Turning moment Diagram and Flywheel problems

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 Chapter 16 : Turning Moment Diagrams and Flywheel 565 16 F F F F Fea ea ea ea eatur tur tur tur tures es es es es T T Tur ur urning ning ning T Tur urning ning 1. Introduction. 2. Turning Moment Diagram for a Single Cylinder Moment Moment Moment Moment Moment Double Acting Steam Engine. 3. Turning Moment Diagram Diagrams Diagrams Diagrams Diagrams Diagrams for a Four Stroke Cycle Internal Combustion Engine. and Flywheel and Flywheel and Flywheel and Flywheel and Flywheel 4. Turning Moment Diagram for a Multicylinder Engine. 5. Fluctuation of Energy. 16.1. 16.1. Intr Introduction oduction 16.1. 16.1. 16.1. Intr Intr Introduction oduction oduction 6. Determination of Maximum Fluctuation of Energy. The turning moment diagram (also known as crank- 7. Coefficient of Fluctuation effort diagram) is the graphical representation of the turning of Energy. moment or crank-effort for various positions of the crank. It is 8. Flywheel. plotted on cartesian co-ordinates, in which the turning moment is taken as the ordinate and crank angle as abscissa. 9. Coefficient of Fluctuation of Speed. 16.2. 16.2. T Tur urning Moment Diagram f ning Moment Diagram for a Single or a Single 16.2. 16.2. 16.2. T T Tur ur urning Moment Diagram f ning Moment Diagram f ning Moment Diagram for a Single or a Single or a Single 10. Energy Stored in a Cylinder Double Acting Steam Engine Cylinder Double Acting Steam Engine Cylinder Double Acting Steam Engine Cylinder Double Acting Steam Engine Cylinder Double Acting Steam Engine Flywheel. A turning moment diagram for a single cylinder 11. Dimensions of the Flywheel double acting steam engine is shown in Fig. 16.1. The vertical Rim. ordinate represents the turning moment and the horizontal 12. Flywheel in Punching Press. ordinate represents the crank angle. We have discussed in Chapter 15 (Art. 15.10.) that the turning moment on the crankshaft,  sin 2 θ  TF=×r sinθ+ P  22  2s n−θin  565 566 Theory of Machines Fig. 16.1. Turning moment diagram for a single cylinder, double acting steam engine. where F = Piston effort, P r = Radius of crank, n = Ratio of the connecting rod length and radius of crank, and θ = Angle turned by the crank from inner dead centre. From the above expression, we see that the turning moment (T ) is zero, when the crank angle (θ) is zero. It is maximum when the crank angle is 90° and it is again zero when crank angle is 180°. This is shown by the curve abc in Fig. 16.1 and it represents the turning moment diagram for outstroke. The curve cde is the turning moment diagram for instroke and is somewhat similar to the curve abc. Since the work done is the product of the turning moment and the angle turned, therefore the area of the turning moment diagram represents the work done per revolution. In actual practice, the engine is assumed to work against the mean resisting torque, as shown by a horizontal line AF. The height of the ordinate a A represents the mean height of the turning moment diagram. Since it is assumed that the work done by the turning moment per revolution is equal to the work done against the mean resisting torque, therefore the area of the rectangle For flywheel, have a look at your tailor’s manual aAFe is proportional to the work done against sewing machine. the mean resisting torque. Notes: 1. When the turning moment is positive (i.e. when the engine torque is more than the mean resisting torque) as shown between points B and C (or D and E) in Fig. 16.1, the crankshaft accelerates and the work is done by the steam. Chapter 16 : Turning Moment Diagrams and Flywheel 567 2. When the turning moment is negative (i.e. when the engine torque is less than the mean resisting torque) as shown between points C and D in Fig. 16.1, the crankshaft retards and the work is done on the steam. 