Workingcycles9At about 120-150°ATDC the exhaust valve opens (EVO),the timing beingchosen to promote a very rapid blow-down of the cylinder gases to exhaust.This is done to (a) preserve as much energy as is practicable to drive the turbo-charger and (b)reduce the cylinder pressure to a minimum by BDC to reducepumping work on the 'exhaust'stroke.The rising piston expels the remainingexhaust gas and at about 70-80° BTDC, the inlet valve opens (IVO) so thatthe inertia of the outflowinggas, and the positive pressure difference, whichusually exists across the cylinder by now, produces a throughflow of air to theexhaustto'scavenge'thecylinder.If the engine is naturally aspirated, the IVO is about 10° BTDC. The cyclenowrepeats.TheTwo-StrokeCycleFigure 1.6 shows the sequence of events in a typical two-stroke cycle,which,asthename implies,isaccomplished inonecompleterevolution of thecrank.Two-stroke engines invariablyhave ports to admit air when uncovered by thedescendingpiston (orair piston where theengine has twopistons per cylinder)The exhaust may be via ports adjacent to the air ports and controlled by thesame piston (loop scavenge)or via piston-controlled exhaust ports or poppetexhaust valves at the other end of the cylinder (uniflow scavenge).The princi-ples of the cycle apply in all cases.Starting at TDC, combustion is already under way and the exhaust opens(EO) at 110-120°ATDC to promote a rapid blow-down before the inlet opens(IO)about 20-30°later (130-150°ATDC).Inthiswaythe inertia of theTDCInjectioncommences=10-20°BTDCFuel valve closes (full load)ExhaustopensExhaustcloses-110-120=110-150°BTDCATDCInletcloses=130-150°BTDCInletopens=130-150°ATDCBDCFIGURE 1.6 Two-stroke cycle
At about 120–150° ATDC the exhaust valve opens (EVO), the timing being chosen to promote a very rapid blow-down of the cylinder gases to exhaust. This is done to (a) preserve as much energy as is practicable to drive the turbocharger and (b) reduce the cylinder pressure to a minimum by BDC to reduce pumping work on the ‘exhaust’ stroke. The rising piston expels the remaining exhaust gas and at about 70–80° BTDC, the inlet valve opens (IVO) so that the inertia of the outflowing gas, and the positive pressure difference, which usually exists across the cylinder by now, produces a throughflow of air to the exhaust to ‘scavenge’ the cylinder. If the engine is naturally aspirated, the IVO is about 10° BTDC. The cycle now repeats. The Two-Stroke Cycle Figure 1.6 shows the sequence of events in a typical two-stroke cycle, which, as the name implies, is accomplished in one complete revolution of the crank. Two-stroke engines invariably have ports to admit air when uncovered by the descending piston (or air piston where the engine has two pistons per cylinder). The exhaust may be via ports adjacent to the air ports and controlled by the same piston (loop scavenge) or via piston-controlled exhaust ports or poppet exhaust valves at the other end of the cylinder (uniflow scavenge). The principles of the cycle apply in all cases. Starting at TDC, combustion is already under way and the exhaust opens (EO) at 110–120° ATDC to promote a rapid blow-down before the inlet opens (IO) about 20–30° later (130–150° ATDC). In this way the inertia of the Working cycles TDC Injection commences �10–20°BTDC Fuel valve closes (full load) Exhaust opens �110–120° ATDC Inlet opens � 130–150° ATDC BDC Exhaust closes �110–150°BTDC Inlet closes � 130–150° BTDC Figure 1.6 Two-stroke cycle
10TheoryandGeneral Principlesexhaust gases—moving at about the speed of sound-is contrived to encour-age the incoming air to flow quickly through the cylinder with a minimum ofmixing,because any unexpelled exhaust gas detracts from the weight of airentrained for the next stroke.The exhaustshould closebeforetheinleton thecompression stroketomaximize the charge, but the geometry of the engine may prevent this if thetwo events are piston controlled.It can be done in an engine with exhaustvalves, but otherwise the inlet closure (IC) and exhaust closure (EC) in a single-piston engine will mirror their opening.The IC may be retarded relative to ECin an opposed-piston engine to a degree which depends on the ability of thedesigner and users to accept greater out-of-balance forces.At all events the inlet ports will be closed as many degrees ABDC as theyopened before it (i.e. again 130-150°BTDC) and the exhaust in the sameregion.Where there are two cranks and they are not in phase, the timing isusually related to that coupled to the piston controlling the air ports.The de-phasing is described as 'exhaust lead.Injection commences at about 10-20° BTDC depending on speed, andcombustion occurs over 3050°, as with the four-stroke engine.HORSEPOWERDespite the introduction of the SI system, in which power is measured in kilo-watts, horsepower cannot yet be discarded altogether. Power is the rate of doingwork.In linear measure it is themean force acting on a