L"PHIUPPINECOPYRIGHTC1993"ByCHIEFENGR._ FERDINANDG. MARCOSoNo part of thisbookmay be reproducedIn any form, by mimeographingorby any meanswithout permissionIn writingfromthe authorPrintedby: SSPMakatlACKNOWLEDGEMENTThisbook was exclusively prepared to help the Marine Engineers while reviewing the differentsubjects in preparation for the government licensure examination conducted by the ProfessionalRegulation Commission.1t is a compilation ofsolutions to the problems encountered during the recent examination.There are also exercise questions including. an outline to examinees, to serve as an instantrefresher on the most fundamentalconcept and principles in accordance with the scope ofthe examination usUally given by the Board of Examiners at thePRe.It is also a complete practical guide to all apprentice cadet, ship personnel and engineersonboard, on the latest technology tobring youthe most up-to-date coverage possible of highstandard onthe job aboardship. .The author gratefully acknowledges andappreciates the support of the staffandstudents ofSEALANE CONSULTANCYAND GENERAL SERVICES, INe.This book is lovingly dedicated to my wife,Terry and my daughters, Sarah Jane, ChristineJoy. andmysons Ferdinand Jr;, Ferdinand Marcos II who have been myconstant inspiration inmy journey tothe port ofsuccess. .~FERDINAND G. MARCOSChief EngineerPMMA, Class 1981Manila, Philippines11/29/96IIICONTENTSPAGEPARTI MATHEMATICS 1 BasicFundamental of Mathematics(Indudes: Algebra, Arithmetic, Physics, Strength of Material) BoardProblemsandAnswers: 1987-93All Ranks(4E, 3E, 2E, CE) BoardMultipleChoice: All Ranks Useful Engineer Formulas ConversionTables- Guide Only.'"--PARTII ELECTRICITYANDELECTRICALLVDRIVENPROPULSION 71 Definitions, Functionsof Electrical Terms BoardQuestionsand Answers: All Ranks Trouble Shooting of Electrical Component Test Equipment andUses SafetyProcedureonElectrical andInterpretation Motor: Operation andMaintenance SwitchboardProtection Electrical FormulasandSymbols BoardProblemSolving: All Ranks BoardMultipleChoice: All RanksPARTIII STEAMBOILER, TURBINESINTERNALCOMBUSTIONENGINESSectIon I: SteamBoilers ~ ....................................... 151 Type, Uses, Classification Boiler Mountings,AccessoriesandFunctions Boiler Terminology, UsesandFunctions SafetyValves Boiler Water Level Gauges MaintenanceOperation Boiler CorrosionWater Treatment Boiler Water Testing 'Procedures WasteHeat Boiler ProblemsandMaintenance Boiler Safetyand Description EmergencyProceduresSection II: Internal CombustionEngine 191 Definitions, Classifications Prindplesof Operation Component PartsandUses ScavengingProcess TUrbochargingProcess Definitionof Terms BoardQuestionsandAnswers Fuel, LubeOil, FreshWater System StandardOperating Procedures, Trouble ShootingSectIon III: SteamTurbines, Engines 222 Definition, Classification, Operation Fittings andFunctions BoardQuestionsandAnswers- All RanksIVCodeof Ethics 58) , ReciprocatingSteamEngine: Definitions, Advantage$, ConstructionandOperations BoardQuestions: MultipleChoice: All RanksPARTIV ANDAIR-CONDITIONINGMACHINERY 261 Definitions, Characteristic, Functionsof Typical Partso safetyDevices Definitions of Technical Terms Operationand MaintenanceSystem Trouble Shooting Guideto RefrigerationProblems BoardProblemSolvings - All Ranks BoardMultipleChoice: All RanksPARTV PRACTICALENGINEERGUIDES . 317 Main EngineIndicator Diagram Main EnginePerformance Test Fuel-LubeOil TankCalculation Inspection, Measurement, Procedures, Cylinder tlner, PistonRings Crankshaft Deflection CheckingClearancesof Main Bearing,Crosshead, CrankpinBearing Reading on EngineCondition Emergency ProceduresinEngineCylinder DrawDiagramsandInterpretation MonthlyReports, MaritimeRegulations, Survey BoardQuestionandAnswers: All Ranks Test Procedure: safetyMaintenanceProgram Principles, OperationandMaintenance:FreshWater Distiller, Air Compressor, Purifier Ordering SpareParts, Safety Bill BasicInstrumentation WeldingSafety andTechniquesPARTVI' D,RAWING 395PART VIISection I: Safetyof lifeat Sea .........................428 FirefightingandPrevention First Aid Survival at Sea Lifeboat HandlingSection II: 011Tanker. Safety ..................................................................474SectIon III : Inert Cias System 487PARTVIIISection I: MachineShop....................503 WeldingTechniques, ToolsandEquipments, SymbolsSection II: PumpTheory, OperationandMalntenance 531Section III: Control Automation: Introduction 553Section IV: Organizationof Engine Department56S Watchkeeping, SafeOperation, BunkeringProcedures BoardExamsRegulations andRequirementsSection V:v..PART IMATHEMATICS...MATHEMATICSInperforming ourdailyduties asshippersonnel, engineers, andcrewaboard-shipweoften solve simple problems involving tank calculations, ship speed,horsepower, con-sumptionsand mathematical calculationwhichneedour basic fundamental learningprocess in solving everyday problems:BASICFUNDAMENTAL OFMATHEMATICSMULTIPLICATION - is the processin whichit is desired to knowhowmuch onenumber is timeanother.Examples:2 x 6=12432 x 19=82080.32 x 0.0046=0.0014723.9472 x 43.16=170.36115DIVISION - this is the process in which it is desired to know howmanytimes onenumber will go intoanother. -Examples:6 3=2 6/3=2 6=281-;-9=9,-18653.18=1036.28-.1121.33/4=298.93ADDITION - adding numbers in similarterms andaddthenumbers in each columnseparately.Examples:81+5+12=986.5 + 3+.5=102a+7b+3c 946.755a-2b+6c 8.427a+5b+9c .00842955.178422 .. ..SUBTRACTION-to subtractnumbers or algebraic terms, change thesignof thetermto be subtracted andthenadd.Examples:92 -12=8012-2.5=9.58x - (-5x)=13x8x-5x=3xOPERATIONS WITHSIGNED NUMBERS1. ADDITIONa. Fornumberswithsamesigns, addtheirabsolutevalues andprefixthecommonsignto thesum.Examples:(+8) +(-8) +(+4)(-4)==+12-12 ./b. Fortwo numbers withdifferentsigns, subtract thelower absolute valuefromthehigherabsolute valueandprefixthesignof thenumber withhigherabsolute valuetothedifference.Examples:(+8) + (-4)(-8) + (+4)==+4-42. SUBTRACTIONa. Anytwonumberswithsamesigns, subtractthelowerabsolute fromthehigherabsolute valueandprefixthe signof the number with higher absolute to the difference.Examples:(+4) - (+2)(-4) - (-2)= +2= -2b. Any two numbers withdifferent signs, addtheabsolute valuesand prefixthesign of the numberwith the higherabsolute valueto the sum.Examples:(+4) -:- ('-2)(-4) - . (+2)==+6-63..3. MULTIPLICATIONa. The product of two numbers having the same signs is always positive.Examples:(+6)(+3) = (+18)(-6) (-3) = (+18)b. The product of two numbers with different signs is always negative.Examples:(+6)(-3) = (-18)(-6) (+3) = (-18)4. DIVISIONa. The quotient of two numbers having the same signs is always positive.Examples:(+9) / (+3)(-9) / (-3)= (+3)= (+3)b. The quotient of two numbers with different signs is always negative.Examples:(+9) / (-3)(-9) / (+3)= (-3)= (-3)4TEMPERATURESCALE CONVERSIONSubtract 32 fromOF and divide remainder by 9 andmultiply by 5.Ex: 212-32 =180+9 =20x5 =100CTo convert 260 CtoOFDivide by 5, mUltiplyby 9 and add 32Ex: 260 + 5 =52x 9 =468 +32 = 500OF....I DECIMALS- a number lessthan awhole number maybeexpressed as a fraction oras a decimal.one tenth=1=0.110onehundredth=1=0.01100onethousandth=1=0'.0011000oneandthree tenths=1...3.-=1.310When decimal number areadded together or subtracted, thedecimal pointmustbe placed onebelow theother.Examples:a) Add 4.3785 to 29.46,4.378529.4633.8385..b) Subtract 3.8648 from 48.8248.82003,864844.9552Conversion of Percent toOeclmalExamples:88%=0.880.35=35%1.58=158%99.34%= 0.99345Conversion of Fraction to Decimal1/2 = 0.55/8 = 0.6253/4 = 0.75POWER - an indexis a short method of expressing a quantity multiplied by itself anumber of times.Examples:=27(adding indices)33(a subtracting indices)26(multiplying indices)ROOTS - is the opposite of a power andthe root symbol is -r":Examples:the square root of 49the cube root of 2762= 32"2= V49==~ 2 7 == 973RATIO-isacomparison ofthe magnitude of one quantitywithanotherquantityof thesamekind; it expresses therelationship of onetotheotherandthereforestatedinfractionalform. Theratio signis the colon:Example:Thelengths of twobarsare250millimeters and2 meters respectively, theratioof oneto another expressed.250:or 1 :120008note: bothquantities must be sameunitsPROPORTION-isanequation of ratios, expresses that ratioof