BALANCING OF RECIPROCATING MASSES

Transcription

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

2 The first ter of the equatio (), i.e. force the secod ter rω cosθ Maxiu value of priary acceleratig force is rω cosθ is called priary acceleratig is called the secodary acceleratig force. rω rω Ad Maxiu value of secodary acceleratig force is Geerally, value is uch greater tha oe; the secodary force is sall copared to priary force ad ca be safely eglected for slow speed egies. I Fig (a), the iertia force due to priary acceleratig force is show.

3 I Fig (b), the forces actig o the egie frae due to iertia force are show. At O the force exerted by the crakshaft o the ai bearigs has two copoets, 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 for a ubalaced shakig couple. The agitude ad directio of these ubalaced force ad couple go o chagig with agle θ. The shakig force produces liear vibratios of the frae i horizotal directio, whereas the shakig couple produces a oscillatig vibratio. h The shakig force F 1 is the oly ubalaced force which ay haper the sooth ruig of the egie ad effort is ade to balace the sae. However it is ot at all possible to balace it copletely ad oly soe odificatios ca be carried out. BALANCING OF THE SHAKING FORCE: Shakig force is beig balaced by addig a rotatig couter ass at radius r directly opposite the crak. This provides oly a partial balace. This couter ass is i additio to the ass used to balace the rotatig ubalace due to the ass at the crak pi. This is show i figure (c).

4 The horizotal copoet of the cetrifugal force due to the balacig ass is rω cosθ ad this is i the lie of stroke. This copoet eutralizes the ubalaced reciprocatig force. But the rotatig ass also has a copoet rω siθ perpedicular to the lie of stroke which reais ubalaced. The ubalaced force is zero at θ or 18 ad axiu at the iddle of the stroke i.e. θ 9. The agitude or the axiu value of the ubalaced force reais the sae i.e. equal to rω. Thus istead of slidig to ad fro o its outig, the echais teds to jup up ad dow. To iiize the effect of the ubalace force a coproise is, usually ade, is of the reciprocatig ass is balaced or a value betwee 1 to. If c is the fractio of the reciprocatig ass, the The priary force balaced by the ass crω cosθ ad The priary force ubalaced by the ass (1 c) rω cos θ Vertical copoet c rω siθ of cetrifuga l force which reais ubalaced I reciprocatig egies, ubalace forces i the directio of the lie of stroke are ore dagerous tha the forces perpedicular to the lie of stroke. Resultat ubalaced force at ay istat [(1 c)rω cosθ] + [ crω siθ] The resultat ubalaced force is iiu whe, 1 c This ethod is just equivalet to as if a revolvig ass at the crakpi is copletely balaced by providig a couter ass at the sae radius diaetrically opposite to the crak. Thus if P is the ass at the crakpi ad c is the fractio of the reciprocatig ass to be balaced, the ass at the crakpi ay be cosidered as c + P which is to be copletely balaced.

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

6 Proble : 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 rp, Stroke 5. If 6 % of the reciprocatig parts ad all the revolvig parts are to be balaced, deterie the, (i) balace ass required at a radius of ad (ii) ubalaced force whe the crak has tured 5 fro the top-dead cetre. Solutio: Give : ass of the reciprocatig parts kg P kg, N 15 rp, L legth of the c 6%, r c stroke 5 (i) Balace ass required at a radius of 5 We have, πn ω 6 L 5 r π x rad/s Mass to be balaced at the crak pi M M c + P.6x + 5 kg ad c r c i.e. Mr c therefore 5 x175 c 9.5 kg Mr r c (ii) Ubalaced force whe the crak has tured 5 fro the top-dead cetre. Ubalaced force at θ 5 [( 1 c) rω cosθ] + [ crω siθ] [( 1.6) xx.175x ( 15.7) cos5 ] +.6xx.175x( 15.7) 88.7 N [ si5 ] 6

