10/29/2018. Chapter 16. Turning Moment Diagrams and Flywheel. Mohammad Suliman Abuhaiba, Ph.D., PE

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1 1 Chapter 16 Turning Moment Diagrams and Flywheel

2 2 Turning moment diagram (TMD) graphical representation of turning moment or crank-effort for various positions of the crank

3 3 Turning Moment Diagram for a Single Cylinder Double Acting Steam Engine

4 4 Turning Moment Diagram for a Single Cylinder Double Acting Steam Engine Turning moment on crankshaft, F P = Piston effort Out stroke = curve abc In stroke = curve cde

5 5 Turning Moment Diagram for a Single Cylinder Double Acting Steam Engine Work done = turning moment angle turned work done = area of turning moment diagram per rev In actual practice, engine is assumed to work against the mean resisting torque Area of rectangle aafe is proportional to work done against the mean resisting torque

6 6 Turning Moment Diagram for a Single Cylinder Double Acting Steam Engine When turning moment is positive (engine torque is more than mean resisting torque) as shown between points B & C or D & E, crankshaft accelerates and work is done by steam. When turning moment is negative (engine torque is less than mean resisting torque) as shown between points C & D, crankshaft retards and work is done on steam.

7 7 Turning Moment Diagram for a Single Cylinder Double Acting Steam Engine T = Torque on crankshaft at any instant T mean = Mean resisting torque Accelerating torque on rotating parts of engine = T T mean If (T T mean ) is positive, flywheel accelerates If (T T mean ) is negative, flywheel retards

8 8 Turning Moment Diagram - 4 Stroke Cycle ICE

9 9 Turning Moment Diagram for a 4 Stroke Cycle ICE Pressure inside engine cylinder is less than atmospheric pressure during suction stroke, therefore a negative loop is formed. During compression stroke, work is done on gases, therefore a higher negative loop is obtained. During expansion or working stroke, fuel burns & gases expand, therefore a large positive loop is obtained. In this stroke, work is done by gases. During exhaust stroke, work is done on gases, therefore a negative loop is formed.

10 10 Fluctuation of Energy

11 11 Fluctuation of Energy Fig. 16.1: TMD for a single cylinder double acting steam engine When crank moves from a to p, work done by engine = area abp energy required = area aabp Engine has done less work (= area aab) than required This amount of energy is taken from flywheel and hence speed of flywheel decreases

12 12 Fluctuation of Energy, Fig From p to q, work done by engine = area pbbcq requirement of energy = area pbcq Engine has done more work than requirement Excess work (area BbC) is stored in flywheel and hence speed of flywheel increases while crank moves from p to q

13 13 Fluctuation of Energy, Fig When crank moves from q to r, more work is taken from engine than is developed Loss of work = area CcD To supply this loss, flywheel gives up some of its energy and thus speed decreases while crank moves from q to r.

14 14 Fluctuation of Energy, Fig As crank moves from r to s, excess energy is again developed given by area DdE and speed again increases. As piston moves from s to e, again there is a loss of work and speed decreases.

15 15 Fluctuation of Energy, Fig Fluctuations of energy: variations of energy above & below mean resisting torque line Areas BbC, CcD, DdE, etc. represent fluctuations of energy. Engine has max speed either at q or at s. Flywheel absorbs energy while crank moves from p to q & from r to s.

16 16 Fluctuation of Energy, Fig Engine has min speed either at p or at r. Flywheel gives out some of its energy when crank moves from a to p and q to r. Max fluctuation of energy: difference between max & min energies

17 17 Determination of Max Fluctuation of Energy

18 18 Determination of Max Fluctuation of Energy Fig. 16.4: TMD for a multi-cylinder engine Line AG: mean torque line a 1, a 3, a 5 = areas above mean torque line a 2, a 4, a 6 = areas below mean torque line

19 19 Determination of Max Fluctuation of Energy Let energy in flywheel at A = E Energy at B = E + a 1 Energy at C = E + a 1 a 2 Energy at D = E + a 1 a 2 + a 3 Energy at E = E + a 1 a 2 + a 3 a 4 Energy at F = E + a 1 a 2 + a 3 a 4 + a 5 Energy at G = E + a 1 a 2 + a 3 a 4 + a 5 a 6 = Energy at A (cycle repeats)

