SEPTEMBER 2018 EXAMINATION OF MARINE ENGINEER OFFICER Function: Controlling the Operation of The Ship & Care for Persons on Board at Management Level NAVAL ARCHITECTURE TIME ALLOWED - 3 HOURS Instructions: - 1. Answer SIX questions, THREE from each section. 2. All Questions carry equal marks 3. Neatness in handwriting and clarity in expression carries weightage 4. Write the full question before attempting to write the answer to same. 5. Illustration of an Answer with clear sketches /diagram carries weightage 6. All unused pages of answer script must be cancelled out by two lines (X) across the page. SECTION – I Q1. With respect to the sacrificial anodes fitted to a ship’s hull.
Transcript
SEPTEMBER 2018
EXAMINATION OF MARINE ENGINEER OFFICERFunction: Controlling the Operation of The Ship & Care
for Persons on Board at Management LevelNAVAL ARCHITECTURETIME ALLOWED - 3 HOURS
Instructions: -
1. Answer SIX questions, THREE from each section.
2. All Questions carry equal marks
3. Neatness in handwriting and clarity in expression carries weightage
4. Write the full question before attempting to write the answer to same.
5. Illustration of an Answer with clear sketches /diagram carries weightage
6. All unused pages of answer script must be cancelled out by two lines (X) across the page.
SECTION – I
Q1. With respect to the sacrificial anodes fitted to a ship’s hull.
A. State the purpose of fitting anodes to a hull structure. B. Describe with the aid of sketch, how anodes may be attached to the hull.
C. Upon inspection in dry- dock it is found that the anodes have not wasted and areas of hull structure have experienced severe corrosion. Explain possible reasons for this situation.
Answer:-(A) Corrosion is the wasting away of a material due to its tendency to return to its natural state, which, in the case of a metal, is in the form of an oxide.
If a metal or alloy is left exposed to 4 damp atmosphere, an oxide will form on the surface. If this layer is insoluble, it forms a protective layer which prevents any further corrosion. Copper and aluminium are two such metals. If, on the other hand, the layer is soluble, as in the cast of iron, the oxidation continues, together with the erosion of the material. When two dissimilar metals or alloys are immersed in an electrolyte, an electric current flows through the liquid from one metal to the other and back through the metals. The direction in which the current flows depends upon the relative position of the metals in the electro chemicals series. For common metals in use in ships, this series is in the following order of electrode _ potential: :
• copper +
lead
tin
iron
chromium
zinc
aluminium
magnesium -
The current flows from the anode to the cathode which is higher in the scale, or more electro-positive. Thus, if copper and iron are joined together and immersed in an electrolyte, a current will flow through the electrolyte from the iron to the copper and back through the copper to the iron (Fig..)
On the ship the galvanic circuit is formed by the propeller which has a large mass of bronze( alloy of copper) which forms the cathode ( positive terminal) and the steel hull forming the anode ( negative terminal), and the sea water surrounding the ship is the electrolyte. The circuit is made from the propeller to the shaft which is connected to the hull through the various supports and the engine bed plate and on the water side from the hull surface back to the propeller through the seawater(electrolyte). which is acidic and so galvanic corrosion is likely to occur in locations where the protective paint coating is damaged or wiped off. To provide an easier path for the galvanic current to flow into the sea water the use of Zinc slabs fitted all around the stern portion and the rudder is resorted to and these are called sacrificial anodes . The zinc slabs have a lower electric potential than steel and so offer less resistance to the flow of the galvanic current . Hence instead of the steel hull corroding the sacrificial anodes corrode and they are renewed at every dry-docking. During inspection of a ship in dry dock , it is observed that almost all the slabs have corroded completely leaving only the bare lugs holding the
slabs , it means that the slabs were of good quality and they have done their job.
(B) 8
(C) The reasons are: * It is possible that at last drydocking the zinc slabs were painted along wiyth the hull surface and the paint on the slabs were not removed( by error)
* The purity of the zinc is doubtful and hence it has not corrioded.
Q2: Give a reasoned opinion as to the validity of the following assertions concerning the ship concerning ship structure: A. Crack propagation in propeller shaft ‘A’ brackets or spectacles frames is indicative of inadequate scantlings and strength; B. The adequate
provision of deck scuppers and freeing ports is as critical to sea worthiness as water tight integrity.
Answer:- (A) The ship is constructed in accordance with classification rules and also monitored during construction by class. To state that the scantlings are in adequate would be wrong because the class rules provide a set of proportional scantlings for the spectacle frame to conform to the shaft diameter and the mass of the propeller. Hence to apportion blame on the inadequacy of the scantlings would not be correct.
