VTU EDUSAT PROGRAMME-17. DYNAMICS OF MACHINES Subject Code -10 ME 54 BALANCING OF RECIPROCATING MASSES ADARSHA H G

Transcription

1 Subject Code -1 ME 5 BALANCING OF RECIPROCATING MASSES Notes Compiled by: ASSOCIATE PROFESSOR DEPARTMENT OF MECHANICAL ENGINEERING MALNAD COLLEGE BALANCING OF ENGINEERING HASSAN -57 OF. KARNATAKA RECIPROCATING Mobile: MASSES vvb@mcehassa.ac.i SLIDER CRANK MECHANISM: PRIMARY AND SECONDARY ACCELERATING FORCE: Acceleratio of the reciprocatig mass of a slider-crak mechaism is give by, a p Acceleratio of pisto cosθ r ω cosθ+ ( ) 1 l Where r Ad, the force required to accelerate the mass m is F i mr ω mr ω cosθ cosθ+ cosθ cosθ + mrω ( ) 1

2 The first term of the equatio (), i.e. force the secod term mrω cosθ Maximum value of primary acceleratig force is mrω cosθ is called primary acceleratig is called the secodary acceleratig force. m rω mrω Ad Maximum value of secodary acceleratig force is Geerally, value is much greater tha oe; the secodary force is small compared to primary force ad ca be safely eglected for slow speed egies. I Fig (a), the iertia force due to primary acceleratig force is show.

3 I Fig (b), the forces actig o the egie frame due to iertia force are show. At O the force exerted by the crakshaft o the mai bearigs has two compoets, h v horizotal F 1 ad vertical F 1. h F 1 is a horizotal force, which is a ubalaced shakig force. v F 1 ad v F 1 balace each other but form a ubalaced shakig couple. The magitude ad directio of these ubalaced force ad couple go o chagig with agle θ. The shakig force produces liear vibratios of the frame i horizotal directio, whereas the shakig couple produces a oscillatig vibratio. h The shakig force F 1 is the oly ubalaced force which may hamper the smooth ruig of the egie ad effort is made to balace the same. However it is ot at all possible to balace it completely ad oly some modificatios ca be carried out. BALANCING OF THE SHAKING FORCE: Shakig force is beig balaced by addig a rotatig couter mass at radius r directly opposite the crak. This provides oly a partial balace. This couter mass is i additio to the mass used to balace the rotatig ubalace due to the mass at the crak pi. This is show i figure (c).

4 The horizotal compoet of the cetrifugal force due to the balacig mass is mrω cosθ ad this is i the lie of stroke. This compoet eutralizes the ubalaced reciprocatig force. But the rotatig mass also has a compoet mrω siθ perpedicular to the lie of stroke which remais ubalaced. The ubalaced force is zero at θ or 18 ad maximum at the middle of the stroke i.e. θ 9. The magitude or the maximum value of the ubalaced force remais the same i.e. equal to m rω. Thus istead of slidig to ad fro o its moutig, the mechaism teds to jump up ad dow. To miimize the effect of the ubalace force a compromise is, usually made, is of the reciprocatig mass is balaced or a value betwee 1 to. If c is the fractio of the reciprocatig mass, the The ad primary force balaced by the mass cmrω The primary force ubalaced by the mass (1 c) mrω Vertical c mrω compoet siθ cosθ cos θ of cetrifuga l force which remais ubalaced I reciprocatig egies, ubalace forces i the directio of the lie of stroke are more dagerous tha the forces perpedicular to the lie of stroke. Resultat ubalaced force at ay istat [(1 c)mrω cosθ] + [ cmrω siθ] The resultat ubalaced force is miimum whe, 1 c This method is just equivalet to as if a revolvig mass at the crakpi is completely balaced by providig a couter mass at the same radius diametrically opposite to the crak. Thus if m P is the mass at the crakpi ad c is the fractio of the reciprocatig mass m to be balaced, the mass at the crakpi may be cosidered as c m+ m P which is to be completely balaced.

