Marine System Design Project

ROJECT #1

PROPELLER SHAFT [WATER COOLED]

The transmission system on a ship transmits power from the engine to the propeller .itis made up of shaft, bearings & finally the propeller itself. Also the thrust from the

propeller is transferred to the ship through the transmission system. The different items in thesystem include the thrust shaft, one or more intermediate shafts and the tail shaft. These shaftsare supported by the thrust block, intermediate bearings & the stern tube bearing. A sealingarrangement is provided at the inboard end of the tail shaft with the propeller & cone completingthe arrangement.

There may be one or more sections of intermediate shafting between the thrust shaft & the tailshaft depending upon the machinery space location. All shafting is manufactured from solid forgedingot steel with the integral flanged coupling. The shaft sections are joined together by solidforged steel bolts. The intermediate shafting has flanges at each end and maybe increased indiameter where it is supported by bearings. The propeller shaft (or tail shaft) has a flanged facewhere it joins the intermediate shafting. The other end will also be threaded to take a nut whichholds the propeller in place.

Basically bearings are of 2 types, shaft bearings & stern tube bearings. Shaft bearings are furtherclassified into 2 types, the aftermost tunnel bearing (the aftermost bearing placed forward of theaft peak bulkhead in the shaft tunnel) and all other bearings. The aftermost tunnel bearing has top& bottom bearing shell. All the other shaft bearings have only the lower half bearing shell.

A stern tube forms the aft bearing for the propeller shaft. This bearing is also a part of the shaftsupport system.

The stern tube bearing being at the end of the shaft is affected by the overhanging weight of thepropeller. The load pulls the outboard end of the shaft downwards so that there is tendency ofedge loading of the stern tube bearing. Thus the stern tube bearings support the overhangingweight of the propeller and the tail shaft .the load on these bearings is transferred onto the sterntube which is supported by the stern frame & the internal part of the ship structure around the aftpeak bulkhead. Two forms of stern tubes are in use, the most commonly fitted having waterlubricated bearings with the aft end open to the sea. In this type, the bearings were traditionallylignum vitae strips and the tail shaft had a brass liner shrunk fit on it. Nowadays phenolic resinmaterials like tufnol are fitted in lieu of lignum vitae in water lubricated stern tube bearings.

P

The other type of stern tube bearing is the oil lubricated type used for oil cooled tail shaft, whichshall be dealt with in the next design

Let us proceed to design project on hand

PROPELLER SHAFT

The following particulars refer to a general cargo vessel of 14900GRT havingsingle screw:

1. Main Engine: 6 cylinder 2 stroke single acting in line reversible dieselengine developing 9900 BHP at 150RPM and directly coupled topropeller shaft through thrust Block and intermediate shaft.

2. Stern tube to be cooled and lubricated by sea water and stern tubebearing to be lined with Lignum vitae. Any portion of the shaft which islikely to come in contact with sea water to be protected with suitableliner shrunk on the shaft.

3. Shaft and coupling bolts to be made from forged steel SF 44 asapproved by Classification Society having the following specifications:

a) Allowable shear stress 27 MN/m2

b) Specified minimum tensile strength 44 Kg/cm2

4. Vessel should conform to LR ICE class 3 which requires 5% increase tothe Rule diameter for propeller shaft.

Calculate:-

1) (a) Intermediate shaft diameter

(D1) = 89 �(�(�/�)|60/(� + 16)|� mm

WhereC=a constant to be determined from the table supplied (Table H 2.5)H= B.H.PR= R.P.M of the shaftT= Specified minimum tensile stress in Kg/cm2

(b) Propeller shaft diameter Ds = 1.236 D1

Ds to be increased as per rules of Ice class 3 vessel

2) (a) Diameter of the propeller shaft at the coupling flangeDst = 1.05 D1

And to be increased as per rules of Ice class 3 vessels.

(b) Diameter of coupling bolts

(�0) = ��(106⁄��.�b).�. |�/�|

WhereB= 4.0 for crankshaft and thrust shaft/ crankshaft couplingB= 3.0 for other shaft couplingsC= a constant to be found from Table H 2.5H= B.H.PR= R.P.M of the propeller shaft

No. of boltsn= 9 for coupling bolts between thrust and intermediate shaft

= 10 for coupling bolts between intermediate and prop. ShaftR= Pitch circle radius of bolts in mm (assume suitable value)Tb= minimum specified tensile strength for bolt material inKg/cm2

3) Thickness of coupling flange:Find out suitable flange thickness from following rulerequirements.

(i)Thickness should be more than coupling bolt diameter(ii)Thickness should not be less than 25% of intermediateshaft diameter

4) Length of the stern tube bush(lignum vitae) considering thefollowing:

(i)Allowable bearing pressure- 2 Kg /cm2

(ii)Mass of propeller and cap nut= 10860 Kg(iii) a) Thickness of liner (shrunk fit) =5.7 meter (approx)

b) Thickness of liner as per classification rules=(Ds+230)/32

c) Density of liner material=8800 Kg/m3

(iv)Density of steel shaft material= 7800Kg/m3

Length of steel shaft to be determined from drawing.Length of lignum vitae should not be less than 4 times theprop. Shaft diameter.

5) Using Guest’s maximum shear stress theory, show that thediameter (D) determined from L.R.S specification isacceptable. Attached graph may be used for requiredinformation.

Equivalent torque to be increased by 20% to take care of stressconcentration.

6) Coupling bolt diameter considering average torque only andcompare the value obtained from L.R.S specification andaccept the greater value.

7) Shrink fit allowance of the liner considering the following

i) Interface of elasticity of shaft material= 11.7 GN/m2

ii) Modulus of elasticity of shaft material= 210 GN/m2

Modulus of elasticity of liner material= 85 GN/m2

iii) Poisson’s ration for both materials=0.25

B. Drawings must be freehand lined sketched.8) Draw the sectional elevation with free hand fairly

dimensional.9) Draw the cross sectional view at XX.

C.

10) Enumerate briefly what are the steps required to be taken forwithdrawal of Tail shaft of this kind. What are the parts needed tobe examined on Tail shaft and its attachments and why?

Solution:-

Calculate (A)

INTERMEDIATE SHAFT DIAMETER

D1 = 89 �� [ ��

��]

�

Where

C = A constant to be determined from table supplied

H = B.H.P

R = R.P.M. of the shaft

T = Specified minimum tensile stress in kg/cm2

This formula for the diameter of intermediate shaft is given as per the rules of the classificationsociety.

C = 1.28 [For 6 cylinder, 2 stroke inline diesel engine from table]

H = 9900 BHP

R = 150 RPM

Tb = specified minimum tensile strength as per classification society rules = 44 kg/cm2

Thereby putting the values in the L.R.S. specified formulae, the diameter of intermediate shaft isfound out

D1 = 89 �� [ ��

��]

�

D1 = 89 �1.28 ��

� [ ��

]�

D1 = 390.518 mmNow allowing a tolerance limit of 1% for machining, the diameter of the intermediate shaft isfound out as

D1 = 390.518 + 1% of 390.518

= 394.423 mmNOTE: - 1] The shaft diameter needs to be increased by 1% with consideration for machiningtolerances. This increased diameter is the diameter with which the shaft is manufactured .The 1%excess compensates the reduction in diameter of the shaft due to machining and surface finishingprocesses to which the shaft is subjected.

For manufacturing considerations, the dimensions are mostly rounded up to the next highermultiple of 10 or sometimes next higher multiple of 5.

D1 = 394.4 mm ≈ 400 mmNOTE: - 2] The sizes of shafting are governed by the formulae adopted by various shipclassification societies. The factors used take into account the strength of the steel used; thequality of steel is controlled by the upper and lower specified limits of the tensile strength. Thehorsepower being transmitted together with the rotational speed of the shaft, the maximumtorque & the respective mass moment of the inertia of the propeller & the flywheel are used in theformula to evaluate intermediate shaft diameter. The mass moment of inertia of the propellerincludes the entrained water.

NOTE: - 3] To ensure that the ships are properly built, equipped and maintained, a number of‘CLASSIFICATION SOCIETIES’ have been formed and have drawn up rules governing theconstruction of ships .the rules of different societies vary somewhat in detail but are mainly verysimilar; they are not arbitrary and other forms of construction are allowed if they give equivalentstrength.

Rules of Lloyd’s Register of Shipping (L.R.S) a British Classification society have been followedthroughout this design project.

NOTE :- 4]Once the diameter of the intermediate shafting has been found, the diameter of thescrew shaft is found by increasing the diameter of the intermediate shaft by some percentage, towhich is added a further amount which takes into account of propeller diameter and the method

by which the shaft is protected from the corrosive action sea water. In practice, the screw shaft isapproximately 15% to 17% larger in diameter than the intermediate shaft. The larger % isassociated with large propeller diameter.

CALCULATE (B)

PROPELLER SHAFT DIAMETER

As per the classification rules; the diameter of the propeller shaft is given by 1.236 times thediameter of the intermediate shaft

i.e. DS = 1.236 D1

Here D1= 390.518 mmThis value is increased as per rules of ICE CLASS 3 vessel of LRS by 5%

Thus Ds = 482.68+ (5/100X482.68)

Ds= 506.814 mmNOTE :- 5]Ships which are specially strengthened for navigating in ice may be given ‘ICE CLASSNOTATION’ .There are 4 ICE CLASSES based on Baltic Ice conditions-Class 1*,Class 1,Class 2,Class3

NOTE: - 6] ICE CLASS 3 ships are those which are intended to navigate in light ship condition.There are also some extra classes (1A to 1C) for vessels with special strengthening, intended totrade in the Northern Baltic.

Allowing a tolerance limit of 1% for machining, the diameter of the propeller shaft is given by

Ds = 506.814 + (1/100x506.814) => [NOTE 1]

= 511.882 mm

≈ 520 mm

This value of Ds looks very odd from the design point of view, thus it can be rounded off to anearest possible value.

The propeller shaft diameter is equal to 520 mm

CALCULATE (C)

DIAMETER OF COUPLING FLANGE OF PROPELLER SHAFT

As per classification society rules, the diameter of the propeller shaft, at coupling flange is given by1.05 times the diameter of the intermediate shaft.

NOTE: - 7] The coupling arrangement can either have a flange forged integral with the shaft orloose, depending on requirements.

Flanges that are forged integral with the shaft have their thickness, determined by classificationsociety rules, at least equal to the diameter of the flange coupling bolts measured, at the couplingflange. By using this ruling the possibility of using tapered or stepped bolts can be accommodated.The flange thickness should not be less than 0.2 times the diameter of the shaft.

The forging of the flange with the shaft must be provided with a fillet radius which is machined inand must not be less than 0.08 times the diameter of the shaft .these fillets must have a smoothfinish

Loose couplings arte usually fitted to the last line of intermediate shafting connecting it to the tailshaft .this method of coupling permits the removal of the tail shaft out through the stern bearingwith common practice to fit them in twin screw vessels. There are 2 types of loose coupling incommon use in current practice.

a) Muff coupling b) keyed and flanged couplings

COUPLING BOLT HOLES

The bolt holes are usually drilled to an approximate size in the workshops. The final dimension isreamed out when the coupling flanges are brought together. The final fit between the bolt and thehole.

Dst=1.05 D1

where D1 =390.518THUS THE DIAMETER Dst IS CALCULATED AS

Dst =1.05 X 390.518

= 410.044 mmThe value of the propeller shaft diameter at coupling flange is to be increased by 5% as per L.R.S.rules for ICE CLASS 3 vessel.

Dst = 410.044+ (5/100X 410.044) => [NOTE 5 & 6]

=430.546 mmAllowing the tolerance of 1% for machining, the dst is given by

Dst =430.546 + (1/100x430.546)

Dst = 434.851 mm

≈ 440 mmThis value looks odd from the design point of view, thus it can be rounded off to the nearestvalue. Thus the diameter of the propeller at the coupling flange is 440 mm

CALCULATE (D)

DIAMETER OF THE COUPLING BOLT IS GIVEN BY

do = B �� (��)

� � � � ��

B= 3 for shaft couplings other than crank shaft

C= 1.28 from table

R= 150 r.p.m

N= 10 for coupling bolts between intermediate and propeller shaft

Tb= minimum specified tensile strength= 44 kg/cm2

r= pitch circle radius of bolts

P.C.D. =1.6 x Dst

=1.6 x 410.044

= 656.070 mm

r= P.C.D/2 = 328.035 mm

NOTE 8:- There are various types of coupling bolts, the diameter of which must comply with theclassification requirements .the material used should have a tensile strength at least equal to thatof the shaft material.

There are 3 types of bolts.

PARALLEL NON –STANDARD BOLTS

This type of bolt has parallel shank and is in common practice .the surface finish is to a higherstandard, giving a good fit in the hole.

STEPPED BOLTS

These are rarely used and they limit the interference distance over which it is necessary to drivethe bolt home when fitting. The cost of machining limits their use.

SPECIAL TYPE BOLTS

Used in high powered installations .each bolt has an accurately bored hole over its full length. Thebolt has an arrangement to take a hydraulic attachment. on fitting the bolt, a high tensile steel rodis inserted into the bolt and using the hydraulic attachment, pressure is exerted, stretching thebolt within its elastic limit .this reduced the diameter of the bolt by 0.5 µm per mm of boltdiameter.

The bolt diameter is inserted into the hole in the stretched condition. On releasing the pressurethe bolt returns to its original duiameter, so exerting radial grip. the nut is fitted before releasingthe hydraulic pressure, thereby transferring the compressive force to the coupling bolt

do = B �� (��)

� � � � ��

do = 3 �� .�� (�� )

�� .��

do = 72.579 mm

≈ 80 mmDiameter of the coupling bolt may be approximately taken as 80 mm

CALCULATE (E)

THICKNESS OF THE COUPLING FLANGE (t)

It has to follow the following conditions.

(i) thickness should be more than coupling bolt diameter

� t > 75 mm

(ii) thickness should not be less than 25% of intermediate shaft diameter

� t >25/100 x D1

� t >25/100 x 390.518

� t >97.629 mm

The value of coupling flange thickness is taken as 100 mm

CALCULATE (F)

LENGTH OF STERN TUBE BUSH

Allowable bearing pressure = 2 kg/cm2

Total load acting on bush = propeller load + shaft load + liner load

Mass of propeller and capnut = 10860 kg

Mass of liner = density of the liner material x length x cross sectional area

= 8800 x 5.7 x�� x [Do

2 – Di2]

Thickness of the liner = (DS +230) / 32

= (482.68 +230) / 32

= 22.271 mmInner diameter of the liner = Diameter of the shaft

Di = 506.814 mm

Liner outer diameter = 506.814 + (22.271 x 2)

= 551.356 mm

Mass of liner = 8800 x 5.7 x � x [551.3562 -506.8142] x 10-6

= 1856.832 kgLoad on shaft = Density x Area x Length

= 7800 x (� / 4 x 506.814 x 10-3)2 x 7.2

= 11329.589 kgTotal load = Propeller load +Shaft load+ Liner load

= 10860 + 11329.589 + 1856.832

= 24046.421 kgLoad = Pressure x Area

24046.421 = 2 x 104 x 552.864 x 10-3 x L

L = 2.174 mLength shall not be less than 4 times the propeller shaft diameter

i.e. 4 x 482.68 mm = 1.93 m

l = 2.174 m > 1.93 mlength of the stern tube bush is taken as 2.174 m

CALCULATE (G)

Engine power = 9900 B.H.P.

= 746 x 9900 /1000

= 7385.4 kW

Power = (2 x � x N X T)/60

T = (60 X 7385.4) / (2 X � X 150) = 470.169 kNm

Bending moment = 100 kN-m => [from graph]Equivalent torque is given by

Teq = √�� + ��

= √100� + 470.169��

= 480.686 kN-mIncrease by 20% to take care of stress concentration

Teq = 480.686 + (20% of 480.686)

= 576.823 kN-mUsing torsion equation to find diameter

�� = ��

�

��.��

= (�� )��

d = 0.4774 m

= 477 mm

≈ 480 mm

Using guest’s maximum shear stress theory ,the minimum allowed diameter of shaft at whichfailure will not occur is 477 mm. the value obtained from LRS specification was 482.68 mm .thevalue being higher the propeller diameter is taken as 482 mm .final value can be approximated as490 mm.

CALCULATE (H)

average torque is the equivalent torque itself and is 470.169 kN-m

Torque = force X pitch circle radius

= F X 0.328

Force = ��.��

= ��.��.��

= 1443.447 kN

Load on each bolt = 1433.447 kN

Shear stress (�) = �� (�)��

�� = �

�� Rdb X n X �

db2 = ��.��

�.��

= 82.2 mm

≈ 85 mmThe diameter of the bolt is 85 mm

Shrink fit allowance

(i) Shrink fittingThe liner is firmly fitted to the shaft by boring it out to a diameter slightly less than thatof the shaft, and after expanding it by heat placing it in a position on the shaft which itgrips as it cools. The shrinking on the brass liner is not an easy operation, as not onlymust the liner be bored out to a size and heated to a temperature that will ensure itsslipping easily over the shaft, but if bored out to too great diameter, it will not grip theshaft sufficiently and if heated too high to a temperature, the properties of the materialwould be adversely affected. The liner should cool out evenly. Should the ends cooldown more rapidly then the middle portion, the metal may be overstressed and crackcircumferentially as it contracts in length. Some cool the middle portion by compressedair to prevent this, but if the liner is allowed to cool down evenly there is no rick ofmetal bring overstressed. The usual shrinkage allowance is 1/1000 inch per inchdiameter. It is not a good practice to pin the liners to the shaft as apart from thepossibility of water leakage past the pins and reaching the shaft, holes drilled into theshaft are liable to start fracture.

Let r3 be the radius of junction when shrinking.�r3 is the difference between the radius of theouter part of inner tube and junction.

�r3 =�r1 + �r2

GIVEN

P’ = 11.7 MN/m2 = 11.7 N/mm2

ES = 210 GN/m2 = 210 x 103 N/mm2

EL = 85 GN/m2 = 85 x 103 N/mm2

�� = 0.25

By applying Lami’s theorem

Radial stress hoop stress

Px =�� − � fx =

��

+ � ( x is the radius)

For outer tube i.e. liner

Px =�� − �1 fx =

�� − �1

At the outer surface of the liner i.e. r1 = 276.5 mm

∴ 0 = ��.��

− �1 … … … … … … … … … … … …. (1)

At the junction of liner and shaft, Px =11.7 N/mm2

Px =�� − �1

11.7 = ��

− �1 … … … … … … … … … … … …. (2)

From (1) & (2)

a1 = 60.18 , b2 = 4601329.65

Circumferential strain for outer tube (liner)

�� = �

�� [ �

�� + �1 + �’

�] … … … … … … … … … … … …. (3)

At the outer surface of the shaft, PX = 11.7 N/mm2

r2 = 253 mm

Px =�� − �2

11.7= �� − �2 … … … … … … … … … … … … . (4)

Hoop stress at the junction

fx =��

+ �2 … … … … … … … … … … … … … . (5)

From (4) & (5)

a2 = 7.65, b2 = 1238574.15Circumferential stress for the shaft

− �� = �

�� [ ��

�� + �2 + �’

�]…………………………………(6)

Subtracting (6) from (3)

�� = ��

� = �

�� [ �

�� + �1 + �’

�] - �

�� [ ��

�� + �2 + �’

�]

�� = �

�� [ ��.��

��.�� + 60.18 + 11.7 � 0.25] -

��

[��.��

+ 7.65 + 11.7 � 0.25]

�� = 1.3048 X 10-3 (r is the radius of the shaft)

�� = 1.3048 � 10��

��3 = 1.3048 � 10�� X 253

��3 = 0.338

��3 ≈ 0.4��

Shrink fit Allowance = 2 X r

= 2 X 0.4

=0.8 mmShrink fit allowance is 0.8 mm

(C).Enumerate briefly the steps required to be taken for withdrawal of tail shaft of this kind.What are the parts needed to be examined on tail shaft and its attachments and why?

Line shafting is examined once every 5 years for classification purposes and unless something hasgone wrong, this is the only time the shafting is looked at. When examined for the classificationsurvey it is usual to open up the main bearings, but it is not necessary to remove coupling bolts ifeverything has been operating satisfactorily. As well as examination of the bearing, the holdingdown arrangement should be looked at. Sometimes the holdings down bolts have slackened back.

At the dry docking of a vessel, the tail shaft wear down reading is recorded. For a tail shaft with acontinuous liner (TCSL), the reading should be less than 6mm. If the reading approaches thisfigure, tail shaft should be with drawn for the bush to be removed.

During dry dock inspection, bearing wear down is measured by poker gauge. Examination, afterremoval of the propeller and inward withdrawal of the propeller shaft may reveal various defects.

A keyway milled in the shaft tape acts as a weakening factor which allows some deformation of thesurface. Transmission of torque from the shaft via the key, to the propeller hub causes adeformation which tends to open the keyway. Grip of the propeller along the side of the keywaydoes this as well. Cracks having been a problem are reduced by the employment of sled type keys,radiused corners and spooning.

The first place to examine is around the forward end of the keyway to see If any fractures havestarted; usually they are found to run in a circumferential direction from the forward end of thekeyway. If the keyway is of the older type, the lower corners of the keyway must be carefullyexamined from the forward end on one side round the radius of the end of the keyway and alongthe forward end of the other side. To assist with the examination, it is normal to use magneticparticle crack detection equipment. In order to find cracks at the corners of the bottom of thekeyway dye-penetrators can be used as they usually show up cracks better in this location. If theshaft is of older design it will not be ‘eased’ at the ends of the liner. Cracks have been found runningin a circumferential direction in older shafts just under the end of the liner in these circumstances

2-3 millimetres is machined of the ends to expose the steel shaft underneath. When this operationis carried out it is usual also to machine a semi-circular groove in each end of the liner. Thisrelieves the herd edge which covers from the shrink fir and prevent it from acting as a stress-raiser.

[ MC GEORGE] The rubber seal sandwiched by the propeller hub and protective bronze liner,prevents ingress of sea water which could act as an electrolyte to promote galvanic corrosion of theexposed part of the shaft. Wastage from corrosion or fretting of the steel shaft beneath the forwardend of the hub or locally under the liner could weaken the shaft surface at the hub Liner notch tocause shaft failure through fatigue or corrosion fatigue. Shaft droop from over changing weight ofthe propeller stretches the upper surface and compress the lower, giving conditions when the shaftis rotating, which are likely to cause failure. The imposed alternating effect is of low frequency andhigh stress.

The shrunk or bronze liner, fitter to protect the steel shaft against ‘black corrosion’ may itself bedamaged by working conditions. Shaft whirl can lead to patched marked by cavitation erosion,scoring occurs in way of the stern gland packing and liner cracking has sometimes penetratedthrough to cause corrosion cracking in the shaft.

The parts of the shaft exposed after machining the ends of the liner are checked for cracks using themagnetic particle r dye penetrator.

The forward end of the shaft, where it is reduced from the large diameter down to the radius whichis swept into the back of the flange is also a location to be examined. The critical region is at thesmallest diameter. Old shafts which may have had the coupling bolt holes enlarged, should alsohave the thinnest section of material examined for cracks running radially outwards from the holeto the outer circumference of the flange. The aft end of the shaft at the reduced section from theend of the thread to the face meeting the smaller end of the cone should also be examined thoughthis area rarely gives trouble. If the shaft liner Is made up of 2 pieces, the joint between the 2sections should be carefully examined. Dye penetrated tests would be carried out on the joint.

Common mistakes

v 1]Neglecting 1% of machining tolerancev 2] Adding of 1% machining tolerance before 5% LR ICE Class 3

Rounding up values in multiples of 10

v 3] Not taking initial value of D1 without machining tolerance for further calculations.v 4] r = pitch circle diameter /2.v 5] Convert B.H.P. into Kw.

Concept review questions

What are parts of the shafting system? what are the bearings in the shafting system? Whatis aftermost tunnel bearing? Why aftermost tunnel bearing has has top and bottom bearingshell? Why the other shaft bearings have only lower half bearing shells?

ANS- The transmission system on a ship transmits power from the engineto the propeller. It is made up of shaft, bearings and finally the propelleritself. The thrust from the propeller is also transferred to the ship through thetransmission system.

The different items in the system include the thrust shaft, one or moreintermediate shafts and the tail shaft. These shafts are supported by thethrust block, intermediate bearings and the stern tube bearing. A sealingarrangement is provided at either end of the tail shaft with the propeller andcone completing the arrangement.

Shaft bearings are of 2 types, the after most tunnel bearing and all others.The aftermost tunnel bearing has a top and bottom bearing shell because itmust counter act the propeller mass and take a vertical upward thrust at theforward end of the tail shaft. The other shaft bearings only support shaftweight and thus have only lower half bearing shells.

What is edge loading of stern tube bearing?

ANS- The stern tube bearing is also part of the shaft support system. Insome of the later designs, the bearing is accessible from the machinery space.The stern tube bearing being at the end of the shaft is affected by theoverhanging weight of the propeller. The load pulls the outer end of the shaftdown so that there is a tendency for edge loading of stern tube bearings. Theforward part of the tail or propeller shaft is tilted upwards. Wear down tendsto make the alignment worse and whirl may give an additional problem.

Is there any substitute for lignum vitae?ANS--The traditional lignum vitae staves are fitted with end grainvertical beneath the shaft for better wear resistance. Staves in the upperpart are cut with grain in the axial direction for economy. The staves areshaped with V or U grooves between them at the surface, to allow accessfor the water. The grooves alsoi accommodate any debris. They are heldin place, in the bronze bush by bronze keys, attached to the bush by the

counter sunk screws. Bearing length is equal to four times shaftdiameter.

Why are lignum vitae strips dove tailed?ANS--Strips of lignum vitae about 2 inch wide and ¾ thick are dovetailed into the bush, water ways being left at 4 or more points to allowsea water to have free access to the rubbing surfaces. The lignum vitae isprevented by rotating with the shaft by 4 projecting longitudinal stripscast on the inside of the bush, while the forward movement is preventedby the lip on the inner end of the bush. To prevent lignum vitae fromworking outboard, a check ring secured by means of tap bolts to theflange of the brass bush is provided.

What is rewooding and when is it required?ANS--Excessive wear down of bearing materials due to vibration orwhirl, poor quality of work when re wooding, inferior materials,presence of sand/ sediments in the water or propeller damage, couldnecessitate early re wooding. Bearing life for vessels with engines aftand particularly tankers and ore carriers which spend long periods inthe ballast has been shot with re wooding being needed in perhaps 18months.

What are radial face seals?ANS--Radial face seals can be used for sea water lubricated stern tubes.These are fitted at the inboard end of the tube with the outboard endopen. However sea water is supplied from the sea water circulatingsystem and runs out through the after end of the tube. The amount ofany sand in water would tend to be less after passing through the pipesystem.

What are the types of seals?Special seals are fitted at the outboard and inboard ends of the tailshaft; they are arranged to prevent the entry of sea water. Older designs,usually associated with sea water lubricated stern bearings, made use ofa conventional stuffing box and gland at the after bulkhead

What is black corrosion?The shrunk on the bronze liner, fitted to protect the steel shaft against“black corrosion” may itself be damaged by working conditions. Shaftwhirl can lead to patches marked by cavitation erosion, screwing occursin the way of stern gland packing and liner cracking has sometimespenetrated through to cause corrosion cracking of the shaft.

PROJECT # 2

HOLLOW TAIL SHAFT (OIL COOLED)

As we have discussed about water cooled propeller shaft earlier, our next project is about hollow oilcooled propeller shaft. This type of propeller shaft is supported by a stern tube closed at both endsand having metal bearing surfaces lubricated by oil.

