Engine Balancing PDF

Engine balancing In an engine, the pressure of gasses inside of the cylinder is used to create torque, through the crank-slide mechanism, which can be referred to as the crank gear. 3.1 Crank gear mechanism The crank gear is the mechanism that transforms the reciprocating motion of the piston in...

Engine balancing In an engine, the pressure of gasses inside of the cylinder is used to create torque, through the crank-slide mechanism, which can be referred to as the crank gear. 3.1 Crank gear mechanism The crank gear is the mechanism that transforms the reciprocating motion of the piston into a rotating motion of the shaft. We have that the piston is a reciprocating mass, the crank a rotating mass, while the conrod is a mass that has a motion in-between a reciprocating and a rotating motion. The forces needed to accelerate and decelerate the masses, are the iner- tia forces. The pressure forces act- ing on the piston are translated in a force which acts perpendicularly to the surface of the piston. This force can be decomposed in a force paral- lel to the conrod (Fconrod ) and one force parallel to the piston surface (FN ). The force acting on the con- rod is then transmitted to the crank as the vector sum of a force tangen- tial to the direction of rotation of the crank (FT ) and a force radial to the direction of rotation (FR ). The tangential force, combined with the radius of the crank gives us (instant by instant) the instantaneous engine torque: Meng = FT · r. The radial force, instead, does not contribute to torque generation and it is com- pensated by the crankshaft bear- ings. Obviously, the engine block has to be able to compensate for Figure 3.1: The crank gear the engine torque, through a reac- tion torque, which is the product of the normal force acting on the piston, 27 multiplied by the distance of the point of application of this force from the centre of rotation of the crankshaft: Mreaction = FN · b. Two forces arise in the engine restraining system in order to compensate for this reaction torque. 3.1.1 Mass balancing of centrifugal forces Rotating masses produce centrifugal forces which are unwelcome. Fortu- nately, they are also fairly easy to balance by using counterweights that produce the same amount of centrifugal force of the rotating masses, but in opposite direction. 3.2 Engine balancing: introduction The aim of engine balancing is to eliminate or at least attenuate the vibra- tions that come from all of the stresses that the engine block is subjected to. Notice that these vibrations are then transmitted through the whole structure. 3.2.1 Forces and moments Forces acting on the engine frame There are several forces that act on the engine frame: we have the sur- face forces that the engine exchanges withe the mounts, the torque that the engine exchanges withe the outside through the crankshaft, the forces ex- changed with engine accessories and forces acting on the engine when it is fitted on the vehicle (clutch,...). Forces acting on the engine The forces due to gas pressure are already compensated by the crank bear- ings, so they are not transferred to the frame. The weight is a constant force and so does not influence vibrations. The last two forces that we need to consider are the reciprocating inertia motions (constant in direction, vari- able in magnitude) and centrifugal (inertia) forces, which are constant in magnitude and varying in direction. Moments along the crankshaft axis We have several moments acting on the axis of the crankshaft. Obviously, we have the engine torque, which is due to the pressure forces of the gas on the piston, summed up withe the inertia forces of the piston itself. We also have the inertial torque, which are dependant from the crank position and is a result of the inertia of the crank and conrod: Mi,e = Ie · !. ˙ Since 28 the conrod has a reciprocating and a rotating part, we calculate its inertia moment by using a schematic representation of the rod. Schematization of the connecting rod The connecting rod has a part (con- nected to the piston) which has a certain mass and a part connected to the crank with another mass. In between there is a distributed mass. The conrod is approximated as two concentrated masses mA and mB , withe the centre of gravity placed as in Figure 3.2. In reality, we need to consider the fact that di↵erent distances from G generate di↵erent inertia moments so, in order to be consistent with the reality, the over- all inertia moment of the conrod will be corrected by adding the term I0 < 0: Irod = mA x2A + mB x2B + I0 Figure 3.2: The conrod The term I0 is called the ”on frame correction of the rod inertia” Moments acting perpendicularly to the crankshaft axis There are also moments which act perpendicularly to the engine. These mo- ments are caused by the inertia forces (of the pistons) and by the centrifugal forces. 