It is possible that the stern tube is lined with lignum vitae as bearing for the propeller shaft and these wear very quickly because of the abrasive action of sand which enters freely with the sea water, When this happens the shaft whirls due excessive wear and the unbalanced mass of the propeller which it supports can cause a fatigue condition on the spectacle frame attachment to the stern structure and that could lead to the propagation of the crack.
(B) This statement is correct. The inadequacy of deck scuppers and freeing ports on a ship with bulwarks will retain the sea water washed onto the main deck causing a high free surface effect leading to loss of transverse stability and the ship may capsize. This has the same criticality as water tight integrity of the maindeck, since both conditions lead to loss of the ship.
If the water collecting on the main deck is unable to be drained out by the inadequacy of freeing ports and scuppers, it leads to a situation of capsizing the ship.
In the case of inadequacy of water tight integrity it leads to loss of reserve buoyancy which will sink the ship. Hence both are equally critical
Q3. (a) What is free surface effect? How can be avoided or reduced.(b) Give the components of ships resistance while vessel is ‘enroute’
Answer:- Any tank containing water or any liquid if not fully pressed up to fill a portion of the air vent, will provide a space for the surface to be free to find its level when the ship heels or trims. The effect of the movement of the surface causes the mass of the liquid to shift to one side in the same direction as the ship heels or trims. This action quantitatively reduces the initial GM of the ship and so proportionately also the range of statical stability of the ship , which is not safe for the ship.
To avoid this situation all ballast tanks when filled up should be pressed so that the free surface is eliminated
This cannot be done in the case of oil tanks used for bunkers or lubricants, but they are comparatively of very small capacity compared to the ballast capacity and hence they do not pose a critical situation. Whereas in tankers which carry oil as cargo maximisation of the cargo space is done by providing a small expansion trunk in each tank so the tank is full without any free surface. The free surface in the expansion trunks all added up is still negligible,
When ballast tanks are emptied they should as far as practical be emptied completely to allow any free surface within the tank. Hence ballast tanks need to be cleaned
during dry docking to remove any accumulated sludge or mud so the the flow of water in the tanks through the drain holes is maintained during pumping out the ballast.
(B) The total resistance to the ships motion is made up of frictional resistance and residuary resistance.
The frictional resistance of a ship depends upon :
1 The speed of the ship .
2The wetted surface area.
3 The LENGTH OF THE SHIP.
4 The roughness of the hull .
5 The density of the water.
The frictional resistance is given by
RF= fxSxVn N
Where : f is coefficient which depends on the length of the ship, the roughness of the hull, and the density of the water.
S is the wetted surface area in m.
V is thespeed of the ship in knots and
N is an index of about 1.825.
The residuary resistances of a ship may be divided into:
(i) Resistance caused by the formation of streamlines round the ship, i.e. due to the change in the direction of the water. If the water changes direction abruptly, such as round a box
barge, the resistance may be considerable, but in modern, well-designed ships should be very small.
(ii) Eddy resistance caused by sudden changes in form. This resistance will be small in a ship where careful attention is paid to detail. The eddy resistance due to fitting rectangular stern-frame and single plate rudder may be as much as 5% of the total resistance of the ship. By streamlining the sternframe and fitting a double plate rudder, eddy resistance is practically negligible.
(iii) Resistance caused by the formation of waves as the ship passes through the water. In slow or medium-speed ships the wave making resistance is small compared with the frictional resistance. At high speeds, however, the wave making resistance is considerably increased and may be 50% or 60% of the total resistance.
Several attempts have been made to reduce the wave making resistance of ships, with varying degrees of success. One method which has proved to be successful is the use of the bulbous bow. The wave produced by the bulb interferes with the wave produced by the stem, resulting in a reduced height of bow wave and consequent reduction in the energy required to produce the wave. The relation between the frictional resistance and the residuary resistances is shown in Fig.
Q4. (a) Draw the mid ships section of an oil tanker with Double Hull & name each part. (b) What is Bow Flare? Why is it so important in Bulk Carriers?
Answer:-
(B) The bow flare is the outward curvature given to the bows to ensure that the wake is symmetrically distributed to both sides of the ship. This ensures directional stability when moving ahead. It is important for all ships for the reasons mentioned above. To some extent it reduces wave making resistance.