Oil cooled propeller stern tube is preferred in many ships with machinery aft, where the short shaftis to be relatively stiff and only small deflections are to be tolerated. Where this patent oillubricated stern tube is fitted, glands are provided at both ends to retain oil and to prevent theingress of water. Besides, white metal (high lead content) bearing surfaces are also provided and oilfor cooling and lubrication is supplied from a reservoir. Progress from sea water to early oillubricated stern tubes involved an exchange of the wooden bearing (LIGNUM VITAE) in its bronzesleeve with a white metal lined cast iron (or sometimes bronze) bush. Oil retention and exclusion ofsea water necessitated the fitting of an external face type seal. The stuffing box was retained inmany early oil lubricated stern tubes, at the inboard end. In oil lubricated bearings the shaft doesnot require a full length protective bronze sleeve.

This is the conventional type of arrangement

Now, without much ado let us proceed to the design project and learn the intricacies andchallenges involved in design, operation and maintenance of this propeller shaft.

HOLLOW PROPELLER SHAFT

The following information relates to a steel propeller shaft for a single screw bulk carrier.

The propeller shaft is hollow and connected to a solid intermediate shaft.

The propeller is fastened to the shaft by 16 pre-stressed internal bolts on P.C.D. of 1m fitted withhollow dowel pins around them. The hollow dowel pins can be regarded to carry the completetorsion load while the bolts take all other loads.

Additional load due to vibrations may be assumed to be acting downwards through the C.G. ofpropeller and is given by

LOAD =��

�� x 0.75

Additional B.M. due to thrust may be taken as a product of Propeller Thrust x Propeller Dia. (in m)

x 0.15 where Propeller Thrust = ��

Shaft RPM = 75

Shaft power = 22400 KW

Propeller dia = 9 m

Propeller mass = 51000 kg

Propeller buoyancy = 80 KN

Max. Shear Stress allowed –

For solid shafting and coupling bolts – 50 MN/m2

For hollow shaft due to welding involved – 25 MN/m2

For hollow dowel pins – 50 MN/m2

Neglecting mass of shafting

Calculate

Dia. ‘D’ of intermediate shaft considering torsion only.Dia. of intermediate shaft coupling bolts.Outside dia. of hollow dowels.The total thrust given that the propeller efficiency 60% and the speed of advance 10 knots.The total B.M. on the shaft.Show that the equivalent torque on the shaft may be rounded to 5680 kN-m and hence,determine the outside dia. of the hollow shaft.Based on dia. so evaluated on hollow shaft, calculate a suitable flange diameter and size ofcoupling bolts placed on possible pitch circle.Mention suitable materials used for each component in design and manufacture, alsodescribe in brief how welding carried out during assembly.Describe the purpose of chrome liner fitted with the shaft seals. What bearing material isused in stern bush bearing of this type?

What is the normal periodicity of tail shaft (OG) survey? What are the specific areas need tobe inspected? Under what conditions this survey may be allowed with extended period?

Solution: -

Calculate

1) Diameter of Intermediate Shaft

Mass of the shafting is neglected

Power, P = 22400KW (given), N=75 rpm

P = ��

Torque,

∴ T = �×��×�×�

= ��×��×�×��

∴ T = 2.85 MN-m

We can determine the diameter of the shaft by applying the torsion equation:

�� = ��

�

��

� =��

�.��×��×��

= ��×��

∴ D = 0.6623m

∴ D = 662.3mm

Considering 1% machining tolerance

D = 662.3 + (0.01×662.3)

∴ D = 668.923

∴ D ≈ 670mm

2) Diameter of intermediate shaft coupling bolts

Torque acting on each bolt (For 12 bolts)

= �� .��

= �.��×��

��

= 237.671 KN-m

Force = ��

= ��.��×��.��

= 306.672 KN

Force = Max. Shear stress× Area

306.672×103 = 50×106×�� × ��

∴ d = 88.37mm

∴ d ≈ 90mm (Bolts connecting intermediate shaft and thrust shaft)

The value is rounded off to 90mm

(THIS DIAGRAM IS ONLY FOR PROPER UNDERSTANDING)

Torque acting on each bolt (for 16 bolts)

= ��

= �.��×��

��

= 178.25 KN-m

Force = ��

= ��.��×��.�×��.��

= 336.42 KN

To find the diameter of the coupling bolts

Force = Max. Shear stress × Area

336.42×103 = 50×106×��×d2

∴ P d = 92.557mm

∴ d ≈ 95mm

(Bolts connecting intermediate shaft and tail shaft)

The value is rounded off to 95mm.

Note:

There is a twist (to baffle students) in the design question (iii) Outside diameter of the hollowdowel pin (do), here we need to use the equivalent torque which can be found in question (vi) andhence (iii) follows (vi)

5) Total B.M. on shaft

Total B.M. = Additional B.M. due to thrust + B.M. due to mass of propeller + B.M. through C.G. ofpropeller - B.M. due to Buoyancy.

Additional B.M. due to thrust,

M = Propeller thrust × Propeller diameter × 0.15

= 2612.752 × 103 × 9 × 0.15

= 3527.216 KN-m.

B.M. due to mass of propeller,

M = 57000 × 9.81 × (�.�� + 0.6 +1.05)

= 500.31 × 103 × 2.1

= 1050.651 KN-m.

Load due to vibrations = ��

× 0.75

= �.�� ×��

� × 0.75

= 237.5 KN

As it acts through C.G. of propeller

B.M.

∴ M = 237.5 × 103 ×2.1

= 498.75 KN-m.

B.M. due to buoyancy,

M = 80× 2.1

M = 168 KN-m.

Total BM,

M = 3527.216 + 1050.651 + 498.75 – 168

∴ M = 4908.617 KN-m.

6) Equivalent Torque is given by,

Te = √�� + ��

=�(4908.617 × 10�)� + (2.85 × 10�)�

∴ Te = 5678.6039 KN-m.

∴ Te ≈ 5680 KNm.

The value of equivalent torque can be rounded off to 5680 KN-m.

To get the tail shaft we apply the torsion and substitute the value of Te in place of T.

Applying torsion equation

�� = ��

P

�

�� ×�� (��.��)

= ��×��

We already know that the internal diameter (bore) of the hollow tail shaft is 1.22m.

1.1563 = ��.��

��

∴ Do4 = 3.627

∴ Do = 1.38m

3) Outside Diameter of hollow dowel pin (do)

Force on each hollow dowel = ��

(here equivalent torque is taken)

Force = �� ×��

�� ×�.� = 710 KN

di = 0.08 m (From diagram )


∴ 710 × 103 = 50 × 106 × �� × (do2 – 0.082)

∴ �� × ��

�� × �� × ��

+ 0.082 = do2

∴ do = 0.156 m

∴ do = 156 mm

∴ do ≈ 160 mm.

The outer diameter hollow dowel pin is rounded off to 160mm.

Equivalent torque is taken because dowel pins are subjected to both

torsion as well as bending.

4) Total Thrust

Propeller thrust = �� ×��

(GIVEN)

= �� ×�� ×�.�� ×�.��

Propeller thrust = 2612.752 KW.

7) Flange diameter is given by 2.2 times shaft diameter (here no. of bolts, n = 16)

∴Do ≈ 1.4m

Flange diameter, d f = 2.2 × Do

df = 2.2 × 1.4

df = 3.08m.

PCD = 1.6 × Do

=1.6 × 1.4 = 2.24 m.

Torque = Force × PCR

5680 × 103 = Force × �.��

∴ Force = 5.071 MN

��

= �.��×��

�� = 316.96 KN.

To find diameter of coupling bolts,


316.96 × 103 = 50 × 106 × �� × dB

∴ dB = 0.0898 m

∴ dB = 89.8mm

∴ dB ≈ 90mm

The coupling bolt diameter can be rounded off to 90mm.

8) Materials used for tail shaft are:

Tail shaft -- The tail shaft in this type of oil cooled arrangement is made of mild steel with highcarbon content. The tail shaft is subjected to shock and fatigue. High carbon content strengthens itagainst such loads.

Hollow Dowel – Hollow dowels used in this type of tail shaft are subjected to torsional loads.To ensure that these dowels work properly they are made up of high carbon content steel fordurability, ductility, and strength.

9) Purpose of Chrome Liner

Chrome lines are fitted in order to safeguard the tail shaft against grooving action of lip seal. Lipseals are provided in order to prevent the ingress of sea water to the engine room and also preventthe lube oil used for tail shaft lubrication from leaking. Hence, chrome liners are fitted on the tailshaft where the lip seal rubs the material by push fit on slide fit.

10) Periodicity of tail shaft (OG) survey and areas to be inspected:

Tail shaft survey—is done within a period of normally every 5 yrs.

Extension is allowed and survey is done in a period up to 7 to 10 yrs, depending on the type ofseals fitted under hydraulic jacking.

This could be the condition of tail shaft if not surveyed within the specified periodCondition – During every possible dry–docking of the vessel :

1) Circulating oil sample from stern tube is tested.

2) Propeller drop is checked.

When withdrawn, normally seals are changed along with chrome liners. Crack testing is done onshaft, meeting dowels, bolts, nuts, etc.

Now let us take a brief glance at Directorate General of SHIPPING (D.G.S.)requirements of items to be inspected during a tail shaft survey

EIGHT SCHEDULE[See rule 73(2) (h) and (i)]EXAMINATION OF PROPELLER SHAFTS

Part 1

The examination required to extend the interval between surveys as permitted shall include.-

(1) An inspection of the bearing oil to establish that it is not contaminated by water or debris.

(2) Measurement of the clearance between the shaft bearing and the shaft to ascertain that the wearis negligible.

(3) Removal of the propeller from the shaft to the extent that a full visual and non-destructive crackdetection inspection of the shaft by the forward end of the keyway can be made, and

(4) An inspection of the shaft sealing arrangements to establish that they will remain efficient forthe extended period.

Part 2


(1)An inspection of the bearing oil to establish that it is not contaminated by water or debris.

(2) Measurements of the clearance between the shaft bearing and the shaft to ascertain that the wearis negligible.

(3) Where the propeller is fitted to a taper on the shaft without a key, a visual and non-destructivecrack detection examination of the forward part of the taper to establish that corrosion or corrosioncracking has not occurred. Alternative methods of ascertaining that sea water has not penetrated theshaft taper/propeller boss bore and that corrosion or corrosion cracking has not occurred may beaccepted by the Chief Surveyor of the Government of India.

(4) Where the propeller is attached to the shaft by a bolted flange, a visual and a non-destructivecrack detection examination of the shaft flange radii and bolt hole bores and recesses.

(5) An inspection of the shaft sealing arrangements, which shall require dismantling the shaft seals,to the extent considered necessary by the Chief Surveyor of the Government of India to establishthat they will remain efficient for the extended period, and

(6) An inspection of the surface of that part of the shaft that normally lies within the aft part of theaft bearing to a distance at least equal to one-half of the shaft diameter.

Concept review questions:Q: SOME SCREW SHAFTS DO NOT HAVE LINERS FITTED, IN SUCH CASES HOW IS THE SHAFT ENDPROTECTED FROM SEA WATER? WHERE IS THE MOST LIKELY POINT OF INGRESS OF SEA WATER?

ANS: Where screw shafts are not fitted with liners, the shaft is oil lubricated and the stern tubebearings are white metal-lined cast iron bushes. In order to retain oil in the stern tube, the inboardend of the shaft is fitted with a mechanical seal which prevents the oil from running out. Seal in theaft end would be discussed later.

Q: HOW IS THE PROCEDURE OF OIL LUBRICATION CARRIED OUT IN STERN TUBE? WHAT ARE THE TWO

TYPES OF HEADER TANKS?

ANS : Oil is pumped to the bush through external axial grooves and passes through holes on eachside into axial passages. The oil leaves from the ends of the bush and circulates back to the pump

and the cooler. Oil pressure within the stern tube is maintained at approximately at the same levelas that of the surrounding sea water by a header tank.

The static lubrication system for vessels with moderate changes in draft, have header tanks placed2-3 m above the maximum load waterline. The small differential pressure ensures that water isexcluded. The cooling of simple stern tubes necessitates keeping the aft peak water level at least 1mabove the stern tube. Tankers and other ships with large changes in draught may be fitted with 2 oilheader tanks for either the fully loaded condition or ballast condition.

Q: WHAT IS THE MATERIAL COMPOSITION OF BEARING BUSH?

ANS: The bearing bush is normally of grey or nodular cast iron, centrifugally lined with white metal.A typical analysis of white metal would be 3% Cu, 7.5% antimony and remainder tin. White metal’sthickness is varies according to classification society specifications. Figures of 3.8 mm for a shaft of300 mm diameter to 7.4 mm for 900 mm diameter shaft have been quoted, with bearingclearances of 0.51 to 0.63 mm and 1.53 to 1.89 mm respectively. (NOTE:-These figures are only forbetter understanding)

Q:WHAT ARE THE TWO MOST COMMONLY USED TYPE OF OIL LUBRICATED STERN TUBE BEARING?WHERE IS THE MOST LIKELY POINT OF INGRESS OF SEA WATER?

ANS : Oil lubricated stern bearings use either lip seals or radial face seals or combination of the two.Lip seals, in which a number of flexible membranes in contact with the shaft are shaped rings ofmaterial with a projecting lip or edge which is held in contact with a shaft to prevent oil leakage orwater entry. Oil is contained within the simplex type stern tube by lip seals .The elastic lip of eachnitrile rubber seal, grips a rubbing surface provided by short chrome steel liners at outboard andinboard ends of the steel propeller shafts. A number of lip seals are usually fitted depending uponthe particular application.

LIP SEALSFace seals use a pair of mating radial faces to seal against leakage. One face is stationary and theother rotates. The rotating face of the after seal is usually secured to the propeller boss .thestationary face of the forward or inboard seal is the after bulkhead. A spring arrangement forcesthe stationary and rotating faces together. The aft end is the most likely point of ingress of sea-water to the screwshaft.

RADIAL FACE SEAL

What are the inbuilt safety precautionary devices fitted in case of St. L.O. P/P failure and sealfailure?

ANS : One of the two header tanks will provide a back pressure in the system and a period of oilsupply in the event of the pump failure. A low-level alarm will be fitted to each header tank. Oilpressure in the lubrication system is higher than the static sea water head to ensure that sea watercannot enter the stern tube in the event of seal failure.

PROTECTION MADE FOR LIP SEALS

What problem can arise out of the chrome liners and how can it be tackled?

ANS : The chrome liners act as rubbing surfaces for the rubber lip seals but grooving from frictionalwear has been a problem. The difficulty has been overcome by using ceramic filler for the groove oralternatively a distance piece to displace axially the seal and ring assembly. Allowance must bemade for the relative movement of shaft and stern tube due to differential expansion. New seals arefitted by cutting and vulcanizing in position.

WHAT KIND OF LUBRICANT CAN BE USED IN STERN TUBES WITH WHITE-METAL LINEDBEARINGS?

ANS : The lubricant used in stern tubes systems must have the ability to maintain alubrication film in the presence of water so that it is not washed away. The lubricant must also havegood affinity for metal surfaces so that it affords good protection of the metal in the stern tube andthe shaft areas against sea water. This affinity for metal surfaces so that it affords good protectionof the metal surfaces and is also necessary when the screw shaft starts to revolve, as boundarylubrication conditions are present at this time.

Compounded oils have these properties: they are blends of mineral oil and fatty oils .thefatty constituent causes water to physically to combine with the oil and form an emulsion. The fattyconstituent may be lanolin or synthetic fatty oil having similar properties.

The lubricant used have a specific gravity at 15.5 0C within the range of 0.92 to 0.95, with aviscosity REDWOOD 1 at 60 0C .The viscosity index of many brands of stern tube lubricant havewide ranges which is unfortunate , as the conditions under which they operate make a highviscosity index desirable

If leakage of sea-water into an oil lubricated screwshaft occurs, what indications will bethere? When is seawater leakage into the system most likely to occur?

ANS : Leakage of sea water into the stern tube is indicated by emulsification of the lubricant.the best time to carry out checks of the vessel is at ports. As the ship loads or discharges cargo, theaft draught of the vessel changes and the static head of sea water above the screw shaft outer sealwill change in a similar manner .if the pressure gauge showing the oil pressure in the stern tubesystem changes with the change of draught it may under many conditions indicate a defective outer

seal. Also when the stern tube system drain cock is opened and any water is drained off. If there is alot of water drained out ,it shows a seal leak

If the ship has a light aft draught and there is no, or only small ,current round the ship ,the sterntube system can be pressurized to some pressure greater than the head of water outside the outerseal.if the seal is defective oil globules will be seen coming to the surfaces from the seal.

How can outer seal leakage can be founded when at sea? What can be done to keep the system safewithout dry docking the vessel?

ANS : Outer seal leakage has often been found by an increase in the consumption of the stern tubelubricant. It is more likely to be noticed when the ship is proceeding in the tropical waters fromcold water areas. Leakage inboard is shown by oil emulsification and reduced oil consumption.

Why is cooling system required and how?

Heat produced by the friction will result in hardening and loss of elasticity of the rubber, shouldtemperature of the seal material exceed 110o C. Cooling at the outboard end is provided by the sea.Inboard seals, unlike those at the outboard end, cannot dissipate heat to the surrounding water. Oilcirculation aided by convection, is arranged to maintain the low temperature of the seals at theinboard end. Connection for circulation, are fitted top and bottom between the two inboard sealsand the small local header tank.

Is oil pressure dependent on shaft speed?

ANS : The requirement of steaming at a slow, economical speed during periods of high fuel prices (orfor other reasons) gives a lower fluid film or hydrodynamic pressure in stern tubes, due to theslower speed. The possibility of bearing damage occurring prompted the installation of forcedlubrication systems to provide a hydrostatic pressure which is independent of shaft speed. Thesupplied oil pressure gives adequate lift to separate shaft and bearing and an adequate oil flow forcooling.

How are such stern tubes supported?

ANS : The later designs of oil lubricated stern tubes are fitted in a stern frame with an elongated bossto provide better support for the white metal lined bearing. A minimum bearing length of twotimes shaft diameter will ensure that bearing load does not exceed 0.8 N/mm2.

The forward part of the stern tube is fabricated and welded direct to the extension of the sternframe boss and into the aft peak bulkhead. The outboard liner additionally protects the steel shaftfrom sea water contact and corrosion.

PROJECT # 3AIR STARTING SYSTEM

WARTSILA-SULZER 12RTA96C ENGINE, built by Doosan Heavy Industries,South Korea (a licensee of WARTSILA)

The picture shows WARTSILA-SULZER 12RTA96C SLOW SPEED 2-STROKE MARINE DIESEL ENGINE. This goliath engine producing power inexcess of 90,000 BHP propels ULTRA LARGE CONTAINER VESSELS like M.V.EMMA MAERSK at speeds of about 25 knots. To start this engine,compressed air at 25 - 30 bar pressure is used.

To discover more about air starting system of such mammoth enginesand how it is designed, take a look at the technical challenge put forth by thisproject.

INTRODUCTIONAir starting is the method used for starting large diesel engines (land

based /marine) by supplying compressed air into the cylinders in a fixedsequence (one after other). The compressed air is supplied by air

compressors, i.e. either diesel or motor driven. After a period of being atstandstill (i.e. stationary), the engine requires to be started with a high initialtorque at low revolutions in order to accelerate the engine rotating andreciprocating masses. The compressed air is then supplied by a large borepipe to a remote-operating non-return or automatic valve and then to thecylinder air start valve. The momentum built in the rotating elements of thecrankshaft will help in smooth starting once the initial inertia has beenovercome. The pressure of the starting air must be sufficient to impartenough speed to the engine pistons to quickly compress the combustion airsufficiently during the compression stroke for it to reach a temperature atwhich the combustion of the injected fuel initiates. The pressurerequirements will also vary according to engine size, design and servicerequirement and mainly working temperature.

An Air Compressor

The starting air is admitted from the air bottle to the engine cylinderthrough a starting air master valve common for all cylinders and cylinderstarting air valve mounted on cylinder head, one for each cylinder. Thedistributor ensures the air is introduced into the relevant cylinder at thecorrect time to achieve starting in the desired direction from any position ofthe engine at rest. The automatic master air starting valve is forced open bymain air acting against a collar and air is admitted to the starting air valves.Air is simultaneously admitted, via the distributor .To pressurize the top sideof piston operating air starting valve on cylinder head while bottom side ofthe piston is vented.

Air compressors preferably multi-stage types are used to provide air athigh pressures required for diesel engine starting. If all the air had beencompressed in a single-stage, it would unfortunately generate compressiontemperatures similar to those in a diesel engine. Multi-stage air compressorunits with various cylindrical configurations and piston shapes are used inconjunction with inter-cooling and after-cooling to provide the nearestpossible approach to the ideal of the isothermal compression. Such inter-cooling and after-cooling is achieved by water cooling or can be air-cooled(i.e. sea water can be used in water cooled type and fin type coolers in aircooled). Air cooling is achieved by multi-tubular heat exchangers where wateris circulated.

Schematic of position of starting air line in marine diesel engines

Air starting system

Technical Challenge:

It is required to design an Air starting system for a 6 cylinder 2-strokereversible diesel engine, consisting of two air storage receivers withhemispherical ends and having necessary fittings and condensers. Thecompressed air supplied through two single acting 2 stage compressors, eachof them capable of coping up with the starting air requirements at anaverage of one start every 3 minutes. As per specification rules, it isnecessary that the capacity of an air bottle should be such that 12 starts maybe given to the engine without replenishing the air bottle. Forprompt starting of the engine it is necessary that air be admitted to twocylinders consecutively for a period of 70o crank rotation at a mean pressureof 20 bars. The compressors are two stage tandem-type with H.P cylinder ontop having an intercooler and after cooler giving perfect cooling.

The following otherdata may be used:

1. Main engine bore 780mm2. Main engine stroke 1400mm3. Maximum air bottle pressure 30bar4. Minimum pressure allowed when the compressor

has to be started20bar

5. External dia. Of air bottles 1.5m6. Maximum allowable stress for air bottle material 90 MN/m2

7. Average engine room temperature 270C8. Compressor bore/stroke ratio(L.P. cylinder) 1.29. Index of compression and expansion.(compressor) 1.310. Rotational speed of compressor 750rpm11. Mechanical efficiency of compressor 85%12. Sea water inlet temperature to compressor 200C13. Sea water outlet temperature from compressor 350C14. specific heat of sea water 4.12 KJ/Kg K15. Compressor L.P suction pressure and temperature 1 bar and 250C16. Clearance volume in both the stages 3% of stroke

volume

I. Calculate

1) Capacity and overall length of air bottle.2) Thickness of air bottle, using thin cylinder theory.3) Capacity of each compressor at free air delivery conditions of

1.013bar and 15oC.4) The motor power required to drive the compressor.5) Compressor cylinder dimensions.6) Capacity of sea water cooling pump in Tonnes/hour.7) A spring loaded safety valve is to be fitted as a mounting to each

of the air receiver to blow off at 10% above maximum pressure.The diameter of valve is 40mm and maximum lift of the valve is10mm.Design a suitable compression spring for the safety valveassuming spring index 6 and stress concentration factor 1.25. Youare required to provide initial compression of spring 30mm.Maximum shear stress in the material of spring wire limited to450 MN/m2.

G=90 MN/m2; spring index=� ��

�

II Draw (hand sketch only)8) A section of the air receiver showing the weld details necessary.9) An outside view showing the mountings.

10) Comment on the material used and welding procedureadopted for construction of the air bottle.What are the safety features in its construction and during use?

The above figure shows a sketch of an air reservoir

GIVEN:

Main engine 6 cylinder, 2 stroke, reversible

2 main air bottles

Main engine bore, d = 780mm

Main engine stroke = 1400mm

Max. air bottle pressure = 30 bar

Minimum pressure allowed when the Compressor has to be started= 20bar

External diameter of air bottles = 1.5m

Max. Allowable Stress for air bottle material = 90MN/m2

Average engine room temperature = 27oC

Compressor bore/stroke ratio (L.P. cycle) = 1.2

Index for compression and expansion (comp.) = 1.3

Rotational speed of compressor = 750 rpm

Mech. Efficiency of compressor = 85%

Sea water inlet temperature to compressor = 20oC

Sea water outlet temperature from compressor = 35oC

Specific heat of sea water = 4.12KJ/kg K

Compressor L.P. Suction pressure & temp. = 1 bar & 25oC

Clearance volume in both the stages = 3% of stroke volume

� 2main air, 2 stage tandem with H.P. cycle on top� 12 starts continuously and one start in every 4 minutes

I. CALCULATE1. Capacity and overall length of the air bottle

We know that

Stroke length = 2 x crank radius

L = 2 x R

R = L / 2 = 1400 / 2 = 700mm

From diagram,

AB = r - OB

= r r cos Ѳ

= r (1 – cos Ѳ)

= 700 (1 - cos70)

= 460.58 mm

AB = 461mm

Volume of 1 cylinder =π� x D2 x AB

= π4 x 0.782 x 0.461

= 0.22 mm3

According to the firing order air is injected in 2 cylinders at start,

Volume = 0.22 x 2 = 0.44 m3

Since 12 starts are required from one air bottle

= 12 x 0.44 = 5.28 m3

Let v be the capacity of the air bottle

30 x V = 20(V+5.28)

V= 10.57m3 ≈10.6m3

2. Thickness of air bottle using thin cylinder theoryAccording to thin cylinder theory,

F= � ��

Where,F = max. allowable stress

P = max. test pressure = 1.5 x max. air bottle pressure= 1.5 x 30 = 45 bar

t= thickness

Substitute the values

90x106 = �� (�.� � ��)� �

40t = 1.5 - 2t

T = 35.7mm

Thickness of cylindrical portion is 35.7 mm

Inside diameter of cylindrical portion of air bottle,= 1.52 x 0.0357

D1 = 1.4286m

Thickness of hemispherical portion, given by formulae

t= Pd1K/2t +0.75x10-3

where,

P = 45x105

K = a constantt = max. allowable stress = 90x106N/m2

t = �� .�� .��

t = 18.6mmInside diameter of the hemispherical end=1.5-2x0.0186=1.4628m

The length of cylindrical portion,

v=π� d1

2 l + 4/3π(d1/2)3

10.56 = 1.6027 x l + 1.6389L = 5.57mThe overall length of air bottle,loverall = l+1.5

= 5.57+1.5

= 7.1m

The overall length of the air bottle is 7.1m

3.) Capacity of each compressor of free air deliveryconditions of 1.013 bar and 15oC.Volume of air following to the cylinder per second,v= 0.44 / 4 x 60 = 1.833 x 10-3 m3/secMass flow rate of air at starting of compressorP V = m R TWhereP = pressure = 20x105N/m2

V = vol. of the air flowing per second to cylinderm= mass flow rateR = gas constantT = temperature = 27oC = 300Km= P V/R T= 20 x 105 x 1.833 x 10-3/ 287 x 300

= 0.0425 kg/sec

Then capacity of each compressor at free air delivery conditionof 1.013 bar and 15oC,Capacity = m R T /P

= 0.0425 x 287 x 288 /1013 x 105

= 0.0347m3/sec= 125m3/hr

The capacity of each air compressor cannot go beyond150m3/hr.