3.2.2 Balancing of inertia and centrifugal actions. Cyclic behaviour of the engine Our goal is now to balance all actions due to inertia and centrifugal forces. We want the forces on the engine mounts to be balanced (resultant equal to zero) and the moments acting on the mount to be balanced (resultant equal to zero) as well. Another important accomplishment in engine balancing is the regularity of engine torque. The engine behaves in a cyclic way. The tangential force that creates the engine torque is in fact a cyclic force that depends from the crank position. In Figure 3.3, we have the tangential pressure (tangential force divided by piston area) as a function of crank angle (the zero is positioned at TDC of combustion phase). 29  Figure 3.3: Cyclic behaviour of the engine 3.3 Fourier analysis of the engine torque A very important tool in the analysis of the engine torque is the Fourier analysis, which allows us to decompose the engine torque in all of its har- monics. The torque produced by a single cylinder can be defined as: Msingle = M0 + ⌃k Mk sin(k!t + k) Where M0 is a constant average torque and Mk is the magnitude of the k th harmonic. 3.3.1 Torque in a multi-cylinder engine In order to get a smooth torque at the crankshaft, we should not have all cylinders firing at the same moment. Actually, we want the engine to be in an evenly spaced cranks. In fact by tweaking the crank position with respect to a reference one, we can get a di↵erent firing order of the cylinders. We have a formula to get the phase shift of each crank with respect to a reference one: m m ' = 2⇡ = 360 i i For example, a four-stroke, four cylinder engine will have a phase shift of 180 for each crank. The resulting moment of the engine is obtained by properly sum up all contributions of all cylinders, taking the phases into account. The phase of the cylinder next to fire with respect of the first to fire is: 'j = '(j 1). The overall torque can be then written as: Mmulti = ⌃j Mj = iM0 + ⌃ij ⌃k Mk sin(k!t + k jk ) 30  Figure 3.4: Tangential pressure for a four cylinder engine With jk = k '(j 1). We can identify the principal harmonics by identifying the harmonics in which all torque vectors sum up (we basically sum up the same harmonic through all cylinders). Another important thing to keep in mind, is the fact that for an in-line engine, with evenly spaced cranks, the torque vectors sum up when the harmonic is a multiple of i/2 and cancel each other out when the harmonic is not a multiple of i/2. For a multi- cylinder engines, we can choose to only consider the principal harmonics (p) and say that: Mmulti = i[M0 + ⌃in/m Mp sin(p!t + p )] Obviously we are supposing that each cylinder has the same torque output for a given crank position, but in reality this is obviously not real. So at all harmonics we will have a certain amount of torque. 3.4 Analysis of centrifugal and inertia forces We are now interested in analyzing the contribution of centrifugal and inertia forces. The centrifugal forces are written as: FC = mc ! 2 r. The inertia forces can be written as a Taylor expansion: Fa = ma ẍ = ma ! 2 r[cos ✓ + cos 2✓ +...], where = r/l is the conrod ratio. 3.4.1 Centrifugal forces, four-stroke engines We can use the so-called star diagrams in order to understand the position of the pistons (up or down) in order to see if the centrifugal forces are balanced or not. We can clearly see that apart from the four-stroke two cylinder engine, the centrifugal forces are automatically balance. It is important 31  Figure 3.5: Star diagrams for two and four stroke engines to notice that even tough the centrifugal forces are always balanced, the moments that they create might not be balanced as well. 3.4.2 Lengthwise layout of cranks and firing order In order to balance the moments produced by the centrifugal forces, we need to make sure to have the firing cylinders in a certain position with respect to the ones that are not firing.We also need to determine the firing order. We can number the cylinders according to standards: they are numbered consecutively, in the order in which they would be intersected by an imagi- nary reference plane. We usually start from the opposite side of the power output. The plane is located to the left and the numerical assignments then proceeds clockwise along the longitudinal axis of the engine. If there are more cylinders on the same plane, they will be numbered starting from the closest to the observer. Regarding the firing order, the rule is to try not to have two consecutive cylinders firing at once (not always possible). Let us see a couple of exam- ples. Four-stroke engine, i even Let us consider a four cylinder engine. The position of the pistons must be a symmetrical lengthwise layout, relative to the middle of the crankshaft. This means having two pistons on opposite sides of the middle of the crankshaft at the same height at the same time.In this way the moment produced by the centrifugal forces acting on the pistons will be zero. Regarding the firing order, we can start from firing cylinder 1, then we need to fire a cylinder which was at BDC (2 or 3), so we choose 3 in order not to have two con- secutive cylinders firing at once. Then we go with 4 and, in the end 2. We restart from one. The order is: 1-3-4-2-1-... Another equivalent order is 1-2-4-3-1-... 