onepairof quantitiesis equal to the ratio of another pair. Theproportion sign is the double colon:Examples:5 10..20 40 ..or 5 10=20 40or 5=20- -10 406Q. A pump takes 55 minutes to deliver 4400liters of water. How long would It taketo deliver 6000 liters?Let X = timein minutes to deliver 6000 liters.Ratio of timestaken Ratio of quantities deliver55 : x :: 4400 6000X x 4400 = 55 x 6000x = 55 x 60004400x = 75 minutes.METHODOF UNITY - deals toproportionproblemsespeciallywith compoundproportion withmorethan two pair quantities.Example:Ashiptravelling at 12knots cancomplete acertain voyage in 16days. Howmanydayswould theshiptaketo dothesamevoyage at aspeedof 15 knots?At a speed of 12 knots, time=16 daysAt a speed of 1 knot, time=16 x 12 daysAt a speed of 15 knots, time=16 x 1215=12.8 daysPERCENTAGE-isanother method of expressing aratio infractional formusing 100as the denominator andsymbol %.Ratio of 4 to 25= 4 in fractional form25= 16 denominator of 100100= 16%in percentage form.FACTORING-isthereverse of multiplying, it istheprocess of finding thenumbersor quantities which, whenmultiplied together will constitute the expression given to befactorized.7Example: 2 x 3+2 x 4 2 x 5= 2 (3+4 5)3x+2xy xz= x (3+2y- z)y2-16= (y+4) (y- 4)EVALUATION - is the processof substituting the numerical valueof thealgebraicsymbolsand workingout the value of the wholeexpression.Examples:====Evaluate 3 x y + X2- 4 Ywhen x =2 and y =33 xy + x2- 4y3 x 2 x 3 +22- 4 x 318 + 4 1210LOGARITHMS - purpose Is to be reduced the amount of labor and time involved inmultiplication and division and the solution of powers and root.Examples:1. Find the value of 0.04218 x 4750Log of 0.04218 = 1.37489Log of 4750 = 3.67669(add)Sum = 2.3018antilog of 2.3018 = 200.4 ans.==2.38023.6380= -1.25780.05524 ans.2. Divide 240 by 4345Log of 240LOg of 4345differenceantilog ot -1.2578 =8POWER - Find the value of (4.189)2log of 4.189 = 0.6221multiply by the power = 11.2442antilogof 1.2442 = 17.55 ans.ROOT - Find the squareroot of 7365log of 7365 = 3.86723.8672 divide by 2 = 1.9336antilogof 1.9336 = 85.82 ans.EQUATION - is an expression consistiJlg of two sides. one side being;equal invalue to the other,4x +4x =4x =x =10 =18 -82181014x - 2x- 5x =8- 15+287x =21x =3Simplify the following equations:a) (a + b)2 b) (a-b)2Ans. a'+b a' - bx a+b x a ba2+ab a2- ab+ab+b2ab+b2a2+2ab+ b2a2- 2ab+ b2b) Find the value of x and y in the followingequations:5x -7x +12x5y =5y ==x =x =9Substitute:5x -5y=505 (15)- 5y=5075 - 5y=505y=50 - 75Y=-25-5Y = 5c) Add the following mixed numbers.5 3/8, 12 1/4,3 3/443 +849 +-4154LCD 43 + 98 + 308= 1718MATHEMATICS (ALGEBRA)or 21.3751. One number Is 8 times another number and their sum Is 45. Find the unknown?Let x=smaller number8x=larger numberx+8x=459x=45x=45=. 5 smaller number-98 (5)=40 bigger number102. If a rectangle Is 4 times as long as Its width and Its perimeter Is 60 ft. Find thelength and the width. .Let x=width 2(4x)+(2x)= 604x=length ax+2x = 6010x= 60x= 6 widththerefore: 4x=4(6)= 24 length3. How 10ngtWlllit take Oscar and Bong, together to plow a field which Oscar cando alone In ~ days, and Bong do the job In 8 days?Let x = number of days Oscar & Bong can plow the field together.1 = work done by Oscar & Bong in one dayx1 = work done by Oscar in one day51 = work done by Bong in one day8Equation:1+1=1-5 a xLCD: 5+a=140 x13=1-40 x13x= 40 x= 4013\ X =3.08days4. In,e sum of two consecutive number Is 26. What are the numbers.Solution: Let x=1st numberx+1=2nd numberx+x+1=262x+1=262x=26-1x=252x=12.51112oQ. When you add 5/8; 7/12 and 11/24, what will be the Sum?~ + . L + . 1 18 12 24 Find: LCD= 2415+ 14+ 11 =~ or1 -224 24 3Q. When you subtract 5/6 from8/15, what will be the difference?~ _ J l LCD=306 1525- 16= .9.or.a30 30 10Q. What Is the product of 5/8 and 4n?.s,x~ = ~x.!= .s8 7 2 7 14Q. The quotient of 13 divided by znIs:13+.3.-.1.3.X.1. .9.Lor 30 .1=7 1 3 3 3Q. Solve for x In the equation 12x+25- 35 = 14x+22x- 212x +25- 35=14x +22x- 212x- 14x- 22x=- 22+35- 2512x- 36 x= 35- 47-24x= -12x =..