4.) The motor power required to drive the compressorWork done by the compressor,W = 2 n / (n-1) x R T1 [(P2/P1)

n-1/n-1]WhereP2= Intermediate pressure= √ (P1P3)= √ (1x30)= 5.47bar

H = 1.3T1 = 273 + 25

= 298KR = Gas constant = 0.287 KJ/KgKP1 = 1 barW = 2x 1.3/1.3-1 x 0.287x298x [(5.47/1)1.3-1/1.3-1]

=355.896KJ/KgW x mass flow

= 355.896x0.0425= 15.125KW

Therefore, the motor power required to drive the compressor,= Work done/mech. Efficiency

= 15.125/0.85= 17.794KW=18KW

5.) Compressor cylinder dimensionsFrom P-V diagramV1-V5 = Swept volume

= (V1-V6) + (V6-V5) ---1V5= clearance volume both stageV5 = 0.03 x (V1-V5) ----2And P1 (V1-v6) = mRT1 ----3WhereM=0.0425 Kg/secR=287 J/ KgKT1= 25oC

= 25+273=298KP1= 1barSubstitute in eqn. 31x105 (V1-V6) = 0.0425 / 750 x 60 x287 x 298V1-V6 = 2.9078x10-3 m3

For low pressure cylinder,Polytropic Process, P5V5

n = P6V6n

[P5/P6]1/n-1 = V6/V5 -1

[5.47/1]1/1.3 -1 = V6-V5/V5

V6-V5 = 2.7 V5

From eqn. 2V6-V5 = 2.7 x 0.03 x (V1-V5)From eqn. 1V1-V5 = 2.9078x10-3 + 0.08102(V1-V5)

V1-V5 = 3.164x10-3m3

Since swept volumeV1-V5 = π/4 d

2l andBore/stoke ratio, d/l = 1.2d= 1.2lV1-V5= π/4 (1.2l)

2 x l = 3.164 x 10-3

L3=2.27x10-3

Stroke of L.P. Cylinder, l = 140.9mm = 141mmFor high pressure cylinder,Swept volume,V2

’-V4 = (V2’-V5

’) + (V5’-V4) ----4

P2 (V2’-V5

’) = mRT2’ ----5

V4 = 0.03 (V2’-V4) ----6

Wherem= 0.0425 Kg/sR = 287 J/Kg KT2/T1 = (P2/P1)

n-1/n

=[5.47/1]1.3-1/1.3

T2=441.26 KAnd T2 = T2

’

And P2’ = P2 = 5.47 bar

Since,P4V4

n = P5‘V5

,n

[P4/P5]1/n – 1 = V5

’/V4 - 1V5

’-V4 = 2.69V4

From equation 4 and 6V2

’- V4 = 7.853 x 10-4 + 0.0807(V2’-V4)

V2’ - V4 = 5.772x10-4 m3

Since, stroke of the H.P. and L.P. cylinder are same

i.e. l= 141 mmSwept volume of the L.P. cylinder5.772 x 10-4 =π/4 d2 x 141D2 = 5.212 x 10-3

Diameter of H.P. cylinder,D= 72.1mm = 75mm

6.) Capacity of the sea water cooling pump intones/hourLetm= mass flow rate = 0.0425Kg/sCp= specific heat at the constant pressure = 1.005H1 = enthalpy of the air at inletH2 = enthalpy of air at outletT1 = temp. of air at suddenT2 = temp. of air at dischargeQ = heat input during polytropic compressionW = Indicated work outputQ = (h2-h1) + W

= mCp(T2 – T1) – 15.125/2= 0.0425 x 1.005(441.25 – 298) – 7.5625= 6.1189 – 7.5625

Q = -1.4435 KJThis is the heat rejected by air which is further received byJCWQ1 = 2 x 1.4435 = 2.887KWFor perfect inter cooling, T2

’ = T1

Q2 = mCp (T2 – T2’)

=0.0425 x 1.005 (441.26 – 298)

=6.118 KWFor perfect after cooling , T3 = T2 & T2

’=T1

Q3 = m Cp(T3 – T2’)

= 0.0425 x 1.005(441.26-298)= 6.118KW

Total heat rejected to sea water,Q = Q1 + Q2 + Q3

=2.887 + 6.118 +6.118=15.14 KW

Heat rejected = m Cw (T2 – T1)Hence, the capacity of sea water cooling pump,15.14 = m Cw (T2-T1)Wherem= mass flow rateCw= sp. Heat of sea water = 4.12 KJ/KgKT2 = seawater outlet temp. from compressor =35oCT1 = seawater inlet temp. to compressor = 200C15.14 = m x 4.12 (35-20)m = 15.14 / 4.12 x 15

=244.98Kg/s

m= 0.881 tonnes/hr is the capacity of seawater cooling pump

7.) A spring loaded safety valve is to be fitted as amounting to each of the air receiver to blow of at 10%above Max. pressure. The diameter of valve is 40mm andthe Max. lift of the valve is 10mm.

Design a suitable compression spring for the safety valve assumingspring index 6 and stress concentration factor 1.25%. You arerequired to provide initial compression of spring 30mm.

Max. shear stress in the material of spring wire limited to 450MN/m2

G= 90 MN/m2; spring index = coil diameter/wire diameter

Diameter of valve, d1 = 40 mm

Max. Lift of the valve, S2 = 10mm

Max. Pressure, P = 3.3N/mm2

Spring index, C = D/d =6

Where,

D= Coil diameter

d= wire diameter

Stress concentration factor, K = 1.25

Initial Compression of spring, S1 = 30 mm

Max. Shear stress in the material of spring wire,

τ= 450MN/m2

Modulus of rigidity, G= 90 GN/m2

Maximum Load, W1 = π/4 d12x3.3

= π/4 x 402 x3.3 = 4146.9N

Max. Compression of the spring

δ max. = δ 1+δ 2

=30 + 10

= 40mm

Load of 4146.9 keeps value on set by providing initial compressionof 30mm, therefore max. load on the spring when the value is openi.e., for max. compression of 40mm.

W= 4146.9/30 x 40

=5529.2N


Max. Shear stress , τ

, τ= K x 8 WC/πd2

450= 1.25 x 8 x 5529.6 x 6/πd2

D2 = 234.66

D = 15.31mm

For consideration the wire diameter is, d = 15.4mm

Mean diameter = spring index x wire diameter

= 6 x 15.4

= 92.4mm

No. of active turns = n

Max. compression of the spring(δmax.)

δmax. = 8WC3n/Gd

40 = 8 x 5529.6 x 62 x n/ 90x103x15.4

N = 5.8

Say 6 turns

Taking the ends of the coil as squared and ground the total no. ofturns.

n’ = n +2 = 6+2 =8

Free length of the spring

LF = n’d + δmax.+ 0.15 δmax.

= 8 x 15.4 + 40 + 0.15 x 40

= 169.2mm

Pitch of the coil = free length / n’-1

= 169.2/8-1

= 24.17mm

Q. 8) A section of the air receiver showing the weld detailsnecessary.

Q9.) An outside view showing the mountings.

Q 10.) Comment on the material used and welding used andwelding procedure adopted for construction of theair bottle. What are the safety features in itsconstruction and during use?

The material used for construction of large or small reservoirsis of good quality mild steel plate having similar specification toboiler plate material. The steel will have an ultimate tensilestrength within the range of 360 MN/m2 to 500MN/m2 will have anelongation of not less than 23% to 25%. Large starting airstorage reservoirs have dish ends; one end has an openingformed within lip to take an elliptical manhole door. The fore ofthe dished end may be either tori-spherical or semi-spherical.

The dished ends are usually made by the spinning process.The edge left by the spinning is machined and tapered down to thethickness of the cylindrical shell. Theends are welded to the cylindrical shell by full penetration welds.The longitudinal seams are machine welded. The circumferentialseams where the dished ends join the cylindrical shell may beeither machine or hand welded. Smaller air bottles havehemispherical ends and inspection holes may be fitted in thecylindrical shell. The air compressor is useon board the ship is of reciprocating-tandem or inline type.Compressor valves are normally plate type annular and springloaded. Air passes through tubes inintercooler, the water jackets being protected from over-pressureby fitting a bursting disc. Compressor air discharge must lead to airreceiver directly. No interconnection allowed with engine starting airline. Relief valves are to be provided in all stages as per stagepressures.

Air receiver should have manhole doors with positive closing.Normally, air bottles are mounted on foundation with inclination toaft. Drains must be provided in each stage of compressor and airreceiver. Relief valve fitted should have its discharge leading toopen deck.

CONCEPT REVIEW QUESTIONS

Q1. What are the uses of compressed air onboard ship?

Ans : It is used for starting main and diesel engines in motor ships and forauxiliary diesel engines of steam ships. Also control air of low pressure isrequired for ships of both categories for control equipment andinstrumentation purposes. Auxiliary boilers and economizers are fitted withsoot blowers which use compressed air .Portable tools such as drillingmachines, impact wrenches, torque wrenches, hard grinders and lifting gearsuse compressed air. Compressed air can also be used for chipping, scalingmachines paint spraying equipments.

Q2.Why multistage compressors preferred over single stage? What is the useof inter cooler and after cooler?

Ans : For a single stage air compressor, acting as isothermal duringcompression by provision of perfect cooling is an ideal situation which is verydifficult to obtain. Moreover the necessary final pressure of compressed airis achieved by the number of stages. For higher pressure, more number ofstages are required .When the air is compressed in stages; it is easier tocontrol the temperatures during its passage through the compressors.

This can be achieved by water jacketing the air compressor cylinders andpassing the air through heat exchangers (intercoolers). As the air leaves eachstage of compressor, it is cooled in the intercooler .This lowers the work donein compressing the air and prevents a lot of mechanical problems which couldarise if the air temperature were uncontrolled.

Apart from practical considerations and the as abovequestion, a 3 stage air compressor is more desirable and requires less

energy or work input than a single stage air compressor when compressingair over the same pressure range.

Q3.What is the basic principle of working of an air compressor?

Ans : Compressor produces high compression temperature approx. equal tothe one in the diesel engine which is sufficient to ignite vaporized oil . Theheat produced in a single stage of compression would be wasted and couldadd energy and produce a resultant rise in pressure apart from the pressurerise expected from the action of piston .However the air cools, the pressurerise due to the heat generated is lost. Only the pressure from compressionremains .The extra pressure due to heat is of no use and actually demandsgreater power for the upward movement of the piston through thecompression stroke.

Q4. Factors affecting volumetric efficiency of an air compressor.

Ans : The factors affecting volumetric efficiency of an air compressor are asfollows :

i. The clearance between the cylinder cover and the end of the pistonwhen the piston is at end of its discharge stroke. The larger theclearance, the less air is discharged per stroke.

ii. Sluggish opening and closing of suction and delivery valves.iii. Leakage past the compressor piston rings.iv. Insufficient cooling water (or) the cooling water inlet temperature is too

high.v. Inlet temperature of the air to the first or low pressure stage

compressor too high.vi. Throttling of air supply to L.P. suction –example dirty air inlet strainers.

Q5. How is clearance volume of air compressor checked?

Ans : This check is made by making up a small ,loosely woven ball of leadwire .This is placed on the top of the piston which will have been moved alittle off the end of the stroke .The cylinder cover is replaced on the cylinderwith a joint in place and tightened down. The compressor is then barredslowly over top centre so that the ball of lead wire is compressed .After it isremoved from the top of the piston it can be measured by a micrometer .Themechanical clearance measured is compared with the compressormanufacture’s recommendations. Adjustments are made by altering the coverjoint thickness or by fitting or removing shims between the foot of theconnecting rod and the bottom end bearing.

Note: The ball of the wire must be placed centrally on the top of the piston; ifthe wire is on one side of the piston, there is a possibility that the piston rodcould bend when the piston is barred over the top centre position.

Q6. How many starting air compressors are fitted on board ships?

Ans. It is usual to have atleast 2 starting air compressors and sometimesthere are more than two .The compressors may be independently driven byan electric motor or steam engine .Also there is an emergency air compressorwhich may be diesel engine driven. In case of air in air bottle and at thesame time crew are facing ‘black out’ condition then the diesel enginecompressor comes into play.

Q7. What are the mountings fitted on air bottle and air compressors?

Ans. Safety valves are normally fitted to the air bottles but in someinstallations the reservoirs are protected against over pressure by those ofthe compressors .There is a requirement that if the safety valves can beisolated from the reservoirs the latter must have fusible plugs fitted torelease the air in the event of fire. Reservoirs are designed ,built and testedunder similar regulations to those for boilers .Many air reservoirs are also

fitted with other outlet valves which have connections with other air systemssuch as auxiliary engine starting systems ,instrument air ,workshop airservices, ship’s whistle, filter cleaning systems ,emergency air supply to boilerfeedback pump.

Excess air pressure is prevented by spring loaded relief valves on airbottle. In some cases the relief valves are fitted on the common dischargeline from the compressors and fusible plugs are fitted on the air reservoirs.

The fusible plugs prevent serious pressure rise if a fire should breakout near the air bottle. Drain valves or cocks are fitted to the bottom of thereservoir to drain off oil or moisture carried over with the air from thecompressor. An independent connection is fitted for the pressure gaugesused to indicate the air pressure within the reservoir.

Q8. Why are the outlets from relief valve and fusible plug from air bottle letoutside the engine room?

Ans : Fusible plugs are fitted if the receiver is isolable from a relief valve.There will always a relief valve on the high pressure side of the compressorso that when the compressor is being used, the bottle is protected. However,this means that the receiver is only protected when the compressor isrunning. A fusible plug therefore offers protection against pressuredevelopment in the event of an engine room fire. The fusible plug thereforeoffers protection against pressure development in the event of an engineroom fire. The fusible plug (lead, bismuth and antimony) softens as itstemperature rises and extrudes from its fir tree type sockets.

Q9. What is the implication of starting air reservoir on fixed fire fighting CO2

system?

Ans : The release of large amount of air in case of engine room fire wouldimpair the effectiveness of any CO2 fixed fire fighting system gas unless extragas is provided in compensation or the air is piped out of the engine room.

Q10. At what position of piston in the cylinder is the starting air introduced?

Ans : Modern practice is to introduce air into the cylinder slightly before TDC(the alignment of piston rod with the connecting rod at this point is such thatlittle if any turning moment is developed). This allows the air to accumulatein the clearance volume ready to force down the piston once it is over TDC.At the same time, another cylinder will be receiving air (because of overlapdiscussed in next question).This unit will be one in which the crank is wellpast TDC so that it generates an adequate turning moment to carry theabove unit over TDC. The first unit, already pressurized, will be able toaccelerate the engine up to the fuel initiation speed. The useful expansion ofthe starting air will cease at the opening of the exhaust to continue airinjection any further would be futile. This limit is normal to 3-cylinder enginesbut is unnecessarily long in engines with more than three units.

A starting air pressure well below the compression of an enginewill be able to turn the engine over against the compression because thecompression pressure is only reached towards the end of the stroke, whereasstarting air is introduced for a much longer period of the stroke.

Q11. In the above project, why is the period of admission of compressed airfor 70 degree of crank rotation?

Ans : In order to conserve starting air, starting valves are designed to closeas early as possible consistent with good starting and some explosion of thestarting air then takes place. The opening and closing of starting air valves iscontrolled by the cam (operating within the starting air distributor) actuatingthe distributor valve.

To enable a propulsion engine to be started from anycrank position, overlap is necessary in the timing of the starting valves of anengine. Overlap is necessary in the timing of the starting valves of an engine.Overlap occurs during the period that any two valves are open, it beingintroduced that one valve will be opening whilst the other valve is closing.One valve will then always open when air is put on the engine to start somemaneuver. If there were no overlaps in the valves it would be possible for theengine to stop in some position where all the valves remained closed whenair was put on the engine.

The amount of overlap is dependent upon the number of cylinders, thetiming of the exhaust opening and so on (the greater the number ofcylinders, the less overlap required)

Over lap calculation:

In the above project there are 6 cylinders, with a obvious total rotation of3600 of crank.

Therefore an equal admission for a period of 600 of crank rotation in eachcylinder, at the same time there needs to be a overlap for atleast 100 of crankrotation for the advantages of overlap stated just above.

In the above example the firing order 1-5-3-6-2-4 indicates the order inwhich these units would be receiving the starting air during the air startingoperation. If for example unit number 5 starting air valve has opened and theengine is rotating in clockwise direction. Before this valve shuts the startingair valve of unit 3 will open, unit 6 will be followed by unit 4 and so on (foranticlockwise rotation the order of air entry will be 1-4-2-6-3-5 i.e. reverse toclockwise timing order).

It is evident from above that no matter in what position the enginestops, there will always be at least one of the cylinders with its starting airvalve opened to admit, on starting, the compressed air to start the engine.

Q12. What are the interlocks in engine, air starting system?

Ans.) The starting interlocks prevent the engine being put on fuel before allthe sequences of the starting system have been completed.

With the systems controlled by the operation of hand lever, the interlocksmay be cams or pins which lock and prevent hand lever movement. In theengines started by hand wheel controls the interlocks ate often slotted discs(fitted on the wheel shafts) and the small levers which engage or clear theslots in the disc.

Some engines have a blocking device connected with the ships E/Rtelegraph which for events the engine being put as when an ahead order is

given and vice versa. Blocking devices are fitted to the engine turning gear sothat the engine cannot be inadvertently started with the turning gear in

In the event of E/R telegraph failure, any interlocking and blockingdevices operated from the telegraph would prevent the engine beingmaneuvered during the time of emergency. It is therefore important to knowhow the interlock and blocking devices may be overridden so that the enginecan be maneuvered under emergency conditions with orders via the bridge toengine room telephone.

Q 13.Why should there be minimum of 12 starts in case of reversible engine?

Ans : Reversibility can be achieved by introducing air into the cylinder wherethe piston is approaching TDC in the direction of rotation in which it wasstopped. During maneuvering the engine is frequently subjected to aheadand astern starts. It is also an important requirement which is checked by asurveyor before entry of vessel in narrow channels like Panama Canal. Theengine should be capable of being started 6 times in ahead direction and 6times in astern direction without refilling the air bottle. This is necessary toavoid the ship being stranded in the canal or locks in case of failure of aircompressor.

Q 14.How does working temperature of the engine affect the starting airpressure?

Ans : An engine at working temperature will start with a lower starting airpressure than a cold engine. The temperature of the lubricating oil suppliedto the engine bearings will also have an influence on the minimum requiredstarting air pressure. When the lubricating oil is at working temperature, theengine will swing more easily than when lubricating oil is cold; hence a lowerair pressure will start an engine with lube oil at working temperature.

During trials of newly built ships, it is a part of the main enginemaneuvering tests to put one starting air reservoir on to the starting systemand continue maneuvers ahead and astern until the pressure falls to thepoint that the engine cannot start. The pressure is recovered with the ship’strial.

One of the principle factor is to overcome the forces of adhesion b/wthe bearing surface due to the presence of cold lubricating oil b/w them. Thewalls of the combustion chamber being cold, there will be a greater heat flowto lower the temperature of the compression at the time of start. An enginewhich has been warmed up at the metallic surface of the combustionchamber, viscosity of lubricating oil lowered, will have the starting speedreached earlier.

Q 15.What is Starting Air Line Explosion? What are the safety devices toavoid it? What are starting line safety devices?

Ans : Explosions can and do occur in diesel engine starting air system. Alsoair start valves and other parts are sometimes burned away withoutexplosion. These problems have been caused by cylinder air start valveswhich have leaked or not closed after operation and have allowed accessfrom the cylinder to the air start system of the flame from combustion.Carbon deposits from burning fuel and oily deposits from compressor areavailable as substances which may be ignited and produce an explosion inthe air start system. If no explosion occurs, the flame from the cylinder andhigh temperature air from compressor can cause carbon deposits in thesystem to burn. Careful maintenance of air start valves, distributors andother parts is vital as is regular cleaning of air start system components toremove deposits.

The lubrication of components is limited as excess lubrication couldcause the air start valves to be stuck by grease which has become hardenedby the heat and oil could accumulate in the pipes from this source. Thedraining of compressor coolers and air receivers is important. Drains on airstart systems are also checked. Flame traps or bursting caps are fitted ateach air start valve.

PROJECT # 5

MARINE STEAM TURBINE UNIT

The world remembers the sinking of R.M.S. TITANIC, but that R.M.S. standsfor Royal Mail Steamer. The 269.1m vessel was equipped with tworeciprocating four – cylinder, triple expansion steam engines and one lowpressure Parson’s turbine each driving a propeller. There were 29 boilers firedby 159 coal burning furnaces that made possible a top speed of 23 knots(43 km/h; 26 mph). On paying special attention to the latter, the followingproject deals with designing of a steam turbine. Why has the triple expansionsteam engine being replaced by steam turbine, the answer is in this project.

To discover more about steam turbines of seagoing vessels and how it isdesigned, take a look at the technical challenge put forth by this project.

INTRODUCTION

A steam turbine is a mechanical device that extracts thermal energyfrom pressurized steam, and converts it into rotary motion. Its modernmanifestation was invented by Sir Charles Parsons in 1884. The speed of the

steam jet is dependent on the rate of expansion, it had long been known andbuilt into reciprocating engines, that a vacuum, or partial vacuum, at theexhaust end of the engine would create a higher pressure ratio betweeninput and output and so impart more energy to the pistons, this vacuum isthe main function of the condenser in which cooling steam is used to create adrop in pressure below that of atmosphere.

Parson’s idea was to reverse that function and create an exhaust thatwas pressurized, but still below that of the feed. This effectively resulted in asmaller pressure drop between feed and exhaust and a slower steam jet. Byrepeating this process a number of times most of the energy from the steamjet can be extracted without the turbine having to destroy itself.

A simple turbine schematic of the Parsons type, rotating and fixedstators alternate and steam pressure drops by a fraction of the total acrosseach pair, the stators grow larger aspressure drops.

There are 2types of turbinesdepending on theiroperation, i.e. animpulse turbine has

fixed nozzles that orient the steam flow into high speed jets. These jetscontain significant kinetic energy, which the rotor blades, shaped likebuckets, convert into shaft rotation as the steam jet changes direction. Apressure drop occurs across only the stationary blades, with a net increase insteam velocity across the stage. As the steam flows through the nozzle itspressure falls from inlet pressure to the exit pressure (atmospheric pressure,or more usually, the condenser vacuum). Due to this higher ratio ofexpansion of steam in the nozzle the steam leaves the nozzle with a veryhigh velocity. The steam leaving the moving blades has a large portion of themaximum velocity of the steam when leaving the nozzle. The loss of energydue to this higher exit velocity is commonly called the "carry over velocity" or

"leaving loss".

In reaction turbine, the rotor blades themselves are arranged to formconvergent nozzles. This type of turbine makes use of the reaction forceproduced as the steam accelerates through the nozzles formed by the rotor.Steam is directed onto the rotor by the fixed vanes of the stator. It leaves thestator as a jet that fills the entire circumference of the rotor. The steam thenchanges direction and increases its speed relative to the speed of the blades.A pressure drop occurs across both the stator and the rotor, with steamaccelerating through the stator and decelerating through the rotor, with nonet change in steam velocity across the stage but with a decrease in bothpressure and temperature, reflecting the work performed in the driving of therotor.

Let U=Blade speedCi= velocity of steam at inlet to blade, i.e. leaving nozzle (giving nozzleangle)Ci rel= velocity of steam relative to the blade (giving blade inlet angle)Co = Velocity of steam at outlet of blade

Parsons Impulse-Reaction

The original blade design was thin section with a convergent path. Thedesigned blades similar to bull nose impulse blades which allowed for aconvergent-divergent path. However due to the greater number of stages thesystem did not find favor over impulse systemsU/Ci = 0.9

If the heat drop across the fixed and moving blades are equal thedesign is known as half degree reaction. Steam velocity was kept small onearly designs; this allowed the turbine to be directly coupled to the propshaft.

Increased boiler pressure and temperature meant that the expansionhad to take place over multiple rotors and gear set. As there is full admissionover the initial stage, blade height is kept low. This feature alone causes adecrease in blade and nozzle efficiency at part loading. In addition, althoughclearances at the blade tips are kept as small as practical, steam leakagecauses a proportionally higher loss of work extracted per unit steam. Bladetip clearances may be kept very tight so long as the rotor is kept at steadystate. Manoeuvring, however, introduces variable pressures andtemperatures and hence an allowance must be made.

End tightening for blades is normally used. This refers to an axialextension of the blade shroud forming a labyrinth. When the rotor is warmedthrough a constant check is made on the axial position of the rotor. Onlywhen the rotor has reached its normal working length may load beintroduced. Alternatively tip tightening may be used referring to the use ofthe tips of the blade to form a labyrinth against the casing/rotor. This systemis requires a greater allowance for loading and is not now generally used. Tokeep annular leakage as small as possible these rotors tend to have a smallerdiameter than impulse turbines. To keep the mass flow the same with theincreasing specific volume related to the drop in pressure requires anincrease in axial velocity, blade height or both -see above. Altering the bladeangle will also give the desired effect but if adopted would cause increasedmanufacturing cost as each stage would have to be individual. Generally therotor and blading is stepped in batches with each batch identical.

The gland at the HP end is subjected to full boiler conditions and issusceptible to rub. The casing must be suitably designed and manufacturedfrom relevant materials.A velocity compounded wheel is often used as the first stage(s) giving a largedrop in conditions allowing simpler construction of casing and rotor andreducing length. Special steels are limited to the nozzle box.

TURBINE CONSTRUCTION

Vertical Casting

Only the bottom part of the ingot isused.

Rough ForgingIt is a requirement that forgings are heavily worked. Any small holes ordefects can become hammer welded together. No forging is carried outbelow the plastic flow temperature as this can lead to work hardening.

Forging will allowcontinuous grain flow.

Ultimate tensile stress andelongation checked. Thismust be near enough equalin all 3 directions.

After roughmachining it is put in for athermal stability test. Forthis final machining is givento the areas indicated. The

end flange is marked at 90' intervals. Then the rotor is encased in a furnace.Pokers are placed onto the machined areas and accurate micrometerreadings taken. The rotor is rotated though 4 positions marked on the flange.The rotor is then heated to 28 0 C above normal operating temperature andslowly rotated.

Measurement is then taken at hourly intervals until 3 consistentreadings are taken (hence the rotor has stopped warping). The rotor is thenallowed to cool and a set disparity allowed.For turbine sets operated at greater than 28'C above their designedsuperheat then run the risk of heavy warping as well as high temperaturecorrosion and creep.Final machining is now given. The rotor is statically balanced and thendynamically balanced and check to ensure homogeneity. The rotor is bladedthen again dynamically balanced.

TECHNICAL CHALLENGE

In a ship’s propulsion steam turbine installation following dataavailable:-

The turbine is a multistage pressure compounded impulse type witheach stage producing equal power. No steam is bled of at any stageof the machine.

1. Turbine rotor speed - 6000 r.p.m.2. Maximum shaft power - 13.5 Mw3. Input steam conditions - 55 bar 5000C

4. Exhaust steam conditions - 2 bar 1500 C5. Steam consumption - 78 tonnes/hr6. Nozzle angle for all stages - 200 C7. Diameter of forged discs for all stages - 540 mm8. Overall length of shaft between inner edges of bearing -

1290 mm9. Density of rotor and blade material - 7856 kg/m3

QUESTIONS:-

1. Calculate velocity of steam at the end of 1st stage nozzle box.

Given Pressure at the outlet of the nozzle = 40 barNozzle efficiency= 93%

2. Calculate the blade angles assuming blades are symmetricaland mean blade speed for the first stage = 175 m/s

3. Calculate power developed per stage considering blade frictionfactor = 0.9

4. Calculate blade efficiency.5. Calculate 1ST stage blade height.6. Calculate number of stages required for developing the

required power.7. Calculate the blade height and radial stress exerted by the

blades in last stage.8. Show that the rotor shaft can be designed as 140 mm

diameter. Take allowable shear stress = 41MN/m2

9. Calculate mass of rotor shaft between the bearing supports,frequency of transverse vibration and first critical speed of theshaft in r.p.s.Assumptions: -

(a) Exclude mass of the disc(b) Rotor shaft mass as uniformly distributed E = 200 X 109

N/m2

10. Determine natural frequency of transverse vibration ofthe loaded shaft and the first critical speed in r.p.s. Hencecalculate the critical speed of the complete rotor system.

