Automobile Engineering - DMX5208 PDF

FACULTY OF ENGINEERING TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING BACHELOR OF TECHNOLOGY HONOURS IN ENGINEERING – LEVEL 5 DMX5208 – AUTOMOBILE ENGINEERING BOOK 1 AUTOMOBILE ENGINEERING BOOK 1...

FACULTY OF ENGINEERING TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING BACHELOR OF TECHNOLOGY HONOURS IN ENGINEERING – LEVEL 5 DMX5208 – AUTOMOBILE ENGINEERING BOOK 1 AUTOMOBILE ENGINEERING BOOK 1 DMX5208 AUTOMOBILE ENGINEERING Published by the Open University of Sri Lanka AUTOMOBILE ENGINEERING Session 1 Firing Interval, Number and Arrangement of Cylinders in Engines 1 Session 2 Dynamics of Crank Mechanisms 24 Session 3 Balancing of Engines 41 Session 4 Combustion chamber design 58 Session 5 Pollution & pollutants 91 Session 6 Diesel Fuel injection systems 109 Session 7 Properties of fuels 142 Session 8 Alternative Fuels 161 Session 9 Electric and Fuel Cell Vehicles 170 Session 10 Computer Controlled Fuel-Injection System [Diesel engine]: 177 Session 11 Computer Controlled Fuel-Injection System [gasoline engine] 184 iii Session 12 Construction of systems used in a hybrid vehicle 194 Session 13 Electro mechanical configuration of ABS systems 202 Session 14 Electro mechanical configuration of traction control systems 210 Session 15 Automatic gear shift control systems 216 iv Session 01 Firing Interval, Number and Arrangement of Cylinders in Engines Contents 1.0 Introduction 1 1.1 Single Cylinder Engine 1 1.2 Necessity for having Multi Cylinders 2 1.3 Firing Order 2 1.4 Firing Interval 2 1.5 Number and Arrangement of Cylinders 3 Aim 23 Objective 23 1.0 Introduction In the previous lesson we observed that the internal combustion engines can be classified very broadly in accordance with the, 1. Mode of operation (spark or compression ignition) 2. Fuel used by the engine (Petrol or Diesel engine) 3. System of sequence (Two or Four stroke operation) In this lesson we study the classification according to the number and arrangement of cylinders in multi cylinder engines. 1.1 Single Cylinder Engine In a single cylinder engine, as the name suggests, there is only one cylinder and these are generally used in motor cycles, lawn mowers etc. In these engine there is only one power impulse in every revolution or two revolutions depending on whether the engine operates on two or four stroke cycles. This leads to difficulty in balancing reciprocating parts resulting in vibrations. The maximum capacity of a high speed single cylinder engine is in the range of 250-300 cc. 1 1.2 Necessity for having Multi Cylinders The power developed by an engine can be increased either by enlarging the size of a single cylinder or by having more cylinders of the same size. Although it appears that a large single cylinder with lesser parts would be the best choice, its advantages are overweighed by the disadvantages. For a given speed and cylinder pressure, the engine power varies with the square of the bore whereas the weight varies as the cube of the bore. Therefore, if the diameter of the cylinder of an engine is increased by two times (doubled), the power would increase by four times and weight by eight times. This indicates a greater rate of increase of weight than power, yielding a lower power to weight ratio at larger diameters. In the case of a single cylinder four stroke engine, having one power stroke for two revolutions, a heavy flywheel is required to smoothen out Torque fluctuations. A two cylinder four stroke engine can be arranged in such a way that there would be one power stroke in every revolution. Therefore, depending on the number of cylinders it is possible to have one power stroke every half revolution or even less. This would require only a lighter flywheel, having less inertia and this results in improved acceleration. The vibrations of a single cylinder engine would be extremely difficult to balance or eliminate, whereas in certain configurations of multi cylinder engines they can be completely balanced and it is another point in favour of multi cylinder engines. Also there are difficulties in cooling large pistons of single cylinder engines than small pistons of multi cylinder engines and the noise produced due to combustion in a single cylinder is difficult to silence or dampen. Since it is economical to produce a single unit of a given size than a number of small units accounting for the same size, one, two and three cylinders are still manufactured. The most popular type and a good compromise for passenger cars is the four cylinder engine. 