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

8 r( ω) Cetrifugal force iduced i the iagiary crak r( ω) Copoet of this force alog the lie of stroke is cosθ Thus the effect of the secodary force is equivalet to a iagiary crak of legth rotatig at double the agular velocity, i.e. twice of the egie speed. The iagiary crak coicides with the actual at ier top-dead cetre. At other ties, it akes 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 ultiplicatio of the secodary force with the distace l of the plae fro the referece plae. r COMPLETE BALANCING OF RECIPROCATING PARTS Coditios to be fulfilled: 1. Priary forces ust balace i.e., priary force polygo is eclosed.. Priary couples ust balace i.e., priary couple polygo is eclosed.. Secodary forces ust balace i.e., secodary force polygo is eclosed.. Secodary couples ust balace i.e., secodary couple polygo is eclosed. Usually, it is ot possible to satisfy all the above coditios fully for ulti-cylider egie. Mostly soe 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 exaples havig four cider i-lie egies. I a reciprocatig egie, the reciprocatig ass is trasferred to the crakpi; the axial copoet of the resultig cetrifugal force parallel to the axis of the cylider is the priary ubalaced force. Cosider a shaft cosistig of three equal craks asyetrically spaced. The crakpis carry equivalet of three uequal reciprocatig asses, the 8

9 Priary force rω cos θ (1) Priary couple rω l cos θ () Secodary force ( ω) r cos θ () Ad Secodary couple r ω r ( ω) l cos θ l cos θ () GRAPHICAL SOLUTION: To solve the above equatios graphically, first draw the r cos θ polygo ( ω is coo to all forces). The the axial copoet of the resultat forces ( F r cosθ ) ultiplied by ω provides the priary ubalaced force o the syste at that oet. This ubalaced force is zero whe θ 9 ad a axiu wheθ. 9

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

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

12 Secodary couple 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 ters of θ The axiu values of these forces ad couples vary istat to istat ad are equal to the values as give by the equivalet rotatig asses at the crak pi. If a particular positio of the crak shaft is cosidered, the above expressios ay ot give the axiu values. For exaple, the axiu value of priary couple is rω l ad this value is obtaied at crak positios ad 18. However, if the crak positios are assued 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 copoets of the force ad couple ca be take to fid the axiu values. For exaple, if the crak positios cosidered as 1 ad, the priary couple ca be obtaied as X copoet r ω 1 r ω l l cos 1 + l cos ( ) Y copoet rω rω l si1 l + l si 18 ( + 1 ) Therefore, Priary couple rω l 1 rω l + rω 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 iddle bearig about which the arrageet is syetrical as the referece plae. Priary force rω [ cosθ + cos(18 + θ) + cos(18 + θ) + cosθ] Priary couple rω l l cosθ + cos(18 + θ) l l + cos(18 θ) cosθ + + Secodary rω cos θ + cos(6 + θ) force + cos(6 + θ) + cos θ rω cosθ rω Maxiu value at θ,18 θ,9,6,18 ad ad 7 5 or Secodary couple rω l l cosθ + cos(6 + θ) l l + cos(6 θ) cosθ + + Thus the egie is ot balaced i secodary forces. 1

15 Proble 1: A four-cylider oil egie is i coplete priary balace. The arrageet of the reciprocatig asses i differet plaes is as show i figure. The stroke of each pisto is r. Deterie the reciprocatig ass of the cylider ad the relative crak positio. Solutio: Give: 8 kg,?, 1 L r crak legth r 59 kg, 8 kg Plae Mass () kg Radius (r) Cet. Force/ω ( r ) kg Distace fro Ref plae Couple/ ω ( r l ) kg 1 8 r 8 r r (RP) r 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 copoets ad take their sus. Su of the horizotal copoets gives i.e., 9 r cos θ + 9 cos θ rcos cos θ Su of the vertical copoets 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 or 19.1 taθ i.e., 1 θ 1968 si(167.9 ) cos (167.9 ) Step : Resolve the forces ito their horizotal ad vertical copoets ad take their sus. Su of the horizotal copoets gives 8 r cos( 1. or cos θ ) + r cos θ + 59 rcos ( ) + 8rcos( ) 16

17 Su of the vertical copoets gives 8 r si( 1. or si θ ) + r si θ + 59 rsi ( ) Squarig ad addig () ad (), we get + 8rsi( ) 7. 1 kg As Dividig () by (), we get 17.9 taθ or θ 8 As Graphical Method: Step 1: Draw the couple diagra takig a suitable scale as show. 17