32  Figure 3.6: Symmetric lengthwise layout Four-stroke engine i=2 This is a special case of the family of configurations discussed above. Since there are only two cylinders, we can balance the momentum by adopting a symmetrical lengthwise layout, but we cannot balance the centrifugal forces. This in why we need to add a counterweight of equal mass of the cylinders whose centrifugal force will balance the one produced by the cylinders. Four-stroke engine, i odd In case of an odd number of cylinders, the best layout in order to minimize (but never fully cancel out) the momentum is the anti-metric lengthwise layout of cranks. The principle of this layout is to position the middle cylinder at the top, then the two close to it and then the outermost ones in swapped positions. In Fig- ure 3.7 we can see an example of this principle for a five cylinders en- gine. If thee were more cylinders, we would swap only the two furthest ones. We can clearly see that the moment resultant is not zero, but it certainly is reduced with respect to the standard layout. A three cylinder engine is the only one that cannot be adjusted with this layout Figure 3.7: Anti-metric layout for a and needs counterweights. Regard- five cylinders engine ing the firing order, we still use the 33 aforementioned rule. 3.4.3 Inertial forces, four-stroke engine The inertial forces can be expressed as Fa = ma ẍ ⇡ ma ! 2 r[cos ✓ + cos 2✓ +...] Since the terms of higher harmonics (we take the coefficient of the ✓ as the order of the harmonic) get smaller and smaller, we will consider only the first 0 two, defined as: Fa = ma ! 2 r cos ✓ which varies periodically synchronous 00 to the shaft speed and Fa = ma ! 2 r cos 2✓, which varies periodically twice per shaft revolution. The first and second order forces display a variable modulus and a constant direction. We can decide to reduce these forces to an equivalent field, made up of two rotating vectors of constant modulus equal to half of the value of the force. The sum of the two vectors is the actual force, since the vectors rotate in opposite directions. In this way we can analyze inertial forces as we did with centrifugal forces. First-order forces 0 At each angle step, the force Fa is the resultant of two symmetrical forces with constant modulus that rotate with ! and ! respectively. The two force fields are equivalent to the cen- trifugal forces rotating at crankshaft speed in the two directions. We 0 can define Fa1 as the first order force rotating n crank direction and 0 Fa2 as the first order force rotating in opposite crank direction (contra- rotating force). For in-line engines, 0 if the system of Fa1 forces cancels 0 out, so will the system of forces Fa2. Figure 3.8: First-order force This is not true for V-engines. Second-order forces 00 As we did for the first order forces, we can call Fa1 the second order force 00 rotating in crank direction and Fa2 the second order forces rotating in the opposite crank directions. The system of forces rotating in the crank direc- tion or opposite to it constitutes a constant rigid system. We can consider the balance of the rigid system of forces in any time instant. Once we have determined the resultant of the system of forces for an instant, in the follow- ing instants the amplitude of the resultant will be the same and the vector will rotate with 2!. 34  Figure 3.9: Balancing of second order inertial forces General considerations Notice how the horizontal component of the two counter-rotating forces will always cancel out and the resultant will always be obtained by summing the vertical components. The first order forces have a period of 2⇡ while the second order ones have a period of ⇡. The overall inertial force acting on a cylinder is the vectorial sum of rotating and oscillating inertial forces of at least first and second order (possibly also higher orders). The resultant inertial forces are not cancelled out in a single cylinder. In order to balance them, we can use two masses, one rotating with ! and the other rotating with ! (this way we balance the first order ones). Multi-cylinder engine We can balance the inertial forces in a multi-cylinder engine as we did for centrifugal forces, with the addition of two counter-rotating shafts. In fact, the resultant of the first order forces is zero, since they depend from ✓ as the centrifugal forces do. Instead second order forces depend from 2✓, so they will all have the same direction, therefore summing up with each other. By adopting a symmetrical lengthwise layout (i even) we can cancel the resultant moment but the resultant force will still be there. In order to balance these second order forces, we use two shaft which rotate with 2! and 2!. These shafts have an o↵-centre mass on them which are located in a normal plane passing through the centre line. The mass can be determined by determining the resultant force RF 00 (same for the counter- a1 rotating forces). The balanced shaft is called the Lanchester shaft. 