1.3 Firing Order The firing order is the order in which the cylinders deliver their power strokes and it is selected as a design parameter. The best design provides a well distributed pattern along the crank shaft. 1.4 Firing Interval The firing interval of a multi cylinder engine is the angle of rotation of crankshaft between any two consecutive firing of cylinders. In general the firing interval of a multi cylinder four stroke engine is given as, FI = 720o/N Where N is the number of cylinders. 2 The firing interval of a multicylinder two stroke engine is, FI = 360o/N However this depends on the configuration or the arrangement of cylinders. Take the case of a four cylinder in line engine, working on four strokes, having a firing order 1-3-4-2. Now let us determine what the firing order is, Since N = 4, FI = 720o/4 = 180o This means the second firing takes place 180 crank shaft degrees (half a revolution) after the first firing. Similarly the third and fourth firing. That is, if the first firing occurs at 0, then, second firing is at 180o + 0o third firing is at 180o + (180o + 0o) i.e. (360o + 0o) fourth firing is at 180o + (180o + 180o + 0o) i.e. (540o + 0o) The firing order gives the sequence of firing which occurs in cylinders. A firing order of 1-3-4-2 implied that, 1st firing occurs in cylinder No. 1 at 0 2nd firing occurs in cylinder No 3. at 180o + 0o 3rd firing occurs in cylinder No. 4. at 360o + 0o 4th firing occurs in cylinder No. 2 at 540o + 0o Also it can be seen that there is a power stroke for every stroke of the four stroke cycle of this four cylinder engine. 1.5 Number and Arrangement of Cylinders In this section we deal with the arrangement of cylinders (crankshaft configurations) in four stroke and two stroke multi cylinder engines. 3 1.5.1 Four Stroke System Various configurations of two, four, six and eight cylinder engines, working on the four stroke operation are outlined below. 1.5.1.1 Two Cylinder Engines 1. In line arrangement with both pistons on the same side and having identical movements. Fig. 1.1 In this arrangement shown in figure 1.1 both pistons 1 and 2, reach Top dead centre positions simultaneously and they move down to the Bottom dead centre position at the same time. A cutaway view of a two cylinder engine is shown in figure 1.2. Fig. 1.2 4 Firing order (FI) is 1-2 and has a regular firing interval of 360o. Fig. 1.3 Fig. 1.3 shows that at, 1,1 – cylinder No. 1 fires for the 1st time, 1,2 – cylinder No. 1 fires for the 2nd time, 2,1 – cylinder No. 2 fires for the 1st time, 2,2 – cylinder No. 2 fires for the 2nd time. Cylinder No. 1 fires first at 0o as shown and its subsequent firing occurs at 720o. Cylinder No. 2 fires first at 360o and its subsequent firing occurs at 1080o. The first firing of cylinder No. 2 occurs 360o after first firing of cylinder No. 1. This cycle of operations repeat for the other revolutions. This has a regular firing interval of 360o. Sequence of Operations Firing Order 1,2 Cylinder Firing Interval – Regular (360o) 1 Power Exhaust Suction Compression 2 Suction Compression Power Exhaust Crank Shaft Rotation 0o 180o 360o 540o 720o 5 2. Inline arrangement with both pistons on the same side and having opposite movements Fig. 1.4 In this arrangement, shown in figure 1.4, the piston of cylinder No. 1 is at TDC, when the piston of cylinder No. 2 is at BDC, and vice versa. FO is 1-2. Fig. 1.5 6 From figure 1.5, it can be clearly seen that, Cylinder No. 1 fires first at 0o Cylinder No. 2 fires first at 180o Cylinder No. 1 fires second at 720o Cylinder No. 2 fires second at 900o and First firing of Cyl. No. 2 occurs 180o after 1st firing of Cyl. No. 1, Second firing of Cyl. No. 1 occurs 540o after 1st firing of Cyl. No. 2, Second firing of Cyl. No. 2 occurs 180o after 2nd firing of Cyl. No. 1, Therefore it is clear that the firing intervals between consecutive firings are not regular and they are 180o and 540o. Sequence of operations Firing Order 1,2 