3. If T = Torque on the crankshaft at any instant, and T = Mean resisting torque. mean Then accelerating torque on the rotating parts of the engine = T – T mean 4. If (T –T ) is positive, the flywheel accelerates and if (T – T ) is negative, then the flywheel retards. mean mean 16.3. Turning Moment Diagram for a Four Stroke Cycle Internal Combustion Engine A turning moment diagram for a four stroke cycle internal combustion engine is shown in Fig. 16.2. We know that in a four stroke cycle internal combustion engine, there is one working stroke after the crank has turned through two revolutions, i.e. 720° (or 4 π radians). Fig. 16.2. Turning moment diagram for a four stroke cycle internal combustion engine. Since the pressure inside the engine cylinder is less than the atmospheric pressure during the suction stroke, therefore a negative loop is formed as shown in Fig. 16.2. During the compression stroke, the work is done on the gases, therefore a higher negative loop is obtained. During the expansion or working stroke, the fuel burns and the gases expand, therefore a large positive loop is obtained. In this stroke, the work is done by the gases. During exhaust stroke, the work is done on the gases, therefore a negative loop is formed. It may be noted that the effect of the inertia forces on the piston is taken into account in Fig. 16.2. 16.4. Turning Moment Diagram for a Multi-cylinder Engine A separate turning moment diagram for a compound steam engine having three cylinders and the resultant turning moment diagram is shown in Fig. 16.3. The resultant turning moment diagram is the sum of the turning moment diagrams for the three cylinders. It may be noted that the first cylinder is the high pressure cylinder, second cylinder is the intermediate cylinder and the third cylinder is the low pressure cylinder. The cranks, in case of three cylinders, are usually placed at 120° to each other. 568 Theory of Machines Fig. 16.3. Turning moment diagram for a multi-cylinder engine. 16.5. Fluctuation of Energy The fluctuation of energy may be determined by the turning moment diagram for one complete cycle of operation. Consider the turning moment diagram for a single cylinder double acting steam engine as shown in Fig. 16.1. We see that the mean resisting torque line AF cuts the turning moment diagram at points B, C, D and E. When the crank moves from a to p, the work done by the engine is equal to the area aBp, whereas the energy required is represented by the area aABp. In other words, the engine has done less work (equal to the area a AB) than the requirement. This amount of energy is taken from the flywheel and hence the speed of the flywheel decreases. Now the crank moves from p to q, the work done by the engine is equal to the area pBbCq, whereas the requirement of energy is represented by the area pBCq. Therefore, the engine has done more work than the requirement. This excess work (equal to the area BbC) is stored in the flywheel and hence the speed of the flywheel increases while the crank moves from p to q. Similarly, when the crank moves from q to r, more work is taken from the engine than is developed. This loss of work is represented by the area C c D. To supply this loss, the flywheel gives up some of its energy and thus the speed decreases while the crank moves from q to r. As the crank moves from r to s, excess energy is again developed given by the area D d E and the speed again increases. As the piston moves from s to e, again there is a loss of work and the speed decreases. The variations of energy above and below the mean resisting torque line are called fluctuations of energy. The areas BbC, CcD, DdE, etc. represent fluctuations of energy. A little consideration will show that the engine has a maximum speed either at q or at s. This is due to the fact that the flywheel absorbs energy while the crank moves from p to q and from r to s. On the other hand, the engine has a minimum speed either at p or at r. The reason is that the flywheel gives out some of its energy when the crank moves from a to p and q to r. The difference between the maximum and the minimum energies is known as maximum fluctuation of energy. 16.6. Determination of Maximum Fluctuation of Energy A turning moment diagram for a multi-cylinder engine is shown by a wavy curve in Fig. 16.4. The horizontal line AG represents the mean torque line. Let a , a , a be the areas above the 1 3 5 mean torque line and a , a and a be the areas below the mean torque line. These areas represent 2 4 6 some quantity of energy which is either added or subtracted from the energy of the moving parts of the engine. Chapter 16 : Turning Moment Diagrams and Flywheel 569 Let the energy in the flywheel at A = E, then from Fig. 16.4, we have Energy at B = E + a 1 Energy at C = E + a – a 1 2 Energy at D = E + a – a + a 1 2 3 Energy at E = E + a – a + a – a 1 2 3 4 Energy at F = E + a – a + a – a + a 1 2 3 4 5 Energy at G = E + a – a + a – a + a – a 1 2 3 4 5 6 = Energy at A (i.e. cycle repeats after G) Let us now suppose that the greatest of these energies is at B and least at E. Therefore, A flywheel stores energy when the supply Maximum energy in flywheel is in excess and releases energy when = E + a energy is in deficit. 