PROJECT # 4

HEAT EXCHANGER

There are three methods employed for water cooled marine diesel engines: direct, keel cooling and heat exchanger cooling. Direct cooling of the cylinders and heads by seawater is unsatisfactory, because the engine which was probably originally designed for radiator cooling will run too cold and the seawater will eventually ruin the cylinder block and heads. Keel cooling is suitable for small boats operating in shallow weedy water, but the need for pipe work external to the hull is a severe limitation.

Heat exchanger cooling is the most common method, the seawater being isolated in components which can be designed to withstand its corrosive effect. The closed fresh-water circuit can be thermostatically controlled so that the engine operates at its design temperature. The tube stack is fully floating, thus minimising thermal stresses, and it can easily be removed should cleaning be necessary. Heat exchanger header tanks prevent aeration of the engine water circuit which must be designed so that the system is self-venting on initial filling. It is usual for all the components in the seawater circuit to be in series, the gearbox oil and engine oil coolers being on the suction side of the seawater pump and the heat exchanger and any seawater cooled exhaust. Manifolds being on the discharge side of the heat exchanger. In the case of turbocharged engines the charge air cooler should receive the seawater first so that the lowest possible air temperature is obtained. The sea-water outlet from the heat exchanger should be from the end cover equipped with the upper connection; this ensures that the tube stack is always full of water. The gearbox cooler size will depend on the type of transmission used, but it will usually be a very size smaller as compared to the engine oil cooler. If preferred, the oil coolers can be fresh water cooled; these will need to be larger owing to the higher water temperature but need not be suitable for sea-water.

HEAT EXCHANGER

A six cylinder single acting 2-stroke Marine Diesel Engine consumes 60 tonnes of fuel oil

of 48 MJ/Kg calorific value per day. It is required to design a cooling water system

consisting of a cooler where fresh water circulated through the engine piston, jacket

and cylinder is cooled by sea water supplied from a single stage circulating pump. The

cooler is a single pass contra-flow type and a fresh water by-pass line is provided for

mixing so that the jacket cooling temperature is maintained at a higher value (see

attached sketch for further details).

Engine Details

1. Specific fuel consumption - 0.25 Kg/KW(brake)/ hr

2. Bore - 0.84m

3. Stroke - 1.6m

4. Mean pressure from indicator card - 11.2 bar

5. Mechanical Efficiency - 83.9 %

Cooler Details

6. Effective length of cooler tubes - 1.47m

7. External diameter of tubes - 20mm

8. Wall thickness of tube - 1mm

For turbulent flow in tube take,

2

St =

1+1.5(Pr)-1/6 (Re)-1/8(Pr-1)

Where f = 0.0791 (Re)-1/4

All fluid properties are to be evaluated at mean bulk temperature. Neglect thermal

resistance of the tube wall.

For flow through tube it can be assumed that flow will be turbulent when Re (Reynolds

number) is greater than 2100.

A. Calculate

(a) Brake power and speed of engine in R.P.M

(b) Mass of fresh water to be circulated through the engine to cope up with 10%

extra fuel burnt for overload running and given that 28% of total heat supplied

to the engine is transferred to the circulating water of 4.18 KJ/KgK specific heat

capacity.

(c) The temperature T at cooler inlet considering that 2/3rd of fuel circulating water

is used for cylinder jacket and cylinder head cooling and 1/3rd for piston cooling.

Determine also the mass flow through the bypass line and hence calculate actual

mass flow of fresh water through the cooler.

(d) Show that 400 tubes of given dimensions will be sufficient for the heat transfer

based on the mean diameter of the tubes given that the velocity of sea water

through the tube is 1.648 m/s for the required mass flow of sea water of 1020

kg/m3 density and 4.2 KJ/KgK specific heat capacity

Use Tables for properties of H2O at near bulk temperature.

(e) Calculate shell diameter considering tube plate area as 1.5 times the area utilized

by the tube holder. Considering thin shell theory, find the allowable thickness of

side shell plating when working pressure does not exceed 5 Kg/cm2.

Ultimate tensile stress of the material is 240MN/m2. Take Factor of Safety as 6.