00 RF 00 = 4Fa1 = 4 · 12 ma ! 2 r , to be balanced by mc (2! 2 )Rx. In the slides a1 35 (PSAV03, 61-68) there are several examples of balancing. 3.4.4 Counterweights in cranks The engine crankshaft has counterweights. In an i-cylinder engine, there are i + 1 main bearing caps for support. The counterweights are needed for reducing the stresses on the supports. We need to make sure that the counterweights produce a balanced field of forces. 3.4.5 Centrifugal and inertia forces, two-stroke engine In two-stroke engines, we can adopt the same lengthwise layout of cranks that we used for four-strokes. If i is odd, we will use the anti-metric length- wise layout (complete balance is not possible) while if i is even, we will have two situations: i 2 is odd: the moment’s resultant is zero; i 2 is even: the complete balance is not possible. In the slides (PSAV3, 71-77) there are several examples of two-stroke engine balancing. 3.5 Balancing of V-engines Before defining what a V-engine is, we need to define two parts of the engine: bank of cylinders: it is a group of cylinders located on the same side of the shaft with their axes in a plane passing through the crankshaft axis; row of cylinders: it is a group of radial cylinders in a plane at right angles to the crankshaft axis and operating on a single crank. Now we can define the V-engine as an engine with two banks of cylinders, with corresponding cylinders in each bank forming a two-cylinder row. The angle between the cylinder planes is called V-angle. Other engines can be the W-engine, which is a V-engine with three banks (the two V-angles are usually equal). We can also have an opposed engine, which is a V-engine with a 180 V-angle. Notice that in a V-engine, we have two connecting rods for each crank pin. If the cylinders form a proper V as well, the engine is called proper V-engine. 36  Figure 3.10: Examples of V-engines 3.5.1 Partial rotation principle In order to obtain a V-engine and to maintain the uniform firing of the cycles, we apply the partial rotation principle to the in-line layout. The partial rotation principle states that ”if we rotate the cylinder and the crank rigidly, the phase of the cycle does not change compared to the other cylinders”. There can be several possible layouts for a V-engine and they depend from considerations linked to inertial and centrifugal forces. The partial rotation does not change the time instant in which we have, for example the TDC of combustion. It only changes the spatial position of the cylinder. In Figure 3.10 we can find some examples of V-engines. 3.5.2 Balancing of proper V-engines In many V-engines, the forces due to inertial and centrifugal forces are bal- anced for each bank of cylinders. There are some cases in which the balance of the engine is excellent even tough the banks are not balanced themselves. In general, the best way to analyze the inertial forces in V-engines is to consider a row of two cylinders that acts on a crank. We can project the forces acting on the pistons on the symmetry axis of the row and on the axis perpendicular to it, in order to find the resultant force acting on the row. If is the V angle, and the crank angle for the left cylinder is ✓, the crank angle of the right cylinder will be 2⇡ ( ✓) or just ✓ if we refer to the crank itself and the direction of the pistons. Both the centrifugal forces and the first order inertial forces that rotate in crank direction are added up and are parallel with the direction of the crank. The contra-rotating first order forces are phase shifted from the other of an angle = 2✓ + 2( ✓) = 2. 37  (a) First order forces (b) Second order forces Figure 3.11: First and second order forces in a row The second order inertial forces that are still rotating in the crank direction are phase shifted of an angle = 2( ✓) ( 2✓) = , while the second order contra-rotating forces are phase shifted from one another of an angle = 2✓ + + 2( ✓) = 3. Depending on the V-angle the forces will either cancel each other out or they will be added together. 38 Models of engine processes The main goal of this chapter is to study the main processes that happen inside of the ICE, especially in terms of fuel conversion efficiency. As we de- Wb scribed in previous chapters, the fuel conversion efficiency is: ⌘f = mf Q LHV. It is convenient to split this efficiency in di↵erent components and to analyze each component individually: Wa f Wi Wb ⌘f = · · = ⌘a f ⌘✓i ⌘m mf QLHV Wa f Wi The efficiencies are respectively: the air-fuel efficiency (takes into account the losses due to the fact that the fluid is real), the internal thermodynamic efficiency (takes into account the losses during the process) and the mechan- ical efficiency (takes into account the losses when we transfer the work from the piston to the shaft). We can also consider an ideal efficiency, linked to and ideal work, which is the efficiency of a cycle with an ideal fluid. During our analysis we will consider three types of cycle: Ideal cycle: no losses, we use an ideal fluid; Air-fuel cycle: we have only the losses due to the fact that the fluid is now considered a real fluid; Indicated cycle: we have a real fluid and all of the losses that happen in reality. 