1 Minimum energy in the flywheel = E + a – a + a – a 1 2 3 4 ∴ Maximum fluctuation of energy, ∆ E = Maximum energy – Minimum energy =(E + a ) – (E + a – a + a – a ) = a – a + a 1 1 2 3 4 2 3 4 Fig. 16.4. Determination of maximum fluctuation of energy. 16.7. Coefficient of Fluctuation of Energy It may be defined as the ratio of the maximum fluctuation of energy to the work done per cycle. Mathematically, coefficient of fluctuation of energy, Maximum fluctuation of energy C = E Work done per cycle The work done per cycle (in N-m or joules) may be obtained by using the following two relations : 1. Work done per cycle = T × θ mean where T = Mean torque, and mean θ = Angle turned (in radians), in one revolution. =2π, in case of steam engine and two stroke internal combustion engines =4π, in case of four stroke internal combustion engines. 570 Theory of Machines The mean torque (T ) in N-m may be obtained by using the following relation : mean PP × 60 T == mean 2πω N where P = Power transmitted in watts, N = Speed in r.p.m., and ω= Angular speed in rad/s = 2 πN/60 2. The work done per cycle may also be obtained by using the following relation : P × 60 = Work done per cycle n where n = Number of working strokes per minute, = N, in case of steam engines and two stroke internal combustion engines, = N /2, in case of four stroke internal combustion engines. The following table shows the values of coefficient of fluctuation of energy for steam engines and internal combustion engines. Table 16.1. Coefficient of fluctuation of energy (C ) for steam and internal E combustion engines. S.No. Type of engine Coefficient of fluctuation of energy (C ) E 1. Single cylinder, double acting steam engine 0.21 2. Cross-compound steam engine 0.096 3. Single cylinder, single acting, four stroke gas engine 1.93 4. Four cylinders, single acting, four stroke gas engine 0.066 5. Six cylinders, single acting, four stroke gas engine 0.031 16.8. Flywheel A flywheel used in machines serves as a reservoir, which stores energy during the period when the supply of energy is more than the requirement, and releases it during the period when the requirement of energy is more than the supply. In case of steam engines, internal combustion engines, reciprocating compressors and pumps, the energy is developed during one stroke and the engine is to run for the whole cycle on the energy produced during this one stroke. For example, in internal combustion engines, the energy is developed only during expansion or power stroke which is much more than the engine load and no energy is being developed during suction, compression and exhaust strokes in case of four stroke engines and during compression in case of two stroke engines. The excess energy developed during power stroke is absorbed by the flywheel and releases it to the crankshaft during other strokes in which no energy is developed, thus rotating the crankshaft at a uniform speed. A little consideration will show that when the flywheel absorbs energy, its speed increases and when it releases energy, the speed decreases. Hence a flywheel does not maintain a constant speed, it simply reduces the fluctuation of speed. In other words, a flywheel controls the speed variations caused by the fluctuation of the engine turning moment during each cycle of operation. Chapter 16 : Turning Moment Diagrams and Flywheel 571 In machines where the operation is intermittent like punching machines, shearing machines, rivetting machines, crushers, etc., the flywheel stores energy from the power source during the greater portion of the operating cycle and gives it up during a small period of the cycle. Thus, the energy from the power source to the machines is supplied practically at a constant rate throughout the operation. Note: The function of a governor in an engine is entirely different from that of a flywheel. It regulates the mean speed of an engine when there are variations in the load, e.g., when the load on the engine increases, it becomes necessary to increase the supply of working fluid. On the other hand, when the load decreases, less working fluid is required. The governor automatically controls the supply of working