pistonmultiplied by thedistance it moves in a given time. The force here is the mean pressure acting onthe piston. This is obtained by averaging the difference in pressure in the cylin-der between corresponding points during the compression and expansion strokesIt can be derived by measuring the area of an indicator diagram and dividing itby its length. This gives naturally the indicated mean effective pressure (imep),also known as mean indicated pressure (mip). Let this be denoted by'p.Mean effective pressure is mainly useful as a design shorthand for the sever-ity of the loading imposed on the working parts by combustion. In that context itis usually derived from the horsepower. If the latter is "brake' horsepower (bhp),themep derived is thebrakemean effectivepressure (bmep);but it should beremembered that it then has no direct physical significance of its own.To obtain the total force, the mep must be multiplied by the area on whichit acts.This in turn comprises the area of one piston,a =d/4, multiplied bythe number of cylinders in the engine, denoted by N. The distance moved percycle by the force is the working stroke (), and for the chosen time unit, thetotal distance moved is theproduct of Ix n,where n is thenumber of workingstrokes in one cylinder in the specified time. Gathering all these factors givesthe well-known'planformula:Power= pX/XaXnxN(1.5) k
10 Theory and General Principles exhaust gases—moving at about the speed of sound—is contrived to encourage the incoming air to flow quickly through the cylinder with a minimum of mixing, because any unexpelled exhaust gas detracts from the weight of air entrained for the next stroke. The exhaust should close before the inlet on the compression stroke to maximize the charge, but the geometry of the engine may prevent this if the two events are piston controlled. It can be done in an engine with exhaust valves, but otherwise the inlet closure (IC) and exhaust closure (EC) in a singlepiston engine will mirror their opening. The IC may be retarded relative to EC in an opposed-piston engine to a degree which depends on the ability of the designer and users to accept greater out-of-balance forces. At all events the inlet ports will be closed as many degrees ABDC as they opened before it (i.e. again 130–150° BTDC) and the exhaust in the same region. Where there are two cranks and they are not in phase, the timing is usually related to that coupled to the piston controlling the air ports. The dephasing is described as ‘exhaust lead’. Injection commences at about 10–20° BTDC depending on speed, and combustion occurs over 30–50°, as with the four-stroke engine. Horsepower Despite the introduction of the SI system, in which power is measured in kilowatts, horsepower cannot yet be discarded altogether. Power is the rate of doing work. In linear measure it is the mean force acting on a piston multiplied by the distance it moves in a given time. The force here is the mean pressure acting on the piston. This is obtained by averaging the difference in pressure in the cylinder between corresponding points during the compression and expansion strokes. It can be derived by measuring the area of an indicator diagram and dividing it by its length. This gives naturally the indicated mean effective pressure (imep), also known as mean indicated pressure (mip). Let this be denoted by ‘p’. Mean effective pressure is mainly useful as a design shorthand for the severity of the loading imposed on the working parts by combustion. In that context it is usually derived from the horsepower. If the latter is ‘brake’ horsepower (bhp), the mep derived is the brake mean effective pressure (bmep); but it should be remembered that it then has no direct physical significance of its own. To obtain the total force, the mep must be multiplied by the area on which it acts. This in turn comprises the area of one piston, a d2 /4, multiplied by the number of cylinders in the engine, denoted by N. The distance moved per cycle by the force is the working stroke (l), and for the chosen time unit, the total distance moved is the product of l n, where n is the number of working strokes in one cylinder in the specified time. Gathering all these factors gives the well-known ‘plan’ formula: Power p l a n N k (1.5)
