Literature review

Modelling and Simulating Ballast Tank

Blowing and Venting Operations in

Manned Submarines ⋆

Roberto Font ∗ Javier Garćıa ∗∗ Diana Ovalle ∗∗∗

∗ Departamento de Matemática Aplicada y Estad́ıstica, ETSI Industriales, Universidad Politécnica de Cartagena, 30202

Cartagena,Spain (e-mail: [email protected]). ∗∗ Departamento de Ingenieŕıa, Navantia S.A., 30205 Cartagena,

Spain. (e-mail: [email protected]) ∗∗∗ Departamento de Matemática Aplicada y Estad́ıstica, ETSI Industriales, Universidad Politécnica de Cartagena, 30202

Cartagena,Spain (e-mail: [email protected])

Abstract: The aim of this work is to obtain mathematical models for both the blowing and venting of ballast tanks, as well as the effect of these operations over the submarine behavior. Although tanks are usually blown only on emergency situations, blowing and venting can potentially be used as a suitable control mechanism for specific manoeuvres, not only in submarines but in any marine system using ballast tanks, like offshore facilities or pontoon docks. We propose mathematical models for blowing and venting that are coupled with a 6 degree– of–freedom model for the equations of motion. All resulting equations are solved altogether thus obtaining a solution for the submarine behavior taking into account the effect of the mass change. In order to explore the potential of these operations as a control mechanism, results of a computer simulation are presented in which the submarine performs a depth change manoeuvre by only blowing and venting its ballast tanks without any use of diving planes.

Keywords: Mathematical models, submarine manoeuvrability, computer simulation.

1. INTRODUCTION

Ballast tanks are distributed along the length of the sub- marine. When filled with water, they contribute with the submarine mass, allowing it to submerge. In this condition, the weight of the submarine is nominally equal to its buoyancy. During an unexpected event or emergence, they act like a dispositive for emerging to the surface: air is blown into the ballast tanks from very high pressure bot- tles, expelling the water out of the tanks. The submarine loss weight, its buoyancy is higher, and it can emerge quicker. In the last years, several works have addressed these emergency rising manoeuvres (see Watt (2001, 2007); Bettle et al. (2009) and the references given there). To fill the tanks with water again, air is vented out of the ballast tanks. A valve located on the top of the tank is opened, air escapes outside, and water flows back into the tanks.

Under certain circumstances, like gathering intelligence missions or special operations, submarines need to perform manoeuvres with very specific requirements. Consider, for example, the case of special forces entering in a submerged submarine: the submarine needs to stay at a fixed depth without any propulsion. In a gathering intelligence mis-

⋆ The first author was supported by an I+D+i grant from Univer- sidad Politécnica de Cartagena. The third author was supported

by projects 2113/07MAE between Navantia S.A. and UPCT, and

08720/PI/08 from Fundación Séneca (Agencia de Ciencia y Tec-

noloǵıa de la Región de Murcia. II PCTRM 2007-10).

sion, the submarine may need to manoeuvre in shallow littoral waters where an accurate control over the ver- tical movement is needed. In cases like the cited above, submarines often perform small blowing and venting op- erations, not because of an emergency, but to slightly modify the buoyancy of the vehicle. This way, blowing and venting become a complementary tool for manoeuvring. This is a concept we would like to expand in this and future works. These manoeuvres are currently performed based exclusively on the operator experience and, due to the high degree of accuracy required, would enormously benefit from the implementation of a control system.

In a previous work, Garćıa et al. (2009), an algorithm capable of providing a set of admissible and optimal trajectories for any manoeuvre (using propeller and diving planes) has been proposed. Our purpose is to expand this algorithm by including blowing and venting as additional control mechanisms. To this end, first, accurate models for these processes are needed. The aim of this work is to provide all the required mathematical models and use them to perform several numerical simulations.

The blowing model we propose is based on the one sketched by Byström (2004). It has several advantages over the one proposed by Watt (2007). Since this last one was developed to study the rising roll instability, it may have some limitations when used for other purposes. We will further discuss these advantages in Section 3.1.

8th IFAC Conference on Control Applications in Marine Systems Rostock-Warnemünde, Germany September 15-17, 2010

978-3-902661-88-3/10/$20.00 © 2010 IFAC 67 10.3182/20100915-3-DE-3008.00029

Up to the knowledge of the authors, venting has not been addressed before.