11.a. Discuss important properties of ideal blade materialb. What is blade erosion and how is it prevented?c. Find the length of the bearing. Allowable pressure in the

bearing = 254KN/m2

SOLUTION:-Notations used:-C a i = velocity absolute at inletC a e = velocity absolute at exitC b = blade velocity = constantC w = whirl velocityβ i = blade inlet angleCf i & Cf e = inlet and exit flow velocity respectivelyCr i & Cr e = inlet and exit resultant velocity respectively

Calculate:-

1)Velocity of steam at the end of first stage nozzle box.Given: pressure at outlet of nozzle = 40 bar

Nozzle efficiency = 93 %

ð Pressure at the outlet of the nozzle = 40 barFrom steam tables,

Enthalpy at nozzle inlet, h1 = 3433.3 KJ/kg (at 55 bar 5000 C)

Enthalpy at nozzle outlet, h2 = 3329.97 KJ/kg & h2s = 3322.2 KJ/kgSince, nozzle efficiency = ��

��h1 - h2 = 0.93 (h1 – h2s)

= 0.93 (3433.33 – 3322.2)h1 – h2 = 103.32 KJ/kg

Velocity of steam at the end of 1st stage nozzle box,

C a i =�2 X (h1 – h2)

= �2 X (103.32 X 10)= 454.58 m/s

2)Calculate blade angles assuming blades aresymmetrical and mean blade speed for 1st stage = 175m/s

ð C b for first stage = 175 m/s & Nozzle angle ∝ = 200

From the combined velocity diagram

Sin ∝ = ��

Cf i = Ca i × Sin ∝

= 454.58 Sin 20

= 155.47 m/sAnd also,

Cos ∝ = ��

Cos 200 = �� .��

BE = 252.07 m/s and then

Tan β i =��

β i = tan�� ( ��

) = tan�� ( ��.��.��

)

β i = 31.67 0

3)Calculate power developed per stage considering bladefriction factor = 0.9

ð We know that ,

Blade friction factor = ��

0.9 x Cr i = Cr e

From the combined velocity diagram,Whirl velocity, Cw = Cr i X cos β i + Cr e X cos β e

Where Cr i =��

��β �

= ��.��.��

Cr i = 296.26 m/s

Cr e = 0.9 x 296.26 = 266.7 m/s

On substituting the values, we getCW = (296.26 X cos 31.67) + (266.7 X cos 31.7)

= 479.23 m/s

Steam consumption per hour = 78 tonnes.

Then steam consumption per sec = ��

= 21.67 kg/s

Power per stage = m (Cb X CW)= 21.67 (175 X 479.23)= 1.81 MW

4)Calculate blade efficiency

ð Blade efficiency = ��

Where, power available = �� m CW

2

= �� X 21.67 X 479.232

= 2.238 MW

Substitute the values in above equationBlade efficiency = �.��

�.��

= 0.8087= 80.87 %

5)Calculate first stage blade height

ð Assuming 25% of the annular circumference is covered,

Coverage of steam area wise = circumference of thenozzle x height of blade

= π D X height

Where D = diameter of forged discCoverage of steam area wise = π x 0.54 x height

Area = 1.696 x height x 0.25

We know that, area x velocity of flow (axial velocity) Cf i = volumeof flow of steam

From steam table at 40 bar, specific volume = 0.08634 m3/kg.

Substituting the value, we get,1.696 x height x 0.25 x 155.47 = 0.08634 x 21.67

Height of blade in first stage = 28.38 mm ≈ 28.5 mm

6)Calculate number of stages required for developing therequired power

ð From the steam tables , hi = 3433.32 KJ/kghe = 2768.5 KJ/kg.

then total power = m (hi - he)

= 21.67 (3433.3 – 2768.5) X 103

= 14.325 MJ/s

Then, the no. of stages = ��

= ��.��.��

≈ 7.933

= 8 stages

7)Calculate the blade height and radial stress exerted bythe blades in the last stage

ð Area = circumference of the nozzle X height X 0.25

= π D X height X 0.25, Where D= 0.54 m

Area = 1.696 x 0.25 x height

Then, area x velocity of flow (Cf e) = volume of flow of steam

Cf e= sin β e X Cr e

= sin 31.67 X 266.7

= 140 m/s - (from combined velocity diagram)

From a steam table at 2 bar and 1500C, the specific volume =0.9595 m3/kg

Therefore, 1.696 x 0.25 x height x 140 = 0.9595 x21.67

Height of the blade in last stage = 350.27 mm ≈ 350.5mm

Radial stress exerted by the blade = � ω� ��

= �� ω� ��

= � � � � �� ω� ��

= h x ρ x ω� r= 0.35027 x 7856 x 628.312 x 0.4451

Where ω = � � � � ��

= � � � � ��

= 628.31 rad/sec

r = disc radius + ½ height of the blade

= 0.54�2 + 0.35027�2 = 0.4451 m

8)Show that the rotor shaft can be designed as 140 mmdiameter. Take allowable shear stress = 41 MN/m2 andtotal mass of the rotor and disc =580 kg

ð Mass of rotor shaft and blade = 580 kg (Given)Length of the shaft between the inner edges of bearingsl = 1290 mm = 12.9 m

Mass per unit length = ��.��

= 449.61 kg/m

Bending moment acting on the shaft = WL� 8�

= ��.�� .�� .��

= 917.48 Nmwe know that,Maximum shaft power P = � � � � � � �

��

Then the torque acting on the rotor shaft T = � � ��

=� � ��.� � ��

� � � � ��

Equivalent torque = √T� + B. M.� = √21.485� + 0.917� = 21.5 kNm

TEQ = 21.5 kNm

Since by the torsion equation,�� = τ

� ………………………. (1)

Where τ = allowable shear stress = 41 MN/m2

J = polar modulus of inertia = � ��

��

r = d2� = diameter of the rotor shaft�2

From equation (1)�� = τ

�

�� = τ� ��

��

∴ T��=τ x � ��

��

∴ 21.5 x 103 = 41 x 106 x � ��

��∴ d3 = 2.67

d = 138.74 mm ≈ 140 mm

Therefore d=140 mm is perfectly suitable.

9)Calculate mass of the rotor shaft of transversevibration, first critical speed of the shaft in r.p.s.

ð Mass of the rotor shaft between bearing support = 7856 x1.29 x �

� x 0.142 = 156 kg

Moment of inertia of rotorIg = �

�� x d4

= π��

x 0.144

= 1.8857 x 10 -5 m4

The frequency of transverse vibration

fn = � � � ��

= ��

�� .�� .�� .��

= 166.69 HzSince we know that, first critical speed of the shaft in r.p.s. =the frequency of transverse vibrations = 166.69 r.p.s.

10) Determine natural frequency of transversevibration of the loaded shaft and the first critical speedin r.p.s. Hence calculate the critical speed of thecomplete rotor system.

ð According to Dunkerley’s formula, natural frequency oftransverse vibration of the blade shaft,

fn = �.��δ��δ��⋯δ��δ/�.��

and the deflection of a simply supported beam subjected to a

point load δ = ��

� � � �

The rotor shaft is subjected to point load, by 8 stage rotor blade,equidistantly placed

Weight of the blade W = (�� – ��) � �.�� = 519.93 N

Since,

i. a = 0.143 m and b = 1.147 m

δ1 = ��

� � � � = ��.�� .�� .��

� � �� .�� .��

= 9.58 x 10-7 m

ii. a = 0.286 m and b = 1.004 m

δ2 = ��

� � � � = ��.�� .�� .��

� � �� .�� .��

= 2.937 x 10-6 m

iii. a = 0.429 m and b = 0.861 m

δ3 = ��

� � � � = ��.�� .�� .��

� � �� .�� .��

= 4.86 x 10-6 m

iv. a = 0.572 m and b = 0.718 m

δ4 = ��

� � � � = ��.�� .�� .��

� � �� .�� .��

= 6 x 106 m

v. a = 0.715 m and b = 0.575 m

δ5 = ��

� � � � = ��.�� .�� .��

� � �� .�� .��

= 6.021 x 10-6 m

vi. a = 0.858 m and b = 0.432 m

δ6 = ��

� � � � = ��.�� .�� .��

� � �� .�� .��

= 4.89 x 10-6 m

vii. a = 1.001 m and b = 0.289

δ7 = ��

� � � � = ��.�� .�� .��

� � �� .�� .��

= 2.98 x 10-6 m

viii. a = 1.144 m and b = 0.146 m

δ8 = ��

� � � � = ��.�� .�� .��

� � �� .�� .��

= 9.937 x 10-6 m

∴� 1 + � 2 +… + � 8 = 9.58 x 10-7 m + 2.937 x 10-6 m + 4.86 x 10-6 m+ 6 x 106 m + 6.021 x 10-6 m + 4.89 x 10-6 m + 2.98 x 10-6 m +9.937 x 10-6 m

= 2.964 x 10-5 m

Hence the natural frequency of transverse vibration of bladeshaft

f n = �.��δ��δ��⋯δ�

= �.��.��

= 91.5 Hz

The first critical speed in r.p.s = 91.5 r.p.s.

Considering the mass of the rotor shaft only, the weight of theshaft per meter

W = �� .��.��

= 1186.32 N

For UDL the deflection is

� 8 = ��

x � ��

� �

= ��

x ��.�� .��

�� .��

= 1.1342 X 10-5 m

By Dunkerley’s formula

fn = �.��δ��δ��⋯δ��δ/�.��

= 80.267 r.p.s.

≈ 81 �.�. �.

11.

a. Discuss important properties of ideal bladematerial.

-> Properties expected out of the blade material depends on portion ofblade in turbine. Good high temperature strength and resistance to creep isrequired at high pressure turbine inlet

The L.P. rotor is high strength to withstand the high centrifugalstresses generated by long last row blades and it also has to avoid anytendency for brittleness at relatively low temperatures which exists at thisexhaust side of machine.

b. What is blade erosion and how is it prevented?

The turbine blade undergoes severe corrosion in lower stages whilsthigh strength is required for highly stressed long L.P. turbine blade. Thematerial used in L.P. turbine blade should resist corrosion although additionalmeasures are also taken to avoid erosion. Manufacturing of blades in variousgrade 12- 13% chrome steel to give required property depending on positionof blade in turbine.

Additional toughness is provided by use of tungsten, molybdenum,steel especially for L.P. turbine. Stellite coating is given on L.P. turbine bladebut if coating is given it is very dangerous. Tungsten is coated to give

toughness for 10000 -15000 hrs running of blade, the blades may bereplaced.

Shield is also provided only for longer durability of blade, no blade canlast forever. Creep will occur, very slowly but suddenly in blade but fatiguenormally doesn’t occur, erosion shield is assisting to fight with erosion.Reheating of steam mechanism

Advantages of reheating are that it increases the work done throughturbine; it increases the efficiency of the turbine, reduced erosion of bladebecause of increase in dryness fraction of steam at exhaust. Amount of waterrequired in condenser of turbine is reduced due to reduction in specific steamconsumption, a slot (special dog tail) design is cut on the wheel and in thisblade is fitted in sliding and they are tied up by tool.

Heat the blades and hammer on sides to remove blades, on it can becut up. Blade root grooves are of various types viz. fir tree, dove tailed, T-grooves. Wire lashing is done on blades together to prevent bending ofblades.

c. Find the length of the bearing. Allowable pressurein the bearing = 254KN/m2

Mass of rotor of blade = allowable pressure X projected area

∴ 9.81 x 580 = 250 X 102 X 0.14 X L

∴ L= 0.16 m = 160 mm


Why has the steam turbine replaced the steam engines?

Ans: Steam turbine has almost completely replaced the reciprocating piston steam engineprimarily because of its greater thermal efficiency and higher power-to-weight ratio. Evenin Electrical power stations use large steam turbines driving electric generators to producemost (about 80%) of the world's electricity. The advent of large steam turbines made

central-station electricity generation practical, since reciprocating steam engines of largerating became very bulky, and operated at slow speeds.Because the turbine generates rotary motion, it is particularly suited to be used to drive anelectrical generator – about 80% of all electricity generation in the world is by use ofsteam turbines. The steam turbine is a form of heat engine that derives much of itsimprovement in thermodynamic efficiency through the use of multiple stages in theexpansion of the steam, which results in a closer approach to the ideal reversible process.The problem was one of basic engineering: the reciprocating engine utilizes the pressureof steam, but the turbine principle uses the speed of steam, and that is fast, 2,000 mph isfairly typical of a moderate power boiler. In order to utilize that energy the turbine bladeshave to rotate at least have the speed of the steam jet. Even by the 1880's it was just notpossible to construct a device that could rotate at those speeds without melting or flyingapart, probably both.

What are the various sizes of a steam turbine?

Ans: Steam turbines are made in a variety of sizes ranging from small <1 hp (<0.75 kW)units (rare) used as mechanical drives for pumps, compressors and other shaft drivenequipment, to 2,000,000 hp (1,500,000 kW) turbines used to generate electricity.

What are the various types of classifying a modern steam turbine dependingupon steam supply and exhaust conditions?

Ans: These types include condensing, non-condensing, reheat, extraction and induction.Non-condensing or back pressure turbines are most widely used for process steamapplications. The exhaust pressure is controlled by a regulating valve to suit the needs of

the process steam pressure. These are commonly found at refineries, district heatingunits, pulp and paper plants, and desalination facilities where large amounts of lowpressure process steam are available.

Condensing turbines are most commonly found in electrical power plants. Theseturbines exhaust steam in a partially condensed state, typically of a quality near 90%, at apressure well below atmospheric to a condenser.

Reheat turbines are also used almost exclusively in electrical power plants. In areheat turbine, steam flow exits from a high pressure section of the turbine and isreturned to the boiler where additional superheat is added. The steam then goes back intoan intermediate pressure section of the turbine and continues its expansion.

Extracting type turbines or Regenerative turbines are common in allapplications. In an extracting type turbine, steam is released from various stages of theturbine, and used for industrial process needs or sent to boiler feed water heaters toimprove overall cycle efficiency. Extraction flows may be controlled with a valve, or leftuncontrolled.

Induction turbines introduce low pressure steam at an intermediate stage toproduce additional power.

What are the casing or shaft arrangements in a steam turbine?

Ans: These arrangements include single casing, tandem compound and cross compoundturbines. Single casing units are the most basic style where a single casing and shaft arecoupled to a generator. Tandem compound are used where two or more casings aredirectly coupled together to drive a single generator. A cross compound turbinearrangement features two or more shafts not in line driving two or more generators thatoften operate at different speeds. A cross compound turbine is typically used for manylarge applications.

What is the principle of operation and design of steam turbine?

Ans: An ideal steam turbine is considered to be an isentropic process, or constant entropyprocess, in which the entropy of the steam entering the turbine is equal to the entropy ofthe steam leaving the turbine. No steam turbine is truly ―isentropic, however, with typical

isentropic efficiencies ranging from 20%-90% based on the application of the turbine. Theinterior of a turbine comprises several sets of blades, or buckets as they are morecommonly referred to. One set of stationary blades is connected to the casing and one setof rotating blades is connected to the shaft. The sets intermesh with certain minimumclearances, with the size and configuration of sets varying to efficiently exploit theexpansion of steam at each stage.

What are the measures to enhance the efficiency of steam turbine?

Ans: To maximize turbine efficiency the steam is expanded, doing work, in a number ofstages. These stages are characterized by how the energy is extracted from them and areknown as either impulse or reaction turbines. Most steam turbines use a mixture of thereaction and impulse designs: each stage behaves as either one or the other, but theoverall turbine uses both. Typically, higher pressure sections are impulse type and lowerpressure stages are reaction type.

Explain the starting procedure of a steam turbine.

Ans: When warming up a steam turbine for use, the main steam stop valves (after theboiler) have a bypass line to allow superheated steam to slowly bypass the valve andproceed to heat up the lines in the system along with the steam turbine. Also, a turninggear is engaged when there is no steam to the turbine to slowly rotate the turbine toensure even heating to prevent uneven expansion. After first rotating the turbine by theturning gear, allowing time for the rotor to assume a straight plane (no bowing), then theturning gear is disengaged and steam is admitted to the turbine, first to the astern bladesthen to the ahead blades slowly rotating the turbine at 10 to 15 RPM to slowly warm theturbine.

What are the problems caused due to incorrect warming up procedure?Ans: The main object of warming through is to ensure straightness of the rotor. To do thisa negligible temperature gradient must exist throughout the rotor. There is a tendency forthe rotor to hog where the steam is introduced( that is to say the rotor bends due totemperature gradient rather than sagging under gravitational forces) with the rotor steamis introduced. Hence the rotor must be rotated.The graph below indicates the importance of this.

The line is the out of balance force due to centrifugal force equal to the mass of the rotor.Hence, the offset at 3000rpm to cause an out of balance equivalent to the mass of therotor is 0.102 mm.

Testing of the engines after shut down ahead and astern should be taken as part ofthe warming through process. Close watch of the relevant nozzle box temperatures is agood indication of the condition of the turbine.

Second object of warming through is to prevent distortion of the casing. Rotation ofthe rotor churns up the steam and provides adequate mixing. With underslung condensersthe temperature gradient is virtually unavoidable, hence separate condensers are better.The third objective is to prevent thermal stresses caused by the temperature gradient inthick materials such as at the bolt flanges. Vertical slots are often provided to helpalleviate this problem, this distortion can also lead to non concentricity of the casing

This is particularly prevalent in open cylinderdesigns such as axial plane or doublecasings. Heat transfer rate is at its greatestwhere the steam is condensing on thesurface of the casing. This in turn isgoverned by the inlet pressure of thewarming through steam. Hence, warming

through in steps providing adequate period to stabilize the temperature at each step.Complete warming through cannot occur until nearly at full power, hence, warming

through much above atmospheric saturation temperature is pointless.

Also as part of the LP turbine runs at lower temperature, warming above 100oC isunnecessary. Protracted warming through periods are unnecessary. A temperature of 82oCat the LP inlet belt in 30 mins is acceptable

Vibration caused by an out of balance of the rotor may be alleviated by running fora short period at reduced engine speed followed by a slow increase in speed.

What are the possible failures and their causes in a modern steam turbine andwhat are the design features to avoid them?

Ans: Problems with modern steam turbines are rare and maintenance requirements arerelatively small. Any imbalance of the rotor can lead to vibration, which in extreme casescan lead to a blade letting go and punching straight through the casing. It is, however,essential that the turbine be turned with dry steam - that is, superheated steam withminimal liquid water content. If water gets into the steam and is blasted onto the blades(moisture carryover), rapid impingement and erosion of the blades can occur leading toimbalance and catastrophic failure. Also, water entering the blades will result in thedestruction of the thrust bearing for the turbine shaft. To prevent this, along with controlsand baffles in the boilers to ensure high quality steam, condensate drains are installed inthe steam piping leading to the turbine.

How is speed control achieved in a steam turbine?

Ans: The control of a turbine with a governor is essential, as turbines need to be run upslowly, to prevent damage while some applications (such as main propulsion unit or thegeneration of alternating current electricity) require precise speed control. Uncontrolledacceleration of the turbine rotor can lead to an overspeed trip, which causes the nozzlevalves that control the flow of steam to the turbine to close. If this fails then the turbinemay continue accelerating until it breaks apart, often spectacularly. Turbines areexpensive to make, requiring precision manufacture and special quality materials. Duringnormal operation in synchronization with the electricity network, power plants aregoverned with a five percent droop speed control. This means the full load speed is 100%and the no-load speed is 105%. This is required for the stable operation of the networkwithout hunting and drop-outs of power plants. Normally the changes in speed are minor.Adjustments in power output are made by slowly raising the droop curve by increasing thespring pressure on a centrifugal governor. Generally this is a basic system requirement forall power plants because the older and newer plants have to be compatible in response tothe instantaneous changes in frequency without depending on outside communication.

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, andultimately split in three and sank, all together resulting in thelargest oil spill of its kind in history to that date. What was thereason of such an catastrophic disaster. The answer is in thisproject.

INTRODUCTION

The direction of the ship iscontrolled by the steering gear. As theship moves through the water, the angleof the rudder at the stern determines thedirection it will move. Modern ships are sobig that moving the rudder necessitatesthe use of hydraulics or electrical power.

The steering starts at the Bridge.The required rudder angle is transmittedhydraulically or electrically from thesteering wheel at the Bridge to thetelemotor at the steering gear, justabove the rudder.

There are a few commonarrangements for using hydraulic power.There are the 4-rams, 2-rams, androtary vane types. The heart of thesehydraulic systems is the variable deliverypump. This type of pump can becontrolled by just moving a spindle. Thepump is driven by an electrical motor atconstant speed.

By moving the control spindleaway from the central point, the pumpstroke increases, and the hydraulic fluidis pumped in one direction. Moving thespindle more from the central point will cause more fluid to be pumped andconsequently more pressure is generatedto drive the rams. Moving the controlspindle back to the original position andthen away in the opposite directioncauses the hydraulic fluid to be pumpedin the reversed direction. The rams willalso move in the reversed direction.

By using a floating lever feedback mechanism, when the rudder stockhas reached the desired angle, the pump control lever moves back to theoriginal position, and the pumping action stops. The rudder is stopped at therequired angle. Moving the steering wheel to the opposite direction will causethe rudder to come back to the original zero position.

TECHNICAL CHALLENGE

A vessel of 10,000 Tonnes displacement has LBP 120m, Breadth15.6m and loaded draught 6.7m, fitted with a semi-balanced rudderhaving a single guide pintle operated by a 2 Ram electro-hydraulicsteering gear (see accompanying sketch) has the followingparticulars. It is required to design a hydraulic steering gear withthe help of information available and using a Rapson slidemechanism.

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 stockbearing 0.4 m

(m)

Horizontal distance of center of press on rudder from leadingedge 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 velocityof water passed in m/sec, and may be assumed to bemaximum speed of the vessel (S) 16 knots for ahead runningand 50% of vessel’s speed for astern running. (� is the anglesubtended 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 pressurein cylinder due to abnormal conditions.

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

(r) Owner requires 10% increase in rudder stock dia. overcalculated 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 tothe force on the rudder. Accepted ram diameter should be amultiple of 10.

(vi) Diameters of tiller arm assuming that the maximum stress occursat 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 forapproximation and then rounding up to a suitable figure.

(viii) Motor power, rpm and shaft diameter given that the shear stress inthe motor shaft is limited to 50MN/m2 and the motor drivingvariable delivery pump is a 4-pole induction motor connected to440v, 3-phase, 50Hz supply mains. Full load slip is 4% with the 0.8lagging Power Factor Motor should run at 85% MCR and availablemotor power ratings are multiple of 5.

(ix) Calculate the capacity of pump considering probable losses whereco-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 theallowance for rudder drop. (Tiller arm and the boss are not to beshown 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 fromdamage 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 ramelectro-hydraulic steering gear.

The steering gear uses Rapson slide mechanism.

a) Max speed of vessel = 16 knotsb) Max rudder angle � = 35°c) Max working pressure on rams ≤ 100 bard) 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 mh) Ram packing thickness = 20 mmi) 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 immersedmiddle plane value at this ratio normally between 60 and70.

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

Area of Rudder = � ��

Where L = LBP in m = 120 mH = Max loaded draught = 6.7 m

A = ��

A = 13.4 m2

Rudder area is 13.4 m2

But �� = 1.5

∴ h = 1.5 bWhere 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 RudderForce 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 thecenter line of the ship for maximum force

= 35°F = 577 X 13.4 X(16 X 0.5144)2

X sin 35

∴ Max Force on Rudder = 300.38 KN

(c) Maximum Bending movement and Torque on the rudderstock with lower pintle.

Since it can be treated as a SSB, when rudder issupported by lower pintle, the max Bending movement isgiven by

F = �4 + �� ℎ� - R ��

� ℎ�

For equilibrium F = R

Max B.M. = �300.38 �0.4 + �� ∗ 4.48��− 300.38 ��

� ∗

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 stockwithout lower pintle.

Max Torque will be same for rudder having lower pintleand without lower pintle

Max. Torque = 120.152 KN-m

Since it is cantilever beam, when it is considered forrudder without lower pintle

∴ Max. B.M. = Force X Distance from rudder stock bearing toC.O.P

= F X L

Where L = 0.4 + ��

h = 0.4 + �� 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 pintleThe equivalent torque is given by

Teq =�(M��)� + (T��)�

As we have already found max B.M. and max Torque withlower pintle

Teq =�(567.718)� + (120.152)�

= 580.29 KN-m

20% increase has to be done for undue forces

Teq =580.29 + ��

X 580.29

= 696.349 KN-m

W.K.T. Teq =��

d3 fs

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

= 75 X 106 N/m2

∴ d3 = 16��

= �� .��

� � ��

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 = �(M��)� + (T��)�

=�(1015.28)� + (120.152)�

= 1022.36 KN-m

As per undue forces, we have increased the eq. torque by20%

∴Teq = 1022.36 X 1.2

= 1226.832KN-m

W.K.T. Teq =��

d3 fs

∴ d3 = 16��

= �� .��

� � ��

d = 0.4367 m

d = 436.7 mm

d ≈ 450 mm

This is the diameter of the rudder stock without lowerpintle.

(e) Diameter of Rams

Torque due to force on Rudder = 120.152 KN-m

W.K.T.

Force = Pressure X Area

= P X �� d2

= 100 X 105 X �� d2

Force = 7853.9816 d2 KN

From Triangle OAB

cos 35 = ��

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 ramfrom rudder stock

= 7853.9816 d2 X 0.8569

= 6730.0768 d2

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

�� = ��

�

��

�=��

d��= ��

� � ��

d��= ��.��

� � ��

��= 145 mm

(g) Thickness of Ram cylinderBy thin cylinder theory, we know thickness

t = ��

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 = ��

= �� .��

t = 14 mm

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

W.K.T.Voltage V = 440v

Current I = 20 AFrequency F = 50 HzNo of poles P = 4

cos� = 0.8 (lag)Motor power = √3 V I cos�

= √3 X 440 X 20 X 0.8Motor power = 12.193 KW

Since power rating should be multiple of 5Motor power = 15 KW

W.K.T.

Synchronous Speed, Ns =��

=��

= 1500 rpm

% slip = ��

X 100

∴ = �� X100

∴ N = 1440 rpm

Shear stress in motor shaft = 50MN/m2

W.K.T.

Power = ��

T = ��

= ��

� � � ��

∴ T = 99.47 N-mBut, torque T = � 3

��d fs

d3 = � � ��

d3 = �� .��π � ��

d = 0.0216 md = 21.6 mm

As the diameter should be multiple of 5d = 25 mm

This the diameter of motor shaft

(i) Capacity of pumpsCd = 0.94 (given)

Cd = ��

QTh = area of flow X velocity of flow

= � ��

� X �.�

�.�

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

= 0.67.5

∴QTh = π X 0.142

4X 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 = �32

d4 (Polar movement of inertia of shaft)

d = Dia. of Shaft

f� = shear stress = 41 X 106 N/m

R = d2

T = Torque acting

T = fs � JR =

fs �� 432dd2

= � � fs � d3

16

Now W.K.T.

Equivalent Torque, Teq = �M2 + T2

M → Bending Movement, T → TorqueAlso given length between bearing = 1290 mmMass of Rotor and drive = 580 KgShaft of turbine is considered as UDL with w/unit length

∴ Wt. per unit length, w = 580�.��

w = 449.612 kg/m

B.M. due to load, M = w�2

8

Which can be calculated as follows:-

Now total load on turbine rotor shaft

= WtLength

� Length

= w X l

Two reaction supports = wl2 (for each)

Now total weight is acting through the center of the shaftdownwards

Now taking moment about center pt. considering left ofreaction

Moment due to reaction = wl2� l

2= w�2

4

Moment due to wt. of half beam = wl2

X l4 = w�2

8

Net B.M. = w�2

4 - w�2

8 = w�2

8

Subs. we get,

w = 449.612 kg/m

l = 1.29

∴ B.M. = 449.612 � 1.292 � 9.818 = 917.48 N-m

W.K.T.