4.1 Ideal heat cycle The ideal heat cycle is the simplest way to describe (approximately) the functioning of an ICE. In fact, we have an ideal fluid, which never changes (so the cycle is closed) and never alters its properties and chemical composition. The combustion is modeled as an heat addition to the gas. The ideal gas can be described with the ideal gas law: pv = RT. The constant R is R= R Mmol. Since the gas is ideal, we consider constant cp and cv. As usual, univ c cp cv = R and = cvp is the isentropic coefficient. During our cycle we will have a quantity of heat that is received qin at high temperature and a quantity that is rejected qout at low temperature. 39 4.1.1 Types of cycles We will consider three main cycles, which are all made out of: 1 ! 2: isentropic compression starting from ambient pressure; 2 ! 3: combustion (heat received, qin ); 3 ! 4: isentropic expansion; 4 ! 1: heat rejection (heat rejected, qout. Notice that the ideal heat rejection phase corresponds to both intake and exhaust phases. In fact the heat addition is actually due to the combustion and the rejection of heat is there in order to account for the spent (hot) charge being replaced with a fresh one. We can model the heat addition as Qin = mf qin = mf QLHV. The term mf QLHV is the energy content of the fuel that we can use during combustion in the real cycle. We can define an ideal efficiency: ⌘id = wqin id = qinqinqout = 1 qqout in. The three cycles that we will consider di↵er in the type of combustion phase: Otto cycle: the combustion is isocoric and instantaneous; Diesel cycle: the combustion is isobaric. Used for big, slow Diesel engines; Sabathè cycle: the combustion is a combination of an isocoric and of an isobaric transformation. Used in smaller, faster Diesel engines (automotive applications. Regarding the isentropic expansion, we would like to prolong it until a point from which we could have a isothermal rejection phase, since this would greatly increase the work and the efficiency of the cycle. The problem is that the cycle would take too long (the frequency of the cycle would approach zero). Another solution could be to have the expansion end in a point from which we could have a rejection phase that is a constant pressure one. This can be done, but for most cases this is not really a good solution, since the displacement of the cylinder would be too large (see PSAV04 10). Lastly, if we consider the fact that in an engine we have an exhaust and intake phase, we have to add two ”phases” that take these things into account. These phases, tough, happen at the same pressure, so they do not contribute to work production. For each cycle the combustion is complete. 4.1.2 Comparison between cycles We can compare di↵erent cycles in terms of efficiency, work and pressure allowed. There are many relationships that we can write for each cycles (PSAV04 18-27). 40  Figure 4.1: Ideal cycles in pV and TS charts Otto cycle We can consider the Otto cycle (very high pressures can be reached after combustion). Its efficiency can be written as a function of the com- 1 pression ratio: ⌘id = 1. It rc 1 is interesting to see how the effi- ciency changes with respect to the composition of the charge and with respect to the compression ratio. As we move from air to mixtures of air and fuel we can notice a change in , which a↵ects the efficiency (low- ers it). The compression ratio is also a factor. When it increases, the ef- ficiency increases as well. We can notice that for larger compression ratios we have an increase of effi- ciency that is not that great. We can also say that increasing at constant rc leads to an increase of Figure 4.2: Efficiency for an Otto cy- maximum pressure, which increases cle the work. We need to make sure that the maximum pressure (p3 ) is not too large, otherwise we will risk to damage the engine. At least we will need a larger and stronger engine. In fact, we would like to have an high imep p3 ratio. 