Dynamical Modelling and Predictive Response
of
Ships Steering Gear
W.G. Peletier
TU Delft
Faculty of Mechanical Engineering and Marine Technology
and
University of Newcastle upon Tyne Department of Marine Technology
PREFACE
With the disappearance of the European borders in the near future, the inhabitants of Europe are preparing themselves to take maximum advantages of it. This disappearance of borders doesn't mean that Europe is going to be one land with one culture and one language. To learn each others culture and language, the EC founded the ERASMUS Scheme (EuRopean community Action Scheme for Mobility of University Students) which provides the opportunity to study at another university for a period no longer than a year.
This project is a result of an ERASMUS exchange program between the Faculty of Mechanical Engineering and Marine Technology of the TU Delft, The Netherlands, and the Department of Marine Technology of the University of Newcastle-upon-Tyne, England.
The project was supervised by Mr. T. Ruxton in Newcastle and Professor Klein Wow]! in Delft.
3
LIST OF SYMBOLS
AA = area on which pressure acts in the actuator [m2
= area of section n = constant factor Cc. = contraction coefficient Cd = discharge coefficient Cr = clearance = velocity coefficient
cint = internal leakage coefficient
d = diameter of tube
Ph = hydraulic diameter
F = Force [N]
g = acceleration of gravity [m\s2]
LTA = inertia of actuator and water
k = spring rate [N/m]
1 = length [m]
m = Mass
n = number of revolutions per second'
pressure' in chamber n
Qin = incoming volumeflow
Cout = outgoing volumeflow
On,, = flow from n to m
R = resistance
Re = Reynolds number
transition Reynolds number
r = radius S = slip T = Torque u = overlap V = velocity V = Volume Volume in chamber n 4
[kg]
[1/s]
[N/m2][m3/s]
[m3/s]
[m3/s]
[m] [Nm] [m][m/s]
[rn3] [m3]5 w
=width
[m]x
= displacement [m] z = height offluid
[m] 13= bulkmodulus
[N/m2] 8 = flow coefficient C = friction factor A = constant factor 11= dynamic
viscosity [Ns/m2] v = kinematic viscosity [m2/s]0 = rotation angle [rad]
8 = angular velocity [rad/s]
e
.
angular acceleration [rad/s2]P = density [kg/m3]
4) = angle of poppet-cone [rad]
6
TABLE OF CONTENTS
2.
LIST OF SYMBOLS 44 be .04 4 54 . 4 r.gCi
'TABLE OF CONTENTS lom V VA 4 .6
SUMMARY
1 INTRODUCTION H., .. " ^.0 4``,0
11
2 LITERATURE ,SEARCH AND SYSTEM DESCRIPTION
13
14
2.1 Historic survey ... _
,
2.2
Steering gear in modem ships...,.
...
Description of the system 16
3 FAULT ANALYSIS
- ...--'-'4'
173.1 Introduction to fault tree analysis.
...
173.2 Single failure point !L.@ ; 17
3.3 Fault Tree Analysis of the steeringsystem 18
34
Fault selection_
.21 4 THEORETICAL PRINCIPLES OF THE MATHEMATICAL MODEL...
4.1 Flow balance . . ,gg mg . m m m gmil g.gumg mgmm ...mg....gm,
4.2 Orifice equation P A. .1 n.11.0 V Pr
21 22 4.2.1 Turbulent orifice flow
-
. . *Iso .1 16, 22'4.2.2 Laminar flow v 24
4.2.3 Leakage. 25
4.3 Pressure drop in conduits . . _ . . . . 27
4.4 Flow calculations based on analogy of electric circuits .
PREFACE 2 3 9 13 2.3 18 28
7 5
6
MATHEMATICAL MODELLING OF DYNAMIC RESPONSE
Si
The actuating vane subsystem 5.1.1 The actuator5.1.2 The safety valves 5.1.3 The locking valves 5.2 The power supply subsystem
5.2.1 The screw pump 5.2.2 The electric motor 5.2.3 The bypass valve 5.2.4 The filter
5.3 The control valve 5.4 The conduits
5.5 Composition of the complete model
5.5.1 Flow calculation between volume A and volume 1 5.5.2 Calculation of relief valve flow
5.5.3 Calculation of the rudder torque SIMULATIONS
6.1 Preliminary simulations
6.1.1 The check valve 6.1.2 The relief valve 6.2 The general complete model
6.2.1 Introduction
6.2.2 Description of the working of the steering gear 6.2.3 General simulations
6.2.4 Simulations made in order to assess the working order of the steering gear system if faults are specially introduced in the system 45 55 31 33 33 36 39 43 43 44 44 44 46 47 48 49 50 51 52 52 53 55 55 56 59 . . .
. . ...
...,....,
The. . . _... .......
... .
. , ....
8
7 EXPERT SYSTEM 67
7.1 Introduction 67
7.2 Specification and requirements of the expert system 67
7.3 Construction of the expert system 69
7.4 Discussion of results 71
8 DISCUSSION 73
9 CONCLUSIONS 75
10 REFERENCES 76
APPENDIX 1 PROJECT DESCRIPTION 78
APPENDIX 2 SIMULATIONS OF CHECK AND RELIEF VALVES . . . 79
APPENDIX 3 SYSTEM SIMULATIONS 91
APPENDIX 4 THE ID3 INFERENCE ALGORITHM 106
APPENDIX 5 DATABASE OF SIMULATION RESULTS 108
. .
.
. . . . .
Steering gear system
Fault tree
analysis
Mathematical
model
Computer
simulation
Simulation
of faults
Expert
system
Figure 1, Procedure followed in order to produce an expert system to assess the condition of a steering gear system.
Operating
Conditions SimulationComputer
Measurement Results Results Database Measurements Database Advisor ._Conclusions
Figure 2, Methodology utilised at the integration of computer simulation in the expert system
SUMMARY
A steering gear system is of vital importance to the safety of a vessel and its
crew.
In this investigation, a dynamical simulation model is produced, which is able to predict the response of the steering gear under various operating conditions. This prediction ability is utilised to increase insight in design parameters at the design stage of the system and to produce an expert system which is able to assess the condition of a steering gear system during operation.
The following procedure has been followed (figure 1):
A mathematical analysis has been carried out of a common steering gear system. Using TUTSIM, a dynamical simulation model has been produced. This model is utilised to simulate the operation of the steering gear in a healthy state, and of the steering gear with specifically implemented faults. These faults are selected by
carrying out a Fault Tree Analysis of the steering gear system. With the results of the simulation, an expert system is produced, that is able to monitor the condition of the steering gear system.
The simulation results of the steering gear with implemented faults are compared with the simulation results of the healthy system. The changes in
performance are assessed, and fed into an inference algorithm, which then generates general rules about changes in performance which occur due to a certain fault. This set of rules forms the basis for the advice given by the expert system.
The methodology given in figure 2 explains the way advice can be given on the condition of the steering gear by integration of a computer simulation:
The measurement results of the steering gear system in actual operation are compared with results of the computer simulation of the steering gear system, operating under the same operating conditions, but assumed to be without faults. The difference between the experimental and the simulation results is now fed into
10
the expert system, which gives advice about the condition of the steering gear system.