fluid to the engine with the varying load condition and keeps the mean speed of the engine within certain limits. As discussed above, the flywheel does not maintain a constant speed, it simply reduces the fluctuation of speed. It does not control the speed variations caused by the varying load. 16.9. Coefficient of Fluctuation of Speed The difference between the maximum and minimum speeds during a cycle is called the maximum fluctuation of speed. The ratio of the maximum fluctuation of speed to the mean speed is called the coefficient of fluctuation of speed. Let N and N = Maximum and minimum speeds in r.p.m. during the cycle, and 1 2 NN + 12 N = Mean speed in r.p.m. = 2 ∴ Coefficient of fluctuation of speed, 2NN − NN − () 12 12 C == s NN +N 12 2 ω−ω ω−ω () 12 12 == ...(In terms of angular speeds) ωω+ω 12 vv − 2vv − () 12 12 == ...(In terms of linear speeds) vv +v 12 The coefficient of fluctuation of speed is a limiting factor in the design of flywheel. It varies depending upon the nature of service to which the flywheel is employed. Note. The reciprocal of the coefficient of fluctuation of speed is known as coefficient of steadiness and is denoted by m. 1 N m== ∴ CN −N s 12 16.10. Energy Stored in a Flywheel A flywheel is shown in Fig. 16.5. We have discussed in Art. 16.5 that when a flywheel absorbs energy, its speed increases and when it gives up energy, its speed decreases. Let m = Mass of the flywheel in kg, k = Radius of gyration of the flywheel in metres, Fig. 16.5. Flywheel. See Art. 16.12. See Chapter 18 on Governors. 572 Theory of Machines I = Mass moment of inertia of the flywheel about its axis of rotation 2 2 in kg-m = m.k , N and N = Maximum and minimum speeds during the cycle in r.p.m., 1 2 ω and ω = Maximum and minimum angular speeds during the cycle in rad/s, 1 2 NN + 12 N = Mean speed during the cycle in r.p.m. = , 2 ω+ω 12 ω = Mean angular speed during the cycle in rad/s = , 2 NN−ω−ω 12 1 2 = or C = Coefficient of fluctuation of speed, S N ω We know that the mean kinetic energy of the flywheel, - 11 222 EI =×.. ω = ×mk.ω (in N-m or joules) 22 As the speed of the flywheel changes from ω to ω , the maximum fluctuation of energy, 1 2 ∆ E = Maximum K.E. – Minimum K.E. 2 22 2 11 1 =×II ω − × ω =×I ω −ω () () () ( )  12 1 2 22 2  1 = ×II ω+ω ω −ω = .ω ω−ω ...(i) ()( ) () 12 1 2 12 2 ω+ω  12 ... ω=   2  ω−ω  2 12 =ω I .  ... (Multiplying and dividing by ω)  ω  2 2 2 2 = I.ω .C = m.k .ω .C ... (∵ I = m.k ) ...(ii) S S 1  2 ... EI =× .ω =2.E.C (in N–m or joules)  ... (iii) S 2  The radius of gyration (k) may be taken equal to the mean radius of the rim (R), because the thickness of rim is very small as compared to the diameter of rim. Therefore, substituting k = R, in equation (ii), we have 2 2 2 ∆ E = m.R .ω .C = m.v .C S S where v = Mean linear velocity (i.e. at the mean radius) in m/s = ω.R Notes. 1. Since ω = 2 π N/60, therefore equation (i) may be written as 2 22 ππ NN  24 ππ N 12 ∆=EI× − = × I× NN− N ()  12 60 60 60 3600  2 π 2 =×mk..N N− N () 12 900  NN − 2 12 π 2 2 ... C = =×mk..N.C  s S N 900  Chapter 16 : Turning Moment Diagrams and Flywheel 573 2. In the above expressions, only the mass moment of inertia of the flywheel rim (I) is considered and the mass moment of inertia of the hub and arms is neglected. This is due to the fact that the major portion of the mass of the flywheel is in the rim and a small portion is in the hub and arms. Also the hub and arms are nearer to the axis of rotation, therefore the mass moment of inertia of the hub and arms is small. Example 16.1. The mass of flywheel of an engine is 6.5 tonnes and the radius of gyration is 1.8 metres. It is found from the turning moment diagram that the fluctuation of energy is 56 kN-m. If the mean speed of the engine is 120 r.p.m., find the maximum and minimum speeds. 3 Solution. Given : m = 6.5 t = 6500 kg ; k = 1.8 m ; ∆ E = 56 kN-m = 56 × 10 N-m ; N = 120 r.p.m. Let N and N = Maximum and minimum speeds respectively. 1 2 We know that fluctuation of energy (∆ E), 2 2 π π 3 2 2 = = 56 × 10 × m.k . N (N – N ) × 6500 (1.8) 120 (N – N ) 1 2 1 2 900 900 = 27 715 (N – N ) 1 2 3 ∴ N – N = 56 × 10 /27 715 = 2 r.p.m. ...(i) 1 2 We also know that mean speed (N), NN + 12 120=+ orNN= 120× 2= 240 r.p.m. ...