Chapter 15. Inertia Forces in Reciprocating Parts

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1 Chapter 15 Inertia Forces in Reciprocating Parts

2 2 Approximate Analytical Method for Velocity and Acceleration of the Piston n = Ratio of length of ConRod to radius of crank = l/r

3 3 Approximate Analytical Method for Velocity and Acceleration of the Piston Velocity of the piston

4 4 Approximate Analytical Method for Velocity and Acceleration of the Piston Velocity of the piston

5 5 Approximate Analytical Method for Velocity and Acceleration of the Piston Velocity of the piston

6 6 Approximate Analytical Method for Velocity and Acceleration of the Piston Acceleration of the piston

7 7 Approximate Analytical Method for Velocity and Acceleration of the Piston Acceleration of the piston At the inner dead center (IDC), q = 0

8 8 Approximate Analytical Method for Velocity and Acceleration of the Piston Acceleration of the piston At outer dead center (ODC), q = 180 Direction of motion is reversed at ODC therefore changing sign of above expression,

9 9 Angular Velocity and Acceleration of the Connecting Rod

10 10 Angular Velocity and Acceleration of the Connecting Rod

11 11 Angular Velocity and Acceleration of the Connecting Rod

12 12 Angular Velocity and Acceleration of the Connecting Rod

13 13 Angular Velocity and Acceleration of the Connecting Rod Since sin 2 q is small as compared to n 2, and 1.0 is small compared to n 2 therefore:

14 14 Example 15.3 If the crank and the connecting rod are 300 mm and 1000 mm long respectively and the crank rotates at a constant speed of 200 rpm, determine: 1. The crank angle at which the maximum velocity occurs 2. Maximum velocity of the piston

15 15 Example 15.4 The crank & ConRod of a steam engine are 0.3 m and 1.5 m in length. The crank rotates at 180 rpm cw. Determine the velocity & acceleration of the piston when the crank is at 40 from the IDC center position. Also determine the position of the crank for zero acceleration of the piston.

16 16 Example 15.5 In a slider crank mechanism, the length of the crank & ConRod are 150 mm & 600 mm respectively. The crank position is 60 from IDC. The crank shaft speed is 450 rpm cw. Using analytical method, determine: 1. Velocity & acceleration of the slider 2. Angular velocity & angular acceleration of the ConRod

17 17 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod

18 18 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod Piston effort, F P : net force acting on the piston or crosshead pin, along the line of stroke m R = Mass of reciprocating parts, piston, crosshead pin or gudgeon pin W R = m R.g

19 19 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod

20 20 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod In a horizontal engine, reciprocating parts are accelerated from rest when piston moves from IDC to ODC It is, then, retarded during the latter half of the stroke when piston moves from ODC to IDC

21 21 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod Inertia force due to acceleration of reciprocating parts, opposes the force on the piston due to difference of pressures in the cylinder on the two sides of the piston Inertia force due to retardation of reciprocating parts, helps the force on the piston.

22 22 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod ve sign is used when piston is accelerated +ve sign is used when piston is retarded

23 23 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod In a double acting reciprocating steam engine, net load on the piston, F L = p 1 A 1 p 2 A 2 = p 1 A 1 p 2 (A 1 a) p 1, A 1 = Pressure & cross-sectional area on back end side of piston p 2, A 2 = Pressure & cross-sectional area on crank end side of piston a = Cross-sectional area of piston rod

24 24 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod In case of a vertical engine, weight of reciprocating parts assists piston effort during the downward stroke (piston moves from IDC to ODC) and opposes during upward stroke of piston (piston moves from ODC to IDC). Piston effort,

25 25 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod Force acting along the ConRod

26 26 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod Crank-pin effort, F T : component of F Q perpendicular to crank

27 27 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod Thrust on crank shaft bearings, F B : component of F Q along crank produces a thrust on crank shaft bearings

28 28 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod Crank effort or turning moment on crank shaft

29 29 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod Crank effort or turning moment on crank shaft

30 30 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod Crank effort or turning moment on crank shaft

31 31 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod Crank effort or turning moment on crank shaft

32 32 Example 15.6 Find the inertia force for the following data of an IC engine: Bore = 175 mm Stroke = 200 mm Engine speed = 500 rpm Length of ConRod = 400 mm Crank angle = 60 from IDC Mass of reciprocating parts = 180 kg

33 33 Example 15.7 Crank-pin circle radius of a horizontal engine is 300 mm. The mass of reciprocating parts is 250 kg. When the crank has travelled 60 from IDC, the difference between driving & back pressures is 0.35 N/mm 2. The ConRod length between centers is 1.2 m & the cylinder bore is 0.5 m. If engine runs at 250 rpm & if the effect of piston rod diameter is neglected, calculate : 1. Piston effort 2. Thrust in ConRod 3. Tangential force on the crank-pin 4. Turning moment on the crank shaft

34 34 Example 15.8 A vertical double acting steam engine has a cylinder 300 mm diameter & 450 mm stroke and runs at 200 rpm. The reciprocating parts has a mass of 225 kg & piston rod is 50 mm diameter. The ConRod is 1.2 m long. When the crank has turned through 125 from IDC, steam pressure above piston is 30 kn/m 2 & below piston is 1.5 kn/m 2. Calculate effective turning moment on the crank shaft.