20 20 Determination of Max Fluctuation of Energy Suppose that greatest of these energies is at B and least at E Max energy in flywheel = E + a 1 Min energy in flywheel = E + a 1 a 2 + a 3 a 4 Max fluctuation of energy, E = Max energy Min energy = (E + a 1 ) (E + a 1 a 2 + a 3 a 4 ) = a 2 a 3 + a 4

21 21 Coefficient of Fluctuation of Energy q = Angle turned in one rev q = 2p in case of steam engine and two stroke ICE q = 4p in case of four stroke ICE

22 22 Coefficient of Fluctuation of Energy n = Number of working strokes per minute n = N, in case of steam engines & two stroke ICE n = N /2, in case of four stroke ICE

23 23 Coefficient of Fluctuation of Energy Table 16.1: C E for steam and ICEs Type of engine Single cylinder, double acting steam engine 0.21 Cross-compound steam engine Single cylinder, single acting, four stroke gas engine 1.93 Four cylinders, single acting, four stroke gas engine Six cylinders, single acting, four stroke gas engine C E

24 24 Flywheel A flywheel serves as a reservoir It stores energy during period when supply of energy is more than requirement It releases it during period when requirement of energy is more than supply

25 25 Flywheel In case of steam engines, ICEs, reciprocating compressors and pumps, energy is developed during one stroke and engine is to run for the whole cycle on energy produced during this one stroke.

26 26 Flywheel In four stroke ICE, energy is developed only during expansion stroke which is much more than engine load and no energy is being developed during suction, compression and exhaust strokes.

27 27 Flywheel Excess energy developed during power stroke is absorbed by flywheel and releases it to crankshaft during other strokes in which no energy is developed, thus rotating crankshaft at a uniform speed.

28 28 Flywheel When flywheel absorbs energy, its speed increases and when it releases energy, speed decreases. A flywheel does not maintain a constant speed It reduces fluctuation of speed

29 29 Flywheel A flywheel controls speed variations caused by fluctuation of engine turning moment during each cycle of operation.

30 30 Flywheel In machines where operation is intermittent like crushers, flywheel stores energy from power source during the greater portion of the operating cycle and gives it up during a small period of the cycle. Energy from power source to machines is supplied practically at a constant rate throughout the operation.

31 31 Governor Governor in engine regulates mean speed of engine when there are variations in load, e.g., when load on engine increases, it becomes necessary to increase supply of working fluid. When load decreases, less working fluid is required. Governor automatically controls supply of working fluid to engine with varying load condition and keeps mean speed of engine within certain limits.

32 32 Flywheel Flywheel does not maintain a constant speed, it simply reduces fluctuation of speed. It does not control speed variations caused by varying load.

33 33 Coefficient of Fluctuation of Speed Max fluctuation of speed: difference between max & min speeds during a cycle Coefficient of fluctuation of speed: ratio of max fluctuation of speed to mean speed N 1 & N 2 = Max & min speeds in rpm N = (N 1 + N 2 )/2

34 34 Coefficient of Fluctuation of Speed

35 35 Coefficient of Fluctuation of Speed C s is a limiting factor in design of flywheel It varies depending upon nature of service to which flywheel is employed. Coefficient of steadiness = reciprocal of C s

36 36 Energy Stored in a Flywheel When a flywheel absorbs energy, its speed increases and when it gives up energy, its speed decreases. m = Mass of flywheel k = Radius of gyration of flywheel I = Mass moment of inertia of flywheel about its axis of rotation = m.k 2

37 37 Energy Stored in a Flywheel

38 38 Energy Stored in a Flywheel Mean kinetic energy of flywheel Maximum fluctuation of energy

39 39 Energy Stored in a Flywheel Radius of gyration (k) may be taken equal to mean radius of rim (R),

40 40 Example 16.1 The mass of flywheel of an engine is 6.5 tons and radius of gyration is 1.8 m. It is found from turning moment diagram that fluctuation of energy is 56 kn.m. If mean speed of engine is 120 rpm, find max & min speeds.

41 41 Example 16.2 Flywheel of a steam engine has a radius of gyration of 1 m and mass 2500 kg. The starting torque of steam engine is 1500 N.m and may be assumed constant. Determine: 1. Angular acceleration of flywheel 2. Kinetic energy of flywheel after 10 seconds from start

42 42 Example 16.3 A horizontal cross compound steam engine develops 300 kw at 90 rpm. The coefficient of fluctuation of energy as found from turning moment diagram is to be 0.1 and fluctuation of speed is to be kept within ± 0.5% of mean speed. Find weight of flywheel required, if radius of gyration is 2 m.