P = 2 ��60

T =260��

P = 13.5 X 106 W

T = 60 � 135 � 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 =� � �� 3

16

�3 = ��.� � ��

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 = � 2� d

dia. of shaft, d = 140 mm

Area, A = �� 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 = � 4��

d = ��

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

Subs we get.

fn=�� .��

��.�� (�.��)� � �.��

fn=��.��

��.��

fn = 53.211 Hz

First critical speed is Nc= 53.211 Hz


Describe the sequence of events in the accident of AmocoCadiz.

En route from the Persian Gulf to Rotterdam, Netherlands, via ascheduled stop at Lyme Bay, Great Britain, the shipencountered stormy weather with gale conditions and highseas while in the English Channel. At around 09:45, a heavywave hit the ship's rudder and it was found that she was nolonger responding to the helm. This was due to the shearing ofWhitworth 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 provedunsuccessful. 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) oflight crude oil from Ras Tanura, Saudi Arabia and Kharg Island,Iran. Severe weather resulted in the complete breakup of theship before any oil could be pumped out of the wreck, resultingin its entire cargo of crude oil (belonging to Shell) and 4,000tons of fuel oil being spilled into the sea.

PROJECT # 2

HOLLOW TAIL SHAFT (OIL COOLED)

As we have discussed about water cooled propeller shaft earlier, our next project isabout hollow oil cooled propeller shaft. This type of propeller shaft is supported by astern tube closed at both ends and having metal bearing surfaces lubricated by oil.

Oil cooled propeller stern tube is preferred in many ships with machinery aft, where theshort shaft is to be relatively stiff and only small deflections are to be tolerated. Wherethis patent oil lubricated stern tube is fitted, glands are provided at both ends to retainoil and to prevent the ingress of water. Besides, white metal (high lead content)bearing surfaces are also provided and oil for cooling and lubrication is supplied from areservoir. Progress from sea water to early oil lubricated stern tubes involved anexchange of the wooden bearing (LIGNUM VITAE) in its bronze sleeve with a whitemetal lined cast iron (or sometimes bronze) bush. Oil retention and exclusion of seawater necessitated the fitting of an external face type seal. The stuffing box wasretained in many early oil lubricated stern tubes, at the inboard end. In oil lubricatedbearings the shaft does not require a full length protective bronze sleeve.

This is the conventional type of arrangement

Now, without much ado let us proceed to the design project and learn the intricaciesand challenges involved in design, operation and maintenance of this propeller shaft.

HOLLOW PROPELLER SHAFT

The following information relates to a steel propeller shaft for a single screw bulkcarrier.

The propeller shaft is hollow and connected to a solid intermediate shaft.

The propeller is fastened to the shaft by 16 pre-stressed internal bolts on P.C.D. of 1mfitted with hollow dowel pins around them. The hollow dowel pins can be regarded tocarry the complete torsion load while the bolts take all other loads.

Additional load due to vibrations may be assumed to be acting downwards through theC.G. of propeller and is given by

LOAD =��

�� x 0.75

Additional B.M. due to thrust may be taken as a product of Propeller Thrust x Propeller

Dia. (in m) x 0.15 where Propeller Thrust = ��

Shaft RPM = 75

Shaft power = 22400 KW

Propeller dia = 9 m

Propeller mass = 51000 kg

Propeller buoyancy = 80 KN

Max. Shear Stress allowed –

For solid shafting and coupling bolts – 50 MN/m2

For hollow shaft due to welding involved – 25 MN/m2

For hollow dowel pins – 50 MN/m2

Neglecting mass of shafting

Calculate

Dia. ‘D’ of intermediate shaft considering torsion only.Dia. of intermediate shaft coupling bolts.Outside dia. of hollow dowels.The total thrust given that the propeller efficiency 60% and the speed ofadvance 10 knots.The total B.M. on the shaft.Show that the equivalent torque on the shaft may be rounded to 5680 kN-m andhence, determine the outside dia. of the hollow shaft.Based on dia. so evaluated on hollow shaft, calculate a suitable flange diameterand size of coupling bolts placed on possible pitch circle.

Mention suitable materials used for each component in design and manufacture,also describe in brief how welding carried out during assembly.Describe the purpose of chrome liner fitted with the shaft seals. What bearingmaterial is used in stern bush bearing of this type?What is the normal periodicity of tail shaft (OG) survey? What are the specificareas need to be inspected? Under what conditions this survey may be allowedwith extended period?

Solution: -

Calculate

1) Diameter of Intermediate Shaft

Mass of the shafting is neglected

Power, P = 22400KW (given), N=75 rpm

P = ��

Torque,

∴ T = �×��×�×�

= ��×��×�×��

∴ T = 2.85 MN-m

We can determine the diameter of the shaft by applying the torsion equation:

�� = ��

�

��

� =��

�.��×��×��

= ��×��

∴ D = 0.6623m

∴ D = 662.3mm

Considering 1% machining tolerance

D = 662.3 + (0.01×662.3)

∴ D = 668.923

∴ D ≈ 670mm

2) Diameter of intermediate shaft coupling bolts

Torque acting on each bolt (For 12 bolts)

= �� .��

= �.��×��

��

= 237.671 KN-m

Force = ��

= ��.��×��.��

= 306.672 KN

Force = Max. Shear stress× Area

306.672×103 = 50×106×�� × ��

∴ d = 88.37mm

∴ d ≈ 90mm (Bolts connecting intermediate shaft and thrust shaft)

The value is rounded off to 90mm

(THIS DIAGRAM IS ONLY FOR PROPER UNDERSTANDING)

Torque acting on each bolt (for 16 bolts)

= ��

= �.��×��

��

= 178.25 KN-m

Force = ��

= ��.��×��.�×��.��

= 336.42 KN

To find the diameter of the coupling bolts


336.42×103= 50×106

×��×d

2

∴ P d = 92.557mm

∴ d ≈ 95mm

(Bolts connecting intermediate shaft and tail shaft)

The value is rounded off to 95mm.

Note:

There is a twist (to baffle students) in the design question (iii) Outside diameter of thehollow dowel pin (do), here we need to use the equivalent torque which can be found inquestion (vi) and hence (iii) follows (vi)

5) Total B.M. on shaft

Total B.M. = Additional B.M. due to thrust + B.M. due to mass of propeller + B.M.through C.G. of propeller - B.M. due to Buoyancy.

Additional B.M. due to thrust,

M = Propeller thrust × Propeller diameter × 0.15

= 2612.752 × 103 × 9 × 0.15

= 3527.216 KN-m.

B.M. due to mass of propeller,

M = 57000 × 9.81 × (�.�� + 0.6 +1.05)

= 500.31 × 103 × 2.1

= 1050.651 KN-m.

Load due to vibrations = ��

× 0.75

= �.�� ×��

� × 0.75

= 237.5 KN

As it acts through C.G. of propeller

B.M.

∴ M = 237.5 × 103 ×2.1

= 498.75 KN-m.

B.M. due to buoyancy,

M = 80× 2.1

M = 168 KN-m.

Total BM,

M = 3527.216 + 1050.651 + 498.75 – 168

∴ M = 4908.617 KN-m.

6) Equivalent Torque is given by,

Te = √�� + ��

=�(4908.617 × 10�)� + (2.85 × 10�)�

∴ Te = 5678.6039 KN-m.

∴ Te ≈ 5680 KNm.

The value of equivalent torque can be rounded off to 5680 KN-m.

To get the tail shaft we apply the torsion and substitute the value of Te in place of T.

Applying torsion equation

�� = ��

P

�

�� ×�� (��.��)

= ��×��

��

We already know that the internal diameter (bore) of the hollow tail shaft is 1.22m.

1.1563 = ��.��

��

∴ Do4 = 3.627

∴ Do = 1.38m

3) Outside Diameter of hollow dowel pin (do)

Force on each hollow dowel = ��

(here equivalent torque is taken)

Force = �� ×��

�� ×�.� = 710 KN

d i = 0.08 m (From diagram )


∴ 710 × 103 = 50 × 106 × �� × (do

2 – 0.082)

∴ �� × ��

�� × �� × ��

+ 0.082 = do2

∴ do = 0.156 m

∴ do = 156 mm

∴ do ≈ 160 mm.

The outer diameter hollow dowel pin is rounded off to 160mm.

Equivalent torque is taken because dowel pins are subjected to both

∴ Do ≈ 1.4m

torsion as well as bending.

4) Total Thrust

Propeller thrust = �� ×��

(GIVEN)

= �� ×�� ×�.�� ×�.��

Propeller thrust = 2612.752KW.

7) Flange diameter is given by 2.2 times shaft diameter (here no. of bolts, n = 16)

Flange diameter, df = 2.2 × Do

df = 2.2 × 1.4

df = 3.08m.

PCD = 1.6 × Do

=1.6 × 1.4 = 2.24 m.

Torque = Force × PCR

5680 × 103 = Force × �.��

∴ Force = 5.071 MN

��

= �.��×��

�� = 316.96 KN.

To find diameter of coupling bolts,


316.96 × 103 = 50 × 106 × �� × dB

∴ dB = 0.0898 m

∴ dB = 89.8mm

∴ dB ≈ 90mm

The coupling bolt diameter can be rounded off to 90mm.

8) Materials used for tail shaft are:

Tail shaft -- The tail shaft in this type of oil cooled arrangement is made of mildsteel with high carbon content. The tail shaft is subjected to shock and fatigue. Highcarbon content strengthens it against such loads.

Hollow Dowel – Hollow dowels used in this type of tail shaft are subjected totorsional loads. To ensure that these dowels work properly they are made up of highcarbon content steel for durability, ductility, and strength.

9) Purpose of Chrome Liner

Chrome lines are fitted in order to safeguard the tail shaft against grooving action oflip seal. Lip seals are provided in order to prevent the ingress of sea water to theengine room and also prevent the lube oil used for tail shaft lubrication from leaking.Hence, chrome liners are fitted on the tail shaft where the lip seal rubs the material bypush fit on slide fit.

10) Periodicity of tail shaft (OG) survey and areas to be inspected:

Tail shaft survey—is done within a period of normally every 5 yrs.

Extension is allowed and survey is done in a period up to 7 to 10 yrs, depending onthe type of seals fitted under hydraulic jacking.

This could be the condition of tail shaft if not surveyed within the specifiedperiod

Condition – During every possible dry–docking of the vessel :

1) Circulating oil sample from stern tube is tested.

2) Propeller drop is checked.

When withdrawn, normally seals are changed along with chrome liners. Crack testingis done on shaft, meeting dowels, bolts, nuts, etc.

Now let us take a brief glance at Directorate General of SHIPPING (D.G.S.)requirements of items to be inspected during a tail shaft survey

EIGHT SCHEDULE[See rule 73(2) (h) and (i)]EXAMINATION OF PROPELLER SHAFTS

Part 1


(1) An inspection of the bearing oil to establish that it is not contaminated by water or debris.

(2) Measurement of the clearance between the shaft bearing and the shaft to ascertain that thewear is negligible.

(3) Removal of the propeller from the shaft to the extent that a full visual and non-destructivecrack detection inspection of the shaft by the forward end of the keyway can be made, and

(4) An inspection of the shaft sealing arrangements to establish that they will remain efficient forthe extended period.

Part 2


(1)An inspection of the bearing oil to establish that it is not contaminated by water or debris.

(2) Measurements of the clearance between the shaft bearing and the shaft to ascertain that thewear is negligible.

(3) Where the propeller is fitted to a taper on the shaft without a key, a visual and non-destructive crack detection examination of the forward part of the taper to establish thatcorrosion or corrosion cracking has not occurred. Alternative methods of ascertaining that seawater has not penetrated the shaft taper/propeller boss bore and that corrosion or corrosioncracking has not occurred may be accepted by the Chief Surveyor of the Government of India.

(4) Where the propeller is attached to the shaft by a bolted flange, a visual and a non-destructivecrack detection examination of the shaft flange radii and bolt hole bores and recesses.

(5) An inspection of the shaft sealing arrangements, which shall require dismantling the shaftseals, to the extent considered necessary by the Chief Surveyor of the Government of India toestablish that they will remain efficient for the extended period, and

(6) An inspection of the surface of that part of the shaft that normally lies within the aft part ofthe aft bearing to a distance at least equal to one-half of the shaft diameter.

Concept review questions:Q: SOME SCREW SHAFTS DO NOT HAVE LINERS FITTED, IN SUCH CASES HOW IS THE SHAFT END

PROTECTED FROM SEA WATER? WHERE IS THE MOST LIKELY POINT OF INGRESS OF SEA WATER?

ANS: Where screw shafts are not fitted with liners, the shaft is oil lubricated and thestern tube bearings are white metal-lined cast iron bushes. In order to retain oil in thestern tube, the inboard end of the shaft is fitted with a mechanical seal which preventsthe oil from running out. Seal in the aft end would be discussed later.

Q: HOW IS THE PROCEDURE OF OIL LUBRICATION CARRIED OUT IN STERN TUBE? WHAT ARE THE

TWO TYPES OF HEADER TANKS?

ANS : Oil is pumped to the bush through external axial grooves and passes through holeson each side into axial passages. The oil leaves from the ends of the bush andcirculates back to the pump and the cooler. Oil pressure within the stern tube ismaintained at approximately at the same level as that of the surrounding sea water bya header tank.

The static lubrication system for vessels with moderate changes in draft, have headertanks placed 2-3 m above the maximum load waterline. The small differential pressureensures that water is excluded. The cooling of simple stern tubes necessitates keepingthe aft peak water level at least 1m above the stern tube. Tankers and other ships withlarge changes in draught may be fitted with 2 oil header tanks for either the fullyloaded condition or ballast condition.

Q: WHAT IS THE MATERIAL COMPOSITION OF BEARING BUSH?

ANS: The bearing bush is normally of grey or nodular cast iron, centrifugally lined withwhite metal. A typical analysis of white metal would be 3% Cu, 7.5% antimony andremainder tin. White metal’s thickness is varies according to classification societyspecifications. Figures of 3.8 mm for a shaft of 300 mm diameter to 7.4 mm for 900mm diameter shaft have been quoted, with bearing clearances of 0.51 to 0.63 mm and1.53 to 1.89 mm respectively. (NOTE:-These figures are only for better understanding)

Q:WHAT ARE THE TWO MOST COMMONLY USED TYPE OF OIL LUBRICATED STERN TUBE BEARING?WHERE IS THE MOST LIKELY POINT OF INGRESS OF SEA WATER?

ANS : Oil lubricated stern bearings use either lip seals or radial face seals or combinationof the two. Lip seals, in which a number of flexible membranes in contact with the shaftare shaped rings of material with a projecting lip or edge which is held in contact with ashaft to prevent oil leakage or water entry. Oil is contained within the simplex typestern tube by lip seals .The elastic lip of each nitrile rubber seal, grips a rubbing surfaceprovided by short chrome steel liners at outboard and inboard ends of the steelpropeller shafts. A number of lip seals are usually fitted depending upon the particularapplication.

LIP SEALSFace seals use a pair of mating radial faces to seal against leakage. One face isstationary and the other rotates. The rotating face of the after seal is usually secured tothe propeller boss .the stationary face of the forward or inboard seal is the afterbulkhead. A spring arrangement forces the stationary and rotating faces together. Theaft end is the most likely point of ingress of sea-water to the screwshaft.

RADIAL FACE SEAL

What are the inbuilt safety precautionary devices fitted in case of St. L.O. P/P failureand seal failure?

ANS : One of the two header tanks will provide a back pressure in the system and aperiod of oil supply in the event of the pump failure. A low-level alarm will be fitted toeach header tank. Oil pressure in the lubrication system is higher than the static seawater head to ensure that sea water cannot enter the stern tube in the event of sealfailure.

PROTECTION MADE FOR LIP SEALS

What problem can arise out of the chrome liners and how can it be tackled?

ANS : The chrome liners act as rubbing surfaces for the rubber lip seals but groovingfrom frictional wear has been a problem. The difficulty has been overcome by usingceramic filler for the groove or alternatively a distance piece to displace axially the sealand ring assembly. Allowance must be made for the relative movement of shaft andstern tube due to differential expansion. New seals are fitted by cutting and vulcanizingin position.

WHAT KIND OF LUBRICANT CAN BE USED IN STERN TUBES WITH WHITE-METAL LINED

BEARINGS?

ANS : The lubricant used in stern tubes systems must have the ability to maintaina lubrication film in the presence of water so that it is not washed away. The lubricantmust also have good affinity for metal surfaces so that it affords good protection of themetal in the stern tube and the shaft areas against sea water. This affinity for metalsurfaces so that it affords good protection of the metal surfaces and is also necessarywhen the screw shaft starts to revolve, as boundary lubrication conditions are presentat this time.

Compounded oils have these properties: they are blends of mineral oil and fattyoils .the fatty constituent causes water to physically to combine with the oil and form anemulsion. The fatty constituent may be lanolin or synthetic fatty oil having similarproperties.

The lubricant used have a specific gravity at 15.5 0C within the range of 0.92 to0.95, with a viscosity REDWOOD 1 at 60 0C .The viscosity index of many brands ofstern tube lubricant have wide ranges which is unfortunate , as the conditions underwhich they operate make a high viscosity index desirable

If leakage of sea-water into an oil lubricated screwshaft occurs, what indicationswill be there? When is seawater leakage into the system most likely to occur?

ANS : Leakage of sea water into the stern tube is indicated by emulsification of thelubricant .the best time to carry out checks of the vessel is at ports. As the ship loads ordischarges cargo, the aft draught of the vessel changes and the static head of seawater above the screw shaft outer seal will change in a similar manner .if the pressuregauge showing the oil pressure in the stern tube system changes with the change ofdraught it may under many conditions indicate a defective outer seal. Also when thestern tube system drain cock is opened and any water is drained off. If there is a lot ofwater drained out ,it shows a seal leak

If the ship has a light aft draught and there is no, or only small ,current round the ship,the stern tube system can be pressurized to some pressure greater than the head ofwater outside the outer seal.if the seal is defective oil globules will be seen coming tothe surfaces from the seal.

How can outer seal leakage can be founded when at sea? What can be done to keepthe system safe without dry docking the vessel?

ANS : Outer seal leakage has often been found by an increase in the consumption of thestern tube lubricant. It is more likely to be noticed when the ship is proceeding in thetropical waters from cold water areas. Leakage inboard is shown by oil emulsificationand reduced oil consumption.

Why is cooling system required and how?

Heat produced by the friction will result in hardening and loss of elasticity of the rubber,should temperature of the seal material exceed 110o C. Cooling at the outboard end isprovided by the sea. Inboard seals, unlike those at the outboard end, cannot dissipateheat to the surrounding water. Oil circulation aided by convection, is arranged tomaintain the low temperature of the seals at the inboard end. Connection forcirculation, are fitted top and bottom between the two inboard seals and the small localheader tank.

Is oil pressure dependent on shaft speed?

ANS : The requirement of steaming at a slow, economical speed during periods of highfuel prices (or for other reasons) gives a lower fluid film or hydrodynamic pressure instern tubes, due to the slower speed. The possibility of bearing damage occurringprompted the installation of forced lubrication systems to provide a hydrostatic pressurewhich is independent of shaft speed. The supplied oil pressure gives adequate lift toseparate shaft and bearing and an adequate oil flow for cooling.

How are such stern tubes supported?

ANS : The later designs of oil lubricated stern tubes are fitted in a stern frame with anelongated boss to provide better support for the white metal lined bearing. A minimumbearing length of two times shaft diameter will ensure that bearing load does notexceed 0.8 N/mm2.

The forward part of the stern tube is fabricated and welded direct to the extension ofthe stern frame boss and into the aft peak bulkhead. The outboard liner additionallyprotects the steel shaft from sea water contact and corrosion.

PROJECT # 3

AIR STARTING SYSTEM

WARTSILA-SULZER 12RTA96C ENGINE, built by Doosan Heavy Industries,South Korea (a licensee of WARTSILA)

The picture shows WARTSILA-SULZER 12RTA96C SLOW SPEED 2-STROKE MARINE DIESEL ENGINE. This goliath engine producing power inexcess of 90,000 BHP propels ULTRA LARGE CONTAINER VESSELS likeM.V. EMMA MAERSK at speeds of about 25 knots. To start this engine,compressed air at 25 - 30 bar pressure is used.

To discover more about air starting system of such mammothengines and how it is designed, take a look at the technical challenge putforth by this project.

INTRODUCTIONAir starting is the method used for starting large diesel engines (land

based /marine) by supplying compressed air into the cylinders in a fixedsequence (one after other). The compressed air is supplied by aircompressors, i.e. either diesel or motor driven. After a period of being atstandstill (i.e. stationary), the engine requires to be started with a highinitial torque at low revolutions in order to accelerate the engine rotatingand reciprocating masses. The compressed air is then supplied by a largebore pipe to a remote-operating non-return or automatic valve and then tothe cylinder air start valve. The momentum built in the rotating elements ofthe crankshaft will help in smooth starting once the initial inertia has beenovercome. The pressure of the starting air must be sufficient to impartenough speed to the engine pistons to quickly compress the combustion airsufficiently during the compression stroke for it to reach a temperature atwhich the combustion of the injected fuel initiates. The pressurerequirements will also vary according to engine size, design and servicerequirement and mainly working temperature.

An Air Compressor

The starting air is admitted from the air bottle to the engine cylinderthrough a starting air master valve common for all cylinders and cylinderstarting air valve mounted on cylinder head, one for each cylinder. The

distributor ensures the air is introduced into the relevant cylinder at thecorrect time to achieve starting in the desired direction from any position ofthe engine at rest. The automatic master air starting valve is forced openby main air acting against a collar and air is admitted to the starting airvalves. Air is simultaneously admitted, via the distributor .To pressurize thetop side of piston operating air starting valve on cylinder head whilebottom side of the piston is vented.

Air compressors preferably multi-stage types are used to provide airat high pressures required for diesel engine starting. If all the air had beencompressed in a single-stage, it would unfortunately generate compressiontemperatures similar to those in a diesel engine. Multi-stage air compressorunits with various cylindrical configurations and piston shapes are used inconjunction with inter-cooling and after-cooling to provide the nearestpossible approach to the ideal of the isothermal compression. Such inter-cooling and after-cooling is achieved by water cooling or can be air-cooled(i.e. sea water can be used in water cooled type and fin type coolers in aircooled). Air cooling is achieved by multi-tubular heat exchangers wherewater is circulated.

Schematic of position of starting air line in marine diesel engines

Air starting system

Technical Challenge:

It is required to design an Air starting system for a 6 cylinder 2-stroke reversible diesel engine, consisting of two air storage receivers withhemispherical ends and having necessary fittings and condensers. Thecompressed air supplied through two single acting 2 stage compressors,each of them capable of coping up with the starting air requirements at anaverage of one start every 3 minutes. As per specification rules, it isnecessary that the capacity of an air bottle should be such that 12 startsmay be given to the engine without replenishing the air bottle.

For prompt starting of the engine it is necessary that air be admittedto two cylinders consecutively for a period of 70o crank rotation at a meanpressure of 20 bars. The compressors are two stage tandem-type with H.Pcylinder on top having an intercooler and after cooler giving perfectcooling.

The following other data may be used:

1. Main engine bore 780mm2. Main engine stroke 1400mm3. Maximum air bottle pressure 30bar4. Minimum pressure allowed when the compressor

has to be started20bar

5. External dia. Of air bottles 1.5m6. Maximum allowable stress for air bottle material 90 MN/m2

7. Average engine room temperature 270C8. Compressor bore/stroke ratio(L.P. cylinder) 1.29. Index of compression and expansion.(compressor) 1.310. Rotational speed of compressor 750rpm11. Mechanical efficiency of compressor 85%12. Sea water inlet temperature to compressor 200C13. Sea water outlet temperature from compressor 350C14. specific heat of sea water 4.12 KJ/Kg K15. Compressor L.P suction pressure and temperature 1 bar and 250C16. Clearance volume in both the stages 3% of stroke

volume

I. Calculate

1) Capacity and overall length of air bottle.2) Thickness of air bottle, using thin cylinder theory.3) Capacity of each compressor at free air delivery conditions of

1.013bar and 15oC.4) The motor power required to drive the compressor.5) Compressor cylinder dimensions.6) Capacity of sea water cooling pump in Tonnes/hour.7) A spring loaded safety valve is to be fitted as a mounting to

each of the air receiver to blow off at 10% above maximumpressure. The diameter of valve is 40mm and maximum lift ofthe valve is 10mm.Design a suitable compression spring for the safety valveassuming spring index 6 and stress concentration factor 1.25.You are required to provide initial compression of spring 30mm.Maximum shear stress in the material of spring wire limited to450 MN/m2.

G=90 MN/m2; spring index=� ��

�

II Draw (hand sketch only)8) A section of the air receiver showing the weld details

necessary.9) An outside view showing the mountings.

10) Comment on the material used and welding procedureadopted for construction of the air bottle.What are the safety features in its construction and during use?

The above figure shows a sketch of an air reservoir

GIVEN:

Main engine 6 cylinder, 2 stroke, reversible

2 main air bottles

Main engine bore, d = 780mm

Main engine stroke = 1400mm

Max. air bottle pressure = 30 bar

Minimum pressure allowed when the Compressor has to bestarted = 20bar

External diameter of air bottles = 1.5m

Max. Allowable Stress for air bottle material = 90MN/m2

Average engine room temperature = 27oC

Compressor bore/stroke ratio (L.P. cycle) = 1.2

Index for compression and expansion (comp.) = 1.3

Rotational speed of compressor = 750 rpm

Mech. Efficiency of compressor = 85%

Sea water inlet temperature to compressor = 20oC

Sea water outlet temperature from compressor = 35oC

Specific heat of sea water = 4.12KJ/kg K

Compressor L.P. Suction pressure & temp. = 1 bar & 25oC

Clearance volume in both the stages = 3% of stroke volume

� 2main air, 2 stage tandem with H.P. cycle on top� 12 starts continuously and one start in every 4 minutes

I. CALCULATE1. Capacity and overall length of the air bottle

We know that

Stroke length = 2 x crank radius

L = 2 x R

R = L / 2 = 1400 / 2 = 700mm

From diagram,

AB = r - OB

= r r cos Ѳ

= r (1 – cos Ѳ)

= 700 (1 - cos70)

= 460.58 mm

AB = 461mm

Volume of 1 cylinder =π� x D2 x AB

= π4 x 0.782 x 0.461

= 0.22 mm3

According to the firing order air is injected in 2 cylinders atstart,

Volume = 0.22 x 2 = 0.44 m3

Since 12 starts are required from one air bottle

= 12 x 0.44 = 5.28 m3

Let v be the capacity of the air bottle

30 x V = 20(V+5.28)

V= 10.57m3 ≈10.6m3

2. Thickness of air bottle using thin cylinder theoryAccording to thin cylinder theory,

F= � ��

Where,F = max. allowable stressP = max. test pressure = 1.5 x max. air bottle pressure

= 1.5 x 30 = 45 bart= thickness

Substitute the values

90x106 = �� (�.� � ��)� �

40t = 1.5 - 2t

T = 35.7mm

Thickness of cylindrical portion is 35.7 mm

Inside diameter of cylindrical portion of air bottle,= 1.52 x 0.0357

D1 = 1.4286m

Thickness of hemispherical portion, given by formulae

t= Pd1K/2t +0.75x10-3

where,

P = 45x105

K = a constantt = max. allowable stress = 90x106N/m2

t = �� .�� .��

t = 18.6mmInside diameter of the hemispherical end=1.5-2x0.0186=1.4628m

The length of cylindrical portion,

v=π� d1

2 l + 4/3π(d1/2)3

10.56 = 1.6027 x l + 1.6389L = 5.57mThe overall length of air bottle,loverall = l+1.5

= 5.57+1.5

= 7.1m

The overall length of the air bottle is 7.1m

3.) Capacity of each compressor of free air deliveryconditions of 1.013 bar and 15oC.Volume of air following to the cylinder per second,v= 0.44 / 4 x 60 = 1.833 x 10-3 m3/secMass flow rate of air at starting of compressorP V = m R T

WhereP = pressure = 20x105N/m2

V = vol. of the air flowing per second to cylinderm= mass flow rateR = gas constantT = temperature = 27oC = 300Km= P V/R T= 20 x 105 x 1.833 x 10-3/ 287 x 300

= 0.0425 kg/secThen capacity of each compressor at free air deliverycondition of 1.013 bar and 15oC,Capacity = m R T /P

= 0.0425 x 287 x 288 /1013 x 105

= 0.0347m3/sec= 125m3/hr

The capacity of each air compressor cannot go beyond150m3/hr.