41  Figure 4.3: Comparison between cycles Diesel and Sabathè cycles 0 1 ⌧ 1 We can write the ideal efficiency of Diesel cycle as ⌘id = 1 1 (⌧ 0 1). rc 0 T3b ⌧ = 3a. Comparison of the cycles As we can see from Figure 4.3, we have very di↵erent maximum pressures between Otto cycle and the other two. The higher maximum pressure makes the Otto cycle the best in terms of work and efficiency. This can be said by fixing the fluid, the compression ratio and the energy inserted in the cycle. Like we said before, having a too high maximum pressure can be dangerous for the engine. In particular, for SI engines, a too high p2 can lead to self- ignition, which generates the knock phenomenon. This is why we usually use Otto cycles for SI engines, since we can limit p2 and have a larger p3 and a large work and efficiency. On the other hand, for CI engines, we do not care about the value of p2 , since we do not have fuel mixed with air during the compression phase. So we usually use Diesel (or Sabathè) cycles for CI engines. The limitation is on p3 , in order to avoid engine damage. We can see from Figure 4.4a, the Otto cycle is the best when we have a fixed rc and no restriction on the maximum pressure. Instead, from Figure 4.4b we can see that the Diesel cycle is the best when the maximum pressure is fixed. Notice how, in this 42  (a) Fixed rc (b) Fixed pmax Figure 4.4: Comparison in di↵erent conditions case, we need to have a lower p2 in the Otto cycle in order not to go above the maximum pressure p3. Notice how we also fixed qin for both cases. Similar considerations can be made by fixing qout (PSAV04 32). 4.1.3 Over-expanded cycles Like we were saying before, we would like to keep the expansion going until a point from which we can then have the heat rejection phase at a con- stant pressure. We also said that in order to implement this solution,we would need an larger engine displacement. Some engineers have actually devised two solutions for this problem. These two were Atkinson and Miller. Atkinson’s engine idea is very complex and we will not consider it in this notes. Instead, Miller’s engine is much simpler and achieves this over-expansion in a very interesting way. Basically we tweak the intake valve opening timing in two ways: either we close it early (early intake valve clos- ing EIVC) or we close it late (late intake valve closing LIVC) with respect to the BDC position of the piston during intake. As we can see from efficiency and impep ratios with the Otto cycle, the Miller cycle can be a valid alter- native. We use Miller’s cycle in large Diesel engines for ships, but also for some niche automotive applications (Mazda Millenia). From the expression of Miller’s cycle efficiency (PSAV04 38) we can see that it is the only cycle which depends from qin in the efficiency expression. Early intake valve closing In this case, we have that we open the valve at the end of the expansion, but we close it before the piston reaches the BDC. This means that the gas 43  (a) EIVC (b) LIVC Figure 4.5: Comparison in di↵erent conditions will be expanded first, as the piston keeps moving downwards (point 7 in Figure 4.5a and then it will be compressed back at the pressure with which it entered the cylinder (point 1) and it will start the compression phase. We obviously have less charge in the cylinder than we would have in a regular Otto cycle. This a↵ects the work, diminishing it but it increases the cycle’s efficiency. Late intake valve closing In this case (Figure 4.5b) we close the intake valve after the piston has reached BDC (normally we would close it at point 5, but we close it at 1). this means that some charge will actually be pushed back inside of the intake manifold. This means that we will have less charge inside of the cylinder, again reducing the work. The efficiency is actually improved. 4.2 Air-fuel cycles Up until now we have considered ideal cycles, in which the charge never changes composition ad it is never replaced (we add some extra strokes to represent its replacement). In the so called air-fuel cycles, we add this phe- nomenon: the gasses change chemical composition during the combustion phase. We will not consider any other loss that is not related to the fuel burning in a ”real” way. The composition of the working fluid changes during the cycle. During in- take and compression, the mixture of air (N2 + O2 +...) and fuel (in vapor form) can be considered frozen, since it does not alter its properties that much. For CI engines, the charge is just air plus some remaining burned 44  (a) Miller/Otto efficiency ratios (b) Miller/Otto imep ratios Figure 4.6: Efficiency and imep ratios gasses. When air and fuel react chemically during the combustion, an equilibrium condition might be reached in which we not only have the products of the combustion, but also some of the original reactants that were obtained by recombination of the products at high temperature (T > 1850K). This phe- nomenon is called dissociation and it is an unwanted phenomenon since it takes some useful heat away in order to recombine the products into the re- actants. As the combustion products cool down, another phenomenon called recombination happens, but it cannot compensate for all of the power lost because of dissociation. During exhaust, we can consider the gas composition as frozen again. The fluid properties are no longer constant (when the mixture is not frozen) and are depending from the temperature (cp (T ), cv (T ) ! (T ). 