B. Draw a section through the center of cooler end cover, tube plate and the cooler

body showing details of a tube fitted in place.

Show also arrangements for anodic protection

C. a) Comment upon sacrificed anode required in coolers of the type as designed:

b) Give the suitable materials required for various components under

consideration.

c) How will you ascertain a leaky coater tube and what action will you propose

to take for continuing the use of cooler?

GIVEN:

Six cylinders, single acting two stroke marine diesel engine consumes 60 tonnes of fuel

oil of 42/MJ calorific value per day

ENGINE DETAILS:

Specific Fuel consumption = 0.25kg/kW (brake)/hr

Daily consumption of F.O. = 60 tonnes

Engine Cylinder Diameter (Bore) = 0.84m

Length of Stroke = 1.6m

Mean pressure from indicator card = 11.2bar

Mechanical efficiency = 83.9%

COOLER DETAILS:

The cooler is a single pass, contra flow type

Effective length of cooler tubes = 1.47m

External diameter of tubes = 20mm

Wall thickness of tube = 1mm

Also, the given calorific value of fuel = 42 MJ/kg

A) CALCULATE:

(a) Brake power and speed of engine in RPM

Fuel consumption per day = 60 tonnes

Fuel consumption per hour = 60

24 tonnes / hour

= 2500 tonnes / hr

Specific fuel consumption =

..

B.H.P = 2500

0.25 = 10,000 kW

mec = MECHANICAL EFFICIENCY = ..

.

INDICATED POWER =

60 X n

Where n = number of cylinders = 6

P = Mean pressure from Indicator Card = 11.2 bar = 11.2 x 105 N/m2

L = Length of stroke = 1.6 m

A = Area of the cylinder bore =

4 X 0.842

N = Engine speed in r.p.m.

= 11.2 X 105 X 1.6 X

4 X 0.842 X N X 6

60

= 99308.5 N

Therefore mec = ..

. =

10000

99308.5 = 0.833

N = 120 RPM

SPEED OF THE ENGINE IS 120 RPM

b) Mass of the fresh water to be calculated through the engine to cope up with

10% extra fuel burnt for overload running and given that 28% of total heat supplied to

the engine is transferred to the circulating water of 4.15KJ/ KG K specific heat

capacity

Since daily fuel consumption = 2500kg/hr

10% of extra fuel consumption = 2500 + 10% of 2500

= 2750kg/hr

= 0.7638kg /sec

Heat generated by the engine = Fuel consumption/second x calorific value

Heat generated = 0.7638 X 42 X 106

= 32.08 MW

As given 28% of the heat generated is transferred to circulating water of 4.18 KJ/kg K

Specific heat capacity

Q = m CW (T2-T1)

8.98 = m X 4.18 (65-55)

m = 214.88 kg/s

This is amount of fresh water circulated to cope up with 10% extra fuel

consumption due to overload.

(c) The temperature at cooler inlet (sketch) considering that 2 3 rd of the total

circulating water in used for cylinder jacket and cylinder head cooling is 1/3 rd of

piston cooling .Determine also the mass flow through the by pass line and hence actual

mass flow of freshwater through the cooler.

SOLUTION: Let,

a = mass of water circulated to jacket cooler

b = mass of water circulated to piston cooler

a = 2

3 m & b =

1

3 m ()

Since from diagram and applying Kirchhoffs law at junction A

65a + 60b = (a + b) T

65 X 2

3 m + 60 X

1

3 m = (

2

3+

1

3) m X T

T = 65 X 2

3 +

60

3

T = 63.33OC or 333.33K

And then applying Kirchhoffs law at B

55a - 50 (a - x) = x T

55 X 2

3 X 214.88 - 50 (

2

3 X 214.88 - x) = 63.33 x

Therefore mass flow through the bypass line x = 53.7 kg/ s

Actual mass flow through cooler

= a + b - x

= 2

3 X 214.88 +

1

3 X 214.888 - 53.7

= 161.17 kg/s

(d) Show that 400 tubes of given dimensions will be sufficient for the heat transfer

based on the mean diameter of the tubes given that the velocity of the sea water

through the tubes 1.648 m/s for the required mass flow of sea water of 1020kg/m3

density and 4.2KJ/Kg k specific heat capacity. Use tables for properties of H20 at mean

bulk temperature.