The rest of the paper is organized as follows. In Section 2 the required mathematical models are proposed. Section 3 is devoted to the results of several computer simulations. Conclusions are given in Section 4.

2. MATHEMATICAL MODELS

The aim of this section is to provide mathematical models for both blowing and venting processes as well as their influence over the behavior of the submarine. Our math- ematical model for the equations of motion is based on Gertler and Hagen (1967) six degree of freedom (DOF) submarine equations of motion, which were revised by Feldman (1979), adapted to the particular characteristics of a prototype developed by the company Navantia S.A. Cartagena Shipyard (Spain). We refer the reader to Garćıa et al. (2009) for a complete description of the resulting model.

2.1 Variable mass model

The equations of motion assume the mass of the submarine to be constant. Since this will not be true while blowing or venting, it is necessary to identify which terms, for- merly constant, will become time dependent due to its dependance with mass. We will need to write the following properties as a function of the amount of water in the tanks:

• Mass (m). • Weight (W). • Moments and products of inertia (Ix, Iy, Iz, Ixy, . . . ). • Location of the center of gravity (xG, yG, zG).

Let us assume there are N ballast tanks with geometrical centers located at points (xbi, ybi, zbi) (where the subscript i denotes the i–th ballast tank). The mass of the submarine is:

m(t) = m0 −

N ∑

i=1

∆mi(t) (1)

where m0 is the initial mass of the submarine (with all tanks filled with water) and ∆mi is the mass loss in the i– th tank. It is 0 when the tank is completely filled with water, and reach its maximum value when it empties. The way to obtain these ∆mi during both blowing and venting will be covered in detail in the next sections, for the moment, let us assume they are known.

In the same manner we can see that the weight of the submarine is:

W(t) = g

(

m0 −

N ∑

i=1

∆mi(t)

)

(2)

with g the acceleration due to gravity.

To find expressions for the location of the center of gravity and moments and products of inertia we will work under the assumption that the mass loss in each tank occurs at a point, (xbi, ybi, zmli(t)), where zmli(t) is the height at which mass loss happens for each tank. It varies from the top of the tank, when it is completely filled, to its geometric center, zbi, when it is completely empty. This

variation is assumed to be linear. Let Ix0, Iy0, Iz0, Ixy0, Ixz0, Iyz0 and xG0, yG0, zG0 be respectively the initial (all tanks completely filled) moments and products of inertia and coordinates of the center of gravity. Then the moments and products of inertia are:

Ix(t) =

∫

(

y2 + z2 )

dm = Ix0 −

N ∑

i=1

(

y2bi + zmli(t) 2 )

∆mi(t) (3)

and the coordinates of the center of gravity:

xG(t) = 1

m0 − ∑

N

i=1 ∆mi(t)

(

m0xG0 −

N ∑

i=1

xbi∆mi(t)

)

(4)

The rest of coordinates and moments and products of inertia are calculated analogously. Substituting (1)−(4) into the equations of motion completes our variable mass model.

2.2 Blowing model

The blowing model is divided into three parts: air flow from pressure bottle, water flow out from ballast tank and evolution of pressure in ballast tank.

Air flow from pressure bottle: The air in the bottle is blown into the tank through a valve that acts as a nozzle. Pressure losses and heat transfer in the tube that connects the bottle and the tank are neglected. Under the above conditions, we need to study the one dimensional steady flow of an ideal compressible gas. This can be found in any classic text on fluid mechanics (see for example Crowe et al. (2004)). At the beginning of blowing the flow will be supersonic due to the high pressure difference between bottle and tank. As air flows out of the bottle, this difference decreases and the flow will become subsonic at a certain time. This transition will happen when pressures in bottle, pF , and tank, pB, are such that:

(

pF

pB

)

=

(

γ + 1

2

)

γ

γ−1

where γ is the isentropic constant. Expressions for the mass flow rate from the bottle can be found for both cases (see Crowe et al. (2004)):

Supersonic flow: In this case, it is easy to calculate the mass flow rate at the nozzle throat, where the Mach number is unity:

ṁF = ρ ∗

v ∗

At = ρ0a0 ρ∗

ρ0

a∗

a0 At (5)

where At is the area in the nozzle throat, a is the speed of sound, v is the air velocity, ρ is the density of air, the asterisk signifies conditions wherein the Mach number is equal to unity and the subscript 0 denotes stagnant conditions. These two conditions are related by:

(

p0

p∗

)

γ−1

γ

=

(

ρ0

ρ∗

)γ−1

= (

a0

a∗

)2

= T0

T ∗ =

γ + 1

2

where T is the temperature.