This investigation demonstrates that this technique is a feasible one, in improving the safety of a steering gear system.
By using simulation, together with an expert system, the physical side of the steering gear is incorporated in the expert system, which makes it an effective tool in assessing the condition of the steering gear system.
The results of the investigation do improve insight in the operation of the system, and the impact of changing of parameters, which can be used to improve the safety of the steering gear.
1./
1 INTRODUCTION
Steering gear systems are of vital importance to a ship. A malfunction of the steering gear can cause a fatal accident which can lead to loss of lives, or large ecological damage, as demonstrated with the loss of the Amoco Cadiz a large crude oil tanker, near the French coast in 1978.
It is therefore important to improve the safety ofa steering gear system in order to prevent those disasters.
To improve the safety of any engineering system, the reliability can be influenced at two stages:
at the design stage at the operational stage
At the design stage, the reliability of a system is influenced by: the use of elements/components with a high reliability building redundancy into the system
eliminating single failure points as much as possible
During the operational stage, the reliability of a system is influenced by: performance of the right amount of maintenance
checking regularly/constantly the working order of the system/components
for hydraulic systems: cleanliness of the hydraulic oil
The purpose of this investigation is to improve the safety ofa steering gear system, by using computer simulation (See project description in appendix 1). This computer simulation will be utilised in two different ways:
11
Cock)
It
NA/).44,1
1) At designing a simple system with a high reliability. Emphasis has been placed
12
and more profound understanding of the system. By using computer simulation in design, it is possible to obtain the real dynamic loads which affect the
components. With this knowledge the reliability of a component can be predicted with more accuracy.
orn--'77)
/-°-2) To build an expert system, that monitors the condition of the steering gear. The computer simulation assists the expert system in its decisions. The expert system is only partially developed, because a complete expert system would be too extensive for this project.
The model utilised in this investigation is a derivation of an existing steering gear. It includes however many estimations by the author, because the steering gear
manufacturer was unwilling or unable to give more detailed information.
In chapter 2, a survey of existing steering gears with their particular properties, is given together with a more detailed description of the steering gear under
investigation.
In chapter 3, the possible faults of this system are investigated by means of a Fault Tree Analysis (FTA), and faults to be investigated later, are selected.
Chapter 4 describes some theoretical principles which are used in the
development of the mathematical model of the system, which is derived in chapter 5.
In chapter 6, the simulations of this model are discussed. These simulations are carried out with, and without the faults selected in chapter 3.
Chapter 7 gives an explanation of the expert system and demonstrates the operation.
In chapter 8, the results of the investigation are discussed. Finally the conclusions are presented in chapter 9.
4/zrn.
-Electric rudder indicator Spring loaded return linkage Bypass valve. _ Figure 2.1,
Arrangement of steering gear system
using Hele-Shaw pumps, andcontrol box (above).
Cylinder isolating valves
/
Electric motor ,Relief :valve blot Donkin helestime type pumaExtended crossbar plates for gyro control and telemotor receiver
Local control
-13
2 LITERATURE SEARCH AND SYSTEM DESCRIPTION
ii
Historic surveyIn the earliest days of the Egyptians, the steering gear was just an oar. The position of the oar was gradually changed, and in the 13th century, the first ship was found with a rudder at the stern.
As ships became larger, the power to move the rudder increased, and this resulted in the early eighteenth century, in the use of a steering wheel. The utilisation of steam for ship propulsion provided steam as a power source for the steering gear.
In the early twentieth century, the ships had such large rudders and hull speeds, that the required torques could not be effectively developed and resisted by steam steering gear. With the introduction of the Hele-Shaw variable displacement hydraulic-pump, a means was found to provide those torques.
The working mechanism of this system is as follows (figure 2.1):
Rotational steering inputs from bridge control shafting, were fed through a rotary differential. This differential mechanism provided an output arm which adjusted the Hele-Shaw pump control lever such that when the bridge-commanded rudder angle equalled the actual rudder angle the pump was then off stroke, and did not cause rudder movement. Any difference between rudder angle or helm angle would cause the output lever to move, thereby putting the pump on stroke to apply the appropriate flow direction and magnitude between helm and rudderangle would be reduced to zero. The output flow of the pump was connected to a single or dual pair of rams which acted through a crosshead (Rapson slide) and caused rotation of the tiller and consequently the rudder.
This steering system was well conceived and has remained largely unchanged for over 70 years.
Figure 2.2a, single loop
system
4../-ELECTRONIC 'FEEDBAG< TE LE MOTOR STAGE
Figure 2.2b, twin
loop systemSTEERING DIAGRAM for Modulated Flow-contrOf
PUMP AND ALAS
1
Figure 2.3, modulated flow
grr, if/ FIR ',STROKE
ACTUATOR MODEL N10-41 Ei
FLOATING LEVER FEEDBAG< MAIN PUMP STAGE
Za
Steering gear in modem shipsThe steering gears which are installed in ships currently under construction can be specified as follows:,
By the way the actuation takes, place:
The Rapson slide system. This system consists of one or two pairs of rams,, which act through a crosshead on the tiller, which acts on the rudder.
The rotary vane system. In this system, a rotary vane actuator works directly on the rudderstocic, and transforms the fluid power, without losses in the
transmission,, to the rudder.
By the pump which provides the oil pressure:i Fixed displacement pump
Variable displacement pump
By the control system which controls the rudder movement (figure 2.2):
Single loop systems. The feedback of the rudder 'position is taken directly, electronically from the rudderstock.
Twin loop systems, which include a telemotor stage from which the electronic, feedback is taken. The output stage employs mechanical feedback to a floating lever pump stroking mechanism.
By the control principle:
Analog control, where the rudder is moved at a speed that is proportional to the error in the servo system.,
Bang-bang control,, where one or more servo loops have discreet levels, on/off, and operates with a deadband.
Modulated flow, the flow is gradually increased tip to full flow in ± 1.5 s. (figure 23). A fixed displacement pump is combined with a control valve. (not
a servo-valve)
I
\10
()f;Lit
ok.
u'vc"- ihri Aft"
15
Advantages/disadvantages of different configurations:
[1],[2],[3] GA-14-6e6?141"
The Rapson slide mechanism result in high side loading, which decreases the theoretical steering gear efficiency and increases wear.
The rotary vane actuator has seal problems at high pressures.
The rotary vane is a more compact unit in the steering compartment and is less vulnerable to accidental damage. Additionally it has a built-in radial and carrier bearing, and weighs about 25% less than an equivalent ram unit.
A variable displacement pump causes many failures due to their mechanical complexity and has low tolerance to small amounts of oil contamination. For high power systems, variable displacement pumps are preferred from the comrnutational point of view. A low flow with a fixed displacement pump and a servo-valve would induce too much heat. An alternative for this configuration is modulated flow. However a fixed displacement pump is cheaper, less noisy and less vulnerable to oil contamination than a variable displacement pump.
Analog control decreases the amount of rudder activity, compared to bang-bang control and improves the fuel economy at certain speed by 1.5 to 2 %. This system outperforms the Bang-Bang type in precision steering systems for special purpose vessels when steering quality is essential for operators.