(ii) 12 2 From equations (i) and (ii), N = 121 r.p.m., and N = 119 r.p.m. Ans. 1 2 Example 16.2. The flywheel of a steam engine has a radius of gyration of 1 m and mass 2500 kg. The starting torque of the steam engine is 1500 N-m and may be assumed constant. Determine: 1. the angular acceleration of the flywheel, and 2. the kinetic energy of the flywheel after 10 seconds from the start. Solution. Given : k = 1 m ; m = 2500 kg ; T = 1500 N-m 1. Angular acceleration of the flywheel Let α = Angular acceleration of the flywheel. We know that mass moment of inertia of the flywheel, 2 2 2 I = m.k = 2500 × 1 = 2500 kg-m ∴ Starting torque of the engine (T), 2 1500 = I.α = 2500 × α or α = 1500 / 2500 = 0.6 rad /s Ans. 2. Kinetic energy of the flywheel First of all, let us find out the angular speed of the flywheel after 10 seconds from the start (i.e. from rest), assuming uniform acceleration. Let ω = Angular speed at rest = 0 1 ω = Angular speed after 10 seconds, and 2 t = Time in seconds. We know that ω = ω + α t = 0 + 0.6 × 10 = 6 rad /s 2 1 574 Theory of Machines ∴ Kinetic energy of the flywheel 11 2 2 =× Iω = × 2500× 6 = 45 000 N-m = 45 kN-m Ans. () 2 22 Example 16.3. A horizontal cross compound steam engine develops 300 kW at 90 r.p.m. The coefficient of fluctuation of energy as found from the turning moment diagram is to be 0.1 and the fluctuation of speed is to be kept within ± 0.5% of the mean speed. Find the weight of the flywheel required, if the radius of gyration is 2 metres. 3 Solution. Given : P = 300 kW = 300 × 10 W; N = 90 r.p.m.; C = 0.1; k = 2 m E We know that the mean angular speed, ω =2 π N/60 = 2 π × 90/60 = 9.426 rad/s Let ω and ω = Maximum and minimum speeds respectively. 1 2 Since the fluctuation of speed is ± 0.5% of mean speed, therefore total fluctuation of speed, ω – ω =1% ω = 0.01 ω 1 2 and coefficient of fluctuation of speed, ω−ω 12 C== 0.01 s ω We know that work done per cycle 3 3 = P × 60 / N = 300 × 10 × 60 / 90 = 200 × 10 N-m ∴ Maximum fluctuation of energy, 3 3 ∆ E = Work done per cycle × C = 200 × 10 × 0.1 = 20 × 10 N-m E Let m = Mass of the flywheel. We know that maximum fluctuation of energy ( ∆ E ), 3 2 2 2 2 20 × 10 = m.k .ω .C = m × 2 × (9.426) × 0.01 = 3.554 m S 3 ∴ m= 20 × 10 /3.554 = 5630 kg Ans. Example 16.4. The turning moment diagram for a petrol engine is drawn to the following scales : Turning moment, 1 mm = 5 N-m ; crank angle, 1 mm = 1°. The turning moment diagram repeats itself at every half revolution of the engine and the areas above and below the mean turning 2 moment line taken in order are 295, 685, 40, 340, 960, 270 mm . The rotating parts are equivalent to a mass of 36 kg at a radius of gyration of 150 mm. Determine the coefficient of fluctuation of speed when the engine runs at 1800 r.p.m. Solution. Given : m = 36 kg ; k = 150 mm = 0.15 m ; N = 1800 r.p.m. or ω = 2 π × 1800/60 = 188.52 rad /s Fig. 16.6 The turning moment diagram is shown in Fig. 16.6. Chapter 16 : Turning Moment Diagrams and Flywheel 575 Since the turning moment scale is 1 mm = 5 N-m and crank angle scale is 1 mm = 1° = π /180 rad, therefore, 2 1 mm on turning moment diagram ππ =×5N =-m 180 36 Let the total energy at A = E, then referring to Fig. 16.6, Energy at B = E + 295 ... (Maximum energy) Energy at C = E + 295 – 685 = E – 390 Energy at D = E – 390 + 40 = E – 350 Flywheel of an electric motor. Energy at E = E – 350 – 340 = E – 690 ...(Minimum energy) Energy at F = E – 690 + 960 = E + 270 Energy at G = E + 270 – 270 = E = Energy at A We know that maximum fluctuation of energy, ∆ E = Maximum energy – Minimum energy 2 =(E + 295) – (E – 690) = 985 mm π =× 985 = 86 N - m= 86 J 36 Let C = Coefficient of fluctuation of speed. S We know that maximum fluctuation of energy (∆ E), 2 2 2 2 86 = m.k ω .C = 36 × (0.15) × (188.52) C = 28 787 C S S S ∴ C = 86 / 28 787 = 0.003 or 0.3% Ans. S Example 16.5. The turning moment diagram for a multicylinder engine has been drawn to a scale 1 mm = 600 N-m vertically and 1 mm = 3° horizontally. The intercepted areas between the output torque curve and the mean resistance line, taken in order from one end, are as follows : 2 + 52, – 124, + 92, – 140, + 85, – 72 and + 107 mm , when the engine is running at a speed of 600 r.p.m. If the total fluctuation of speed is not to exceed ± 1.5% of the mean, find the necessary mass of the flywheel of radius 0.5 m. Solution. Given : N = 600 r.p.m. or ω = 2 π × 600 / 60 = 62.84 rad / s ; R = 0.5 m Fig. 16.7 Since the total fluctuation of speed is not to exceed ± 1.5% of the mean speed, therefore ω – ω =3% ω = 0.03 ω 1 2 576 Theory of Machines and coefficient of fluctuation of speed, ω−ω 12 0.03 C== s ω The turning moment diagram is shown in Fig. 16.7. Since the turning moment scale is 1 mm = 600 N-m and crank angle scale is 1 mm = 3° = 3° × π/180 = π / 60 rad, therefore 2 1 mm on turning moment diagram = 600 × π/60 = 31.42 N-m Let the total energy at A = E, then referring to Fig. 16.7, Energy at B = E + 52 ...