5 Problem 1: A sigle cylider reciprocatig egie has a reciprocatig mass of 6 kg. The crak rotates at 6 rpm ad the stroke is mm. The mass of the revolvig parts at 16 mm radius is kg. If two-thirds of the reciprocatig parts ad the whole of the revolvig parts are to be balaced, determie the, (i) balace mass required at a radius of 5 mm ad (ii) ubalaced force whe the crak has tured 5 from the top-dead cetre. Solutio: Give: m mass of the reciprocatig parts 6 kg N 6 rpm, L legth of thestroke mm m kg, c, rc P (i) Balace mass required at a radius of 5 mm We have, ad πn πx6 ω π rad/s 6 6 L r 16 mm 5 mm Mass to be balaced at thecrak pi M M c m + m x6 + 8 kg P m c r c i.e. m Mr c 8 x16 5 therefore m c 6.57 kg Mr r (ii) Ubalaced force whe the crak has tured 5 from the top-dead cetre. Ubalaced force at θ 5 [( 1 c) mrω cosθ] + [ cmrω siθ] 1 x6x.16x 9.9 N c ( π) cos5 + x6x.16x( π) si5 5

6 Problem : The followig data relate to a sigle cylider reciprocatig egie: Mass of reciprocatig parts kg Mass of revolvig parts kg at crak radius Speed 15 rpm, Stroke 5 mm. If 6 % of the reciprocatig parts ad all the revolvig parts are to be balaced, determie the, (i) balace mass required at a radius of mm ad (ii) ubalaced force whe the crak has tured 5 from the top-dead cetre. Solutio: Give : m mass m P kg, c 6%, of r c the N 15 rpm, mm reciprocatig (i) Balace mass required at a radius of 5 mm We have, ad πn π x15 ω 15.7 rad/s 6 6 L 5 r 175 mm M c m + mp.6x + 5 kg parts kg Mass to be balaced at the crak pi M m c r c i.e. m Mr c 5 x175 therefore m c 9.5 kg L legth of the stroke 5 mm Mr r (ii) Ubalaced force whe the crak has tured 5 from the top-dead cetre. Ubalaced force at θ 5 [( 1 c) mrω cosθ] + [ cmrω siθ] [( 1.6) xx.175x ( 15.7) cos5 ] +.6xx.175x( 15.7) 88.7 N c [ si5 ] 6

7 SECONDARY BALANCING: cosθ Secodary acceleratio force is equal to mrω ( 1) Its frequecy is twice that of the primary force ad the magitude magitude of the primary force. cosθ The secodary force is also equal to mr( ω ) ( ) 1 times the Cosider, two craks of a egie, oe actual oe ad the other imagiary with the followig specificatios. Actual Imagiary Agular velocity ω ω Legth of crak r r Mass at the crak pi m m Thus, whe the actual crak has tured through a agle would have tured a agle θ ω t θ ω t, the imagiary crak 7

8 mr( ω) Cetrifugal force iduced i the imagiary crak mr( ω) Compoet of this force alog the lie of stroke is cosθ Thus the effect of the secodary force is equivalet to a imagiary crak of legth rotatig at double the agular velocity, i.e. twice of the egie speed. The imagiary crak coicides with the actual at ier top-dead cetre. At other times, it makes a agle with the lie of stroke equal to twice that of the egie crak. The secodary couple about a referece plae is give by the multiplicatio of the secodary force with the distace l of the plae from the referece plae. COMPLETE BALANCING OF RECIPROCATING PARTS Coditios to be fulfilled: 1. Primary forces must balace i.e., primary force polygo is eclosed.. Primary couples must balace i.e., primary couple polygo is eclosed.. Secodary forces must balace i.e., secodary force polygo is eclosed.. Secodary couples must balace i.e., secodary couple polygo is eclosed. Usually, it is ot possible to satisfy all the above coditios fully for multi-cylider egie. Mostly some ubalaced force or couple would exist i the reciprocatig egies. BALANCING OF INLINE ENGINES: A i-lie egie is oe wherei all the cyliders are arraged i a sigle lie, oe behid the other. May of the passeger cars such as Maruti 8, Ze, Satro, Hoda-city, Hoda CR-V, Toyota corolla are the examples havig four cider i-lie egies. I a reciprocatig egie, the reciprocatig mass is trasferred to the crakpi; the axial compoet of the resultig cetrifugal force parallel to the axis of the cylider is the primary ubalaced force. Cosider a shaft cosistig of three equal craks asymmetrically spaced. The crakpis carry equivalet of three uequal reciprocatig masses, the r 8