1 = 50 - 12 = 38 C

2 = 63.3 - 32 = 31.3 C

Then

M = 12

(12

) =

3831.3

38

31.3

m = 34.54 C

Stantum Number St = {

2

1+ 1.5()

16 ()

18(1)

}

Where f = 0.0791()1/4

All fluid properties are to be evaluated at mean bulk temperature. Neglect thermal

resistance to the tube wall for flow through tube, it can be assumed that flow will be

turbulent when Re (Reynoldss number) is greater than 2100

Re = ..

Where = density = 1020 kg/ m

d = inner dia of tube =18mm

= 979 X 10-6 N / m2

Hence,

Re = 1020 X 1.648 X 0.018

979 X 106 = 30906.312

Since Re is greater than 2100, the flow is turbulent

f = 0.0791 (Re)-1/4

= 0.791 X (30906) - 1/4

= 5.95 X 10-3

Pr = 6.778 [Given]

Finally = 5.965 X 103

1+1.5(6.778)

16(3096.312)

18(6.7781)

= 1.092 X 10 -3

w.k.t

HEAT supplied to the engine, Q = ha X Qm

=

..

Where C = velocity

c = calorific value

1.092 X 10-3 =

1020 1.648 4.2 103

h = 7709 55 KW / m2h

Therefore

Q = Om

where a = projected area of tube = X d x l X h

where d = 0.019m

l = 1.47m

n = no. of tubes

Om = 34.54

Substitute in the above equation

8.98 X 106 = 7709.55 X X 0.019 X 1.47 X n X 35.54

n = 384.32 < 400

Therefore, the heat exchanger should have a minimum of 385 tubes

Since, it has been provided with 400 tubes which will be sufficient for heat transfer

(e) Calculate shell diameter considering tube plate areas as 1.5 times the area

utilised by tube holes

Answer: Area utilised by tube holes =

4 X d X n

=

4 X 0.022 X 400

= 0.125m2

Tube plate area = 1.5 X area utilised by tube holes

= 1.5 X 0.1256

= 0.18849m2

Hence, tube plate area =

4 X d 2

Where d = shell diameter

0.18849 =

4 d 2

d 2 = 0.24

d = 489.89mm

Diameter of the shell is 490 mm

(f) Considering thin shell theory find the allowable thickness of side shell plating

where working pressure does not exceed 5kg/ cm2. Ultimate tensile stress of the

material 240MN/m2.Take factor of safety is 6

SOLUTION: According to the thin shell theory

f =

2 Where,

P = Working pressure = 0.5 MN/ m2

we know that,

working stress =

=

240

6 = 40MN

On substituting the values

40 X 106 = 0.5 X 106 X 0.49

2 X

t = 3.06 mm

Since the thickness obtained is very small, for a safer consideration of value, we

assume it to be 5mm.

C)

A) Comment upon the sacrificial anodes required in cooling of this type as designed

Answer: There are two ways of protecting the sea water cooling side of heat

exchanger by coating and by cathodic protection systems. The fresh water side of the

heat exchanger is protected by addition of chemical additives to cooling water.

Coating varying to bitumen based paints through the epoxy coatings

of various forms can be based to protect water boxes and water box covers. In other

cases rubber sheet materials may be bonded to the mild steel or steel or cast iron so

that sea water cannot make contact with steel or cast iron. The steel or cast iron is

then, in effect electrically insulated from the sea water which is in contact with other

metals of heat exchanger.

Cathodic protection systems using sacrificial anodes are also used

within the sea water spaces of heat exchangers. The metals used for sacrificial anodes

must be such that they are higher in the electromotive or galvanic series of metals then

the metals which they have to protect. These metals or alloys are referred to being

active or anodic. In descending order of the galvanic series, they are cadmium,

commercially pure aluminium, zinc and magnesium alloys. In the presence of sea water

and with continued electrical continuity, the anode waste away and so protect the other

metals within the heat exchanger. In some cases a protective or passive film is

deposited on the metals being protected.