=12-24x = ..12Q. Awire is to becut so that one piece is shorten than the other by 8 meters. Howlong are the pieces If their combined length Is 24 meters.x + x 82x - 82xxxx- 8=16- 8CONVERSION FACTORS= 24= 24= 24 + 8= 322=16meterslongerwire= 8metersshorter wireTemperature SCale:1. Convert 1oocofand212FDCAns.OF=9 C+32DC=2..- (OF 32)-59=-i... x100+ 32 =5 (212 - 32)-5 9=900+32 =...E- (180)--59=180+32 =900--OF=2129DC=1002. Convert 3000CofAnswer: 572.oF4000FDC 204.44 OC800Cof176FMeasurements:1. Convert the followings:a) 60 milhr. - FtiSec, meter/sec.b) 6.56 kmlhr. FtiSec.c) 375hp- wattsd) 700 mm - feet13Answer: a) 60 ml x 1.6 km x1000 m x 3.28 ft. x 1 hr. x 1 min.hr 1 ml. 1 km. 1 m 60 min. 60 sec.= 314880 =87.47 ftIsec.3600= 87.46 ftIsec =26.67 mlsec3.28ft1mb) 6.56 kmlhr x 1000m1 kmx 3.28 ft. x1m1 hr3600 sec.=21516.8 ft.3600 sec=5.98 ft./sec.c) 375hp x 746 watts1hp= 279.750 watts.d) 700 mm x 1 cm x10mm.= 2.30 ft.1 Inch x2.54cm. x1ft.-12 In.Q. Atruck's speed Increases uniformlyfrom36 km/hr. to 108km/hr.ln 20seconds,determine the:a) average speed (velocity)b) the acceleration In meterlsec.c) the distance Sin meter covered during this period.14Solution: 36 krnlhr =10 mlseca) V =VI + Vo2=30+ 102= 20 mlsec.V =36+ 1082=72 km./hr.108 kmlhr =30 mlsec.b) a =VI - Vot=30 - 1020= 1 mlsec.2c) S =Vt= 20 (20)S = 400m.PYTHAGOREANTHEOREM'or RIGHT ANGLE TRIANGLEI1. The baseof a triangleIs 5ft. andaltitudeIs 8ft. WhatIs thehypothenuse of thegiven triangle?Figurea =8 ft.b=5 ft.Formula:C2 A2+ b2~ "== 82+ 52C="54+25= Vi9C =9.43 ft.2. Findthevaluesof thethreetrigonometricfunctions of anangleAIf Its sineIs315.Ba=3Cb =?Note: SOH-CAH-TOASine=opposite=ale=3/5hypothenuseCosine=adjacent=b/c=4/5hypothenuse.Tan=opposite=alb=3/4adjacentBy Phythagorean Theorem:A'-------'C2= a2+ b2b2 = C2-a2b ="52- 32b ="25- 9b =V'16b = 4Q. StatethePhytagoreanTheorem-expressedthat thehypothenuseIsequal tothesumof the squareof the two legs. It Is also called right angle whOse formula ca =at + b2; and the angle sides areopposite, adjacent and base.163. The dlamiteraf a'rounditeel bar IsSOmm. What Is the biggestnut that can be madefrom the bar.By: PhythBgoreBn Theorem.aa=d2= (50)2= 25002="1250= 35.3555 mm.sizeof thesquarea2=1250 m24. What Is the circumference of a circle whose radius Is 71/2 meters?Formula: Circumference of a circle=21rrwhere r=71/2= 7.5C=2{3.1416) (7.5)=(6.2834) (7.5)=47.13 meters5. What Is the lateral area of a sphere whose diameter is 10ft.Solution:Area of sphere=1102=3.1416 (10)2=3.1416 (100)=314.16166. Find the area of a circle whose diameter Is 3 ft. What Is the area ,In. millimeter?Ans. A = 1i' d24= 3.1416 (3)24= .7854 (9)= 7.068 sq. ft.= .7854 (914.4 mm)2= 656694.42 sq. mm.7. Find theheight of a cylinder tank which hold 250 gallons and dla. 24 Inches?Volume=.785402h1 gal.=231 cu. in.height=volume.785402=250 (231).7854 (24)2=127.6' approx. 101/2ft.8. A cylindrical tank 12 ft. longholds 2600 gals when fUll,what Is the diameter ofthe tank?VolumeDiameter= .7854 (0)2H= Volume.7854 (H)1 gal. = 231 cu. in.02= 2,600 x 231.7854 (12x 12)02= 600600113.09o ="5310.81incheso =72.87 inches179. Ahexagonof equal sides Is InscribedInacircle whosecircumferenceIs95 em.What arethe length of the sides of the hexagon?Area=2nR2Circumference=2 ITrRadius=C21T=952 (3.1416)=956.2832=15.12em.Sincehexagon has 6 equal sidestherefore length equal is 15.12em.10. Theservicetank of acontainer ship Is 15 ft. Indiametera n ~ 7 metershigh. Howmuchfuel 011 canIt accommodate If thespecific gravity of afuel Is 0.95assuming novolume expansion.Given:h=7m=22.96ft. Sp. gr. =0.95dia.=15ft.Solution:Volume=areaof base x height=.7854 d2h=.7854(15}2 x (22.96)=.7854 (225) (22.96) (0.95)Volume=8354.5 ft.311. WhatIsthe minimumdiameterof aroundstock necessaryto makeasquarekey5" on eachside?By: Phythagorean Theorem.x =V(5)2 + (5)2="25 inc.2+ 25 inc.2=V50 inc.2x = 7.07inches..1812. FInd the volume of the givencylinder and Its content In metric ton of fuel 011whose specific gravity Is 0.96 ?dla = 5 ft. h=5.25m where5 ft.