Torque11Thevalue of the constant k depends onthe units used.Theunitsmustbeconsistent as regards force, length and time.If, for instance, SI units are used (newtons,metres and seconds),k will be1000and thepowerwill begiven inkilowatts.If imperial units are used (pounds, feet and minutes),k will be 33000 andthe result is in imperial horsepower.Onboard ship the marineengineer's interest in the above formula will usu-ally be to relate mep and power for the engine with which he or she is directlyconcerned. In that case I, a,n become constants as well as k and(1.6)Power=p×CxNwhere C=(lxaxn)/kNote that in an opposed-piston engine'I totals the sum of the strokes of thetwopistons in each cylinder.Applying theformulaeto double-acting enginesis somewhatmorecomplex since,for instance,allowancemustbemadeforthepiston rod diameter. Where double-acting engines are used, it would be advis-ableto seek the builder's advice about the constants to beused.TORQUEThe formula for power given in Equations (1.5) and (1.6) is based on the move-ment of the point of application of the force on the piston in a straight line.The inclusion of the engine speed in the formula arises in order to take intoaccount the total distance moved by the force(s) along the cylinder(s): that is,thenumberofrepetitions ofthe cycle in unittime.Alternatively,power can bedefinedinterms ofrotation.If F is the effective resulting single force assumed to act tangentially atgiven radius fromthe axis of the shaft about which it acts, r the nominatedradius at which Fis reckoned, and n revolutions per unit time of the shaft spec-ified,thenthe circumferential distance moved bythetangential force inunittime is 2rn.Hence,F×2rmPower =KFr×2元n(1.7)KThe value of K depends on the system of units used, which, as before, mustbe consistent.In this expressionF× r=T,the torqueacting on the shaft,andis measured (in SI units) in newton metres.Note that Tis constant irrespective of the radius at which it acts. If n is inrevolutions per minute, and power is in kilowatts, the constant K=100o,so that
The value of the constant k depends on the units used. The units must be consistent as regards force, length and time. If, for instance, SI units are used (newtons, metres and seconds), k will be 1000 and the power will be given in kilowatts. If imperial units are used (pounds, feet and minutes), k will be 33 000 and the result is in imperial horsepower. Onboard ship the marine engineer’s interest in the above formula will usually be to relate mep and power for the engine with which he or she is directly concerned. In that case l, a, n become constants as well as k and Power p C N (1.6) where C (l a n)/k. Note that in an opposed-piston engine ‘l’ totals the sum of the strokes of the two pistons in each cylinder. Applying the formulae to double-acting engines is somewhat more complex since, for instance, allowance must be made for the piston rod diameter. Where double-acting engines are used, it would be advisable to seek the builder’s advice about the constants to be used. Torque The formula for power given in Equations (1.5) and (1.6) is based on the movement of the point of application of the force on the piston in a straight line. The inclusion of the engine speed in the formula arises in order to take into account the total distance moved by the force(s) along the cylinder(s): that is, the number of repetitions of the cycle in unit time. Alternatively, power can be defined in terms of rotation. If F is the effective resulting single force assumed to act tangentially at given radius from the axis of the shaft about which it acts, r the nominated radius at which F is reckoned, and n revolutions per unit time of the shaft specified, then the circumferential distance moved by the tangential force in unit time is 2rn. Hence, Power F rn K Fr n K 2 2 π π (1.7) The value of K depends on the system of units used, which, as before, must be consistent. In this expression F r T, the torque acting on the shaft, and is measured (in SI units) in newton metres. Note that T is constant irrespective of the radius at which it acts. If n is in revolutions per minute, and power is in kilowatts, the constant K 1000, so that Torque 11
12TheoryandGeneralPrinciplespower×60×1000T=2元130000×power(1.8)(Nm)TnIf the drive to the propeller is taken through gearing, the torque acts at thepitch circle diameter of each of the meshinggears.If thepitch circle diametersof the input and output gears are, respectively, d, and d2 and the speeds of thetwo shafts are n and nz,the circumferential distance travelled by a tooth oneither of these gears must be d, × ny and d × n2, respectively. But sincetheyaremeshedd,n,=den2Therefore=dn2The tangential force F on two meshing teeth must also be equal.Thereforethe torque on the input wheelT=Fd2and the torque on the output wheelT =Fd2orI-4d2If thereis morethanonegeartrain,thesameconsiderations applyto eachIn practice there is a small loss of torque at each train due to friction, etc.Thisis usually of the order of 1.5-2 per cent at each train.MEANPISTONSPEEDThis parameter is sometimes used as an indication of how highly rated anengine is.However, although in principle a higher piston speed can implyagreater degree of stress and wear, in modern practice, the lubrication of pis-tonringsand liner,as well as other rubbing surfaces,has becomemuch morescientific. It no longer follows that a high'piston speed is of itself more detri-mentalthanalowerone in a well-designed engine