Chemical additives in fresh water are known as inhibitors. There

action creates a protective film of passive material on the metallic surfaces which they

protect. They not only protect the heat exchangers but other parts of cooling system in

which they circulate.

B) Give the suitable materials required for various components under construction

ANSWER: Tubes _ Aluminium brass

Anode - Zinc and Aluminium

End cover and water boxes - Cast iron & fabrication from mild steel

Shell - Gun metal

Baffle plates - Monel metal

Gasket - Nitrile rubber

Although there is an abundance of free sea water available, marine diesel engines do not use it directly to keep the hottest parts of the engine cool. This is because of the corrosion which would be caused in the cooling water spaces, and the salts which would be deposited on the cooling surfaces interfering with the heat flow.

Instead, the water circulated around the engine is fresh water (or better still, distilled water) which is then itself cooled using sea water. This fresh water is treated with chemicals to keep it slightly alkaline (to prevent corrosion) and to prevent scale formation. Of course, if distilled water, which some ships can make from sea water using evaporators, is used then there is a reduced risk of scale formation.

The cooling water pump which may be engine driven or be a separate electrically driven pump pushes the water around the circuit. After passing through the engine, where it removes the heat from the cylinder liners, cylinder heads, exhaust valves and sometimes the turbochargers, it is cooled by seawater and then returns to the engine. The temperature of the cooling water is closely controlled using a three way control valve. If the water is allowed to get too cold then it will cause thermal shocking which may lead to component failure and will also allow water and acids to condense on the cylinder bores washing away the lubricating film and causing corrosion. If it gets too hot then it will not remove the heat effectively causing excessive wear and there is a greater danger of scale formation. For this reason the cooling water outlet temperature is usually maintained at about 78-82C. Because it is at a higher temperature than the cooling water used for other purposes (known as the LT cooling), the water for cooling the engine is known as the HT (High Temperature) cooling water.

Cooling can be achieved by using a dedicated cooler or by mixing in some of the water from the LT cooling circuit. The LT cooling water is then cooled in the sea water coolers. The temperature is controlled using cascade control which monitors both the inlet and outlet temperatures from the engine. This allows a fast response to any change in temperature due to a change in engine load.

To make up for any leaks in the system there is a header tank, which automatically makes up any deficiency. Vents from the system are also led to this header tank to allow for any expansion in the system and to get rid of any air (if you are familiar with a domestic central heating system then you will see the similarities). The header tank is relatively small, and usually placed high in the engine room. It is deliberately made to be manually replenished, and is fitted with a low level alarm. This is so that

any major leak would be noticed immediately. Under normal conditions, the tank is checked once per watch, and if it needs topping up, then the amount logged.

The system will also contain a heater which is to keep the cooling water hot when the engine is stopped, or to allow the temperature to be raised to a suitable level prior to starting. Some ships use a central cooling system, whereby the same cooling water is circulated through the main engine(s) and the alternator engines. This system has the advantage whereby the engines which are stopped are kept warm ready for immediate starting by the engines which are running.

A fresh water generator (FWG) which is used to produce fresh water from sea water is also incorporated.

A drain tank has been included. This is for when the engine is drained down for

maintenance purposes. Because of the quantities of water involved and the chemical

treatment, it is not economically viable or environmentally responsible to dump the

treated water overboard each time. This way the water can be re used.

Project #6

STEERING GEAR

Amoco Cadiz was a very large crude carrier (VLCC), owned by Amoco, that ran aground on Portsall Rocks, 5 km (3.1 mi) from the coast of Brittany, France, on 16 March 1978, and ultimately split in three and sank, all together resulting in the largest oil spill of its kind in history to that date. What was the reason of such an catastrophic disaster. The answer is in this project.

INTRODUCTION

The direction of the ship is controlled by the steering gear. As the ship moves through the water, the angle of the rudder at the stern determines the direction it will move. Modern ships are so big that moving the rudder necessitates the use of hydraulics or electrical power. The steering starts at the Bridge. The required rudder angle is transmitted hydraulically or electrically from the steering wheel at the Bridge to the telemotor at the steering gear, just above the rudder.