=1.52mIVolume=Area of base x height=.7854 (1.52)2(5.25)=.7854 (2.31) (5.25)=9.52 m3x .96s.g.Volume=9.145 MTPUMPSPROBLEMS1. A single acting power pump making 200 rpm has dimensions 5" x 6" x 4". SlipIs 4.5%. What Is Its actual discharge in gallons per minute? (G.P.M.)1 gallon = 231 cu. in.GPM=vol. of cyl. x no. of strokes x Efficiency231=.7854 x 62x 4 x 200 x 0.955231=21601.64231GPM=93.52. A double bottom tank holds 6530gallons. A duplex double-acting pump 8" x 6"x 10" makes 35 double strokes per minute. Leakage 10%. How long will it take topump out the tank?GPM=vel, of cyl. x no. of strokes x Efficiency231=.7854 x 62x 10x(35 x 4) x .90231=35625.74231GPM= 154.224therefore 6530 +154.224=42.34 minutes19------...--__43. A duplex double acting pump 4"x 6"x 6" makes 25 RPMslip 4%. What Is Itsactual discharge In G.P.M?4. Aship covers242.6 actual milesIn aday. Findthepitchof thepropeller Ifefficiency Is 87% and speed Is 98 RPM?Formula:Pitch =6080 x observed milesN x 60 x 24 x E= 6080 x 242.698 x 60 x 24 x 0.87= 1475008122774.4=12 ft.5. A ship travels 5742 miles In 26 days, 16 hour and 8 minutes. Find the averagespeedIn knots for the entire voyage.No. of min. per voyage=[(26 x 24)] + 16 x 60 + 8= 640 x 60 + 8=38400 + 8=38408 min.Mile per minute = 5742miles =0.149538408 mins.Mile per hour = 60 x 0.1495= 8.97or 9 knots.6. A ship makes an observed speed of 17knots per hour. The engine speed Is 17.5knots. What Is the propeller slip In % and how many nautical miles the ship makesIn 24 hours?20'If% slip =ES - OS x 100ES=17.5 - 17 x 10017.5=0.02857 x 100=2.857%Nautical miles = observed speed x 24 hours= 17 x 24 hours= 408 knots or= 408 NM7. A merchant shipnavigated a distance of 7,200miles In 22 days, 12hours and 30 minutes. Compute the average speed for the whole voyage.,Ans. SttAve. speed=7200 NM= 22 days x 24 + 12 + .5=540.5 hours=distance =7,200 mi.time 540.5 hrs.=13.32 knots.8. A ship crane lifts a 1,500 Ibs. steelbeam to a height of 44 ft. In 10 sec. Find thepower developed.Given: F=1500 lb. x 1 kg=681.82 kg.2.2 lb.d=44 ft.=13.415 Mt=10 sec.Power=work donetime ellapsed=force x distancetime=681.82 kg. x 9.8 m/sec.2x 13.415 m.10 sec.= joule per sec. or watts9. Awire 120 Inch long;with across section of 0.125In2hangvertically when a loadof 450 Ibs. Isappliedtothewire it streches0.015 inch. FindYoungModulus ofElasticity. ..21y = Stress=F/A-Strain AULy = 450 Ibs.10.125 in.20.015 in. 1120 inch.y = 3600 psi0.000125Y = 2.88x107 psi or28.800.000 psi10. Aship left port with 12000barrel of fuel 011 on boardat 18 knots, the consumptionIs 400 barrel per day, after the vessel hastravelled 2,000 miles, what Is the steamingradius?Speed =18 knots x 24432 milesCons. of oil per 1 mile =400 -:- 432=.9259 barrelFuel cons. at 2000 mile =2000 x .9259=1851.8 barrelFuel on board =12000 - 1851.8=10148.2barrelRemainingsteaming radius = 10148.2.9259=10.960 miles11. Aships make 320 mllday at 70RPM withpropeller pitch of 21 ft. What Is thepropeller efficiency?Propeller Eft. = ED x 6080 ft.P x RPM x Time= 320 x 608021 x 70 x 1440= 19456002.116.800= .9191 x 100% = 91.91'12. Your engineconsumes 130grams of fuel per BHP-HR. Howmanygallonsof fuelwill your engine developing 12000 BHP, consume dally with specific graVity of fuelat .92?22cons.ldaycons/day in gal= 130gr.lBHP-HR x 12000BHP x 24.92 x 1000.000 grms.= 31449000920.000= 40.7 rna= 40.7 x ~ x 0.2642 gal.~ 1L= 10.752.9413. Your dallyusefuel tank hasor diameter of 7ft. Every 4hourswatchthe heightlevel goes down15Inches. Wl)at Is your average hourly consumptionIn liters?Given:Tankdia. =7 ft. = 84 inchesHeightdiff.=15 inches,..,.Consumption Vol. =.!! 02h4=.7854(84)2(15)=83126.736 in3per 4 hrs.=20781.684 in3per hr.Cons. in liters =20781.684 x 1 liter61.0128 in3=340.611Litersper hourCOMPUTATION FORFUELCONSUMPTION ONBOARD1. MYDonaEvelynconsumes20MY/daysailingwhosefuel specificgravityat 15C=.9730 and correction factor at 85C heated Is .9542. Find the cons. In liters, perwatchhour andminutes.Solution: 1. 20 MT x 1000L =20,000 Liters1 Ton2. 29000 L.9730(sg) 15003. 21,541L6 watch/day= 20.554 =21.541 Liters.9542 (CF) at 8500=3590 Uwatch=897L/Hr.= 15Umin.232. Find the fuel consumption In GRMS-BHP/HR whose cons. per day Is 27.10 MT(metric ton) and actualBHP Is 7109.52.Solution:=27.10MT x 1000 kg=27.100 kg/day1 ton=27100kg/day=1,129 kglhr.24 hr/day=1,129 kg x 1000 grms = 1.129.166 grmslhrHr. 