Cylinder Firing Interval – Regular (180o, 540o ) 1 Power Exhaust Suction Compression 2 Compression Power Exhaust Suction Crank Shaft Rotation 0o 180o 360o 540o 720o 3. “V” Configuration Fig. 1.6 In the configuration shown in figure 1.6. the piston of cylinder No. 1 comes out while the piston of cylinder No. 2 moves in, when the crank is gradually rotated in the clockwise 7 direction. This is designed in such a way that the piston of cylinder No.2 reaches outer dead centre 90o after piston of Cyl. No. 1 has reached outer dead centre. FI is 1-2. Fig. 1.7 From figure 1.7. it is clear that, Cyl. No. 1 fires for the first time at 0o Cyl. No. 2 fires for the first time at 90o Cyl. No. 1 fires for the second time at 720o Cyl. No. 2 fires for the second time at 810o In this case too the firing interval are not regular and firing occurs at intervals of 90 and 630. Sequence of operations Firing Order 1,2 Cylinder Firing Interval – Irregular (90o, 630o ) 1 Power Exhaust Suction Compression 2 Power Exhaust Suction Compression Crank Shaft Rotation 0o 90o 180o 270o 360o 450o 540o 630o 720o 8 4. Inline but Opposed pistons Fig. 4.8 Fig. 1.8 In this configuration shown in figure 1.8 the pistons have been positioned on the opposite sides of the crankshaft and both the pistons move in together and move out together. When piston 1 moves out on the compression stroke, piston two moves out on the exhaust stroke. Fig. 1.9 From figure 1.9, can be seen that, Cyl 1 fires first at 0o Cyl. 2 fires first at 360o Cyl. 1 fires for the second time at 720o Cyl. 2 fires for the second time at 1080o 9 The firing of alternate cylinders occur at regular intervals of 360o Sequence of operations Firing Order 1,2 Cylinder Firing Interval – Regular (360o ) 1 Power Exhaust Suction Compression 2 Suction Compression Power Exhaust Crank Shaft Rotation 0o 180o 360o 540o 720o 1.5.1.2 Four Cylinder Engines 1. Conventional form with three bearings Fig. 1.10 In this configuration shown in figure 1.10, the pistons of cylinders 1 and 4 are at TDC positions when the pistons of cylinders 2 and 3 are at BDC positions. Firing order : 1-3-4-2 When the piston of cylinder 1 is at TDC, at the end of the compression stroke, a) the piston of cylinder 4 is at TDC, at the end of the exhaust stroke. b) the piston of cylinder 2 is at BDC at the end of the power stroke. c) the piston of 3 is at BDC at the end of the suction stroke. 10  Fig. 1.11 It can be seen from figure 1.11, that firing interval is regular and is 180o. Sequence of operation Firing Order 1,3,4,2 Cylinder Firing Interval - 180o 1 Power Exhaust Suction Compression 2 Exhaust Suction Compression Power 3 Compression Power Exhaust Suction 4 Suction Compression Power Exhaust Firing Order : 1-2-4-3 See Figure 1.12 Fig. 1.12 11 2. More rigid arrangement with five main bearings. Fig. 1.13 Figure 1.13 shows a more rigid arrangement with five main bearings. This is a more rigid version of the configuration explained under (1) above. This too can have two firing orders, 1-3-4-2 or 1-2-4-3, depending on the design, with regular firing intervals of 180o. 3. V - configuration In this configuration shown in figure 1.14 there are two banks of cylinders which are at an angle of 90o. Each bank has two cylinders. Fig. 1.14 12 The pistons marked 1 and 2 are on one bank while those marked 3 and 4 are on the other bank. A cutaway view of a V-4 engine is shown in figure 1.15. Fig. 1.15 Firing order 1-3-4-2 Fig. 1.16 13 In this arrangement shown in figure 1.16 when piston 1 of the left bank is at the outer dead centre position, a) Piston 2 of left bank would be at inner dead centre position, b) Piston 3 of right bank would be at inner dead centre position c) Piston 4 of right bank would be at outer dead centre position. This has a regular firing interval of 180o as can be seen from figure 1.16. 