43 43 Example 16.4 Turning moment diagram for a petrol engine is drawn to following scales: Turning moment, 1 mm = 5 N.m; crank angle, 1 mm = 1. Turning moment diagram repeats itself at every half revolution of engine and areas above and below mean turning moment line taken in order are 295, 685, 40, 340, 960, 270 mm 2. Rotating parts are equivalent to a mass of 36 kg at a radius of gyration of 150 mm. Determine coefficient of fluctuation of speed when engine runs at 1800 rpm.

44 44 Example 16.4

45 45 Example 16.5 TMD for a mult-icylinder engine has been drawn to a scale 1 mm = 600 N.m vertically and 1 mm = 3 horizontally. The intercepted areas between output torque curve and mean resistance line, taken in order from one end, are as follows : + 52, 124, + 92, 140, + 85, 72 and mm 2, when engine is running at a speed of 600 rpm. If total fluctuation of speed is not to exceed ± 1.5% of the mean, find the necessary mass of flywheel of radius 0.5 m.

46 46 Example 16.5

47 47 Example 16.6 A shaft fitted with a flywheel rotates at 250 rpm and drives a machine. The torque of machine varies in a cyclic manner over a period of 3 revolutions as shown. Determine the power required to drive the machine and percentage fluctuation in speed, if driving torque applied to shaft is constant and the mass of flywheel is 500 kg with radius of gyration of 600 mm.

48 Example

49 49 Example 16.7 During forward stroke of the piston of double acting steam engine, turning moment has max value of 2000 N.m when crank makes an angle of 80 with IDC. During backward stroke, max turning moment is 1500 N.m when crank makes an angle of 80 with ODC. Turning moment diagram for the engine is shown. If crank makes 100 rpm and radius of gyration of flywheel is 1.75 m while keeping the speed within ± 0.75% of mean speed, find 1. the coefficient of fluctuation of energy 2. mass of flywheel 3. crank angle at which speed has its min & max values

50 50 Example 16.7

51 51 Example 16.8 A three cylinder single acting engine has its cranks set equally at 120 and it runs at 600 rpm. Torque-crank angle diagram for each cycle is a triangle for the power stroke with max torque of 90 N.m at 60 from IDC of corresponding crank. Torque on return stroke is sensibly zero. Determine: 1. power developed 2. coefficient of fluctuation of speed, if mass of flywheel is 12 kg and has a radius of gyration of 80 mm 3. coefficient of fluctuation of energy 4. maximum angular acceleration of flywheel

52 52 Example 16.8

53 53 Example 16.8

54 54 Example 16.9 A single cylinder, single acting, four stroke gas engine develops 20 kw at 300 rpm. Work done by gases during expansion stroke is three times work done on gases during compression stroke, work done during suction and exhaust strokes being negligible. If total fluctuation of speed is not to exceed ± 2 % of mean speed and TMD during compression and expansion is assumed to be triangular in shape, find moment of inertia of flywheel.

55 Example

56 56 Example TMD for a 4-stroke gas engine may be assumed to be represented by four triangles, areas of which from line of zero pressure are as follows: Suction stroke = m 2 ; Compression stroke = m 2 ; Expansion stroke = m 2 ; Exhaust stroke = m 2. Each m 2 of area represents 3 MN.m of energy. Assuming resisting torque to be uniform, find mass of rim of a flywheel required to keep speed between 202 & 198 rpm. Mean radius of rim is 1.2 m.

57 Example

58 58 Example Turning moment curve for an engine is represented by the equation, T = ( sin 2q 5700 cos 2q ) N.m, where q is angle moved by crank from IDC. If resisting torque is constant, find: 1. Power developed by engine 2. Moment of inertia of flywheel in kg-m 2, if total fluctuation of speed is not exceed 1% of mean speed which is 180 rpm 3. Angular acceleration of flywheel when crank has turned through 45 from IDC

59 59 Example 16.11

60 60 Example A certain machine requires a torque of ( sinq ) N.m to drive it, where q is angle of rotation of shaft measured from certain datum. The machine is directly coupled to an engine which produces a torque of ( sin 2q) N.m. The flywheel and other rotating parts attached to engine has a mass of 500 kg at a radius of gyration of 0.4 m. If mean speed is 150 rpm, find: 1. fluctuation of energy 2. total percentage fluctuation of speed 3. maximum and minimum angular acceleration of the flywheel and the corresponding shaft position.