1
re I
-1ainesioa.
wig
0
ing Valve 'Control Valve I 0 Cto LOW LEVEL ALARM EXPANSIONTAN
Po el. steering control
itsisol;Thtt
117
._\
PEV/SEELie. -86eOrrespondIng arrangement drawing. no
21186, 2118$
1
}to el. seeering control
1 1 SAFETY VALVE 8 Eft PUMP 7
OIL PUMP &pe tiotweut 6 ELECTRIC MOTOR
S PUMP UN!! tole P11.11:4-4 . I 1 STEERING nor& Am.!, Or 411".. - ! ---w .it_ Ma On Mt 0., P4 i II. re
.es
1 Mitnie -'--VA, --I... ..-kw./ 4/ .. .HYDR. CIRCUIT DIAGRAM FOR
STEEPING _ow I
Ia
' .4...., Ie.,
G.r
-f-al I^..-.. -.... IIP"Isr. floes' 2 POMP LINOS IM e PT( I ISTEERING NOWA. LOCK VALVE
i i I 7; --II
MEW
11 Relief Valve -1511911:141. II II Check Valvesa===,:..
0-10
I Bypass Valve Car--1
It
Li,23
Description of the systemThe system which is investigated, is shown in figure 2.4. It is a system that is both modern and simple.,
it consists of one rotary vane actuator(1), with two chambers divided by the rotor of the actuator. The chambers are connected by safety valves(8). When the pressure in one chamber is too high, the valves open, and pressure is released to the non-pressurized side of the chamber. The inlet and outlet of the actuator are locked by two locking valves. They only open when pressure is applied, otherwise, the actuator remains closed.
The actuator is powered by two independent "powerpacks"(5,6). It is however possible to power the actuator with only one powerpack. The speed of operation of the rudder will then be half the speed as when two units are in use. A powerpack consists of a pump of the screw type powered by an electric motor. The hydraulic oil flow delivered by the pump is directed by the control valve to the actuator, or to the pump inlet without pressure being built up. No oil cooler is necessary.
The control valve is either a two-stage valve, or a three-stage valve. In the first case, there is only one speed at which the rudder is moved. In the latter case, there is a possibility to manipulate the flow at two different flow rates. Le. for the first 1.5
_
seconds, the flowrate is low, to allow the pressure to be gradually built up, and afterwards the flowrate is maximum.
In this investigation, a simplified model of modulated flow is utilised, which is the three-stage configuration (section 5.3)
The oil pressure in the system is maintained by the position of the reservoir. It is usually placed approximately 2m above the pump, in order to achieve a suction pressure.
I
FAULT ANALYSISIn this chapter, a fault tree analysis of the steering gear system is made. A selection of faults is made which are used for the development of a fault diagnosis expert system.
j.
Introduction to fault tree analysisFault Tree Analysis (ETA) is a method to investigate the logical development of the contributing failures which may result in an accident.
It breaks the main failure, the top event, down to its respectivecauses, until the smallest event is reached. This smallest event is called a basic event.
When the whole tree has been build from the top event until all the basic events, an analysis can be made of the minimal cut sets (MCS). Each MCS is the smallest set of basic events that is sufficient to cause the top event if all the failures in that MCS occur simultaneously.
This is the ETA, it displays the logical relationships between the basic events and the top event.
32
Single failure pointIf a MCS consist of only one basic event, then this is called a single failure point. Single failure points are to be avoided, for they are the weak spots in the system.
After the loss of the Amoco Cadiz in 1978, where a tanker and cargo were lost
as
result of a single failure (flange rupture) in the steering system, regulations fortankers were changed according to 'single failure criterion'. 17
WPA:Mv SPRNG WOKEN BYPASS vALvf OPEN =NOWT FITTING RUPTURE POwER SYSTDA FNLURE POWER STSTEIA FAALURE RELIEF VALvt °POI RELIEF VALVE BuoCKED TOTAL LOSS OF STEERING
Figure 3.1, Fault tree of steering gear system
FREE OIL MOVENENT ACTUA1OR vgivAlvEASS WRONG BYPASS V. SETTTNG TOO LOW CAPACITY REUEF V. TOO LOW
0
AN OR Ot. CoNTA-LANAIXAN 1 Dcassw wA rota18
This means that a steering gear's power actuating and remote control systems must be duplicated. No single failure in one of the two systems must be able to destroy steering ability. After a failure it must be possible to regain steering within 45 seconds.
Loss of hydraulic fluid shall be detected, and the defective system automatically isolated, so that half of the system will remain fully operational.
1,3 Fault Tree Analysis of the steering system
The FTA of the steering system under investigation is derived from a ETA of a four ram steering gear.[4](flgure 3.1)
The top event is total loss of steering. This occurs when either the control system, the hydraulic system or the mechanical structure fails.
In this investigation is assumed that only the hydraulic system can fail. The hydraulic system fails, if either no fluid is fed to the system, or the oil is free to move in the actuator.
The power system is duplicated, and for simplicity of the fault tree, only one is analyzed, the second system is identical to the first power system. If one system fails, the other one is designed to take over immediately.
The fault tree is self explanatory, the names of the different components mentioned in the fault tree are given in (figure 2.4).
4 Fault selection and failure modes of selected faults [5]
To select the faults for the development of the expert system, it is important to know which kind of faults cause total failure of the system, the so called single failure points. From these faults, it is important to know their failure modes, in order to prevent the fault, if during inspection a particular failure mode is found.
19 The single failure points for hydraulic failure are:
low oil level sump
relief valve open when no pressure applied check valve open
safety valve between chamber 1 and chamber 2 open safety valve between chamber 2 and chamber 1 open
seals between actuator chambers damaged by wear due to oil contamination
low oil level in the oil reservoir can mean an external leakage in the system. An alarm in the tank (low level). An alarm sounds when the level in the oil reservoir drops below a certain level. In this situation, the integrity of the system can be restored by changing over from one power system to the other, and refilling the tank.
relief valve is open. This can occur due to different causes. E.g. the return spring is broken, the piston is stuck in its seating, or the clearance between piston and seating has increased due to wear.
check valve open. This is nearly the same case as with the relief valve. The spring might be broken, or the clearance between poppet and seat has increased due to wear.
4 and 5) as relief valve
6) oil contamination can cause failure of components which operate with a small
clearances, e.g. seals, pumps and valves. It can affect components by wear, increasing the clearances and thus preventing correct working of the component.
20
Increased valve leakage can result from abrasive particles wearing away metal surfaces by severe cutting or milder abrasive wear mechanisms.
If particles of about the same size of the static clearance of a valve lodge in this clearance, this can lead to increased actuating forces, and eventual seizure.
Another failure mode is sticking. The valve tends to stick to its seat.
The most common types of seal failure are by fatigue-like embrittlement, abrasive removal of material, and corrosion. Hard particles can become embedded in soft elastomeric and metal surfaces, leading to abrasion of the harder mating
surfaces forming the seal. It is found that the wear rate of the seal increases at an accelerating rate with reduced quality of environmental contamination.
As oil contamination is likely to occur [1][6], and the effects of oil
contamination do affect some single failure points, the faults used in the development of the expert system are:
leakage of the seal between the two pressurised chambers of the rotary vane leakage of the relief valve
sticking or jamming of the safety valve pump wear
Other effects used in the development of the system, which are not related to one component are:
air in the system
lower kinematic viscosity
4 THEORETICAL PRINCIPLES OF THE MATHEMATICAL MODEL: In this chapter theoretical principles, which are used in the development of the dynamical model, are specified.