4. Flat engine low built configuration Fig. 1.17 In this arrangement shown in figure 1.17, pistons 1 and 2 are on one side whereas 3 and 4 are on the other side. (compare with a V engine – 180 angle between rows of cylinders) Firing order can be either 1-3-4-2 or 1-2-4-3, depending on the design. The firing interval is 180o in either case. Figure 1.18 shows a modern horizontally opposed four cylinder engine. Fig. 1.18 14 1.5.1.3 Six Cylinder Engines 1. In line engine Fig. 1.19 In this configuration shown in figure 1.19, the crank throws of each set of two cylinders are in the same radial plane. In other words crank throws of cylinders 1 and 6 are in one radial plane, 2 and 5 in another and 3 and 4 in a third plane. This implied that two outer cranks (1 and 6) are parallel and are at an angle of 240o to the middle pair. Similarly cranks 2 and 5 too are parallel and they are at an angle of 120 from the middle pair. With this arrangement, in every 120 (1/3 of a revolution), a pair of pistons will be at the TDC. Ex. At 0o crank angle, pistons 1 and 6 will be at TDC At 120o crank angle, pistons 2 and 5 will be at TDC At 240o crank angle, pistons 3 and 4 will be at TDC Out of the two pistons of a pair at TDC, one is at the compression stroke and the other is at the exhaust stroke. There are 6 power impulses during two rotations of the crank shaft or one power stroke in every 120o. 15 The possible firing orders are, Fig. 1.20 From figure 1.20, it can be seen that the firing intervals are regular and equal to 120o. 2. V - configuration Fig. 1.21 In this arrangement shown in figure 1.21, there are two banks of three cylinders each and they are at an angle of 60. The pistons 1,2 and 3 are in one bank whereas pistons 4,5 and 6 are in one another bank. As in the case of the previous six cylinder engine, this too has a power impulse in every one third of a revolution. Firing order : 1-5-6-3-2 Firing interval : 120o (see figure 1.22) 16  Fig. 1.22 3. Flat engines Flat engines are compact and low built and they are often air cooled. Fig. 1.23 Here too the cylinders are in two banks on opposite side as shown in figure 1.23. Pistons 1, 2 and 3 are in one bank and pistons 4, 5 and 6 are in the other bank. This arrangement too has six power impulses in every two revolutions and has a regular firing interval of 120o. The firing order is 1-5-3-6-2-4. 17 1.5.2 Two Stroke System The different arrangement of cranks in two, four and six cylinder engine working on two stroke operation are described below. 1.5.2.1 Two Cylinder Engine Fig. 1.24 You have already studied the operation of engines working on two stroke cycle and you know that it takes only one revolution to complete a cycle. In the arrangement shown in figure 1.24, the piston 1 is at TDC, at the end of the compression stroke while the piston 2 is at BDC, at the beginning of the suction stroke. Fig. 1.25 From figure 1.25 it can be seen that, cylinder 1 fires first at 0o cylinder 2 fires first at 180o 18 cylinder 1 fires for the second time at 360o cylinder 2 fires for the second time at 540o This has a regular firing interval of 180o Firing order is 1-2 Sequence of operations Firing Order 1,2 Cylinder Firing Interval – Regular (180o ) 1 Power Suction Power Suction Exhaust Compression Exhaust Compression 2 Suction Power Suction Power Compression Exhaust Compression Exhaust Crank Shaft Rotation 0o 180o 360o 540o 720o S.A.Q A three cylinder engine having a crank configuration as shown below operates on a two stroke cycle. The firing order is 1-2-3. Tabulate the sequence of operations. Answer 19 In this configuration, the piston 2 reaches the TDC position, 120o after piston 1 and piston 3 reaches TDC position, 120o after piston 2 has reached TDC. As this is operating on a two stroke cycle, each cylinder produces a power impulse per revolution of the crank shaft. From above diagram it is clear that, Cylinder 1 fires for the first time at 0o Cylinder 2 fires for the first time at 120o Cylinder 3 fires for the first time at 240o Cylinder 1 fires for the second time at 360o Cylinder 2 fires for the second time at 480o Cylinder 3 fires for the second time at 600o Also it can be seen that the firing interval is regular throughout and it is equal to 120o Firing Order 1-2-3 Cylinder Firing Interval – Regular 120o 1 Power Suction Power Exhaust Compression Exhaust 2 Power Suction 20  Exhaust Compression 3 Power Suction Exhaust Compression Crank Shaft Rotation 0o 120o 180o 240o 300o 360o 420o 480o 540o 600o 1.5.2.2 Four Cylinder Engine Fig. 1.26 Figure 1.26 shows the arrangement of a four cylinder engine. This system has a non symmetrical crankshaft. When piston 1 is at TDC, the piston 4 would be at BDC