61 61 Example 16.12

62 62 Example The equation of turning moment curve of a three crank engine is ( sin 3q ) N.m, where q is crank angle in radians. The moment of inertia of flywheel is 1000 kg.m 2 and mean speed is 300 rpm. Calculate: 1. power of the engine 2. max fluctuation of speed of flywheel in percentage when resisting torque is constant resisting torque is ( sinq ) N.m.

63 63 Example 16.13

64 64 Dimensions of Flywheel Rim D = Mean diameter of rim R = Mean radius of rim A = x-sectional area of rim r = Density of rim material N = Speed of flywheel, rpm w = Speed of flywheel, rad/s v = Linear velocity at mean radius = w R s = hoop stress due to centrifugal force

65 65 Dimensions of Flywheel Rim Volume of small element = A R.dq Total vertical upward force tending to burst the rim across diameter X Y

66 66 Dimensions of Flywheel Rim

67 67 Example TMD for a multi-cylinder engine has been drawn to a scale of 1 mm to 500 N.m torque and 1 mm to 6 of crank displacement. The engine is running at 800 rpm. The engine has a stroke of 300 mm and fluctuation of speed is not to exceed ± 2% of mean speed. Determine a suitable diameter and cross-section of flywheel rim for a limiting value of the safe centrifugal stress of 7 MPa. The material density may be assumed as 7200 kg/m 3. The width of rim is to be 5 times thickness.

68 68 Example 16.14

69 69 Example A single cylinder double acting steam engine develops 150 kw at a mean speed of 80 rpm. The coefficient of fluctuation of energy is 0.1 and fluctuation of speed is ± 2% of mean speed. If mean diameter of flywheel rim is 2 m and the hub and spokes provide 5% of the rotational inertia of flywheel, find mass and cross-sectional area of flywheel rim. Assume density of flywheel material (cast iron) as 7200 kg/m 3.

70 70 Example A multi-cylinder engine is to run at a speed of 600 rpm. On drawing the turning moment diagram to a scale of 1 mm = 250 N.m and 1 mm = 3. The speed is to be kept within ± 1% of mean speed of engine. Calculate necessary moment of inertia of flywheel. Determine suitable dimensions of a rectangular flywheel rim if breadth is twice its thickness. The density of cast iron is 7250 kg/m 3 and its hoop stress is 6 MPa. Assume that rim contributes 92% of flywheel effect.

71 71 Example 16.16

72 72 Example TMD of a four stroke engine may be assumed to be represented by four triangles in each stroke. Areas of these triangles are as follows: Suction stroke = m 2 ; Compression stroke = m 2 ; Expansion stroke = m 2 ; Exhaust stroke = m 2. All areas excepting expansion stroke are negative. Each m 2 of area represents 14 MN.m of work. Assuming resisting torque to be constant, determine moment of inertia of flywheel to keep the speed between 98 & 102 rpm. Also find size of a rim-type flywheel based on min material criterion, given that density of flywheel material is 8150 kg/m 3 ; allowable tensile stress of flywheel material is 7.5 MPa. The rim cross-section is rectangular, one side being four times the length of the other.

73 Example

74 74 Example An otto cycle engine develops 50 kw at 150 rpm with 75 explosions per minute. The change of speed from commencement to end of power stroke must not exceed 0.5% of mean on either side. Find mean diameter of flywheel and a suitable rim cross section having width four times depth so that hoop stress does not exceed 4 MPa. Assume that flywheel stores 16/15 times the energy stored by rim and work done during power stroke is 1.40 times work done during the cycle. Density of rim material is 7200 kg/m 3.

75 Example

Flywheel. 776 A Textbook of Machine Design

Flywheel. 776 A Textbook of Machine Design 776 A Textbook of Machine Design C H A P T E R Flywheel 1. Introduction.. Coefficient of Fluctuation of Speed. 3. Fluctuation of Energy. 4. Maximum Fluctuation of Energy. 5. Coefficient of Fluctuation