Source: https://www.scribd.com/document/112009795/CouserIJME04OnTheEffectOfTankFreeSurfacesOnVesselStaticStability
Timestamp: 2019-04-26 16:36:19+00:00

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1 CONDITION A The reduction of GZ. δVCG = Hence FSM s ρs (12) We shall now ﬁnd the shift of the ﬂuid CG in the ship coordinate system. the horizontal shift of the tank CG in the earth-ﬁxed axis system is then easily found by simply rotating the coordinate system. bh 2 bh T 2 T T bh 2 bh V 2 V = −A × Tem + A × Timm = = −b2 tan φ −b b2 tan φ b × + × 8 3 8 3 2 b tan φ 6h (9) Condition A inverted π − tan −1 = −A × Vem + A × Vimm = −b2 tan φ × 8 b2 tan φ + × 8 b2 tan2 φ = 12h h b tan φ − 2 6 h b tan φ + 2 6 (10) h b ≤φ≤π (2) Condition B (increased angle of heel) tan−1 h b ≤ φ ≤ π − tan−1 h b V (3) In the derivation presented below. which is equal to the horizontal shift of the vessel CG in the earth-ﬁxed axis system. 2. The area of the immersed and emerged triangles is given by Equation 4 A= 1 b b tan φ b2 tan φ × × = 2 2 2 8 (4) An equivalent free surface moment. the shift of the tank centre of gravity (CG) is calculated in the ship axis system. the equivalent FSM can be calculated from Equation 14: δGZ = FSM = δGZ s ρs sin φ (14) Vimm = h 1 b tan φ h b tan φ + × = + 2 3 2 2 6 (6) Substituting δGZ from Equation 11 into Equation 14 relates the horizontal shift of tank CG (δy) to the equivalent FSM – Equation 15. can be calculated by considering the effect of the tank CG shift to be equivalent to a virtual rise in CG (this is the traditional approach). Timm b 2 b = × = 3 2 3 (5) FSM sin φ (13) s ρs Rearranging Equation 13. FSM. δy. FSM = δy t ρt sin φ (15) Similarly. the centroid of the emerged triangle is given by Equations 7 and 8 respectively. Condition A is true for the heel angle range given by Equation 1. In this case. δys s ρs = δy = δy t ρt t ρt s ρs δys = δGZ (11) Consider the situation as shown in Figure 2. δGZ = δVCG sin φ and the virtual rise in the CG is given by Equation 12. The transverse and vertical components of the centroid of the immersed triangle. δGZ. where the ﬂuid in the tank has moved due to the vessel heeling to an angle φ. can be calculated from the horizontal shift of the tank CG in the earth-ﬁxed axis system.ure 1 and the remaining condition is the inverted case of Condition A. measured from the upright tank CG (in the ship axis system). δys . are given by Equations 5 and 6 respectively. by taking moments about the original vessel CG – as shown in Equation 11. in the ship coordinate system. Condition A Inverted is valid for the heel angle range given by Equation 2 and Condition B is true for the middle heel angle range given by Equation 3. can be found by taking moments about the original upright tank CG – Equations 9 and 10. Tem 2 −b −b = × = 3 2 3 (7) Now the horizontal shift of the tank CG in the earth-ﬁxed axis system is given by Equation 16. δy = T cos φ + V sin φ b2 tan φ b2 tan2 φ = cos φ + sin φ 6h 12h Vem h 1 b tan φ h b tan φ = −× = − 2 3 2 2 6 (8) δy (16) c Royal Institution of Naval Architects 2004 . Condition A (moderate angle of heel) 0 ≤ φ ≤ tan−1 h b (1) The CG of the ﬂuid in the heeled tank.