There are a few common arrangements for using hydraulic power. There are the 4-rams, 2-rams, and rotary vane types. The heart of these hydraulic systems is the variable delivery pump. This type of pump can be controlled by just moving a spindle. The pump is driven by an electrical motor at constant speed.

By moving the control spindle away from the central point, the pump stroke increases, and the hydraulic fluid is pumped in one direction. Moving the spindle more from the central point will cause more fluid to be pumped and consequently more pressure is generated to drive the rams. Moving the control

spindle back to the original position and then away in the opposite direction causes the hydraulic fluid to be pumped in the reversed direction. The rams will also move in the reversed direction. By using a floating lever feedback mechanism, when the rudder stock has reached the desired angle, the pump control lever moves back to the original position, and the pumping action stops. The rudder is stopped at the required angle. Moving the steering wheel to the opposite direction will cause the rudder to come back to the original zero position.

TECHNICAL CHALLENGE

A vessel of 10,000 Tonnes displacement has LBP 120m, Breadth

15.6m and loaded draught 6.7m, fitted with a semi-balanced rudder

having a single guide pintle operated by a 2 Ram electro-hydraulic

steering gear (see accompanying sketch) has the following

particulars. It is required to design a hydraulic steering gear with

the help of information available and using a Rapson slide

mechanism.

S.No PARTICULARS VALUE

(a) Maximum speed of vessel (S) 16 knots

(b) Maximum rudder angle () 35

(c) Maximum working pressure on rams 100 bar

(d) Allowable Shear stress in rudder stock 75MN/m2

(e) Allowable bending stress in tiller arm 120MN/m2

(f) Allowable hoop stress in ram cylinders 50MN/m2

(g) Stroke of rams from mid-ship to hard-over 0.6m

(h) Time taken from mid-ship to hard-over 7.5 sec

(i) Ram packing thickness 20mm

(j) Variable delivery pump efficiency 70%

(k) Motor efficiency 90%

(l) Vertical distance of rudder top from rudder stock bearing

0.4 m

(m)

Horizontal distance of center of press on rudder from leading edge of rudder is given by x = (0.195 + 0.305 Sin ) b,

where b is the breadth of the rudder & = rudder angle

(n)

Force acting on the rudder is given by F = 577 X A X V2 Sin Newtons, where A is the area of the rudder in m2 , V is velocity of water passed in m/sec, and may be assumed to be maximum speed of the vessel (S) 16 knots for ahead running and 50% of vessels speed for astern running. ( is the angle subtended between rudder and center line of the ship)

(o) Rudder area for fast ships and slow ships are 1/60th or 1/70th of middle line area respectively.

(p) Suitable relief valves to be provided to prevent high pressure in cylinder due to abnormal conditions.

(q) Height: breadth of the rudder is 1.5:1

(r) Owner requires 10% increase in rudder stock dia. over calculated value.

Calculate:-

A.

(i) Find rudder area and dimensions;

(ii) Maximum force on the rudder;

(iii) Maximum Bending movement and Torque on the rudder stock with

lower pintle in place and without lower pintle.

(iv) Calculate rudder stock dia. allowing 20% increase for undue forces

and compare the values with diameter without lower pintle.

(v) Diameters of the rams so that the component of the force acting

on the tiller arm is sufficient to counter balance the torque due to

the force on the rudder. Accepted ram diameter should be a

multiple of 10.

(vi) Diameters of tiller arm assuming that the maximum stress occurs

at the junction of the tiller arm to the boss for rudder stock

(600mm. from center of rudder stock).

(vii) Thickness of ram cylinder using Thin Cylinder Theory for

approximation and then rounding up to a suitable figure.

(viii) Motor power, rpm and shaft diameter given that the shear stress in

the motor shaft is limited to 50MN/m2 and the motor driving

variable delivery pump is a 4-pole induction motor connected to

440v, 3-phase, 50Hz supply mains. Full load slip is 4% with the 0.8

lagging Power Factor Motor should run at 85% MCR and available

motor power ratings are multiple of 5.