(Maximum energy) Energy at C = E + 52 – 124 = E – 72 Energy at D = E – 72 + 92 = E + 20 Energy at E = E + 20 – 140 = E – 120 ...(Minimum energy) Energy at F = E – 120 + 85 = E – 35 Energy at G = E – 35 – 72 = E – 107 Energy at H = E – 107 + 107 = E = Energy at A We know that maximum fluctuation of energy, ∆ E = Maximum energy – Minimum energy =(E + 52) – (E – 120) = 172 = 172 × 31.42 = 5404 N-m Let m = Mass of the flywheel in kg. We know that maximum fluctuation of energy (∆ E ), 2 2 2 2 5404 = m.R .ω .C = m × (0.5) × (62.84) × 0.03 = 29.6 m S ∴ m = 5404 / 29.6 = 183 kg Ans. Example 16.6. A shaft fitted with a flywheel rotates at 250 r.p.m. and drives a machine. The torque of machine varies in a cyclic manner over a period of 3 revolutions. The torque rises from 750 N-m to 3000 N-m uniformly during 1/2 revolution and remains constant for the following revolution. It then falls uniformly to 750 N-m during the next 1/2 revolution and remains constant for one revolution, the cycle being repeated thereafter. Determine the power required to drive the machine and percentage fluctuation in speed, if the driving torque applied to the shaft is constant and the mass of the flywheel is 500 kg with radius of gyration of 600 mm. Solution. Given : N = 250 r.p.m. or ω = 2π × 250/60 = 26.2 rad/s ; m = 500 kg ; k = 600 mm = 0.6 m The turning moment diagram for the complete cycle is shown in Fig. 16.8. We know that the torque required for one complete cycle = Area of figure OABCDEF = Area OAEF + Area ABG + Area BCHG + Area CDH 11 =× OF OA+ × AG× BG+ GH× CH+ × HD× CH 22 Chapter 16 : Turning Moment Diagrams and Flywheel 577 1 6 750 3000 750 2 3000 750 =π× + ×π() − + π() − 2 1 +×π 3000− 750 () 2 = 11 250 π N-m ...(i) If T is the mean torque in N-m, then torque required for one complete cycle mean = T × 6 π Ν-m ...(ii) mean From equations (i) and (ii), T = 11 250 π /6 π = 1875 N-m mean Fig. 16.8 Power required to drive the machine We know that power required to drive the machine, P = T × ω = 1875 × 26.2 = 49 125 W = 49.125 kW Ans. mean Coefficient of fluctuation of speed Let C = Coefficient of fluctuation of speed. S First of all, let us find the values of LM and NP. From similar triangles ABG and BLM, LM 3000 − 1875 LM BM == 0.5 = or or LM = 0.5 π π− 3000 750 AG BG Now, from similar triangles CHD and CNP, NP CN NP 3000 − 1875 = or or NP = 0.5 π == 0.5 HD CH π− 3000 750 From Fig. 16.8, we find that BM = CN = 3000 – 1875 = 1125 N-m Since the area above the mean torque line represents the maximum fluctuation of energy, therefore, maximum fluctuation of energy, ∆ E = Area LBCP = Area LBM + Area MBCN + Area PNC 11 =× LM× BM+×+ MN BM × NP× CN 22 578 Theory of Machines 11 =× 0.5π× 1125+ 2π× 1125+ × 0.5π× 1125 22 = 8837 N - m We know that maximum fluctuation of energy (∆ E), 2 2 2 2 8837 = m.k .ω .C = 500 × (0.6) × (26.2) × C = 123 559 C S S S 8837 C== 0.071 Ans. S 123559 Flywheel of a pump run by a diesel engine. Example 16.7. During forward stroke of the piston of the double acting steam engine, the turning moment has the maximum value of 2000 N-m when the crank makes an angle of 80° with the inner dead centre. During the backward stroke, the maximum turning moment is 1500 N-m when the crank makes an angle of 80° with the outer dead centre. The turning moment diagram for the engine may be assumed for simplicity to be represented by two triangles. If the crank makes 100 r.p.m. and the radius of gyration of the flywheel is 1.75 m, find the coefficient of fluctuation of energy and the mass of the flywheel to keep the speed within ± 0.75% of the mean speed. Also determine the crank angle at which the speed has its minimum and maximum values. Solution. Given : N = 100 r.p.m. or ω = 2π × 100/60 = 10.47 rad /s; k = 1.75 m Since the fluctuation of speed is ± 0.75% of mean speed, therefore total fluctuation of speed, ω – ω =1.5% ω 1 2 and coefficient of fluctuation of speed, ωω – 12 C== 1.5%= 0.015 S ω Coefficient of fluctuation of energy The turning moment diagram for the engine during forward and backward strokes is shown in Fig. 16.9. The point O represents the inner dead centre (I.D.C.) and point G represents the outer dead centre (O.D.C). We know that maximum turning moment when crank makes an angle of 80° (or 80 × π / 180 = 4π/9 rad) with I.D.C., ∴ AB = 2000 N-m Chapter 16 : Turning Moment Diagrams and Flywheel 579 and maximum turning moment when crank makes an angle of 80° with outer dead centre (O.D.C.) or 180° + 80° = 260° = 260 × π /180 = 13 π / 9 rad with I.D.C., LM = 1500 N-m Let T = EB = QM = Mean resisting torque. mean Fig. 16.9 We know that work done per cycle = Area of triangle OAG + Area of triangle GLS 11 =× OG× AB+ × GS× LM 22 11 = × π× 2000 + × π× 1500 = 1750π N-m ...