1 Kg=fuel cons. in grms=1.129.166actualBHP 7109.52@=158.82 gms-bhp/hr.Q. Findthecylinder 011 Ingrm-bhp/hr. whoseconsumption189.36liters/day,maximum BHP8200; Specific gravity =.95; ave.rpm=141.30and shop trial rpm =150Formula:Cyl. oil cons.=Ne x Vd xVax 1000whereNe= shop rpmN x nex 24 N = actualrpm=150 x 189.36 x .95 x 1000 ne= ratedoutput141.30 x 8200 x 24 Vd = cyl. oil cons.=26.983.800vo=specific27.807,840 gravityCyl. oil cons.=0.970 grms-bhp/hr.TANKS CALCULATIONS WHEN BUNKERINGQ. Your fuel tank on board capacity Is 1500 M3 (cubicmeter) at 100%full. Howmanymetric tons are you required of fuel whose specific gravity is .9768 at 15C If tank tobe filled up to 95% full of fuel whose temperature Is 45C, coefficient of expansion Is.000720 given data:24Formula:Net vol.M3FirstSecondTo be bunker======(T2- T1x coef. of expansionx vol.m3)1500 m3x 95% = 1425M31425 - (45-15x .000720 x 1425)1425-30.781394.22 m3at 15C (0.9768)1361.87 MTQ. Fuel Consumption per voyage distance?How much fuel be consumed to cover the distance of 7,000 miles?Given datas: Bore =680 mm: stroke =1250 mm; 6 cylinder mechanical efficiency =85%, MEP=8.5 Kg/cm2;RPM=140; Pitch =3.15 M, F.O. cons. gr-BHP/Hr =156, F.O.S.G. at 15DC=0.9700, Heating temp. of F.O. + 85DC,propeller slip =5%.where: Bore =68 em ; Stroke =1.25 ; Area =3,631.689Solutions:1. cylinder constant =L x A = 1.25 x 3,631.684500 4500=1.00882. Prop. dist. = 3.15M x 3.28Ft. =10.33Ft. =.00169736 Mile6080FtlMi.3. BHP =cyl. constant x RPM x P. x ME x no. of cyl.100=1.0088 x 140 x 8.5 x 85 x 6100BHP =61224. F.O. Cons.lHr. =6122 x 156 gms=955,032 -7- 1,000,000=0.95503 MTProp. distance/min. = .00169736 x 140 RPM =0.23762 mi.Prop. distance.hour = 0.23762 x 60 =14.25 mi.Prop. distancelhr. with5% slip= 14.25 x .95 =13.54 mi.= 7000 miles13.54 mi/hr.= 5 1 6 . ~ 8 Hours.Total consumption = (0.95503MT/HR)(516.98)= 493.74MTQ. A 1,500 HP turbine operating at full load for an entire day requires the burningof 6.5tons offuel 011. Calculate the fuel consumption in pounds per horsepower hour.Given: Fuel cons. = 6.5 tonsHP of turbine = 1,5006.5 tons x 2000Ibs. = 13,000 Ibs.1 tonFuel cons. = 13,000 Ibs.1500 HP= 8.666lb. per horsepower-hr.25Q. A ship travels 5700 miles In 26 days, 16 hours and 8 minutes. Find the averagespeed In knots for the entire voyage.Given:distancetimeAve. speed= 5700 mi.= 26 days, 16 hr. and 8 mins.=26 x 24 + 16 + .133=640.133 hrs.=distance travelledtime ellapsed= 5700 miles640.133 hrs.=8.90 knotsQ. Arevolutioncounter reads69,985at 8:00amat 11:00amtheclockwasadvanced 17 minutes and at noon the counter reads 87, 319. What was the averagespeed onthe 8 to 12 o'clock watch?Formula:Ave. SpeedAve. speed= advanced in counter readingminutes in watch= 87,319 - 69,9853 hrs. (60) + 43 mins.= 17334223= 77.73 RPM.Q. Afuel 011 has a specific gravity of 0.948 at 24C. What is Its specific gravity at15C? Correction coefficient is .00063 per1DC.Given:0.94824.5C15C0.00063===SG =T1T2corr. coeff.Solution:a. T1- T2= 24.5 -15 = 9.5Cb. 9.5Cx .00063 = .005985c. 0.948 + .005985 =0.9539 SG at 15C26Q. Specific gravity of diesel ollis 0.865 at 30F. What Is Its gravity at 84F? S. G.correction Is .00037 per 1of.Given:0.86530F84F0.00037 per 1F===SG =T1T2corr. coeff.Solution:a. T2- T1= 84-30 =54Fb. 54 x 0.00037 = 0.01998c. 0.865 - 0.01998 =0.8450 at 84FQ. DurIng Bunkering, how much shlpownner will lose If F.O. supplier supply youF.OI1 at $80per MT. The supplier figures on the deliveryreceipt are S.G0.9785at 15C;pumpingtemp. 25C; total volume515m3 Yourrequirement Is500MT.Beforebunkering hydrometer test shows: S.G0.9525at 35Cafter bunkering sounding wastaken and found 511m3at 40Cafter applying ship trim correction.NOTE: Ship owner will lose if you use supplier figure, wJII not if you use interpolatedhydrometer figure:Solution:a. Supplier figure in Metric Tons:MT =SG .9785 x 511 M3- [(40 - 15x .000720 x 511)]= .9785x(511-9.198)=.9785 (501.802)MT= 491.01b. Using Ship Figure by hydrometer test: S.G .9525 at 35CS.G at 15C = .9525 + (35 - 15x .000720)= .9525 + 0.0144S.G = 0.9669 .MT = .9669 x 511 - (40 - 15x .000720 x 511)= .9669x511-9.198= .9669 x 501.802MT= -485.19- ShipFigureTherefore: 491.01 - Supplier figure485.195.82 MT short of delivery5.82 MT x $90m = $523.80 Losses27c. Using All Suppliers figures:MT = .9785 x 515 - (25 - 15x .000720 x 515)=.9785 x 515 - 3.708=.9785 x 511.292 m3MT =500.30OwnerLosses = 500.30 - MTSupplier- 485.19 - Shipfigure15.11 MT x $90 = $1,359.90Formula of fuel mixed with specific gravity:MIXED S.G.