9 Primary force mrω cos θ (1) Primary couple mrω l cos θ () Secodary Ad Secodary ( ω) force mr cos θ () couple GRAPHICAL SOLUTION: m r ω m r ( ω) l cos θ l cos θ () To solve the above equatios graphically, first draw the m r cos θ polygo ( ω is commo to all forces). The the axial compoet of the resultat forces ( F r cosθ ) multiplied by ω provides the primary ubalaced force o the system at that momet. This ubalaced force is zero whe θ 9 ad a maximum wheθ. 9

10 If the force polygo ecloses, the resultat as well as the axial compoet will always be zero ad the system will be i primary balace. The, F P h ad F P V To fid the secodary ubalace force, first fid the positios of the imagiary secodary ( ω ) craks. The trasfer the reciprocatig masses ad multiply the same by or to get the secodary force. I the same way primary ad secodary couple ( m r l ) polygo ca be draw for primary ad secodary couples. Case 1: IN-LINE TWO-CYLINDER ENGINE Two-cylider egie, craks are 18 apart ad have equal reciprocatig masses. ω 1

11 Takig a plae through the cetre lie as the referece plae, Primary Primary force couple m rω m rω [ cosθ + cos(18 + θ) ] Maximum values are mrω l at θ ad 18 Secodary force Maximum values are m rω l l cosθ + cos(18 + θ) m rω mrω whe m rω l cosθ [ cos θ + cos(6 + θ) ] cosθ θ or θ, 18, 9, 6,18 ad 5 ad 7 11

12 Secodary couple m rω l cosθ + l cos(6 + θ) ANALYTICAL METHOD OF FINDING PRIMARY FORCES AND COUPLES First the positios of the craks have to be take i terms of θ The maximum values of these forces ad couples vary istat to istat ad are equal to the values as give by the equivalet rotatig masses at the crak pi. If a particular positio of the crak shaft is cosidered, the above expressios may ot give the maximum values. For example, the maximum value of primary couple is mrω l ad this value is obtaied at crak positios ad 18. However, if the crak positios are assumed at 9 ad 7, the values obtaied will be zero. If ay particular positio of the crak shaft is cosidered, the both X ad Y compoets of the force ad couple ca be take to fid the maximum values. For example, if the crak positios cosidered as 1 ad, the primary couple ca be obtaied as X compoet Y compoet l m r ω cos 1 1 m r ω l mrω mrω Therefore, Primary couple l si1 l mrω l 1 mrω l l + cos l + si 18 ( ) ( + 1 ) + mrω l Case : IN-LINE FOUR-CYLINDER FOUR-STROKE ENGINE 1

13 This egie has tow outer as well as ier craks (throws) i lie. The ier throws are at 18 to the outer throws. Thus the agular positios for the craks are θ for the first, 18 + θ for the secod, 18 + θ for the third ad θ for the fourth. 1

14 FINDING PRIMARY FORCES, PRIMARY COUPLES, SECONDARY FORCES AND SECONDARY COUPLES: Choose a plae passig through the middle bearig about which the arragemet is symmetrical as the referece plae. Primary Primary Secodary force couple m rω force m rω [ cosθ + cos(18 + θ) + cos(18 + θ) + cosθ] m rω m rω θ l l cosθ + cos(18 + θ) l l + cos(18 θ) cosθ + +,9 cos θ + cos(6 + θ) + cos(6 + θ) + cos θ cosθ m rω Maximum value at θ,18 Secodary couple,6,18 ad ad 7 5 mrω Thus the egie is ot balaced i secodary forces. or l l cosθ + cos(6 + θ) l l + cos(6 θ) cosθ + + 1