4.) The motor power required to drive thecompressorWork done by the compressor,W = 2 n / (n-1) x R T1 [(P2/P1)

n-1/n-1]WhereP2= Intermediate pressure= √ (P1P3)= √ (1x30)= 5.47bar

H = 1.3T1 = 273 + 25

= 298KR = Gas constant = 0.287 KJ/KgKP1 = 1 barW = 2x 1.3/1.3-1 x 0.287x298x [(5.47/1)1.3-1/1.3-1]

=355.896KJ/KgW x mass flow

= 355.896x0.0425= 15.125KW

Therefore, the motor power required to drive thecompressor,

= Work done/mech. Efficiency= 15.125/0.85= 17.794KW=18KW

5.) Compressor cylinder dimensionsFrom P-V diagramV1-V5 = Swept volume

= (V1-V6) + (V6-V5) ---1V5= clearance volume both stageV5 = 0.03 x (V1-V5) ----2And P1 (V1-v6) = mRT1 ----3WhereM=0.0425 Kg/secR=287 J/ KgKT1= 25oC

= 25+273=298KP1= 1bar

Substitute in eqn. 31x105 (V1-V6) = 0.0425 / 750 x 60 x287 x 298V1-V6 = 2.9078x10-3 m3

For low pressure cylinder,Polytropic Process, P5V5

n = P6V6n

[P5/P6]1/n-1 = V6/V5 -1

[5.47/1]1/1.3 -1 = V6-V5/V5

V6-V5 = 2.7 V5

From eqn. 2V6-V5 = 2.7 x 0.03 x (V1-V5)From eqn. 1V1-V5 = 2.9078x10-3 + 0.08102(V1-V5)V1-V5 = 3.164x10-3m3

Since swept volumeV1-V5 = π/4 d

2l andBore/stoke ratio, d/l = 1.2d= 1.2lV1-V5= π/4 (1.2l)

2 x l = 3.164 x 10-3

L3=2.27x10-3

Stroke of L.P. Cylinder, l = 140.9mm = 141mmFor high pressure cylinder,Swept volume,V2

’-V4 = (V2’-V5

’) + (V5’-V4) ----4

P2 (V2’-V5

’) = mRT2’ ----5

V4 = 0.03 (V2’-V4) ----6

Wherem= 0.0425 Kg/sR = 287 J/Kg KT2/T1 = (P2/P1)

n-1/n

=[5.47/1]1.3-1/1.3

T2=441.26 KAnd T2 = T2

’

And P2’ = P2 = 5.47 bar

Since,P4V4

n = P5‘V5

,n

[P4/P5]1/n – 1 = V5

’/V4 - 1V5

’-V4 = 2.69V4

From equation 4 and 6V2

’- V4 = 7.853 x 10-4 + 0.0807(V2’-V4)

V2’ - V4 = 5.772x10-4 m3

Since, stroke of the H.P. and L.P. cylinder are samei.e. l= 141 mmSwept volume of the L.P. cylinder5.772 x 10-4 =π/4 d2 x 141D2 = 5.212 x 10-3

Diameter of H.P. cylinder,D= 72.1mm = 75mm

6.) Capacity of the sea water cooling pump intones/hourLetm= mass flow rate = 0.0425Kg/sCp= specific heat at the constant pressure = 1.005H1 = enthalpy of the air at inletH2 = enthalpy of air at outletT1 = temp. of air at sudden

T2 = temp. of air at dischargeQ = heat input during polytropic compressionW = Indicated work outputQ = (h2-h1) + W

= mCp(T2 – T1) – 15.125/2= 0.0425 x 1.005(441.25 – 298) – 7.5625= 6.1189 – 7.5625

Q = -1.4435 KJThis is the heat rejected by air which is further received byJCWQ1 = 2 x 1.4435 = 2.887KWFor perfect inter cooling, T2

’ = T1

Q2 = mCp (T2 – T2’)

=0.0425 x 1.005 (441.26 – 298)=6.118 KW

For perfect after cooling , T3 = T2 & T2’=T1

Q3 = m Cp(T3 – T2’)

= 0.0425 x 1.005(441.26-298)= 6.118KW

Total heat rejected to sea water,Q = Q1 + Q2 + Q3

=2.887 + 6.118 +6.118=15.14 KW

Heat rejected = m Cw (T2 – T1)Hence, the capacity of sea water cooling pump,15.14 = m Cw (T2-T1)Wherem= mass flow rateCw= sp. Heat of sea water = 4.12 KJ/KgK

T2 = seawater outlet temp. from compressor =35oCT1 = seawater inlet temp. to compressor = 200C15.14 = m x 4.12 (35-20)m = 15.14 / 4.12 x 15

=244.98Kg/s

m= 0.881 tonnes/hr is the capacity of seawater coolingpump

7.) A spring loaded safety valve is to be fitted as amounting to each of the air receiver to blow of at 10%above Max. pressure. The diameter of valve is 40mm andthe Max. lift of the valve is 10mm.

Design a suitable compression spring for the safety valveassuming spring index 6 and stress concentration factor 1.25%.You are required to provide initial compression of spring 30mm.

Max. shear stress in the material of spring wire limited to 450MN/m2

G= 90 MN/m2; spring index = coil diameter/wire diameter

Diameter of valve, d1 = 40 mm

Max. Lift of the valve, S2 = 10mm

Max. Pressure, P = 3.3N/mm2

Spring index, C = D/d =6

Where,

D= Coil diameter

d= wire diameter


Initial Compression of spring, S1 = 30 mm

Max. Shear stress in the material of spring wire,

τ= 450MN/m2

Modulus of rigidity, G= 90 GN/m2

Maximum Load, W1 = π/4 d12x3.3

= π/4 x 402 x3.3 = 4146.9N

Max. Compression of the spring

δ max. = δ 1+δ 2

=30 + 10

= 40mm

Load of 4146.9 keeps value on set by providing initialcompression of 30mm, therefore max. load on the spring whenthe value is open i.e., for max. compression of 40mm.

W= 4146.9/30 x 40

=5529.2N


Max. Shear stress , τ

, τ= K x 8 WC/πd2

450= 1.25 x 8 x 5529.6 x 6/πd2

D2 = 234.66

D = 15.31mm

For consideration the wire diameter is, d = 15.4mm

Mean diameter = spring index x wire diameter

= 6 x 15.4

= 92.4mm

No. of active turns = n

Max. compression of the spring(δmax.)

δmax. = 8WC3n/Gd

40 = 8 x 5529.6 x 62 x n/ 90x103x15.4

N = 5.8

Say 6 turns

Taking the ends of the coil as squared and ground the total no. ofturns.

n’ = n +2 = 6+2 =8

Free length of the spring

LF = n’d + δmax.+ 0.15 δmax.

= 8 x 15.4 + 40 + 0.15 x 40

= 169.2mm

Pitch of the coil = free length / n’-1

= 169.2/8-1

= 24.17mm

Q. 8) A section of the air receiver showing the welddetails necessary.

Q9.) An outside view showing the mountings.

Q 10.) Comment on the material used and welding used andwelding procedure adopted for construction of theair bottle. What are the safety features in itsconstruction and during use?

The material used for construction of large or smallreservoirs is of good quality mild steel plate having similarspecification to boiler plate material. The steel will have anultimate tensile strength within the range of 360 MN/m2 to500MN/m2 will have an elongation of not less than 23% to25%. Large starting air storage reservoirs have dish ends; oneend has an opening formed within lip to take an ellipticalmanhole door. The fore of the dished end may be either tori-spherical or semi-spherical.

The dished ends are usually made by the spinning process.The edge left by the spinning is machined and tapered down tothe thickness of the cylindrical shell.

The ends are welded to the cylindrical shell by fullpenetration welds. The longitudinal seams are machine welded.The circumferential seams where the dished ends join thecylindrical shell may be either machine or hand welded.

Smaller air bottles have hemispherical ends and inspectionholes may be fitted in the cylindrical shell.

The air compressor is use on board the ship is ofreciprocating-tandem or inline type. Compressor valves arenormally plate type annular and spring loaded.

Air passes through tubes in intercooler, the water jacketsbeing protected from over-pressure by fitting a bursting disc.

Compressor air discharge must lead to air receiver directly.No interconnection allowed with engine starting air line.

Relief valves are to be provided in all stages as per stagepressures.

Air receiver should have manhole doors with positive closing.Normally, air bottles are mounted on foundation with inclinationto aft. Drains must be provided in each stage of compressor andair receiver. Relief valve fitted should have its discharge leadingto open deck.


Q1. What are the uses of compressed air onboard ship?

Ans : It is used for starting main and diesel engines in motor ships and forauxiliary diesel engines of steam ships. Also control air of low pressure isrequired for ships of both categories for control equipment andinstrumentation purposes. Auxiliary boilers and economizers are fitted withsoot blowers which use compressed air .Portable tools such as drillingmachines, impact wrenches, torque wrenches, hard grinders and liftinggears use compressed air. Compressed air can also be used for chipping,scaling machines paint spraying equipments.

Q2.Why multistage compressors preferred over single stage? What is theuse of inter cooler and after cooler?

Ans : For a single stage air compressor, acting as isothermal duringcompression by provision of perfect cooling is an ideal situation which isvery difficult to obtain. Moreover the necessary final pressure ofcompressed air is achieved by the number of stages. For higher pressure,more number of stages are required .When the air is compressed instages; it is easier to control the temperatures during its passage throughthe compressors.

This can be achieved by water jacketing the air compressor cylinders andpassing the air through heat exchangers (intercoolers). As the air leaveseach stage of compressor, it is cooled in the intercooler .This lowers thework done in compressing the air and prevents a lot of mechanicalproblems which could arise if the air temperature were uncontrolled.

Apart from practical considerations and the as abovequestion, a 3 stage air compressor is more desirable and requires lessenergy or work input than a single stage air compressor when compressingair over the same pressure range.

Q3.What is the basic principle of working of an air compressor?

Ans : Compressor produces high compression temperature approx. equalto the one in the diesel engine which is sufficient to ignite vaporized oil .The heat produced in a single stage of compression would be wasted andcould add energy and produce a resultant rise in pressure apart from thepressure rise expected from the action of piston .However the air cools, thepressure rise due to the heat generated is lost. Only the pressure fromcompression remains .The extra pressure due to heat is of no use andactually demands greater power for the upward movement of the pistonthrough the compression stroke.

Q4. Factors affecting volumetric efficiency of an air compressor.

Ans : The factors affecting volumetric efficiency of an air compressor are asfollows :

i. The clearance between the cylinder cover and the end of the pistonwhen the piston is at end of its discharge stroke. The larger theclearance, the less air is discharged per stroke.

ii. Sluggish opening and closing of suction and delivery valves.iii. Leakage past the compressor piston rings.iv. Insufficient cooling water (or) the cooling water inlet temperature is

too high.v. Inlet temperature of the air to the first or low pressure stage

compressor too high.vi. Throttling of air supply to L.P. suction –example dirty air inlet

strainers.

Q5. How is clearance volume of air compressor checked?

Ans : This check is made by making up a small ,loosely woven ball of leadwire .This is placed on the top of the piston which will have been moveda little off the end of the stroke .The cylinder cover is replaced on thecylinder with a joint in place and tightened down. The compressor is thenbarred slowly over top centre so that the ball of lead wire is compressed.After it is removed from the top of the piston it can be measured by amicrometer .The mechanical clearance measured is compared with thecompressor manufacture’s recommendations. Adjustments are made byaltering the cover joint thickness or by fitting or removing shims betweenthe foot of the connecting rod and the bottom end bearing.

Note: The ball of the wire must be placed centrally on the top of thepiston; if the wire is on one side of the piston, there is a possibility that thepiston rod could bend when the piston is barred over the top centreposition.

Q6. How many starting air compressors are fitted on board ships?

Ans. It is usual to have atleast 2 starting air compressors and sometimesthere are more than two .The compressors may be independently drivenby an electric motor or steam engine .Also there is an emergency aircompressor which may be diesel engine driven. In case of air in air bottleand at the same time crew are facing ‘black out’ condition then the dieselengine compressor comes into play.

Q7. What are the mountings fitted on air bottle and air compressors?

Ans. Safety valves are normally fitted to the air bottles but in someinstallations the reservoirs are protected against over pressure by those ofthe compressors .There is a requirement that if the safety valves can beisolated from the reservoirs the latter must have fusible plugs fitted torelease the air in the event of fire. Reservoirs are designed ,built andtested under similar regulations to those for boilers .Many air reservoirs arealso fitted with other outlet valves which have connections with other airsystems such as auxiliary engine starting systems ,instrument air,workshop air services, ship’s whistle, filter cleaning systems ,emergencyair supply to boiler feedback pump.

Excess air pressure is prevented by spring loaded relief valves on airbottle. In some cases the relief valves are fitted on the common dischargeline from the compressors and fusible plugs are fitted on the air reservoirs.

The fusible plugs prevent serious pressure rise if a fire should breakout near the air bottle. Drain valves or cocks are fitted to the bottom of thereservoir to drain off oil or moisture carried over with the air from thecompressor. An independent connection is fitted for the pressure gaugesused to indicate the air pressure within the reservoir.

Q8. Why are the outlets from relief valve and fusible plug from air bottle letoutside the engine room?

Ans : Fusible plugs are fitted if the receiver is isolable from a relief valve.There will always a relief valve on the high pressure side of the compressorso that when the compressor is being used, the bottle is protected.

However, this means that the receiver is only protected when thecompressor is running. A fusible plug therefore offers protection againstpressure development in the event of an engine room fire. The fusible plugtherefore offers protection against pressure development in the event of anengine room fire. The fusible plug (lead, bismuth and antimony) softens asits temperature rises and extrudes from its fir tree type sockets.

Q9. What is the implication of starting air reservoir on fixed fire fightingCO2 system?

Ans : The release of large amount of air in case of engine room fire wouldimpair the effectiveness of any CO2 fixed fire fighting system gas unlessextra gas is provided in compensation or the air is piped out of the engineroom.

Q10. At what position of piston in the cylinder is the starting airintroduced?

Ans : Modern practice is to introduce air into the cylinder slightly beforeTDC (the alignment of piston rod with the connecting rod at this point issuch that little if any turning moment is developed). This allows the air toaccumulate in the clearance volume ready to force down the piston once itis over TDC. At the same time, another cylinder will be receiving air(because of overlap discussed in next question).This unit will be one inwhich the crank is well past TDC so that it generates an adequate turningmoment to carry the above unit over TDC. The first unit, alreadypressurized, will be able to accelerate the engine up to the fuel initiationspeed. The useful expansion of the starting air will cease at the opening ofthe exhaust to continue air injection any further would be futile. This limitis normal to 3-cylinder engines but is unnecessarily long in engines withmore than three units.

A starting air pressure well below the compression of an enginewill be able to turn the engine over against the compression because the

compression pressure is only reached towards the end of the stroke,whereas starting air is introduced for a much longer period of the stroke.

Q11. In the above project, why is the period of admission of compressedair for 70 degree of crank rotation?

Ans : In order to conserve starting air, starting valves are designed to closeas early as possible consistent with good starting and some explosion ofthe starting air then takes place. The opening and closing of starting airvalves is controlled by the cam (operating within the starting airdistributor) actuating the distributor valve.

To enable a propulsion engine to be started from anycrank position, overlap is necessary in the timing of the starting valves ofan engine. Overlap is necessary in the timing of the starting valves of anengine. Overlap occurs during the period that any two valves are open, itbeing introduced that one valve will be opening whilst the other valve isclosing. One valve will then always open when air is put on the engine tostart some maneuver. If there were no overlaps in the valves it would bepossible for the engine to stop in some position where all the valvesremained closed when air was put on the engine.

The amount of overlap is dependent upon the number of cylinders,the timing of the exhaust opening and so on (the greater the number ofcylinders, the less overlap required)

Over lap calculation:

In the above project there are 6 cylinders, with a obvious total rotation of3600 of crank.

Therefore an equal admission for a period of 600 of crank rotation in eachcylinder, at the same time there needs to be a overlap for atleast 100 ofcrank rotation for the advantages of overlap stated just above.

In the above example the firing order 1-5-3-6-2-4 indicates the orderin which these units would be receiving the starting air during the air

starting operation. If for example unit number 5 starting air valve hasopened and the engine is rotating in clockwise direction. Before this valveshuts the starting air valve of unit 3 will open, unit 6 will be followed byunit 4 and so on (for anticlockwise rotation the order of air entry will be 1-4-2-6-3-5 i.e. reverse to clockwise timing order).

It is evident from above that no matter in what position the enginestops, there will always be at least one of the cylinders with its starting airvalve opened to admit, on starting, the compressed air to start the engine.

Q12. What are the interlocks in engine, air starting system?

Ans.) The starting interlocks prevent the engine being put on fuel before allthe sequences of the starting system have been completed.

With the systems controlled by the operation of hand lever, theinterlocks may be cams or pins which lock and prevent hand levermovement. In the engines started by hand wheel controls the interlocksate often slotted discs (fitted on the wheel shafts) and the small leverswhich engage or clear the slots in the disc.

Some engines have a blocking device connected with the ships E/Rtelegraph which for events the engine being put as when an ahead orderis given and vice versa. Blocking devices are fitted to the engine turninggear so that the engine cannot be inadvertently started with the turninggear in

In the event of E/R telegraph failure, any interlocking and blockingdevices operated from the telegraph would prevent the engine beingmaneuvered during the time of emergency. It is therefore important toknow how the interlock and blocking devices may be overridden so that theengine can be maneuvered under emergency conditions with orders via thebridge to engine room telephone.

Q 13.Why should there be minimum of 12 starts in case of reversibleengine?

Ans : Reversibility can be achieved by introducing air into the cylinderwhere the piston is approaching TDC in the direction of rotation in which itwas stopped. During maneuvering the engine is frequently subjected toahead and astern starts. It is also an important requirement which ischecked by a surveyor before entry of vessel in narrow channels likePanama Canal. The engine should be capable of being started 6 times inahead direction and 6 times in astern direction without refilling the airbottle. This is necessary to avoid the ship being stranded in the canal orlocks in case of failure of air compressor.

Q 14.How does working temperature of the engine affect the starting airpressure?

Ans : An engine at working temperature will start with a lower starting airpressure than a cold engine. The temperature of the lubricating oil suppliedto the engine bearings will also have an influence on the minimum requiredstarting air pressure. When the lubricating oil is at working temperature,the engine will swing more easily than when lubricating oil is cold; hence alower air pressure will start an engine with lube oil at working temperature.

During trials of newly built ships, it is a part of the main enginemaneuvering tests to put one starting air reservoir on to the startingsystem and continue maneuvers ahead and astern until the pressure fallsto the point that the engine cannot start. The pressure is recovered withthe ship’s trial.

One of the principle factor is to overcome the forces of adhesion b/wthe bearing surface due to the presence of cold lubricating oil b/w them.The walls of the combustion chamber being cold, there will be a greaterheat flow to lower the temperature of the compression at the time of start.An engine which has been warmed up at the metallic surface of thecombustion chamber, viscosity of lubricating oil lowered, will have thestarting speed reached earlier.

Q 15.What is Starting Air Line Explosion? What are the safety devices toavoid it? What are starting line safety devices?

Ans : Explosions can and do occur in diesel engine starting air system. Alsoair start valves and other parts are sometimes burned away withoutexplosion. These problems have been caused by cylinder air start valveswhich have leaked or not closed after operation and have allowed accessfrom the cylinder to the air start system of the flame from combustion.Carbon deposits from burning fuel and oily deposits from compressor areavailable as substances which may be ignited and produce an explosion inthe air start system. If no explosion occurs, the flame from the cylinder andhigh temperature air from compressor can cause carbon deposits in thesystem to burn. Careful maintenance of air start valves, distributors andother parts is vital as is regular cleaning of air start system components toremove deposits.

The lubrication of components is limited as excess lubrication couldcause the air start valves to be stuck by grease which has becomehardened by the heat and oil could accumulate in the pipes from thissource. The draining of compressor coolers and air receivers is important.Drains on air start systems are also checked. Flame traps or bursting capsare fitted at each air start valve.

PROJECT # 4

HEAT EXCHANGERThere 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 isunsatisfactory, because the engine which was probably originally designed for radiator coolingwill run too cold and the seawater will eventually ruin the cylinder block and heads. Keel coolingis suitable for small boats operating in shallow weedy water, but the need for pipe workexternal to the hull is a severe limitation.

Heat exchanger cooling is the most common method, the seawater being isolated incomponents which can be designed to withstand its corrosive effect. The closed fresh-watercircuit 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 removedshould cleaning be necessary. Heat exchanger header tanks prevent aeration of the enginewater circuit which must be designed so that the system is self-venting on initial filling. It isusual for all the components in the seawater circuit to be in series, the gearbox oil and engineoil coolers being on the suction side of the seawater pump and the heat exchanger and anyseawater cooled exhaust. Manifolds being on the discharge side of the heat exchanger. In thecase of turbocharged engines the charge air cooler should receive the seawater first so that thelowest possible air temperature is obtained. The sea-water outlet from the heat exchangershould be from the end cover equipped with the upper connection; this ensures that the tubestack is always full of water. The gearbox cooler size will depend on the type of transmissionused, 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 watertemperature 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 oilof 48 MJ/Kg calorific value per day. It is required to design a cooling water systemconsisting of a cooler where fresh water circulated through the engine piston, jacketand cylinder is cooled by sea water supplied from a single stage circulating pump. Thecooler is a single pass contra-flow type and a fresh water by-pass line is provided formixing so that the jacket cooling temperature is maintained at a higher value (seeattached sketch for further details).

Engine Details

1. Specific fuel consumption - 0.25 Kg/KW(brake)/ hr2. Bore - 0.84m

3. Stroke - 1.6m4. Mean pressure from indicator card - 11.2 bar5. Mechanical Efficiency - 83.9 %

Cooler Details

6. Effective length of cooler tubes - 1.47m7. External diameter of tubes - 20mm8. Wall thickness of tube - 1mm

For turbulent flow in tube take,

��

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 thermalresistance of the tube wall.

For flow through tube it can be assumed that flow will be turbulent when Re (Reynold’snumber) 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 suppliedto the engine is transferred to the circulating water of 4.18 KJ/KgK specific heatcapacity.

(c) The temperature T at cooler inlet considering that 2/3rd of fuel circulating wateris 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 actualmass flow of fresh water through the cooler.

(d) Show that 400 tubes of given dimensions will be sufficient for the heat transferbased on the mean diameter of the tubes given that the velocity of sea waterthrough the tube is 1.648 m/s for the required mass flow of sea water of 1020kg/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 utilizedby the tube holder. Considering thin shell theory, find the allowable thickness ofside 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 coolerbody 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 fueloil 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 =��

tonnes / hour

= 2500 tonnes / hr

Specific fuel consumption =��

�.�.�

B.H.P = ��.��

= 10,000 kW

ηmec = MECHANICAL EFFICIENCY =�.�.��.�

INDICATED POWER =��

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 = �� X 0.842

N = Engine speed in r.p.m.

=11.2 X 105 X 1.6 X

�

4X 0.842 X N X 6

��

= 99308.5 N

Therefore η mec =�.�.��.� =

��.� � � = 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 with10% extra fuel burnt for overload running and given that 28% of total heat supplied tothe engine is transferred to the circulating water of 4.15KJ/ KG K specific heatcapacity

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 fuelconsumption 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 ofpiston cooling .Determine also the mass flow through the by pass line and hence actualmass 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 = �� m & b = �

� m ⟶ (��)

Since from diagram and applying Kirchhoff’s law at junction A

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

65 X �� m + 60 X �

� m = (�

� + �

�) m X T

T = �� + ��

�

T = 63.33OC or 333.33K

And then applying Kirchhoff’s law at B

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

55 X �� X 214.88 - 50 ( �

� 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

= �� X 214.88 + �

� X 214.888 - 53.7

= 161.17 kg/s

(d) Show that 400 tubes of given dimensions will be sufficient for the heat transferbased on the mean diameter of the tubes given that the velocity of the sea waterthrough the tubes 1.648 m/s for the required mass flow of sea water of 1020kg/m3density and 4.2KJ/Kg k specific heat capacity. Use tables for properties of H20 at meanbulk temperature.

θ1 = 50 12 = 38˚ C

θ2 = 63.3 32 = 31.3 ˚ C

Then

θ M = ��(��

)= ��.�

�� .�

θm = 34.54 ˚C

Stantum Number St = {��

�� .�(��)�� (��)�

��(��)

}

Where f = 0.0791(��)��/�

All fluid properties are to be evaluated at mean bulk temperature. Neglect thermalresistance to the tube wall for flow through tube, it can be assumed that flow will beturbulent when Re (Reynolds’s number) is greater than 2100

Re =�.�.�

µ

Where δ = density = 1020 kg/ m

d = inner dia of tube =18mm

µ = 979 X 10-6 N / m2

Hence,

Re =�� .�� .��

�� = 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 �� = �.��

��.�(�.��)��(��.��)�

��(�.��)

�� = 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 =��

��.�� .� � ��

h = 7709 55 KW / m2h

Therefore

Q = ℎ� SOm

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 areautilised by tube holes

Answer: Area utilised by tube holes = �� X d X n

= �� 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 = �� X d 2

Where d = shell diameter

0.18849 = �� 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 platingwhere working pressure does not exceed 5kg/ cm2. Ultimate tensile stress of thematerial 240MN/m2.Take factor of safety is 6

SOLUTION: According to the thin shell theory

f = ��

Where,

P = Working pressure = 0.5 MN/ m2

we know that,

working stress = ��

= �� = 40MN

On substituting the values

40 X 106 = �.� � �� .��

t = 3.06 mm

Since the thickness obtained is very small, for a safer consideration of value, weassume 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 heatexchanger by coating and by cathodic protection systems. The fresh water side of theheat exchanger is protected by addition of chemical additives to cooling water.

Coating varying to bitumen based paints through the epoxy coatingsof various forms can be based to protect water boxes and water box covers. In othercases rubber sheet materials may be bonded to the mild steel or steel or cast iron sothat sea water cannot make contact with steel or cast iron. The steel or cast iron isthen, in effect electrically insulated from the sea water which is in contact with othermetals of heat exchanger.

Cathodic protection systems using sacrificial anodes are also usedwithin the sea water spaces of heat exchangers. The metals used for sacrificial anodesmust be such that they are higher in the electromotive or galvanic series of metals thenthe metals which they have to protect. These metals or alloys are referred to beingactive or anodic. In descending order of the galvanic series, they are cadmium,commercially pure aluminium, zinc and magnesium alloys. In the presence of sea waterand with continued electrical continuity, the anode waste away and so protect the othermetals within the heat exchanger. In some cases a protective or passive film isdeposited on the metals being protected.