(ix) Calculate the capacity of pump considering probable losses where

co-efficient discharge of pump is 0.94.

B.

With the help of isometric drawing supplied draw:-

Elevation in section at the center line of the rams with rams,

trunnion and the right ram cylinder in place clearly showing the

allowance for rudder drop. (Tiller arm and the boss are not to be

shown in this view.)

C.

(i) Why the rudder stock is preferred to be more than 230mm dia.?

What are the requirements in SOLAS pertaining to above criteria?

(ii) What arrangement is required to protect the steering gear from

damage due to jumping in rough sea?

(iii) How wear down of rudder carrier bearing is measured?

Steering Gear

Displacement = 10000 tonnes.

LBP = 20m

Breadth = 15.6m

Loaded Draught = 6.7m

Semi-balanced rudders with single guide pintle operated by 2 ram

electro-hydraulic steering gear.

The steering gear uses Rapson slide mechanism.

a) Max speed of vessel = 16 knots

b) Max rudder angle = 35

c) Max working pressure on rams 100 bar

d) Allowable shear stress in rudder stock = 75 MN/m2

e) Allowable bending stress in tiller arm = 120 MN/m2

f) Allowable hoop stress in ram cylinders = 50 MN/m2

g) Stroke of ram from mid-ship to hard over = 0.6 m

h) Ram packing thickness = 20 mm

i) Variable delivery pump efficiency = 70%

j) Motor efficiency = 90%

k) Vertical distance of rudder top from rudder stock bearing = 0.4

mm

A

(a) Rudder Area and Dimensions

The area of rudder is added to the area of the immersed

middle plane value at this ratio normally between 60 and 70.

Since speed of vessel is 60 knots, we can use area formula as

Area of Rudder =

Where L = LBP in m = 120 m

H = Max loaded draught = 6.7 m

A = 120 x 67

60

A = 13.4 m2

Rudder area is 13.4 m2

But

= 1.5

h = 1.5 b

Where h = height of rudder in meters

b = breadth of rudder in meters

Area of rudder = h X b

13.4 = 1.5 b X b

b = 2.98 m

h = 1.5 X 2.98

h = 4.48 m

These are the dimensions of the rudder

(b) Maximum force on Rudder

Force acting on the rudder is given by

F = 577 X A X V2 X sin

Where A = area of rudder in m2 = 13.4 m2

V = velocity of water passed in m/s

= 16 knots

= (16 X 0.5144) /

= Angle subtended between rudder and the center

line of the ship for maximum force

= 35

F = 577 X 13.4 X(16 X 0.5144)2X sin 35

Max Force on Rudder = 300.38 KN

(c) Maximum Bending movement and Torque on the rudder stock

with lower pintle.

Since it can be treated as a SSB, when rudder is supported by

lower pintle, the max Bending movement is given by

F = (4 + 2

3 ) - R (

1

3 )

For equilibrium F = R

Max B.M. = [300.38 (0.4 + (2

3 4.48)) 300.38 (

1

3 4.48)]X

103

= 567.718 KN-m

Also Max torque = Force X Distance from axis

= F X a

Where a = x - 0.7

= (0.195 + 0.305sin ) b 0.7

b = 2.98 m

a = (0.195 + 0.305sin 35) X 2.98 - 0.7

= 1.1 - 0.7

a = 0.4 m

Max Torque = F X a

= 300.38 X 103 X 0.4

= 120.152 KN-m

Maximum Bending movement and Torque on the rudder stock

without lower pintle.

Max Torque will be same for rudder having lower pintle and

without lower pintle

Max. Torque = 120.152 KN-m

Since it is cantilever beam, when it is considered for rudder

without lower pintle

Max. B.M. = Force X Distance from rudder stock bearing to

C.O.P

= F X L

Where L = 0.4 + 2

3 h = 0.4 +

2

3 X 4.48

= 3.38 m

B.M. = 300.38 X 3.38 X 103

= 1015.28 KN-m

(d) Diameter of Rudder stock with lower pintle

The equivalent torque is given by

Teq =(Mmax)2 + (Tmax)2

As we have already found max B.M. and max Torque with

lower pintle

Teq =(567.718)2 + (120.152)2

= 580.29 KN-m

20% increase has to be done for undue forces

Teq =580.29 + 20

100 X 580.29

= 696.349 KN-m

W.K.T. Teq =

16 d3 fs

fs = allowable shear stress in rudder stock = 75 MN/m2

= 75 X 106 N/m2

d3 = 16 Teq

fs=

16 X 696.349 X 103

X 75 X 106

d = 0.3616 m

d = 361.6 mm

d 370 mm

This is the diameter of rudder stock with lower pintle.

