Chapter 15 Inertia Forces in Reciprocating Parts
2 Approximate Analytical Method for Velocity & Acceleration of the Piston n = Ratio of length of ConRod to radius of crank = l/r
3 Approximate Analytical Method for Velocity & Acceleration of the Piston Velocity of piston
4 Approximate Analytical Method for Velocity & Acceleration of the Piston Velocity of piston
5 Approximate Analytical Method for Velocity and Acceleration of the Piston Velocity of the piston
6 Approximate Analytical Method for Velocity & Acceleration of the Piston Acceleration of piston
7 Angular Velocity & Acceleration of ConRod
8 Angular Velocity & Acceleration of ConRod
9 Angular Velocity and Acceleration of the Connecting Rod
10 Angular Velocity & Acceleration of ConRod
11 Angular Velocity & Acceleration of ConRod Since sin 2 q is small as compared to n 2, and 1.0 is small compared to n 2 therefore:
12 Example 15.3 If the crank and ConRod are 300 mm & 1000 mm long respectively and the crank rotates at a constant speed of 200 rpm, determine: 1. Crank angle at which max velocity occurs 2. Max velocity of piston
13 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 1. velocity & acceleration of piston when the crank is at 40 from IDC 2. position of the crank for zero acceleration of the piston
14 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 slider 2. Angular velocity & angular acceleration of ConRod
15 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod
16 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod Piston effort, F P : net force acting on piston or crosshead pin, along line of stroke m R = Mass of reciprocating parts, piston, crosshead pin W R = m R.g
17 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod
18 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 stroke when piston moves from ODC to IDC
19 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod Inertia force due to acceleration of reciprocating parts, opposes force on 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 force on piston.
20 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
21 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod In a double acting reciprocating steam engine, net load on 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
22 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 downward stroke (piston moves from IDC to ODC) and opposes during upward stroke of piston (piston moves from ODC to IDC). Piston effort,
23 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod Force acting along ConRod
24 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod Crank-pin effort, F T : component of F Q perpendicular to crank
25 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
26 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod Crank effort or turning moment on crank shaft
27 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod Crank effort or turning moment on crank shaft
28 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod Crank effort or turning moment on crank shaft
29 Forces on Reciprocating Parts of an Engine, Neglecting Weight of ConRod Crank effort or turning moment on crank shaft
30 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
31 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 crank-pin 4. Turning moment on crank shaft
32 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.
33 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
34 Example 15.10 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
35 Example 15.11 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
Example 15.12 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 36
37 Compound Pendulum compound pendulum: a vertically suspended RB oscillating with a small amplitude under action of gravity m = Mass of pendulum W = m g k G = Radius of gyration about G
38 Compound Pendulum If pendulum is given a small angular displacement q, then couple tending to restore pendulum to equilibrium position OA,
39 Compound Pendulum Angular acceleration of pendulum
40 Compound Pendulum Compare this equation with equation of simple pendulum Equivalent length of a simple pendulum, which gives same frequency as compound pendulum, is
41 Equivalent Dynamical System To determine motion of a rigid body (RB), it is convenient to replace RB by two masses placed at a fixed distance apart such that: 1. Sum of their masses = total mass of RB 2. Center of gravity (CG) of two masses coincides with that of RB 3. Sum of mass moment of inertia of masses about their CG = mass moment of inertia of body
42 Equivalent Dynamical System k G = Radius of gyration about G
43 Equivalent Dynamical System
44 Equivalent Dynamical System When k G is not known, then position of 2 nd mass may be obtained by considering body as a compound pendulum Length of simple pendulum which gives same frequency as RB (compound pendulum) is
45 Equivalent Dynamical System 1 st mass is situated at center of oscillation of body
46 Graphical Determination of Equivalent Dynamical System of Two Masses
47 Example 15.15 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 dynamical equivalent two mass system of ConRod if one of masses is located at small end center.
48 Example 15.16 A ConRod is suspended from a point 25 mm above the center of small end, and 650 mm above its CG, 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 small end center.
Example 15.17 The following data relate to a ConRod 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 ConRod is suspended from small end = 1.83 s; Time of oscillation when ConRod is suspended from big end = 1.68 s. 49
Example 15.17 50 Determine: 1. Radius of gyration of rod about an axis passing through CG & perpendicular to plane of oscillation 2. Moment of inertia of rod about same axis 3. Dynamically equivalent system for ConRod, constituted of two masses, one of which is situated at small end center.
51 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: Moment of inertia condition is not possible to satisfy
52 Correction Couple to be Applied to Make Two Mass System Dynamically Equivalent Consider two masses, one at A & the other at D be placed arbitrarily
53 Correction Couple to be Applied to Make Two Mass System Dynamically Equivalent I 1 = New mass moment of inertia of two masses k 1 = New radius of gyration k G = Radius of gyration of a dynamically equivalent system
54 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,
55 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.
56 Correction Couple to be Applied to Make Two Mass System Dynamically Equivalent
Example 15.18 A ConRod has a mass of 2 kg and the distance between 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 CG perpendicular to plane of rotation is 110 mm. 57
Example 15.18 58 1. Find the equivalent dynamical system if only one of masses is located at gudgeon pin. 2. If ConRod is replaced by two masses, one at gudgeon pin and the other at crank pin and the angular acceleration of rod is 23000 rad/s 2 cw, determine the correction couple applied to the system to reduce it to a dynamically equivalent system.
59 Analytical Method for Inertia Torque Mass of ConRod (m C ) is divided into two masses: One at crosshead pin P & the other at crankpin C CG of the two masses coincides with CG of rod G
60 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
61 Analytical Method for Inertia Torque Mass of ConRod at P Mass of reciprocating parts (m R ) is acting at P Total equivalent mass of reciprocating parts acting at P
62 Analytical Method for Inertia Torque Total inertia force of equivalent mass acting at P,
63 Analytical Method for Inertia Torque Corresponding torque exerted on crank shaft,
64 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.
65 Analytical Method for Inertia Torque To compensate for it, a correcting torque is necessary whose value is given by
66 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
67 Analytical Method for Inertia Torque
68 Analytical Method for Inertia Torque Equivalent mass of rod acting at C, Torque exerted on crank shaft due to mass m 2,
69 Analytical Method for Inertia Torque Total torque exerted on crankshaft due to the inertia of the moving parts = T I + T C + T W
70 Example 15.20 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 CG of ConRod from crank pin = 500 mm radius of gyration of ConRod about an axis through CG = 650 mm Determine magnitude & direction of torque exerted on crankshaft when crank has turned through 30 from IDC.
71 Example 15.21 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.
Example 15.21 72 For the position shown When the crank is at 40 from TDC & the piston is moving downwards, 1. Calculate radius of gyration of ConRod about an axis through its mass center. 2. find acceleration of piston and angular acceleration of ConRod 3. Find the inertia torque exerted on the crankshaft.