Chemical additives in fresh water are known as inhibitors. Thereaction creates a protective film of passive material on the metallic surfaces which theyprotect. They not only protect the heat exchangers but other parts of cooling system inwhich 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 donot use it directly to keep the hottest parts of the engine cool. This is because of thecorrosion which would be caused in the cooling water spaces, and the salts whichwould be deposited on the cooling surfaces interfering with the heat flow.

Instead, the water circulated around the engine is fresh water (or better still, distilledwater) which is then itself cooled using sea water. This fresh water is treated withchemicals to keep it slightly alkaline (to prevent corrosion) and to prevent scaleformation. Of course, if distilled water, which some ships can make from sea waterusing 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 electricallydriven pump pushes the water around the circuit. After passing through the engine,where it removes the heat from the cylinder liners, cylinder heads, exhaust valvesand sometimes the turbochargers, it is cooled by seawater and then returns to theengine. The temperature of the cooling water is closely controlled using a three waycontrol valve. If the water is allowed to get too cold then it will cause thermalshocking which may lead to component failure and will also allow water and acids tocondense on the cylinder bores washing away the lubricating film and causingcorrosion. If it gets too hot then it will not remove the heat effectively causingexcessive wear and there is a greater danger of scale formation. For this reason thecooling water outlet temperature is usually maintained at about 78-82°C. Because itis at a higher temperature than the cooling water used for other purposes (known asthe LT cooling), the water for cooling the engine is known as the HT (HighTemperature) cooling water.

Cooling can be achieved by using a dedicated cooler or by mixing in some of thewater from the LT cooling circuit. The LT cooling water is then cooled in the seawater coolers. The temperature is controlled using cascade control which monitorsboth the inlet and outlet temperatures from the engine. This allows a fast responseto 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 automaticallymakes up any deficiency. Vents from the system are also led to this header tank toallow for any expansion in the system and to get rid of any air (if you are familiarwith a domestic central heating system then you will see the similarities). The headertank is relatively small, and usually placed high in the engine room. It is deliberatelymade 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 ischecked 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 whenthe engine is stopped, or to allow the temperature to be raised to a suitable levelprior to starting. Some ships use a central cooling system, whereby the same coolingwater is circulated through the main engine(s) and the alternator engines. Thissystem has the advantage whereby the engines which are stopped are kept warmready for immediate starting by the engines which are running.

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

A drain tank has been included. This is for when the engine is drained down formaintenance purposes. Because of the quantities of water involved and the chemicaltreatment, it is not economically viable or environmentally responsible to dump thetreated water overboard each time. This way the water can be re used.

PROJECT # 5

MARINE STEAM TURBINE UNIT

The world remembers the sinking of R.M.S. TITANIC, but that R.M.S.stands for Royal Mail Steamer. The 269.1m vessel was equipped with tworeciprocating four – cylinder, triple expansion steam engines and one lowpressure Parson’s turbine each driving a propeller. There were 29 boilersfired by 159 coal burning furnaces that made possible a top speed of 23knots (43 km/h; 26 mph). On paying special attention to the latter, thefollowing project deals with designing of a steam turbine. Why has thetriple expansion steam engine being replaced by steam turbine, the answeris in this project.

To discover more about steam turbines of seagoing vessels and how it isdesigned, take a look at the technical challenge put forth by this project.

INTRODUCTION

A steam turbine is a mechanical device that extracts thermal energyfrom pressurized steam, and converts it into rotary motion. Its modernmanifestation was invented by Sir Charles Parsons in 1884. The speed ofthe steam jet is dependent on the rate of expansion, it had long beenknown and built into reciprocating engines, that a vacuum, or partialvacuum, at the exhaust end of the engine would create a higher pressureratio between input and output and so impart more energy to the pistons,this vacuum is the main function of the condenser in which cooling steamis used to create a drop in pressure below that of atmosphere.

Parson’s idea was to reverse that function and create an exhaust thatwas pressurized, but still below that of the feed. This effectively resulted ina smaller pressure drop between feed and exhaust and a slower steam jet.By repeating this process a number of times most of the energy from thesteam jet can be extracted without the turbine having to destroy itself.

A simple turbine schematic of the Parsons type, rotating and fixedstators alternate and steam pressure drops by a fraction of the total acrosseach pair, the stators grow larger as pressure drops.

There are 2 types of turbines depending on their operation, i.e. animpulse turbine has fixed nozzles that orient the steam flow into highspeed jets. These jets contain significant kinetic energy, which the rotorblades, shaped like buckets, convert into shaft rotation as the steam jetchanges direction. A pressure drop occurs across only the stationaryblades, with a netincrease in steamvelocity across thestage. As the steamflows through thenozzle its pressure fallsfrom inlet pressure tothe exit pressure(atmospheric pressure,or more usually, thecondenser vacuum).Due to this higher ratio of expansion of steam in the nozzle the steamleaves the nozzle with a very high velocity. The steam leaving the movingblades has a large portion of the maximum velocity of the steam whenleaving the nozzle. The loss of energy due to this higher exit velocity iscommonly called the "carry over velocity" or "leaving loss".

In reaction turbine, the rotor blades themselves are arranged toform convergent nozzles. This type of turbine makes use of the reactionforce produced as the steam accelerates through the nozzles formed by therotor. Steam is directed onto the rotor by the fixed vanes of the stator. Itleaves the stator as a jet that fills the entire circumference of the rotor.The steam then changes direction and increases its speed relative to thespeed of the blades. A pressure drop occurs across both the stator and therotor, with steam accelerating through the stator and decelerating throughthe rotor, with no net change in steam velocity across the stage but with adecrease in both pressure and temperature, reflecting the work performedin the driving of the rotor.

Let U=Blade speedCi= velocity of steam at inlet to blade, i.e. leaving nozzle (giving nozzleangle)Ci rel= velocity of steam relative to the blade (giving blade inlet angle)

Co = Velocity of steam at outlet of blade

Parsons Impulse-Reaction

The original blade design was thin section with a convergent path.The designed blades similar to bull nose impulse blades which allowed for aconvergent-divergent path. However due to the greater number of stagesthe system did not find favor over impulse systemsU/Ci = 0.9

If the heat drop across the fixed and moving blades are equal thedesign is known as half degree reaction. Steam velocity was kept small onearly designs; this allowed the turbine to be directly coupled to the propshaft.

Increased boiler pressure and temperature meant that the expansionhad to take place over multiple rotors and gear set. As there is fulladmission over the initial stage, blade height is kept low. This feature alonecauses a decrease in blade and nozzle efficiency at part loading. Inaddition, although clearances at the blade tips are kept as small aspractical, steam leakage causes a proportionally higher loss of workextracted per unit steam. Blade tip clearances may be kept very tight solong as the rotor is kept at steady state. Manoeuvring, however, introducesvariable pressures and temperatures and hence an allowance must bemade.

End tightening for blades is normally used. This refers to an axialextension of the blade shroud forming a labyrinth. When the rotor iswarmed through a constant check is made on the axial position of therotor. Only when the rotor has reached its normal working length may loadbe introduced. Alternatively tip tightening may be used referring to the useof the tips of the blade to form a labyrinth against the casing/rotor. Thissystem is requires a greater allowance for loading and is not now generallyused. To keep annular leakage as small as possible these rotors tend tohave a smaller diameter than impulse turbines. To keep the mass flow thesame with the increasing specific volume related to the drop in pressurerequires an increase in axial velocity, blade height or both -see above.Altering the blade angle will also give the desired effect but if adoptedwould cause increased manufacturing cost as each stage would have to beindividual. Generally the rotor and blading is stepped in batches with eachbatch identical.

The gland at the HP end is subjected to full boiler conditions and issusceptible to rub. The casing must be suitably designed and manufactured

from relevant materials.A velocity compounded wheel is often used as the first stage(s) giving alarge drop in conditions allowing simpler construction of casing and rotorand reducing length. Special steels are limited to the nozzle box.

TURBINE CONSTRUCTION

Vertical Casting

Only the bottom part of the ingot isused.

Rough Forging

It is a requirement that forgings are heavily worked. Any small holes ordefects can become hammer welded together. No forging is carried out

below the plastic flowtemperature as this canlead to work hardening.Forging will allowcontinuous grain flow.

Ultimate tensile stressand elongation checked.This must be near

enough equal in all 3 directions.

After rough machining it is put in for a thermal stability test. For thisfinal machining is given to the areas indicated. The end flange is marked at90' intervals. Then the rotor is encased in a furnace. Pokers are placedonto the machined areas and accurate micrometer readings taken. Therotor is rotated though 4 positions marked on the flange.The rotor is then heated to 28 0 C above normal operating temperature andslowly rotated.

Measurement is then taken at hourly intervals until 3 consistentreadings are taken (hence the rotor has stopped warping). The rotor isthen allowed to cool and a set disparity allowed.For turbine sets operated at greater than 28'C above their designedsuperheat then run the risk of heavy warping as well as high temperaturecorrosion and creep.Final machining is now given. The rotor is statically balanced and thendynamically balanced and check to ensure homogeneity. The rotor isbladed then again dynamically balanced.

TECHNICAL CHALLENGE

In a ship’s propulsion steam turbine installation following dataavailable:-

The turbine is a multistage pressure compounded impulse typewith each stage producing equal power. No steam is bled of atany stage of the machine.

1. Turbine rotor speed - 6000 r.p.m.2. Maximum shaft power - 13.5 Mw3. Input steam conditions - 55 bar 5000C4. Exhaust steam conditions - 2 bar 1500 C5. Steam consumption - 78 tonnes/hr6. Nozzle angle for all stages - 200 C7. Diameter of forged discs for all stages - 540 mm8. Overall length of shaft between inner edges of bearing -

1290 mm9. Density of rotor and blade material - 7856 kg/m3

QUESTIONS:-

1. Calculate velocity of steam at the end of 1st stage nozzlebox.

Given Pressure at the outlet of the nozzle = 40 barNozzle efficiency= 93%

2. Calculate the blade angles assuming blades are symmetricaland mean blade speed for the first stage = 175 m/s

3. Calculate power developed per stage considering bladefriction factor = 0.9

4. Calculate blade efficiency.5. Calculate 1ST stage blade height.6. Calculate number of stages required for developing the

required power.7. Calculate the blade height and radial stress exerted by the

blades in last stage.8. Show that the rotor shaft can be designed as 140 mm

diameter. Take allowable shear stress = 41MN/m2

9. Calculate mass of rotor shaft between the bearing supports,frequency of transverse vibration and first critical speed ofthe shaft in r.p.s.Assumptions: -(a) Exclude mass of the disc(b) Rotor shaft mass as uniformly distributed E = 200 X 109

N/m2

10. Determine natural frequency of transverse vibration ofthe loaded shaft and the first critical speed in r.p.s. Hencecalculate the critical speed of the complete rotor system.

11.a. Discuss important properties of ideal blade materialb. What is blade erosion and how is it prevented?

c. Find the length of the bearing. Allowable pressure inthe bearing = 254KN/m2

SOLUTION:-Notations used:-C a i = velocity absolute at inletC a e = velocity absolute at exitC b = blade velocity = constantC w = whirl velocityβ i = blade inlet angleCf i & Cf e = inlet and exit flow velocity respectivelyCr i & Cr e = inlet and exit resultant velocity respectively

Calculate:-

1)Velocity of steam at the end of first stage nozzle box.Given: pressure at outlet of nozzle = 40 bar

Nozzle efficiency = 93 %

ð Pressure at the outlet of the nozzle = 40 barFrom steam tables,

Enthalpy at nozzle inlet, h1 = 3433.3 KJ/kg (at 55 bar 5000 C)Enthalpy at nozzle outlet, h2 = 3329.97 KJ/kg & h2s = 3322.2 KJ/kgSince, nozzle efficiency = ��

��h1 - h2 = 0.93 (h1 – h2s)

= 0.93 (3433.33 – 3322.2)h1 – h2 = 103.32 KJ/kg

Velocity of steam at the end of 1st stage nozzle box,

C a i =�2 X (h1 – h2)

= �2 X (103.32 X 10)= 454.58 m/s

2)Calculate blade angles assuming blades aresymmetrical and mean blade speed for 1st stage =175 m/s

ð C b for first stage = 175 m/s & Nozzle angle ∝ = 200

From the combined velocity diagram

Sin ∝ = ��

Cf i = Ca i × Sin ∝

= 454.58 Sin 20

= 155.47 m/sAnd also,

Cos ∝ = ��

Cos 200 = �� .��

BE = 252.07 m/s and then

Tan β i =��

β i = tan�� ( ��

) = tan�� ( ��.��.��

)

β i = 31.67 0

3)Calculate power developed per stage consideringblade friction factor = 0.9

ð We know that ,

Blade friction factor = ��

0.9 x Cr i = Cr e

From the combined velocity diagram,Whirl velocity, Cw = Cr i X cos β i + Cr e X cos β e

Where Cr i =��

��β �

= ��.��.��

Cr i = 296.26 m/s

Cr e = 0.9 x 296.26 = 266.7 m/s

On substituting the values, we getCW = (296.26 X cos 31.67) + (266.7 X cos 31.7)

= 479.23 m/s

Steam consumption per hour = 78 tonnes.

Then steam consumption per sec = ��

= 21.67 kg/s

Power per stage = m (Cb X CW)

= 21.67 (175 X 479.23)= 1.81 MW

4)Calculate blade efficiency

ð Blade efficiency = ��

Where, power available = �� m CW

2

= �� X 21.67 X 479.232

= 2.238 MW

Substitute the values in above equationBlade efficiency = �.��

�.��

= 0.8087= 80.87 %

5)Calculate first stage blade height

ð Assuming 25% of the annular circumference is covered,

Coverage of steam area wise = circumference of thenozzle x height of blade

= π D X heightWhere D = diameter of forged disc

Coverage of steam area wise = π x 0.54 x height

Area = 1.696 x height x 0.25

We know that, area x velocity of flow (axial velocity) Cf i = volumeof flow of steam

From steam table at 40 bar, specific volume = 0.08634 m3/kg.

Substituting the value, we get,1.696 x height x 0.25 x 155.47 = 0.08634 x 21.67

Height of blade in first stage = 28.38 mm ≈ 28.5 mm

6)Calculate number of stages required for developingthe required power

ð From the steam tables , hi = 3433.32 KJ/kghe = 2768.5 KJ/kg.

then total power = m (hi - he)

= 21.67 (3433.3 – 2768.5) X 103

= 14.325 MJ/s

Then, the no. of stages = ��

= ��.��.��

≈ 7.933

= 8 stages

7)Calculate the blade height and radial stress exertedby the blades in the last stage

ð Area = circumference of the nozzle X height X 0.25

= π D X height X 0.25, Where D= 0.54 m

Area = 1.696 x 0.25 x height

Then, area x velocity of flow (Cf e) = volume of flow of steam

Cf e= sin β e X Cr e

= sin 31.67 X 266.7

= 140 m/s - (from combined velocity diagram)

From a steam table at 2 bar and 1500C, the specific volume =0.9595 m3/kg

Therefore, 1.696 x 0.25 x height x 140 = 0.9595 x21.67

Height of the blade in last stage = 350.27 mm ≈ 350.5mm

Radial stress exerted by the blade = � ω� ��

= �� ω� ��

= � � � � �� ω� ��

= h x ρ x ω� r= 0.35027 x 7856 x 628.312 x 0.4451

Where ω = � � � � ��

= � � � � ��

= 628.31 rad/sec

r = disc radius + ½ height of the blade

= 0.54�2 + 0.35027�2 = 0.4451 m

8)Show that the rotor shaft can be designed as 140 mmdiameter. Take allowable shear stress = 41 MN/m2

and total mass of the rotor and disc =580 kg

ð Mass of rotor shaft and blade = 580 kg (Given)Length of the shaft between the inner edges of bearingsl = 1290 mm = 12.9 m

Mass per unit length = ��.��

= 449.61 kg/m

Bending moment acting on the shaft = WL� 8� P

= ��.�� .�� .��

= 917.48 Nmwe know that,Maximum shaft power P = � � � � � � �

��

Then the torque acting on the rotor shaft T = � � ��

=� � ��.� � ��

� � � � ��

Equivalent torque = √T� + B. M.� = √21.485� + 0.917� = 21.5kNm

TEQ = 21.5 kNm

Since by the torsion equation,�� = τ

� ………………………. (1)

Where τ = allowable shear stress = 41 MN/m2

J = polar modulus of inertia = � ��

��

r = d2� = diameter of the rotor shaft�2

From equation (1)�� = τ

�

�� = τ� ��

��

∴ T�� = τ x � ��

��

∴ 21.5 x 103 = 41 x 106 x � ��

��∴ d3 = 2.67

d = 138.74 mm ≈ 140 mm

Therefore d=140 mm is perfectly suitable.

9)Calculate mass of the rotor shaft of transversevibration, first critical speed of the shaft in r.p.s.

ð Mass of the rotor shaft between bearing support = 7856 x1.29 x �

� x 0.142 = 156 kg

Moment of inertia of rotorIg = �

�� x d4

= π��

x 0.144

= 1.8857 x 10 -5 m4

The frequency of transverse vibration

fn = � � � ��

= ��

�� .�� .�� .��

= 166.69 HzSince we know that, first critical speed of the shaft in r.p.s.= the frequency of transverse vibrations = 166.69 r.p.s.

10) Determine natural frequency of transversevibration of the loaded shaft and the first criticalspeed in r.p.s. Hence calculate the critical speed ofthe complete rotor system.

ð According to Dunkerley’s formula, natural frequency oftransverse vibration of the blade shaft,

fn = �.��δ��δ��⋯δ��δ/�.��

and the deflection of a simply supported beam subjected to a

point load δ = ��

� � � �

The rotor shaft is subjected to point load, by 8 stage rotorblade, equidistantly placed

Weight of the blade W = (�� – ��) � �.�� = 519.93 N

Since,

i. a = 0.143 m and b = 1.147 m

δ1 = ��

� � � � = ��.�� .�� .��

� � �� .�� .��

= 9.58 x 10-7 m

ii. a = 0.286 m and b = 1.004 m

δ2 = ��

� � � � = ��.�� .�� .��

� � �� .�� .��

= 2.937 x 10-6 m

iii. a = 0.429 m and b = 0.861 m

δ3 = ��

� � � � = ��.�� .�� .��

� � �� .�� .��

= 4.86 x 10-6 m

iv. a = 0.572 m and b = 0.718 m

δ4 = ��

� � � � = ��.�� .�� .��

� � �� .�� .��

= 6 x 106 m

v. a = 0.715 m and b = 0.575 m

δ5 = ��

� � � � = ��.�� .�� .��

� � �� .�� .��

= 6.021 x 10-6 m

vi. a = 0.858 m and b = 0.432 m

δ6 = ��

� � � � = ��.�� .�� .��

� � �� .�� .��

= 4.89 x 10-6 m

vii. a = 1.001 m and b = 0.289

δ7 = ��

� � � � = ��.�� .�� .��

� � �� .�� .��

= 2.98 x 10-6 m

viii. a = 1.144 m and b = 0.146 m

δ8 = ��

� � � � = ��.�� .�� .��

� � �� .�� .��

= 9.937 x 10-6 m

∴ � 1 + � 2 +… + � 8 = 9.58 x 10-7 m + 2.937 x 10-6 m + 4.86 x10-6 m + 6 x 106 m + 6.021 x 10-6 m + 4.89 x 10-6 m + 2.98 x 10-

6 m + 9.937 x 10-6 m

= 2.964 x 10-5 m

Hence the natural frequency of transverse vibration of bladeshaft

f n = �.��δ��δ��⋯δ�

= �.��.��

= 91.5 Hz

The first critical speed in r.p.s = 91.5 r.p.s.

Considering the mass of the rotor shaft only, the weight of theshaft per meter

W = �� .��.��

= 1186.32 N

For UDL the deflection is

� 8 = ��

x � ��

� �

= ��

x ��.�� .��

�� .��

= 1.1342 X 10-5 m

By Dunkerley’s formula

fn = �.��δ��δ��⋯δ��δ/�.��

= 80.267 r.p.s.

≈ 81 �.�. �.

11.

a. Discuss important properties of ideal bladematerial.

-> Properties expected out of the blade material depends on portionof blade in turbine. Good high temperature strength and resistance tocreep is required at high pressure turbine inlet

The L.P. rotor is high strength to withstand the high centrifugalstresses generated by long last row blades and it also has to avoid anytendency for brittleness at relatively low temperatures which exists at thisexhaust side of machine.

b. What is blade erosion and how is it prevented?

The turbine blade undergoes severe corrosion in lower stages whilsthigh strength is required for highly stressed long L.P. turbine blade. Thematerial used in L.P. turbine blade should resist corrosion althoughadditional measures are also taken to avoid erosion. Manufacturing ofblades in various grade 12- 13% chrome steel to give required propertydepending on position of blade in turbine.

Additional toughness is provided by use of tungsten, molybdenum,steel especially for L.P. turbine. Stellite coating is given on L.P. turbineblade but if coating is given it is very dangerous. Tungsten is coated togive toughness for 10000 -15000 hrs running of blade, the blades may bereplaced.

Shield is also provided only for longer durability of blade, no bladecan last forever. Creep will occur, very slowly but suddenly in blade butfatigue normally doesn’t occur, erosion shield is assisting to fight witherosion. Reheating of steam mechanism

Advantages of reheating are that it increases the work done throughturbine; it increases the efficiency of the turbine, reduced erosion of bladebecause of increase in dryness fraction of steam at exhaust. Amount ofwater required in condenser of turbine is reduced due to reduction inspecific steam consumption, a slot (special dog tail) design is cut on thewheel and in this blade is fitted in sliding and they are tied up by tool.

Heat the blades and hammer on sides to remove blades, on it can becut up. Blade root grooves are of various types viz. fir tree, dove tailed, T-grooves. Wire lashing is done on blades together to prevent bending ofblades.

c. Find the length of the bearing. Allowablepressure in the bearing = 254KN/m2

Mass of rotor of blade = allowable pressure X projected area

∴ 9.81 x 580 = 250 X 102 X 0.14 X L

∴ L= 0.16 m = 160 mm


Why has the steam turbine replaced the steam engines?

Ans: Steam turbine has almost completely replaced the reciprocating piston steamengine primarily because of its greater thermal efficiency and higher power-to-weightratio. Even in Electrical power stations use large steam turbines driving electricgenerators to produce most (about 80%) of the world's electricity. The advent of largesteam turbines made central-station electricity generation practical, since reciprocatingsteam engines of large rating became very bulky, and operated at slow speeds.Because the turbine generates rotary motion, it is particularly suited to be used to drivean electrical generator – about 80% of all electricity generation in the world is by use ofsteam turbines. The steam turbine is a form of heat engine that derives much of itsimprovement in thermodynamic efficiency through the use of multiple stages in theexpansion of the steam, which results in a closer approach to the ideal reversibleprocess. The problem was one of basic engineering: the reciprocating engine utilizesthe pressure of steam, but the turbine principle uses the speed of steam, and that isfast, 2,000 mph is fairly typical of a moderate power boiler. In order to utilize thatenergy the turbine blades have to rotate at least have the speed of the steam jet. Evenby the 1880's it was just not possible to construct a device that could rotate at thosespeeds without melting or flying apart, probably both.

What are the various sizes of a steam turbine?

Ans: Steam turbines are made in a variety of sizes ranging from small <1 hp (<0.75kW) units (rare) used as mechanical drives for pumps, compressors and other shaftdriven equipment, to 2,000,000 hp (1,500,000 kW) turbines used to generateelectricity.

What are the various types of classifying a modern steam turbine dependingupon steam supply and exhaust conditions?

Ans: These types include condensing, non-condensing, reheat, extraction and induction.Non-condensing or back pressure turbines are most widely used for process steamapplications. The exhaust pressure is controlled by a regulating valve to suit the needsof the process steam pressure. These are commonly found at refineries, district heatingunits, pulp and paper plants, and desalination facilities where large amounts of lowpressure process steam are available.

Condensing turbines are most commonly found in electrical power plants.These turbines exhaust steam in a partially condensed state, typically of a quality near

90%, at a pressure well below atmospheric to a condenser.

Reheat turbines are also used almost exclusively in electrical power plants. Ina reheat turbine, steam flow exits from a high pressure section of the turbine and isreturned to the boiler where additional superheat is added. The steam then goes backinto an intermediate pressure section of the turbine and continues its expansion.

Extracting type turbines or Regenerative turbines are common in allapplications. In an extracting type turbine, steam is released from various stages of theturbine, and used for industrial process needs or sent to boiler feed water heaters toimprove overall cycle efficiency. Extraction flows may be controlled with a valve, or leftuncontrolled.

Induction turbines introduce low pressure steam at an intermediate stage toproduce additional power.

What are the casing or shaft arrangements in a steam turbine?

Ans: These arrangements include single casing, tandem compound and crosscompound turbines. Single casing units are the most basic style where a single casingand shaft are coupled to a generator. Tandem compound are used where two or morecasings are directly coupled together to drive a single generator. A cross compoundturbine arrangement features two or more shafts not in line driving two or moregenerators that often operate at different speeds. A cross compound turbine is typicallyused for many large applications.

What is the principle of operation and design of steam turbine?

Ans: An ideal steam turbine is considered to be an isentropic process, or constantentropy process, in which the entropy of the steam entering the turbine is equal to theentropy of the steam leaving the turbine. No steam turbine is truly ―isentropic,however, with typical isentropic efficiencies ranging from 20%-90% based on theapplication of the turbine. The interior of a turbine comprises several sets of blades, orbuckets as they are more commonly referred to. One set of stationary blades isconnected to the casing and one set of rotating blades is connected to the shaft. Thesets intermesh with certain minimum clearances, with the size and configuration of setsvarying to efficiently exploit the expansion of steam at each stage.

What are the measures to enhance the efficiency of steam turbine?

Ans: To maximize turbine efficiency the steam is expanded, doing work, in a number ofstages. These stages are characterized by how the energy is extracted from them andare known as either impulse or reaction turbines. Most steam turbines use a mixture ofthe reaction and impulse designs: each stage behaves as either one or the other, butthe overall turbine uses both. Typically, higher pressure sections are impulse type andlower pressure stages are reaction type.

Explain the starting procedure of a steam turbine.

Ans: When warming up a steam turbine for use, the main steam stop valves (after theboiler) have a bypass line to allow superheated steam to slowly bypass the valve andproceed to heat up the lines in the system along with the steam turbine. Also, a turninggear is engaged when there is no steam to the turbine to slowly rotate the turbine toensure even heating to prevent uneven expansion. After first rotating the turbine by theturning gear, allowing time for the rotor to assume a straight plane (no bowing), thenthe turning gear is disengaged and steam is admitted to the turbine, first to the asternblades then to the ahead blades slowly rotating the turbine at 10 to 15 RPM to slowlywarm the turbine.

What are the problems caused due to incorrect warming up procedure?

Ans: The main object of warming through is to ensure straightness of the rotor. To dothis a negligible temperature gradient must exist throughout the rotor. There is atendency for the rotor to hog where the steam is introduced( that is to say the rotorbends due to temperature gradient rather than sagging under gravitational forces) withthe rotor steam is introduced. Hence the rotor must be rotated.The graph below indicates the importance of this.

The line is the out of balance force due to centrifugal force equal to the mass of therotor. Hence, the offset at 3000rpm to cause an out of balance equivalent to the massof the rotor is 0.102 mm.

Testing of the engines after shut down ahead and astern should be taken as partof the warming through process. Close watch of the relevant nozzle box temperatures isa good indication of the condition of the turbine.