Assuming that in this particular case stagnant conditions are the bottle conditions, using the above relations, the

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ideal gas law ρ0 = p0

RgT0 and a0 =

√

γRgT0, the mass flow

rate is:

ṁF (t) = At · pF (t) √

RgTF (t)

√

√

√

√

γ

(

2

γ + 1

)

γ+1

γ−1

(6)

where Rg is the gas constant for air and TF is the temperature in the bottle.

Subsonic flow:

In this case, the mass flow rate is:

ṁF = ρeMeaeA = ρ0a0 ρe

ρ0

ae

a0 MeA (7)

where A is the exit area, M is the Mach number and the subscript e denotes conditions at the exit.

The stagnant and static conditions are related by:

T0

T =

(

ρ0

ρ

)γ−1

=

(

p0

p

)

γ−1

γ

= 1 + γ − 1

2 M

2 .

Particularizing for p0 = pF and p = pe = pB, the Mach number at the exit is given by:

Me =

√

√

√

√

2

γ − 1

[

(

pF

pB

)

γ−1

γ

− 1

]

(8)

Substituting (8) into (7) and proceeding as in the super- sonic case, the mass flow rate is:

ṁF (t) = A · pF (t) √

RgTF (t)

√

√

√

√

2γ

γ − 1

[

(

pB(t)

pF (t)

) 2

γ

−

(

pB(t)

pF (t)

)

γ+1 γ

]

(9)

As the pressure in the bottle drops the temperature will increase. If the heat transmission is neglected, however, the process can be considered to be adiabatic. Let mF (t), pF (t) and TF (t) be respectively the air mass, pressure and temperature in the bottle, mF 0, pF 0 be the initial mass and pressure in the bottle and VF the bottle volume. Under the above assumptions the momentary pressure and temperature will be:

pF (t) =

(

mF (t)

mF 0

)γ

· pF 0, TF (t) = pF (t) · VF RgmF (t)

(10)

Thus the final equations for the mass flow rate are: 





















































































ṁF (t) = A

C ·

(

γ

(

2

γ + 1

)

γ+1 γ−1 pF 0

m γ

F 0 VF

· (mF (t)) γ+1

)

1

2

if pF (t)

pB(t) ≥

(

γ + 1

2

) γ

γ−1

ṁF (t) = A ·

(

pF 0mF (t) γ+1

m γ

F 0 VF

2γ

γ − 1

)1

2

·

·





(

pB(t)

pF 0 (

mF (t) mF 0

)γ

)2

γ

−

(

pB(t)

pF 0 (

mF (t) mF 0

)γ

)

γ+1 γ





1

2

if

(

γ + 1

2

)

γ γ−1

≥

pF (t)

pB(t) ≥ 1

(11)

where C = A/At. Although an analytic solution exists for the first expression, it is not the case for subsonic flow.

Since the initial mass flow rate depends only on the initial conditions in the bottle, it can be considered as a constant with value:

ṁF (0) = A

C ·

(

γ

(

2

γ + 1

)

γ+1

γ−1 pF 0mF 0

VF

)

1

2

(12)

This initial mass flow rate has been measured for several blowing intensities. Let ṁF max be this measured maxi- mum mass flow rate. Instead of using the numerical value of area A when computing (11), we obtain it from (12) and the measured value for ṁF max:

A = C · ṁF max

(

(

2

γ + 1

)

− γ+1

γ−1 VF

γpF 0mF 0

)

1

2

(13)

By doing so, we ensure that the initial mass flow rate calculated coincides with the real measured value. This way, although pressure losses are not considered in the model, they are indirectly taken into account. We plan to expand the model by including both major and minor pressure losses in future work.