Equation 19. Ttri imm = 2 h h × = 3 2 tan φ 3 tan φ h 1 h 5h + × = 4 3 2 12 (22) where δyCond. given by Equation 19: δy δy = = h sin φ − δyCond. Considering the triangles ﬁrst: the area of the immersed and emerged triangles is given by Equation 21. it may be shown that the horizontal shift in tank CG. Atri = h h2 1 h × × = 2 2 2 tan φ 8 tan φ (21) In this condition. into Equation 15 yields the equivalent FSM – Equation 17 which simpliﬁes to Equation 18. FSM = t ρt b 2 sin φ (17) 2.A is the earth-ﬁxed. where the heel angle has been increased.A 2 b2 tan φ b2 tan2 φ h sin φ − cos φ − sin φ 2 6h 12h (19) The transverse and vertical components of the centroid of immersed triangle are given by Equations 22 and 23 respectively. yields the equivalent FSM – Equation 20. horizontal shift for Condition A – Equation 18.2 CONDITION A INVERTED Now we shall consider the situation as shown in Figure 3. FSM = b tan φ b tan φ cos φ + sin φ 6h 12h 2 2 2 t ρt Substituting for δy.3 6h 3h2 tan2 φ −1− b2 2 (20) CONDITION B FSM = t ρt b 2 6h 1+ tan2 φ 2 (18) 2.Condition A Condition B Figure 1: Two conditions arise depending on the heel angle Figure 2: Shift of ﬂuid for 0 ≤ φ ≤ tan−1 (h/b) Thus substituting for δy. The immersed and emerged areas can be divided into a triangle and a rectangle. Equation 16. in the earth-ﬁxed axis system is. Vtri imm = (23) c Royal Institution of Naval Architects 2004 . into Equation 15 and simplifying.
the centroid of the emerged triangle is given by Equations 24 and 25 respectively. the centroid of the emerged rectangle is given by Equations 29 and 30 respectively. in the ship coordinate system. can be found by taking moments about the upright CG – Equations 31 and 32 Vtri em h −h 2 h = + × = 4 3 2 12 Now consider the rectangles: the area of the immersed and emerged rectangles is given by Equation 26. Trec em = = −h 1 h − b− 2 tan φ 4 tan φ −1 h b+ 4 tan φ Vrec em = 0 V (29) (30) Again the horizontal shift of the tank CG in the earth-ﬁxed c Royal Institution of Naval Architects 2004 . Arec = h 1 h × × b− 2 2 tan φ = 1 4 bh − h2 tan φ (26) The transverse and vertical components of the centroid of immersed rectangle are given by Equations 27 and 28 respectively. Trec imm = = h 1 h + b− 2 tan φ 4 tan φ 1 h b+ 4 tan φ Vrec imm = h 2 T (27) V (28) V bh 2 bh 2 = −Atri × Vtri em − Arec × Vrec em +Atri × Vtri imm + Arec × Vrec imm h2 h 1 h2 = − × − bh − 8 tan φ 12 4 tan φ 2 5h 1 h2 h × + bh − + 8 tan φ 12 4 tan φ 2 h h = − 4 6b tan φ ×0 × h 2 (32) Similarly. Ttri em = 2 −h −h × = 3 2 tan φ 3 tan φ (24) T (25) T bh 2 bh 2 = −Atri × Ttri em − Arec × Trec em +Atri × Ttri imm + Arec × Trec imm h2 −h = − × 8 tan φ 3 tan φ 1 h2 −1 h − bh − × b+ 4 tan φ 4 tan φ h2 h + × 8 tan φ 3 tan φ 1 h2 1 h + bh − × b+ 4 tan φ 4 tan φ h2 b − (31) = 4 12b tan2 φ The CG of the ﬂuid in the heeled tank.Figure 3: Shift of ﬂuid at tan−1 (h/b) < φ < 2π − tan−1 (h/b) Similarly.