35 35 Example 15.9 Crank & ConRod of a petrol engine, running at 1800 rpm are 50 mm & 200 mm respectively. The diameter of piston is 80 mm & mass of reciprocating parts is 1 kg. At a point during the power stroke, pressure on piston is 0.7 N/mm 2, when it has moved 10 mm from IDC. Determine : 1. Net load on gudgeon pin 2. Thrust in ConRod 3. Reaction between piston & cylinder 4. Engine speed at which the above values become zero

36 36 Example During a trial on steam engine, it is found that the acceleration of piston is 36 m/s 2 when the crank has moved 30 from IDC. The net effective steam pressure on piston is 0.5 N/mm 2 and frictional resistance is equivalent to a force of 600 N. The diameter of piston is 300 mm & mass of reciprocating parts is 180 kg. If the length of crank is 300 mm & ratio of ConRod length to crank length is 4.5, find: 1. Reaction on guide bars 2. Thrust on crank shaft bearings 3. Turning moment on crank shaft

37 37 Example A vertical petrol engine 100 mm diameter & 120 mm stroke has a ConRod 250 mm long. The mass of piston is 1.1 kg. The speed is 2000 rpm. On the expansion stroke with a crank 20 from IDC, the gas pressure is 700 kn/m 2. Determine: 1. Net force on piston 2. Resultant load on gudgeon pin 3. Thrust on cylinder walls 4. Speed above which, other things remaining same, the gudgeon pin load would be reversed in direction

38 Example A horizontal steam engine running at 120 rpm has a bore of 250 mm & a stroke of 400 mm. The ConRod is 0.6 m & mass of reciprocating parts is 60 kg. When the crank has turned through an angle of 45 from IDC, the steam pressure on cover end side is 550 kn/m 2 & that on crank end side is 70 kn/m 2. Considering the diameter of piston rod equal to 50 mm, determine: 1. Turning moment on crank shaft 2. Thrust on bearings 3. Acceleration of flywheel, if the power of engine is 20 kw, mass of flywheel 60 kg & radius of gyration 0.6 m 38

39 4.8. Compound Pendulum When a RB is suspended vertically, and it oscillates with a small amplitude under the action of force of gravity, body is known as compound pendulum m = Mass of pendulum W = m g k G = Radius of gyration about G 39

40 4.8. Compound Pendulum If pendulum is given a small angular displacement q, then couple tending to restore pendulum to equilibrium position OA, 40

41 Compound Pendulum Angular acceleration of pendulum

42 Compound Pendulum Compare this equation with equation of simple pendulum Equivalent length of a simple pendulum, which gives the same frequency as compound pendulum, is

43 43 Equivalent Dynamical System To determine motion of a rigid body (RB), it is usually convenient to replace RB by two masses placed at a fixed distance apart, in such a way that: 1. Sum of their masses = total mass of RB 2. Center of gravity (CG) of the two masses coincides with that of the RB 3. Sum of mass moment of inertia of the masses about their CG = mass moment of inertia of the body

44 44 Equivalent Dynamical System k G = Radius of gyration about G

45 45 Equivalent Dynamical System

46 46 Equivalent Dynamical System When k G is not known, then position of 2 nd mass may be obtained by considering the body as a compound pendulum Length of simple pendulum which gives same frequency as RB (compound pendulum) is

47 47 Equivalent Dynamical System 1 st mass is situated at center of oscillation of body

48 48 Determination of Equivalent Dynamical System of Two Masses by Graphical Method

49 49 Example The ConRod of a gasoline engine is 300 mm long between its centers. It has a mass of 15 kg & mass moment of inertia of 7000 kg.mm 2. Its CG is at 200 mm from its small end center. Determine the dynamical equivalent twomass system of the ConRod if one of the masses is located at the small end center.

50 50 Example A ConRod is suspended from a point 25 mm above the center of small end, and 650 mm above its center of gravity, its mass being 37.5 kg. When permitted to oscillate, time period is found to be 1.87 seconds. Find dynamical equivalent system constituted of two masses, one of which is located at the small end center.