Water flow out from ballast tank: The difference between the tank and outside pressure will force the water to flow out from the tank trough the flood port located at the bottom. A detailed analysis of a draining tank filled with an ideal fluid can also be found in Crowe et al. (2004). Let ρ be the density of water, pB(t) the pressure in the ballast tank and pSEA the outside pressure. If losses at the flood port are neglected, then the Bernoulli equation applied at both sides of the port leads to a volume flow from the ballast tank:

qB = CnAhvh = CnAh

√

2(pB − pSEA)

ρ (14)

with:

pSEA = patm + ρg(z + zh − xb sin θ)

where patm is the atmospheric pressure, z is the vehicle depth, zh is the vertical distance between the origin of the body–fixed frame and the outlet hole, and the term xb sin θ models the tank height variation with the vehicle pitch. Thus, (z + zh − xb sin θ) is the depth at which this outlet hole is located. Ah is the outlet hole area and Cn is a coefficient that takes into account that, since the outlet hole is actually a grid, the effective area is smaller than Ah.

Pressure in ballast tank: The air is blown into the tank at a very high velocity, rapidly mixing with water. This promotes good heat transfer from the water to the expanding air and we can work under the assumption that the air will immediately adopt the temperature in the tank. The process can then be considered to be isothermal. Let TB, VB and mB be respectively the temperature, the volume and mass of air in the tank. The ideal gas law can be applied to the air in the tank: pBVB = mBRgTB.

Taking the derivative with respect to time in the above equation and considering the process isothermal (ṪB = 0):

ṗB(t) = ṁB(t)RgTB − V̇B(t)pB(t)

VB(t)

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69

The volume occupied by air increases (and in the same quantity) as the volume occupied by water decreases. Thus

the rate at which it changes, V̇B, will be the volume flow out of the ballast tank, qB, given by (14). Even when the tanks are completely filled with water, a small amount of air, mB0, will always be present. The mass of air in the tank will then be the sum of this initial mass of air, mB0, and the difference between the initial and the momentary mass of air in the bottle. Thus mB(t) = mB0+ (mF 0 − mF (t)), ṁB(t) = −ṁF (t), and the variation in tank pressure is then:

(mB0 + mF 0 − mF (t))ṗB(t) + pB(t)ṁF (t) =

= − CnAh

RgTB p 2 B(t)

√

2(pB(t) − pSEA(t))

ρ

(15)

Mass change while blowing: As discussed in Section 2.1, we need to find expressions for the mass loss at each tank. The volume of water that has left the tank is equal to the volume occupied by air except for the initial air volume in the tank, VB0, which depends on the initial mass of air in the tank, mB0, and the initial depth. The mass loss in the i − th tank during blowing is then:

∆mi(t) = ρ

(

(mB0 + mF 0 − mF i(t)) RgTB

pBi(t) − VB0

)

(16)

We now have two new variables, mF i and pBi, and a set of two new equations, (11) and (15), for each ballast tank. Since ∆mi(t) depends on mF i and pBi and it is widely present in the equations of motion, as discussed in Section 2.1, these equations and the equations of motion are highly coupled and need to be solved altogether.

2.3 Venting model

The venting model is similar to the one proposed for the blowing process and is divided in two parts: water flow into the tank and air flow out from the tank.

Water flow into the tank: We can model the water flow into the tank using an approach analogous to that used to model the water flow out from the tank. Bernoulli equation applied at both sides of the outlet hole leads to:

vh(t) =

√

2

ρ (pSEA(t) − pB(t) − ρgs(t)) (17)

where s(t) is the momentary height of the water column in the tank and the rest of the terms have the same meaning as in Section 2.2.

Let Atk be an averaged cross–section area for the tank (tanks usually does not have regular shapes) and vs be the velocity at which the water column in the tank raises. Then, we can use the continuity equation to relate vs with the velocity at the flood port given by (17):

vs(t) = vh(t) CnAh

Atk . Since vs is the velocity at which the

water column raises, it is immediate to see that vs(t) = ṡ(t). Using the above relations and (17), the equation for the water column height is:

ṡ(t) = CnAh

Atk

√

2

ρ (pSEA(t) − pB(t) − ρgs(t)) (18)

Air flow out from the tank: Equations for the air flow from the tank are the same obtained for the air flow from the bottle in Section 2.2. Since the pressure differences are now much lower, only subsonic flow will happen. Let pext be the pressure outside the vent valve pipe system and Av the section in the vent valve. The mass flow rate out from the tank is then:

ṁB(t) = Av · pB(t) √

RgTB(t)

√

√

√

√

2γ

γ − 1

[

(

pext(t)

pB(t)

) 2

γ

−

(

pext(t)

pB(t)

)