4i
Flow balance:Flow in - flow out = increase in volume + flow storage (continuity equation).
Qin clout dV + V dP
dt
13 dt with:Qin = incoming flow [m3/s]
°out = outgoing flow
[m3/s] V = volume [m3]= pressure [N/m2]
A P = pressure difference [N/m2] AV = compressed volume [m3]
i3 = effective bulk modulus [N/m2]
The bulk modulus is defined as the relation between the pressure difference needed to compress(negative) a volume V an amount AV, (relative compression).
For the effective bulk modulus, the structural elasticity of the components, and the effect of entrapped air in the oil are taken into account.
It has been shown, that in order to investigate the dynamic performance of a hydraulic system, it is important to incorporate the bulk modulus in the system, as it relates to the stiffness of the liquid.[7] It is up to the model developer to judge ifa model has to be considered dynamically.
21
(4.1)
If
Ao
0 2
'EN
42
Orifice equation (figure 4.1):4.2,1 Turbulent orifice flow:
The flow between the points 1 and 2 is considered as streamline or potential flow, and experience justifies the use of Bernoulli's equation in this region [7].
Bernoulli: v2 2
+ zg = c
(4.3)
= pressure = density of fluid [kg/m31 = velocity of fluid [m/s]= height of fluid (negligible) [m] = constant [m2/s2]
= acceleration of gravity
Bernoulli's equation applied to an orifice neglecting gravity:
+
in-
at-- 4),4k'c tit
i
)Pi v12 p2 v22
Pt 2 P2 2
r.... Ce-.
1, to.pi
rvw,j,i,
4 . 4)/1,0 cif . ' ISC-01.0.-.-221 i-,-0- I
LA. lei liCA-4--% (ftn/CA.A..") 60 2.61_
J,.i ra v.,-.4
(7
L.)c-416 ti) ,VAAv-L.. c-0....,
c-r-A--Continuity equation for incompressible flow: 22 P(1 or 2) pressure section 1 or 2 [N/m2] v(1 or 2) = velocity section 1 or 2 [m/s]
P1 = p2 = p = density fluid
[kg/m3] = A2 v2 = Q A1 = area of section 1 [m2]A2 = area of section 2, where the flow area is minimum [m2] = flow through section [m3/s]
Ao = orifice area [m2]
(4.5)
p
=
23
= CcA0
(4.6)
Cc = contraction coefficient, dimensionless
(4.4) and (4.5) give, together with a coefficient to take into account the velocity loss due to viscous friction (Cv):
C. = velocity coefficient (-4.98) [7] This leads with (4.6) to:
Q = CctA0 NI (P1 P2) with: C,Cc.
- C,Cc for Ao << Ai
\11 A 2 Cc2 ( ) Cd = discharge coefficientFor turbulent flow, sharp-edged orifices and Ao << A1, is shown experimentally that: Cd 0.61 [7] 2 (Pi P2)
(4.7)
(4.8)
(4.9)
A2,0+
4.2.2 Laminar flow:
Laminar flow exists at low Reynolds numbers, i.e. under the transition Reynolds number [8]
0.611 2
)
V flow DOrif lee
V ik
Re, = Reynolds number
Rt = transition Reynolds number = flow coefficient
-=' kinematic viscosity fm2/'s1
6 is dependent on the geometry of the orifice. Wuest [91 has theoretically determined expressions for laminar flaw through orifices:
for a circular orifice: .6 0.2 for a rectangular slit 6' = 0.157' The laminar flow can be described as:
2624A0 (
/A = dynamic viscosity [Ns/m2)
Dh = hydraulic diameter [ml
For a circular orifice:', Dh = ,diameter of orifice K4.10)
(4.11t
S v(4.12)
25
For a rectangular slit orifice of width 'w' and height 'b' where w >> b, the hydraulic diameter is defined by:
Dh
42.3 Leakage:
There are three types of leakage, which are used in the simulation: leakage through a slit orifice
leakage between two plates
leakage between a piston and the housing
The leakage is assumed to be proportional to the pressure difference, as leakage is nearly always a laminar flow.
1) In paragraph 4.2.2, the laminar flow through a slit orifice is explained: 4 wb
2b
2 (b+w) 2 IS 2DAo Q (Pi P2) I. where: 6 = 0.157 and Dh hydraulic diameter = 2b = height of slit orifice(4.13)
2) The leakage between two plates: 3 wc Q
(Pl-P2)
12 where:= contact length of plate = width of plate
Cr = clearance between plates
3) leakage between piston and housing:
For a piston, the surface between piston and housing may be treated as two parallel plates, of which: w = Trd:
26
it dc
3r
(P -P)
1211.t, 1 2
= diameter of piston
Cr = clearance between piston and housing
1 = length of piston
(4.12a)
Li
Pressure drop in conduits:The empirical equation, giving the pressure drop for fully developed turbulent flow is:
AP
CD2
(4.14)
= friction factor
II = length of the pipe
I) = diameter of the pipe = speed of flow
C is dependent on the Reynolds number, the type of flow, and the condition of the surface of the pipe.
For laminar flow you get the Hagen-Poiseuille law:
c 64 Re
For turbulent flow, Re < 100.000 and smooth pipe, Blasius:
0.316
Re° 2 5
27
(4.15)
(4.16)
4A Flow calculations based on analogy of electric circuits:[10]
Flow is caused by a pressure difference, according to the following experimental law:
R is the resistance to the flow, influencing the flow generated by a pressure difference.
If at the other hand, a certain flow is given which flows through a certain resistance, the pressure loss is as follows:
AP = Q2R2
(4.18)
To calculate the resistance of a network of resistors, the following rules are derived (Figure 4.2):
For resistors parallel:
Q T = Q1 4- Q2
with (4.17), (4.19) gives: 1-A-7c
1/L-F2-R R1 R2
VAPT 11,&131 IIAP2
(4.20) + (4.21) give: 1 1 + 1 R T R1 R2 28
(4.17)
(4.19)
(4.20)
(4.21)
(4.22)
= =For resistors in series:
APTLP1 +
P2 with (4.18), (4.23) gives: QT2 RT2 = 2 Ri 2 + 02 2 R2 2 QT = 01 = Q2 1:4.24) + (4.25) give RT2 = R12 + R22R
-- Turbulent flow through conduits:
with: =
\ir121±2_
D 2
29(4.23)
(4.24)
(4.25)
(4.26)
(4.28)
(4.29)
Calculation of resistor values:
- Turbulent flow through orifice:
R-
(4.17)
30 It.D2 142 -v M11. 8Q2
(4.30)
4 2 (nD2)2 and (4.17), gives:R=
J81p
(4.31)
Tc2D5- using the classical equation for pressure drop:
AP = Kp v2
(4.32)
2 V2R=
Kp(4.33)
vA P-
2 with (4.30) gives: R Kp8(4.34)
(rD2)2
1
control Valve subsystem
L
-locking valves
actuator
safety valves
rn
--li 1 I 1actuating Vane subsystem
iii i il! L
r
Jr
L
L
31
MATHEMATICAL MODELLING OF DYNAMICRESPONSE
In this chapter, first the different components of the steering gear are modeled separately. In paragraph 5.5, the models of the different components are integrated in one model, describing the response of the entire steering gear system.