15 Problem 1: A four-cylider oil egie is i complete primary balace. The arragemet of the reciprocatig masses i differet plaes is as show i figure. The stroke of each pisto is r mm. Determie the reciprocatig mass of the cylider ad the relative crak positio. Solutio: Give: m 8 kg, m?, m 1 L r crak legth r 59 kg, m 8 kg Plae Cet. Distace Mass (m) Radius (r) Force/ω Couple/ ω from Ref ( m r l ) kg m (m r ) plae kg m kg m m 1 8 r 8 r r (RP) m r m r 59 r 59 r r 8 r 8 r r 15

16 Aalytical Method: VTU EDUSAT PROGRAMME-17 Choose plae as the referece plae adθ.. Step 1: Resolve the couples ito their horizotal ad vertical compoets ad take their sums. Sum of the horizotal compoets gives i.e., 9 r cos θ + 9 cos θ rcos cos θ Sum of the vertical compoets gives 9 r si θ i.e., 9 si θ rsi 1968 si θ Squarig ad addig (1) ad (), we get rcos θ rsi θ ( 1) ( ) ( 9) ( cos θ ) + ( 1968 si θ ) i.e., ( 9) ( 165) + x165x1968cos θ + ( 1968 cos θ ) + ( 1968 si θ ) O cos θ solvig we get,.978 ad θ Choosig oe value, say θ Dividig () by (1), we get taθ i.e., 1 Step : θ 1 1. or si(167.9 ) cos (167.9 ) 7.8 Resolve the forces ito their horizotal ad vertical compoets ad take their sums. Sum of the horizotal compoets gives 8 r cos( 1. or m cos θ ) + m r cos θ + 59 rcos ( ) + 8rcos( ) 16

17 Sum of the vertical compoets gives 8 r si( 1. or m si θ ) + m r si θ + 59 rsi ( ) Squarig ad addig () ad (), we get m 7. 1 kg Dividig () by (), we get As + 8rsi( taθ or θ 8 As ) Graphical Method: Step 1: Draw the couple diagram takig a suitable scale as show. 17

18 1,m ad m This diagram provides the relative directio of the masses. Step : Now, draw the force polygo takig a suitable scale as show. This gives the directio ad magitude of mass m. The results are: m θ m 168 r, θ 1 7r 1, θ or m 8 7 kg As 18

19 Problem : Each crak of a four- cylider vertical egie is 5 mm. The reciprocatig masses of the first, secod ad fourth craks are 1 kg, 1 kg ad 1 kg ad the plaes of rotatio are 6 mm, mm ad mm from the plae of rotatio of the third crak. Determie the mass of the reciprocatig parts of the third cylider ad the relative agular positios of the craks if the egie is i complete primary balace. Solutio: Give: r 5 mm m 1 kg, m 1 kg ad m 1 kg 1 Plae Cet. Distace Mass (m) Radius (r) Force/ω Couple/ ω from Ref ( m r l ) kg m (m r ) plae kg m kg m m (RP) m.5.5 m

20 Aalytical Method: VTU EDUSAT PROGRAMME-17 Choose plae as the referece plae ad θ. 1 Step 1: Resolve the couples ito their horizotal ad vertical compoets ad take their sums. Sum of the horizotal compoets gives i.e., 8. 1 cos θ i.e., 1. 5 cos 8. 1 cos θ 8. 1 cos θ cos θ cos θ Sum of the vertical compoets gives i.e., 1. 5 si 8. 1 si θ 8. 1 si θ si θ Squarig ad addig (1) ad (), we get (8.1) (6.75 cos θ cos θ 5.56(cos θ cos θ i.e., 18.5 cos θ cos θ ( 1) si θ 18.5 cos θ ( ) 1.5) + (6.75 siθ ) + si θ ) 18.5 cos θ si θ Therefore, 16. cosθ ad θ Dividig () by (1), we get taθ i.e., θ As 6.75 si(7.1 ) cos (7.1 ) Step : Resolve the forces ito their horizotal ad vertical compoets ad take their sums. Sum of the horizotal compoets gives

21 i.e., i.e.,.5m VTU EDUSAT PROGRAMME-17.5 cos( ) + 7 cos ( ) +.5m cosθ m cosθ Ad sum of the vertical compoets gives i.e., i.e., m.5m siθ cosθ ().5 si( ) + 7 si ( ) +.5m siθ siθ Squarig ad addig () ad (), we get (.5) i.e., m ( 17.55) m kg 1 kg As Dividig () by (), we get (.518) ().518 taθ or θ 9.5 As +.5 cos(7.1 ) +.5 si(7.1 ) 1