and pistons 2 and 3 would be at halfway positions moving in opposite directions. Firing order is 1-3-4-2 Firing interval is 90o and it is regular as can be seen from figure 1.27. Fig. 1.27 21 1.5.2.3 Six Cylinder Engine Fig. 1.28 This arrangement shown in figure 1.28, too has a non symmetrical crankshaft. The firing interval is regular and is equal to 60o (See figure 1.29). The firing order is 1-5-3-6-2-4 Fig. 1.29 22 Aim To study the configuration and operation of two, three, four, six cylinder engines. Objective At the end of the lesson, the student should be aware of the different Firing orders, Firing intervals corresponding to various crankshaft configurations of multi cylinder engines and the regularity of power impulses. 23 Session 02 Dynamics of Crank Mechanisms Contents 2.1 Velocity and Acceleration of the Piston of a Reciprocating Engine 24 2.2 Forces and Couples Transmitted in Direct Engine Mechanism (Trunk Piston Type) 28 2.3 Balancing of Engines 31 2.4 Why Balancing is Necessary 31 2.5 Static and Dynamic Balancing 32 2.6 Balancing of Rotating Masses 33 2.7 Balancing of Reciprocating Masses 34 2.8 Primary Balancing 37 2.9 Secondary Balancing 39 2.10 Condition of Balance in Various Crank Types 39 Aim 40 Objective 40 2.1 Velocity and Acceleration of the Piston of a Reciprocating Engine Fig. 2.1 24 In the figure 2.1. AB represents the connecting rod and CB represents the crank of a simple engine mechanism of a reciprocating Engine. Let  be the constant angular velocity with which the crank rotates and let  be the crank angle measured from the inner dead centre. Take X axis to be the line of stroke and  the inclination of the connecting rod to this axis. If we take the inner dead centre position of the gudgeon pin as the datum, then the displacement of the piston at a crank angle  is given by, x = r + L – r cos  - L cos  Where r is the crank radius and L is the length of the connecting rod. Now L sin  = r sin  and hence, sin  = r/L sin  = sin  / n where n = L/r Therefore, x = r (1 + n – cos  - n cos ) = r (1 + n – cos  - n 2  sin 2  )   1  sin 2   2 x  r 1  n  cos   n1      n 2   cos(2x) = 1 – 2sin2(x) =. sin2(x) = (1-cos(2x)/2).   sin 2    r 1  n  cos  n1     n 2    1  2 cos 2   r 1  n  cos  n1     2n 2  Approximately.  1  or x  r 1  cos   sin   2n   1 (1  cos 2)   r 1  cos     2n 2 25   1 1   r 1   cos   cos 2  4n 4n  Differentiating with respect to time t and substituting d/dt =  (angular velocity), we have piston velocity, dx  sin  cos     r  sin    dt  n 2  sin 2   Piston acceleration,   d 2x 2  n 2 cos 2  sin 4     r cos    dt 2   3  n 2  sin 2  2  At the inner dead centre, when  = 0, the  1 piston acceleration = 2 r 1   n  and at the outer dead centre, when  = ,  1 piston acceleration = -2 r 1    n These two positions are indicated by points M and N in figure 2.1. Now let us determine the position of the piston when its acceleration is zero. d2x This can be done by equating 2 to zero. dt That is,  n 2 cos 2  sin 4   2 r cos  0   n  sin  2 2  3 2   cos  (n2 – sin2 )3/2 = - (n2 cos 2 + sin4) Squaring both sides and simplifying we obtain the equation giving the crank angle  radians for which the piston acceleration is zero. sin6  - n2 sin4 - n4 sin2 + n4 = 0 26 The value of n is usually 5 or 6 in practice and when n is greater than 4 the equation is very nearly satisfied by a value of,  = tan-1 (n) The identity tan  = n = L/r implies than the connecting rod and the crank are at right angles. Fig. 2.2 The velocity of the piston when the piston acceleration is zero can be found by substituting tan  = n in the expression, dx  sin  cos     r sin    dt  n 2  sin 2   r  Relevant piston velocity = 1 n2 n Figures 5.2 shows the variation of piston velocity and piston acceleration with crank angle. Now let us try to obtain approximate expressions for piston velocity and acceleration.   1  sin 2   2 x  r 1  n  cos   n1      n 2     sin 2    r 1  n  cos   n1     2n 2  27 Approximately.  1  or x  r 1  cos   sin   2n   1 (1  cos 2)   r 1  cos     2n 2  1 1   r 1   cos   cos 2  4n 4n  Differentiating as before with respect to t and substituting d /dt = , dx  sin 2   r sin   dt  n  d 2x  cos 2    2 r cos   dt 2  n  In general this first approximation is sufficient for all practical purposes. 