γ+1 γ

]

(19)

As discussed above for the case of the bottle, the process can be considered adiabatic. Thus the following relation will hold for the air pressure in ballast tank:

(

pB0

pB(t)

)

γ−1

γ

= TB0

TB(t)

where TB0 and pB0 are respectively the temperature and pressure in ballast tank before the starting of venting. TB(t) can be obtained from the ideal gas law and the volume occupied by air can be put as: VB(t) = Vtk − Atk · s(t), where Vtk is the total tank volume. Using the above relations the momentary tank pressure is:

pB(t) =

(

mB(t)RgTB0 Vtk − Atk · s(t)

)γ

·

(

1

pB0

)γ−1

(20)

Mass change while venting: The volume of water missing in the tank is in this case equal to the difference between the maximum volume available in the tank and the volume of water in the tank. The mass loss in the i–th tank during venting is then:

∆mi(t) = ρ(Vtki − VB0 − Atki · si(t)) (21)

3. RESULTS

The above models have been implemented in a Matlab code capable of simulating a wide variety of manoeuvres. Since the equations vary depending on which operation is being performed, the code has been designed to identify the situation and switch to the appropriate ODEs system. The system is then solved using the Matlab functions ODE45 and ODE15s. We refer to Shampine and Reichelt (1997) for detailed information about both solvers.

3.1 Model analysis

We have tested our model by comparing its results with those obtained with previous models. Simulations in Watt (2007), page 21, have been reproduced and the air volume in each ballast tank has been calculated using both our model and the one proposed there. For geometrical data required by our model but not used and thus not provided in Watt (2007), we have used the values for Navantia Submarine P–650, to which we are applying our models. It is easy to see, from figures 1 and 2, that results obtained by both models are extremely similar. However, Figure 3 highlights some of the advantages of our model. Starting at a depth of 100 m, ballast tanks are blown at t = 5 s. Ballast tank pressure is plotted for three different scenarios: dash/dot line shows tank pressure considered to be ocean pressure as in Watt (2007), solid line shows

CAMS 2010 Rostock-Warnemünde, Germany, Sept 15-17, 2010

70

Fig. 1. Comparison between air volume in ballast tanks ob- tained by our model (above) and the model proposed by Watt (2007) (below), while blowing at a constant depth of 50 m.

ballast tank pressure given by (15), and dashed line corresponds to ballast tank pressure given by (15) when the outlet hole area, Ah, is reduced to half its actual value.

It is easy to see, from these results, the importance of modelling the evolution of ballast tank pressure as a result of blowing process. The increase in ballast tank pressure can be high enough to break the tanks if not taken into account. Comparison between dashed and solid lines shows the importance of considering all geometrical parameters involved, specially during the preliminary state design, when all these parameters need to be adjusted to meet the design requirements.

Once the blowing and venting models were coupled with the 6–DOF model for the equations of motion, several emergency rising manoeuvres were simulated and the re- sults examined by experienced submariners. The agree- ment between the simulation results and the behavior of a real vehicle was found satisfactory.

Unfortunately, there are currently no unclassified experi- mental data available for comparison. However, the com- pany Navantia S.A. plans to perform sea tests for the upcoming S–80 class submarines that will provide us with all the experimental data required for model validation.

3.2 Depth change manoeuvre

In order to explore the potential of blowing and venting as a control mechanism, we have simulated a depth change manoeuvre in which the submarine goes from a depth of

Fig. 2. Comparison between air volume in ballast tanks obtained by our model (above) and the model pro- posed by Watt (2007) (below), while blowing at a depth given by z(t) = 100 − 1.5t.

Fig. 3. Ocean pressure (dash/dot line) and ballast tank pressure with actual Ah value (solid line) and half the actual value (dashed line).

400 to 50 meters without any use of diving planes. Of course, performing such a manoeuvre is not a standard practice in real operation, since the same result can be achieved faster by using diving planes. Details for this manoeuvre are:

• Initial depth is 400 meters. • The submarine starts at straight and level flight. This condition is maintained during one minute to let the system stabilize.

• At t = 60 s, 4 of the 7 main ballast tanks are blown. The blowing stops when the pressure in any of the

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71

bottles falls under a prefixed value (20 bar). (This happens at t = 298 s.)

• After the blowing is stopped the submarine keeps rising. Expanding air in the tanks causes water to keep flowing out of the tanks.