22 Problem : The craks of a four cylider marie oil egie are at agular itervals of 9. The egie speed is 7 rpm ad the reciprocatig mass per cylider is 8 kg. The ier craks are 1 m apart ad are symmetrically arraged betwee outer craks which are.6 m apart. Each crak is mm log. Determie the firig order of the cyliders for the best balace of reciprocatig masses ad also the magitude of the ubalaced primary couple for that arragemet. Aalytical Solutio: Note: Give: m 8 kg, mrω N 7 rpm, 8 x. x(7.) r. m, πn ω 7. rad/s 6 There are four craks. They ca be used i six differet arragemets as show. It ca be observed that i all the cases, primary forces are always balaced. Primary couples i each case will be as uder. Takig 1 as the referece plae,

23 C C C C C C p1 p6 p p5 p p mrω 761 Nm C C p1 mrω p VTU EDUSAT PROGRAMME-17 ( l ) + ( l l ) ( 1. 8) + (. 8. 6) 761 Nm oly, sice l 795 mrω 198 C p ad l are it erchaged ( l ) + ( l l ) (. 6) + ( ) Nm 795 Nm oly, sice l ad l are it erchaged ( l ) + ( l l ) (. 8) + ( ) Nm 198 Nm oly, sice l ad l are it erchaged Thus the best arragemet is of rd ad th. The firig orders are 1 ad 1 respectively. Ubalaced couple 198 N m. Graphical solutio:

24 Case : SIX CYLINDER, FOUR STROKE ENGINE Crak positios for differet cyliders for the firig order 165 for clockwise rotatio of the crakshaft are, for First Third Fifth 1 θ 1 Fourth θ Secod θ Sixth 5 θ θ 1 θ 6 Sice all the force ad couple polygos close, it is iheretly balaced egie for primary ad secodary forces ad couples. Ad m m m m m m 1 1 r r r r r r

25 Problem 1: Each crak ad the coectig rod of a six-cylider four-stroke i-lie egie are 6 mm ad mm respectively. The pitch distaces betwee the cylider cetre lies are 8 mm, 8 mm, 1 mm, 8 mm ad 8 mm respectively. The reciprocatig mass of each cylider is 1. kg. The egie speed is 1 rpm. Determie the out-of-balace primary ad secodary forces ad couples o the egie if the firig order be 165. Take a plae midway betwee the cyliders ad as the referece plae. Solutio: Give: We have, r 6mm, l coectig rod legth mm, m reciprocatig mass of each cylider 1. kg, N 1 rpm ω πn 6 πx rad/s 5

26 Plae VTU EDUSAT PROGRAMME-17 Mass (m) kg Radius (r) m Cet. Force/ω (m r ) kg m Distace from Ref plae m Couple/ ω ( m r l ) kg m Graphical Method: Step 1: Draw the primary force ad primary couple polygos takig some coveiet scales. Note: For drawig these polygos take primary craks positio as the referece NO UNBALANCED PRIMARY FORCE NO UNBALANCED PRIMARY COUPLE 6

27 Step : Draw the secodary force ad secodary couple polygos takig some coveiet scales. Note: For drawig these polygos take secodary craks positio as the referece Problem : The firig order of a six cylider vertical four-stroke i-lie egie is 165. The pisto stroke is 8 mm ad legth of each coectig rod is 18 mm. The pitch distaces betwee the cylider cetre lies are 8 mm, 8 mm, 1 mm, 8 mm ad 8 mm respectively. The reciprocatig mass per cylider is 1. kg ad the egie speed is rpm. Determie the out-of-balace primary ad secodary forces ad couples o the egie takig a plae midway betwee the cyliders ad as the referece plae. Solutio: Give: We have, L 8 r mm, l coectig rod legth 18 mm, m reciprocatig mass of each cylider 1. kg, N rpm ω πn 6 πx rad/s NO UNBALANCED SECONDARY FORCE NO UNBALANCED SECONDARY COUPLE 7