Q5. (a) Considering the vessel as a compound beam define Bending moment shearing force. Which is the point of Maximum Bending Moment? (b) Sketch and Describe Hatch coaming of a large bulk carrier.
Answer;- A ship may be regarded as non-uniform beam, carrying non- uniformly distributed weights and having varying degrees of support along its length.
(a) Still water bending
Consider a loaded ship lying in still water. The upthrust at any one meter length of the ship depends upon the immersed cross- sectional area of the ship at that point. If the values of upthrust at different positions along the length of the ship are plotted on a base representing the ship’s length, a buoyancy curve is formed (Fig. 2-1). This curve increases from zero at each end to a maximum value in way of the parallel midship portion. The area of this curve represents the total upthrust exerted by the water on the ship. The total weight of a ship consists of a number of independent weights concentrated
over short length of the ship, such as cargo, machinery, accommodation, cargo handling gear, poop and forecastle, and a number of items which form continuous material over the length of the ship, such as decks, shell and tank top. A curve of weight is shown in Fig. 2.1. the difference between the weight and buoyancy at any point is the load at that point. In some cases the load is an excess of weight over buoyancy and in other cases an excess of buoyancy over weight. A load diagram formed by these differences is shown in the figure. Since the total weight must be equal to the total buoyancy, the area of the load diagram above the base line must be equal to the area below the base line. Because of this unequal loading, however, shearing forces and bending moments are set up in the ship. The maximum bending moment occurs about midships.
SECTION – II
Q6. A. Describe how the force on the ship’s bottom and the GM vary when grounding takes place. B. A ship of 8,000 tonnes displacement takes the ground on a sand bank on a falling tide at an even keel draft of 5.2 metres. KG 4.0 metres. The predicted depth of water over the sand bank at the following low water is 3.2 metres. Calculate the GM at this time assuming that the KM will then be 5.0 metres and that mean TPC is 15 tonne. 2017/SR11 2018/SR03 2018/SR0
Answer:-Repeat Question
Q7. A. List the precautions necessary before an inclining experiment is carried out.B. A box shaped vessel, 50 metres long X 10 metres wide, floats in salt water on an even keel at a draft of 4 metres. A centre line longitudinal watertight bulkhead extends from end to end and for the full depth of the vessel. A compartment amidships on the starboard side is 15 metres long and contains cargo with permeability 30%. Calculate the list if this compartment is bilged. KG = 3m 2017/SR11 2018/SR03 2018/SR09
Answer:- Repeat Question.
Q8. A. Define longitudinal centre of gravity (LCG) and longitudinal centre of buoyancy (LCB). B. A ship 120m long floats has draughts of 5.50m forward and 5.80 forward and 5.80m aft; MCTI cm 80 tonne m, TPC 13, LCF 2.5m forward of midships. Calculate the new
draughts which a mass of 110 tonne is added 24m aft of midships. 2017/SR11 2018/SR07 2018/SR0
Answer:-Repeat question.
Q9. With reference to international load line statutory certification, A. State the reasons for the freeboard requirements; B. Explain the term conditions of assignments; C. List the items that may be examined during a related survey after major repairs in the drydock. 2018/SR04 2018/SR07 2018/SR09
Answer:- Repeat Question.
Q10. With reference to fixed pitch propellers: a. Explain Propeller Slip and Propeller Thrust.
b. The shaft power of a ship is 3000 KW, the ship’s speed V is 13.2 knot. Propeller rps is 1.27. Propeller pitch is 5.5m and the speed of advance is 11 Knots. Find:i. Real Slipii. Wake fractioniii. Propeller thrust, when its efficiency, η = 70% 2018/SR09
Answer:- Pitch P:If the propeller is assumed to workin sn unyielding fluid , then in one revolution of the shaft the propeller will move forward a distance which is known as the pitch.
Propeller slip has two values . When considered in relation to the ship speed itg is called apparent slip and is given by
( VT-V)/ VT, Where VT is the theoretical speed of the ship and given by P*n where P is the pitch in meters per rev and n is rps( revs per sec. V is the ship speed in m/sec.
When considered with the speed of advance of the propeller it is called true slip and is given by (VT—VA,)/ VT
where VT is the theoretical speed and VA, is the sped of advance
(B) The theoretical speed in knots is :
(1.27x5.5x3600)/ 1000 KM/ Hr =25..146 KM /Hr
Hence 25.146 x 0.54 =13.57 knots
( 1) Real slip is (13.57-11)/ 13.57= 0.189
(2) The wake fraction is (V—VA)/V = (13.2-11)/13.2 = 0.167