• Venting begins when the submarine reaches a prefixed depth of 100 meters. (This happens at t = 353 s.)

• Venting continues until all tanks are completely filled. (This happens at t = 381 s.)

• Simulation ends when straight and level flight at a depth of approximately 50 m is achieved.

Results for this simulation are shown in Figure 4. As noted in Section 2, the submarine state variables are calculated using the 6–DOF model for the equations of motion proposed in Garćıa et al. (2009) and the variable mass model in Section 2.1.

4. CONCLUSIONS AND FUTURE WORK

Mathematical models for both blowing and venting pro- cesses have been proposed, and some advantages over previous models have been noted. Preliminary analysis of simulation results suggest that the models are accurate enough for this stage of the work.

A complete model of the blowing system and all pressure losses present will be considered in future works. However, at this moment, our main objective is to demonstrate the feasibility of blowing and venting of ballast tanks as a control mechanism. Results in Section 3.2 show that an underwater vehicle can effectively manoeuvre by means of this mechanism, without any use of diving planes. Design of a control algorithm based on blowing and venting of ballast tanks will be the subject of future research.

REFERENCES

Bettle, M.C., Gerber, A.G., and Watt, G.D. (2009). Un- steady analysis of the six DOF motion of a buoyantly rising submarine. Computers and Fluids, 38, 1833–1849.

Byström, L. (2004). Submarine recovery in case of flood- ing. Naval Forces, XXV.

Crowe, C.T., Elger, D.F., and Roberson, J.A. (2004). Engineering Fluid Mechanics, 8th edition. John Wiley & sons.

Feldman, J. (1979). Revised standard submarine equations of motion. Report DTNSRDC/SPD-0393-09. David W. Taylor Naval Ship Research and Development Center, Washington DC.

Garćıa, J., Ovalle, D., and Periago, F. (2009). Nonlinear optimization tool for the analysis of the manoeuvre capability of a submarine. http://www.dmae.upct.es/ ∼fperiago/archivos investigacion/ovalle periago garcia.pdf.

Gertler, M. and Hagen, G.R. (1967). Standard equations of motion for submarine simulation. NSRDC Report 2510.

Shampine, L. and Reichelt, M. (1997). The matlab ODE suite. SIAM Journal on Scientific Computing, 18, 1–22.

Watt, G. (2001). A quasy-steady evaluation of submarine rising stability: the stability limit. In RTO-AVT sym- posium on advanced flow management. Loen, Norway.

Watt, G. (2007). Modelling and simulating unsteady six degrees-of-freedom submarine rising maneuvers. DRDC Atlantic TR 2007-08.

50 100 150 200 250 300 0

100

200

300

Time (s)

B o

tt le

p re

s s u

re (

b a

r)

50 100 150 200 250 300 0

100

200

300

Time (s)

M a

s s o

f a

ir i n

b o

tt le

( k g

)

100 200 300 0

50

100

Time (s)

F lo

o d

e d

v o

lu m

e p

e rc

e n

ta g

e

0 200 400 600 0

500

1000

1500

2000

Time (s)

x p

o s it io

n (

m )

0 200 400 600 0

5

10

15

20

Time (s)

y p

o s it io

n (

m )

0 200 400 600 0

100

200

300

400

Time (s)

D e

p th

( m

)

0 200 400 600 −0.1

−0.05

0

Time (s)

φ (

ro ll a

n g

le )

(d e

g re

e s )

0 200 400 600 −15

−10

−5

0

Time (s)

θ (

p it c h

a n

g le

) (d

e g

re e

s )

0 200 400 600 0

0.5

1

1.5

2

Time (s)

ψ (

y a

w a

n g

le )

(d e

g re

e s )

0 200 400 600 1

1.5

2

2.5

3

Time (s)

u (

s u

rg e

v e

lo c it y )

(m /s

)

0 200 400 600 −5

0

5 x 10

−4

Time (s)

v (

s w

a y v

e lo

c it y )

(m /s

)

0 200 400 600

−2

−1

0

Time (s)

w (

h e

a v e

v e

lo c it y )

(m /s

)

Fig. 4. Simulation results for submarine state variables and bottle pressure, air mass in the bottle and flooded volume percentage in main ballast tank 2.

CAMS 2010 Rostock-Warnemünde, Germany, Sept 15-17, 2010

72