= (Qty. Before Loading:ms>.lS.G.) +'Oty.Received m3xS.G.Qty.Bef. Loading + Qty. Received m3= 100 m3(.950) + 200m3(.960)100 m3+ 200 m3= 95 + 192300S.G. = 0.956Q. A hydraulic Is fitted with a raised reservoir to prevent cavitation and gives a 6meter column of all at specific gravity 910 kg/cm3 Determine the pressure workedat the pump Intake port.Pressure =Solution:I. ForceII.=6 x 910 x 9.81= 53,562 NForce =53,562Area 1 m2=53,562 Pa=53.56 KPaQ. In aforce multiplication systemthe area ratio Is 100:1. The large piston diameterIs 150 mm and It move through a distance of 130 mm. If the small piston stroke 400times. What,dlstance does It travel per.stroke.Solution:I. Volume to displace large piston in volume displace by a small piston:A = 'ii'D2h428= 0.7854 (0.15)(0.15)(0.13)= .0022972 m3II. Area of Small Piston= 1xO.15xO.15xO.7854100=.0001767 m2III. Total Stroke = Volume = 2.2972 x10-3 m2Area 0.1767 x 10-3 m2= 13 meterIV. Single Stroke = 13 m = 32.5 x 10-3m400=32.5 mmBOYLESLAW:1. An accumulator In a hydraulicsystemIs precharged to 900 KPaand Is then filledwith hydraulic fluid until the gas pressure shows 2,700 KPa. Howmuch 011 has beenpumped In, If the accumulator volume Is 0.4 m3V, =P, =P2=Formula:Y.1.=V20.4m3900 + 101.3 KPa2,700 + 101.3 KPaP, V2=V, P,PIP2=1001.3 KPa x 0.4 m32801.3KPa=0.143 m3CHARLES LAW:2. A rubber gas reservoir has a volume of 0.1 m3at -14OC.ltstemperature Is raisedto 90GC.What Is It volume Increase If the pressure remains the same?V2= V,x T2 =0.1.m3x (90 + 273)T,-14 + 273=0.1 m3(363)259=0.14 m3Q. Two days after a tank was filled with arrival ballast you check the 011 content Inthe tank and found 0.5 em. of 011 on top of the water. Dimension of the tank L =43 mB =21 mi d =22 m. Is It okey to maritime regulation the amount of 011 to discharged29overboard? If no. What shall you do?NOTE: By Regulation: Required_1of total oil volume by parts can be discharged30,000'itsea.Solution:I. Total volume of the tank = Length x Breadth x Depth= 43 x 21 x 22= 19,866 m3II. Total volume of oil in the tank =43 x 21 x 0.005 m=4;515 m3III. By Regulation volume can bedischarged~ , ~.'FOURTH ENGINEER - January 1989=19,866 m330,000=0.6622 m31. A trapezoidal planefigure with sides in meters measuring 8 3/4, 105/8,53/4and211/4. Find the perimeter. Give your answer in mixed number. What Is the area of theabove figure? The parallel sides are the 10 5/8 and the 21 1/4.Solution:abPerimeter =21 1/4+83/4+105/8+53/4=85+35 +85+23------4 4 8 4=170+70+ 85+468P =371 or 463/88Area = 1/2 (a +b) (d)=(21 1/4 + 105/8) (53/4)230= 85+8523- -4842=85+85 23- --8 16 4=170+85 2316 4- 586564A =91 41 m2642. At the start of your 4-hour watch,the readingof therevolution counter of themain engine Is 996,430. At the end of your watch the reading Is 026,430. What Is theaverage rpm. If the time will be advanced 20minutes, during the watch, What will i)ethe reading at the end of the watch? "Given: Previous reading =996430End of watch =026430Advanced 20 mins.Solution:Revolution before the counter set to 0 is 1,000,0001,000,000 - 996,430 =3570Total revolution after the watch =026430 + 3570=30,000 rev.=30,000 rev.240 - 20 mins, advancedRPM =136.363. A lookout looking towards the bow of the ship is standing In the bridgewith thelevel of his eyes about 50 meters above the water line. The distance of the bowfromthebridge Is 160 meters andits height fromthelookout be able tosee a floatingobject? What Is the distance of the floating object fromthe bridge?SOm31By Similar Triangle:50160 + dSOdSOd50 d= 30d= (160 + d) 30= 4800 + 30 d= 30 d = 480020 d = 4800d = 480020d = 240 m distance from bow.The distance of object from bridge:= 240 + 160= 400 M4. A 12-knot ship eensumes 125MTof fuel 011 per day. How many days will It takeher to navigate a distance of 6,280nautical miles and how many metric tons of fuel(who) will she consume? If the unpumpable fuel Is about 3% and the allowance for.delay