Some components however, are not as essential as others and in this case, assumptions have been made about their influence on the system, to simplify modelling of those components. This has been done for the control valve and the electric motor.
The steering gear system consists of two power packs, which under normal conditions are both in use. As they are identical, only one power pack is modeled.
The steering gear system can be divided into three major subsystems (fig. 5.1):
5.1 Actuating vane subsystem 5.2 Power supply subsystem 5.3 Control valve subsystem
5.1) The actuating vane subsystem consists of three major components:
5.1.1 actuator
5.1.2 safety valves, these valves are of the relief valve type, designed to prevent excessive pressures in one of the chambers of the actuator.
32
5.1.3 locking valves, these valves "lock" the hydraulic fluid in the actuator, and can only be opened when the pressure from the incoming flow into the actuator is higher than the pressure setting (P-set) of the valve, placed in the output line of the actuator, and thus the outgoing flow is replaced by the incoming flow, and a certain pressure is maintained in the actuator.
5.2) The power supply system consists of four major components:
5.2.1 hydraulic screw type pump 5.2.2 electric motor
5.2.3 bypass valve 5.2.4 filter
5.3) The control valve subsystem directs the flow from the power supply subsystem (the power pack) to the actuating vane subsystem, and consists of two
elements:
5.3.1 solenoid valve 5.3.2 pilot valve
The connections between the different subsystems, are made by conduits, which are modeled in paragraph 5.4.
Li
The actuating vane subsystem51.1 The actuator:
The actuator (fig 5.2) consists of four chambers, which are connected two by two. To simplify of the model, it is assumed that there are only two chambers, as the pressures in the two connected chambers will be nearly identical. The consequence of this assumption it that it is only possible to include seal leakage between the two
modeled chambers. The seal leakage between the other chambers mustbe .disregarded. The volumes of the two chambers depend on the rudder angle..
The flow balance can be described as:
dv
sV2 (0) dip].dt
1(3dt
dVV (9) dP2
QB; 2 + Ql. 2
dt
2-
2dt
Qi,2 = cint ( Pc P2 )1
(5.3)
mt = internal leakage coefficient QA,B the volume flow from A to R
The leakage flow is assumed to have a low Reynolds number, and therefore can be assumed to be laminar, and thus proportional to the pressure difference between the chambers.
33.
QA,
- °1,2 -
(5.1)
(5.2)
It is considered as flow between two parallel plates: Q 3 W C, (PI -P2)
121k
12 1 p.w = width of two plates
er = clearance between two plates
1 = contact length
V2 = v
- ce
dVdt
- -ce
dV,
dt
= ce
relation between rudder angle and volume, swept by the actuator Vmax = the maximum volume achievable in the chamber.
the angle over which the actuatorturns.
Equation of motion of actuator plus rudder:
+ c
= A(P1 P2 ) r
Trudder(5.6)
damping coefficient representing friction and hydraulic damping resultant radius arm of the actuating pressure on the actuator
moment of inertia of actuator
+ rudder + 30 % (due to
entrained water effect) rudder torque 34(4.12a)
(5.4)
(5.5)
Trudder +Figure 5.3, Blockdiagram of the actuator
Pr
35
The damping coefficient as a result of viscous friction can be found by considering the movement of the seals as the movement of two parallel plates:
A .
= F friction Cr
Ap. damping
A surface of the seals
clearance between the seals
Figure 5.4, Relief valve
Fret
fFigure 5.5, Blockdiagram of the relief valve
leak
°
ae,1_ ottiv`tuu
act IP1 PD51.2 The safety valves;
The safety valve is a relief valve and is modeled as in (figure 5.4). The block diagram is shown in (figure 5.5). The safety valve setting can be changed by changing the spring assembled load by turning the screw, to compress the spring.[7],[13]
Continuity flow equation:
V2 dP2
Qin - Qv - Qc = T at
Orifice equation valve outflow:
Qv = Cd42 P2-Pd) Qleak
A2 = TC D X
Continuity equation control flow:
Q, = CdAiNi (P2
-Pi)
36(5.7)
(5.8)
(5.9)
(5.10)
QC QaCCdt
°act = A .k , = Vo + AxEquation of motion of the valve:
Mt + ct + .icx =
- Pd) A Fpreset-A1
= area of
control orifice0
i= diameter of the piston= mechanical and hydraulic spring rate.
= damping factor, representing viscous friction
,F = spring assembled pre-load = valve mass + th of spring mass,
(= equivalent mass ofmoving system) .37
(5.11)
''The damping factor due to Viscous friction is:
Cdamping
n Drip,
D '= diameter of the piston -- length of the piston
cr F clearance between piston and housing
In The non-linearities of thesafety valve are:
can only be positive
overlap of piston:
A2 = IrD(x - overlap) instead of IrDx
leakage:
Leakage in the valve takes place through the clearance between the piston and the cylinder. Because the overlap is only very small, the leakage flow through the clearance is assumed to be as through a slit orifice:
262DhA0 Clleak2,D (P2 - PD) Ii 4 62x Dcr2 Cleak 38 where: 5 = 0.157 (see paragraph 4.2.2) Dh = hydraulicdiameter = 2 Cr
Cr = clearancebetween piston and cylinder
AO
=irDb
= length of piston
In the model where all the components are assembled, the relief valve is modeled as a resistance in the flow:
=
2
CdA2NI
(4.12)
This resistance varies as the valve opening A2 varies. The valve movement is a
function of the pressure before and after the valve (see section 5.5). A2 is taken from (equation 5.9); the dynamics of the valve are included.
(4.17)
(4.28)
R 1
Relief
Valve
Check
Valve
Figure 5.6, locking valves
5.1.3 The locking valves:
This system of valves, which are called the locking valves (figure 5.6) consists of two paired components, each of them incorporating a relief valve and a check
valve.
5.1.3.1. The check valve (figure 5.7): [141
Continuity flow equation:
dP
in -
Qvalve-
--2 10
dt
V2 dP2 Qvalve - Qout- ---
13dt
Turbulent orifice flow:
Qvalve = Ce.210\1-2-p (IDI - p2)
A0 = d it x sin 4)
The velocity at the vena contracta is:[11]
_ Qvalve C,A0 39 2 CdAo\ (p3_- P2) C,A0
(5.12a)
(5.12b)
(5.15a)
v -
Qvalve ArealFigure 5.8, Bipckdiagram. of the check valve.
