG 4 m and KM 4.5 m. A double- bottom tank in the ship 20 m long and 10 m wide is partly full of salt-water Calculating the effect of free surface of liquids (FSE) 203 ballast. Find the moment of statical stability when the ship is heeled 5 degrees. 8 A box-shaped vessel has the following data. Length is 80 m, breadth is 12 m, draft even keel is 6 m, KG is 4.62 m. A double bottom tank 10 m long, of full width and 2.4 m depth is then half-®lled with water ballast having a density of 1.025 t/m 3 . The tank is located at amidships. Calculate the new even keel draft and the new transverse GM after this water ballast has been put in the double bottom tank. Chapter 21 Bilging and permeability Bilging amidships compartments When a vessel ¯oats in still water it displaces its own weight of water. Figure 21.1(a) shows a box-shaped vessel ¯oating at the waterline WL. The weight of the vessel (W) is considered to act downwards through G, the centre of gravity. The force of buoyancy is also equal to W and acts upwards through B, the centre of buoyancy. b  W. Now let an empty compartment amidships be holed below the waterline to such an extent that the water may ¯ow freely into and out of the compartment. A vessel holed in this way is said to be `bilged'. Figure 21.1(b) shows the vessel in the bilged condition. The buoyancy provided by the bilged compartment is lost. The draft has increased and the vessel now ¯oats at the waterline W 1 L 1 , where it is again displacing its own Fig. 21.1(a) weight of water. `X' represents the increase in draft due to bilging. The volume of lost buoyancy (v) is made good by the volumes `y' and `z'. ; v  y z Let `A' be the area of the water-plane before bilging, and let `a' be the area of the bilged compartment. Then: y  z  Ax Àax or v  xA Àa Increase in draft  x  v A À a i.e. Increase in draft  Volume of lost buoyancy Area of intact waterplane Note. Since the distribution of weight within the vessel has not been altered the KG after bilging will be the same as the KG before bilging. Example 1 A box-shaped vessel is 50 metres long and is ¯oating on an even keel at 4 metres draft. A compartment amidships is 10 metres long and is empty. Find the increase in draft if this compartment is bilged. See Fig. 21.1(c). x  v A À a  l  B  B L À lB let B  Breadth of the vessel then x  10  B  4 50  B À 10 ÂB  40B 40B Increase in draft  1 metre Bilging and permeability 205 Fig. 21.1(b) Example 2 A box-shaped vessel is 150 metres long Â24 metres wide Â12 metres deep and is ¯oating on an even keel at 5 metres draft. GM  0.9 metres. A compartment amidships is 20 metres long and is empty. Find the new GM if this compartment is bilged. Old KB  1 2 Old draft  2X5m Old BM  B 2 a12d  24  24 12  5  9X6m Old KB 2X5m Old KM  12X1m Old GM À0X9m KG  11X2m This KG will not change after bilging has taken place. x  v A À a  20  24  5 150  24 À 20 Â24  2400 130  24 Increase in draft  0X77 m Old draft  5X00 m New draft  5X77 m  say draft d 2 206 Ship Stability for Masters and Mates Fig. 21.1(c) New KB  1 2 New draft  d 2 2  2X89 m New BM  B 2 a12d 2  24  24 12  5X77  8X32 m  New KB  2X89 m New KM  11X21 m À As before, KG  11X20 m Ans. New GM  0X01 m This is ve but dangerously low in value!! Permeability k Permeability is the amount of water that can enter a compartment or tank after it has been bilged. When an empty compartment is bilged, the whole of the bouyancy provided by that compartment is lost. Typical values for permeability m are as follows: Empty compartment m  100% Engine room m  80% to 85% Grain ®lled cargo hold m  60% to 65% Coal ®lled compartment m  36% approx Filled water ballast tank (when ship is in salt water) m  0% Consequently, the higher the value of the permeability for a bilged compartment, the greater will be a ship's loss of bouyancy when the ship is bilged. The permeability of a compartment can be found from the formula: m  Permeability  Broken Stowage Stowage Factor  100 per cent The broken stowage to be used in this formula is the broken stowage per tonne of stow. When a bilged compartment contains cargo, the formula for ®nding the increase in draft must be amended to allow for the permeability. If `m' represents the permeability, expressed as a fraction, then the volume of lost buoyancy will be `mn' and the area of the intact waterplane will be `A À mn' square metres. The formula then reads: x  mn A À ma Example 3 A box-shaped vessel is 64 metres long and is ¯oating on an even keel at 3 Bilging and permeability 207 metres draft. A compartment amidships is 12 m long and contains cargo having a permeability of 25 per cent. Calculate the increase in the draft if this compartment be bilged. x  mn A À ma  1 4  12 ÂB  3 64  B À 1 4  12 ÂB  9B 61B Ans. Increase in draft  0X15 m Example 4 A box-shaped vessel 150 m  20 m  12 m is ¯oating on an even keel at 5 metres draft. A compartment amidships is 15 metres long and contains timber of relative density 0.8, and stowage factor 1.5 cubic metres per tonne. Calculate the new draft if this compartment is now bilged. The permeability `m' must ®rst be found by using the formula given above. i.e. Permeability  BS SF  100 per cent  ` m' The stowage factor is given in the question. The broken stowage per tonne of stow is now found by subtracting the space which would be occupied by one tonne of solid timber from that actually occupied by one tonne of timber in the hold. One tonne of fre sh water occupies one cubic metre and the relative density of the timber is 0.8. ; Space occupied by one tonne of solid timber  1 0X8  1X25 cubic metres Stowage Factor  1X50 cubic metres ; Broken Stowage  0X25 cubic metres Permeability `m'  BS SF  100 per cent  0X25 1X50  100 per cent  100a6 per cent ; `m'  1a6 or 16.67 per cent Increase in draft  x  mn A À ma  1a6  15  20 Â5 150  20 À 1a6 Â15  20  250a2950  0X085 m 208 Ship Stability for Masters and Mates [...]... of 4 .5 metres Find the new drafts if a forward compartment 5 metres long is bilged (a) First let the vessel sink bodily Fig 21.4 w ˆ x  B  d1  1X0 25 tonnes ˆ 5  10  4X5  1X0 25 w ˆ 230X63 tonnes WPA L2  B ˆ 97X56 97X56 70  10 ˆ 97X56 TPC ˆ 7X1 75 TPC ˆ 212 Ship Stability for Masters and Mates Increase in draft ˆ w/TPC ˆ 230.63 /7. 1 75 ˆ 32X14 cm ˆ 0X321 m ‡ Old draft ˆ New mean draft ˆ 4X500 m... d2 (b) Now consider the trim W ˆ L  B  d1  1X0 25 tonnes BML ˆ IL aV ˆ W ˆ 3 459 tonnes BL3 2 12V ˆ ˆ 75  10  4X5  1X0 25 10  70 3 12  75  10  4X5 BML ˆ 84X7 metres W  BML 100L 3 459  84X7 ˆ 100  75 MCTC 9 ˆ 39X 05 tonnes m per cm Change of trim ˆ Moment changing trim MCTC where dˆ LBP 75 ˆ ˆ 37X5 m ˆ lever from new LCF 2 2 230X6  37X5 ˆ 39X 05 ˆ 221X4 cm by the head After bilging, LCF has moved... 3X53 ˆ S1 Dynamical stability 221 Fig 22.3 h ˆ 10 10 radians ˆ common interval CI hˆ 57 X3 1 The area under the stability curve ˆ  CI  S1 3 1 10  3X53 ˆ  3 57 X3 ˆ 0X2 053 metre-radians Dynamical stability ˆ W  Area under the stability curve ˆ 50 00  0X2 053 Ans Dynamical stability ˆ 1026 .5 metre tonnes Example 2 A box-shaped vessel 45 m  10 m  6 m is ¯oating in salt water at a draft of 4 m F and. .. 