Second object of warming through is to prevent distortion of the casing. Rotationof the rotor churns up the steam and provides adequate mixing. With underslungcondensers the temperature gradient is virtually unavoidable, hence separatecondensers are better. The third objective is to prevent thermal stresses caused by thetemperature gradient in thick materials such as at the bolt flanges. Vertical slots areoften provided to help alleviate this problem, this distortion can also lead to nonconcentricity of the casing

This is particularly prevalent in opencylinder designs such as axial plane ordouble casings. Heat transfer rate is at itsgreatest where the steam is condensingon the surface of the casing. This in turnis governed by the inlet pressure of thewarming through steam. Hence, warming

through in steps providing adequate period to stabilize the temperature at each step.Complete warming through cannot occur until nearly at full power, hence,

warming through much above atmospheric saturation temperature is pointless.Also as part of the LP turbine runs at lower temperature, warming above 100oC isunnecessary. Protracted warming through periods are unnecessary. A temperature of82oC at the LP inlet belt in 30 mins is acceptable

Vibration caused by an out of balance of the rotor may be alleviated by runningfor a short period at reduced engine speed followed by a slow increase in speed.

What are the possible failures and their causes in a modern steam turbineand what are the design features to avoid them?

Ans: Problems with modern steam turbines are rare and maintenance requirements arerelatively small. Any imbalance of the rotor can lead to vibration, which in extremecases can lead to a blade letting go and punching straight through the casing. It is,however, essential that the turbine be turned with dry steam - that is, superheatedsteam with minimal liquid water content. If water gets into the steam and is blastedonto the blades (moisture carryover), rapid impingement and erosion of the blades canoccur leading to imbalance and catastrophic failure. Also, water entering the blades willresult in the destruction of the thrust bearing for the turbine shaft. To prevent this,along with controls and baffles in the boilers to ensure high quality steam, condensatedrains are installed in the steam piping leading to the turbine.

How is speed control achieved in a steam turbine?

Ans: The control of a turbine with a governor is essential, as turbines need to be run upslowly, to prevent damage while some applications (such as main propulsion unit or thegeneration of alternating current electricity) require precise speed control. Uncontrolledacceleration of the turbine rotor can lead to an overspeed trip, which causes the nozzlevalves that control the flow of steam to the turbine to close. If this fails then the turbinemay continue accelerating until it breaks apart, often spectacularly. Turbines areexpensive to make, requiring precision manufacture and special quality materials.During normal operation in synchronization with the electricity network, power plantsare governed with a five percent droop speed control. This means the full load speed is100% and the no-load speed is 105%. This is required for the stable operation of thenetwork without hunting and drop-outs of power plants. Normally the changes in speedare minor. Adjustments in power output are made by slowly raising the droop curve byincreasing the spring pressure on a centrifugal governor. Generally this is a basicsystem requirement for all power plants because the older and newer plants have to becompatible in response to the instantaneous changes in frequency without dependingon outside communication.

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, andultimately split in three and sank, all together resulting in thelargest oil spill of its kind in history to that date. What was thereason of such an catastrophic disaster. The answer is in thisproject.

INTRODUCTION

The direction of the ship iscontrolled by the steering gear. As theship moves through the water, the angleof the rudder at the stern determines thedirection it will move. Modern ships areso big that moving the ruddernecessitates the use of hydraulics orelectrical power.

The steering starts at the Bridge.The required rudder angle is transmittedhydraulically or electrically from thesteering wheel at the Bridge to thetelemotor at the steering gear, justabove the rudder.

There are a few commonarrangements for using hydraulic power.There are the 4-rams, 2-rams, and rotaryvane types. The heart of these hydraulicsystems is the variable delivery pump.This type of pump can be controlled byjust moving a spindle. The pump isdriven by an electrical motor at constantspeed.

By moving the control spindleaway from the central point, the pumpstroke increases, and the hydraulic fluidis pumped in one direction. Moving thespindle more from the central point willcause more fluid to be pumped andconsequently more pressure is generatedto drive the rams. Moving the control

spindle back to the original position and then away in the opposite directioncauses the hydraulic fluid to be pumped in the reversed direction. The ramswill also move in the reversed direction.

By using a floating lever feedback mechanism, when the rudder stockhas reached the desired angle, the pump control lever moves back to theoriginal position, and the pumping action stops. The rudder is stopped at therequired angle. Moving the steering wheel to the opposite direction will causethe rudder to come back to the original zero position.

TECHNICAL CHALLENGE

A vessel of 10,000 Tonnes displacement has LBP 120m, Breadth15.6m and loaded draught 6.7m, fitted with a semi-balanced rudderhaving a single guide pintle operated by a 2 Ram electro-hydraulicsteering gear (see accompanying sketch) has the followingparticulars. It is required to design a hydraulic steering gear withthe help of information available and using a Rapson slidemechanism.

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 stockbearing 0.4 m

(m)

Horizontal distance of center of press on rudder from leadingedge 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 velocityof water passed in m/sec, and may be assumed to bemaximum speed of the vessel (S) 16 knots for ahead runningand 50% of vessel’s speed for astern running. (� is the anglesubtended 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 pressurein cylinder due to abnormal conditions.

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

(r) Owner requires 10% increase in rudder stock dia. overcalculated 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 tothe force on the rudder. Accepted ram diameter should be amultiple of 10.

(vi) Diameters of tiller arm assuming that the maximum stress occursat 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 forapproximation and then rounding up to a suitable figure.

(viii) Motor power, rpm and shaft diameter given that the shear stress inthe motor shaft is limited to 50MN/m2 and the motor drivingvariable delivery pump is a 4-pole induction motor connected to440v, 3-phase, 50Hz supply mains. Full load slip is 4% with the 0.8lagging Power Factor Motor should run at 85% MCR and availablemotor power ratings are multiple of 5.

(ix) Calculate the capacity of pump considering probable losses whereco-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 beshown 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 fromdamage 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 ramelectro-hydraulic steering gear.

The steering gear uses Rapson slide mechanism.

a) Max speed of vessel = 16 knotsb) Max rudder angle � = 35°c) Max working pressure on rams ≤ 100 bard) 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 mh) Ram packing thickness = 20 mmi) 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 immersedmiddle 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 mH = Max loaded draught = 6.7 m

A = ��

A = 13.4 m2

Rudder area is 13.4 m2

But �� = 1.5

∴ h = 1.5 bWhere 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 RudderForce 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 centerline of the ship for maximum force

= 35°F = 577 X 13.4 X(16 X 0.5144)P

2X sin 35

∴ Max Force on Rudder = 300.38 KN

(c) Maximum Bending movement and Torque on the rudder stockwith lower pintle.

Since it can be treated as a SSB, when rudder is supported bylower pintle, the max Bending movement is given by

F = �4 + �� ℎ� - R ��

� ℎ�

For equilibrium F = R

Max B.M. = �300.38 �0.4 + �� ∗ 4.48��− 300.38 ��

� ∗ 4.48�� RX

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 stockwithout lower pintle.

Max Torque will be same for rudder having lower pintle andwithout lower pintle

Max. Torque = 120.152 KN-m

Since it is cantilever beam, when it is considered for rudderwithout lower pintle

∴ Max. B.M. = Force X Distance from rudder stock bearing toC.O.P

= F X L

Where L = 0.4 + ��

h = 0.4 + �� 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 pintleThe equivalent torque is given by

Teq =�(M��)� + (T��)�

As we have already found max B.M. and max Torque withlower pintle

Teq =�(567.718)� + (120.152)�

= 580.29 KN-m

20% increase has to be done for undue forces

Teq =580.29 + ��

X 580.29

= 696.349 KN-m

W.K.T. Teq =��

d3 fs

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

= 75 X 106 N/m2

∴ d3 = 16��

= �� .��

� � ��

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 = �(M��)� + (T��)�

=�(1015.28)� + (120.152)�

= 1022.36 KN-m

As per undue forces, we have increased the eq. torque by20%

∴Teq = 1022.36 X 1.2

= 1226.832KN-m

W.K.T. Teq =��

d3 fs

∴ d3 = 16��

= �� .��

� � ��

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 � 2� d

= 100 X 105 X � 2� d

Force = 7853.9816 d2 KN

From Triangle OAB

cos 35 = ��

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 fromrudder stock

= 7853.9816 d2 X 0.8569

= 6730.0768 d2

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

r Distance= 141.138 X 103 X 0.256

= 36.643 KN-m

�� = ��

�

��

�=��

d��= ��

� � ��

d��= ��.��

� � ��

��= 145 mm

(g) Thickness of Ram cylinderBy thin cylinder theory, we know thickness

t = ��

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 = ��

= �� .��

t = 14 mm

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

W.K.T.Voltage V = 440vCurrent I = 20 A

Frequency F = 50 HzNo of poles P = 4

cos� = 0.8 (lag)Motor power = √3 V I cos�

= √3R X 440 X 20 X 0.8Motor power = 12.193 KW

Since power rating should be multiple of 5Motor power = 15 KW

W.K.T.

Synchronous Speed, Ns =��

=��

= 1500 rpm

% slip = �� 100��

RX

∴ = ��

RX100

∴ N = 1440 rpm

Shear stress in motor shaft = 50MN/m2

W.K.T.

Power = ��

T = ��

= ��

� � � ��

∴ T = 99.47 N-mBut, torque T = � 3

��d fs

d3 = � � ��

d3 = �� .��π � ��

d = 0.0216 md = 21.6 mm

As the diameter should be multiple of 5d = 25 mm

This the diameter of motor shaft

(i) Capacity of pumpsCd = 0.94 (given)

Cd = ��

QTh = area of flow X velocity of flow

= � ��

� X �.�

�.�

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

= 0.67.5

∴QTh = π X 0.142

4X 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 = �32

d4 (Polar movement of inertia of shaft)

d = Dia. of Shaft

f� = shear stress = 41 XR 106 N/m

R = d2

T = Torque acting

T = fs � JR =

fs �� 432dd2

= � � fs � d3

16

Now W.K.T.

Equivalent Torque, Teq = �M2 + T2

M → Bending Movement, T → TorqueAlso given length between bearing = 1290 mmMass of Rotor and drive = 580 KgShaft of turbine is considered as UDL with w/unit length

∴ Wt. per unit length, w = 580�.��

w = 449.612 kg/m

B.M. due to load, M = w�2

8

Which can be calculated as follows:-

Now total load on turbine rotor shaft

= WtLength

� Length

= w X l

Two reaction supports = wl2 (for each)

Now total weight is acting through the center of the shaftdownwards

Now taking moment about center pt. considering left ofreaction

Moment due to reaction = wl2� l

2= w�2

4

Moment due to wt. of half beam = wl2

X l4 = w�2

8

Net B.M. = w�2

4 - w�2

8 = w�2

8

Subs. we get,

w = 449.612 kg/m

l = 1.29

∴ B.M. = 449.612 � 1.292 � 9.818 = 917.48 N-m

W.K.T.

P = 2��60

T =260��

P = 13.5 X 106 W

T = 60 � 135� 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 =� � �� 3

16

�3 = ��.� � ��

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 = � 2� d

dia. of shaft, d = 140 mm

Area, A = ��

RX 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 = � 4��

d = ��

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

Subs we get.

fn=�� .��

��.�� (�.��)� � �.��

fn=��.��

��.��

fn = 53.211 Hz

First critical speed is Nc= 53.211 Hz


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

En route from the Persian Gulf to Rotterdam, Netherlands, via ascheduled stop at Lyme Bay, Great Britain, the ship encounteredstormy weather with gale conditions and high seas while in theEnglish Channel. At around 09:45, a heavy wave hit the ship'srudder and it was found that she was no longer responding to thehelm. This was due to the shearing of Whitworth thread studs inthe 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 longermanoeuvrable" and asking other vessels to stand by wastransmitted at 10:20, no call for tug assistance was issued until11:20.

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

All cargo ships are required to be registered with concerned Administration as per Merchant Shipping Act.The Registrar issues the Certificate of Registry. In India ships can be registered in Mumbai, Chennai andKolkata. The Principal Officer of M.M.D in each above individually acts as the Registrar of Indian Shipsunder Act.

STATUTORY CERTIFICATES AND DOCUMENTS FOR MERCHANT SHIPS

Cargo Ships Foreign going (all in original)CERTIFICATES VALIDITY CONVENTION SURVEY ISSUANCE

International TonnageCertificate (1969)

Valid for lifetime unlessmajor change inconstruction.

International TonnageConvention 1969

Surveyed by Class,measurements aretaken, computed.

Issued by Registrar orothers assigned for suchjob as per M.S Act

International Loadline Certificate (1966)

Valid five years withannual and intermediatesurveys

ILLC 1966 andprotocol of 1988

Surveyed by Class Issued by Head quartersof ClassificationSociety

International Loadline ExemptionCertificate

Same as above Same as above Surveyed by Class Same as above

Minimum SafeManning document

Valid for lifetime unlessmajor change in constrn.

SOLAS 1974 (1989amendments)

M.M.D Issued by The Principalofficer M.M.D

Certificates ofmasters, officers andratings

As applicable undercompetence STCW 1995 code

M.M.D Issued by The Principalofficer M.M.D

TonnageComputation Booklet

Valid for lifetime unlessmajor change inconstruction.

International TonnageConvention 1969

Surveyed by Class,measurements aretaken, computed

Checked by M.M.D andfinally approved byD.G.S (Indian Ships)

Intact StabilityBooklet with damagecalculations

Valid for lifetime unlessmajor change inconstruction

SOLAS 1974regulation II-1/22including calculationspart B 25 regn 1-10

Same as above Same as above

For oil tankers above 150 GRT and other ships above 400 GRT

IOPP Certificate +Record of constructionas per Regulation


MARPOL 73/78annex I regulation 5

Surveyed by Class Issued by the Principalofficer (Registrar) ofM.M.D

Oil Record Book Continuous recorddocument

MARPOL 73/78annex I regulation 20

Same as above As approved by theFlag State

Shipboard Oil PollutionEmergency Plan(3copies) SOPEP

As governed by IOPP MARPOL 73/78annex I regulation 26

Same as above Same as above

In addition to all above cargo ships including tankers must have following in original

Safety ConstructionCertificate (SAFECON)


SOLAS 1974 asamended and GMDSS

Surveyed by Class Issued by the Principalofficer (Registrar) ofM.M.D

Safety EquipmentCertificate (500 GRTand above)

Valid Five years &annual (HSSC effectivesince year 2000)

SOLAS 1974Chapter II-2, III andCOLGEG 1972

Surveyed by M.M.DSurveyors

Issued by the Principalofficer of concernedM.M.D

Safety Equipment Planand Record of SafetyEquipment

Record issued everyFull term survey

Same as aboveScrutiny and check byM.M.D Surveyors

Plan approved byD.G.S and Recordissued with SafetyEquipment Certificate.

Safety RadioTelegraphy/Telephonyor GMDSS

Valid Five years &annual (HSSC effectivesince year 2000)

SOLAS Chapter IVas amended

Surveyed by RadioInspector of M.M.D

Issued by the Principalofficer of concernedM.M.D

Exemption certificateSafety Equipment

If required as per SECand valid same term

SOLAS 1974Regulation I/12

Surveyed by M.M.DSurveyor

Issued by the Principalofficer of M.M.D

DOC Specialrequirement for shipswith dangerous cargo

As and when requiredto carry dangerouscargo

SOLAS 1974Regulation II/54

Dangerous goodsManifest

Stowage Plan asrequired

SOLAS 1974Regulation VII,MARPOL annex III/4

Document ofauthorization forcarriage of Grain

As required where grainis required to be carried

SOLAS 1974 asamended Chapter VIRegulation 9

Certificate of Insuranceor other financialsecurity in respect ofcivil liability

As per terms ofagreement with regardto oil pollution damageTOVALOP

CLC 1969 article VII

Enhanced SurveyReport file

SOLAS 1974 ChapterXI/2, MARPOL annexI Regulation 13G

In addition to all above ships carrying noxious liquid chemicals in bulk shall carry

International pollutionprevention certificatefor carriage of noxiousliquid in bulk (NLS)

Valid five years subjectto annual andintermediate surveys asapplicable

MARPOL 73/78Annex IIRegulation 12/12a

Surveyed by ClassSurveyor

Certificate issued byThe Registrar of ship

Certificate of fitness forcarriage of NLS in bulk

Same as above BCH code Section 1.6 Same as above Same as above

OR Internationalcertificate of fitness

Same as above IBC code Section 1.5 Same as above Same as above

Cargo Record Book In support of above MARPOL Annex II/9 Same as above Same as above

For Gas Carriers

Certificate of fitness forcarriage of liquefiedgasses in bulk

Valid five years subjectto annual andintermediate Surveys

GC code Section 1.6Surveyed by ClassSurveyor

Certificate issued byThe Registrar of Ship

OR InternationalCertificate of fitness Same as above IGC code Section 1.5 Same as above Same as above

For passenger ships

Passenger Ship SafetyCertificate

Valid one year only SOLAS/MARPOL/ILLCall combined as required

Surveyed by MMD andClass Surveyors

Certificate issued byThe Registrar of ship

Passenger ShipExemption Certificate Same as above

SOLAS 1974 RegulationI/2 as amended

Surveyed by MMDSurveyors Same as above

Special TradePassenger ShipCertificate

Same as aboveSTP agreement 1971Regulation 6 Same as above Same as above

Special TradePassenger Ship SpaceCertificate

Same as aboveSSTP agreement 1973Regulation 5 Same as above Same as above

For high-speed crafts as defined by M.S Rules requires a SAFETY CERTIFICATEin compliance with Chapter X of SOLAS as amended

With reference to Chapter IX of SOLAS as amended ships to which ISM appliesDocument ofcompliance (Certifiedcopy) DOC

Valid five years subjectto annual andintermediate audits

SOLAS 1974 ChapterIX + ISM provisos asamended

Audit conducted bycompetent leadauditor/s as approvedby the flag state

Flag State or acompetent authority onbehalf of Flag State

Safety ManagementCertificate SMC Same as above Same as above Same as above Same as above

All such ships must possess elaborate SAFETY MANNUAL as document on board

International Maritime Organization (IMO) is the nucleus for origination of all shipping Regulations byway of holding conventions. Various maritime countries all over the globe are signatory members to IMO.These countries usually send representatives to IMO’s conventions when held. All such valid signatorymembers of IMO are also termed as Flag States. For example, India is signatory member of IMO, so Indiais a Flag State. All Indian ships, obviously will fly Indian flag whether sailing high seas or in port.

Each country has administrative body set up by the Government of that country to look after Shipping. Forexample, Government of India has Ministry of Shipping and Transport, a division under which theDirectorate General of Shipping (DGS) has been established to control merchant shipping operation. Forthe purpose of legal proceedings M.S ACT 1958 has been formulated as key Instrument for merchantshipping and its alliances in India.

Class Society (Classification Society) is an autonomous body established for the benefits of shipping, andpromoted by the concerned Flag State for the activities in the field of shipping industry. Every Flag Statecan have one or more Classification Societies for the purpose of conducting ship related Surveys andmonitoring internationally. Names of a few major Classification Societies: -

CLASS SOCIETY HEADQUARTERS SCOPE OF SURVEYS/ JOBLloyds Register of Shipping LRS United Kingdom (London) Ships, Industry, Testing house,

Construction, ISO, ISM, Plan approval,Researches, Seminars

American Bureau of Shipping ABS United States (New Jersey) Same as aboveDet Norske Veritas DNV Norway (Oslo) Same as aboveBureau Veritas BV France (Paris) Same as aboveIndian Register of Shipping IRS India (Mumbai) Same as above

Classification Society is a non-profit organization and operates under the vested guidelines of IACS(International Association of Classification societies) in liaison with Flag States. Merchant shipindividually needs to be classed with any of international Class Societies as stated above right from thebuilding yard. Ship-owners at their option can switch over to another Classification Society if they cansatisfactorily prove their reason to do so. Whenever a merchant ship is proposed to be taken under class theSociety arranges Its surveyor/s to board the ship for a First entry survey which is very exhaustive and maytake a longer time depending on condition of ship and her equipments. After completion of first entrysurvey satisfactorily the ship is assigned with class notation like 100A1 and machinery class LMC (both arefor Lloyds, + is added if the ship was originally built to LRS specifications). From that particular day whenthe ship is assigned with class notation, the ship’s class surveys start. All hull items are normally surveyedas per Special Survey sequences which is of 5 years interval having Intermediate Survey 2.5 + 0.5 yrsand Annual Survey having window period 3 months either side. All items need to be surveyed by 5 years.Machinery items are coded in a survey book and they are surveyed in continuous basis called CSM(continuous survey of machinery).

1

4. BASIC OUTLINES OF PROPELLER DESIGN

Propellers are designed to absorb minimum power and to give maximum efficiency,minimum cavitation and minimum hull vibration characteristics.

The above objectives can be achieved in the following stages:

1. Basic design2. Wake adaptation3. Design analysis

a) Propeller Design Basis

The term propeller design basis refers to:

i. Powerii. Rotational speediii. Ship speed.

that are chosen to act as the basis for the design of the principal propeller geometricfeatures.

Resistance and Power Estimation

A ship owner usually requires that the ship will achieve an average speed in servicecondition (fouled hull in full displacement and rough weather), VSERVICE, at a certainengine power. Initial acceptance will be the basis of demonstration of a higher speedon trial condition (clean ship usually in light displacement), VTRIAL, at some power,i.e.

� VVV SERVICETRIAL Γ

where ΓV is the speed increase due to the fouled hull, rough weather and other effects,and is usually taken as 1 knot.

ETRIALESERVICE x PP (1 � )

where (1+x) is the sea margin that the ship resistance is increased usually by 10 to20% in average service conditions.

2

b) The Use of Standard Series Data in Design

Propeller design diagrams of Standard Series, such as Wageningen-B, can be used infive design options as:

Known variables Required1. Known power (PD), rotation rate (N) and advanced velocity (VA) Dopt

In this case the unknown variable optimum diameter, Dopt, can be eliminated from thediagrams by plotting KQ/J5 versus J instead of KQ vs J. KQ/J5 can be written as:

5

35

525 ( )AA

Q

VQN

VND

DNQ

JK

ΥΥ

since P QND Σ2

pA

D

A

DQ BV

NPVNP

JK

| 2.5

2/1

5

2

5 2ΣΥ

This parameter Bp is plotted againstVA

NDJ

Γ1 and these diagrams are called Bp-Γ

diagrams. On these diagrams optimum Κ0 and Γ are read off at the intersection ofknown Bp value on the optimum efficiency line. Dopt is then calculated.

2. Known power (PD), diameter (D) and advanced velocity (VA) Nopt (required)

If the delivered power PD and the diameter D are known a diagram of KQ/J3 can beobtained.

32323

523 2)(

A

D

AA

Q

D VP

VDQN

VND

DNQ

JK

ΣΥΥΥ

From these diagrams Nopt is calculated on the optimum efficiency line.

3. Known thrust (T), diameter (D) and advanced velocity (VA) Nopt(required)

If the resistance values of the ship are available, thrust T can be calculated using thethrust deduction factor t:

T= RT/(1-t)

T and D are known optimum rate of rotation Nopt can be eliminated using KT/J2.

222

422 N( )

V DT

VND

DT

JKT

AA ΥΥ

3

4. Known thrust (T), rotation rate (N) and advanced velocity (VA) Dopt(required)

T and N are known and optimum diameter Dopt can be eliminated similarly usingKT/J4.

4

24

424 ( )AA

T

VTN

VND

DNT

JK

ΥΥ

5. Determination of Optimum RPM and Propeller Size (Diameter) (GeneralCase)

To determine the propeller diameter D and rate of rotation N for a propeller whenabsorbing certain delivered power PD in association with the ship speed VS.

i) Propeller type is chosen depending on the ship type, initial installationcoast, running costs, maintenance requirements.

ii) Number of blades is determined by the need to avoid harmful resonantfrequencies of the ship structure and the machinery.

iii) BAR is initially determinediv) First it is necessary to determine a mean design Taylor wake fraction (wT)

from experience, published data or model test results.v) Advance propeller speed VA can be determined as VA=(1-wT)VSvi) Diameters of behind hull and open water are calculated as Dmax is assumed

to be usually % of the draught

aTDD max B

a<0.65 for bulk carriers and tankersa<0.74 for container ships

where DB and T are the behind hull diameter and draught of the ship, respectively.

4

When the diameter is determined the diameter should be as large as the stern of hullcan accommodate, to obtain the maximum propeller efficiency. The typical figures ofthe clearances of propeller-hull, propeller-rudder and propeller-baseline should be:

X 5% to 10% of DY 15% to 25% of DZ up to 5% of D

vii) Open water diameter D0 is then calculated by increasing the DB by 5% and3% for single and twin screws respectively.

for single screw propellers1 0.050

�

DBD

viii) This diameter D0 should absorb the delivered power for trial condition atthe optimum RPM which would correspond to the maximum propellerefficiency.

ix) From the Power-Speed diagram (PE vs. VS) PETRIAL is read off at theVSTRIAL

0

2000

4000

6000

8000

10000

12000

12 13 14 15 16 17 18 19 20

VS (Knots)

P E(k

W) Trial Condition

Service Condition

x) Propulsive efficiency ΚD is obtained by iteration for the optimum RPM, N.Initially ΚD is assumed (i.e. ΚD(assumed)=0.7) and the delivered power PD iscalculated as:

)(

)E (TRIAL)D(TRIAL

D assumed

PP

Κ

xi) Bp and Γ values are calculated for a range of N (e.g. N=80 to 120)

5

xii) From the Bp-Γ diagram open water propeller efficiency, Κ0 is read off atthe corresponding Bp-Γ (D0)

xiii) N (RPM) vs. Κ0 diagram is plotted and Κ0max and N values are read offfrom the diagram

0.604

0.606

0.608

0.61

0.612

0.614

0.616

0.618

0.62

0.622

0.624

80 90 100 110 120 130

N (RPM)

Κ0

xiv) ΚD is calculated

00)( 11

Κ ΚΚΚΚΚ D calculated H R w R� t�

xv) The difference between the ΚD(calculated) and ΚD(assumed) is calculated as:

)(D(calculated ) ΚD assumedΚΗ �

iterate if requiredif Η value > Ηthreshold, go back to step (x) and assume ΚD(assumed)=ΚD(calculated)

if Η value δ Ηthreshold, ΚD(calculated) is converged

and ΚD is the latest calculated ΚD(calculated)

xvi) Based upon the latest value of ΚD break power in trial condition PB(TRIAL) iscalculated as:

S

D

SD

TRIALETRIALB

PP (P ( ΚΚΚ)

)

xvii) Installed Maximum Continuous Power is taken as

0.85B(TRIAL)P

6

xviii) PD = PBΚS

xix) 1.158 2.5 and 3.28082/1

A

BB

A

Dp V

NDV

NPB Γ are calculated

xx) PB/DB is read off at the calculated (Bp, ΓB) from the Bp-Γ diagram and themean face pitch is calculated.

3. Engine Selection

xxi) We have optimum RPM (latest), brake power in trial condition PBTRIAL andinstalled maximum continuous power. For engine selection refer to enginelayout diagrams provided by the manufacturers.

4. Prediction of Performance in Service

Prediction of the ship speed and propeller rate of rotation in service condition with theengine developing 85% of MCR

5. Determination of Blade Surface Area and BAR (Cavitation Control)

Cavitation control is carried out for trial condition. This is due to fact that the shipwill have the maximum speed in trial condition

If the calculated BAR is less than the selected BAR, the design stage is completed.

If the calculated BAR is greater than the selected BAR, a new BAR greater than thecalculated BAR is chosen. All the calculations are performed for the new BAR.

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