Fpreset
c4A0
P2 t
V
40
IJ
+ ct
OPi-P2) 4d2 - Fprase C - X ( PI. -P2) X.where:. Fpreset = constant
This leads to the block diagram in (figure 5.8) with:
= Camping
e2 k c3 d Tr sin 4(5.15b)
(5.18)
with: Cd= Ccig,Cv velocity coefficient, a reduction in theoretical velocity caused by friction
and turbulence
The momentum force on the poppet is:
momentum PiQvalve:( V out,COS V in)
yin =- inlet velocity
out = exit velocity of the flow through the valve
Since yin is very small compared to vow, yin is neglected. The equations (5.13),(5.14),(5.15)and (5.16) lead to
F
ntwa = A (P1- P2) x, = n dCdCvSin240 C5.17)The equation of motion of thevalve is:
+ kx =
=
2 P1-
)26 2D hAo
P2)
cleak
In the model, where all the components are assembled, the check valve is modeled as a resistance in the flow:
Leakage: the leakage which can occur in the check valve, is leakage due to wear between poppet and seat. It is modeled as a slit orifice clearance:
1 Is where:
8 = 0.157
Dh = 2 cr
cr = clearance between piston and cylinder
AO
=irDb
= width of poppet seat
j_erCvs. L'L LA AN-2
(4.12)
(4.28)
C Ad p Q =This resistance varies as the valve opening Ao varies. The valve movement is a
function of the pressure before and after the valve (see section 5.5). Ao is taken from (equation 5.14); the dynamics of the valve are included.
4 627cDcE2 (3,,A
tr
41 at vt-,n
The non-linearities of the check valve are:
x can only be positive AA^' vi
Q =
(4.17)
53.12
The relief valveThe relief valve incorporated in the locking valve is modeled as in paragraph 5.12. The difference is that the flow Qc is not connected to the flow in the
depressurised line of the actuator (in which the valve is placed), but is connected to the flow from the pump to the pressurised chamber of the actuator as shown in figure
5.6.
52
The power supply subsystemThe power supply subsystem (figure 5:9) consists of an electric motor, a pump, a filter and a bypass valve. The pump is continuously running, and is of the screw
type.
5.2.1 The screw pump'
Since a screw pump is a positive displacement pump, the theoretical delivery is Proportional to the speed of revolution.
Pth i(m3/s)i
D F Displacement per radianfm311
n = radians, per second
[1/4
In practice the pump is not 100 % efficient, and there are several reasons 'which reduce the pump capacity. The reduction in capacity is represented by the slip
'S.
= oth
Ecif
CS. 20
Slip is a function of the clearance, theliquid viscosity and the .differential pressure. When there is no differential pressure, the slip is negligible.
M =
liquid viscosity AP = pressure difference? 43 1(5.2141 = D n(5.19)
Q SThe flow equation is:
Qout = Oth Qslip
(5.22)
The electric motor
The electric motor in the simulation is assumed to be perfect, ie. to deliver the desired torque at a constant speed, without being disturbed by load changes. This is assumed because the pump is a screw pump, aconstant volume delivery pump, where
- -
-
--the output pressure does not affect the delivered volume flow. (except for the
44
c.\lealcage)
5.2.3 The bypass valve
The bypass valve is a relief valve, identical to the safety valve, modeled in paragraph 5.1.2. The only difference is that the movement of the poppet depends on
P1, thus this valve opens due to an absolute instead of arelative pressure difference.
5.2.4 The filter:
In the system, the filter is placed before the pump. The pressure here is the atmospheric pressure plus the weight of an oil column with a height of the vertical distance between filter and expansion tank, as this tank is placed above the pump.
The filter is not included in the model. 5.22
53
The control valveThe control valve is a directional valve. Its function is to direct the flow to either side of the actuator or back to the reservoir, in order to turn clockwise, anti-clockwise, or to remain stationary. Because the assumption is made that the control valve does not fail, this valve is modeled in the general model only as a resistance in
the flow, which is high if it is supposed to be closed, and low if the valve is supposed to be open.
5.3.1 Modulated Flow
This control valve system has got 2 phases:
Modulated flow control Full rudder rate
During modulated flow, the oil flow delivered by the pump is partially bypassed, and directed to the suction side of the pump. The smaller oil volume is
directed gradually to the actuator. This will cause less rudder rate.
After phase 1, which is 1 - 1,5 s., the full flow is directed to the actuator. This control valve is modeled as a linear increase in flow directed to the actuator; from 0 to 100 % in 1,5 s.
5A The conduits
The flow between the different subsystems, flows through conduits. This flow is turbulent flow, which is caused by a pressure drop over the conduit.
The resistance of the conduit to the flow consists of friction. In paragraph 4.4, the value for the resistance of a conduit is derived:
R = n2D5
(4.31)
47
Si
Composition of the complete modelThe models of all the different basic elements of the steering gear are assembled in the complete model (figure 5.10).
The relations between the different elements are found by assuming four major volumes (see fig 6.1):
volume 1 = the volume of chamber 1 in the actuator volume 2 = the volume of chamber 2 in the actuator
volume A = the volume of the conduit leading from pump pressure side to the pressurised actuating chamber
volume B = the volume of the conduit leading from the depressurised actuating chamber to the pump suction
The flow between the volumes is generated by a pressure difference between two volumes, in contact with each other.
The flow is dependent on the pressure difference and the resistance encountered. The valves are incorporated in the model as a restriction in the main flows (QA.1,Q2_B).The size of the restriction depends on the valve and the position of the poppet or piston. The poppet or piston movement is a function of the pressure before and after the valve, which is calculated according to the pressures in one of the four
major volumes, and the pressure losses already encountered from that volume to the valve opening or outlet.
5.5.2 Calculation of relief valve flow
There are different relief valves incorporated in the system, each for a different purpose. Their flow can be calculated by assuming turbulent flow between the two volumes which are connected by the valve, and by considering the valve opening a restriction (equation 4.17).
The value of the resistance is dependant on the valve.
5.
VPA -PBk.ey
Qbypassd,
cryr
1 R3 Rsafety,1-2 //kAA/,)t)J
/6--°/)my-r
sa fety, 1-2 ' )1 Pi P2
Rsafety, 1-2 Qsafe Cy, 2 -1 Rbypass V P2 Pi Rsafety, 2-1 R2 = Rbypass 49 R4 = Rsafety,2-1The value of the different constants has been derived in the relevant paragraphs.
-n
-0
Figure 5.11, typical torque versus rudder angle of attack curve
L.; CC !-D 0 0 U.! > _ F-(71 0 C._ I SO -20 -10 0 +10 +20 ANGLE OF 4.:.! ATTACK o w n 0 cc 0 w > ).-= cc co LLJ Z
50 5.5,3 Calculation of the rudder torque [12]
The rudder torque is influenced by different factors. The main factors are: ships speed, Vs
angle of attack, a area of rudder, A
In case of a trailing rudder, the rudder torque would be: Trudder =
CV: ASilla
In order to minimise the rudder torque required, most rudders are not designed as trailing rudders, but the nidderstock is placed near to the centre of pressure of the rudder. The centre of pressure however, moves aft, if the angle of
attack increases.(The angle of attack is 5/7 of the rudder angle) _ ce cv it.arzio re4)
The location of the rudderstock will therefore be determined on the basis that the hydrodynamic torque should be zero at anangle of attack of about 10-15 degrees. This angle is chosen in the interest ofminimising the power required for routine steering and course keeping, which on most ships seldom requires more than 10 to 15 deg. of rudder angle.
Thus a typical torque versus angle of attack curve is shown in (figure 5.11). In this figure, one can see that the position of a=0 deg. is an unstable position. If it were freed, the rudder would move from 0 to plus or minus 15 degrees.