2.2 Forces and Couples Transmitted in Direct Engine Mechanism (Trunk Piston Type) A – Gudgeon pin centre B – Crank pin centre C – Crankshaft centre Fig. 2.3 The schematic arrangement of the connecting rod, crank and the piston of a single acting internal combustion engine is shown in figure 2.3. 28 P – Effective thrust on the piston in the direction of the line of stroke. Q – Thrust in the connecting rod N – Normal reaction of cylinder (Neglecting friction between piston rings/cylinder walls) Since Q is the resultant of two forces P and N, we can write, P = Q cos  N = Q sin  Turning Moment The turning moment is the moment of the connecting rod thrust Q acting at the centre crank pin centre, about the crankshaft centre C, and it is given by Turning Moment = Q X CZ = Q X CF cos  = Q cos  X CF = P X CF Crank pin Effort The crank pin effort R, shown in figure 2.3. is define as the connecting rod thrust Q in a direction tangential to the crank pin circle. Turning moment = R X BC = P X CF Resistance to the motion of the crankshaft This is equivalent to a force Q acting at the crank pin centre B in a direction opposing the connecting rod thrust. This force of resistance is equivalent to, (a) a resting couple Q X CZ, and (b) a reacting Q at the crankshaft bearing C (See figure 5.4). 29  Fig. 2.4 This reaction will have two components namely Q cos  (=P) and Q sin  (= N) at C. The component Q sin  will produce a couple with the equal and opposite force N acting at A and it is equivalent to N X AC acting in the clockwise direction. Also it can be shown that, this couple N X AC is equivalent to the turning moment since, N X AC = N X CF / tan  = N / tan  X CF = P X CF Fig. 2.5 For a single acting internal combustion engine shown in figure 2.5. the equation of motion of the piston can be written as, pA – Q cos  = m1 d2x / dt2 where p – the gas pressure 30  A – area of cross section of the piston Q – thrust in the connecting rod x – displacement of the piston from the inner dead centre position. Piston Effort P = Q cos   d2x  = pA  m 2   dt  The crank pin effort R, shown in above figures, is defined as the connecting rod thrust Q in a direction tangential to the crank pin circle. The turning moment = R x BC = P X CF The resistance to the motion of the crankshaft is equivalent to force Q acting at the crank pin centre B in a direction opposing the connecting rod thrust. This force is equivalent to, 1. a resisting couple Q X CZ 2. a reaction Q at the crankshaft bearing C. 2.3 Balancing of Engines Balancing of Engines can be categorized under two headings namely. (i) Balancing of rotating masses (ii) Balancing of reciprocating masses 2.4 Why Balancing is Necessary Consider the arrangement of the shaft shown in figure 5.6. Fig. 2.6 31 The shaft has an eccentric mass m at a radius r attached to the shaft and it is mounted on two bush bearings. As the shaft rotates at an angular speed w, a centrifugal force equivalent to mrw2 is produced and it acts radially outwards. This centrifugal forces causes, (a) Steady stress in the revolution shaft, and (b) Alternating stresses in supports which give rise to vibrations. These forces tend to wear only one side of the shaft and the whole circumference of the bearings. The dynamic forces which arise from any lack of balance, not only increases the bearing loads, but also sets up unpleasant and even dangerous vibrations. Therefore in view of these, it is very essential to balance rotating moments in order to arrest vibrations and to reduce wearing of supporting bearings. 2.5 Static and Dynamic Balancing In the analysis of balancing of engines it is necessary to understand what is meant by static and dynamic balancing. When a shaft carries a series of masses such that the centre of gravity of the combined system is on the axis of rotation it is said to be “statically” balanced. Such a system may nevertheless apply dynamic forces to the bearings, when it is rotated. In the case of static balancing, the part to be balanced is mounted on a true spindle (or its own journals), and placed on straight leveled knife edges. If the part is not properly balanced, it will roll on the knife edges until the heaviest side comes to the bottom. B attaching small counterweights, it can be made to remain in any