18 1, ad This diagra provides the relative directio of the asses. Step : Now, draw the force polygo takig a suitable scale as show. This gives the directio ad agitude of ass. The results are: θ 168 r, θ 1 7r 1, θ or 8 7 kg As 18

19 Proble : Each crak of a four- cylider vertical egie is 5. The reciprocatig asses of the first, secod ad fourth craks are 1 kg, 1 kg ad 1 kg ad the plaes of rotatio are 6, ad fro the plae of rotatio of the third crak. Deterie the ass of the reciprocatig parts of the third cylider ad the relative agular positios of the craks if the egie is i coplete priary balace. Solutio: Give: r 5 1 kg, 1 kg ad 1 kg 1 Plae Mass () kg Radius (r) Cet. Force/ω ( r ) kg Distace fro Ref plae Couple/ ω ( r l ) kg (RP)

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 copoets ad take their sus. Su of the horizotal copoets gives i.e., 8. 1 cos θ i.e., 1. 5 cos 8. 1 cos θ 8. 1 cos θ cos θ cos θ Su of the vertical copoets 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 cos θ ( 1) si θ 18.5 cos θ ( ) 1.5) + (6.75 siθ ) + si θ ) 18.5 cos θ si θ i.e., 18.5 cos θ Therefore, 16. cosθ ad θ As Dividig () by (1), we get taθ 6.75 si(7.1 ) 6.75 cos (7.1 ) i.e., θ Step : Resolve the forces ito their horizotal ad vertical copoets ad take their sus. Su of the horizotal copoets gives

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

22 Proble : The craks of a four cylider arie oil egie are at agular itervals of 9. The egie speed is 7 rp ad the reciprocatig ass per cylider is 8 kg. The ier craks are 1 apart ad are syetrically arraged betwee outer craks which are.6 apart. Each crak is log. Deterie the firig order of the cyliders for the best balace of reciprocatig asses ad also the agitude of the ubalaced priary couple for that arrageet. Aalytical Solutio: Give: 8 kg, rω 8 N 7 rp, r., x. x(7.) ω πn 6 7. rad/s Note: There are four craks. They ca be used i six differet arrageets as show. It ca be observed that i all the cases, priary forces are always balaced. Priary couples i each case will be as uder. Takig 1 as the referece plae,

23 C C C C p1 p6 p p5 rω 761 N C 761 p1 795 N C 795 rω p VTU EDUSAT PROGRAMME-17 ( l ) + ( l l ) ( 1. 8) + (. 8. 6) N oly, sice l ad l are it erchaged ( l ) + ( l l ) (. 6) + ( ) N oly, sice l ad l are it erchaged C C p p rω 198 C p ( l ) + ( l l ) (. 8) + ( ) N 198 N oly, sice l ad l are it erchaged Thus the best arrageet is of rd ad th. The firig orders are 1 ad 1 respectively. Ubalaced couple 198 N. 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 Ad 1 1 r r r r r r Sice all the force ad couple polygos close, it is iheretly balaced egie for priary ad secodary forces ad couples.

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

26 Plae VTU EDUSAT PROGRAMME-17 Mass () kg Radius (r) Cet. Force/ω ( r ) kg Distace fro Ref plae Couple/ ω ( r l ) kg Graphical Method: Step 1: Draw the priary force ad priary couple polygos takig soe coveiet scales. Note: For drawig these polygos take priary 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 soe coveiet scales. Note: For drawig these polygos take secodary craks positio as the referece NO UNBALANCED SECONDARY FORCE Proble : NO UNBALANCED SECONDARY COUPLE The firig order of a six cylider vertical four-stroke i-lie egie is 165. The pisto stroke is 8 ad legth of each coectig rod is 18. The pitch distaces betwee the cylider cetre lies are 8, 8, 1, 8 ad 8 respectively. The reciprocatig ass per cylider is 1. kg ad the egie speed is rp. Deterie the out-of-balace priary ad secodary forces ad couples o the egie takig a plae idway betwee the cyliders ad as the referece plae. Solutio: Give: We have, L 8 r, l coectig rod legth 18, reciprocatig ass of each cylider 1. kg, N rp ω πn 6 πx rad/s 7