6 SIMULATIONS
In this chapter, simulations are carried out of the steering gear system for normal and fault conditions.
First some simulations of valves are made, in order to design them in the right proportions for proper functioning in the model. Next, some simulations are carried out of the healthy system. Finally simulations of the system with specifically
introduced faults are made.
6.1 Preliminary simulations
6.1.1 The check valve
The function of the check valve in the model is to ensure that the flow is supplied to the actuator, and contained. The capacity must be at least the pump
capacity (=3.0 1/s). The resistance must be as low as possible, and the valve must close virtually instantaneously if a negative pressure is applied. The normal pressure after the valve is 40 bar, cause the normal pressure in the actuator is set at 40 bar.
c-y/
ch,_a7)
The first simulation (1.1) (appendix 2), is one where the valve opening Ao is shown against the poppet diameter. The size of the valve opening Ao is directly related to the valve resistance. The larger the valve opening, the lower the resistance. As one can see, the valve opening increases if the poppet diameter is increased. This means that valve resistance is lower if a valve with a larger diameter is utilised.
In the second simulation (1.2), the pressure difference between chamber 1 and chamber 2 in the check valve is plotted when varying the poppet diameter. This can be considered as the pressure loss over the valve.
As a small pressure loss over the valve is needed, the diameter of the poppet is set at d = 24 mm (as a larger poppet diameter would not ensure much difference in the pressure loss).
6.1.2 The relief valve
In the steering gear, the relief valve model is used to represent three different functions:
safety valve relief valve bypass valve
However every relief valve has its own characteristics:
1) Safety valve. This valve opens if the pressure difference between P1 and P2
exceeds 50 bar. The capacity of this valve mustbe at least 10 % higher than the capacity of all pumps added together.
Relief valve. The locking valve is part of the system for two reasons: to maintain a certain pressure level in the actuator
to build up a braking action against negative loads
The requirements for the first function, is that it opens at the desired pressure, and closes if this pressure drops below the setting pressure.
The requirements for the second function, are that enough braking action is induced, in order to slow down the rudder movement if it is affected by negative loads.
Bypass valve. This valve prevents excessive high pressures after the pump. The capacity must be 10 % larger than the pump capacity. The pressure setting is at 100 bar absolute pressure. The regulation for the safety valves in a steering gear is, that the hydraulic pressure in the steering gear system should not exceed 10 % of the setting pressure. This can only be measured in the
simulation of the complete steering gear, as many more factors than the safety valve alone influence the pressure in the steering gear system.
53
Simulations:
Simulation 2.1:
This simulation has been made in order to asses the result of a change in piston diameter to the pressure before the valve. It is found that the larger the diameter, the higher the pressure peak, but also the sooner the pressure is stabilised.
A diameter of 25 mm is chosen as an optimum between maximum damping of the piston movement and minimum pressure peak before the valve.
Simulation 2.2:
This simulation has been made to investigate the effect of the orifice area of the relief valve on the pressure before the valve. As the orifice area is increased, the pressure peak decreases.
An orifice area of 10 * 10-6 is chosenas an acceptable value.
Simulation 2.3:
The simulation is the same as simulation 2.2, except for a setting pressure of loo, bar.
Figure 6.1, parameters used in the simulation.
6.2 The general complete model
6.2.1 Introduction
The simulations of the whole steering gear system can best be described by first explaining the general operation of the steering gear system, and then explaining the different simulations made.
The simulations can be divided into two parts:
general simulations, to assess the normal operating performance of the model. simulations made in order to assess the performance of the steering gear system if faults are specially introduced in the model.
In appendix 5, the results of the simulations are given in table format. This is done because for these simulations the output was generated in numerical format, and it was impossible to make plots from these numerical files due to a fault in the graphical software package. The value of the parameters after each part of the simulation is given in the tables.
6.2.2 Description of the working of the steering gear (figure 6.1)
Hydraulic oil flows from the reservoir and is pumped into volume A. Depending on the position of the control valve, a connection is made between
chamber 1, chamber 2 or volume B. Volume B is directly connected to the reservoir, and the pressure there approximately equal to the reservoir pressure.
If no rudder movement is required, the flow is directed directly to volume B, and the only pressure loss is the loss in the control valve. If some fault (eg. excessive pressure) might occur here, the bypass valve will open (at 100 bar), and the fluid will flow to the reservoir.
55
56
When rudder movement is required, the control valve directs the fluid to one of the actuator chambers, via the locking valves. The check valve opens at the
pressurised side. If the setting pressure of the locking valve in the return line is reached, this valve will open, and fluid will be able to flow from the depressurised chamber to chamber B, the output. The pressure is built up because of the flow to a rigid chamber. If the pressure difference between the two chambers, times their radius and area, are higher than the torque produced by the water on the rudder, then the rudder will be accelerated. By accelerating the rudder, the volume of the pressurised chamber is increased, and the result is a pressure drop. In case of
stationary movement of the rudder, the volume increment is equal to the pump flow, and the pressure in the chambers remains constant.
When the required rudder angle is reached, the flow is redirected to volume B, the reservoir. The pressures in the conduits to the valves will drop, and the locking valves will close; The pressure in the actuator is maintained.
If due to a sudden wave force, a high torque is induced on the rudder, and the pressure in one of the chambers exceeds the safety pressure, a safety valve will open,
the rudder will move, but the steering system will remain intact, as no harm is done to the individual components.
6.2.3 General simulations
Simulations have been carried out to investigate the proper working order of the model (appendix 3).
1) The classification societies prescribe that the rudder has to be able to be moved from neutral to 35 degrees in 15 seconds. This test has been carried out in simulation 3.1, 3.1.a and simulation 3.2.
The first 1.5 second, the rudder movement is unstable, but is quickly stabilised. This phenomenon is enlarged in simulation 3.1.a (a simulation of the first second of rudder movement). It can be seen that at t=0, a pressure is induced in chamber 1, due to the
57
volume flow from the pump. The pressure difference between chamber 1 and
chamber 2, together with the water torque accelerate the rudder, but the acceleration is too high, and this results in an unstable rudder movement. The rudder movement stabilises after ± 1 s.
This is because of a sudden full flow which is generated by the pump. If we compare this with simulation 3.2, where the flow has been built-up during the first 1.5 s, from 0 to 100%, the rudder movement can be considered to be smooth.
A control valve is supplied, which regulates the flow from 0 to 100% in 1.5 s,
to ensure better rudder control.
Between 0 and 10 s, P2 is higher than P1. Chamber 1 is the pressurised chamber, but between 0 and 21 degrees (=0-15 deg. angle of attack), the rudder is
instable because the rudder is not of the trailing type (in order to minimise the rudder torque). (paragraph 5.53). The locking valve generates a braking pressure of 40-45 bar. After e=21 deg. P1 rises, in order to overcome the rudder torque which
now acts on the rudder in a direction opposite of the rudder velocity.
To test the working order of the safety valves, the simulation is started at e =35 deg., hard over. After reaching its rudder stops the pressure difference between chamber 1 and chamber 2 should be limited to the P-set (=50 bar).
Simulation 3.3:
After an initial pressure build-up, the rudder is moved a small amount, against its rudder stops. The pressure rises again, until a pressure difference of P-set is
reached. The pressure difference is 50 bar( = P-set).