Rudder stock diameter without lower pintle

Eq. Torque, Teq = (Mmax)2 + (Tmax)2

=(1015.28)2 + (120.152)2

= 1022.36 KN-m

As per undue forces, we have increased the eq. torque by

20%

Teq = 1022.36 X 1.2

= 1226.832KN-m

W.K.T. Teq =

16d3 fs

d3 = 16 Teq

fs=

16 X 1226.832 X 103

X 75 X 106

d = 0.4367 m

d = 436.7 mm

d 450 mm

This is the diameter of the rudder stock without lower pintle.

(e) Diameter of Rams

Torque due to force on Rudder = 120.152 KN-m

W.K.T.

Force = Pressure X Area

= P X

4 d2

= 100 X 105 X

4 d2

Force = 7853.9816 d2 KN

From Triangle OAB

cos 35 = OA

AB

AB = 0.6 m = stroke of ram

OA = 0.6 X cos 35= 0.8569 m

Torque due to force = F X perpendicular distance of ram from

rudder stock

= 7853.9816 d2 X 0.8569

= 6730.0768 d2

As torque due to force on Rudder = torque due to tiller force

on ram

120.152 X 103 = 6730.0768 d2

d = 0.135 m

d = 135 mm

d 140 mm

[ dia should be a multiple of 10]

Force = 7853.9181 X 0.1352

= 143.138 KN

(f) Diameter of tiller arm (d)

Length of tiller arm = 0.856 0.6

= 0.256 m

B.M. on tiller arm = Max Force X r Distance

= 141.138 X 103 X 0.256

= 36.643 KN-m

M

I =

fs

R

M

64 d1

4=fsd12

d13=

M X 32

X fs

d13=

36.643 X 103 X 32

X 120 X 106

= 145 mm

(g) Thickness of Ram cylinder

By thin cylinder theory, we know thickness

t = Pd

2

Where P = max. Working pressure on rams

= 100 bar = 100 X 105X N/m2

d = diameter of ram = 0.14 m = 140 mm

f = allowable hoop stress in ram cylinder

= 50 MN/m2

t = Pd

2 =

100 X 105 X 0.14

2 X 50 X 106

t = 14 mm

(h) Motor power, rpm, and shaft dia.

W.K.T.

Voltage V = 440v

Current I = 20 A

Frequency F = 50 Hz

No of poles P = 4

cos = 0.8 (lag)

Motor power = 3 V I cos

= 3 X 440 X 20 X 0.8

Motor power = 12.193 KW

Since power rating should be multiple of 5

Motor power = 15 KW

W.K.T.

Synchronous Speed, Ns = 120 f

=120 X 50

4

= 1500 rpm

% slip = Ns N

NsX 100

= 1500 N

1500X100

N = 1440 rpm

Shear stress in motor shaft = 50MN/m2

W.K.T.

Power = 2

60

T = 60 X P

2N=

60 X 15 X 103

2 X X1440

T = 99.47 N-m

But, torque T =

16d3 fs

d3 = T X 16

X fs

d3 = 16 X 99.47

50 X 106

d = 0.0216 m

d = 21.6 mm

As the diameter should be multiple of 5

d = 25 mm

This the diameter of motor shaft

(i) Capacity of pumps

Cd = 0.94 (given)

Cd = Qact

QTh

QTh = area of flow X velocity of flow

= D4

4X

0.6

7.5

Velocity of flow = Stroke of ram from midship to hard over

Time taken from midship to hard over

= 0.6

7.5

QTh = X0.142

4 X

0.6

7.5

= 1.23 X 10-3 m3/s

Actual discharge, Qact = Cd X QTh

= 0.94 X 4.43

= 4.16 m3/s

Now considering probable losses.