one having a square cross-section (b/h = 1) and the other having a rectangular cross-section (b/h = 1. expressed as percentage of FSM calculated from the upright tank waterplane. the Hydromax ﬂuid simulation. it is relatively easy to compute the actual position of the ﬂuid in the tank as the vessel heels and trims. Figure 4 shows a comparison of the theoretical FSM derived in this paper with The traditional FSM based on the upright tank geometry is given by Equation 36 FSMupright = It ρt (36) c Royal Institution of Naval Architects 2004 . with any level of ﬂuid. 3. Although similar equations could be derived for other crosssection shapes. this is rather a cumbersome approach.Figure 4: Comparison of theoretical effective FSM with Hydromax ﬂuid simulation. that are not necessarily prismatic. Using this (or similar software) it is possible to investigate arbitrarily shaped tanks. expressed as percentage of FSM calculated from the upright tank waterplane. With a computer model.1 EFFECT OF VOLUME OF FLUID IN TANK FSM = t ρt b h2 h + − 4 4 tan φ 6b tan φ 1+ 1 2 tan2 φ (35) 3 STATIC SIMULATION OF FLUID The equations derived above hold only for prismatic tanks with rectangular cross-section. Hydromax has then been used to investigate the effect of tank ﬂuid level on the FSM in the following section. (33) and hence the equivalent FSM is given by Equation 34. expressed as percentage of FSM calculated from the upright tank waterplane. axis system is given by Equation 33: δy δy = T cos φ + V sin φ h2 b cos φ − = 4 12b tan2 φ h h2 + − sin φ 4 6b tan φ Figure 5: Effect of tank ﬂuid volume on effective FSM. This calculation method is available in Formation Design Systems’ Hydromax stability software. which simpliﬁes to Equation 35 FSM FSM δy t ρt sin φ b h2 = − cos φ 4 12b tan2 φ h h2 t ρt + − sin φ (34) 4 6b tan φ sin φ = Figure 6: Effect of tank breadth:height ratio (b/h) on effective FSM (broad tanks).722). Results are expressed as a percentage of the FSM as computed from the upright waterplane (It ρt ) and as expected tend to 100% of the upright FSM at zero heel. for a 1m×1m×1m cubic tank. half-ﬁlled with ﬂuid. The results are presented for two prismatic tanks.
where large angles of vessel heel are expected. It can be seen that Equation 36 is independent of the volume of ﬂuid in the tank. These results were computed for a 1m×1m×1m cubic tank.3 – Values for coefﬁcient k for calculating free surface correction1 k= sin φ 12 1+ tan2 φ 2 b/h The effect of tank breadth:height ratio (b/h) on FSM has been investigated and the results presented in Figures 6 and 7. have a reduced FSM in the upright condition. this has been done and the results are shown in Figure 8. b/h ≥ 1.0. In this case the broad tanks have signiﬁcantly higher FSM upright. 3.) It can be seen that the maximum effect occurs when the tank is half-full (50%). but the FSM diminishes quite rapidly as heel angle increases and becomes negative once the vessel is heeled above approximately 100 ◦ . especially at large heel angles. The intermediate values are determined by interpolation. Hence at heel angles above about 80 ◦ . Figure 5 shows the actual effect of the ﬂuid movement in the tank as an effective FSM for different levels of ﬂuid in the tank. for the upright tank and ρt is the density of the ﬂuid in the tank. on the other hand. but the FSM increases with heel angle. expressed as percentage of FSM calculated from the upright tank waterplane. in m tonnes v is the total tank capacity. the case could be made for broad tanks. These results are for a half-full tank with rectangular cross-section. It can be seen that there is a dramatic effect for tall. the IMO code on intact stability.2 EFFECT OF TANK BREADTH:HEIGHT RATIO In §3. the broad tanks have reduced FSM compared with the narrow tanks (for the same capacity). in m3 b is the tank maximum breadth. (The same effect is observed if the tank is x% full or 100 − x% full. Table 3. b/h > 3. in reality this is not the case and the effective FSM does depend on the volume of ﬂuid in the tank. The narrow tanks. the relevant section of the code is reproduced below: 3.0.. the effective FSM reduces rapidly compared with the upright FSM at heel angles above 30 ◦ .3. describes how ﬂuid free surfaces in slack tanks should be taken into account.3. tanks are shown in Figure 6.ferent breadth:height ratio but constant total capacity. in m ρ is the mass density of liquid in the tank. The values are expressed as a percentage of the FSM calculated from the upright waterplane.8 The values of Mfs for each tank may be derived from the formula: √ Mfs = vbρhk δ where: Mfs is the free surface moment at any inclination. which despite reducing initial stability would be less detrimental as heel angle increases.3 according to the ratio b/h. in tonnes/m3 δ is equal to v/blh (the tank block coefﬁcient) h is the tank maximum height. where It is the transverse second moment of area of tank waterplane. not the entire table. c Royal Institution of Naval Architects 2004 . However.3 Effect of free surfaces of liquids in tanks . The results for tanks with. narrow tanks (Figure7). However. about its centroid. For the broader tanks. perhaps a more useful comparison can be made by examining the actual effective FSM for tanks of dif- where cot φ ≤ b/h cos φ tan φ 1+ 8 b/h cos φ cot2 φ − 1+ 2 2 12 (b/h) k = where cot φ > b/h 1 Only the equations for k are reproduced here. 4 COMPARISON WITH IMO Figure 7: Effect of tank breadth:height ratio (b/h) on effective FSM (tall tanks).. This could be an important consideration for self-righting craft or craft that are expected to operate in severe conditions. In some situations. in m l is the tank maximum length. 3. amended by . with a signiﬁcant jump at approximately 90 ◦ . in m k is the dimensionless coefﬁcient to be determined from table 3.3.