12 Theory and General Principles T n n power power (N m) 60 1000 2 30 000 π π (1.8) If the drive to the propeller is taken through gearing, the torque acts at the pitch circle diameter of each of the meshing gears. If the pitch circle diameters of the input and output gears are, respectively, d1 and d2 and the speeds of the two shafts are n1 and n2, the circumferential distance travelled by a tooth on either of these gears must be d1 n1 and d2 n2, respectively. But since they are meshed d1n1 d2n2. Therefore n n d d 1 2 2 1 The tangential force F on two meshing teeth must also be equal. Therefore the torque on the input wheel T Fd 1 1 2 and the torque on the output wheel T Fd 2 2 2 or T T d d 1 2 1 2 If there is more than one gear train, the same considerations apply to each. In practice there is a small loss of torque at each train due to friction, etc. This is usually of the order of 1.5–2 per cent at each train. Mean piston speed This parameter is sometimes used as an indication of how highly rated an engine is. However, although in principle a higher piston speed can imply a greater degree of stress and wear, in modern practice, the lubrication of piston rings and liner, as well as other rubbing surfaces, has become much more scientific. It no longer follows that a ‘high’ piston speed is of itself more detrimental than a lower one in a well-designed engine
Vibration13Ixn(1.9)Mean piston speed is simply30This is given in metres per second if /=stroke inmetresandn=revolutions perminute.FUELCONSUMPTIONIN24HIn SI units:w×kW×24W=(1.10)10001000W(1.11)Wkw×24where wis thefuel consumption rate (kg/kWh)W = total fuel consumed per day in tonnes(1 tonne= 1000kg)VIBRATIONMany problems have their roots in, or manifest themselves as,vibration.Vibrationmay be in any linear direction, and it may be rotational (torsional). Vibration maybe resonant, atone of its naturalfrequencies,orforced.It may affect anygroup ofcomponents,orany one.It can occurat anyfrequency uptothosewhich aremorecommonly called noise.That vibration failures are less dramatic now than formerlyis dueto theadvances in our understanding of vibration during the last 80 years. It can becontrolled, oncerecognized,by design and correct maintenance,byminimizingat source, by damping and by arranging to avoid exciting resonance. Vibrationis a very complex subject and all that will be attempted here is a brief outlineAnyelasticallycoupled shaft orothersystemwill haveoneormorenaturalfrequencies,which,if excited, can build up to an amplitudewhich is perfectlycapable of breaking crankshafts.'Elastic'in this sense means that a displace-ment or a twistfrom rest creates a force or torque tending to return the systemto its position of rest, and which is proportional to the displacement. An elas-tic system, once set in motion in this way, will go on swinging, or vibrating,about its equilibrium positionforever,in the theoretical absence of any damp-ing influence.Theresulting time/amplitude curve is exactly represented byasine wave; that is, it is sinusoidal.In general, therefore, the frequency of torsional vibration of a single masswill be as follows:1J cycles per second(1.12)2元V1
Mean piston speed is simply l n 30 (1.9) This is given in metres per second if l stroke in metres and n revolutions per minute. Fuel consumption in 24 h In SI units: W w kW 24 1000 (1.10) w 1000 24 W kW (1.11) where w is the fuel consumption rate (kg/kWh) W total fuel consumed per day in tonnes (1 tonne 1000 kg). Vibration Many problems have their roots in, or manifest themselves as, vibration. Vibration may be in any linear direction, and it may be rotational (torsional). Vibration may be resonant, at one of its natural frequencies, or forced. It may affect any group of components, or any one. It can occur at any frequency up to those which are more commonly called noise. That vibration failures are less dramatic now than formerly is due to the advances in our understanding of vibration during the last 80 years. It can be controlled, once recognized, by design and correct maintenance, by minimizing at source, by damping and by arranging to avoid exciting resonance. Vibration is a very complex subject and all that will be attempted here is a brief outline. Any elastically coupled shaft or other system will have one or more natural frequencies, which, if excited, can build up to an amplitude which is perfectly capable of breaking crankshafts. ‘Elastic’ in this sense means that a displacement or a twist from rest creates a force or torque tending to return the system to its position of rest, and which is proportional to the displacement. An elastic system, once set in motion in this way, will go on swinging, or vibrating, about its equilibrium position forever, in the theoretical absence of any damping influence. The resulting time/amplitude curve is exactly represented by a sine wave; that is, it is sinusoidal. In general, therefore, the frequency of torsional vibration of a single mass will be as follows: f q I 1 2π cycles per second (1.12) Vibration 13