Pump efficiency = 70%

Qact= 4.16 X 0.7

= 2.917 m3/s

Qact 3 m3/hr.

Actual discharge of pump is 3 m3/hr.

Rotor shaft dia.

W.K.T. Torsional eq. of shaft.

T

J=

f

J =

32 d4 (Polar movement of inertia of shaft)

d = Dia. of Shaft

f = shear stress = 41 X 106 N/m

R = d

2

T = Torque acting

T = fs J

R =

fs

32d

4

d

2

= X fs X d

3

16

Now W.K.T.

Equivalent Torque, Teq = M2 + T2

M Bending Movement, T Torque

Also given length between bearing = 1290 mm

Mass of Rotor and drive = 580 Kg

Shaft of turbine is considered as UDL with w/unit length

Wt. per unit length, w = 580

1.29

w = 449.612 kg/m

B.M. due to load, M = wl2

8

Which can be calculated as follows:-

Now total load on turbine rotor shaft

= Wt

Length Length

= w X l

Two reaction supports = wl

2 (for each)

Now total weight is acting through the center of the shaft

downwards

Now taking moment about center pt. considering left of

reaction

Moment due to reaction = wl

2

l

2 =

wl2

4

Moment due to wt. of half beam = wl

2X

l

4 =

wl2

8

Net B.M. = wl2

4 -

wl2

8=

wl2

8

Subs. we get,

w = 449.612 kg/m

l = 1.29

B.M. = 449.612 1.292 X 9.81

8= 917.48 N-m

W.K.T.

P = 2

60

T = 60

2 P = 13.5 X 106 W

T = 60 X 135 X 106

2 6000= 21.48 KN-m

Eq. Torque, Teq = T2 + M2

Teq = (21.48)2 + (0.917)2

Teq = 21.5 KN-m

Teq = f

3

16

3 = 21.5 X 106 X16

X 41 X106

d = 0.1387 m

d = 138.7 mm = dia. of rotor shaft

d 140 mm

Mass of Rotor

Mass of rotor = density X volume

= e X l X A

Area of cross-section of shaft = A =

4 d2

dia. of shaft, d = 140 mm

Area, A =

4 X 0.142

A = 0.01539 m2

e = density of shaft material = 7856 kg/m3

Length = 1.29 m

Mass of rotor = 156 kg

Mass of rotor/m = 156/1.29

= 120.934 kg/m

W.K.T. freq. of transverse vibration

fn=

2

E I

M l4

E = mod. Of elasticity of shaft material = 200 X 109 N/m2

I = Moment of Inertia of shaft

I =

64 d4=

64 (0.14)4 = 1.885 X 10-5 m4 X

Subs we get.

fn=

2

200 X 109X 1.885 X 105

120.934 X(1.29)4 X 9.81

fn=

2

3.77 X 106

3285.309

fn = 53.211 Hz

First critical speed is Nc= 53.211 Hz

CONCEPT REVIEW QUESTIONS

Describe the sequence of events in the accident of Amoco Cadiz.

En route from the Persian Gulf to Rotterdam, Netherlands, via a scheduled stop at Lyme Bay, Great Britain, the ship encountered stormy weather with gale conditions and high seas while in the English Channel. At around 09:45, a heavy wave hit the ship's rudder and it was found that she was no longer responding to the helm. This was due to the shearing of Whitworth thread studs in the Hastie four ram steering gear, built under licence in Spain, causing a loss of hydraulic fluid. Attempts to repair the damage

were made but proved unsuccessful. While the message "no longer manoeuvrable" and asking other vessels to stand by was transmitted at 10:20, no call for tug assistance was issued until 11:20.

Amoco Cadiz contained 1,604,500 barrels (219,797 tons) of light crude oil from Ras Tanura, Saudi Arabia and Kharg Island, Iran. Severe weather resulted in the complete breakup of the ship before any oil could be pumped out of the wreck, resulting in its entire cargo of crude oil (belonging to Shell) and 4,000 tons of fuel oil being spilled into the sea.