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

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

30 Proble : The stroke of each pisto of a six-cylider two-stroke ilie egie is ad the coectig rod is 8 log. The cylider cetre lies are spread at 5. The craks are at 6 apart ad the firig order is 156. The reciprocatig ass per cylider is 1 kg ad the rotatig parts are 5 kg per crak. Deterie the out of balace forces ad couples about the id plae if the egie rotates at rp. Priary 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 priary forces ad couples: Total ass at the crak pi 1 kg + 5 kg 15 kg Plae Mass () kg Radius (r) Cet. Force/ω ( r ) kg Distace fro Ref plae Couple/ ω ( r l ) kg

31 1

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

33 Two Cylider V-egie: BALANCING OF V ENGINE A coo 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 oved by the crak fro the X-axis. Deteriatio of Priary force: Priary force of 1 alog lie of stroke OB 1 rω cos( θ α ) ( 1) Priary force of 1 alog X - axis rω cos( θ α ) cosα ( ) Priary force of alog lie of stroke OB rω cos( θ+α ) ( ) Priary force of alog X-axis rω cos( θ+α )cosα ( ) Totalpriary force alog X -axis rω rω rω rω cosα cosα cos [ cos(θ α) + cos(θ+ α) ] [ cosθcos α+ siθsiα+ cosθcos α siθsiα] cosα xcosθcosα α cosθ (5)

34 Siilarly, Totalpriary force alog Z-axis rω rω rω rω [ cos(θ α)siα-cos(θ+ α)siα] [(cosθcosα+ siθsiα) (cosθcosα siθsiα] siα Resultat Priary force siα x siθsiα si α siθ (6) ( rω cos α cosθ) + ( rω si α siθ) ( cos α cosθ) + ( si α siθ) (7) rω ad this resultat priary force will be at agle β with the X axis, give by, si α taβ cos α siθ (8) cosθ If α 9, the resultat force will be equal to rω rω ( 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 copletely balaced by a ass at a suitable radius diaetrically opposite to the crak, such that, r r r r -----(11) For a give value of α, the resultat force is axiu (Priary force), whe ( cos α cosθ) + ( si α siθ) or axiu ( cos α cos θ + si α si θ ) is axiu 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 priary force is axiu whe θ. Deteriatio of Secodary force: Secodary force of 1 alog lie of stroke OB 1 is equal to rω cos( θ α ) ( 1) rω Secodary force of 1 alog X - axis cos( θ α ) cosα ( ) Secodary force of alog lie of stroke OB rω cos( θ+α ) ( ) rω Priary force of alog X-axis cos( θ+α )cosα ( ) Therefore, Total secodary force alog X-axis rω cosα [ cos (θ α) + cos(θ + α) ] rω cosα rω cosα cosθcosα (5) [(cosθcosα+ siθsiα) + (cosθcosα siθsiα] 5

36 Siilarly, Total secodary force alog Z-axis rω rω 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, r ω siθ rω Secodary force siθ ( 9) ' ' Ad taβ ad β 9 (1) i.e., the force acts alog Z- axis ad is a haroic force ad special ethods are eeded to balace it. Proble 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 log ad the crak radius is 1. The total rotatig ass is equivalet to kg at the crak radius ad the reciprocatig ass is 1. kg per pisto. A balace ass is also fitted opposite to the crak equivalet to. kg at a radius of 15. Deterie the axiu ad iiu values of the priary ad secodary forces due to iertia of the reciprocatig ad the rotatig asses if the egie speed is 8. Solutio: Give: reciprocatig ass of each pisto 1. kg M equivalet rotatig ass kg C balacig ass. kg, l coectig rod legth 6 r crak radius 1 N 8 rp We have, ω πn 6 πx8 6 r C rad/s ad l r

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

38 Cetrifuga l force due to rotatig ass alog Z axis Mrω siθ (6) Cetrifuga l force due to balacig ass alog Z axis C r C ω siθ (7) Therefore total ubalace force alog Z axis (5) + (6) + (7) That is Total Ubalace force alog Z- axis rω si α siθ+ Mrω siθ siθ ω siθ [ r si α + Mr - ] C rc ( 8.78) siθ[ x1.x.1xsi + x.1.x.15] ( 8.78) siθ[ ] -16. siθ N (8) C r C ω Resultat Priary force ( 88.1 cosθ) + (-16. siθ) cos cos This is axiu, whe θ si θ θ (9) θ ad iiu, whe θ 9 Maxiu Priary force, i.e., whe θ N (1) Ad Miiu Priary force, i.e., whe θ cos N (11) Secodary force: The rotatig asses do ot affect the secodary forces as they are oly due to secod haroics of the pisto acceleratio. 8