(i) 22 We also know that work done per cycle = T × 2 π N-m ...(ii) mean From equations (i) and (ii), T = 1750 π / 2 π = 875 N-m mean From similar triangles ACD and AOG, CD OG = AE AB OG OG π CD=× AE= AB – EB= 2000− 875= 1.764 rad or () () AB AB 2000 ∴ Maximum fluctuation of energy, 1 ∆= E = Area of triangle ACD× CD× AE 2 11 CD AB – EB 1.764 2000 875 992 N-m =×()= × () − = 22 We know that coefficient of fluctuation of energy, Max.fluctuation of energy 992 C=== 0.18 or 18% Ans. E Work done per cycle 1750 π 580 Theory of Machines Mass of the flywheel Let m = Mass of the flywheel. We know that maximum fluctuation of energy (∆ E), 2 2 2 2 992 = m.k .ω .C = m × (1.75) × (10.47) × 0.015 = 5.03 m S ∴ m = 992 / 5.03 = 197.2 kg Ans. Crank angles for the minimum and maximum speeds We know that the speed of the flywheel is minimum at point C and maximum at point D (See Art. 16.5). Let θ and θ = Crank angles from C D I.D.C., for the minimum and maximum speeds. From similar triangles ACE and AOB, CE AE = Flywheel of small steam engine. OB AB AE AB – EB 2000−π 875 4 π CE=× OB= × OB= × = rad or AB AB 2000 9 4 4ππ 7π 7π 180 θ− = = rad= × = 35° ∴ Ans. C 9 4 36 36 π Again from similar triangles AED and ABG, ED AE = BG AB AE AB – EB ED=× BG= OG− OB or () AB AB 2000−π 875 4 2.8π =π−= rad  2000 9 9  4ππ 2.8 6.8π 6.8π 180 ∴ θ+ = = rad= × = 136° Ans. D 99 9 9 π Example 16.8. A three cylinder single acting engine has its cranks set equally at 120° and it runs at 600 r.p.m. The torque-crank angle diagram for each cycle is a triangle for the power stroke with a maximum torque of 90 N-m at 60° from dead centre of corresponding crank. The torque on the return stroke is sensibly zero. Determine : 1. power developed. 2. coefficient of fluctuation of speed, if the mass of the flywheel is 12 kg and has a radius of gyration of 80 mm, 3. coefficient of fluctuation of energy, and 4. maximum angular acceleration of the flywheel. Solution. Given : N = 600 r.p.m. or ω = 2 π × 600/60 = 62.84 rad /s; T = 90 N-m; max m = 12 kg; k = 80 mm = 0.08 m Chapter 16 : Turning Moment Diagrams and Flywheel 581 The torque-crank angle diagram for the individual cylinders is shown in Fig. 16.10 (a), and the resultant torque-crank angle diagram for the three cylinders is shown in Fig. 16.10 (b). Fig. 16.10 1. Power developed We know that work done/cycle 1 =× Area of three triangles = 3×π× 90= 424 N-m 2 Work done / cycle 424 and mean torque, T=== 67.5 N - m mean Crank angle / cycle 2π ∴ Power developed = T × ω = 67.5 × 62.84 = 4240 W = 4.24 kW Ans. mean 2. Coefficient of fluctuation of speed Let C = Coefficient of fluctuation of speed. S First of all, let us find the maximum fluctuation of energy (∆ E). From Fig. 16.10 (b), we find that 1 aA=× Area of triangleaB =AB×Aa 1 2 1 π =6 ××7.5−45=5.89N-m= a () 7 ...(∵ AB = 30° = π / 6 rad) 26 1 a=× Area of triangle BbC = BC× bb ' 2 2 1 π =×− 90 67.5= 11.78 N - m () ...(∵BC = 60° = π/3 rad) 23 = a = a = a = a 3 4 5 6 Now, let the total energy at A = E, then referring to Fig. 16.10 (b), Energy at B = E – 5.89 Energy at C = E – 5.89 + 11.78 = E + 5.89 Energy at D = E + 5.89 – 11.78 = E – 5.89 Energy at E = E – 5.89 + 11.78 = E + 5.89 Energy at G = E + 5.89 – 11.78 = E – 5.89 Energy at H = E – 5.89 + 11.78 = E + 5.89 Energy at J = E + 5.89 – 5.89 = E = Energy at A 582 Theory of Machines From above we see that maximum energy = E + 5.89 and minimum energy = E – 5.89 ∴ Maximum fluctuation of energy, ∆ E =(E + 5.89) – (E – 5.89) = 11.78 N-m We know that maximum fluctuation of energy (∆ E), 2 2 2 2 11.78 = m.k .ω .C = 12 × (0.08) × (62.84) × C = 303.3 C S S S ∴ C = 11.78 / 303.3 = 0.04 or 4% Ans. S 3. Coefficient of fluctuation of energy We know that coefficient of fluctuation of energy, Max. fluctuation of energy 11.78 C=== 0.0278= 2.78% Ans. E Work done/cycle 424 4. Maximum angular acceleration of the flywheel Let α = Maximum angular acceleration of the flywheel. We know that, 2 T – T = I.