28 Plae Mass (m) kg Radius (r) m Cet. Force/ω (m r ) kg m Distace from Ref plae m Couple/ ω ( m r l ) kg m Graphical Method: Step 1: Draw the primary force ad primary couple polygos takig some coveiet scales. Note: For drawig these polygos take primary craks positio as the referece Note: No primary ubalaced force or couple 8

29 Step : Draw the secodary force ad secodary couple polygos takig some coveiet scales. Note: For drawig these polygos take secodary craks positio as the referece Note: No secodary ubalaced force or couple 9

30 Problem : The stroke of each pisto of a six-cylider two-stroke ilie egie is mm ad the coectig rod is 8 mm log. The cylider cetre lies are spread at 5 mm. The craks are at 6 apart ad the firig order is 156. The reciprocatig mass per cylider is 1 kg ad the rotatig parts are 5 kg per crak. Determie the out of balace forces ad couples about the mid plae if the egie rotates at rpm. Primary craks positio Relative positios of Craks i degrees Firig order θ 1 θ θ θ θ 5 θ Secodary craks positio Relative positios of Craks i degrees Firig order θ 1 θ θ θ θ 5 θ Calculatio of primary forces ad couples: Total mass at the crak pi 1 kg + 5 kg 15 kg Plae Cet. Distace Mass (m) Radius (r) Force/ω Couple/ ω from Ref ( m r l ) kg m (m r ) plae kg m kg m m

31 1

32 Calculatio of secodary forces ad couples: Sice rotatig mass does ot affect the secodary forces as they are oly due to secod harmoics of the pisto acceleratio, the total mass at the crak is take as 1 kg. Plae Mass (m) kg Radius (r) m Cet. Force/ω (m r ) kg m Distace from Ref plae m Couple/ ω ( m r l ) kg m

33 Two Cylider V-egie: BALANCING OF V ENGINE A commo crak OA is operated by two coectig rods. The cetre lies of the two cyliders are iclied at a agle α to the X-axis. Let θ be the agle moved by the crak from the X-axis. Determiatio of Primary force: Primary force of 1 alog lie of stroke OB 1 mrω cos( θ α ) ( 1) Primary force of 1 alog X - axis mrω cos( θ α ) cosα ( ) Primary force of alog lie of stroke OB mrω cos( θ+α ) ( ) Primary force of alog X-axis mrω cos( θ+α )cosα ( ) Totalprimary force alog X -axis mrω mrω mrω mrω cosα cosα cos [ cos(θ α) + cos(θ+ α) ] [ cosθcos α+ siθsiα+ cosθcos α siθsiα] cosα xcosθcosα α cosθ (5)

34 Similarly, Totalprimary force alog Z-axis mrω mrω mrω mrω [ cos(θ α)siα-cos(θ+ α)siα] [(cosθcosα+ siθsiα) (cosθcosα siθsiα] siα Resultat Primary force siα x siθsiα si α siθ (6) ( mrω cos α cosθ) + ( mrω si α siθ) ( cos α cosθ) + ( si α siθ) (7) mrω ad this resultat primary force will be at agle β with the X axis, give by, si α siθ taβ (8) cos α cosθ If α 9, the resultat force will be equal to mrω mrω ( cos 5 cosθ) + ( si 5 siθ) (9) ad si 5 siθ taβ taθ (1) cos 5 cosθ i.e., β θ or it acts alog the crak ad therefore, ca be completely balaced by a mass at a suitable radius diametrically opposite to the crak, such that, m r r r mr -----(11) For a give value of α, the resultat force is maximum (Primary force), whe ( cos α cosθ) + ( si α siθ) or maximum ( cos α cos θ + si α si θ ) is maximum is