4.2.1 Cycle description As we can see in Figure 4.7 we can see some examples of air-fuel cycles which di↵er from one another in the way in which intake and exhaust take place. The compression is a reversible adiabatic, the combustion is complete with no heat loss, the expansion is a reversible adiabatic and the exhaust/intake are considered ideal. Usually the power in a SI engine is controlled by controlling the pressure in the inlet manifold. If the pressure is less than the environment one, the engine is said to be throttled, while if the pressure is more or less the atmospheric one, the engine is unthrottled. If the inlet pressure is higher than the atmospheric one, the engine is supercharged. 45  Figure 4.7: Air-fuel cycles Ideal exhaust and intake As we said before, the intake and exhaust phases are considered as ideal. We will have adiabatic reversible processes, valve events will occur instantaneously at TDC and BDC, with no change in cylinder volume while the pressure di↵erence across the valve falls to zero. In the un- trhottled cycle, we have an isocoric fall of the pressure to environment pressure. The remaining gasses fol- low the transformation (4 ! 5). If the external pressure is not equal the one in the cylinder, we will have some fresh charge or some burned gas flowing into the intake mani- fold until the pressure is equalized. Since there is no heat transfer, the residual fraction of spent gasses is fixed, because the state of the re- maining gas at point 6 (Figure 4.8) Figure 4.8: Detail of intake/exhaust is the same of the state at point 5, at the end of the isentropic expansion of the gasses in the cylinder at the exhaust pressure. 46  Figure 4.9: Comparison with an ideal cycle 4.2.2 Air-fuel cycle efficiency As we said at the beginning of this chapter, the efficiency of the air-fuel cycle is defined as: Wa f ⌘a f = mf QLHV This efficiency is a↵ected by a few factors: dependence of the specific heats from the temperatures (PSAV04 55); dissociation; variation of the specific gas constant before and after combustion be- cause of a variation of number of moles and of molar mass (usually R increases). We can see from Figure 4.9 the real and air-fuel cycles are the same during compression but start to di↵er greatly during combustion and expansion. The efficiency is obviously lower. 4.2.3 Dissociation Definition of the subtangent of a thermodynamic transformation We can define a useful concept called subtangent of a thermodynamic transformation. We can start from a point A and trace a line tangent to the transformation line in point P. We can write: T dS = dQ + dWw = cdT dT T = dS c 47  Figure 4.10: Subtangent of a thermodynamic transormation dT T = tan = dS AB We can say that AB = c is the subtangent of the polytropic curve in P. If the point was somewhere else, we would have a di↵erent tangent line but the same subtangent. Dissociation (a) E↵ect of dissociation (b) Quantity of dissociation Figure 4.11: Dissociation 48  The dissociation of products into reactants can occur at temperatures higher than 1850K and if the mixture is in the right proportions of air and fuel. Usually dissociation does not occur in lean mixtures, since the temper- ature reached is not high enough. It also does not occur in rich mixtures, since the presence of CO in the burnt gasses (rich mixtures have a great quantity of CO) suppresses the dissociation of CO2. The phenomenon of dissociation is most evident in mixtures which have stoichiometric propor- tions. We can define a mixture richness through the degree of richness: degr = ' 1, where ' = 1/. 4.2.4 Air-fuel cycle analysis The first part of the analysis concerns the variation of the specific gas con- stant. In fact, due to the fact that the number of molecules changes espe- cially during combustion, the constant R is modified as well. Actually, this is a phenomenon that benefits the efficiency of the air-fuel cycle, since the 0 constant is reduced and pp32 = RR TT32 means that for the same p2 , the maxi- mum pressure achievable in the air-fuel cycle is greater than the one for the ideal one, meaning a larger cycle area and a larger work. Summary The overall efficiency of the air-fuel cycle is smaller than the one of the ideal cycle. The variation of the specific heats and the dissociation process contribute to the lowering of efficiency. The variation of R slightly increases the efficiency of the air-fuel cycle, but it is not enough to win over the other two phenomena. Influence of the air-fuel ratio The air to fuel ratio determines the temperature T3 at the end of com- bustion and therefore influences the fuel-air efficiency. In a lean mix- ture, we will have lower tempera- tures because of the dilution of the fuel in a greater mass of air. This means that the phenomena related to large temperature variations like specific heat change and dissocia- tion, will be lower. This means that with an increase in ↵, we will have a slight increase in efficiency. For a rich mixture, instead, we will have higher temperatures, so we will see Figure 4.12: E↵ect of ↵ 49 phenomena like the dissociation be- ing very present. So, smaller values of ↵ lead to smaller values of effi- ciency. Also, the fuel that is not oxidized is ”wasted”, meaning that we do not use its energy. We can see the influence of ↵ in a graph. 50