If the torque, acting on the rudder is in the same direction as the required rudder movement, the following problem occurs:
The water torque accelerates the rudder, but because no pressure is built up in chamber 1, the locking valve is closed. Because the fluid in chamber 2 is compressed, a pressure is build up. Because chamber 1 can not expand a pressure is build up due to fluid flow into chamber 1. If the setting pressure of the locking valve has been
ftekAA
(Jot
c"-0-74-7
,,zek
58
reached, the locking valve opens, and fluid flows from chamber 2 to volume B. The rudder now accelerates, which results in an increase in the volume of chamber 1. If this volume increment is higher than the pump capacity, P1 will drop. If P1 drops under P-set, the locking valve will close, P2 will rise again, the rudder acceleration
stops, and finally reverses.
This whole movement is unstable if the volume change due to rudder movement exceeds the pump capacity. A solution to this problem is found in incorporating a restriction in the return line; by setting the capacity of the locking valve (at a certain pressure), equal to the pump capacity. This 'certain' pressure must be high enough to induce a braking action on the rudder movement, which limits the rudder speed (if converted into volume increment) to a value equal to the pump
capacity.
To show the necessity of the braking action in the locking valves, the following simulations are made:
In simulation 3.5, the effect of a locking valve without enough braking action, on the movement of the rudder is demonstrated. The rudder is moved from e = -35 deg. at ships speed Vs = 15 knots (=Vmax, this corresponds with the highest
available negative torque) to neutral position. With the diameter of the piston of the locking valve = 9 mm, not enough braking action is induced, and this results in an unstable rudder movement as shown in the simulation.
If a locking valve which induces enough braking action is utilised (dpiston = 5 ram), the movement of the rudder stabilises much quicker (simulation 5.4).
I-71r c
fa*
59
6.2.4 Simulations made in order to assess the working order of the steering gear system if faults are specially introduced in the system
The simulations which are carried out in order to assess the working order of the steering gear system consist of three movements:
rudder movement 0-15 deg. no rudder adjustment for 2 s. rudder movement 15-0 deg.
The explanation for utilising these specific movements is given in paragraph
7.4.
In order to obtain as much information as possible, many parameters are monitored. As one full computer simulation (3 movements) takes about 7 hours, and TUTSIM only allows 4 parameters to be monitored at one simulation, only 7
parameters are monitored (the 8th proved to be unnecessary). PA P1 132 QA-1 02-B
de/dt
In the following discussion, first the simulation of the healthy system will be described. Next, the simulations of the system with implemented faults will be described in relation to the response of the healthy system. The results of the
62.4.1 Healthy system
The response of the healthy steering gear system tothe rudder movement is discussed in this paragraph.
Rudder movement 0 - 15 deg.
After initial instability, the rudder movement stabilises and reaches its final value of 15 deg. after ± 6 s. The movement is induced by the pressure difference between chamber 1 and chamber 2, and the torque from the water on the rudder. The torque of the water is cooperating with the pressurised chamber of the actuator. Therefore P2 is higher than Pi, and the pressure difference between Pi and P2
induces a braking action to the acceleration of the rudder by the water.
The flow from the pump to chamber 1 (QA-1) is constant at ± 3.0 1/s. The flow from chamber 2 to the reservoir (02-B) is unstable in the first 1,5 s., but stabilises then and remains constant.
No rudder adjustment for 2 S.
P1 and P2 drop considerably. This is due to the clearance in the locking valves, where it is still possible for the fluid to escape.
This pressure drop can be explained by calculating the average leakage over the locking valves. QL c AP (laminar flow) 4 52 Tc dcr2 1,78 * 10-11 ms Ns 60 (4.12) =
um'
,vu
/vAkA Qt,
AV 2 * 2 * QL (2 locking valves for 2 seconds) V 90 * 10-3 ml (contents of actuator)
/3 7 * 108 N
m2
This results in a pressure drop of:
AP = 28 bar
For a healthy system, this pressure drop is considerable, and the clearance must be smaller for a good working component. For a pressure drop of 1 bar, a
clearance of 2 Am would be needed, instead of the 10 Am utilised in the simulation.
The rudder angle increases to 15.23 deg. This has been made possible due to the force acting on the rudder and leakage. The rudder velocity however is very small.
Rudder movement 15 - 0 deg.
When returning from 15 deg. to 0 deg., the original pressures are restored, they are even increased, because the work now has to be carried out by the steering gear system. Both QA-1 and Q2-13 are constant again and with an angular velocity of 0.05 rad/s, the neutral position is reached.
61
AP = 50 bar (average pressure in the actuator) 8,9 * 10-5 M3
With equation (4.2):
-
A =_
...
,
62.42
Fault 1, seal leakageThe seal between both actuator chambers has a certain clearance. If the sealis worn, that clearance is increased to an unacceptable level. In the simulation the seal clearance is increased from 10 Am to 20 Am.
The difference compared to the healthy system is a higher angular velocity of the rudder during the first movement and the 2 s. without adjustment hold. During the third movement, the angular velocity is less than thatof the healthy system,
except at the end, where the torque of the water is nil; the angular velocity is equal to the healthy system at that rudder position.
The oil flows from the chamber with maximum pressure to a chamber with minimum pressure. If the rudder is moved from 0-15 deg., P2 is higher, and fluid will flow to chamber 1; the rudder will move faster. The leakage flow helps the pump flow in this case. If the rudder is moved from 15-0 deg., P2 is still higher, and more flow is pumped into chamber 1, and the rudder will move slower, because now this leakage
flow works against the pump flow.
If the actuator is divided into four chambers, more seals can be worn. With this model however, it is not possible to diagnose these faults.
Fault 2, locking valve leakage
Both locking valves have got a larger clearance than in healthy condition. This is because both valves are subject to the same type of wear, erosion due to oil
contamination. Although it is possible that wear is not symmetrical, this supposed to be the case.
The clearance between housing and piston is enlarged from 10 Am to 50 Am. This can occur due to contamination of the fluid and erosion. As the valves are open during the rudder movement, no difference in working order is to be expected during
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the first and the third movement. This fault shows itself when supposed to 'lock' the fluid in the actuator. (during 2 s. without rudder adjustment for 2 s.).
QA-1 and Q2_B are higher than in the healthy system, flow escapes from the actuator. The result is a drop in P1 and P2. This pressure drop is considerably higher than in the case of the healthy system.
During the third movement, this simulation gives the same result as the simulation of the healthy system, because leakage in the locking valve is now negligible when compared to the full flow through the valves.
62.4.4 Fault 3, safety valve leakage
The leakage through this valve is because of sticking of the valve. It is assumed to be stuck at an opening of x=1 mm.
The result is qualitatively the same as the result of the seal leakage, as the same process of oil flowing from one chamber to another (when not supposed to), takes place. This specific leakage is more pronounced, and results not only in a higher actuator velocity, but also in a measurably higher rudder angle after the first
movement.
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Fault 4, the worn pumpThe pump is worn due to contamination or some other type of wear. The clearance is enlarged, resulting in a lower pump efficiency.
This means a lower capacity, slower rudder movement and less pressure is built up, as the flow rate through the locking valves is lower, which implies a lower