39 Resultat Secodary rω. VTU EDUSAT PROGRAMME-17 force ( cosα cosθcos α) + ( siα siθsi α) ( cos cosθcos6 ) 5 + ( si siθsi6 ) [.1875 (cosθθ (siθθ ] (1) x1. x.1x(8.78) This is axiu, whe θ ad iiu, whe θ 18 Maxiu secodary force, i.e., whe θ. [.1875 (cos ) (si ) ] N (1) Ad Miiu secodary force, i.e., whe θ 18. [.1875 (cos18 ) (si18 ) ] N (1) BALANCING OF W ENGINE BALANCING OF W, V-8 AND V-1 ENGINES I this egie three coectig rods are operated by a coo crak. Total priary force alog X-axis rω cosθ ( cos α + 1 ) (1) Total priary force alog Z-axis will be sae as i the V twi egie, (sice the priary force of alog Z axis is zero) rω rω si Resultat Priary force α siθ () [ cosθ( cos α + 1) + ( si αsiθ) ] () ad this resultat priary 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 Priary force rω (5) ad taβ taθ (6) i.e., β θ or it acts alog the crak ad therefore, ca be copletely balaced by a ass at a suitable radius diaetrically opposite to the crak, such that, r r -----(7) r r Total secodary force alog X-axis r ω cosθ cosαcosα + 1 (8) Total secodary force alog Z directio will be sae as i the V-twi egie. Resultat secodary force rω [ cosθ( cosαcosα + 1) + ( siαsiαsiθ) ] (9) ta β ' siα siθsi α cosθ cosαcosα ( + 1) (1) If α 6, Secodary force alog X-axis rω Secodary force alog Z-axis cosθ (11) r ω siθ (1) It is ot possible to balace these forces siultaeously

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 arrageet 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 arrageet of cyliders i each bak. If the craks of the six cyliders o oe bak are arraged like the copletely balaced six cylider, four stroke egie the, there is o ubalaced force or couple ad thus the egie is copletely balaced. BALANCING OF RADIAL ENGINES: It is a ulticylider egie i which all the coectig rods are coected to a coo crak. 1

42 Direct ad reverse crak ethod of aalysis: I this all the forces exists i the sae plae ad hece o couple exist. I a reciprocatig egie the priary force is give by, the lie of stroke. rω cosθ which acts alog I direct ad reverse crak ethod of aalysis, a force idetical to this force is geerated by two asses as follows. 1. A ass /, placed at the crak pi A ad rotatig at a agular velocity ω i the couter clockwise directio.. A ass /, placed at the crak pi of a iagiary crak OA at the sae agular positio as the real crak but i the opposite directio of the lie of stroke. It is assued to rotate at a agular velocity ω i the clockwise directio (opposite).. While rotatig, the two asses coicide oly o the cylider cetre lie. The copoets of the cetrifugal forces due to rotatig asses alog the lie of stroke are, Due to ass at Due to ass at A r ω cosθ A rω cosθ ' Thus, total force alog the lie of stroke rω cosθ which is equal to the priary force. At ay istat, the copoets of the cetrifugal forces of these asses oral 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 iagiary crak rotatig i the opposite directio is kow as the reverse crak. Now, Secodary acceleratig force is rω r cosθ r( ω) ( ω) cosθ cosθ

43 This force ca also be geerated by two asses i a siilar way as follows. r 1. A ass /, placed at the ed of direct secodary crak of legth ad rotatig at a agular velocity ω i the couter clockwise directio.. A ass /, 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 at a agle -θ The copoets of the cetrifugal forces due to rotatig asses alog the lie of stroke are, Due to ass at C r rω (ω) cosθ cosθ Due to ass at C' r (ω) cosθ rω cosθ Thus, total force alog the lie of stroke r rω 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 Liited.. Kieatics ad Dyaics of Machiery by R. L. Norto, First Editio i SI uits, Tata McGraw Hill Educatio Private Liited.