position on the knife edge (no rolling takes place to bring the heaviest side to the bottom), and hence the error can be ascertained. This error can be corrected by adding balance masses or removing excess metal. This brings the part to the state of static balancing and the centre of mass of the system lies in the axis of rotation. But this method of balancing may not add balance masses in the same plane as the eccentric mass. Fig. 2.7 32 In the configuration shown in figure 2.7 the eccentric mass m1 can be statically balanced by attaching a mass m2 (equal to m1) at a plane B-B, quite different from plane A-A. When this system rotates the centrifugal forces created by m1and m2 act in opposite directions and they constitute a couple, which tends to rock the shaft in its bearings. Now we say that the system is not dynamically balanced oven through it is statically balanced. Therefore in order to dynamically balance the inertia effects of eccentric mass m, it may be necessary to use two additional balancing masses. The three masses including the eccentric mass must be arranged in such a way that the resultant dynamic force and the couple on the shaft are zero. This requires that the lines of action of three inertia forces shall be parallel and that the algebraic sum of their moments about any point in the same plane shall be zero. Therefore we can summarise that two conditions are required to ensure complete dynamic balancing of a rotating system. 1. The centre of mass of the system should lie on the axis of rotation. 2. The resultant couple due to dynamic forces, when rotating must be zero. 2.6 Balancing of Rotating Masses A general analysis, of balancing of rotating masses, where the masses are in the same axial plane is given under the section on static and dynamic balancing. However to refresh your memories of balancing of masses, studied in your course on Mechanics of Machines, a very brief outline of the balancing of rotating masses which are not in the same axial plane is given below. Fig. 2.8 The figure 2.8 shows a system of masses m1, m2 and m3 at radii r1, r2 and r3 revolving about an axis OX with an angular velocity w. These masses revolve in planes Z1, O1Y1, Z2 O2Y2 and Z3 O3 Y3 at distances x1, x2 and x3 from reference plane ZOY. The angular 33 positions are denoted by 1, 2 and 3, measured clockwise from the reference plane Z O Y. It can be shown that, 1. for static balance, the vectorial sum of the forces must be zero. m1 r1 sin 1 + m2 r2 sin 2 + …………………….. = 0 m1 r1 cos 1 + m2 r2 cos 2 + …………………….. = 0 2. for dynamic balance the vectorial sum of the couples must also vanish. m1 r1 x sin 1 + m2 r2 x sin 2 + …………………….. = 0 m1 r1 x cos 1 + m2 r2 x cos 2 + …………………….. = 0 2.7 Balancing of Reciprocating Masses The moving parts of a direct engine mechanism of a reciprocating engine are subjected to either pure rotation as in the case of crankshaft or pure translation as in the case of the piston with the single exception of the connecting rod. The motion of the connecting rod is a combination of both rotation and translation. The mass of the connecting rod is considered as the connecting rod is considered as dynamically equivalent to two masses concentrated at the two ends of the connecting rod. The magnitude of these masses are inversely proportional to the distances from the centre of gravity of the rod. Based on this assumption, let m1 be the mass of the reciprocating parts and m2 be the mass of the rotating parts The mass m1 is assumed to be concentrated at the gudgeon pin centre whereas the mass m2 to be concentrated at the crank pin centre. The mass m1 includes the mass of the piston and that portion of the connecting rod which is considered to be subject to reciprocatory motion. We know that the acceleration of the reciprocating parts is given by, d2x 2  w 2 rcos   cos 2 / n  dt Therefore the force required to accelerate the reciprocating mass is given by, F = m1 r w2 (cos  + cos 2/n) 34 where m1 is the mass of the reciprocating masses r is the crank radius w is the angular speed of rotation of the crank and  is the angle the crank makes with the line of stroke of the piston Therefore Inertia force = m1rw2 (cos  + cos 2/n) = m1rw2 cos  + mrw cos 2 /n = Fp + Fs where, Fp is the primary component of the inertia force and Fp = m1rw2 cos , and Fs is the secondary component of the inertia force and Fs = m1rw2 cos 2/n = m1 (2w)2 (r/4n) cos 2 From this analysis the following observations can be made. 1. The secondary force has twice the frequency of the primary force. 