35 Or d dθ i.e., ( cos α cos θ + si α si θ ) -cos α x cosθ siθ + si α xsiθ cosθ i.e., i.e., -cos siθ α xsiθ + si [ si α -cos α ] α x siθ As α is ot zero, therefore for a give value of α, the resultat primary force is maximum whe θ. Determiatio of Secodary force: Secodary force of 1 alog lie of stroke OB 1 is equal to mrω cos( θ α ) ( 1) mrω Secodary force of 1 alog X - axis cos( θ α ) cosα ( ) Secodary force of alog lie of stroke OB mrω cos( θ+α ) ( ) mrω Primary force of alog X-axis cos( θ+α )cosα ( ) Therefore, Total secodary force alog X-axis mrω cosα [ cos (θ α) + cos(θ + α) ] mrω cosα mrω cosα cosθcosα (5) [(cosθcosα+ siθsiα) + (cosθcosα siθsiα] 5

36 Similarly, Total secodary force alog Z-axis mrω mrω Resultat Secodary siα siθsiα (6) force ( cosα cosθcosα) + ( siα siθsi α) (7) Ad siα siθsi α ta β ' (8) cosα cosθcos α If α 9 or α 5, mr ω siθ mrω Secodary force siθ ( 9) ' ' Ad taβ ad β 9 (1) i.e., the force acts alog Z- axis ad is a harmoic force ad special methods are eeded to balace it. Problem 1: The cyliders of a twi V-egie are set at 6 agle with both pistos coected to a sigle crak through their respective coectig rods. Each coectig rod is 6 mm log ad the crak radius is 1 mm. The total rotatig mass is equivalet to kg at the crak radius ad the reciprocatig mass is 1. kg per pisto. A balace mass is also fitted opposite to the crak equivalet to. kg at a radius of 15 mm. Determie the maximum ad miimum values of the primary ad secodary forces due to iertia of the reciprocatig ad the rotatig masses if the egie speed is 8 mm. Solutio: Give: ADARSHA m reciprocatig mass of each pisto 1. kg M equivalet rotatig mass kg m C balacig mass. kg, l coectig rod legth 6 mm r crak radius 1 mm N 8 rpm We have, H G ω πn 6 πx8 6 r C 15 mm 8.78 rad/s ad l r

37 Primary Force: Total primary force alog X - axis Mrω mrω cos Cetrifuga l force due to rotatig mass alog X axis m r cosθ () Cetrifuga l force due to balacig mass alog X axis C C ω cosθ () Therefore total ubalace force alog X axis (1) + () + () That is Total Ubalace force alog X axis mrω cos α cosθ+ Mrω cosθ m α cosθ (1) cosθ ω cosθ [ mr cos α + Mr -m ] C rc ( 8.78) cosθ[ x1.x.1xcos + x.1.x.15] ( 8.78) cosθ[ ] 88.1 cosθ N () C r C ω Total primary force alog Z - axis mr ω si α siθ (5) 7

38 Cetrifuga l force due to rotatig mass alog Z axis Mrω siθ (6) Cetrifuga l force due to balacig mass alog Z axis m C r C ω siθ (7) Therefore total ubalace force alog Z axis (5) + (6) + (7) That is Total Ubalace force alog Z- axis mrω si α siθ+ Mrω siθ m siθ ω siθ [ mr si α + Mr -m ] C rc ( 8.78) siθ[ x1.x.1xsi + x.1.x.15] ( 8.78) siθ[ ] -16. siθ N (8) Resultat Primary force ( 88.1 cosθ) + (-16. siθ) cos cos This is maximum, whe Maximum θ si C r C ω θ θ (9) θ ad miimum, whe Primary force, i.e., whe θ θ N (1) Ad Miimum Primary force, i.e., whe θ cos N (11) Secodary force: The rotatig masses do ot affect the secodary forces as they are oly due to secod harmoics of the pisto acceleratio. 8

39 Resultat Secodary force mrω. VTU EDUSAT PROGRAMME-17 ( cosα cosθcos α) + ( siα siθsi α) ( cos cosθcos6 ) 5 + ( si siθsi6 ) [.1875 (cosθθ (siθθ ] (1) x1. x.1x(8.78) This is maximum, whe Maximum. Ad Miimum. θ ad miimum, whe secodary force, i.e., whe θ θ 18 [.1875 (cos ) (si ) ] secodary force, i.e., whe θ 18 [.1875 (cos18 ) (si18 ) ] N (1) BALANCING OF W, V-8 AND V-1 ENGINES BALANCING OF W ENGINE I this egie three coectig rods are operated by a commo crak. Total primary force alog X-axis Total primary mrω mrω cosθ si ( cos α + 1 ) (1) (sice the primary force of alog Z axis is zero) Resultat Primary force mrω α siθ () N (1) force alog Z-axis will be same as i the V twi egie, [ cosθ( cos α + 1) + ( si αsiθ) ] () ad this resultat primary force will be at agle β with the X axis, give by, 9