123/20 .5 ˆ 6 cm TPC ˆ 20X5 tonnes Increase in draft ˆ 0X06 m Fig 21 .5 BML ˆ IL BL 3 B  L3 L2 ˆ ˆ ˆ V 12V 12  LÂBÂd 12  d1 BML ˆ W ˆ L  B  d1  1X0 25 100  100 12  6 ˆ 100  20  6  1X0 25 W ˆ 12 300 tonnes BML ˆ 139 metres MCTC 9 W  BML 12 300  139 ˆ 100  L 100  100 214 Ship Stability for Masters and Mates MCTC ˆ 171 tonnes m per cm Trimming moment ˆ W  B1 B2 ˆwÂd Trimming moment ˆ 123  45 tonnes... compartment 4 metres long and situated amidships is bilged (a) Original condition before bilging Find the KG I V LB 3 B2 ˆ ˆ 12V 12  d1 8Â8 ˆ 12  3 BM ˆ 1X78 m KB ˆ ‡1X50 m KM ˆ 3X28 m GM ˆ À1X00 m KG ˆ 2X28 m d1 2 ˆ 1X5 metres KB ˆ BM ˆ Bilging and permeability 2 15 (b) Vessel's condition after bilging Find the New Draft Lost buoyancy ˆ 4  8  3  1X0 25 tonnes WPA 36  8 ˆ 97X56 97X56 Lost buoyancy Increase... drafts 216 Ship Stability for Masters and Mates Exercise 21 1 2 3 4 5 6 7 8 Bilging amidships compartments (a) De®ne permeability, `m' (b) A box-shaped vessel 100 m long, 15 m beam ¯oating in salt water, at a mean draft of 5 m, has an amidships compartment 10 m long which is loaded with a general cargo Find the new mean draft if this compartment is bilged, assuming the permeability to be 25 per cent... permeability 25 per cent Find the new draft if this compartment is bilged 25  20  10  4  1X0 25 tonnes Buoyancy lost ˆ 100 ˆ 2 05 tonnes Extra mass of water displaced ˆ TPC  X tonnes 210 Ship Stability for Masters and Mates ; X ˆ w/TPC ˆ 205a20 Increase in draft ˆ 10X 25 cm ˆ 0X10 25 m plus the old draft ˆ 6X0000 m Ans New draft ˆ 6X10 25 m Note: The lower the permeability is the less will be the changes... the common interval must be expressed in radians 57 X3 ˆ 1 radian 1 1 ˆ radians 57 X3 or x x ˆ radians 57 X3 Therefore to convert degrees to radians simply divide the number of degrees by 57 .3 Example 1 A ship of 50 00 tonnes displacement has righting levers as follows: Angle of heel GZ (metres) 10 0.21 20 0.33 30 0.40 40 0.43 Calculate the dynamical stability to 40 degrees heel GZ SM 0 0.21 0.33... the curve ˆ 18 45  0X0389 Ans Dynamical stability ˆ 71 X 77 m tonnes Dynamical stability 223 Exercise 22 1 2 3 4 5 A ship of 10 000 tonnes displacement has righting levers as follows: 20 30 40 Heel 10 GZ (m) 0.09 0.21 0.30 0.33 Calculate the dynamical stability to 40 degrees heel When inclined, a ship of 8000 tonnes displacement has the following righting levers: 30 45 60 Heel 15 GZ (m) 0.20... 0. 252 0X668 ˆ S1 222 Ship Stability for Masters and Mates Fig 22.4 At 10 heel: GZ ˆ GM  sin y ˆ 0X6  sin 10 GZ ˆ 0X104 m At 20 heel: GZ ˆ …GM ‡ 1 BM tan 2 y† sin y 2 ˆ …0X6 ‡ 1  2X08  tan 2 20 † sin 20 2 ˆ …0X6 ‡ 0X138† sin 20 ˆ 0X738 sin 20 GZ ˆ 0X 252 m 1 Area under the curve ˆ  CI  S1 3 1 10  0X668 ˆ  3 57 X3 Area under the curve ˆ 0X0389 metre radians Dynamical stability ˆ W  Area . d 1  1X0 25 tonnes  75 Â10  4X5  1X0 25 W  3 459 tonnes BM L  I L aV  BL 3 2 12V  10  70 3 12  75  10 Â4X5 BM L  84X7 metres MCTC 9 W  BM L 100L  3 459  84X7 100  75  39 X 05 tonnes. tonnes TPC  WPA 97X56  L 2  B 97X56  70  10 97X56 TPC  7X1 75 Bilging and permeability 211 Fig. 21.4 Increase in draft  w/TPC  230.63 /7. 1 75  32X14 cm  0X321 m  Old draft  4X500 m  draft. 0X 77 m Old draft  5X00 m New draft  5X 77 m  say draft d 2 206 Ship Stability for Masters and Mates Fig. 21.1(c) New KB  1 2 New draft  d 2 2  2X89 m New BM  B 2 a12d 2  24  24 12  5X 77 