2. As n becomes larger the secondary effects of the inertia forces can be neglected. Applying the method of representative cranks, the primary component is the projection on the line of stroke, of the inertia force due to a mass m1 located at the crank pin centre. (see figure 2.9). Fig. 2.9 35 Also the secondary component is the projection on the line of stroke of the inertia force due to a mass m1/4n, located at the extremity of an imaginary crank of length r rotating at angular velocity 2 rad/sec in the plane of the main crank. (see figure 2.10). Fig 2.10 S.A.Q. Determine a dynamically equivalent system, consisting of two masses concentrated at points and connected rigidly together, which can replace a rigid body. Answer For a two mass system to be dynamically equivalent to a rigid body, the behaviour of the equivalent system, when subjected to a given system of forces should be exactly the same as that of the rigid body. For achieving this the following conditions have to be satisfied by the equivalent two mass system. a) The total mass must be equal to that of the rigid body b) The centre of gravity must coincide with rigid body c) The total moment of inertia about an axis through the centre of gravity must be equal to that of the rigid body. 36 The rigid body shown at (A) has a mass m and the radius of gyration about an axis through G is k. Let this be replaced by a dynamically equivalent system shown at (B). m1 and m2 are the masses of the equivalent system and a1, a2 are the distance of m1 and m2 respectively from G. The three conditions stated above lead to the following equations. m1 + m2 =m (1) m1a1 = m2a2 (2) m1 a 12 + m2 a 22 = m k2 (3) Substituting for m a from (2) in (3), m1 a 12 + m2 a 22 = m2 k2 Therefore m1 = mk/a1 (a1 + a2) From (1) and (2), m1 = m a2 / (a1 + a2) Therefore mk2 / a1 (a1 + a2) = ma2 / (a1 + a2) k2 = a1a2 Also, m1 = m a2 / (a1 + a2) = m a2 / L and m2 = m a/ / L Therefore m1/m2 = a2/a1 The magnitudes of the two masses are inversely proportional to the distances from the centre of gravity. 2.8 Primary Balancing The primary component of the inertia or disturbing force is given by Fp = m1 rw2 cos  The representative crank method assumes that the mass m1 to be located at the crank pin centre as shown in figure 5.9 and it will produce inertia forces, m1rw2 cos  in the plane of stroke (parallel to the line of stroke) m1 rw2 sin  in a transverse plane 37 Therefore it is clear that the magnitude of the primary component of the unbalanced disturbing force is equivalent to the magnitude of the component parallel to the line of stroke of the centrifugal force produced by an equal mass attached to and revolving with the crank pin. Fig. 2.11 With reference to figure 2.11 let us suppose that a weight mo is fixed as shown at a radius ro directly opposite the crank. The component of the centrifugal force created by this mass mo in a direction parallel to the line of stroke is mo ro w2 cos . This is in a direction opposite to that of the primary component of the disturbing force and hence the resultant force in this direction is (m1 r1 w2 – mo ro w2) cos . Therefore if m and r are chosen in such a way that mo ro = m1 r1, there would not be any disturbing force parallel to the line of stroke but the vertical component of the centrifugal force due to mass mp having a magnitude mo ro w2 sin  remains. This force acts in a transverse plane perpendicular to the line of stroke. As the c*rank revolves, this force also goes through he same variations of the magnitude as the original primary force of the reciprocating parts, but its maximum values will occur when the crank is at right angles to the line of stroke. Now it is clear that the introduction of the revolving balance mass mo, has merely served to change the direction of the disturbing force. It has been changed from a direction parallel to the line of stroke to a direction perpendicular to the line of stroke. It is preferable to have mo ro = c m1 r1 where c