40 si α siθ taβ cosθ cos ( α + 1) () If α 6, Resultat Primary force taβ mrω (5) taθ (6) i.e., β θ or it acts alog the crak ad therefore, ca be completely balaced by a mass at a suitable radius diametrically opposite to the crak, such that, r mr -----(7) m r r Total secodary force alog X-axis mr ω cosθ ad cosαcosα + 1 (8) Total secodary force alog Z directio will be same as i the V-twi egie. Resultat secodary force ta β ' If mrω [ cosθ( cosαcosα + 1) + ( siαsiαsiθ) ] (9) siα siθsi α cosθ cosαcosα α 6, ( + 1) Secodary force alog X-axis mrω (1) Secodary force alog Z-axis cosθ (11) mr ω siθ (1) It is ot possible to balace these forces simultaeously

41 V-8 ENGINE It cosists of two baks of four cyliders each. The two baks are iclied to each other i the shape of V. The aalysis will deped o the arragemet of cyliders i each bak. V-1 ENGINE It cosists of two baks of six cyliders each. The two baks are iclied to each other i the shape of V. The aalysis will deped o the arragemet of cyliders i each bak. If the craks of the six cyliders o oe bak are arraged like the completely balaced six cylider, four stroke egie the, there is o ubalaced force or couple ad thus the egie is completely balaced. BALANCING OF RADIAL ENGINES: It is a multicylider egie i which all the coectig rods are coected to a commo crak. 1

42 Direct ad reverse crak method of aalysis: I this all the forces exists i the same plae ad hece o couple exist. I a reciprocatig egie the primary force is give by, the lie of stroke. mrω cosθ which acts alog I direct ad reverse crak method of aalysis, a force idetical to this force is geerated by two masses as follows. 1. A mass m/, placed at the crak pi A ad rotatig at a agular velocity ω i the couter clockwise directio.. A mass m/, placed at the crak pi of a imagiary crak OA at the same agular positio as the real crak but i the opposite directio of the lie of stroke. It is assumed to rotate at a agular velocity ω i the clockwise directio (opposite).. While rotatig, the two masses coicide oly o the cylider cetre lie. The compoets of the cetrifugal forces due to rotatig masses alog the lie of stroke are, Due to mass at Due to mass at m A r ω cosθ A m rω cosθ ' Thus, total force alog the lie of stroke mrω cosθ which is equal to the primary force. At ay istat, the compoets of the cetrifugal forces of these masses ormal to the lie of stroke will be equal ad opposite. The crak rotatig i the directio of egie rotatio is kow as the direct crak ad the imagiary crak rotatig i the opposite directio is kow as the reverse crak. Now, Secodary acceleratig force is mrω r m cosθ mr( ω) ( ω) cosθ cosθ

43 This force ca also be geerated by two masses i a similar way as follows. r 1. A mass m/, placed at the ed of direct secodary crak of legth ad rotatig at a agular velocity ω i the couter clockwise directio. at a agle θ. A mass m/, placed at the ed of reverse secodary crak of legth ad rotatig at a agular velocity ω i the clockwise directio. r at a agle -θ The compoets of the cetrifugal forces due to rotatig masses alog the lie of stroke are, m r mrω Due to mass at C (ω) cosθ m r Due to mass at C' (ω) Thus, total force alog the lie of stroke mrω cosθ cosθ cosθ m r mrω x (ω) cosθ cosθ which is equal to the secodary force. Refereces: 1. Theory of Machies by S.S.Ratta, Third Editio, Tata McGraw Hill Educatio Private Limited.. Kiematics ad Dyamics of Machiery by R. L. Norto, First Editio i SI uits, Tata McGraw Hill Educatio Private Limited.