dueto badweather that maybeencounteredIs20%.What Isthefuelrequirement to complete the voyage?Given: Ship speed=12 knotsFuel cons. = 125 MT/dayDistance=6280 n. miles3%=allowance for unpumpable20%=allowance for delay and weatherSolution:Days to navigate ===Totalallowance =Fuel consumed for the voyage ==326280 miles x 1 day12 n.mlhr. 24 hrs.6280 miles288 hrs.21.8 days3+20 = 23%-21.8x125 1.233351.75MTTHIRD ENGINEER- January 1989Q. What size circular bar Is required to make ahexagonal nut of 16mmsidesalongthe circumferences? .16mma= b=16mm Sin 30= 8d=2xa a=2 x 16 a=8=32mm sin 30a=16mm2. The specific fuel consumption of the main engine rated at 12000 metric brakehorsepower Is 155 g/Bhp - hr. What Is the fuel consumption in metric tons to makeavoyageof 6,280nautical milesat thespeedof 14knots?Allow10% fortheunpumpable In the fuel storage tank. CGiven:BhpSp. fuel coDistanceShip speed10%Solution:Fuel oil cons.Day the shipsFuel oil cons.= 12000= 155 g-bhplhl= 6,280 nm- 14 knots= IIowance= x 12000bhp x 24 hr/day1,000,000 grrnlton= 44,640,0001,000,000= 44.64M.T./day= 6,280 n.m.. x1 day14 knots 24 hrs.= 18.69days= 44.64 x 18.69(1+ 10%)= 44.64 x 18.69x 1.1= 917.98MT.333. The revolution counter reading at the beginning of a 4 hour watch Is 996,430. Attheend of the thecounter reading026,430. During the watch the time wasretarded by 20 minutes. What Is the average rpm of the main engine?996,430= 026,430previousreading =end of watchretarded 20 min.,Given:Solution:\The counter reset to 0 @ 1,000,000Rev. before the counter reset to 0 =, 1,000,000 - 996,430= 3570revEnd of watch eng. rev. = 026430 + 3570= 30,000RPM = 30,000rev.240 + 20 min. retardRPM = 115.384. The main engineIsan8 cylindersingle2-strokecyclediesel with acylinder of 650 mm bore x 1,350 mm stroke. What Is the cylinder constant. What Isthe Indicated horsepower If the indicated pressure Is 11" kglsq.jcm. at 110 rpm?" \Given:No. of cyl.= 8 cyl. 2 strokeBore = 650mm = 65cmStroke= 1350 mm= 1.35 mMEP=11 kg/sq. cm.Rpm = 110Solution:cyl. constant=Ii' 02X L-44500=Ii' (65)2 (1.35)4 4500=.7854 (4225) (1.35)4500cyl. constant=0.9954IHP=MEP x cyl. constant x Rpm x no. of cyl.=(11kg/cm2) (.9954) (110) (8)IHP=9635.47HP.345. The pitch of the propeller of an ocean-going ship is3600 mm. What isthe enginemileage in 24 hours if the propeller makes 118 rpm? If the apparent slip Is minus 3%What Is the observed speed?Given: Pitch = 3600mm= 3.6 mRPM = 118Slip = -3%Solution:Eng. Speed=Pitch x RPMx 601852=3.6 m x 118 x 601852=13.76 knotsSlip=Eng. Speed - O. SpeedEng. Speed-0.03=13.76 - OS13.76- 0.03 (13.76)=13.76 -O.S.OS=13:76 + (0.03) (13.76)=13.76 + 0.41Observed Speed= 14.17 knots.SECOND ENGINEER-January 19891. A cylindrical water tank has a diameter of 3 meters at the base and 4 1/2 metershigh. How many metric tons of fresh water Is to be pumped into the tank In order tohave an ullage of 1 meter? If fuel oil of 0.86specific gravity Is to be pumped Into thetank, how many metric tons are required tohave the same ullage?J.. ,..- ....,1m35Solution:= 1000 kgm3For a F.W. Sp. gr.a) Volume of tank =1102h4For an ullage of 1 m; h = 3 1/2 mVol. of tank @ 3 1/2 m height =1i' (3)2 (3.5 m)4= .7854 (9) (3.5)= 24.74 m3= 1 tonF.W. to be pumped=24.74 m3x 1 MT/m3=24.74 MTb) Sp. gr. of oil=0.86M.T. of fuel oil=24.74 m3x 0.86 (S.G.)=21.28 MT.2. A vessel makes an observed speedof 12 knots with an apparent slip of plus120". The propeller turns 110 rpm. What Is the pitch of the propeller In mm?Given:Ships speed=12 knotsSlip=12%RPM=110Solution:Engine Speed=PxRPMx601852% Slip=ES-SEx100ES0.12=ES-12 knotsESE.S. (0.12)=ES-12E.S. (0.12) - E.S. =-12- 0.88E . S ~=-12E.S.=-12-0.88E.S.=13.63 knotsE.S.=P x RPMx601852 m3613.63 knots=Px110x601852 mP=13.63 (1852)110 x 60P=25242.766600P=3.83 m=3.830 mm.3. Aship'sprovisionisloadedonboardfromamotorlaunchbymeans of amanually operated winch Which work on the same principle as the wheel and axlemachine. The revolving drum of the winch is 30 cm diameter and the crank attachedtotheend ofthedrumIs 40 cm.longfromthe center ofthedrum.What force Isrequired to11ft the provision weighing 300 kg?D ~ F300 I[

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