51 Example The following data relate to a connecting rod of a reciprocating engine: Mass = 55 kg; Distance between bearing centers = 850 mm; Diameter of small end bearing = 75 mm; Diameter of big end bearing = 100 mm; Time of oscillation when the connecting rod is suspended from small end = 1.83 s; Time of oscillation when the connecting rod is suspended from big end = 1.68 s. 51

52 Example Determine: 1. Radius of gyration of the rod about an axis passing through center of gravity and perpendicular to plane of oscillation 2. Moment of inertia of the rod about the same axis 3. Dynamically equivalent system for the connecting rod, constituted of two masses, one of which is situated at the small end center.

53 53 Correction Couple to be Applied to Make Two Mass System Dynamically Equivalent When two masses are placed arbitrarily, then the following conditions will only be satisfied: The Moment of inertia condition is not possible to satisfy.

54 54 Correction Couple to be Applied to Make Two Mass System Dynamically Equivalent Consider two masses, one at A and the other at D be placed arbitrarily

55 55 Correction Couple to be Applied to Make Two Mass System Dynamically Equivalent I 1 = New mass moment of inertia of the two masses k 1 = New radius of gyration k G = Radius of gyration of a dynamically equivalent system

56 56 Correction Couple to be Applied to Make Two Mass System Dynamically Equivalent Torque required to accelerate the body, Torque required to accelerate the twomass system placed arbitrarily,

57 57 Correction Couple to be Applied to Make Two Mass System Dynamically Equivalent Correction couple: difference of torques T This couple must be applied, when the masses are placed arbitrarily to make the system dynamical equivalent.

58 58 Correction Couple to be Applied to Make Two Mass System Dynamically Equivalent

59 Example A ConRod has a mass of 2 kg and the distance between the center of gudgeon pin & center of crank pin is 250 mm. The CG falls at a point 100 mm from the gudgeon pin along the line of centers. The radius of gyration about an axis through the CG perpendicular to the plane of rotation is 110 mm. 59

60 Example Find the equivalent dynamical system if only one of the masses is located at gudgeon pin. 2. If the ConRod is replaced by two masses, one at the gudgeon pin and the other at the crank pin and the angular acceleration of the rod is rad/s 2 cw, determine the correction couple applied to the system to reduce it to a dynamically equivalent system.

61 61 Analytical Method for Inertia Torque Mass of ConRod (m C ) is divided into two masses. One at crosshead pin P and the other at crankpin C CG of the two masses coincides with CG of rod G.

62 62 Analytical Method for Inertia Torque Inertia force due to mass at C acts radially outwards along crank OC mass at C has no effect on crankshaft torque

63 63 Analytical Method for Inertia Torque Mass of ConRod at P Mass of reciprocating parts (m R ) is also acting at P Total equivalent mass of reciprocating parts acting at P

64 64 Analytical Method for Inertia Torque Total inertia force of equivalent mass acting at P,

65 65 Analytical Method for Inertia Torque Corresponding torque exerted on crank shaft,

66 66 Analytical Method for Inertia Torque In deriving the above equation of the torque exerted on the crankshaft, it is assumed that one of the two masses is placed at C and the other at P. This assumption does not satisfy the condition for kinetically equivalent system of a rigid bar.

67 67 Analytical Method for Inertia Torque To compensate for it, a correcting torque is necessary whose value is given by

68 68 Analytical Method for Inertia Torque Correcting torque T' may be applied to the system by two equal & opposite forces F Y acting through P & C

69 69 Analytical Method for Inertia Torque

70 70 Analytical Method for Inertia Torque The equivalent mass of rod acting at C, Torque exerted on crank shaft due to mass m 2,

71 71 Analytical Method for Inertia Torque Total torque exerted on the crankshaft due to the inertia of the moving parts = T I + T C + T W

72 72 Example The following data refer to a steam engine: Diameter of piston = 240 mm Stroke = 600 mm length of ConRod = 1.5 m mass of reciprocating parts = 300 kg mass of ConRod = 250 kg speed = 125 rpm center of gravity of ConRod from crank pin = 500 mm radius of gyration of ConRod about an axis through the center of gravity = 650 mm Determine magnitude and direction of torque exerted on the crankshaft when the crank has turned through 30 from IDC.

73 73 Example A vertical engine running at 1200 rpm with a stroke of 110 mm, has a ConRod 250 mm between centers and mass 1.25 kg. The mass center of the ConRod is 75 mm from the big end center and when suspended as a pendulum from the gudgeon pin axis makes 21 complete oscillations in 20 seconds.

74 Example For the position shown When the crank is at 40 from TDC & the piston is moving downwards, 1. Calculate the radius of gyration of ConRod about an axis through its mass center. 2. find the acceleration of the piston and the angular acceleration of the ConRod 3. Find the inertia torque exerted on the crankshaft.

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