however. half-full of ﬂuid. Some interesting results were found for tall. the effective FSM was generally less than upright and for tanks between 20% and 80% full.3 (Equations 18 and 35) shows that the equations are similar except for the sin φ term. This implies that Mfs is not in fact the free surface moment but the reduction in ship righting moment due to the tank ﬂuid free surface. It is worth pointing out that if the equations in [1.0. An investigation into the effect of tank breadth:height ratio (b/h) has also been made. A comparison with the IMO treatment of ﬂuid free surfaces in slack tanks[1.1 and 2. it is recommended that calculation of a vessel’s large angle. rectangular tank.8]. §3. the FSM varies dramatically from the FSM in the upright condition. §3. These equations have been used to verify the static simulation of c Royal Institution of Naval Architects 2004 .3. narrow tanks b/h < 1. the effective FSM reduces rapidly compared with the upright FSM at heel angles above 30 ◦ . §3. b/h ≥ 3. the IMO equation for Mfs (quoted above) may be rearranged as follows: Condition A: cot φ ≥ b/h Mfs = vt b2 ρt 6h 1+ tan2 φ 2 sin φ (37) ﬂuid movement in Formation Design Systems’ hydrostatics program Hydromax.3] and [2. static stability should accurately account for the effects of ﬂuid free surfaces in tanks by modelling the actual position of the ﬂuid in the tank at any arbitrary angle of heel and trim rather than approximating this effect Condition B: cot φ < b/h Mfs = vt ρt b h + 4 4 tan φ h2 1 − 1+ 6b tan φ 2 tan2 φ sin φ (38) Comparison of these equations (Equations 37 and 38) with the equations derived in §2.8] are used verbatim to calculate the FSM. even for the relatively simple case of a half-ﬁlled. v = 2 t ).0.3. these documents state that: “Mfs is the free surface moment at any inclination. This indicated that there is a typographical error in [1. 2] has been made. Now assuming a rectangular tank. and that the tank is half-full (i.0. For these reasons and given the reality that virtually all vessel hydrostatics are calculated by computer. It has been shown that. 100kg total capacity. 5 CONCLUSIONS The effect of ﬂuid movement in slack tanks on static stability has been derived from ﬁrst principles for tanks with rectangular cross-section. the effective FSM was found to increase dramatically as the heel angle passes through 90 ◦ . below 20% and above 80% full. Under many conditions. For broad tanks. The 20% and 80% full conditions were reasonably close to the FSM calculated from the upright tank free surface. the effective FSM was generally greater than upright.3. It should also be noted that Equation 38 is not valid for all angles where cot φ < b/h. Numerical experiments with Hydromax indicated that the half-full condition was the worst condition for a cubic tank. the FSM for the vessel in the upright condition will be zero resulting in no reduction to GM. for which δ = 1.Figure 8: Effect of tank breadth:height ratio (b/h) on effective FSM (kg m). but is limited by the range as described by Equation 3. rectangular cross-section tank. examination of the equations provided in these documents indicate that Mfs is in fact the reduction in GZ due to the free surface of the tank in question. for a half-ﬁlled. the FSM is far from constant as the vessel heels.e. in m tonnes”.
7. 1999. Resolution A.com. IMO.from the tank free surface in the upright condition.formsys.3.75(69).  Hydromax manual. Resolution MSC.749(18). c Royal Institution of Naval Architects 2004 . Formation Design Systems.  Amendments to the Code on Intact Stability for all types of ships covered by IMO instruments. It is acknowledged that this practice is accepted by some authorities including IMO[2. 2004. §3.1]. http://www. 1995. IMO.2. References  Code on Intact Stability for all types of ships covered by IMO instruments.

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