α = m.k .α max mean 2 90 – 67.5 = 12 × (0.08) × α = 0.077 α 90 − 67.5 2 ∴ α= = 292 rad / s Ans. 0.077 Example 16.9. A single cylinder, single acting, four stroke gas engine develops 20 kW at 300 r.p.m. The work done by the gases during the expansion stroke is three times the work done on the gases during the compression stroke, the work done during the suction and exhaust strokes being negligible. If the total fluctuation of speed is not to exceed ± 2 per cent of the mean speed and the turning moment diagram during compression and expansion is assumed to be triangular in shape, find the moment of inertia of the flywheel. 3 Solution. Given : P = 20 kW = 20 × 10 W; N = 300 r.p.m. or ω = 2π × 300/60 = 31.42 rad/s Since the total fluctuation of speed (ω – ω ) is not to exceed ± 2 per cent of the mean speed 1 2 (ω), therefore ω – ω =4% ω 1 2 and coefficient of fluctuation of speed, ω−ω 12 C== 4%= 0.04 S ω The turning moment-crank angle diagram for a four stroke engine is shown in Fig. 16.11. It is assumed to be triangular during compression and expansion strokes, neglecting the suction and exhaust strokes. Since the area above the mean torque line represents the maximum fluctuation of energy, therefore maxi- mum fluctuation of energy, ∆ E = Area Bbc = Area DdE = Area Ggh 1 π =× (90–67.5)=11.78N-m 23 Chapter 16 : Turning Moment Diagrams and Flywheel 583 We know that for a four stroke engine, number of working strokes per cycle, n = N/2 = 300 / 2 = 150 3 ∴ Work done/cycle =P × 60/n = 20 × 10 × 60/150 = 8000 N-m ...(i) Fig. 16.11 Since the work done during suction and exhaust strokes is negligible, therefore net work done per cycle (during compression and expansion strokes) W 2 E == WWW––= W ... ( ∵ W = 3W ) ...(ii) E C EC E E 33 Equating equations (i) and (ii), work done during expansion stroke, W = 8000 × 3/2 = 12 000 N-m E We know that work done during expansion stroke (W ), E 11 12 000== Area of triangle ABC× BC× AG=×π× AG 22 ∴ AG = T = 12 000 × 2/π = 7638 N-m max and mean turning moment, Work done/cycle 8000 TF==G = = 637 N-m mean Crank angle/cycle 4π ∴ Excess turning moment, T = AF = AG – FG = 7638 – 637 = 7001 N-m excess Now, from similar triangles ADE and ABC, DE AF AF 7001 = or DE=× BC= ×π= 2.88 rad BC AG AG 7638 Since the area above the mean turning moment line represents the maximum fluctuation of energy, therefore maximum fluctuation of energy, 11 ∆=EA Area of∆D=E ×D×EA=F × 2.88× 7001= 10081 N-m 22 The mean turning moment (T ) may also be obtained by using the following relation : mean 3 P = T × ω or T = P/ω = 20 × 10 /31.42 = 637 N-m mean mean 584 Theory of Machines 2 Let I = Moment of inertia of the flywheel in kg-m . We know that maximum fluctuation of energy (∆ E), 2 2 10 081 = I.ω .C = I × (31.42) × 0.04 = 39.5 I S 2 ∴ I = 10081/ 39.5 = 255.2 kg-m Ans. Example 16.10. The turning moment diagram for a four stroke gas engine may be assumed for simplicity to be represented by four triangles, the areas of which from the line of zero pressure are as follows : –3 2 –3 2 Suction stroke = 0.45 × 10 m ; Compression stroke = 1.7 × 10 m ; Expansion stroke –3 2 –3 2 2 = 6.8 × 10 m ; Exhaust stroke = 0.65 × 10 m . Each m of area represents 3 MN-m of energy. Assuming the resisting torque to be uniform, find the mass of the rim of a flywheel required to keep the speed between 202 and 198 r.p.m. The mean radius of the rim is 1.2 m. –3 2 –3 2 –3 2 Solution. Given : a = 0.45 × 10 m ; a = 1.7 × 10 m ; a = 6.8 × 10 m ; 1 2 3 –3 2 a = 0.65 × 10 m ; N = 202 r.p.m; N = 198 r.p.m.; R = 1.2 m 4 1 2 The turning moment crank angle diagram for a four stroke engine is shown in Fig. 16.12. The areas below the zero line of pressure are taken as negative while the areas above the zero line of pressure are taken as positive. ∴ Net area = a – (a + a + a ) 3 1 2 4 –3 –3 –3 –3 –3 2 = 6.8 × 10 – (0.45 × 10 + 1.7 × 10 + 0.65 × 10 ) = 4 × 10 m 2 6 Since the energy scale is 1 m = 3 MN-m = 3 × 10 N-m, therefore, –3 6 3 Net work done per cycle = 4 × 10 × 3 ×10 = 12 × 10 N-m . . . (i) We also know that work done per cycle, = T × 4π N-m . . . (ii) mean From equations (i) and (ii), 3 T = FG = 12 × 10 /4π = 955 N-m mean Fig. 16.12 Work done during expansion stroke –3 6 3 = a × Energy scale = 6.8 × 10 × 3 × 10 = 20.4 × 10 N-m ...(iii) 3

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